multilinear algebra greub

Werner Greub

MultilinearIgenra

2nd Edition

Springer-VerlagNew York Heidelberg Berlin

a I

Un iversitext

Werner Greub

Multiinear Algebra2nd Edition

Springer-VerlagNew York Heidelberg Berlin

Werner Greub

Department of MathematicsUniversity of TorontoToronto M5S 1A1Canada

AMS Subject Classifications: 15-01, 15A75, 15A72

Library of Congress Cataloging in Publication Data

Greub, Werner Hildbert, 1925-Multilinear algebra,

(Universitext)Includes index.1. Algebras, Linear. I. Title.

QA 184.G74 1978 512'.5 78-949ISBN-13:978-0-387-90284-5 e-ISBN-13:978-1-4613-9425-9DO!: 10.1007/ 978-1-4613-9425-9

All rights reserved.

No part of this book may be translated or reproduced in anyform without written permission from Springer-Verlag.

© 1967 by Springer-Verlag Berlin Heidelberg© 1978 by Springer-Verlag New York Inc.

987654321

Preface

This book is a revised version of the first edition and is intended as asequel and companion volume to the fourth edition of Linear Algebra(Graduate Texts in Mathematics 23).

As before, the terminology and basic results of Linear Algebra arefrequently used without reference. In particular, the reader should befamiliar with Chapters 1-5 and the first part of Chapter b of that book,although other sections are occasionally used.

In this new version of Multilinear Algebra, Chapters 1-5 remain essen-tially unchanged from the previous edition. Chapter b has been completelyrewritten and split into three (Chapters b, 7, and 8). Some of the proofshave been simplified and a substantial amount of new material has beenadded. This applies particularly to the study of characteristic coefficientsand the Pf of f ian.

The old Chapter 7 remains as it stood, except that it is now Chapter 9.The old Chapter 8 has been suppressed and the material which it con-tained (multilinear functions) has been relocated at the end of Chapters 3,S, and 9.

The last two chapters on Clifford algebras and their representations arecompletely new. In view of the growing importance of Clifford algebrasand the relatively few references available, it was felt that these chapterswould be useful to both mathematicians and physicists.

In Chapter 10 Clifford algebras are introduced via universal propertiesand treated in a fashion analogous to exterior algebra. After the basicisomorphism theorems for these algebras (over an arbitrary inner productspace) have been established the chapter proceeds to a discussion offinite-dimensional Clifford algebras. The treatment culminates in the com-plete classification of Clifford algebras over finite-dimensional complexand real inner product spaces.

v

vi Preface

The book concludes with Chapter 11 on representations of Cliffordalgebras. The twisted adjoint representation which leads to the definitionof the spin-groups is an important example. A version of Wedderburn'stheorem is the key to the classification of all representations of the Cliffordalgebra over an 8-dimensional real vector space with a negative definiteinner product. The results are applied in the last section of this chapter tostudy orthogonal multiplications between Euclidean spaces and the ex-istence of orthonormal frames on the sphere. In particular, it is shown thatthe (n -1)-sphere admits an orthonormal k-frame where k is theRadon-Hurwitz number corresponding to n. A deep theorem of F. Adamsstates that this result can not be improved.

The problems at the end of Chapter 11 include a basis-free definition ofthe Cayley algebra via the complex cross-product analogous to the defini-tion of quaternions in Section 7.23 of the fourth edition of Linear Algebra.Finally, the Cayley multiplication is used to obtain concrete forms of someof the isomorphisms in the table at the end of Chapter 10.

I should like to express my deep thanks to Professor J. R. Vanstone whoworked closely with me through each stage of this revision and who madenumerous and valuable contributions to both content and presentation. Ishould also like to thank Mr. M. S. Swanson who assisted ProfessorVanstone and myself with the proof reading.

Toronto, April 1978 W. H. Greub

Table of Contents

Chapter 1 Tensor Products 1

Chapter 2 Tensor Products of Vector Spaces withAdditional Structure 41

Chapter 3 Tensor Algebra 60

Chapter 4 Skew-Symmetry and Symmetry in the Tensor Algebra 84

Chapter 5 Exterior Algebra 96

Chapter 6 Mixed Exterior Algebra 148

Chapter 7 Applications to Linear Transformations 174

Chapter 8 Skew and Skew-Hermitian Transformations 193

Chapter 9 Symmetric Tensor Algebra 209

Chapter 10 Clifford Algebras 227

Chapter 11 Representations of Clifford Algebras 260

Index 291

vii

Tensor Products

Throughout this chapter except where noted otherwise all vector spaces will be defined over afixed, but arbitrarily chosen, field T.

Multilinear Mappings

1.1. Bilinear Mappings

Suppose E, F and G are any three vector spaces, and consider a mapping

p : E x F -+ G.

'p is called bilinear if it satisfies the conditions

p(t x 1 + µx 2 , y) = p(x 1, y) + µup(x2 , y)

'p(x, Ay 1 + µY2) _ A p(x, Y 1) + µup(x, Y2)

x1, x2EE, yeF,A,µeF,

xeE,y1,y2eF.

Recall that if G = F, then 'p is called a bilinear function.The set S of all vectors in G of the form '(x, y), x e E, y e F is not in

general a vector subspace of G. As an example, let E = F and G be re-spectively 2- and 4-dimensional vector spaces. Select a basis a 1, a2 in Eand a basis c (v = 1, ... , 4) in G and define the bilinear mapping 'p by

co(x, Y) = 'i 'c1 + 1i2C2 + 2i 'c3 + 2ii2C4

where x = c 1a 1 + 2a2 and y = ri' a 1 + r12a2. Then it is easy to see thata vector

z = VC

of G is contained in S if and only if the components satisfy the relation

- = o.

1

2 1 Tensor Products

Since the vectors z 1 = 2c 1 + 2c2 + c3 + c4 and z2 = c 1 + c3 satisfy thiscondition, while the vector z = z 1 - z2 = c 1 + 2c2 + c4 does not, it followsthat S is not a subspace of G.

We shall denote by Im p the subspace of G generated by S.Now consider the set B(E, F; G) of all bilinear mappings of E x F into

G. By defining the sum of two bilinear mappings p1 and P2 by

('p1 + P2)(x, y) = P1(x, y) + P2(x, y)

and the mapping (hp) by

(A p) (x, y) = p(x, y) x e E, y e F, F,

we can introduce a vector space structure in the set B(E, F; G). The spaceB(E, F; F) of all bilinear functions in E x F will be denoted simply byB(E, F).

1.2. Bilinear Mappings of Subspaces and Factor Spaces

Given a bilinear mapping p : E x F -+ G and a pair of subspaces E 1 c E,F 1 c F, a bilinear mapping p 1: E 1 x F 1 -+ G is induced by

p1(x1, Y1) = P(x1, Y1) x1 e E1, Y1 e F1.

p 1 is called the restriction of 'p to E 1 x F 1.Let E = a Ea and F = F be two direct decompositions of E and F

respectively and assume that for every pair (a, /3) a bilinear mappingpa : Ea x Ffl -+ G is given. Then there exists precisely one bilinear mappinggyp: E x F -+ G whose restriction to Ea x Ffl is In fact, if ?Ca: E -+ Ea andpfl : F -+ Ffl are the canonical projections, define 'p by

'(x, y) = p x y e F.a,13

Then the restriction of 'p to Ea x F13 is p.Now let '1 and '2 be two bilinear mappings of E x F into G whose

restrictions to Ea x F13 are Then it follows that

('p1 - '2)(x, y) = 'p1(x, y) - P2(xry)

= x, p13 Y) - x, p13 Y) = 0a, f3 a, f3

whence Pi = 'p2 .

If E 1 c E and F 1 c F are subspaces and 'pr: E 1 x F 1 -+ G is a bilinearmapping then there exists a (not uniquely determined) bilinear mappingcP : E x F -+ G whose restriction to E 1 x F 1 is 'pr. To prove this choosetwo subspaces E2 c E and F2 c F such that

E=E1QE2, F=F1G F2

Multilinear Mappings 3

and define the bilinear mappings cpi; : E. x F; -+ G (i, j = 1,2) by P 1 1 = Piand cpt; = 0, (i, j) (1, 1). In view of the above remark there exists a bilinearmapping p: E x F -+ G whose restriction to E. x F; is cot;.

Now suppose that p: E x F -+ G is bilinear, and for some subspacesE 1 c: E and G 1 c: G c(x 1, y) E G 1 for every x1 E E 1, y e F. Let p: E -+ E/E 1and it : G -+ G/G 1 be the canonical projections, and define a bilinear mapping

x F -+ G/G1

by

gP(Px, y) = irpCx, y) px e E/E 1, y e F.

It is clear that P is a well-defined bilinear mapping. We say that P is thebilinear mapping induced by gyp.

If also for some subspace F 1 c: F, c(x, y1) E G 1 for each x e E, y1 E F 1,then cp(px, y1) = 0 for each px e E/E 1, y1 E F 1. Denoting the canonicalprojection of F onto F/F1 by a we see that p induces a bilinear mapping

P : E/E 1 x F/F 1 -+ G/G 1

defined by

x i x, xeEE cry eMCP cry) _ C Y) P / 1 Y 1

1.3. Multilinear Mappings

Suppose we are given p + 1 vector spaces Et (i = 1. , p), G. A mappinggyp: E1 x x Ep -+ G is called p-linear if for every i (1 <_ i <_ p)

(p(x 1, ... , x_ 1, ),x1 + µyi, xi + 1, ... , xp) = p(x 1, ... , xt , ... , xp)

+ µup(x1, ..., y1, ... , x,,) x,, yt e E., A,, t e F.

If G = F, then p is called a p-linear function.As in the case p = 2 the subspace of G generated by the vectors

cp(x 1, ... , xp), xt E Et will be denoted by Im gyp. Let L(E 1, ..., E,,; G) be theset of all p-linear mappings p: E 1 x x Ep -+ G. Defining the linearoperations by

(co + I) (x 1, ... , x,,) = (p(x 1, ... , x p) + lU(x 1, ... , x,,)

and

(A p)(x1, ... , xp) = A p(x1, ... , xp)

we obtain a vector space structure in L(E 1, ... , Es,, G). The space of all p-linear functions in E 1 x x Ep will be denoted by L(E 1, ..., E,,).

4 1 Tensor Products

PROBLEMS

1. Establish natural isomorphisms

B(E, F; G) L(E; L(F; G)) L(F; L(E; G)).

2. Given a bilinear mapping (p : E x F --> G, define a mapping li : E x F -- G by

fiz = cp(ir1 z, ire z) z E E x F

where ir1: E x F -- E and ire : E x F --> F are the canonical projections. Show that& satisfies the relations

'Y(z1 + z2) + 'Y(z1 - z2) = 2'&(z1) + 2'&(z2)

and

() _3. Let E and F be arbitrary. Show that the mapping f3: L(E; F) x E -- F defined by

(gyp, x) px is bilinear. Prove that Im f3 = F.

4. Let E, F, G be finite-dimensional real vector spaces with the natural topology.Show that every bilinear mapping p : E x F --> G is continuous. Conclude that themapping L(E; F) x E -- F defined by (gyp, x) --> px is continuous.

5. Given a bilinear mapping qP : E x F -- G define the null-spaces N 1(cp) c E andN2(q,) = F as follows:

N 1(qp) : {x; q (x, y) = 0) for every y e F

and

N2(gp) : {y; q (x, y) = 0) for every x e E

(a) Consider the induced bilinear mapping

P : E/N 1(q,) x F/N2(q) -- G

(cf. Section 1.2). Show that N 1(rp) = 0 and N 2(rp) = 0.Given a linear map f : G -- H consider the bilinear mapping ii : E x F -- H

defined by li(x, y) = f ip(x, y). Show that

N 1(q,) c N and N2(ip) c N2(/i)

(b) Conversely, let ci : E x F -- H be a bilinear mapping such that N 1(ip) c N 1(ci)and N2(gp) c N2(ci). Prove that there exists a linear map f : G -- H such that

li(x, y) = fp(x, y)

Consider the space L of linear maps f : G -- H satisfying this condition. Establish alinear isomorphism

1

L -* L(G/Im gyp; H).

Conclude that f is uniquely determined by ii if and only if Im p = G.

The Tensor Product 5

6. Let E be a vector space and F be the space of all functions h : E -- r. Define a bilinearmapping (p : L(E) x L(E) --> F by

(p(f, 9)(x) = f(x)g(x) x e E.

Show that N1(q) = 0 and N2((p) = 0.

7. Let E, E* be a pair of dual spaces and assume that 4: E* x E -- r is a bilinear func-tion such that

D(T*- lx*, Tx) = D(x*, x)

for every pair of dual automorphisms. Prove that iD(x*, x) = A<x*, x> where A is ascalar.

The Tensor Product

1.4. The Universal Property

Let E and F be vector spaces and let ® be a bilinear mapping from E x Finto a vector space T. We shall say that ® has the universal property, if itsatisfies the following conditions:

®l : The vectors x ® y (x e E, y e F) generate T, or equivalently,Im ®= T.

®2. If p is a bilinear mapping from E x F into any vector space H,then there exists a linear map f:T -+ H such that the diagram

commutes.T

The two conditions above are equivalent to the following single con-dition

®: To every bilinear mapping p: E x F -+ H there exists a uniquelinear map f:T -+ H such that Diagram (1.1) commutes.

In fact, assume that ®1 and ®2 hold and let T -+ H and 12: T -+ Hbe linear maps such that

p(x, y) = fl(x ®x y) and p(x, y) = f2(x ®x y)

Then we have

fi(x8y)=f2(x®y) xEE,yeF.

Now ® 1 implies that f, = f2 and so f is uniquely determined by cp.

6 1 Tensor Products

Conversely, assume that Q holds. Then QZ is obviously satisfied. Toprove ®, let Ti be the subspace of T generated by the vectors x Q y withx e E and y e F. Then Q determines a bilinear mapping gyp: E x F -+ Tisuch that

icp(x, y) = x Q y xeE, yeF,

where i : Ti -+ T denotes the inclusion map. By Q there is a linear mapf : T -- Ti such that

p(x, y) = f (x Q y) x e E, y e F.

Applying i to this relation we obtain

(i°f)(x®y) _ x®yOn the other hand, clearly,

l(x® y) = xQ y xeE,yeFwhere us the identity map of T. Now the uniqueness part of Q implies thati of = 1. Thus i is surjective and so Ti = T This proves Q l .

EXAMPLE. Consider the bilinear mapping F x F -+ F given by Q y = Ay.Since 1 Q y = y this map satisfies condition ®. To verify OZ, let gyp: F xF -+ H be any bilinear mapping and define a linear map f : F -+ H by setting

f(y) _ (p(l, y)Then we have for E F and y e F

y) _ A p(1, y) _ f (y) = f (;y) = f ( O y)

and so OZ is proved.

1.5. Elementary Properties

Before proving existence and uniqueness of bilinear mappings with the uni-versal property we shall derive a few properties which follow directly fromthe definition.

Thus we assume that Q : E x F -+ T is a bilinear mapping with the uni-versal property.

Lemma 1.5.1. Let at (i = 1, ..., r) be linearly independent vectors in E andlet b. (i = 1, ..., r) be arbitrary vectors in F. Then the relation

atQbt = 0

implies that bt = 0 (i = 1, ..., r).


PROOF. Since the at are linearly independent we can choose r linear functionsf t in E such that

f'(a) = b (i, j = 1, ..., r).

Now consider the bilinear functionr

(x, y) _ x e E, y e Fi= 1

where the g` are arbitrary linear functions in F. In view of QZ , there existsa linear function h in T such that

h(x O Y) _ f `(x)g`(Y)t

Then

h a; O b; _ f `(a;)g`(b;) _ g`(bt)

Since L a; Q b; = 0, we obtain

g`(bt) = 0.

But the g` are arbitrary and so it follows that

b= O 0 (i = 1, ... , r). D

Corollary. If a 0 and b 0, then a Q b 0.

Lemma 1.5.2. Let be a basis of E. Then every vector z e T can bewritten in the form

z = > ea Q ba , ba E F,a

where only finitely many ba are different from zero. Moreover, the ba areuniquely determined by z.

PROOF. In view of ®, z is a finite sum

z = x O Yv xv e E, Yv e F.v

Now write

ea JaEF.a

Then we have

where

z = >e®yv = >e®yv = eaObaav, c v, cc

aba = v Yvv

8 1 Tensor Products

To prove uniqueness assume that

ea ® ba = ea ® ba b, ba e F.a a

Then

ea ® (ba - ba) = 0a

and so Lemma 1.5.1 implies that ba = b'.

Lemma 1.5.3. Every nonzero vector z e T can be written in the formr

z = > xl ® yl xleE, yIeF,i=1

where the x, (1, ... , r) are linearly independent and the yl (i = 1. , r) arelinearly independent.

PROOF. Choose a representation z = xl ® yl where r is minimized.If r = 1, it follows from bilinearity that x 1 0 and y1 0.

Now consider the case r > 2. If the vectors x, are linearly dependent wemay assume that

r- 1ixr =

i= 1

Then we haver-1 r-1 r-1 r-1

>x, !!',z = > x1®y+ jL!x1®y,. = ® (y i + `yr) = xi®yii=1 i=1 i=1 i=1

which contradicts the minimality of r. Thus the vectors x, are linearly in-dependent. In the same way it follows that the vectors yl are linearly in-dependent as well.

1.6. Uniqueness

Suppose that ® : E x F -+ T and c : E x F -+ T are bilinear mappingswith the universal property. Then there exists a linear isomorphism f:T Tsuch that

f(x®y)=xy xeE,yeF.In fact, in view of ®2, we have linear maps

f:T-+T and g: T -+ T

such that

f(x ® y) = x y


and

g(x ®y) = x Q y x e E, y e F.

These relations imply that

gf (x O Y) = x ® y and fg(x O Y) = x O Y

Now ®, shows that

gof=i and fig=i.Thus f and g are inverse linear isomorphisms.

1.7. Existence

To prove existence consider the free vector space C(E x F) generated bythe set E x F (see Section 1.7 of Linear Algebra). Let N(E, F) denote thesubspace of C(E x F) generated by the vectors

(x + µx2, y) - (x 1, y) - µ(x2 , y)

and

(x, Y + µ)72) - (x, Y 1) - µ(x, Y2)

Set

T = C(E x F)/N(E, F)

and let it : C(E x F) - T be the canonical projection.Now define a set map ® : E x F - T by

x O Y = ic(x, y).

We shall show that Q is a bilinear mapping and has the universal property.In fact, since

ic( tx 1 + µx2, y) = A r(x 1, y) + µir(x2 , y),

it follows that

(x1 + ,1x2) Q y = Tr(x 1 + µx2, y)= Ait(x 1, y) + µir(x 2 , y) = x1 O y + µx 2 ® y.

In the same way it is shown that Q is linear in y.To prove ®,, observe that every vector z e T is a finite sum

z = t vµ(Yµ) xeE,yµeF.v, µ

It follows that

vµxv 0 yµ yµ) _ it vµ(xv, yµ) = z.v,µ v,µ v,µ

10 1 Tensor Products

To verify 02, consider a bilinear mapping U of E x F into a third vectorspace H. Since the pairs (x, y) x e E, y e F form a basis for C(E x F) thereis a uniquely determined linear map

g:C(ExF)--Hsuch that

g(x, y) = U(x, Y)

The bilinearity of i'/ implies that N(E, F) c ker g. In fact, if

z = (x 1 + µx 2 , y) - (x1, y) - µ(x 2 , y)

is a generator of N(E, F), then

g(z) = g(tx 1 + µx2, y) - 2g(x 1, y) - ug(x2, y)

= U(x 1 + µx2, y) - A,4i(x 1, y) - µ4(x2 , y)

=0.

In a similar way it is shown that

g[(x, Y1 + µY2) - Ax, Y1) - µ(x, Y2)] = 0

and it follows that N(E, F) c ker g. Hence g induces a linear map

f : C(E x F)/N(E, F) -+ H

such that

foj=g.In particular, it follows that

(f ° ®)(x, y) = f ir(x, y) = g(x, y) _ U(x, Y)

This shows that the bilinear mapping Q has the universal property.

Definition. The tensor product of two vector spaces E and F is a pair (T, ®),where Q : E x F -+ T is a bilinear mapping with the universal property. Thespace T, which is uniquely determined by E and F up to an isomorphism, isalso called the tensor product of E and F and is denoted by E Q F.

Now we show that the tensor product is commutative in the sense that

EOF FOE.In fact, consider the bilinear mappings

cp:E x F-+FQE and t/i:F x E-+EQFgiven by

p(x, y) = Y O x and U(y, x) = x Q y.


In view of 02, they induce linear maps f: E Q F -+ F O E and g: F Q E -+E Q F such that

y Q x= f (x Q y) and x Q y= g(y O x)

for all x e E and y e F. These relations imply, in view of Ol, that g o f = iand f o g = t. Thus f and g are inverse isomorphisms.

1.8. Reduction of Bilinear Mappings to Linear Maps

Fix E and F and let G be a third vector space. Then a linear isomorphism

L(EQF;G)=-B(E,F;G)is defined by

1(f) = f° O f e L(E Q F; G).

In fact, Q2 implies that 1 is surjective, since it states that any bilinear map-ping gyp: E x F -+ G may be factored over the tensor product. To show that1 is injective assume that f o Q = 0 for a certain linear map f : E Q F -+ G.In view of Q 1 the space E Q F is generated by the products x Q y and henceit follows that f = 0.

The correspondence between the bilinear mappings p: E x F -+ G and thelinear maps f : E Q F -+ G which is obtained by the above result is expressedby the following commutative diagram:

E x F -- G

Proposition 1.8.1. Let cp: E x F -+ G be a bilinear mapping and f: E Q F -+ Gbe the induced linear map. Then f is surjective if and only if p satisfies OQ l .

Moreoverf is injective [and only if p satisfis ®2.

PROOF. The first part follows immediately from the relation

Imp=ImfTo prove the second part assume that f is injective. Then the pair (Im gyp, 'p)is a tensor product for E and F. Hence every bilinear mapping i'/ : E x F -+ Kinduces a linear map g :Im p -+ K such that

U(x, y) = y)1f f is an extension of g to a linear map f : G -+ K it follows that

/i(x, y) = f p(x, y)

and hence p has the property 02.


Conversely, assume that p satisfies ®2. Then the bilinear mappingE x F -+ E® F induces a linear map h: G -+ E Q F such that

x O y = hto(x, y).

On the other hand, we have p(x, y) = f(x ® y) and it follows that

x®y=hf(x®y).Hence h o f = i and so f is injective.

PROBLEMS

1. Consider the bilinear mapping f3 : r" x rm -- M" x m defined by

(b 1, ... ,x ( 1, ... ,xn 1 .. ,bb bb /

Prove that the pair (M" x m, f3) is the tensor product of r" and rm.n

2. Show that the bilinear mapping r" x E --> QE defined by

(b 1 . ... ,") ® x = ( 1 x, ... ,"x)is the tensor product.

3. Let S and T be two arbitrary sets and consider the vector spaces C(S) and C(T)generated respectively by S and T (cf. Section 1.7 of Linear Algebra). Show thatC(S x T) is isomorphic to C(S) ® C(T).

4. Assuming that a® b 0, a e E, be F prove that

a®b=a'®b'if and only if

a' = A.a and b' = A-1 b A E r, A 0.

5. Let A be a subfield of r and consider a vector space Eo over A. Then r ® Eo is againa A-vector space. Define a scalar multiplication r x (r ® Eo) -- r ® Eo by

A,aer,xeEo(a) Prove that this multiplication makes r ® Eo into a r-vector space E.(b) Show that the restriction of this multiplication to A x Eo coincides with thescalar multiplication in E.(c) If {e2} is a basis of Eo prove that { 1 ® e2} is a basis of E.(d) Let r = C and A = Ft Prove that E is isomorphic to the vector space E x Econstructed in Problem 5, §1, Chapter 1 of Linear Algebra.

6. With the notation of Problem 5 let po be a linear transformation of E.(a) Prove that pr = i is a linear transformation of E.(b) For any polynomial f e r[t] prove that

(.f (q,e))r = .f (qpr)

(c) Find the minimum polynomial of pr in terms of the minimum polynomial of p.(d) Show that pr is semisimple (nilpotent) if po is semisimple (nilpotent) and henceconstruct the decomposition of pr into semisimple and nilpotent parts.

Subspaces and Factor Spaces 13

Subspaces and Factor Spaces

1.9. Tensor Products of Subspaces

Suppose that the bilinear mapping Q : E x F -+ T has the universal propertyand consider two subspaces E 1 E and F 1 F. Let Q' denote the restric-tion of Q to E 1 x F 1 and set T1 = Im Q'. We shall show that (T1, ®') isthe tensor product of E 1 and F 1.

Property Q i is immediate from the definitions. To verify ®2' let P1 : E 1 xF 1 -+ H be a bilinear mapping. Extend P1 to a bilinear mapping p: E x F -+ H.Since O has the universal property, there is a linear map

f:T -+ H

such that

f(x O Y) = p(x, y) x e E, y e F.

This relation implies that

f(x1 0 Y1) = P(x1, Y1) = p1(x1, Y1) x1 E E1, Y1 E 1,

and so P1 factors over 0.

1.10. Tensor Product of Factor Spaces

Again let E 1 E and F 1 c F be subspaces and set

T(E1, F1) = E1 ®F + E O F1Define a bilinear mapping /3: E x F -+ (E Q F)/T (E 1, F 1) by

/3(x, y) = ic(x O Y),

where it denotes the canonical projection.Since /3(x1, y) = 0 if x 1 E E 1, y e F and /3(x, y 1) = 0, if x e E, y l E F 1, /3

induces a bilinear mapping

J3:E/E1 x F/F1 -+ (E Q F)/T(E1, F1)

such that

/3(x, y) = /3(x, y) x E E/E 1, y e F/F 1.

To prove that fi satisfies Q first notice that

Im fi = Im /3 = Im m = (E Q F)/T (E 1, F 1)

and so property Q i follows. To check ®2, let

E/E 1 x F/F 1 -+ H

be any bilinear mapping. Define a bilinear mapping p: E x F -+ H by setting

p(x, y) = (x, Y)


Then there is a linear map f : E Q F -+ H such that

p(x, y) = f (x Q y) x e E, y e F.

Moreover,

f (x 1 0 Y) = p(x 1, y) _ /i(0, j) = 0

and similarly,

x1eE1,yeF

f(x®Y1) = 0 xeE,yeF1.Hence T (E 1, F 1) c: ker f, and so f induces a linear map

f:(E O F)/T(E1, F1) -+ H

such that

fo=fIt follows that

/ (x, j) _ p(x, Y) = f (x O Y) = fir(x O Y)

= f13(x, Y) = ffl(, y) x E E/E 1, y E F/F 1

whence ,/, = f o J3. Thus JJ satisfies condition Q 2 and so the proof is complete.The result obtained above shows that there is a canonical isomorphism

E/E1 0 F/F 1 =- (E O F)/(E 1 0 F+ E O F 1)

PROBLEM

Let E1 and E2 be subspaces of E such that E = E1 + E2 and set E1 n E2 = F. Establishan isomorphism

E/E1 OO E/E2 (E1 Ox E2)/(F OO E2 + E1 OO F).

Direct Decompositions

1.11. Tensor Product of Direct Sums

Assume that two families of linear spaces E, a e 1 and Ffl ,13 E J are given andthat for every pair (a, 13), (Ea Q Ffl , Q) is the tensor product of Ea and Ffl .Then a bilinear mapping p of E = ®Ea and F = O Ffl into the direct sum(= EE a, fl(Ea Q Ffl) is defined by

(p(x, Y) _ i (ica x O p fl Y)a, fJ

where

ica : E -+ Ea , pfl : F -+ Ffl

Direct Decompositions 15

are the canonical projections and

Ea O F -+ G

are the canonical injections. It will be shown that the pair (G, gyp) is the tensorproduct of E and F.

Condition Q 1 is trivially fulfilled. To verify Q 2 let :E x F -+ H be anarbitrary bilinear mapping. Define : Ea x Ffl -+ H by

y) = iU(la x, j y),

where

is : Ea -+ E and j : Ffl -+ F

are the canonical injections. Then induces a linear map fag : Ea Q Ffl -+ Hsuch that

U (x, y) = O Y)

Define a linear map f : G -+ H by

xeEa,yeFf.

f=

where

-' Ea Q Ffl

are the canonical projections. Then it follows that

(f ° 'p)(, j) = f x O P Y)a, fJ

_ x O P Y)a, fJ

= Pfl Y)

=a

= Y)

Hence f ° P = i/i and so 02 is satisfied.

1.12. Direct Decompositions

Assume that the pair (E Q F, Q) is the tensor product of the vector spaces Eand F and that two direct decompositions E = a Ea and F = Ffl aregiven. It will be shown that E Q F is the direct sum of the subspaces Ea Q Ffl,

EOF = >Ea®Ff. (1.2)a, fJ


In view of ®l the space E Q F is generated by the products x Q y;xeE, yeF.

Since x = axa, xa a Ea and y = yp E F it follows that

x®y = xa®x ys.a, Ii

This equation shows that the space E Q F is the sum of the subspacesEa Q F.

To prove that the decomposition (1.2) is direct consider the direct sumsE = Qa Ea, F = Q F and G = pa, Ea Q F and let the injections is, J1,ia and the projections ?Ca, pp, ira be defined as in the previous section. Thenif gyp: E x P -+ G is the bilinear mapping given by

p(i, Y) = iap(ia x O p Y)a, Ii

we have shown (in the previous section) that the pair (G, gyp) is the tensorproduct of E and F.

Now consider the linear isomorphisms

f:E-+E and g:F-+Fdefined by

f x = la xa and g y = >jpYp'a Ii

where

x = xa , xa E Ea and y = Yp, Yp E F.a p

Define a bilinear mapping i/i : E x F -+ G by

/i(x, y) = P(f x, g y)-

In view of the factorization property there exists a linear map h : E Q F -+ Usuch that

h(x O y) = Ji(x, y)and hence

h(x O y) = P(f x, g y).

If x e Et and y e Fa it follows from the definition off, g, and p that

h(x O y) = p(f x, g y) = p(it x, jay)

= l43(ia it x O ppja y) = lta(x 0 y)a, p

But this equation shows that h maps every subspace Ea Q F of E Q F intothe subspace Q of G. Since the decomposition

G = Oa, p

Direct Decompositions 17

is direct, the decomposition

EQF = Ea®Ffa, fi

must be direct. This completes our proof.Conversely, suppose that direct decompositions

E = > Ea , F = > Ffl , G = > Gafla 13 a, f3

and bilinear mappings Q : Ea x F13 -+ Gay are given such that the pairis the tensor product of Ea and F13. Define a bilinear mapping

p : E x F -+ G by

p(x, y) = , xa O Y'

where

x = xa and y = y13.a 13

Then the pair (G, gyp) is the tensor product of E and F.The condition Q 1 is obviously satisfied. To prove 02 let U : E x F -+ H

be an arbitrary bilinear mapping and consider the restriction of U toEa x F13 . Then there exists a linear map fag : Gay -+ H such that O

y13). Define a linear map f : G -+ H by

f (z) = L, f(z)

where z = zap, zap e G.Then

a,f

(f o p)(x, y) = f o(x, y)= Ox y13)

a, fi

= y13) = U(x, y)a, fi

whence f o P = U. Thus 02 is satisfied and the proof is complete.

1.13. Tensor Product of Basis Vectors

Suppose that (aa)a El and (b13) 13 E J are, respectively, bases of vector spaces E andF. Then the products (aa O b13)a El, 1 E J form a basis of E Q F. To prove this,let Ea, F13 denote the one-dimensional subspaces of E and F generated by asand b13 respectively. Then E = Ea, F = in view of the result of theprevious section it follows that

EQF = >Ea®F13.a, 13


Now it was shown in Section 1.5 that as 0, bf 0 implies as Q bf 0.

On the other hand, Q 1 applied to Ea, Ffl and Ea Q Ffl gives that Ea Q Ffl isspanned by the single element as Q bfl . Thus E Q F is the direct sum of theone-dimensional subspaces generated by the products as Q bfl, and hencethese products form a basis of E Q F (see Lemma 1.5.1).

In particular, it follows from these remarks that if E and F have finitedimensions, then E Q F has finite dimension, and

dim(E Q F) = dim E dim F. (1.3)

1.14. Application to Bilinear Mappings

Let E and F be vector spaces with bases (x 3, E I and (yfl) E J respectively. Thensince xa Q y is a basis of E Q F, it follows that every set map of (xa Q intoa third vector space G can be extended in a unique way to a linear map

f:E ® F - G

and every linear map f : E Q F - G is obtained in this way. In view of theisomorphism

L(E Q F ; G) B(E, F ; G),

it follows that every set map

(x«, yfi) - G

can be extended in a unique way to a bilinear mapping p : E x F - G andevery bilinear mapping p is obtained in this way. In particular, the spaceIm 'p is generated by the vectors p(xa, yfl). This result implies that if E and Fhave finite dimension, then

dim Im ' dim E dim F.

It is now easy to construct a basis of the space B(E, F; G) provided that thedimensions of E and F are finite. Let xt (i = 1, ..., n), y; (j = 1, ..., m) and(zy)y E K be bases of E, F, and G respectively. Then the products xt Q y; forma basis of E Q F and hence the linear maps f yl (k = 1, ..., n; l = 1, ... , m;y e K) defined by

f'(xi Ox y;) = 5 5 zy

form a basis of L(E Q F; G). Consequently, the bilinear mappings o givenby

p '(xi , y,) = bkb; zy

form a basis of B(E, F; G).If G has finite dimension as well, it follows that

dim B(E, F; G) = dim E dim F dim G

Direct Decompositions

and so in particular

dim B(E, F) = dim E dim F.

1.15. Intersection of Tensor Products

19

Let E 1 and E2 be two subspaces of a vector space E. Then if F is a secondvector space,

(E1 QF)n(E2QF) = (E1 nE2)QF.

Clearly,

(E1 nE2)OFc (E1 OF)n(E20F)To prove the inclusion in the other direction, let z be an arbitrary vector of(E1 Q F) n (E2 Q F). If (bfl)fE, is a basis of F we can write

z = u Q bfl, ufi E E1 and z = u Q bfl, u E E2.13

This yields

13

(u - v13) x0 b = oe

and since the b13 are linearly independent we obtain u13 = v. Hence u13 EE1 n E2 and z e (E1 n E2) Q F. This completes the proof of (1.4).

Next consider two subspaces E 1 c E and F 1 c F. Then

(E1 O F) n (E O F1) = E1 O F1

To prove (1.5), choose a subspace F' c F such that

F=F10+F'.Then we have in view of Section 1.12 that

E1QF=E1QF1E E1QF'.Intersecting with E Q F 1 and observing that E 1 Q F 1 c E Q F 1 we obtain

(E1 O F) n (E O F1) = (E1 O F1) O+ [(E1 O F') n (E1 O F1)] (1.6)

Now Formula (1.4) yields

(E1 Q F') n (E1 O F1) = E1 Q (F' n F1) = E1 Q 0 = 0

and thus (1.5) follows from (1.6).


Finally let E 1, E2 and F 1, F2 be subspaces respectively of E and F. Then

(E 1 O F 1) n (E2 O F2) _ (E 1 n E2) O (F 1 n F2)

Clearly,

(E1 n E2) OO (F1 n F2) C (E1 Q F1) n (E2 Q F2).

Moreover, we have in view of (1.4) that

(E1 ® F1) n (E2 O F2) c: (E1 O F) n (E2 O F) _ (E1 n E2) O F.

In the same way it follows that

(E1 ®F1) n (E2 O F2) c: (E O F1) n (E O F2) = E 0 (F1 n F2).

Now formula (1.5) yields

(E 1 O F 1) n (E2 O F2) (E 1 n E2) O (F 1 n F2)

This completes the proof of (1.7).

PROBLEMS

1. Let z e E O F, z 0, be any vector. Show by an explicit example that in general thereexist several representations of the form

z = x Qx y x E E, y E F

where the and the are respectively linearly independent. Given two suchrepresentations

r s

z= >x®y and z= xJOyJi=1 j=1

prove that r = s.

2. Let p : E x F --> G be a bilinear mapping. Show that the following property is equiv-alent to ®2 Whenever the vectors xa E E and yp e F are linearly independent then soare the vectors cp(xa, yp).

3. Let A 0 be an algebra of finite dimension and assume that the pair (A, /3) is a tensorproduct for A and A where /3 denotes the multiplication. Prove that dim A = 1.

4. Let E be an arbitrary vector space and F be a vector space of dimension m. Establisha (noncanonical) isomorphism

EE ...Q+ E=+EQx F.m

5. Let E, E* and F, F* be two pairs of dual spaces of finite dimension. Consider thebilinear mapping

fl: E x F -- B(E*, F*)

Linear Maps

given by

IJ,y(x*, Y*) _ <x*, x><Y*, y>

Prove that the pair (B(E*, F*), /3) is the tensor product of E and F.

Linear Maps

1.16. Tensor Product of Linear Maps

Given four vector spaces E, E', F, F' consider two linear maps

cP : E -- E' and t/i : F -- F'.

Then a bilinear mapping E x F -+ E' Q F' is defined by

(x, y) -' cpx Q U y.

In view of the factorization property there exists a linear mapx:EQF-+E'®F'

such that

21

x(x O y) _ cpx O ty (1.8)

and this map is uniquely determined by (1.8). The correspondence ((p, U) -+ xdefines a bilinear mapping

f3: L(E; E') x L(F; F') -+ L(E Q F; E' Q F').

Proposition 1.16.1. Let L(E; E') Q L(F; F') be the tensor product of L(E; E')and L(F; F'). Then the linear map

f:L(E;E')QL(F;F')-+L(EQF;E'QF')induced by the bilinear mapping fi is injectiue.

PROOF. Let w be an element such that f (w) = 0. If w 0 we can writer

w= >Jp® Pi E L(E, E'), thii E L(F, F'),i= 1

where the co and i/ii are linearly independent.Then

f(w) = f3((p, th'1)

and hence f (w) = 0 implies thatr

q (x) O i(y) = 0i= 1

for every pair x e E, y e F. Now choose a vector a e E such that p 1(a) 0.


Let p >_ 1 be the maximal number of linearly independent vectors in theset p 1(a), ... , cpr(a). Rearranging the cot we can achieve that the vectorsP 1(a), ... , p ,,(a) are linearly independent. Then we have

P

= a) J = p + 1, ... , r

and Relation (1.9) yields

i=1 j=p+ 1 \i=1

i= 1

p r p

(a) O i(Y) + ji Pi(a) O (Y) = 0 yeF;

p r

(a)® (Y) + Y' i(Y) = 0 y E F.i=1 j=p+ 1

Since the vectors pi(a) (i = 1, ... , p) are linearly independent it follows that

r

i(Y) + Y) = 0 i = 1, ... , pj=p+ 1

for every y e F, i.e. i/ i + = p + 1 = 0. This is in contradiction to ourhypothesis that the maps l//; are linearly independent and hence f is injective.

Corollary I. The pair (Im /3, l3) is the tensor product of L(E; E') and L(F; F').

Corollary II. The bilinear mapping /3: L(E) x L(F) - L(E Q F) given by

fl(f, g)(x O Y) = f(x)g(y)

is such that (Im /3, /3) is the tensor product of L(E) and L(F).

Corollary III. If E and F have finite dimension the elements f3(cp, U) generate thespace

L(EQF;E'QF)as will be shown in Section 1.27 and so the pair (L(E Q F; E' Q F'), /3) is thetensor product of L(E; E') and L(F; F').

In general, however, this is not the case as the example below will show.Nevertheless, by an abuse of language, we call U) the tensor product ofthe linear maps p and l// and write U) = p Q U. Then formula (1.8)reads

('p O U)(x O Y) = 'x O UY

In particular, if E = F = E' = F' and 'p = ui = t, then t Q l = 1.

Linear Maps 23

1.17. Example

Let E = F be a vector space with a countable basis and put E' = F' = F.Then we have

L(E; E') = L(F; F') = L(E), L(E Q F; E' Q F') = L(E Q E)

and the bilinear mapping /3 is given by

f3: (f g) -'f . g,where

(f . g) (x O y) = f (x)g(y)

It will be shown that /3 does not satisfy the Condition Q l .We associate with every linear function h in E Q E a subspace Eh c: E

(called the nullspace of the corresponding function) in the following way:A vector xo E E is contained in Eh if and only if h(xo Q y) = 0 for every y e E.Now assume that h e Im /3. Then the factor space E/Eh has finite dimension.In fact, h can be written as a finite sum of the form

r

h(x O y) _ f (x)gt(y)-i= 1

It follows that for any x e n ker f we have

h(x O y) = 0 y E F;

i.e.,r

Eh D flker.i= 1

Now a result of Section 2.4 of Linear Algebra implies that dim E/Eh < r.On the other hand, consider the linear function w on E Q E given by

w(xx0

y) _ cv11v

V

where c and iv (v = 1, 2, ... ) are the components of x and y with respect to abasis of E. It is easy to see that the nullspace Ew = 0 and hence dim E/Ew =dim E = oo. Consequently, w is not contained in Im /3.

1.18. Compositions

Consider four linear maps

p : E --> E' gyp' : E' --> E"

F -+ F' t/i' : F' -+ F".

Then it is clear from the definition that

((P' O /J') 0 ((P O IJ) _ ((P' 0 (p) O (c/i' 0 i/i) (1.10)


Now assume that 'p and U are injective. Then there exist linear maps 0: E' -p Eand ,J: F' -+ F such that

(po(p=i and iiol'/=i.Hence, formula (1.10) yields

(® ° (' O i/i) = (o gyp) O (c/iO i/i) = p _showing that (p Q U is injective.

1.19. Image Space and Kernel

It follows immediately from the definition of (p Q U that

ImcppImiji.

In particular, if 'p and U are surjective, then so is 'p Q U. Now the formula

ker((ppU)=kercp® F+EpkerU

will be established. Consider the induced injective linear maps

i:E/ker cp -> E' and F/ker t/i -> F'.

Thenrp®E/ker'QF/keri//- E'QF'

is injective as well (Section 1.18). Let

m, :E -+ E/ker gyp, ir2 : F -+ F/ker U,

and

(1.12)

U)

T(ker gyp, ker U) = ker 'p Q F + E Q ker 1i

be the canonical projections. According to Section 1.10 there exists a linearisomorphism

g: E/ker cp Q F/ker 1i E p F/T(ker p, ker U) (1.13)

such that

9(irx O it2 Y) = ir(x O Y)

Now define a linear map

x = (i O ) ° 9 -'

Linear Maps 25

Clearly x is injective. Moreover, if x e E and y e F are arbitrary we obtain

(x ° ir)(x O y) = (iii O /)g -1 g(irx O ice y)

= (pic x O i2 y

= (pXQi/iy

whence

x is injective, it follows that

ker((p Q U) = ker Tr = T(ker (p, ker U) = ker p Q F + E Q ker U.

PROBLEMS

1. Consider two linear maps (p : E - E' and /i : F - F'.(a) Prove that p Q i/i is injective if and only if both mappings (p and /.i are injective.(b) Assume that E and F have finite dimension. Prove that

r(q Qx /1) = r(ip)r(i/i)

where r denotes rank (see Section 2.34 of Linear Algebra).

2. Let E, F be two vector spaces of dimension n and m respectively and let p : E -p E,/i : F - F be two linear transformations. Prove that

tr(q Q /i) = tr p tr /1

and

det(gp Q Ii) = (det q)m(det /,)n.

3. Consider a vector space E of dimension n. Given two linear transformations a and /3of E let

L(E ; E) - L(E ; E)

be the linear transformation defined by

Q = a o Q o/ Q E L(E; E).

Show that

tr D = tr a tr /3

and

det D = det(a o f3)".

4. Let E and F be two finite-dimensional vector spaces and E', F' be arbitrary vectorspaces. Prove that the bilinear mapping

j3: L(E; E') x L(F; F') -> L(E Q F; E' Q F')

defined by l(q, 1i) = P Qx i/i is a tensor product.


5. Let E be a finite-dimensional vector space and consider two commuting linear trans-formations (p, /i of E. Prove that

((P OO ')s = GPs OO 1s

and

(APO I)N=BPSOx'N+cPNOx 1N

(cf. Section 13.24 of Linear Algebra). Conclude that the tensor product of commutingtransformations is semisimple if and only if both transformations are semisimple.

Tensor Product of Several Vector Spaces


Let Et (i = 1, ... , p) be any p vector spaces and let

Q:E1 x...xEp-Tbe a p-linear mapping. This mapping is said to have the universal property ifit satisfies the following conditions:

Q l : The vectors x 1 ® O xp, (x, e Et) generate T.®2: Every p-linear mapping p: E 1 x x E p -p H (H any vector space)

can be written in the form

p(x1, ... , x,,) = f (x 1 0 ... ® x,,)

where f : T -p H is a linear map.

The existence and uniqueness theorems are proved in the same way as inthe case p = 2 and we shall not repeat them.

Definition. The tensor product of the spaces Et (i = 1, ... , p) is a pair (T, Q)where Q : E1 x x Ep -p T is a p-linear mapping with the universalproperty. T is also called the tensor product of the spaces Et and is denotedby E 1 O ... O Ep .

If H is any vector space, then the correspondence 'p -+f expressed by thecommutative diagram

E1 x xEp- H

E1O...OEp

determines a linear isomorphism

L((E 1Q ... O E r); H) 4 L(E 1, ..., E; H).

Tensor Product of Several Vector Spaces 27

Proposition 1.20.1. Given three arbitrary spaces E 1, E2, E3 there exists a linearisomorphism

f:E1 OE2 OE34(E1 OE2)OE3such that

f(x®y®z) = (x®x y)Qx z.

PROOF. Consider the trilinear mapping

E 1 x E2 x E3 - (E1 0 E2) O E3

defined by

(x, y, z)-(x®y)Oz.In view of the factorization property, there is induced a linear map

f:E1 OE20E3 -(E1 OE2)OE3such that

f (x Q y Q z) = (x xQ y) xQ z. (1.14)

On the other hand, to each fixed z e E3 there corresponds a bilinear mappingI3Z : E 1 x E2 - E 1 Q E2 Q E3 defined by

IZ(x, y) = x Q y Q z.

The mapping f3Z induces a linear map

gZ:EIOE2-E1OE20E3such that

gZ(x®y) = x®y®z.Define a bilinear mapping

fr: (E 1 0 E2) x E3 - E 1 0 E2 O E3

by

(1.15)

,&(u, z) = gZ(u) u e E 1 Q E2, z E E3 . (1.16)

Then /i induces a linear map

g:(E1 OE2)OE3-E1 OE20E3such that

U(u, z) = g(u Q z) u e E 1 Q E2, z e E3. (1.17)

Combining (1.17), (1.16), and (1.15) we find

g((xOy)Oz) = U(xOy,z) = gZ(xOy) = xQyQz. (1.18)

Equations (1.14) and (1.18) yield g f (x O y 0 z) = x O y O z andfg((x O y) O z) = (x O y) O z showing that f is a linear isomorphism ofE 1 Q E2 Q E3 onto (E 1 Q E2) Q E3 and g is the inverse isomorphism.


In the same way a linear isomorphism h : E 1 Q E2 Q E3 --> E 1 Q (E2 Q E3)can be constructed such that

h(x Q y Qx z) = x 0 (y O z).

Hence, h o f -' is an isomorphism of (E1 Q E2) Q E3 onto E1 Q (E2 Q E3)such that (x®y)QzHx®(y®z).

More generally, if Et (i = 1. , p + q) are p + q vector spaces then thereexists precisely one vector space isomorphism

f:(E1 ...Ep) (Ep+1 O ... O Ep+q) E1 O ... O Ep+q

such that

f ((x 1 ®... ® xp) ® (xp + 1 ... O x,+)) = x 1 ®... ® x p + q -

It follows that there is a uniquely determined bilinear mapping

f3:(E1 Q - - - Q Ep) x (Ep+ 1 O ... Q Ep+q) -+ E1 O ... O Ep+q

such that

fJ(xl ®" . ® xp, xp+ 1 ® .. ® xp+q) x1 Qx . Qx xp+q

and that (E 1 O O Ep + q , /3) is the tensor product of E 1 ®... O Ep andEp+ 1 ® ... ® Ep+q-

The theory developed for the case p = 2 carries over to the general casein an obvious way, and the reader will have no difficulty in making (andproving) the generalization himself. In particular, he should verify that ifais a basis for Et (i = 1, ... , p) then the products ail Q - Q a p form abasis for E1 O - - O Es,, and then generalize the results of Section 1.14,obtaining in the finite-dimensional case that

dim L(E1, ... , Ep; G) = dim E1 - - dim G

dim L(E 1, ..., E,,) = dim E 1 - - dim E.

The tensor product of several linear maps can be defined in the same wayas in the case p = 2. If Bpi : Ei -+ Fi (i = 1, ... , p) are linear maps then thereexists precisely one linear map

such that

x(xl ® ... ® xp) = (Plxl ®... O opxp xt e E,.

As in the case p = 2 we shall write x = P 1 Q - .. Q 'pr. As in that case, themapping (p1, ..., cP p) H P 1 Q Q cop induces an injection

L(E1; F1) O ... O L(Ep; Fp) -+ L(E1 O ... O Es,; F1 O - . O Fr).

If

Gi (i = 1, ... , p)

Dual Spaces 29

is another system of linear maps it follows from the above definition that

('l1 ®---®'l p)°(®---® i ) = ('l1° 1)®'--®(4ip°

An argument similar to the one given for p = 2 shows that

O---® pp)= Q...QIm(pP

and

P

ker(rpi O - - O Pp) _ Ei O - - - ker Pt Q O Ep-i= 1

PROBLEM

Let EL be p vector spaces.(a) Prove that x 1 Q Q xp = 0 if and only if at least one x = 0.(b) Assuming that x 1 O . O xp 0 prove that

x1O...Oxp=y1O...Oyp

if and only if y _ (i = 1, ... , p) and Al Ap = 1.

Dual Spaces

1.21. Bilinear Mappings

Let two triples of vector spaces E, E', E" and F, F', F" be given and considertwo bilinear mappings

p : E x E' -+ E" and tai : F x F' -+ F".

Then there exists precisely one bilinear mapping

x:(E®F) x

such that

x(xOy,x'Oy')_ co(x,x')O'l,(y,y') xeE, x'EE', yeF, y'EF'. (1.19)

Since the spaces E Q F and E' Q F' are generated by the products x Q yand x' Q y' respectively it is clear that if x exists it is uniquely determined byp and ,,Ii. To prove the existence of x consider the linear maps

f:E®E'-E" and g:FQF'-F"induced by p and /, respectively. Then f Q g is a linear map of (E Q E') Q(F Q F') into E" Q F". Now let

S:(EQF)O(E'OF')4(EOE')O(FOF')


be the linear isomorphism defined by

S:(x®Y)O(x'OY')-+(x®x')O(YOY')

and define a bilinear mapping x by

x(u, u) _ (f ® g)S(u ® u) u e E ® F, u e E' ® F'.

Then it follows that

x(x®Y, x'®Y') _ (f®g)((x®x')®(Y®Y'))=f(x®x')®g(Y®Y')_ P(x, x') ® U(Y, Y')

We shall denote x by p ® U, the justification for this notation being given inproblem 1 at the end of Section 1.26.

1.22. Bilinear Functions

In particular, every pair of bilinear functions 1 and 'P in E x E' and F x F'induces a bilinear function 1 ® 'P in (E ® F) x (E' ® F') such that

(I ® Ì')(x ® Y, x' ® Y') _'(x, x')Ì'(Y, y').

We shall show that 1 ®'P is nondegenerate if and only if and 'P are bothnondegenerate.

Consider the linear maps

q: E -+ L(E') l//: F -+ L(F') x : E ® F -+ L(E' ® F')

which are determined by

Pa x' _ I (a, x') IbY' _ t'(b, y') x z' _ (I ® Ì') (c, z') (1.20)

(Here sp(a), fi(b), and x(c) are denoted by Pa' 'i6, and xe). Then we havep ® U : E ® F -+ L(E') ® L(F'). On the other hand, L(E') ® L(F') may beconsidered as a subspace of L(E' ® F') (Corollary III to Proposition 1.16.1).

It will be shown that

(1.21)

where i denotes the injection of L(E') ® L(F') into L(E' ® F'). By definitionwe have

('P ='Pa ® I'6 aEE,bEFand thus

1('P ® tIJ)a ®6(x' ® Y') _ 'Pa x' ' b Y' _ 't'(a, x')Ì'(b, y').

On the other hand it follows from (1.20) that

xa ®6(x' ® Y') _ ('Ia ® Ì') (a ® b, x' ® Y') _ 't'(a, x')Ì'(b, y')

Dual Spaces

and so we obtain

l(p ® 41)a®b = xa®b aeE,beF

whence (1.21).Since i is an injection, it follows from (1.12) that

31

ker x = ker p ® ker SIi.

Hence, the nullspaces NE(I), NF(tP), and NE ®F(' ®'F) (see Section 2.21of Linear Algebra) are connected by the formula

NE ®F('t' ® "F) = NE(B) ® F + E O NF('f). (1.22)

In the same way it is shown that

NE, ®F,(t' ® "F) = F' + E' ® NF-("F). (1.23)

Formulas (1.22) and (1.23) imply that 1 ® "F is nondegenerate if and onlyif 1 and "F are nondegenerate.

Suppose now that E*, E and F*, F are two dual pairs and let both scalarproducts be denoted by <p>. Then the above results show that there existsprecisely one bilinear function <,> in E* ® F*, E ® F such that

<x* O Y*, x O Y> = <x*, x> <Y*, y> (1.24)

and this bilinear function is again nondegenerate. In other words, if E*, Eand F*, F are dual pairs, then a duality between E* ® F* and E ® F isinduced.

Next, consider the special case F = E* and F* = E. Then we obtain ascalar product in the pair E* ® E, E ® E* defined by

<x* ® x, y ® y*> = <x*, Y> <Y*, x>.

Since the mapping x* ® x F-* x ® x* is an isomorphism of E* ® E ontoE ® E*, this scalar product determines a scalar product in the pair E ® E*,E ® E* given by

<x O x*, Y ®Y* = <x*, . ,> 3,*, x>. (1.25)

Hence the space E ® E* can be considered as dual to itself. It is clear, more-over, that this scalar product is symmetric.

Now suppose that E*, Et (i = 1, ... , p) are pairs of dual vector spacesand let all the scalar products be denoted by <,>. As in the case p = 2 a scalarproduct is induced between E1 ® ® Ep and E1 ® ® Ep such that

<x* 1 ® ... ® x*p, x 1 ®... ® X,> = <x* 1, X1> ... <x*p, xp>,

32

1.23. Dual Mappings

1 Tensor Products

Let Et, E* and Ft, F* (i = 1,2) be four pairs of dual vector spaces, and let

co:E1-+E2 q*:Ei E2

and

i/i : F 1 -+ F2 ,* : F i -- F2

be two pairs of dual mappings. Then the mappings

p®iji:E1 ®F1 -'E20F2

and

are dual with respect to the induced scalar products.In fact, let x 1 E E 1, y 1 E F 1, 4 E E, and y2 E F2 be arbitrarily chosen

vectors. Then we have

<4 O y, (px 1 O 'Y 1> = <4 (px 1> <YZ , '/'Y 1

= <(p*x2 xi><I1f*YZ, Y1

= <cP*x2 O U*YZ , xi O Y1

whence

((p O /j)* _ (p* O /j*.

1.24. Example

Consider the dual spaces E* = L(E) and F* = L(F). Then the inducedscalar product in L(E) Q L(F) and E Q F is given by

<f 0 9, x 0 Y> = f(x)9(y).

On the other hand, the space L(E Q F) is dual to E Q F with respect to thescalar product

< h, x Q y> = h(x Q y) h e L(E Q F). (1.26)

Now consider the injection

i : L(E) Q L(F) -+ L(E Q F)defined by

i(f O 9) (x O Y) = f(x)9(y).

Then formulas (1.26), (1.27), and (1.24) yield the relation

< i(f O 9), (x O )> = f(x)g(y) = <f O 9, x O Y

showing that the injection i preserves the scalar products.

(1.27)

Dual Spaces

1.25. Inner Product Spaces

33

An inner product in a vector space E is a nondegenerate symmetric bilinearfunction ( , ) in E. So in particular, every Euclidean space is an inner productspace.

Now let E and F be inner product spaces and denote both inner productsby (, ). In view of Section 1.22, there is precisely one bilinear function ( , ) inE Q F satisfying

(x1 Ox Y1 x2 O Y2) = (x1, x2) . (Y1' Y2)

Clearly, this rbHinear function is again symmetric. Moreover, it is non-degenerate as follows from Section 1.22. The inner product space E Q Fso obtained is called the tensor product of the inner product spaces E and F.

Now assume that E and F are Euclidean spaces of dimensions n and mrespectively. Choose orthonormal bases a (v = 1, ..., n) and bµ (u = 1, ... , m)in E and F. Then we have

(a Q bµ, a Q bK) = Sv S/lK .

Thus the products a O bµ form an orthonormal basis of E Q F. In particular,E Q F is again a Euclidean space.

1.26. The Composition Algebra

Let E*, E be a pair of dual vector spaces. Define a multiplication in the spaceE* Q E by setting

(x* O x) ° (Y* O Y) = <x*, Y>(Y* O x)

It is easy to verify that this multiplication makes E* Q E into an associative(noncommutative) algebra called the composition algebra.

Next, consider the linear map T : E* Q E - L(E; E) given by

T(a* Q b)x = <a*, x>b x E E.

Since

T [(ai O b1) ° (a2 O b2)] = T(ai O b1) ° T(a2 O b2),

T is an algebra homomorphism. We show that T is injective. In fact, assumethat T(z) = 0, z e E* Q E. Choose a basis {ej of E. Then z is a finite sum

r

z = a* Q e where a* E E*v= 1

(see Lemma 1.5.2). Then, for every x e E,

r

<a*, x>e,, = 0

34

whence

1 Tensor Products

a *, x> = 0 x E E.

This implies that a* = 0 (v = 1, ..., r) and so z = 0. Hence T is injective.For the further discussion we distinguish two cases:

Case 1. dim E < oo. Set dim E = n. Then dim(E* Q E) = n2 = dim L(E; E)and so T is a linear isomorphism (since it is injective). Thus, = T -'(l) isthe unit element of the composition algebra. It is called the unit tensor of E.

To obtain a more explicit expression for the unit tensor let {e*v},(v = 1, ..., n) be a pair of dual bases for E* and E and consider the element

i e* Q Then we have

n n

T e*v Q e) (x) = x>e = x x E E,v= 1 v= 1

whence

n

Oev=.

v=1

In particular, the sum e* Q e is independent of the choice of the dualbases {e*v},

Next observe that the spaces E* Q E and L(E; E) are self-dual with respectto the scalar products given by

<x* Qx x, y* OO Y> = <x*, y> <Y*, x>

and

a, i> = tr(a o /3) a, /3 E L(E ; E)

respectively. A simple calculation shows that

< T(x* Q x), T (Y* O y)> = x* ® x, y* ® Y>

and so T preserves the scalar products.

Case 2. dim E = oo. Then the composition algebra does not have a unitelement. To prove this, assume that a is a unit element of E* U E. Let {ejbe a basis of E. Then a is a finite sum

r

e = a* Q e with a* E E*.

For x* E E* and x e E,

v= 1

r

e o (x* xQ x) = (a* x0 o (x* O x)v= 1

r r

_ <a* x> (x* O ev) _ >2v(x* O ev)v= 1 v= 1

Finite-Dimensional Vector Spaces

where V = <a*, x>. Since a is a unit element, it follows thatr

v(x* O ev) = x* O x,

whence

v= 1

r

x =v= 1

35

Thus the vectors e1, ..., er generate E and so E has finite dimension.It follows from the result above that, whenever dim E = oo, the linear

map T is not an isomorphism and hence not surjective.

PROBLEMS

1. Given six vector spaces E, E', E" and F, F', F" consider the bilinear mapping

y:B(E,E';E") x

defined by

/i) : (x O Y, x' O Y') H q (x, x') O /i(Y, y').

Show that the pair (Im y, y) is the tensor product of B(E, E'; E") and B(F, F'; F").

2. Let E, E* and F, F* be two pairs of dual vectors spaces and consider subspacesE 1 E and F 1 c F.(a) Show that a nondegenerate bilinear function is induced in

(E* Qx F*)/T(Ei, F±) and E1 Ox F1,

where

T(Ei, Fi) = Ei Ox F* + E* Ox Fi

by the scalar product in E* Q F*, E Q F.(b) Prove the relation

(E1 Ox F1)-L = Ei Ox F* + E* Ox Ff.

3. Given a pair of dual transformations q : E -- E, p* : E* E* prove that the lineartransformation (p Q (p* is self-dual.

Finite-Dimensional Vector SpacesFor the remainder of this chapter all vector spaces will be assumed to havefinite dimension.

1.27.

Let E and F be vector spaces of dimension n and m respectively. Then E Q Fhas dimension nm (see Section 1.13).


Proposition 1.27.1. Let gyp: E x F -+ T be a bilinear mapping where dim T =nm. Then conditions ®i and ®2 for p are equivalent.

PROOF. Consider the induced linear map

f : E ®F -+ T.

Then, if 'p satisfies ®l, f is surjective, (Proposition 1.8.1). Since dim T =nm = dim(E ® F), it follows that f is an isomorphism and so 'p satisfies

®2.

On the other hand, if 'p satisfies ®2, then f is injective (Proposition 1.8.1),and hence it must be a linear isomorphism. Thus 'p satisfies ®l.

Next, let p: E -+ E' and l//: F -+ F' be linear maps where dim E' = n'and dim F' = m'. It has been shown in Section 1.16 that the bilinear mapping

f3:L(E;E') x L(F;F`)-+L(EQF;E'QF')

given by 'p x i/i -+ 'p ® i/i satisfies ®2. In the finite-dimensional case wehave

dim L(E Q F; E' ® F') = (nm) (n'm') = (nn') (mm')= dim L(E ; E') dim(F ; F').

Hence, by the proposition above, /3 satisfies ®1 as well. Thus /3 has the uni-versal property and we may write

L(E ®F; E' ®F') = L(E; E') ®L(F; F').

This yields for E' = F' = r

L(E ® F; r) = L(E; r) ® L(F; r),

that is,

(E ® F)* = E* ® F*.

Thus the tensor product of linear functions f and g in E and F is the linearfunction in E ® F given by

(f O g)(x O y) = f (x) . g(y) x e E, y e F.

1.28. The Isomorphism T

Let E* be a dual space of E and consider the linear map

T : E* ®F-+ L(E; F)

given by

T(a* ® b)x = <a*, x>b x e E.

Finite-Dimensional Vector Spaces 37

We show that T has the universal property. In fact, we have the commutativediagram

E* Q F

L(E ; F)

L(E) Q L(r; F)where a: E* - L(E) is the canonical isomorphism, /3F -+ L(r; F) is theisomorphism given by f3() = y, y e F, e r, and O is the map defined inSection 1.16. Since the bilinear mapping Q has the universal property(Proposition 1.27.1), the same is true for T. Thus we may identify E* Q F withL(E; F) under T.

A straightforward computation shows that

/joT(a*®b)= T(a* Q t/ib) i/i e L(F; E)

and

T(a* Q b) o = T(t/i*a* Q b) U e L(F; E).

In particular, if a e E, a* e E*, b e F, b* e F*, then

T(b* O a) o T(a* Q b) = <b*, b>T (a* Q a). (1.28)

Finally, consider the trace form tr : L(E, F) x L(F, E) -+ r given by(gyp x U) -+ o U). We show that the operator T satisfies

tr(T(b* O a) o T (a* O b)) = <a* Q b, b* Q a> = <a*, aX b*, b>. (1.29)

In fact, Formula (1.28) yields

tr(T(b* Q a) o T(a* Q b)) = <b*, b>tr T(a* Q a).

But since the linear map T(a* O a) is given by T(a* Q a)x = <a*, x>a,we have

tr T(a* O a) = <a*, a> (1.30)

and so (1.29) follows.Formula (1.29) implies in particular, that the trace form is nondegenerate

(cf. Section 1.22).

1.29 The Algebra of Linear Transformations

To simplify notation we use the isomorphism T of the last section to identifya* Q a with the corresponding linear transformation.

Consider the associative algebra A = A(E; E) of the linear transforma-tions p : E --> E. A linear map cZ : A Q A -+ L(A ; A) is defined by (a, /3) HcZ(a Q /3) where cZ(a Q /3) is the transformation defined by

cZ(oc O /3)(P = a ° P ° /3 (1.31)


Proposition 1.29.1. cZ is an isomorphism.

PROOF. We recall that the space A is dual to itself with respect to the traceform <U, p> = tr(1i o gyp). Now 1'et F : A x A -+ L(A ; A) be the linear mapdefined by

F(a O <a, p>/3

Then the mappings F and cZ are connected by the relation

S = F o Q, (1.32)

where Q is the linear automorphism of L(A Q A) defined by

Q((a * O a) O (b* O b)) = (a * O b) O (b* O a)

To prove (1.32) it is sufficient to show that

O a) O (b* O b)) = F((a* O b) O (b* O a)). (1.33)

Let p : E -+ E be an arbitrary linear transformation. Then we have, in viewof the results of Section 1.28, that

O a) O (b* O (a* O a) ° (P ° (b* O b) = (a* O a) ° (b* O cob)= <a*, cpb>b* Q a

= <<p*a*, b>b* Q a

and

F((a* O b) O (b* O a))co = <a* O b, p>b* Q a= <<p*a*, b>b* Q a

whence (1.33).In view of Section 1.28, F is a linear isomorphism. Since Q is a linear

automorphism of A Q A, relation (1.32) implies that S is an isomorphism.

Corollary. Suppose c and f3. (i = 1, ..., r) are elements of A such that the care linearly independent. Then the relation

al P f3i = 0 for every (p E A

implies that f3i = 0 (i = 1, ..., r).

1.30. The Endomorphisms of A

Every linear automorphism a of E induces an endomorphism ha 0 of thealgebra A given by

hp =ao(poa-'.It will now be shown that conversely, every nonzero endomorphism of thealgebra A is obtained in this way. In other words, every endomorphism

Finite-Dimensional Vector Spaces 39

h 0 of the algebra A can be written in the form hcp = a ° P ° x' where ais a regular linear transformation of E.

Since the pair (L(A; A), SZ) is the tensor product of A and A, we can writer

hcp= at°cp°fJ1 ai,f1EA,i=1

where the ai and the /3i are linearly independent.Now the relation

h(cp ° U) = hcp ° hi/i

implies that

°p°(i/i°f3i - fi°j°i/j°fj) = 0.i j

Since the ai are linearly independent and p is an arbitrary element of Ait follows that (see the Corollary to Proposition 1.29.1)

- f3i°a;°l//°i3. = 0

This can be written in the form

- fi°a.)°ii°fj = 0

(i= 1, ... , r).

(1= 1, ... , r).

Now the linear independence of the elements fl; implies that

For j i we obtain from (1.34)

(i,j = 1,...,r). (1.34)

and for j = i

fi ° ai = 1.

These relations are compatible only if r = 1. Denoting a1 by a we obtain

hcp = ao (p°a-'.

It is easy to show that the element a is uniquely determined by h up toa constant factor.

Our result shows, in particular, that every endomorphism h 0 of thealgebra A preserves the scalar product <iU, 'p> = tr(i1i ° gyp). In fact, for everytwo elements (p, l// E A we have

<h(, hi/i> = tr(h( ° h U)

= trh(cp°I,I/) = tr(a°(°I,Ii°a-')

= tr((° iU) _

whence

<'p,'/'> (p, I/(E A.


PROBLEMS

1. Let a* E E* and b e F be two fixed vectors and consider the linear maps a* Q b : E --> Fand b Q a * : E* F*. Prove that

b Q a* _ (a* Q b)*.

2. Let E, F be Euclidean spaces and consider the induced inner product in E Q F. Giventwo linear transformations p : E -+ E, : F -+ F; prove that(a) p Q i/i is a rotation if and only if p = AtE and _ A-1iF where 'CE and cF arerotations of E, F and A 0 is a real number.(b) p Q i/i is selfadjoint if and only if both transformations p and l.i are selfadjoint orskew.(c) p Q i/i is skew if and only if precisely one of the transformations is selfadjoint andthe other one is skew.(d) p Q /i is normal if and only if both transformations are normal.

3. Let E be a real n-dimensional vector space and consider two regular transformationsp and /i of E. Given an orientation in E Q E prove that(a) if n is even then p Q Ii preserves the orientation.(b) if n is odd, then p Q /i preserves the orientation if and only if both mappings pand c1' are orientation preserving or orientation reversing.

Tensor Products of Vector Spaceswith Additional Structure

Tensor Products of Algebras

2.1. The Structure Map

If A is an algebra, then the multiplication A x A -> A determines a linearmap µA:A Qx A -> A such that

µA(x O Y) = xy. (2.1)

µA is called the structure map of the algebra A. (In this chapter the symbol µAwill be reserved exclusively for structure maps. Since such a notation appearsin no other chapters, there is no possibility of confusion.) Conversely, if A is avector space and µA: A Q A -p A is a linear map, a multiplication is inducedin A by

xY = µA(x O Y) (2.2)

and so A becomes an algebra. The above remark shows that there is a 1-1correspondence between the multiplications in A and the linear mapsµA:AOA-A.

Now let B be a second algebra and µB : B Q B -+ B be the correspondingstructure map. If gyp: A -+ B is a homomorphism we have

(ThµA(x O Y) = I B((PX O SPY) = I B((P O pXx O Y)

whence

'P ° µA = µB ° ('P O 'P) (2.3)

Conversely, every linear map gyp: A -p B which satisfies this relation is ahomomorphism.

41

42 2 Tensor Products of Vector Spaces with Additional Structure

2.2. The Canonical Tensor Product of Algebras

Let A and B be two algebras with structure maps µA and µB respectively.Consider the flip-operator

S:(AQB)Q(AQB)--(AQA)Q(BQB)

defined by

S(x1 O Y1 ® x2 ® Y2) = x1 ® x2 ® Y1 ® Y2

Then a linear map

µA®B.(AQB)Q(AQB)-+AQB

defined by

µA ®B - (µA ®µB) ° S

determines an algebra structure in A ® B. The algebra A ® B is called thecanonical tensor product of the algebras A and B. It is easily checked that themultiplication in A ® B satisfies

(x1 ® Y1)(x2 ® Y2) = xlx2 ® Y1Y2 (2.5)

This formula shows that a canonical tensor product of two associative(commutative) algebras is again associative (commutative). If A and B haveunit elements IA and IB respectively then IA ® IB is the unit element of A ® B.If B has a unit element IB we can define an injective linear mapping (p : A -+AQBby

(px = x Q I B x e A.

It follows from (2.5) that

(p(xx') = xx' ® I B = (x ®I B)(x' ® I B) = (ox (px', x, x' e A,

i.e., (p preserves products and so it is a monomorphism.

2.3. Tensor Product of Homomorphisms

Let A 1, B 1, A2, B2 be algebras and suppose that

(P1 Al -' B1 (p2:A2 -+ B2

are homomorphisms. Then (i 1 ®(p2 is a homomorphism of the algebraA 1 ® A2 into the algebra B 1 ® B2. In fact, since

((p l ®1 ®(P2 ®(p 2) ° S = S ° ((p 1 ®(P2 ®(Pl ® 'P2)

Tensor Products of Algebras 43

it follows that

0 cP2) ° µA1 ®A2 = [(cP1 ° µA1) 0 (cP2 ° µA2)] ° S

= [(µB1 ° O P1)) O µB2 ° O P2)] ° S

= [(µB1 O µB2) ° S] ° ((P1 ®42 ® Pi ®(P2)

= µB1 ®B2 ° [('P1 0 (P2) 0 0 (P2)]

This equation shows that P1 0 P2 is a homomorphism.Now consider two involutions WA and WB in A and B respectively. Then

the mapping

W A ®B = WA 0 WB

is an involution in the algebra A 0 B.

2.4. Antiderivations

Let WA be an involution in the algebra A and SZA be an antiderivation withrespect to WA; i.e.,

SZA(x . y) = AX y + WAX A Y (2.6)

In terms of the structure map, (2.6) can be rewritten as

A ° µA = µA ° 0 l + WA 0 SZA) (2.7)

To simplify notation we write

A O l+ WA 0 A = A ® A

and then (2.7) reads

A°µA =

Now let B be a second algebra and SZB be an antiderivation of B with respectto an involution WB,

B ° µB = µB ° B ® B

Consider the linear maps

A®B 0 lA®B + WA®B 0 A®B

and

(A ®A) ®(B ®B) = SZA ®B 0 1B ®B + WA ®A 0 SZB ®B . (2.12)


On the assumption that

So (A®B)®(A®B) = °S (2.13)

(this always holds if WA = 1A, WB = 'B), Equations (2.10), (2.11), (2.12), and(2.13) imply that

µA ® B ° (A ® B) ® (A ® B) = (h A ® µB) ° S ° SZ(A ® B) ® (A ® B)

C UA ® µB) ° (A ® A) ®(B ® B) ° S

CUA ® JIB) ° (AA®A ® lB®B + WA®A S

CUA ° A®A ® µB + µA ° WA®A ® µB ° B®B) ° S= (cZA ° µA ®µB + WA ° JA QX cZB ° µB) ° S

= (cZA Q lB + WA Q SZB) ° (hA ® JIB) ° S

A®B °µA®B

This relation shows, in view of (2.8), that SZA ® B is an antiderivation in thealgebra A Q B with respect to the involution WA ® B

Tensor Products of G-Graded Vector Spaces

2.5. Poincare Series

Let E _ > E, a e G and F = F, fi e H be respectively G- and H-graded vector spaces. Then a (G O H)-gradation is induced in the spaceEQFby

EQF= Ea®Ff. (2.14)a, fJ

If H = G, then (2.14) is a G-bigradation. The corresponding (simple) G-gradation is given by

EQF = (EQF)y, (EQF)y = Ea®FJ. (2.15)y

The space E Q F, together with its G-gradation is called the tensor product ofthe G-graded spaces E and F.

It follows from (2.15) that for every two homogeneous elements x e Ea,y e Ffl the element x Q y is homogeneous and

deg(x Q y) = deg x + deg y.

In particular, the linear isomorphism f : E Q F -+ F Q E given by

f(x® y) = y ® x

is homogeneous of degree zero. Moreover, we have as well that each (E Q F)yis linearly generated by homogeneous decomposable elements of the formx®y,xeEa,yeFf,a+/3= y.

Tensor Products of G-Graded Vector Spaces 45

Let E, E', F, F' be G-graded vector spaces, and consider homogeneouslinear maps

cP : E -- E' and /i : F -- F'

of degrees k and l respectively. Then (p ® U : E ® F -+ E' ® F' is homo-geneous of degree k + 1. In fact, if x and y are homogeneous elements ofdegree a and /3 respectively it follows that

deg((p ® t/ixx ® y) = deg((px ® ,y)

= deg cpx + deg ty

=a+k+f3+1=(a+/3)+(k+1)

and hence p ® U is homogeneous of degree k + 1.Now assume that G = 7l and that the gradations of E and F are positive

and almost finite. Then the Poincare series of E ® F is given by

PE ®F(t) = dim(E ® F)k tk.k

Since

dim(E ® F)k = dim E. ® Fj = dim Et dim Fj,i+j=k i+j=k

the above formula reads

PE ® F(t) = dim Ei t` dim Ej tj = PE(t) PF(t)k i+ j=k

showing that the Poincare series of E ® F is the product of the Poincareseries of E and F.

2.6. Tensor Products of Several G-Graded Vector Spaces

Let E. = a E", a e G be G-graded vector spaces. Then a G p-gradation isinduced in the space E = E 1 ® ® Ep by assigning the degree (a 1, ... , a p)to the elements of E;1 ® ® E 7p. The corresponding simple G-gradation isgiven by E = a E G Ea where

the sum being extended over all p-tuples (a 1, ... , a p) such that cc 1 ++ ap = a. The space E together with this gradation is called a tensor product

for the G-graded spaces E,. It follows from the definitions that

deg(x 1 O ... O xp) = deg x i + ... + deg xp


for every p-tuple of homogeneous elements x1. As another immediate con-sequence of the definition we note that the isomorphism f: (E1 Q E2) Q E3 -+E1 Q (E2 Q E3) defined by

f:(x1 0x2)Ox3 -+x1 0(x2 Ox3)is homogeneous of degree zero.

It is easy to verify that if El, F1(i = 1, ... , p) are graded spaces, and if1: E1 -+ F1 are homogeneous of degree kl then the map p1 O O co is

homogeneous of degree p=1 kl.Suppose now that G = 7l and that all the gradations of the E1 are positive

and almost finite. Then clearly the induced gradation in E is again positiveand almost finite. Moreover, the Poincare series of E is given by

PE(t) = PE1(t)... P(t).

The proof is similar to that given for p = 2.

2.7. Dual G-Graded Spaces

Let E = >JizeG Ea, E* = >jzeG Ea and F = >fleG Ff, F* = >fleG F be twopairs of dual G-graded vector spaces and consider the spaces

a, fJ

E* Q F* = Ea Q Fa, fJ

as G-bigraded vector spaces. Then the induced scalar product between E Q Fand E* Q F* respects the G-bigradations. In fact, for any vectors x e Eat,x * E E, y e Ff 1, y* E Fwe have

<x* O y*, x O y> = <x*, x><y*, y> = 0

unless a1 = a2 and f1 = I2 . As an immediate consequence we have that theG-graded spaces E Q F, E* Q F* are dual G-graded spaces.

2.8. Anticommutative Tensor Products of Graded Algebras

Let A = L Ap and B = >q Bq be two graded algebras. Consider the anti-commutative flip operator

Q:(AOB)O(AOB)-'(AOA)O(BOB)defined by

Q(x®y®x'Oy')=(-1)' x®x'Oy®y',


where deg x' = p' and deg y = q. Then the linear map

µA®B:(AQ B)Q(AQB)-AQB

defined by

µA ®B =(µA OO 1 B) ° Q

determines an algebra structure in the graded vector space A Q B. Theresulting algebra, A Q B, is called the anticommutative tensor product or theskew tensor product of A and B. The multiplication in A Q B is given by

(x O Y)(x' O Y') = (-1)1 'qxx' Q yy' p' = deg x', q = deg y. (2.16)

If A and B are algebras without gradation, then by the tensor product ofA and B we shall mean the canonical tensor product (Section 2.2). If, in thischapter, A and B are graded algebras, then by the tensor product of A and Bwe shall mean the anticommutative tensor product. Observe that the under-lying vector spaces of the algebras A Q B and A Q B coincide.

Now it will be shown that A Q B is a graded algebra. In fact, if x 1 E AP 1,

x2 E AP2 , y 1 E Bq 1, Y2 E Bq2 are arbitrary we have

(x1 O y1)(x2 O Y2) = (- 1)P2glx1x2 O YlY2

Since A and B are graded algebras, it follows that

deg(xlx2) = pl + p2 deg(Y1Y2) = q1 + q2

In view of the definition of the gradation in A Q B (Section 2.5) we obtainthat (x 1 O y 1)(x2 O Y2) is homogeneous of degree Pi + P2 + q1 + q2 andhence A Q B is a graded algebra.

It is easy to verify that if C is a third graded algebra then the linear mapf:(AOB)QC-AQ(BQC)given by

f:(x®y)Ox z - xOx (y Ox z)

preserves products and hence is an isomorphism.The anticommutative tensor product of two anticommutative graded

algebras is again an anticommutative algebra. In fact, let x E AP, y E Bq,x' E A,,, and y' E Bq- be homogeneous elements. Then we have

(x O Y)(x' O Y') = (- 1)1qxx' O yy'

( 1)P'q+P'P+q"x'x O y'y

= (- 1)P'q + PP' + qq' + Pq'(x' ® y')(x O y)

= (, 1)(P + q')(x' O Y')(x O Y)

The reader should observe that the canonical tensor product of anticom-mutative graded algebras is not in general an anticommutative algebra.


2.9. Homomorphisms and Antiderivations

Let C = >r Cr and D = > DS be two more graded algebras and assume that

cP : A - C, t/i : B - D

are homomorphisms homogeneous of even degree k and l respectively. Then(p ® U is a linear map homogeneous of degree k + 1. Moreover, we have

as is easily checked. By the same argument as that used in Section 2.3 itfollows that (p ® U is a homomorphism of A ® B into C ® D.

In particular, if WA and WB are involutions in A and B, homogeneous ofdegree zero, then WA® B = WA ®wB is an involution of degree zero in A ® B.As a special case, suppose that WA and WB are the canonical involutions inA and B. Then WA ® B is the canonical involution in A ® B (see Section 6.6 ofLinear Algebra).

Proposition 2.9.1. Let SZA and SZB be homogeneous antiderivations of odd degreek. Then the mapping

AB = A ® 1B + WA

(where WA is the canonical involution) is an antiderivation, homogeneous ofdegree k, in the graded algebra A ® B. If eA and eB are derivations of evendegree k then

eA ®B = eA OO LB + A ®0B

is a homogeneous derivation of degree k in A ® B.

PROOF. Clearly, SZA ® B is a homogeneous linear map of degree k. To show thatSZA ® B is an antiderivation we verify that

Q ° (A®B)®(A®B) = (A®A)®(B®B) ° Q (2.17)

(see Section 2.4 for the notation). Then the same argument as that used inSection 2.4 proves that SZA ® B is an antiderivation in A ® B. Similarly, toprove the second part of the proposition we need only establish the formula

Q ° e(A®B)®(A®B) = e(A®A)®(B®B) ° Q

Now we proceed to the verification of (2.17) and (2.18).Let

(2.18)

xeAp, yEBq, x'EAp-, y'EBq-


be homogeneous elements. Then we obtain

Q (A ® B) ®(A ® B)(x OO Y OO X' O Y')

= Q[QAXOYOX'OY' + (-1)Px®QBYOx'OY'+ (_ 1)P+,x ® y ® QA x' ® y' + (-1)P+q+P'x O y O x` O QBY'](- 1)'Q x O x' O y O y' +(-1)P+q+(k+P')q.aG Q Ax' Q y Q y'

+ (_ 1)P+P (q+k)x O X' O BY O Y'

+ (-1)P+q+P,+P'qx

O x' O y O By'= (-1)" {QA x O x' + (_ 1)P + (k + 1)qx O A x'} O Y O Y'

+ (- 1)'(x O x') O {( -1)P+kP QBy Q y' + (-1)P+q+P y Q QBy'}.

Since k is odd we havex O x' + (_ 1)P +(k + 1)qx O A x') = x O x' + (- 1)"x O A x')

= QA®A(x O x')

and

(- 1)P+kP BY O Y' + (- 1)""y ® BY=

(-1)P+P [ BY O Y' + (-1)y O BY]= (-1)P + P QB ®B(Y O Y')

Hence it follows that

Q A ®B) ®(A ®B)(x OO Y OO x' O Y')

= (- 1)qP' LEA ®A(x O x')® Y O Y' + (-1)P +q (x O x')® B ®B(Y O Y')]

= (- ®A) ®(B ®B)(x OO x' O Y O Y')

= SZ(A ® A) ®(B ® B) Q(x OO Y Ox X' O Y')

whence (2.17).Now let eA and eB be homogeneous derivations of even degree k. Then

we have

Q (A ® B) ® (A ® B)(x Ox Y OO X' OO Y')

= Q(OAx®YOx'OY' + XOOBYOX'®Y'

+ XOYOOAX'OY' + XOYOX'OOBY')= (-1)P'q[OAXQx'Q y®y' + x®x'®y®0BY']

+ (-1)q(P' + k)x ® eA x' O y O Y' + (-1)P' (q + k)x O X' O 0B Y O Y'

= (-1)P'q[OAx®x'Oy®y' +XOOAX'®y®Y'+ XOX'OOBYOY' + XOx'OYOOBY'1

= (-1)P qe(A ®A) ®(B ®B)(x Ox x' Ox Y Ox Y')

= e(A ®A) (B B) Q(x OO Y OO x' OO Y')

whence (2.18).


Tensor Products of Differential Spaces

In Sections 2.10-2.17 the notations BE, B(E); ZE, Z(E); and HE, H(E) for theboundary, cycle, and homology spaces of a differential space will be used inter-changeably.

2.10. Tensor Products of Differential Spaces

Suppose that (E, aE) and (F, aF) are differential spaces. We wish to makeE ® F into a differential space. In order to do so we shall need an involutionw of E such that

Suppose we are given such an involution. Define D = E ®F by

aF.aE®F = aE ®l + (U®

Then we have

(2.19)

aE ®F(x ® Y) = aE x ® y + wx ® 'FY . (2.20)

From (2.19) we obtain that

aE®F=aE®l +waE®aF+aEw®aF+ l®aF

= (waE + aE w) ®aF

=0

and so (E ® F, E ®F) is a differential space. Formula (2.20) implies that

ZE ® ZF C ZE®F.

Moreover

(2.21)

BE ® ZF C BE ®F and ZE ® BF BE ®F ' (2.22)

In fact, if aE x e BE and y e ZF are arbitrary elements, then

EX ® Y = aE ®F(x ® Y)

Similarly, if x e ZE and aF y e BF, then

aE ®F(wx ® Y) = w2X ® 'FY = x ®aFY

It follows from Relations (2.21) and (2.22) that the bilinear mapping (E x F)- E ® F induces a bilinear mapping p: HE x HF -+ HE ® F such that

(p(lrE Z 1, F z2) = E ®F(Z 1 ® Z2) Z1 e ZE, Z2 e ZF, (2.23)

where ?rE, ?CF, and irE ®F are the canonical projections of the cycle spaces ontothe homology spaces.

Tensor Products of Differential Spaces 51

It is the purpose of this section to show that the pair (HE ® F' 'p) is thetensor product of HE and HF. We first establish the formulas

ZE ®F = ZE ® ZF + BE ®F (2.24)

(ZE ® ZF) n BE ®F = BE Q ZF + ZE Q BF. (2.25)

Consider the linear operators D1: E Q F -+ E Q F and D2 : E Q F -+ E Q Fgiven by

D1 = aE Q and D2 = a Q aF.

Then

(2.26)

D1 =D2=0, D1D2+D2D1 =0, (2.27)

and

D1+D2=D.It follows from (2.26) and (1.11) that

ImDI=BEQF, ImD2=EQBF,and

Im(D 1 D2) = BE ® BF

whence, in view of (1.7),

Im(D1D2) = Im D1 n Im D2.

The kernels of D 1 and D2 are given by

ker D l= ZE Q F ker D 2= E Q ZF

(cf. (1.12)) and so we obtain

ker D1 n ker D2 = ZE Q ZF.

Suppose now that z e ZE ® F is arbitrary. Then D 1 z = - D2 z and so

DIzeImD1 nImD2 = ImD1D2.

Let x e E O F be a vector such that D 1 D2 x= D 1 z. Then setting

y = z - (D1x + D2x)

we obtain

D1y = D1z - D1D2x = 0

and

(2.28)

D2y = D2z - D2D1x = -D1z + D1D2x = 0.


Thus

y E ker D 1 n ker D2 = ZF. ®ZF .

It follows that

Z=Dx+ yEBE®F+ZE®ZF

whence

ZE®F C ZE ® ZF + BE®F

Inclusion in the other direction is a consequence of (2.21) and so (2.24) isproved.

Next, note that every element of BE ® F n (ZE ® ZF) can be written in theform

Dx=Dix+D2x.

Then we obtain from (2.26) and (2.27)

D2(D1x) = D2Dx = 0

and

Hence

and

D1(D2x) = D1Dx = 0.

Dix e ker D2 n Im D1 = BE ® ZF

D2xekerD1 nImD2 = ZE®BF.

It follows that Dx E BE ® ZF + ZE ® BF ; i.e.,

BE ®F n (ZE ® ZF) C BE ® ZF + ZE ® BF

The inclusion in the other direction follows from (2.22) and hence (2.25)is proved as well.

Now we are ready to prove that the pair (HE ® F, rp) is the tensor productof HE and HF.

Let Q be the restriction of the canonical projection 1rE ®F . ZE ®F -' HE ® Fto the subspace ZE ® ZF. Then we have, in view of (2.24) and (2.25),

Im Q = HE®F (2.29)

and

ker Q = BE ® ZF + ZE ® BF. (2.30)

Consequently Q induces a linear isomorphism

Q : (ZE ® ZF)/Te(E, F) 4 HE ®F,


where Ta(E, F) = BE ® ZF + ZE ® BF. Hence,

Qop=Q

where p denotes the canonical projection

p . ZE ® ZF -' (ZE ® ZF)/Te(E, F)

Consider the bilinear mapping

fJ: HE x HF -' (ZE ® ZF)/T,(E, F)

defined by

(2.31)

I (irE Z 1, F Z2) = p(Z 1 ® Z2) Z 1 E ZE, z2 E ZF. (2.32)

Then Formulas (2.32), (2.31), and (2.23) yield

h'/(?rEZI, FZ2) = ap(Z1 ® Z2)

= Q(Z1 ® Z2)

= ?LE ®F(Z 1 ® Z2)

(p(?rEZ 1 , F Z2)

and so it follows that Th/i = rp. Hence we have the commutative diagram

Since the pair (ZE ® ZF F), i'/) is the tensor product for HE and HF(see (1.13)) and a is a linear isomorphism, it follows that the pair (HE ®F, co)is also the tensor product of HE and HF. The result is restated in the

Kunneth theorem: Let (E, aE) and (F, aF) be two differential spaces and(E ® F, aE ® F) be their tensor product. Then the pair (HE ® F, cp) is the tensorproduct of HE and HF, where p : HE x HF -+ HE ® F denotes the bilinearmapping induced by the bilinear mapping E x F -+ E ® F.

In view of the above theorem we may denote the mapping p by ®.Then we have the relations

E®F(Z1 ® Z2) = EZ1

and

HE ®F = HE ® HF.


2.11. Tensor Products of Dual Differential Spaces

Suppose that (E, SE), (E*, aE) and (F, SF), (F*, aF) are two pairs of dif-ferential spaces dual with respect to scalar products <, > and suppose furtherthat w, w* is a pair of dual involutions in E and E* respectively. Then theinduced differential operators D, D* in E O F and E* O F* are given by

aF

and

D* = a* i + w* Qx aF .

It follows from Section 1.23 that D and D* are again dual, and hence(E O F, D) and (E* O F*, D*) are dual differential spaces.

As an immediate consequence we have that there is induced a bilinearform 1 in HE* ® F* x HE ® F such that

1(pz*, irz) = <z*, z> z* E ZE* ®F*, z E ZE ®F, (2.33)

where

P ZE* ®F* -+HE* ®F* and 7r : ZE ®F -+HE ® F

are the canonical projections. On the other hand, consider the bilinearfunctions I 1 and 2 in HE* x HE and HF* x HF defined by

1 1(Plu*, ?r1u) = <u*, u> u * E ZE*, u e ZE

and

s,2(P2 U*, ?r2 U) = <U*, v) U* E ZF*, v E ZF.

For the tensor product of the bilinear functions I 1 and '12 (see Section 1.22)we obtain

(I O s2)(P1u* ® P2 u*, ir1u ® ir2 U) _ 11(P1u*,

i 1u)12(P2U*,

m2 U)

_

= (u* O u*, u O u>. (2.34)

On the other hand, Formula (2.33) shows that

CP1u* O P2 u*, lt1u O ir2 u) = (p(u* O U*), ir(u O u)) (2.35)

= <u* OQ u*, u Ox v>.

Comparing (2.34) and (2.35) we find that

2.

In particular, since I 1 and 1 2 are nondegenerate (see Section 1.22) so is 1.In view of Section 6.9 of Linear Algebra the nondegeneracy of 1 is also deriv-able from the duality of D and D*.


2.12. Graded Differential Spaces

Consider two graded differential spaces E = i E. and F = L F. Weshall assume that the operators aE and aF are homogeneous of odd degree k.Then the canonical involution, w, defined by

wx=(-1)"x xeEp

satisfies Condition (2.19). In fact,

(SEw + w aE)x = (- 1)p 1)P+k 3X = 0.

The induced differential operator D in E Q F is given by

D(x Q y) = aE x Q y+(-1)"x Q aF y x e E, y e F. (2.36)

Clearly D is again homogeneous of degree k. In general, the induceddifferential operator in the tensor product of graded differential spaces willmean the operator defined with respect to the canonical involution, w, in E.

The gradations of E and F induce gradations in H(E) and H(F) respec-tively defined by

H(E) = H.(E) H.(E) = ic1Z1(E) (2.37)i

and

H(F) = H (F) H (F) = ir2 Z j(F). (2.38)J

Similarly, we have (in the induced simple gradation)

H(E O F) = Hk(E O F) Hk(E Q F) = irZk(E Q F). (2.39)k

Formulas (2.39), (2.37), and (2.38) yield in view of the Kunneth theorem

Hk(E O F) = H1(E) O Hj(F)

= Hi(E) xQ H (F). (2.40)k i+j=k

Since

H.(E) O H(F) c H+ (Ej(E O F)

we obtain from (2.40) the Kunneth formula for graded differential spaces

Hk(E O F) = H,(E) O H (F). (2.41)i+j=k

In particular, the gradation in H(E Q F) determined by the gradationin E Q F coincides with the gradation obtained by identifying H(E Q F)with the tensor product of the graded spaces H(E) and H(F).


Now assume that E and F are almost finite positively graded spaces.Then so are H(E), H(F) and H(E Q F). Moreover, if H(E)' H(F) andPH(E ®F) are the corresponding Poincare polynomials, then

PH(E ®F) - PH(E) PH(F)

2.13. Dual Graded Differential Spaces

Suppose now that E* = (>t Et, 3) and F* = (L F;, 3) are two gradeddifferential spaces which are dual to the graded differential spaces E and Frespectively. Then aF, aE will be of degree - k and the canonical involutionw* of E* is dual to the canonical involution of E, (k = deg aE = deg SF).

It follows that the differential operator

D*=& ®i+(0*®a*coincides with the differential operator in the graded space E* Q F* definedin Section 2.11. Hence, (E Q F, D) and (E* Q F*, D*) are dual differentialspaces and graded spaces as well. Moreover, we have from Section 2.7that E Q F and E* Q F* are dual graded spaces (i.e., the scalar productrespects the gradation). Thus these spaces are dual graded differential spaces.

PROBLEM

Let i? be a differential operator in a finite-dimensional vector space E. Define dif-ferential operators i1 and 2 in the space L(E; E) by

O1(p= 9 q) and a2(P = qP°a (pEL(E;E)

and let H 1, H2 be the corresponding homology spaces. Prove that

H1 E* Qx H(E) and H2 H(E*) ® E

(H(E) and H(E*) are the homology spaces of E and E*).

Tensor Products of Differential Algebras

2.14. The Structure Map of the Homology Algebra

Suppose (A, &4) is a differential algebra with respect to an involution cvA(see Section 6.12 of Linear Algebra). Then the differential operator aA is anantiderivation with respect to WA; i.e.,

aA(XY) = aA x y + WAX aA y x, y e A. (2.42)

Tensor Products of Differential Algebras 57

Introducing the differential- space (A ® A, aA ® A) we can rewrite (2.42) inthe form

µA aA ®A = aA µA ,

where µA denotes the structure map of A. Now consider the homologyalgebra H(A) (see Section 2.10). The equation

AZ1 AZ2 = A(Z1Z2)

shows that the structure map of H(A) is given by

µH(A) ° (?rA 7tA ° µZ(A)

z i, z2 E Z(A)

(2.43)

where irA : Z(A) -p H(A) denotes the canonical projection and µZ(A) is thestructure map of the algebra Z(A). Since Z(A) is a subalgebra of A it isclear that µZ(A) is the restriction of the structure map µA to the subspaceZ(A) ® Z(A) of A® A.

2.15. Tensor Products of Differential Algebras

Suppose now that (A, '3A) and (B, aB) are two differential algebras withrespect to involutions WA and c0B . Then the induced differential operator inthe space (A ® B, aA ® B), is given by

aA®B-aA®l +WA® 3B.

Now consider the canonical tensor product A ® B. Recall that WA ® B is aninvolution in A ® B and that aA®B is an antiderivation with respect toWA ® B Hence (A ® B, aA ® B) is a differential algebra, and so an algebrastructure is induced in H(A ® B). The structure map, µH(A ® B)' of H(A ® B)is given by

µH(A ® B) (irA ® B ® irA ® B) = irA ® B ° µZ(A ® B)' (2.44)

where irA ® B : Z(A ® B) -+ H(A ® B) denotes the canonical projection, andµZ(A ® B) is the structure map for the algebra Z(A ® B) (cf. (2.43)).

2.16. The Algebra H(A) p H(B)

It follows from the Ki nneth formula that the vector space H(A ® B) maybe considered to be the tensor product of the spaces H(A) and H(B),

H(A ® B) = H(A) ® H(B).

In this section it will be shown that H(A ® B) as an algebra is the canonicaltensor product of the algebras H(A) and H(B).


The structure map of the algebra H(A) ® H(B) is given by

µH(A) ®H(B) = (µH(A) ® µH(B))SH, (2.45)

where SH denotes the flip operator for the pair H(A), H(B) (see Section 2.2).It has to be shown that

µH(A ®B) = µH(A) ®H(B)

To simplify notation we set

µH(A) = Q, µH(B) = ' µH(A ® B) = P

Then we obtain from (2.43) that

(Q ® t) (mA ® 1A ® ThB ® ThB) = Q(mA ® 2A) ® t(ThB

ThA µZ(A) ® T B µZ(B)

(ThA ® ThB) (µZ(A) ® / Z(B))

Next we observe that

(2.46)

SH(mA ® 2 B ® ThA ® ThB) _ (?LA ® 2 A ® 2 B ® TB)SZ, (2.47)

where SZ is the flip operator for Z(A) and Z(B). The preceding two equationsyield

(Q ® t)SH(mA ® 2B ® ThA ® ThB) = (ThA ® ThB) (µZ(A) ® µZ(B))SZ . (2.48)

But

CUz(A) 0 µZ(B))'SZ = µZ(A) ®Z(B)

and so (2.48) implies that

(Q ® t)SH(mA ® 2B ® ThA ® ThB) = (ThA ® ThB)µZ(A) ® Z(B)

On the other hand it follows from (2.44) that

P(2tA ® B ® ThA ® B) = ThA ® B µZ(A ® B)

(2.49)

Restricting this relation to the subspace Z(A) ® Z(B) ® Z(A) ® Z(B), andobserving that the restriction of ThA ®B to Z(A) ® Z(B) is ThA (see(2.23)), we obtain

P(2tA ® ThB ® ThA ® ThB) = (ThA ® ThB)µZ(A) ®Z(B) (2.50)

Combining (2.49) and (2.50), we find that

((Q ®r)SH P)(ThA ® ThB ® ThA ® ThB) = 0. (2.51)

But

ThA ® ThB ® ?LA ® ?LB : Z(A) ® Z(B) ® Z(A) ® Z(B)

- H(A) ® H(B) ® H(A) ® H(B)

Tensor Products of Differential Algebras

is a surjective mapping, and so it follows from (2.51) that

(cr ® i)SH = P'

i.e.,

C UH(A) ® UH(B))SH = 1 H(A ® B)

Relations (2.45) and (2.52) yield (2.46).

2.17. Graded Differential Algebras

59

(2.52)

Let A = A, aA) and B = (>q Bq,'3B) be two graded differential alge-bras and assume that aA and aB are both antiderivations of odd degree k.Consider the anticommutative tensor product A Q B. Then the structuremap of the algebra A Q B is given by

µA ® B = C UA ® µB) ° Q,

where Q denotes the anticommutative flip operator for A and B. It followsfrom Section 2.8 that A Q B is a graded anticommutative algebra and, sincek is odd, A © B is an antiderivation of degree k. Hence (A Q B, B) isa graded differential algebra. It will be shown that H(A Q B) is the anti-commutative tensor product of the graded algebras H(A) and H(B).

Let QH and QZ be the anticommutative flip operators for the pairsH(A), H(B) and Z(A), Z(B). Then we have that

QH ° (irA ® ?CB ® irA ® iB) = (iA ® irA ® irB ® iB) ° QZ

where irA : Z(A) -+ H(A) and 1rB. Z(B) -+ H(B) denote the canonical pro-jections. With this formula the proof coincides with the proof for theanalogous result in Section 2.16.

3Tensor Algebra

In this chapter except where noted otherwise all vector spaces will be defined over an arbitraryfield r.

Tensors

3.1.

Definition. Let E be a vector space and consider for each p >_ 2 the pair(®P E, 0 p) where

OpE=EQ...QE.n

We extend the definition of Op E to the case p = 1 and p = 0 by settingQ 1 E = E and 0° E = F. The paif (0" E, 0") is called the pth tensorialpower of E. The space Qp E is also called the pth tensorial power of E andits elements are called tensors of degree p.

A tensor of the form x1 ® O xp, p 1, and tensors of degree zero arecalled decomposable.

For every pair (p, q) there is a unique bilinear mapping

/3:Qp E X ®"E-+®E+q

f3(x1 0 ... OO xp, xp+ 1 OO ... OO xp+q) = x1 0 ... 0 xp+q x, E E. (3.1)

Moreover, the pair (Op+q E, /3) is the tensor product of Op E and Oq E(see Section 1.20). Hence we may write u Q u instead of /3(u, u) for u e Op E.Then (3.1) reads

(x1 Ox ... Ox xp) Ox (xp + 1 Ox ... Ox xp + q) = x 1 Ox ... Ox x p + q . (3.2)

60

Tensors 61

The tensor u ® v is called the product of the tensors u and v. The product(3.2) is associative, as follows from the definitions.

However, it is not commutative except for the case dim E = 1. (In fact, ifx e E and y e E are linearly independent vectors, then the products x ® y andy ® x are also linearly independent and hence x ® y y ® x). Finally,we notice that the product ® z (e ®° E = r, z e Op E) is the vectorin Op E obtained by multiplying the vector z by the scalar i In particular,the scalar 1 acts as identity.

Let {eV}VEI be a basis of E. Then the products e l)form a basis of ®p E. (see Section 1.20). In particular, if E has finite dimensionand eV (v = 1, ..., n) is a basis of E then the products eVl ® ®(v, = 1, ..., n) form a basis of ®p E and dim Op E = np (n = dim E).Every tensor z e Op E can be uniquely written as a sum

z = :V1 ...., Vpev 1 O ... O e.(v)

The coefficients .VP are called components of z with respect to the basis eV.

3.2. Tensor Algebra

Suppose that (Q p E, Q p) is a pth tensorial power of E (p = 0, 1,...) andconsider the direct sum

OCR

®E= Q ®p E.p=o

The elements of ® E are the sequences

(zo , z 1, ...) zp e ®p E (p = 0, 1, ...)

such that only finitely many zp are different from zero in each sequence.If ip: Qp E -+ ® E denotes the canonical injection we can write

® E = i, E.p=o

Since the pair (ip p p E, i, Q p) is again a pth tensorial power of E we denotethe p-linear mapping i,® p by ®" and the vector space i, O Y' E by ®" E.Then the above equation reads

®E= ®"E.p=0

By assigning the degree p to the elements of Op E we obtain a positivegradation in the space ® E.

62

We now define a bilinear mapping

(u,u)-+uu u, u,uue®E

by

ull = uP Qx Vq U = UP, U = Vq.

P9 P q

3 Tensor Algebra

This multiplication makes ® E into an associative (but noncommutative,if dim E > 2) algebra in which the sequence (1, 0, ...) acts as a unit element.It is clear from the definition that ® E is a positively graded algebra. ®Eis called a tensor algebra over the vector space E. From now on we shallidentify ®° E with r and ®' E with E. Then r and E are subspaces of® E, and the elements of E, together with the scalar 1 generate (in the alge-braic sense) the algebra ® E.

If E has finite dimension the gradation of ® E is almost finite and thePoincare series of the graded space ® E is given by

P(t) = nptp = 1 - t'nP = °n = dim E.

Remark : The reader should observe that if E 0 the bilinear mapping/3:(® E) x (® E) -+ ® E which is defined by the multiplication is not atensor product. In fact, if /3pq denotes the restriction of /3 to (®P E) x (®q E),we have

Im f3Pq = Im f3gP = ® p + q E.

Set E 1 = ®p E and F 1 = ®q E, p q. Then

E1 n F1 = 0,

while j3(E 1 x F 1) = j3(F 1 x E 1). Hence if 3 were a tensor product it wouldfollow that E 1 = 0 or F 1 = 0, whence E = 0.

3.3. The Universal Property of ®E

Consider an arbitrary associative algebra A, with unit element e, and alinear map ri : E -+ A. Then there exists precisely one homomorphismh : ® E -+ A such that h(1) = e and h o i = ri ; i.e., such that the diagram

it

nA

is commutative, where i denotes the injection of E into ® E. To define hconsider the p-linear mapping

1

ExxE-+A

Tensors

given by

(x 1, ..., x,,) - rix 1 ... ,ixP .

In view of 02 there exists a linear map h: OP E - A such that

hP(x 1 O ... O x,,) = yix 1 ...P

We extend the definition of hP to the cases p = 1 and p = 0 by setting

h1 = ri and ho(t) = to for all e r.

Then a homomorphism h : Q E - A is given by

hu = hp uP u, e 0,, E, u = uP.P P

63

In fact, if u, v e Q E are decomposable, then it is clear that

h(uv) = hu hv.

Since every element in Q E is the sum of decomposable tensors, and his linear, it follows that h preserves products.

To show that h is uniquely determined by ri, we notice that the conditionsh o i=, and h(1) = e determine h in Q 1 E= E and in Q° E= r. But Q Eis generated by 1 and the vectors of E ; consequently, h is uniquely deter-mined in Q E.

3.4. Universal Pairs

Now let U be an associative algebra with unit element 1 and E : E - U be alinear map. We shall say that the pair (E, U) has the universal tensor algebraproperty with respect to E if the following conditions are satisfied:

T1: The space Im E together with the unit element 1 generates U (in thealgebraic sense),

T2: If ri is a linear map of E into an associative algebra A with unitelement a then there is a homomorphism h : U - A such that h(1) = eand the diagram

E

E

W

is commutative.

The properties T1 and T2 are equivalent to the property:

T: If ri is a linear mapping of E into an associative algebra A with unitelement, then there exists a unique homomorphism h : U - A such thath(1) = e and Diagram (3.3) is commutative.

64 3 Tensor Algebra

It is clear that Tl and T2 imply T. Conversely, assume that the pair (E, U)satisfies the condition T. Then T2 follows immediately. To prove Tl considerthe subalgebra V of U generated by Im E and the unit element. Then E can beconsidered as a linear map of E into V (which, to avoid confusion, we denoteby EV) and hence there is a homomorphism h : U V such that h(1) = 1 andho E = Ey . If j : V -+ U denotes the inclusion map we have E = J o Ey andhence it follows that E = (j o h) o E. In diagram form we have

and hence the diagram

U

is commutative. On the other hand, we have the commutative diagram

E U

where us the identity map of U. Since j o h is an endomorphism of the algebraU, the uniqueness part of T implies that

joh = z.

Consequently, j is an onto map, and hence U = V. This proves T1.We shall now prove the following

Uniqueness theorem. Let (E, U) and (E; U') be two universal pairs for E. Thenthere exists precisely one isomorphism f : U -+ U' such that

fo =E'.

PROOF. In view of T there exist unique homomorphisms f : U -+ U' andg : U' -+ U such that

f o = ' and goE'=E.Hence, g of is an endomorphism of U which reduces to the identity in Im E.

Tensors 65

Since the space Im E generates U, it follows that g of = i. In the same way itis shown that f o g = i', the identity map of U'. Hence, f is an isomorphism ofU onto U' and g = f -' . The uniqueness theorem is thereby proved.

Since (i, ®E) is a universal pair for E it follows from the above uniquenesstheorem that for every pair(, U) there is precisely one isomorphism f : ®E -+ Usuch that f o i = E.

Since ®E is a graded algebra, a gradation is induced in U by the iso-morphism f. The algebra U furnished with this gradation is a graded algebraand f : ®E -+ U is a homogeneous isomorphism of degree zero.

In view of the uniqueness theorem, the universal algebra U is usuallycalled the tensor algebra over E and is denoted by ®E.

3.5. The Induced Homomorphism

Let p: E -+ F be a linear map from the vector space E to a second vectorspace F. Then p extends in a unique way to a homomorphism

-+ ®F

such that 'p®(1) = 1. In fact, consider the linear map ri : E -+ ®F given byri = j o p (where j denotes the inclusion map) and apply the result of Section3.3.

Clearly, the homomorphism'p® is homogeneous of degree zero. It followsfrom the definition of 'o that

'pO (x 1 ® ... ® x,,) = (px 1 O ... O cpx p xt e E.

Now let G be a third vector space, ® G be a tensor algebra over G andU: F -+ G be a linear map. Then it is clear from the definitions that

° p)® = o ° (3.4)

If F = E and us the identity map then ro is the identity map of ®E,

ro=l. (3.5)

It follows from (3.4) and (3.5) that 'o is injective (surjective) whenever 'p isinjective (surjective). In fact, if 'p is injective there exists a linear map i/i : E - Fsuch that l// o 'p = u. Formulas (3.4) and (3.5) now imply that

7"0 ° 'p® = l® = l

and hence 'o is injective.It is easy to see that

Im 'o = ®(Im 'p).

Hence 'o is surjective whenever 'p is.

66 3 Tensor Algebra

3.6. The Derivation Induced by a Linear Transformation

Let p be a linear transformation of E. Then p can be extended in a uniqueway to a derivation e®(cp) in the algebra ®E. To construct we noticethat for each p >_ 2 a p-linear mapping

E x x E- ®"E

is defined by

p

P

i= 1

This p-linear mapping induces a linear map QPE - QPE such thatP

eo(p)(x 1 O ... O xp) = >x1 ®... ®(pxt ®... ® x.i= 1

We extend this definition to the case p = 1 and p = 0 by setting

Op) = p and eo('p) = 0

and define to be the linear transformation of ®E into itself whichextends the Op).

It remains to be shown that is a derivation, i.e., that

v) = v + u e®(p)v. (3.6)

For the proof, we may assume (as before) that u and u are decomposable sothat

and xieE.

Let us first assume that p >_ 1 and q >_ 1. We then obtain

e ®(p)(u u) = e ®(p)(x 1 ®. ® x p + q)p+q

xl ®...®(pxi®x ...Ox xP+qi= 1

P x1 O ... O ® O xp+qi= 1

p+q

+ x 1 0 ... O xp O ... O cpxi O ... O xp + qi=p+ 1

= e®(p)u U + u e®(p)u.

Formula (3.6) remains true if p = 0 or q = 0. In the case p = 0, for instance,we have

W®(p)U = e®(p)L v+ e®(p)u.

Tensors 67

Since the algebra ®E is generated by the elements of E and the unitelement, it follows that the extension of p into a derivation in the algebra ®Eis unique (see Section 5.6 of Linear Algebra).

Let U be a second linear transformation of E. Then

o®(2p + µt/i) _ Ao®(p) + uO®(i,U) A, µ E r (3.7)

and

o®( ° - t,U° q') = o®C(p) ° o®(I,U) - o®(U) ° o®(p) (3.8)

Formula (3.7) follows immediately from the definition of o®. To prove(3.8) we remark first that the operator on each side is a derivation (see Section5.6 of Linear Algebra). Hence it is sufficient to show that

p) = eo('P) ° eo(n) - eo(U) °

But this is immediately clear since

e®('p) _ p and O®(U) _ fi.

3.7. Tensor Algebra Over a G-Graded Vector Space

Suppose E _ a Ea is a G-graded vector space and let ®E _ L ®"E be atensor algebra over E. Then a G-gradation is induced in each ®"E (p > 1) by

pE _ (®"E), where (®"E)a = Ea 1 ... O E.a al

It will be convenient to extend this gradation to ®°E = r by assigning thedegree 0 to the elements off. The induced G-gradation in the direct sum ®Eis given by ®E _ (®E)a where (®E)a = L (®"E). Then ®E becomesa G-graded algebra. In fact, let

u = xa 1 0 ... O xap E (®E)a and v = yfl 1 0 ... O Yflq e (O E)

be two decomposable elements. Then

uv = xa1O ... Q xap O Y 1 0 ... O Yflq E (O E)a +

and thus we have

deg(uv) = deg u + deg v.

The algebra ®E together with this gradation is called the G-graded tensoralgebra over E. If G = 7l and all the vectors of E are of degree one then theinduced gradation of ®E coincides with the gradation defined in Section 3.2.

6g

PROBLEMS

3 Tensor Algebra

1. Let u 1 = a 1 Q b 1 and u2 = a2 Q b2 be two decomposable tensors and assume thatu 1 0. Prove that u1 + u2 is decomposable if and only if a2 = a1 or b2 = µb 1(A, µ e r).

2. Assume that a 1 O O ap 0. Prove that

a1®...®ap=b1Q...Qbp

if and only if

b1 _ i = 1, ... , .. /., ... , 4,, = 1 (A E r).

3. Use Formula (3.8) to prove that

tr(q Q /i) = tr (p tr /i.

Tensors Over a Pair of Dual Spaces

3.8.

Definition. Suppose that E and E* ire vector spaces, dual with respect to ascalar product <, >, and let QE _ ®"E and ®E* _ ®PE* be tensoralgebras over E and E*, respectively. According to Section 1.22 there isinduced between x®PE and x®PE* for each p >_ 1 a unique scalar productsuch that

<x* 1 O ... Q X*P, X1-® ... QX XP> _ <X* 1, X1>" . <X*P, XP>. (3.9)

We extend the definition of <, > to the case p = 0 by setting

<A,, µ> _ Aµ u E Q °E = r.

The scalar products between ®E and ®PE* can be extended in a uniqueway to a gradation-respecting scalar product <, > between the spaces QE and®E* (see Section 6.5 of Linear Algebra) and this scalar product is given by

<u* U\ _<u*P UP>, u* = u*P, V = VP

P P P

Now assume that E has finite dimension and let a*v be a pair of dualbases for E and E*. Then the scalar product between the induced basisvectors O O eVp and a*µ1 O O e*µp is given by

<e*I O . . . O e*µp' ev1 O O ev,> _ (3.10)

This formula shows that the bases Q Q and a*µ1 O O e*µp areagain dual. It follows from (3.10) that the scalar product of two tensors

u = cv1.....VpeV1 O O and u* _ µp a*µ1 ®... O e*µ"

(v) (µ)

Tensors Over a Pair of Dual Spaces 69

is given by

<u* u> _ V1, ..., Vp1

(V)


Suppose that F, F* is a second pair of spaces dual with respect to a scalarproduct (again denoted by <, >). Let p : E -+ F, p* : E* - F* be a dual pair oflinear maps. The homomorphism (gyp*)®: ®E* - ®F* induced by p* willbe denoted by p ®. Then we have

p®(y* 1 ®... O y*p) = (p* y* 1 ®... ® (p* y*P y*i e F*. (3.11)

Now it will be shown that the homomorphisms

cP®: OE -+ OF and p®: ®E* - QF*

form a dual pair,

(3.12)

Let u e ®E and u* e ®F* be arbitrary elements. It has to be shown that

<U*, (p®u> = <(p®V*, u>

Since 'p® and 'p ® are both homogeneous of degree zero we may assume thatu and u* are both homogeneous of the same degree p. Moreover, we mayassume that u and v* are decomposable, u = x 1 ® ® xp and u* =y* 1 ® ® y*p. Then Formulas (3.11), (3.8) and (3.9) yield

<'P®v*, u> = <(P®(y*1 x0 ... x0 y*p), x1 ®... x0 xp>

= <(P* y* 1 ®... ® 'p* y*p, x 1 0 ... o xp>= <<p*y* 1, x1> ... <(p*y*P, xp> = <y* 1, 'x 1 > ... < y*p, cpxp>

_ < y* 1 ®... ® y*p, 'x 1®... ® oxp> = <v*, o®u>

whence

<v*, P® u> = <(p®V*, u>

If G, G* is a third pair of dual spaces and

iU: F -+ G, t/,* : F* - G*

is a second pair of dual mappings, we have in view of (3.12) and (3.4) that

( ° )® = [(/j° )*]® _ ((p* ° /J*)®_ (p*)®°(/,*)® ='p® ° 1//®

whence

70

3.10. The Induced Derivation

3 Tensor Algebra

Consider again a pair of dual mappings, p : E - E, p* : E* E- E*. The deriva-tion in the algebra ®E* which is induced by p* will be denoted by 8®(p),O®(p) = Then

P

e®(p)(x* 1 0 ... O x*p) => x* 1 p ... P*x*i ... O x*P. (3.13)i= 1

The derivations

and

again form a dual pair,

As in Section 3.9 we have to prove that

<e®(cp)v*, u> = <v*, 8®(cp)u> v* E QE*, u e QE

and we may assume that v* and u are of the form

v* = x* 1 QX ... QX x*P u = x 1 QX ... QX xp .

Then we obtain from (3.13) that

Pu> = x*1 Q ...*x* ... ® x*P, x1 ® ... ® xp

i= 1

P

<x*1 xl> ... xi> ... <x*p, xp>i= 1

whence

P<x* 1, x1 > ... <x*i, Pxi> ... <x*p, xp>

i= 1

P(x*i QX ... QX x*P, x1 O ... (pxi ... O xpi= 1

= <v*, O®(p)u>

(3.14)

<e®(p)v*, u> = <v*, 8®(p)u>

If i'/ : E - E, /i* : E* E- E* is a second dual pair of linear maps, then itfollows from (3.14) and (3.8) that

/,'p) = e®(/,*p* - (p*/,*) = e®(/,*)e®(p*) -

e®(i/i)e®(p) - e®(p)e®(i/i)

Mixed Tensors

whence

PROBLEM

71

Let E, E* be a pair of dual n-dimensional vector spaces, and consider a tensoru e ®"E. Define a subspace c E* as follows: a vector x* e E* is to be contained inEif and only if

<u, x* Q v*> = 0 for every v* e ®"- lE*.

Show that u is decomposable if and only if dim E= p - 1.

Mixed Tensors

3.11.

Definition. Let E*, E be a pair of dual vector spaces and consider, for every pair(p, q), p > 1, q >_ 1, the space

®(E*, E) _ (®PE*) O (®"E).

Extend the definition of ®(E*, E) to the cases q = 0 and p = 0 by setting

®(E*, E) = ®PE* and Qq(E*, E) = QqE.

The elements of ®(E*, E) are called mixed tensors over the pair (E*, E) andare said to be homogeneous of bidegree (p, q). The number p + q is called thetotal degree.

A tensor of the form

w = xi Qx ... xp Qx xl Qx ". xq x*EE*, x;EE,

is called decomposable.The scalar product between E* and E induces a scalar product between

®(E*, E) and ®(E*, E) determined by

<u* p v, v* p u> _ <u*, u><v*, v> (3.15)

(see Section 1.22). Thus any two spaces ®(E*, E) and ®(E*, E) are dual.In particular, every space ®(E*, E) is self-dual. Finally note that

<zl, z2> = <z2, z1> z1e®(E*, E), z2 e ®(E*, E)

as follows from the definition.

72

3.12. The Mixed Tensor Algebra

3 Tensor Algebra

The mixed tensor algebra over the pair E*, E is defined to be the canonicaltensor product of the algebras ®E* and QE (see Section 2.2). It will bedenoted by ®(E*, E),

®(E*, E) = (OE*) O (OE).

Thus ®(E*, E) is an associative (noncommutative) algebra with 1 Q 1 asunit element. It is (algebraically) generated by the elements 1 p 1, x* Q 1,and 1 p x with x* E E* and x e E.

Now let

iP. ®PE * - ®E*, iq : ®E - Q E

and

iq : (p PE *) p (p qE) - ®(E*, E)

be the inclusion maps and identify the spaces ®)E*, ®"E, and ®(E*, E)with their images under these maps. Then we have the direct decomposition

®(E*, E) = (®PE*) O (®"E).P, q

3.13. Contraction

Assume that p >_ 1 and q >_ 1. Fix a pair (i, j) with 1 <_ i <_ p and 1 < j < qand consider the (p + q)-linear mapping

x x E* x E x x E- ® : (E*,E)

given byq

..,xp,x1 ...xq)= <xr, x;>x l 0 ... 0 x O ... 0 xP* 0 x1 0 ... 0 x OX ... OX x.

1 q

In view of the universal property, 'I determines a linear map

C: ®(E*, E) - OP- i (E*, E).

C; is called the contraction operator with respect to the pair (i, j) and thetensor FXw) is called the contraction of w with respect to (i, j). In particular,

Ci(x* Q x) = <x*, x> x* E E*, x e E.

Now assume that E has finite dimension and let {e*v}, {ev} be a pair ofdual bases of E* and E. Then the products

evls ..., Vp = e*vl O... O e*vp O e O.. O eµ1s ..., µ9 µl µ9

Mixed Tensors 73

form a basis of ®(E*, E). Thus every tensor w e ®(E*, E) can be writtenin the form

w = µ l s . µy eV l s : s Vp µ l s : µy E IT.Vls ' VP µ1s s µq V1s Vp

(v)(µ)

The scalars , ; vp are called the components of w with respect to the basis{eV}. Since

Ci.eV1s ...s Vp _ Vi eV1s ...s Vis ...s VP1 µ1....,µg µj µl,..., µj ..... µq

it follows that the components of the contraction C; w are given byn

µls sµq- 1 = rµls sµ1.- 1Aµ1.. sµq- 1vls....Vp-1Vls....VJ-1AVjs....Vp-1

A=1

3.14. Tensorial Maps

Let E be an n-dimensional vector space. Then every linear automorphism aof E determines for each pair (p, q) a linear automorphism 7 of ®(E*, E)given by

T2(u* Q u) = (a®)-1u* Q a®u.

It is easily checked that

T02= T ° Ta

and

T,= i.A linear map

D : ®(E*, E) --, ®(E*, E)

is called Censorial, if it satisfies

Fo Ta = Tao

for all linear automorphisms a of E.As an example consider the contraction operator C. For the sake of

simplicity we let i = j = 1 and write C i = C. Then we have for z = x i pOxpOx10...Ox9

7 (z) = X* -1 x i O ... p a* -1 xp p Xx 1 p ... p axq

whence

(Cl )(z) = <a*-1xr, ax1>a*-1x2 ®... Q a*-1xp Q axe Q ... Q axq

= <xl,xl>a*-1x2 O...O«*-1xp pocx2 p...Oaxq= (Ta C) (z).

Thus C is a tensorial map.

74 3 Tensor Algebra

PROBLEMS

1. Let (v = 1, ... , n) be two bases of E and consider a tensor w E ®(E, E*).Assume that the bases and the dual bases {e*v}, {e*v} are connected bythe relations

e aye a*vA A

Prove the following transformation formulas for the components of w:

V1, ... VP - fJV1 VP K1 . .

(A),(K)

2. Using the formula in problem 1 show explicitly that the components of a contractedtensor satisfy the same transformation formula.

3. Show that <u*, u> = (C1)"(u Q u*) for all u e ®"E, u* E QpE*.

4. Assume that

c1: ®(E, E*) - ®(E, E*) ®(E, E*) ®5(E, E*)

is a pair of dual mappings and that 1 is tensorial. Prove that 1* is also tensorial.

5. Show that sum; composition, and tensor product of tensorial mappings is againtensorial.

6. Let

: ®(E, E*) - ®(E, E*)

be a nonzero tensorial mapping. Prove that

r- p=s - q.

7. Consider the linear map

(a) - a O z,where a e E Q E* is a fixed tensor. Show that qi(a) is tensorial if and only if a = ) t,

where t is the unit tensor.

8. Prove that every tensorial mapping c1: E Q E* + r is of the form

where C is the contraction operator.

9. Verify the relations

if i<k,j<1,C;oC = if i<k,j>1,

if i>k,j<1.

Tensor Algebra Over an Inner Product Space 75

10. Consider the bilinear mapping

®(E, E*) x ®(F, F*) -+ ®(E O F, E* O F*)

defined by

(x1O...Oxp®x*l(x1 0Y1)0...0(xp0 yp)0(x*1 0y*l)0...0(x*q®y*q).

(a) Show that this mapping is a tensor product(b) Prove that

CJ(u Q v) = CX u) ® CJ(v) u e ®(E, E*), v e ®(F, F*)

11. A tensor u e ®(E, E*) is called invariant if T (u) = u for every linear automorphism(see Section 3.14).(a) Show that if u 0 is an invariant tensor then p = q.(b) If E has finite dimension show that a tensor u e E Q E* is invariant if and

only if u = ) t where t is the unit tensor.(c) If E has infinite dimension show that every invariant tensor u e E Q E* must

be zero.(d) Assume that E has finite dimension. Show that a tensor u is invariant if and

only if the components of u are the same with respect to all pairs of dual bases.

Tensor Algebra Over an Inner Product Space

3.15. The Induced Inner Product

Let E be an inner product space and consider the pth tensorial power®pE. The inner product in E induces an inner product in ®pE such that

(x1 Ox ... Q xp, vi (xi, vi) .. (xp, Yp)

(see Section 1.25). In particular, if E is an n-dimensional Euclidean spaceand if ev (v = 1, ..., n) is an orthonormal basis of E, then the productsev1 p p evp form an orthonormal basis of ®"E and so ®"E becomes aEuclidean space as well.

The inner products in the spaces &J E determine an inner product in ®Egiven by

(u, v) _ (up, vp) u, v e ®E,P

where

u= u p and v= vp u p, v p e Q PE.P P

76 3 Tensor Algebra

3.16. The Isomorphism -r®

Let E be an n-dimensional inner product space with dual space E*. Then theinner product in E determines a linear isomorphism -r: E - E* given by

<ix, y> = (x, y) x, y e E.

This isomorphism -r in turn induces a linear isomorphism

i®: ®E - ®E*.

This isomorphism satisfies the relation

<i®u, v> = (u, v) u, y e ®E. (3.16)

In fact, let

u=x1O...Oxp and v=Y1O...OYp

be decomposable tensors. Then we have

<i®u, v> = <ix1 ®... ®ixp, Y1 ®... O Yp

= < x1, Y1 ... <ixp, Yp> = (x1, Y1) ... (xp, Yp)= (x1 ®... O xp, Y1 ®... ® Yp) = (u, v)

Relation (3.16) shows that the restriction of -r® to ®"E coincides with theisomorphism ppE ppE* induced by the inner product in ppE. Finallynote that

since r* = -r.

3.17. The Metric Tensors

Let E be an n-dimensional Euclidean space. Choose an orthonormal basis{ev} (v = 1, ..., n) and set

g = ev ® ev.

v

Since an orthonormal basis is self-dual (with respect to the inner product inE), it follows that the tensor g is independent of the choice of the orthonormalbasis {ev}. It is called the contravariant metric tensor of E. Similarly, thecovariant metric tensor is defined by

g* = e*v O e*v,

v

where {e*v} (v = 1, ..., n) is an orthonormal basis of E* with respect to theinduced inner product.

Tensor Algebra Over an Inner Product Space 77

The inner product of two vectors x and y can be expressed in the form

(x, y) = <g *, x O y> .

In fact, write

x =vey and y = vev .v v

Then we have

x O y> = <e*v O e*v, x O y> = e*v x><e*v, y>v

= v;7v = (x, y)v

V

The same argument shows that

(x*, y*) = <x* O y*, g> x*, y* E E*.

PROBLEMS

1. Given a Euclidean space E prove that

(u1 0 v1, u2 0 v2) = (u1, u2)(v1, v2) u1, u2 e ®E, v1, v2 e QX qE.

2. Let E be a Euclidean space and E* be a dual space. Consider the metric tensorsg E E Q E and g* E E* ®E*.(a) Show that

Ci(g OO g*) = t,

where t is the unit tensor.(b) Prove that the metric tensor of ®"E is given by g O O g

3. Verify the formula

p

(µ(t)u, µ(t)v) = n(u, v) u, v e Q E,

where t is the unit tensor of E, n = dim E (see Section 1.26).

4. Let E be a Euclidean space and E* be a dual space. Consider the space L(E; E) =E* Q E (see Section 1.27).(a) Show that the adjoint of a linear transformation p = a* Q b is given by

P = ib Q i -1 a*.(b) Show that the induced inner product in L(E; E) is given by

rN

(`p, /') = tr(q ° 1'i).

5. Given a Euclidean space E and an arbitrary basis (v = 1, ... , n) show that

(u, v) = 9V11 9V 1...., Up7M1...., NPµvpup

(v)(u)

where

u = U1, ..., vpevl 0 ... 0 and v = ul ..., µpeul 0 ... 0 eµp .(v) (u)

78 3 Tensor Algebra

6. Show that the components of the metric tensor g* with respect to an arbitrarybasis x (v = 1, ..., n) of E are the inner products (xv, xµ).

7. Let E be a Euclidean space and E* be a dual space. Given a pair of dual bases{x*°} (v = 1, ... , n) show that

9* = (xv, xµ)x*V O x

and

v, µ

g = >(x*v,x*M)xv®xM.v, µ

8. Let E, E* be a pair of dual spaces and assume that E is a Euclidean space. Considerthe linear isomorphism i : E -+ E* defined by

<tx, y> = (x, y) x, y e E.

Prove that the isomorphism i®: ®"E -+ ®PE* coincides with the linear isomorphismQ pE + ®PE* which is induced by the inner product in ®"E. Show that i ® = i®and that (i®)2 =z.

The Algebra of Multilinear FunctionsIn Sections 3.18-3.23 E denotes a finite-dimensional vector space.

3.18. The Algebra T(E)

Consider for each p >_ 1 the space Tp(E) of p-linear functions

p

In particular T 1(E) = E*. It will be convenient to extend the definition ofTp(E) to p = 0 by setting T °(E) = F. The product of a p-linear function Fand a q-linear function 'P is the (p + q)-linear function 'F 'P given by

(F `P)(x1, ..., xp+q) = F(x1, ... , xp).'P(xp+ 1, ..., xp+q). (3.17)

In the cases p = 0 or q = 0 we define the multiplication to be the ordinarymultiplication by scalars. This multiplication makes the direct sum

00

T(E) = Tp(E)p=0

into an associative (noncommutative) algebra with the scalar 1 as unitelement.

A linear map p: E -+ F induces a homomorphism p* : T '(E) E- T (F) givenby

((p*'P)(x 1, ..., xp) =((px ..., (pxp) 'P E T "(F)

as follows directly from the definitions.

The Algebra of Multilinear Functions 79

Moreover, a linear transformation co of E determines a derivation oT (cp)in the algebra T(E) given by

P

1, ... , xP) = (x1, ..., xvv=1

3.19. The Substitution Operators

Fix a vector h e E and consider the operators iv(h) : T(E) -* T(E) given by

(iv(h)F) (x 1, ..., x,_1) = F(x 1, ..., h, ..., xP _ 1).

iv(h) is called the with substitution operator in T(E) corresponding to thevector h. Clearly,

iv(h) (iF + 'Y) = iv(h)F + iv(h)`J! F, Y E T'(E).

Moreover, it follows from the definition that for F E TP(E)

i h 'F = iv(h) F 'F v <_ p,

v _ p + 1.

This relation implies that for fv e E*

iv(h)(fi ... fP) = ... fv ... fP.

Now define operators iA(h) and is(h) on T(E) by

P

iA(h) = (_ 1)v -1 iv(h)v= 1

and

P

is(h) = iv(h) (p = 1, 2, ...).v= 1

Then we obtain from (3.18) the formulas

iA(h)((''F) = iA(h)(''F + (-1)M ' iA(h)'

and

is(h) (i ' 'F) = is(h)( ' 'F + ( ' is(h)'

(3.18)

for all (J) E TP(E), 'F E T'(E).

80 3 Tensor Algebra

3.20. The Isomorphism ppE* - T"(E)

In this section we shall establish an isomorphism between the space T"(E)and the pth tensorial power over E*. Consider the p-linear mapping

Tp(E)

given byn

P(f1, ... , fp) = f1 ... fp fV E E*.

It has to be shown that this map has the universal property. Fix a basis{e1, ..., in E and let {e*1, ..., e*"} be the dual basis.

We show that the products a*v1 e*vp (v, = 1, ..., n) form a basis ofTp(E). In fact, every vector x e E can be written in the form

x= e*v(x)ev e*v e E*v.V

Now let F E T "(E). Then we have

xp) _ e*V(xl)ev, ..., e*v(xp)eVv v

_ e*V1(x1) . . . e*VP(xp)F(eV1, . . . , eVP).

(v)

This equation can be written in the form

_ F(eyl, . . . , ev )e*vl . . . e*VPP

(v)

and so it shows that the products a*v1 e*VP generate T"(E).On the other hand, assume a relation

e*vl . . . e*VP = 0

Then

(v)

vP<e*V1, x1>... <e*VP, xp> = 0 xi E E.(v)

Now fix a p-tuple (o, ..., ocp) and set x, = e21(i = 1, ..., p). Then the relationabove implies that dal, ..., aP = 0. Thus the products a*V1 e*Vp form abasis of T"(E). In particular,

dim T "(E) = np n = dim E.

Now consider the linear map a: ppE* - T"(E) induced by co. Then wehave the commutative diagram

E* x x E* ( T(E)


In particular

a(e*v1 O . . . O e*' ) = e*v1 . . . e*vP-

Since the products a*v 1 e*''P form a basis of Tp(E) it follows that a is anisomorphism. Finally observe that, since oc(u* p v*) = oc(u*) oc(v*) a isan algebra isomorphism

a: QE* + T'(E).

3.21. The Algebra T.(E)

Let Tp(E)(p >_ 1) denote the space of p-linear functions in the dualspace E* and set T0(E) = F. Observe that the space T1(E) is canonicallyisomorphic to E under the correspondence a H fa given by

fa(x*) _ <x*, a> a e E.

Applying the results of Section 3.18 with E replaced by E* we obtain amultiplication between the spaces T,(E)(p >_ 0) which makes the direct sum

T,(E) _ Tp(E)p=o

into an associative algebra.A linear map p: E -+ F induces a homomorphism co: * T,(E) -+ T, (F) given

by

t)(yi, ... , yp) _ p*yi, ..., p*yp) Fe 7,(E)

and a linear transformation p of E determines a derivation in T,(E) given byP

(x l , ..., xp) _ (xi, ..., (p*x*, ..., x).v=1

Finally note that T.(E) is isomorphic to the tensor algebra over E (seeSection 3.20).

3.22. The Duality Between T"(E) and T(E)

Fix a pair of dual bases {e*v}, {ev} in E and E* and consider the bilinearfunction <,>: Tp(E) x Tp(E) -+ F given by

<IF, 'I'> _ F(ev1 , .. ., evP)%Tt(e*v 1, . . . , e*vP).(v)

Then we have in particular

<F, xl ' ... xp> _ F(eyl, ..., eyP)<e*v1, x1> ... <e*vp, xp>(v)

_ <e*vl, xl>e <e*vP, xp>vl, . . ., evP .

vl vp

82

Since

><e*v,x>ev=x x E E,v

3 Tensor Algebra

it follows that

<IF, x 1 ... xp> _ iF(x 1, ... , x p) F e T"(E), xv e E.

Similarly

< fi ... fp Y> _ tY(fi,... , fp) Y E Tp(E), fv e E*.

These relations imply that the bilinear function <, > does not depend onthe choice of the dual bases {ev}, {e*v} (v = 1, ..., n) and that it is non-degenerate. Thus it defines a scalar product in the spaces T"(E) and Tp(E).

On the other hand, we have a scalar product between the spaces ppE*and ppE (see Section 3.8). A simple computation shows that the isomorph-isms ®)E* T (E) and ®" E = Tp(E) preserve the scalar products.

3.23. The Algebra T(E)

Denote by T (E) the space of (p + q)-linear functions

f.

In particular,

n 'i

To (E) = T"(E) and T q (E) = 7(E)

(see Section 3.18).The product of two multilinear functions F E T (E) and Y E Ts(E) is

defined by

* * * *_ (x 1, ... , xp, x 1, ... , xq)(x i, ... , xr, x i , ... , xs ).

Form the direct sum T(E) of the spaces T (E) and identify each T (E)with its image under the inclusion map. Then the multiplication abovemakes T(E) into an associative algebra.

Now consider the bilinear mapping T "(E) x 7(E) - T (E) defined bythis multiplication. It follows from Section 1.20 that this bilinear mappinghas the universal property. Thus,

T (E) T (E) Q 1(E)

and

T (E) T (E) Q T.(E).


Moreover, since for F1, F2 E T(E) and 'F1, 'F2 E T,(E)

(b 'F2) = (b . b2)

T(E) is the canonical tensor product of the algebras T(E) and T.(E).Finally, let a : E 3 F be a linear isomorphism. Then an induced iso-

morphism 7: T(E) * T(F) is explicitly given by

for all F E T (E), y E F, y*' E F*.

4 Skew-Symmetry and Symmetryin the Tensor Algebra

All vector spaces in this chapter are defined over a field of characteristic zero.

Skew-Symmetric Tensors

4.1. The Space Np(E)

Given a vector space E consider the pth tensorial power QpE and let Sp denotethe permutation group on p letters. Then every permutation 6 E Sp determinesa linear automorphism of QpE (also denoted by 6) defined by

6(x 1 O ... O xp) = xQ -1(1) Q ... O x y - 1 p xy E E.

As an immediate consequence of the definition we have the formulas

(t6)u = ti(6u) 6, E Sp, u e QpE

and

iu=u

(i being the identity permutation).Now consider the subspace Np(E) of QpE generated by all products

x 1 O O xp such that x, = x for at least one pair i j. Clearly Np(E) isstable under 6 for each 6 E S.

It will be shown that for each u e QpE and each 6 E Sp

U - EQ 6u E Np(E)

For the proof we may assume that u is decomposable, u = x1 Q Q x.

84

Skew-Symmetric Tensors 85

Consider first the case of a transposition i : i ± j. Then we have

u-;tu=x1Qx ...Qxi®...Qx;Q...Qxp+x1Q...QX x;0...0xi0...0xp

= x 1 QX ... QX (xi + x;) QX ... QX (xi + x;) QX ... QX xp- x1 O ... O xi O ... O xi O ... O xp- x 1 O ... 0 x; 0...0x; 0 ... QX xp E Np(E).

Now assume that u - EQ 6u e N"(E) for all permutations 6 which are productsof m transpositions and consider the permutation p = r6 where is atransposition and 6 is a product of m transpositions. By hypothesis we have

u - EQ 6u e N"(E).

Since N"(E) is stable under it follows that

iu - EQ i6u e N(E),

whence

; tu - EtQ t6u e N(E).

On the other hand we have

u - ; iu e N"(E).

Adding these relations we obtain

u - EtQ i6u e N(E).

Now (4.1) follows by induction.

4.2. The Alternator

A tensor u e QpE is called skew-symmetric if

6u = EQ u for every 6 E Sp.

The skew-symmetric tensors of degree p form a subspace X"(E) of QpE.The alternator in QpE is the linear map m: QpE -p QpE given by

1

P Q

It follows from the definition that for e Sp

1 1TGA i = EQ 6't = Et EQt 6

P Q P Q

1

= Et p P = Et TGA .p P

4 Skew-Symmetry and Symmetry in the Tensor Algebra

TGA o 'L = Et TGA ieSp.

t 0 TGA = Et TGA t E Sp .

Next we establish the relations

ker mA = N"(E)

and

Im ThA = X "(E). (4.5)

In fact, let u = x 1 ®. p xp be a generator of N"(E). Then, for sometransposition -r, -ru = u and so Formula (4.2) implies that mA u = - ThA uwhence mA(u) = 0. This shows that N"(E) c ker mA.

On the other hand, it follows from the definition of mA that for u e ®"E

1ThA u - u = (EQ 6u - u) E N"(E).

P Q

Hence, if ThA u = 0, then u e N"(E). Thus (4.4) is established.To prove Formula (4.5) observe that, in view of (4.3), Im ThA c X"(E).

On the other hand, if u e X "(E) then JCA u = u and so X "(E) c Im ThA.Next note that Formula (4.3) implies that

211A = A

and so ThA is a projection operator. Hence we obtain from Relations (4.4) and(4.5) the direct decomposition

QpE = X "(E) Q N(E). (4.6)

Thus every tensor u e QpE can be uniquely decomposed in the form

u= v+ w v e X (E), w e N(E).

The tensor v = mA u is called the skew part of u.

4.3. Dual Spaces

Suppose now that E, E* is a pair of dual spaces, and let mA be the alternatorfor QpE*, p > 2. If x* 1 Q Q x*p and x1 O O xp are two decomposabletensors in QpE* and QpE respectively, we have, for any 6 E Si,,

O x*p, 6(x1 QX ... Q xp)> = <x*1, XQ- 1(1)> ... <x*p, X1 (p)>

x1>... <x*p), x p>

_ <cr 1(x*1 O ... O x*p), x1 OX ... QX xp>

Skew-Symmetric Tensors 87

and hence we obtain the relation

<u*, 6u> = <6- 1u*, u> U* E QPE*, U E QPE,

which shows that 6 and 6-1 are dual operators.Since

E-1= 1 E6 and it A=1

E6= 1 6

it follows that iA and i A are dual operators as well, i.e.,

<U*, TGAU> = <TGAU*, u> u* E QPE*, u e QPE. (4.7)

The duality of iA and iA implies that the restriction of the scalar product tothe subspaces Im iA = XP(E) and Im iA = XP(E*) is again nondegenerateand hence a duality is induced between XP(E) and XP(E*).

Suppose now that

u=x*1x®...x®x*P and

are decomposable tensors in QPE* and QPE respectively. Then we obtainfrom (4.7) the formula

<JCA(x* 1 ®... XO x*P), JCA(xl ®... ® xp)>

= <x* 1 O ... 0 x*P, TGA(x1 0 ... ® xp)>

<x*1 ®... 0 x*P, 1GA(x1 XQ ... ® xp)>

1 *= EQ<x1, XQ- 1(1)> ...<x*p,

XQ- 1(p)>,p Q

whence

1

<mA(x*1 O ... O x*P), ThA(xl ®... O xp)> = i det(<x*`, x;>).p

4.4. The Skew-Symmetric Part of a Product

Let QE = QPE be a tensor algebra over E and consider the subspacesNP(E) C QPE, p > 2. It will be convenient to extend the definition to thecases p = 1 and p = 0 by setting N 1(E) = N°(E) = 0. Accordingly we defineiA to be the identity on ®'E and Q °E and then the previously establishedformulas continue to hold in the cases p = 1 and p = 0.

It follows from the definition of NP(E) that

NP(E) O O9E C NP + 9(E) p>Oq>0 (49)-O PE O N9(E) C N ° + 9(E) - "

88 4 Skew-Symmetry and Symmetry in the Tensor Algebra

Now let u e QpE and v e ®"E be two arbitrary tensors. Then we canwrite

u= TGA u+ u 1 u 1 E N"(E)

and

V= TGA v+ v 1 v 1 E Nq(E),

whence

UQV=TGAUQTGAV+TGAUQV1 +U1QTGAV+U1Qv1.

Applying the projection mA to this equation and observing Relations (4.9)and (4.4), we obtain the formula

TA(U Q v) = TtA(TCA U Q TGA v). (4.10)

Since mA is a projection operator, it follows that

TGA(TGA U Q v) = mA(u Q v) = TGA(U Q mA V). (4.11)

PROBLEM

Show that the mapping 6: QpE -+ QpE is tensorial (see Section 3.14), where QpEis considered as a subspace of Q(E, E*). If the dimension of E is finite, prove that 6 isgenerated by the operators µ(t) and C, t being the unit tensor for E and E*.

The Factor Algebra ®E/N(E)

4.5. The Ideal N(E)

Consider the direct sum

N(E) _ N"(E).n

Formulas (4.9) imply that N(E) is a graded ideal in the graded algebra ®E.Suppose now that u e QpE and v e ®"E are two arbitrary tensors. Then

we have

u Q v-(-1)pgv Q u e Np + q(E). (4.12)

In fact, if 6 is the permutation given by

it follows that

6(u Q v) = v Q u and EQ = (- 1)I"

The Factor Algebra ®E/N(E) 89

and thus Formula (4.1) yields

u Q v-(-1)Pgv Q u= u Q v- EQ 6(u Q v) E NP + q(E).

Applying the operator ThA to (4.12) we obtain the formula

ThA(u Q v) = (-1)P%(v Q u) u e ®E, y e ®E. (4.13)

4.6. The Algebra ®E/N(E)

Consider the canonical projection

m: ®E -4 Q E/N(E). (4.14)

Since N(E) is an ideal in QE, a multiplication is induced in QE/N(E) by

ma mb = m(a Q b) a, be QE. (4.15)

It follows from (4.15) that this multiplication is associative and that m(1) is aunit element. Since the ideal N(E) is graded in the graded algebra QE, agradation is induced in the factor algebra ®E/N(E) by

®E/N(E) = m(® "E)P

and so ®E/N(E) becomes a graded algebra. Since N 1(E) = N°(E) = 0, wehave in particular that m(®1 E) and m(® °E) are isomorphic to Q 'E = Eand ®°E = IT respectively. Consequently, we shall identify m(®1 E) andm(p°E) with E and IT respectively.

From (4.13) we obtain the commutation relation

uv = (-1)Pgvu, (4.16)

for every two homogeneous elements of degree p and q in the algebraQ E/N(E).

4.7. Skew-Symmetric Tensors

Define the subspace X(E) c QE by

X (E) = X P(E)P

and extend the projection operators mA: ®"E - ®"E (ThA = i in ®'E andQ°E) to a linear map mA : QE -p QE. Then we have

ker ThA = N(E)

and

Im ThA = X (E).


Moreover, itA is a projection operator and

Q E = N(E) Q X (E).

If p denotes the restriction of the projection is to the subspace X(E) then

p : X (E) --> ®E/N(E)

is a homogeneous linear isomorphism of degree zero. Let mx : ®E -- X (E)be the restriction of Tx to QE, X(E). Then we have the following commutativediagram:

Q E r X (E)

irl

4.8. The Induced Scalar Product

(4.17)

Let E, E* be a pair of dual vector spaces. Then a scalar product is induced inQE, ®E* (see Section 3.8). It follows from (4.7) that the restriction of thisscalar product to the subspaces X (E) and X (E*) is again nondegenerate.Since p : X(E) =- QE/N(E) is a linear isomorphism, a scalar product <, > inthe pair Q E/N(E), Q E*/N(E*) is induced by

<pu*, pu> = p !<u*, u> u* E Xp(E*), u e XP(E). (4.18)

Clearly the scalar product (4.18) respects the gradation. Moreover it followsfrom (4.17) and (4.18) that

<mu*, mu> = <pm Xu*, pmx u>

= <pm a*, p7cA u> = p !<m u*, mA u> u* E QX pE*, u e ®"E.

(4.19)

Now assume that u and u* are decomposable,

u = x1 QX ... QX xp, u* = x*1 0X ... QX x*I.

Then formulas (4.19) and (4.8) yield

<m(x*1 O ... O x*p), m(x1 O ... Q xp)> = det(<x*`, xi>). (4.20)

Suppose now that E is an inner product space. Then E is dual to itselfwith respect to the inner product and hence we may set E* = E. It followsfrom Section 3.15 that the induced scalar product in QE is again non-degenerate and, hence, so is its restriction to the subspace X(E). Hence an inner

Symmetric Tensors 91

product is determined in the factor space ®E/N(E) such that

(mu, itv) = p !(u, v) u, v e X (E)

(cf. Formula 4.19). Formula (4.20) yields the relation

(m(x1 ®... Q xp), m(Y O ... O Yp)) = det(xi, y) xi E E, y; E E.

PROBLEM

Define a multiplication in X (E) such that the linear map p : X (E) - ®E/N(E)becomes an isomorphism. (This multiplication is necessarily uniquely determined.)Prove that

?LA u ?LA U = ?LA (u Ox U) u, v e Q E.

Symmetric Tensors

4.9. The Space Mp(E)

Consider the subspace M(E) of ppE generated (linearly) by the tensorsu - iu where u e QpE and i is a transposition. The space Mp(E) is stableunder every transposition. In fact, if v = u - iu is a generator of Mp(E) andt' is a transposition we have

= (i'u - u) - (iu - u) + (iu - ftu) E Mp(E).

The same argument as in Section 4.1 shows that

u - 6u e M(E)

for every u e QpE and every permutation 6.

4.10. The Symmetrizer

A tensor u e QpE is called symmetric, if

6u=u 6ESp.

The symmetric tensors form a subspace Y"(E) of ppE.Next consider the linear map Its: QpE -+ QpE given by

1TGS=

p Q

An argument similar to the one given in Section 4.2 shows that

ker ms = MP(E)

(4.21)

(4.22)

(4.23)


and

Im 7s = YP(E). (4.24)

Moreover, 7s is a projection operator,

7s = 7s (4.25)

Thus we have the direct decomposition

Q pE = Y(E) 8 M"(E). (4.26)

The operator 7s is called the symmetrizer in ppE andsymmetric part of u.

4.11. Dual Spaces

7s u is called the

Suppose now that E, E* is a pair of dual spaces and let is be the symmetrizerfor ppE* (p >_ 2). The same argument as that used in Section 4.3 shows that7s and is are dual operators,

<u *, s u > = <Su *, u > u * E ®P*, u e ppE. (4.27)

It follows from (4.27) that the restriction of the scalar product <, > to thesubspaces Yp(E*), YP(E) is again nondegenerate.

Now let

and

be decomposable tensors. Then we obtain from (4.27) and (4.25) that

<lrs(x * 1 O ... Ox x *p), TGs(xl ® ... ® xp)= <x* 1 O ... O x*p, TGS (x 1 OX ... Q xp)>

= <x * 1 O ... O x *p, TGS(x 1 OX ... ® xp)

= 1 x* i x ... x*p xP Q

Introducing the permanent of a p x p-matrix cx by

perm(x) _ aQc 1 ...

we can rewrite (4.28) in the formQ

(4.28)

1<lrs(x*l ®... p x*p), 1s(xi O ... O xp)> = - perm(<x*`, x;>). (4.29)

P

4.12. The Symmetric Part of a Product

Let ®E be a tensor algebra over E, and consider the subspaces M"(E) cppE. As before we set M°(E) = M1(E) = 0, and define 1s to be the identityon ®°E and p 'E. Then the formulas developed above continue to hold.

The Factor Algebra QE/M(E) 93

Now let v = u - -cu be any generator of MP(E), p > 2, and let w e ®"Ebe an arbitrary tensor. Then we have

v®w=u®x w-iu®x w=upw-where -r' E SP+q denotes the transposition given by

'i(v) l< v< p,

i (v) =v 1<v<p + _ _ p +q.

It follows that

MP(E) Q ®E c MP + q(E). (4.30a)

Similarly we obtain

®"E Ox M(E) C MP + q(E). (4.30b)

Now by the same argument as in Section 4.4 it is shown that

1s(u O v) _ u O 7s v)_ 1S(u Q 1C v) _ lcs(lcsu Q v) u E ®E, v e ®E. (4.31)

PROBLEMS

1. Show that a bilinear mapping p: E x E -+ G can be uniquely written in the form

P = (p i +

where 1i is symmetric and p2 is skew-symmetric.

2. For each p > 2 prove that

X (E) n Y(E) = 0.

3. Let p: E x . x E -+ F be a symmetric p-linear mapping such that

p(x,... , x)=0 x E E.

Show that p = 0.

The Factor Algebra ®E/M(E)

4.13. The Ideal M(E)


M(E) _ MP(E).P

Formulas (4.30a) and (4.30b) imply that M(E) is a graded ideal in the gradedalgebra ®E.


Suppose now that u e ®E and v e ®E are two arbitrary tensors. Acalculation similar to that in Section 4.5 shows that

u 0O v- v Qx u e M(E). (4.32)

4.14. The Algebra QE/M(E)

Consider the canonical projection

it : Q E - Q E/M(E).

Since M(E) is an ideal in ®E, a multiplication is induced in ®E/M(E) suchthat

it(a Q b) = ma itb a, b e ®E. (4.33)

It follows from (4.33) that this multiplication is associative, and that it(1) isa unit element. (4.32) implies that the multiplication is commutative as well.

Since M(E) is graded, a gradation is induced in the factor algebra by

®E/M(E) = ir(O PE)P

and so ®E/M(E) becomes a graded algebra. Since M 1(E) = M°(E) = 0,the restriction of it to Q 'E = E and Q °E = IT is an isomorphism. Conse-quently we identify it(Q 1 E) with E and it(Q °E) with F.

4.15. Symmetric Tensors

Let Y(E) c ®E be the space defined by

Y(E) = YP(E).P

Extend the projection operators its: QPE - QPE (its = l in Q 'E and ®°E)to a linear map its : ®E - ®E. Then

ker is = M(E)

Im ms = Y(E)

and

Q E = M(E) Q Y(E).

The restriction 6 of it to the subspace Y(E) is a homogeneous linear iso-morphism

6: Y(E) - QE/M(E)

The Factor Algebra QE/M(E) 95

of degree zero. If icY : ®E -- Y(E) denotes the restriction of its to ®E, Y(E) wehave the following commutative diagram :

O E Y(E)

irl

4.16. The Induced Scalar Product

(4.34)

Let E and E* be a pair of dual vector spaces and consider the induced scalarproduct in ®E and ®E*. According to Section 4.11 the restriction of thisscalar product to the subspaces Y(E) and Y(E*) is again nondegenerate.Consequently, a scalar product is induced in the factor spaces QE/M(E),QE*/M(E*) such that

<6u*, 6u> = p ! <u*, u> u* E Yp(E*), u e Yp(E). (4.35)

Clearly, the scalar product (4.35) respects the gradations. Moreover, itfollows from (4.35) and (4.34) that

<itu*, mu> = p !<msu*, msu> u* E QE*, u e QE. (4.36)

Now let u* and u be decomposable,

u = xi Q...Qxp, u* = x*1 Q...Qx*p.

Then formulas (4.36) and (4.29) yield

O ... O x*p), m(xi O ... O xp)> = perm(<x*i, x;>).

5Exterior Algebra

For this chapter E denotes a vector space over a field of characteristic zero.

Skew-Symmetric Mappings

5.1. Skew-Symmetric Mappings

Let E and F be two vector spaces and let

P

be a p-linear mapping. Then every permutation 6 E SP determines anotherp-linear mapping hip given by

6(p(.7C 1, ... , xP) = P(aCQ(1) , ... , .7CQ(P)).

It follows immediately that

(#r6)q, =

for every two permutations, and that

icp = co,

where i is the identity permutation. A p-linear mapping p is called skew-symmetric if

6(p = EQ (p

for every permutation 6 where EQ = + 1 or -1 if the permutation is, respec-tively, even or odd.

A p-linear mapping co is skew-symmetric if and only if

96

Skew-Symmetric Mappings 97

for every transposition -r. In fact, since a transposition is an odd permutationit follows that every skew symmetric mapping satisfies (5.1). Conversely,assume that co is a p-linear mapping which satisfies (5.1) and let 6 be anarbitrary permutation. Now 6 is a product of m transpositions where m iseven (odd) if 6 is an even (odd) permutation. It follows that co satisfies6cp = EQ co, and hence co is skew-symmetric.

This result implies that a p-linear mapping, gyp, is skew-symmetric if andonly if

p(x1, ..., xp) = 0, (5.2)

whenever x, = x; for at least one pair i j. In fact, suppose that co is skew-symmetric and assume that x, = x; (i j). Let -r be the transposition i ± j.Then

p(x 1, ..., xp) = - -rp(x 1, ..., x p) = - q (x 1, ..., x p),

whence p(x1, ..., xp) = 0. Conversely, assume that co satisfies (5.2). Thenif -r: i ± j is any transposition, it follows that

+ p(x1,...,x,..= p(x1,...,x, + x,...,x, + x;,...,xp)

cP+icp=0.

xi

Hence p is skew-symmetric.Since every transposition -r is a product of an odd number of trans-

positions of the form i ± i + 1, it follows that a p-linear mapping p is skew-symmetric if and only if

p(x 1, ..., xp) = 0, whenever x, = x, + 1 1 <i <p - 1.Formula (5.2) implies that a p-linear mapping, gyp, is skew-symmetric if

and only if

p(x 1, ... , xp) =0, (5.3)

whenever the x, are linearly dependent. In fact, it is clear that this conditionimplies that p is skew-symmetric. Conversely, let co be a skew-symmetricmap. Then if x 1, ... , xp are any linearly dependent vectors, we have

x; = t1x, (for some j).

Without loss of generality, we may assume j = p. It follows thatp-1

i

P(x l ... , xp) = ) p(x 1, ... , xp _ 1, x,) = 0,i=1

which proves (5.3).

98 5 Exterior Algebra

From every p-linear mapping co we can obtain a skew-symmetric p-linearmapping A by setting

1A = EQ 6cp.P Q

To show that A is indeed skew-symmetric let p be an arbitrary permutation.Then it follows that

p ! P(Aq,) = EQ EQ(P6)Q Q

= EP Ep Et P = Ep P !t

whence p(Aq,) = Ep A.The mapping A is called the skew-symmetric part of co, and the operator

A : p - Ac1, is called the antisymmetry operator. If co is skew-symmetric itselfwe have 6cp = EQ co for every 6 and hence it follows that

1-=cp.P Q

This equation shows that a skew-symmetric mapping coincides with itsskew-symmetric part. Since Ac1, is skew-symmetric it follows that A 2 = A.

Proposition 5.1.1. Let

cp:E x x E

p

be a p-linear mapping , where F is an arbitrary vector space, and letf : ®"E - F be the linear map induced by co. Then p is skew-symmetricif and only if N"(E) c ker f.

PROOF. co is skew-symmetric if and only if tp(x1, ... , xp) = 0, wheneverx = x for some pair (i , j), i j. But

p(x 1, ... , xp) = f (x 1 © ... ® xp)

and so co is skew-symmetric if and only if f is zero on the generators ofN"(E); i.e., if and only if N"(E) c ker f.

As an example of a p-linear skew-symmetric mapping, consider the p-linear mapping

p

Exterior Algebra 99

defined by

°

where ThA is the alternator (see Section 4.2). Since ker mA = N"(E), it followsfrom the proposition that /'A is skew-symmetric.

PROBLEMS

1. Denote by Lp(E; F) the space of all p-linear mappings (p : E x x E - F and byAp(E; F) the subspace of skew-symmetric mappings. Assume that

T:Lp(E;F) - Lp(E;F)

is a linear map such that

T cp = (p (p E A p(E ; F)

and

T(6p) = EQT((p) 6 E Si,, (p e Lp(E; F).

Show that T is the antisymmetry operator.

2. Let E be an n-dimensional vector space and suppose that A 0 is a determinantfunction in E.(a) Given an n-linear skew-symmetric mapping (p : E x x E -+ F show that

there is a unique vector b e F such that

(P(x 1, ... , xp) = O(x 1, ... ,

(b) Show that every (n - 1)-linear skew-symmetric function b can be written inthe form

(x1, ... , x,,_ 1) = O(x 1, ... , x,,_ 1, app),

where a E E is a fixed vector. Hint: Consider the n-linear mapping (p defined by

(-1)"-1b(xl,...,X.,...,x,,);.=1

Exterior Algebra


Let

np:E x x E- Ap

be a skew-symmetric p-linear map from E to a vector space A. We shall saythat n p has the universal property (for skew-symmetric maps) if it satisfiesthe following conditions:

A 1: The vectors A "(x1, ... , xp) (x, E E) generate A, or equivalently,Im A p = A.


A 2 : If p is a skew-symmetric p-linear mapping from E into any vectorspace H, then there exists a linear map f : A -p H such that thediagram

'HP

n°I

commutes. A

Conditions A 1 and A 2 are equivalent to the following condition (theproof being the same as in Section 1.4)

n : If

cp:E x x

P

is a skew-symmetric p-linear mapping, there is a unique linear mapf : A - H such that the diagram above commutes.

The skew-symmetry of A p implies that

whenever the vectors x1, ..., xp are linearly dependent. On the other hand,if the vectors x 1, ..., xp are linearly independent, then A "(x1, ... , xp) 0.In fact, assume that A "(x1, ..., x,,) = 0. Then, by A2, /i(x 1, ..., x,,) = 0for every skew-symmetric p-linear mapping i'/. In particular,

JCA(x 1 p= 0.Thus, by Section 4.2, the vectors xv are linearly dependent.

5.3. Uniqueness and Existence

Suppose that

A and

are two p-linear mappings with the universal property. Then there are linearmaps

f: A - A and g: A - A

such that

f o A"= A" and g o n p= A ".

Exterior Algebra 101

Now condition n 1 implies that

gof= i and f i g= i.

Thus f and g are inverse isomorphisms.To prove existence, set

A "E = Qx "E/N"(E)

(see Section 4.1) and let A" denote the p-linear mapping

P

defined by

A "(x 1, ..., xp) = it(x 1 Q ... Q xp), (5.4)

where it denotes the projection (see Section 4.6). In view of Proposition5.1.1, A "is skew-symmetric. Property A 1 follows directly from the definition.To verify A 2' let

q:E x x

P

be any p-linear skew-symmetric mapping. Then p determines a linear maph : p "E - H such that

h(x 1 Q ... Q xp) = p(x 1, ..., xp) xy e E (5.5)

(see Section 1.20). Since p is skew-symmetric, it reduces to zero in thesubspace N"(E) and so it induces a linear map f : A "E -p H such that

fom=h.

Combining this relation with (5.5) we obtain

cp(x 1, ... , xp) = f TG(x 1 Qx ... QX x p) = f A "(x1,..., x p)

whence co = f o A".

Definition. The pth exterior power of E is a pair (A, A p), where

A":E x x

P

is a skew-symmetric p-linear mapping with the universal property. Thespace A (which is uniquely determined up to an isomorphism) will also becalled the pth exterior algebra of E and is denoted by A "E.

The elements of A "E are called p-vectors.


Next we shall give a description of the pth exterior power of E in termsof the subspace X "(E) c ®"E of skew-symmetric tensors. Consider the skew-symmetric p-linear mapping p: E x x E -+ ®"E given by

P(x 1 ..., xp) _ iA(x1 Q ... Q xp) xy e E,

where itA denotes the alternator (see Section 4.2). By the universal propertyit induces a linear map r: n pE -+ ®"E. We show that the diagram

QpE

AE - , opE

commutes. In fact,

rpc(x 1 Q ... Q xp) = i n "(x 1, ..., xp) = p(x 1, ... , xp)_ iA(x 1 Qx ... ® xp)

and so

r O it _ TGA . (5.6)

This relation implies that Im i = Im ltA (since it is surjective). But Im itA =X "(E) and so we have

Im r = X "(E).

On the other hand, it is easy to check the relation

i i determines a linear isomorphism fromn pE onto X "(E).

r: n pE 3 X "(E).

Remark. Since i is injective, the formula i o 7 = iA implies that ker iA =ker it. Thus we have the relation ker iA = N"(E) which was proved inSection 4.2 in a different way.

5.4. Exterior Algebra

Extend the definition of n pE to the cases p = 1 and p = 0 by settingn 1E = E and A °E = IT. Consider the direct sum, n E, of the spaces n pE(p = 0, 1,...) and identify each n pE with its image under the inclusionmap. Then we can write

AE _ ARE.p=o

Exterior Algebra 103

The projections it : QpE -+ A "E (with kernel N"(E)) determine a pro-jection

ir:QE--, AE

with kernel N(E) = >J N"(E). Thus we have a linear isomorphism

f : Q E/N(E) =- AE.

Now recall from Section 4.6 that ®E/N(E) is an associative algebra. Hencethere is a unique multiplication in A E, denoted by A, such that f becomesan algebra isomorphism. Thus we have

u A v = m(u Qx v) uE AE, vE AE,

where u e ®E, v e ®E are elements such that mu = u and iv = v. Thismultiplication makes A E into an associative algebra with the scalar 1 asunit element. It is called the exterior algebra over E. It is generated (as analgebra) by the vectors x e E and the scalar 1.

Formula (5.4) can now be written in the form

A "(x 1, ... , xp) = x 1 A ... A xp x E E.

From (4.16) we obtain the relation

uAv=(-1)"vAu ueARE, veAgE.

In particular,

u A v= v A u if p or q is even. (5.7)

and

u AU =0 if p is odd.

The kth exterior power of an element u e A E is defined by

1uk = u A A u

k.k>1 ,

k

It follows that

k+lUk A U1 =

k)uk+1 u E A E.

Now let u e A "E and v e A qE be arbitrary and assume that p or q iseven. Then it follows from (5.7) that u A v = v A u. This yields the binomialformula

(u + v)k = u` A vji+ j=k

for u e A "E, v e A qE, p or q even.


Now we shall describe the exterior algebra n E in terms of the subspacesXP(E) ®"E (see Section 4.2). Consider the direct sum

X (E) = X "(E)P

and define a multiplication in X(E), denoted by r, by setting

a r b= TCA(a Q b) a, b e X (E),

where mA denotes the alternator. Recall from Section 5.3 the linear isomorph-isms ri : n "E 3 X (E). We shall show that ri preserves products,

ri(u n v) = ri(u) r i(v) u e A "E, v e n qE.

In fact, write

u=TCU, v=m15 u,vEQE.

Then we have, in view of the commutative triangle in Section 5.3 and Formula(4.10)

ri(u A v) = ri(TC u A miS) =rim (u Q v)

= TCA(u OX v) = TCA(TCA u OX TCA v)

N N N N= TCA u n TCA v = riTCU r r7TCV = riu r riv.

Thus ri is an algebra isomorphism ri : n E X (E).

5.5. The Universal Property of A E

Let A be an associative algebra with unit element a and let

h:AE - A

be a homomorphism. Then a linear map o : E - A is defined by

'p=h°i,where i is the injection of E into n E. It follows from (5.7) that

(rpx)2 = 0 x E E. (5.8)

Conversely, assume that 'p : E - A is a linear map satisfying (5.8).Then there exists precisely one homomorphism h : n E - A such that

h(1) = e and

'p = ho i.

For the proof we note first that (5.8) implies that

cpx coy + coy cpx = 0 x, y e E.

Exterior Algebra

In fact, if x, y e E are arbitrary elements, we have

cpx coy + coy cpx = q (x + y). q (x + y) - cpx cpx - coy coy = 0.

To define h consider, for every p >_ 2, the p-linear mapping

oc:E x x E-+A

defined by

p

105

a(x 1, ... , x p) = cpx, ... cpx p .

Then it follows from (5.9) that a is skew-symmetric and hence there exists alinear map h": n pE -+ A such that

hP(x 1 n ... n xp) = (px, ... cpxp p > 2.

Define h 1 and h° by h 1 = co and h°(1) = e and let h : n E -+ A be the linearmap whose restriction to n PE is equal to h", p >_ 0.

To prove that h is a homomorphism, let

u = xl n n xp and v = x,+, n ... n xp+q

be two decomposable elements. Then we have

h(u n v) = h(xl n ... A xp+q) = (x1 ... (pxp+q

= (px, ... (px p) (px p +, ... (px p + q)

= h(x, n ... A xp) . h(xP+ 1 n ... A x p +q) = hu hv.

The uniqueness of h follows from the fact that the algebra n E is generated bythe vectors of E and the scalar 1.

If A is a positively graded associative algebra, A = >P Ap, and co is a linearmap of E into A 1 it follows that the extending homomorphism h is of degree

zero.Let U be an associative algebra with unit element 1 and let E : E -+ U

be a linear map with the following property : If o : E -+ A is a linear mapinto any associative algebra A with unit element a such that 0,

x e E, then there exists precisely one homomorphism h : U -+ A such that

h(1)=e and h o E _ cp.

Then we say that the pair (U, E) has the universal exterior algebra property.An argument completely analogous to that found in Section 3.4 shows

that if the pairs (U, E) and (U', E') have the universal exterior algebra propertythen there exists a unique isomorphism f : U -+ U' such that f o E = E'.

It follows from the results of this section that the pair (n E, i) has theuniversal exterior algebra property, where i : E -+ n E is the inclusion map.Now the above uniqueness theorem implies that for every universal pair(U, E) there exists a unique isomorphism f : n E -+ U such that f o i = E.

106

5.6. Exterior Algebra Over Dual Spaces

5 Exterior Algebra

Let E, E* be a pair of dual vector spaces, and consider the exterior algebrasover E and E*. In view of the induced isomorphisms

f: ®E/N(E) n E and g: Q E*/N(E*) 3 E*

it follows from Section 4.8 that a scalar product <, > may be defined betweenn E and A E* such that

x*1 n ... n x*P, xi n ... A xP> = det(<x*`, xj>) p 1

A,,1u E lT (5.10)if pq.

Condition A 1 implies that the scalar product <, > is uniquely determinedby (5.10). We also have that the restriction of < , > to the pair n PE*, A PEis nondegenerate for each p, and so induces a duality between these spaces.In particular, the restriction of < , > to n 1E* = E* and A 'E = E is justthe original scalar product.

Now expanding the determinant in (5.10) by the ith row we obtain theformula

Plx* 1 n ... A x*P, x 1 n ... n XP> _ (- 1)i + j<x*i, xi>,\

j=1

<x* 1 n ... n x*J n ... n x*P, x i n ... n xi n ... A XP> p> 2.(5.11)

5.7. Exterior Algebra Over a Vector Space of Finite Dimension

Suppose now that E is a vector space of dimension n and let {ev} (v = 1, ... , n)be a basis of E. Then the products

ev 1 A ... n evp , (v1 < ... < vi,) (5.12)

form a basis for n PE. In fact, it follows immediately from A 1 that the pro-ducts (5.12) generate n PE. To prove the linear independence, let E* be adual space of E. If a*v (v = 1, ..., n) is the dual basis in E* we have, in viewof (5.10),

<e*''1 n n e*vp, eµ1 n .. A eµp> = det(<e*vi, det(b). (5.13)

This formula shows that , the products ev1

n n evp and e*v 1 A A e*vp(v1 < < vP) form dual bases of n PE and A PE*. Hence, the dimensionof n PE is given by

dim AE _ n (0 < p < n). (5.14)P

Exterior Algebra 1 07

For the dimension of the exterior algebra n E we obtain from (5.14)it

dimAE= n =2",p=o \PJ

while the Poincare polynomial of the graded vector space n E is given by

" nP(t) = t" = (1 + t)". (5.15)

p=o PEvery p-vector u can be uniquely represented in the form

u = cv1, ..., .Pev1 A ... A evP

<

where the symbol < indicates that the indices (v 1, ... , vp) are subject tothe condition v 1 < < vp . The coefficients V1 ° . °

VP are called the com-ponents of the p-vector u with respect to the basis {ev} of E. Formula (5.13)implies that the scalar product of two p-vectors

U=V1s ..., VPev1 A . A

eyP

and

is given by

e*v1 A ... A e *Vp

<u*, /u\ = V1, ..., vP'1 vt, ,vP

Inner product spaces. Suppose now that E is an inner product space and setE* = E. Let A E be an exterior algebra over E. The isomorphism

f : QE/N(E) -- AE

(see Section 5.6) determines an inner product in n E such that

(x1 A ... A xp, Y1 A ... A yp) = det(x1, y).

If E is Euclidean and {eV} (v = 1, ... , n) is an orthonormal basis of E, itfollows that the products

eV 1 A ... A evP v1 < ... < vp

form an orthonormal basis of n "E.

PROBLEMS

1. Let E, E* be a pair of n-dimensional dual vector spaces and consider the subspaceA c L(E*; E) consisting of the transformations satisfying p* = - 'p. Define abilinear mapping 'p : E x E - A by

<

cpa,bx* = <x*, a>b - <x*, b>a.

Prove that the pair (A, 'p) is a second exterior power of E.


2. Show that a 2-vector z is decomposable if and only if z n z = 0.

3. Let E be a vector space of dimension n. Show that a 2-vector is decomposable ifand only if the matrix of its components (with respect to a basis of E) has rank 1.

4. Establish the general Lagrange identity

n n

v1 v1

n n1 ... bvnpv p v

b11 ...iP '11 ... niP

= det )det(:ZP1 ... 'pP yIpl ... IpP

v1 v1

Hint: Employing a pair of dual bases {e*v}, (v = 1, ... , n), consider the vectors

x = Jev and y*` = '1 e*v

Evaluate the scalar product

<x i n ... n xp , y* 1 n ... n

(i = 1, ... , P)

y*p>

in two different ways.

5. Assume that E has finite dimension, and consider a differentiable mapping(J - n pE, t H u(t). Establish the formula

dUk = _U n Uk - 1.

dt dt

6. Let E be any vector space and n E be an exterior algebra over E. Define a newmultiplication in the space n E by

u n v = + q) U A V u E n pE, v E n qE.P!q!

Prove that the resulting algebra n E is again an exterior algebra over E.

7. Consider the subspace _ o n 21'E of n E. Show that this subspace is a commutativealgebra which is algebraically generated by 1 together with a set of elementssatisfying w = 0. This algebra is called a Nolting algebra over E.

Homomorphisms, Derivationsand Antiderivations


Let A E and A F be exterior algebras over the vector spaces E and F andassume that a linear map co : E -p F is given. Then co can be extended in aunique way to a homomorphism cP A , n E - n F such that PA(l) = 1. Toprove this consider the linear map rl : E - n F defined by ri = i o co where i

Homomorphisms, Derivations and Antiderivations 109

denotes the injection of F into A F. Then we have for every x e E

ri(x) A ri(x) = cpx A cpx = 0 (since cpx e F).

Now it follows from the universal property of A E (see Section 5.5) that ri(and hence co) can be extended in a unique way to a homomorphism

A E - A F. Clearly 'P A is homogeneous of degree zero. Since 'Ppreserves products we have

'A(u A v) = 'Au A (PAv u,ve AE. (5.16)

In terms of the multiplication operator (see Section 5.13) this can be written as

'PA oµ(U) = n u) ° 'P n u e AE. (5.17)

From (5.16) we obtain the formula

'Pn(x1 A...Axp)=(Px1 A...Acpxp x,EE. (5.18)

It follows immediately from (5.18) that Im 'P A = A (Im 'P). In particular,if 'P is an onto map then so is 'P A The kernel of 'P A will be discussed inSection 5.24.

For the identity map i : E - E we have obviously

= l (5.19)

while the homomorphism (- z),, is given by (- i) u = (-1)"u, u e A "E;hence (- z),, is the canonical involution of the algebra A E.

If ,Ii is a linear map of F into a third vector space G and A G is an exterioralgebra over G, it is clear that

(/' °'P) = '/' A °'A (5.20)

Formulas (5.20) and (5.19) imply that P A is injective whenever (p is injective.In fact, if 'P is injective, there exists a linear map /i : E - F such that /' ° 'P = 1.Now formulas (5.20) and (5.19) yield

/'A ° 'PA = (`Y ° 'P) = l n = l

and hence 'P n is injective.In particular, if E 1 is a subspace of E, then the injection i : E 1 - E induces

a monomorphism i A: A E 1 - A E. Hence, A E 1 can be identified with asubalgebra of A E.

5.9. Dual Mappings

Suppose now that E*, E and F*, F are two pairs of dual spaces and that twodual mappings

and 'P*:E*4 F*


are given. Consider the induced mappings

(P A: A E - AF and ((p*) : A E* E- AF*.

The mapping (gyp*) A will be denoted by p ^ . Then co A and co ^ are againdual,

SPA= (SPA)* (5.21)

In fact let u e A E and v* E A E* be two arbitrary elements. Since co A andP ^ are both homogeneous of degree zero we may assume that u and v*are homogeneous of the same degree, say p. Furthermore, we may assumethat the p-vectors u and v* are both decomposable

u= x 1 A ... A xp , U* = y* l A ... A y*P.

Then relations (5.18) and (5.10) yield

<v*, (p, u> = < y* 1 A ... A y*P, (px1 A ... A Pxp

= det(<Y*`, px;>) = det(< p*Y*`, x;>)

= A ... A (p*y*P, x1 A ... A xp> = <A v*, u>whence

<v*, (PA u> = <rp^v*, u> v* E AE*, u e A E.

If G, G* is a third pair of dual spaces with exterior algebras A G and A G*and if l//: F -+ G, /i* : F* E- G* is another pair of dual mappings, we have

( ° P) = ((/I° q)*) = ((P* ° II*)A = (I*)A = (P^ o//whence

( ° P) = (PA ° ,A (5.22)

5.10. The Induced Derivation

Let p: E -+ E be a linear transformation. Then p can be extended in a uniqueway to a derivation, 0 A (gyp), in the algebra A E. The uniqueness of 0A(p)follows immediately from the fact that the algebra A E is generated by thevectors of E and the unit element 1.

To prove existence, consider the p-linear mapping

t/ip:E x x E--> APE

defined byn

P

Y' (x1, ... , xp) = x 1 A ... A (pxi A ... A xp p > 2i= 1

I'1=(P and /io=0.


If for some i <j, xi = x; = x, then

+x1 A...AXA".A(pXA AXP

=(-1x1 A...AxAgxA...AxP+(-1}'-1-`x1 A...AxAgxA...AxP

=0.

Hence P is skew-symmetric. It follows that there is a linear map,8 A (gyp) : A E -+ A E such that

P

e n (x 1 A ... A xp) = x 1 A ... A (pxi A A XPi= 1

and such that

8 A (cp)x = cpx x E E,

p > 2 (5.23)

e o a. e F.

Clearly OA(P) is a homogeneous (of degree 0) linear map extending 'p. Toprove that 8 A ('p) is a derivation let u = x1 A A xP and v = Y1 A A Yq

be arbitrary decomposable p- and q-vectors. Then

e n (co) (u A v) = e n('p) (x 1 A ... A x p A Y 1 A ... A Yq)

P_ (x1 A "A 'pXA . A xP AYlA...AYq

i= 1

q

+ (x 1 A ... A xp) A (Y 1 A ... A 'Yj A ... A Yq)i= 1

= 8 A (cp)u A v+ u A 8 A (cp)v.

Now the linearity of 8 ('p) gives

8 A ('p) (u A v) = 8 A (cp)u A v+ u A 8 A (cp)v u, y e AE

and hence 8 A ('p) is a derivation. In terms of the multiplication operator thisformula reads

8 n ('p) ° z(u) = µ(e n ('p)u) + z(u) ° e n ('p). (5.24)

For the identity map we obtain that

e A (l)u = pu u E n PE.

If l//: E -+ E is a second linear transformation, then we have the relation

8 A (q // - /,'P) = 8 A ('p)e A (//) - 8 (//)e A ('P). (5.25)


For the proof, we notice first that the operation on each side of (5.25) is aderivation in A E (see Section 5.6 of Linear Algebra) and consequently itis sufficient to consider the restriction of these operators to E. But in this case(5.25) is trivial.

Now suppose that cP : E - E, q : E* F-- E* is a dual pair of linear maps,and consider the induced derivations OA(P) and 8 ^ (cp) = 8 ^ (gyp*). It willbe shown that 8^ (gyp) and 8 A (p) again form a dual pair

(5.26)

Let u = x1 A A x1 ,e APE and u* = x* 1 A A x' e APE*. Then, inview of Formula (5.11), we have

P

<u*, 0A ()u> = <.x* 1 A ... A x*P, x 1 A ... A A ... A XP>i= 1

p

(- 1)<x*j, (Pxi> <x* 1 A ... A x*1 A ... A x*p,i, j = 1

P

whence

( 1)1+'<q*x, xi> <x*1 Ai, j = 1

P*1_ <x n n

j= 1

= <e ^ (co)u*, u>,

P*xA

x 1 A ... A 4xi A ... A XP>

/NAx*jA...AX*P,x1A...A4XIA...AX>

... A x*P x 1 A ... A xP

<u*, 8 ^ (cp)u> = <8 ^ (cp)u*, u> u* e A E*, u E AE. (5.27)

If /i : E - E and /i* : E* F- E* is a second pair of dual mappings, thenformulas (5.25) and (5.26) yield

e ^ /,'p) = e ^ (//* p* - */,*) = e ^ (//*)e ^ (p*) - e ^

e^(/,)e^(p) - e^(p)e^(/i)

whence

8 i/np) = 8 ^(/i)e 8 (/i). (5.28)

5.11. Antiderivations

Let w be a homogeneous involution of degree zero in A E and let be a (notnecessarily homogeneous) antiderivation with respect to w. If co denotes therestriction of S to E, we have

A x)=(px A x+wx A (px,


whence

(px A x+ wx A cpx = 0 x e E. (5.29)

Conversely, every linear map co : E -+ A E which satisfies (5.29) can beextended in a unique way into an antiderivation p) (with respect to w)of A E. Since A E is generated by E and the scalar 1, the uniqueness followsat once. To prove the existence of (co) the p-linear mapping

t/i p: E x x E -- A E

defined by

p

P

... , xp) _ wx 1 A . A wxv _ 1 A (pxv A xv + 1 A .. A x p p 2v=1

1 = p and /io = 0. (5.30)

It will be shown that p is skew-symmetric. In view of Section 5.1 it issufficient to verify that //(X1, ..., xi , xi + 1, ... , xp) = 0 whenever xi = xi + 1

Since wx A (Dx = 0 and x A x = 0, we obtain from (5.30) that

... , x, x, ..., xp) = wx 1 A . . A wxi _ 1

whence, in view of (5.29),

l'/p(xl,...,X,X,...,xp) = 0.

The skew-symmetric mappings p induce a linear map cp) : A E -+ A Esuch that

P

SZ((p) (x 1 A A x p) _ wx 1 A ... A wxv _ 1 A (pxv A xv + 1v=1

(5.31)

Now it will be shown that (co) an antiderivation with respect to theinvolution co. Let u = x 1 A A xp and v = xp + 1 A A x p + q be twodecomposable elements. Then

(co) (u A U) = (x 1 A ... A x p + q)

P

wx 1 A ... A wxv _ 1 A (pxv A xv + 1v=1

A."AxP+q

q

+ wxl A ... A wxv_ l A (pxv A xv+ l A ... A xp+qv=p+ 1

= A v + (DU A

which completes the proof.


It is clear that p) is homogeneous of degree k if and only if Im o cn k+ 'E. Assume now that is in fact homogeneous of degree k.

The following two cases are of particular importance:1. w = i A (derivations) : Condition (5.29) reduces to

(pxnx+xn(px=0or, equivalently,

[1 + (- 1)k+ l]q x n x = 0. (5.32)

If k is even (5.32) always holds: any linear map cp : E -+ n 2P + 'E can beextended in a unique way to a homogeneous derivation in n E.

Now assume that k is odd. Then equation (5.32) reads

(px n x=0 xeE,whence

(px n y=- spy n x x, y E E.

Since k + 1 is even we have

(pynx=xn(py.It follows that

(px n y = -x A py.


n ... n xp) = i[1 + (-1)p+ n x2 n ... n xp. (5.33)

It follows in particular that the restriction of cp) to every subspace n 2Eis zero.

2. w = (- i) A (antiderivations) : Condition (5.29) becomes

(pxnx=xncpx,i.e.,

[1 + (-1)k] cpx n x = 0.

If k is odd this condition is always fulfilled: any linear map p:E -+ n 2Ecan be extended in a unique way to an antiderivation (with respect to thecanonical involution) in n E.

Now assume that k is even. Then the above equation implies that

px n x=0,whence

ox^y= -coy nxor equivalently

cpx n y = x n cp y.


Formula (5.31) now yields

n ... n xp) = 2 [1 + (- 1)P + i ] (px 1 n x2 n ... n xp. (5.34)

It follows again that the restriction of to every subspace n 2E is zero.

5.12. a-Antiderivations

Suppose now that F is a second vector space, WF is a homogeneous involutionof degree zero in AF and that (p : E -+ AE, /i : F -+ AF are homogeneousmappings satisfying the conditions

(px n x+WEx n (px=0 xEE,

/iYAY+(DFYA/iY=0 yeF.

Assume further that a : E -+ F is a linear map such that

WF a A = a A WE and /ia = a (p. (5.35)

Then we have

= OC n ° E((P) (5.36)

In fact, it is easy to verify that the operators a A and a A °are a-antiderivations (see Section 5.8 of Linear Algebra). Relation (5.35)implies that the restrictions of these a-antiderivations to E coincide and50 (5.36) follows.

PROBLEMS

1. Let E and F be two vector spaces, QE, QF tensor algebras and n E, A F exterioralgebras over E and F respectively. Consider the projections ltE : Q E -+ n E andF: ®F - n F defined by

2E(x 1 0 ... QX x,,) = X1 A ... n x p

and

2F(Y 1 0 ... 0 Yq) = Y 1 n ... n Yq .

(a) If p: E - F is a linear map prove that

.= 'PA °E

(b) If p: E - E is a linear map show that

=

2. Let a e n kE be a fixed element, where k is odd. Define a linear map 9: n E -+ n E by

eu =JaAU u E n "E, p odd,

0 u e n "E, p even.

Prove that 0 is a derivation in the algebra n E.

116

3. Suppose w is an involution of degree zero in n E such that

WX n x=0 xEE.

Prove that w = l ^ or w = (- i) .

5 Exterior Algebra

4. Show that there does not exist an involution w in n E such that (5.29) holds for everylinear map

p:E - E.

Hint: First show that for such an w the relation wx n x = 0 must hold.

5. Let E and F be vector spaces of finite dimension and consider a linear map p: E -+ Fof rank r.Prove that

r(q) = 2'.

6. Let A E be the exterior algebra over E and consider an antiderivation S2 in A Eof degree 1. Define a new multiplication n in n E by setting

u n v= p+q)UAV uE npE, vE nyE.

p

Show that the operator C defined by S u = p S2u, u e A "E is an antiderivation withrespect to this multiplication.

7. Let d : ®E -+ ®E be a homogeneous linear map of degree 1 such that

d(u Q v) = du Q v + 6(u Q dv),

where 6 is a fixed permutation such that EQ = (-1)deg u. Assume that

ltA dltA = 1tA d,

where 7tA is the alternator (see Section 4.2). Define the operator S by S = 7rAd.Prove that S is an antiderivation in the graded algebra X(E) (cf. the problem inSection 4.8). Prove that S is a differential operator if and only if 7rA d 2 = 0.

8. Let l//: E -+ E be a linear transformation of an n-dimensional space E. Prove thatthere exist uniquely determined transformations iii (i = 0, ... , n) of n E such that

n

('Y - Al) = tIiA"-` AE F.i=o

If cii") denotes the restriction of t ii to n" E, show that

=ai i=O,...,n,tr /4'

where ai is the ith characteristic coefficient of cli (see Section 4.19 of Linear Algebra).Prove that

o = (-1)nl, li1 = (-1)"- II' =

The Operator i(a) 117

The Operator i(a)

5.13.

Fix an element a e A E and consider the linear transformation µ(a) of n Egiven by left multiplication with a,

µ(a)u = a n u u E n E.

Since the algebra n E is associative, we have the relation

µ(a n b) = µ(a) ° µ(b) a, b e A E. (5.37)

Now consider the dual map

i(a) : n E* F- AE*.

It is determined by the equation

(i(a)u*, v> = <u*, a n v> v e A E.

In particular,

i(t)u* = tu* e F.

Now suppose that a is homogeneous of degree p. Then i(a) restricts tolinear maps

n rE* _ n r pE*, r> pand reduces to zero in n rE* if r <p. For u* E A PE* we have

i(a)u* = <u*, a>.

Dualizing formula (5.37) we obtain

i(a n b) = i(b) ° i(a) a, b e A E. (5.38)

In particular,

i(a n b) = (-1)' i(b n a) a e n "E, b E ME. (5.39)

Next, let F be a second vector space and let p: E -p F, p* : E* F- F* be apair of dual maps. Since co is a homomorphism,

'PA oµ(a) µ((PAa) °'PA

Dualizing this relation yields

aE AE.

i(a) ° 'P " = 'P " ° A a) a E A E. (5.40)

Finally, let 'P and (p* be a pair of dual linear transformations of E, and con-sider the induced derivations H A ((P) and 8 A (p) Then Formula (5.24) impliesthat

i(a) ° 8 " (rp) = i(8 A (sp)a) + 8 ^ (rp) ° i(a).


5.14. The Operator i(h)

In this section we shall consider the operator i(h) for the special case h e E.This operator is homogeneous of degree -1. Formula (5.39) implies that

i(h) o i(k) + i(k) o i(h) = 0 h, k e E.

In particular,

i(h)2 = 0.

Proposition 5.14.1. The operator i(h) is an antiderivation in the algebra n E*,

i(h)(u* A v*) = i(h)u* A v* +(-1)pu* A i(h)v* u* E n PE*, v* E A E*.

PROOF. Consider the linear map tph : E* -+ IT given by

(ph x* = <x*, h>, x* E E*.

It follows from Section 5.11 that c°h extends to an antiderivation h of degree-1 in n E* (with respect to the canonical involution). We have to show that

i(h),

v> = <u*, h n v> u* E A E*, v e A E. (5.41)

We may assume that u* and v are decomposable, u* = x* n n xp andv = x 1 n n xq . If p q + 1, both sides of (5.41) are zero and so onlythe case p = q + 1 has to be considered. Then we have, in view of (5.11),

= <Qh(x* n ... A xp ), x 1 A ... A X,_1>P n

= (- 1)v- 1<x*, h> <x* n ... n x* A ... Av= 1

xp, x1 n ... A X_1>

and so Formula (5.41) follows.

Corollary I:

i(h) o 1u(h*) + µ(h*) o i(h) = <h*, h>l he E, h* E E*.

PROOF. Apply the proposition with u* = h*.

Corollary II:P n

i(h)(x* n ... n xp) _ (_ 1)v -1 <x* n ... n x* n ... n 4v=1

Corollary III: Let F be a subspace of E and let Fl be its orthogonal complement.Identify A F and A Fl with subalgebras of n E and A E* respectively. Then

i(a)(u* n v*) =(-1)pqu* n i(a)v* a e A "F, u* E n qFl, v* E A E*.

The Operator i(a) 119

PROOF. We may assume that a = yl n n yp, yv e F. Since i(y) is an anti-derivation, we have for y e F

i(y)(u* n v*) = i(y)u* n v* + (=1)qu* n i(y)v*= (_1)u* n i(y)v*.

It follows that

i(a) (u* n v*) = i(y1 n ... n yp) (u* n v*)= i(yp) ... i(yi)(u* n v*)= (_ 1)1 i * n i(yp) ... i(y i )v*= (-1)P u* n i(a)v*.

Proposition 5.14.2. If an element u* E A PE* (p > 1) satisfies the equationi(h)u* = O for every h e E, then u* = 0.

PROOF. Let v e A E be arbitrary. Since p >_ 1, we can write

v = by n vy by e E, vy e A 'E.V

It follows that

= <u*, by n vy> = <l(hy)u*, vy> = 0v v

whence u* = 0.

PROBLEMS

1. Let u* e n PE*, p > 1 be an element such that i(a)u* = 0 for every a e n kE, wherek < p is a fixed integer. Prove that u* = 0.Hint: Use the duality of the operators i(a) and µ(a).

2. Let E, E* be a pair of n-dimensional dual spaces and {e*°} be a pair of dualbases. Given a linear transformation p: E -+ E show that

e ('P) = µ(coev)i(e*")

O\(p) =

3. Let E, E* be a pair of n-dimensional dual spaces. Prove the following relations:

(a) µ(x)i(x*) + i(x*)u(x) = <x*, x>i x* e E*, x e E

(b) µ(e ji(e)u = pu u e n "E

(c) i(e* (n - p)u u e n "E.


4. Let E, E* be a pair of dual spaces of dimension n = 2m. Prove the formula

h*Ai(h)UjAU1J= 1

he A2E* (j = 1,...,m).

5. Show that the operator

1(a): n E* - n E* a E A "E

is not an antiderivation unless p = 1.

6. Let a E A "E be arbitrary and assume that p < q. Prove the formula

i(a)(x*1 A A x*q)

= (- 1(v -1) la, x*V1 A A x*Vp>x*Vp+ 1 A A x*Vq,v1<...<vp

where (vp+ 1, ... , vq) denotes the complementary ordered (q - p)-tuple.

Exterior Algebra Over a Direct Sum

5.15.

Let E and F be vector spaces and consider the anticommutative tensorproduct of the graded algebras n E and A F (see Section 2.8.) On the otherhand, we have the exterior algebra over the direct sum E Q F.

Theorem 5.15.1. There is a canonical isomorphism between the graded algebrasAE Q n F and n (E Q F).

PROOF. Let

i1:E-+E +QF and +QF

be the inclusion maps. They extend to homomorphisms

(i 1) A : n E - n (E Q F) and (i2) A : n F - n (E p F).

Hence a linear map

f:AEQ A F- A(E®F)

is given by

f (u Q v) = (i 1) A u n (2)A v. (5.42)

Exterior Algebra Over a Direct Sum 121

We show that f is an algebra homomorphism. In fact, let u e A E, v e n qF,u' E n rE, v' E A F. Then, if the multiplication in the algebra n E Q A F isalso denoted by A, we have

f [(u O v) n (u' O v')] _ (-1)qr f{(u n u') Q (v n v')]

_ (u n u') n (i2) A (v n v')

n (il) n u' n (i2) v n (i2) A v'

= (ll)n U A (l2)A v A (ll)n U' A (l2)A v'

= f (u ® v) n f (u' ® v')

To show that f is an isomorphism we construct an inverse homomorphism.Consider the linear map ri : E +O F -+ AE Q AF given by

ii(z) _ it1 z 0 1+ 1 Q ic2 z z e E +0 F, (5.43)

where

it 1: E +Q F -- E and it2 : E +Q F -- F

are the canonical projections. Then we have

ii(z) n ii(z) _ (itlz O 1+ 1 O it2 z) n (itlz O 1+ 1 O it2 z)

_ (it1z n it1z) p 1+ it1z O it2 z- it1z Q it2 z+ 1 Q (it2 z n

Hence r extends to a homomorphism

h: n (E Q F) -- AE Q AF.

Relations (5.42) and (5.43) imply that

hf(x®1)=h(i1x)=ri(ilx)=x xeE,

hf(1Oy)=h(i2y)=rl(i2y)=y yeF,

it2z)=0.

f h(z) = f (TG 1 Z xO 1 + 1 Q TG2 z) = i 1 T G 1 z + l2 TG2 z = z z E E Q F.

Since the vectors x Q 1 and 1 Q y together with the scalar 1 generate thealgebra n E Q A F and the vectors x +0 y together with 1 generate thealgebra n (E Q F) it follows that

hof=l and foh=i.Thus f is an isomorphism and h is the inverse isomorphism.

Corollary. Let E = E1 Q E2 be a direct decomposition of E into two sub-spaces. Then A E A E1 p n E2.

122

5.16. Direct Sums of Linear Maps

5 Exterior Algebra

Suppose that E', F' is another pair of vector spaces and that linear mapsP : E - E', /i : F - F' are given. Then the linear map p Q /i : E Q F -E' Q F' is given by

P +O = ii°(P°lt1 + i2°//°1t2, (5.44)

where

and

denote the canonical injections. It will be shown that

+O/) = ®I1From (5.44) we obtain

(' +O /,) ° it = ii °' and (gyp //) ° i2 = i2 ° /i

(5.45)

whence

°('P +O') A ° (i 1) A = ('l) A ° A and ('P +O') A ° (2)A = ('2)A

Now let u e A E and v e A F be two arbitrary elements. Then we have

+O ) (u O v) = +O Y') n ((i l) n u A (12)A v)

= O+ u] n O+ v]

= ('I) A 'PA U A ('2)A `/' A 1,

=('PA OY1n)(UOv)

whence (5.45).Now consider a linear map ' : E - F and suppose that two direct

decompositions

E=E1 f3E2 and F=F1 f3F2

are given such that 'Ei c F1 and 'E2 c F2. Then it follows from (5.45)that

'PA = ('P1) O ('P2) ' (5.46)

where 'P1: E1 -p F1 and 'P2 : E2 -p F2 are the restrictions of 'P to E1 and E2.Formula (5.46) yields, in view of (1.12) and (1.11), that

ker 'P = (ker ('Pa) A) O A E2 + n E 1 O (ker ('P2) A)

and

Im 'P A = Im ('P 1) A O Im ('P2) A

Exterior Algebra Over a Direct Sum 123

5.17. Derivations

Suppose p: E -+ E and l// : F -+ F are two linear maps and consider theinduced derivations OA(P). A E -+ n E and 8 A (li) . A F -+ n F. Then

0A(Y' o /') = 0A(') o f+ Z o eA(/'). (5.47)

In fact, Proposition 2.9.1 implies that the mapping on the right hand side of(5.47) is a derivation. Hence, it is sufficient to verify that the restrictions ofthe operators to E Q F coincide.

Let z e E +O F be an arbitrary vector. Then

(e (gyp) ®i + l 0 e (/i))z = (e (q) O l+ l 0 e (/i)) (i z O 1+ 1 0 lt2 z)

_ (p7L1Z Q 1 + 1 Q ,/i7c2 z = il(p1L1Z + i2l'/TG2 z

_ (gyp +O i)z = e n ('p +O i) (z)

and so Formula (5.47) follows.

5.18. Direct Sums of Dual Spaces

Consider the dual pairs E, E* and F, F* of vector spaces. Then the inducedscalar product in E Q F and E* Q F* is given by

<z*, z> _ <x*, x> + <Y*, y> z = (x, y), z* _ (x*, Y*)

The multiplication operator in the algebra n (E Q F) is given by

µ(a Ox b) = µ(a) o wq Q µ(b) a E A E, b E n qF, (5.48)

where wq denotes qth iterate of the canonical involution of the graded algebran E (see Section 6.6 of Linear Algebra).

Dualizing (5.48) we obtain the relation

i(a Q b) = wq o i(a) Q i(b). (5.49)

In particular it follows that

i(h Q 1) = i(h) Q i h e E

and

i(1 Q k) = w Qx i(k) k e E

whence

i(h Q k) = i(h) Q i + w Q i(k). (5.50)


Since the vector spaces E Q F and E* Q F* are dual, a scalar product(, ) is induced in n (E Q F) and A (E* Q F*). It will now be shown thatthis scalar product coincides with the induced scalar product if n (E Q F)and A (E* +O F*) are considered as the tensor products n E p n F andA E* p n F*. In other words, it will be proved that

(u* O v*, u O v) = <u*, u> <v*, v> (5.51)

for all u e A E, v e A F, u* E A E*, v* E A F*. Without loss of generality wemay assume that all the elements in (5.51) are homogeneous,

u e ARE, y e n rF, u* E n qE*, v* e A SF*.

If p + r q + s, both sides of (5.51) are zero and hence only the casep + r = q + s remains to be considered. Then we have

(u* O v*, u 0 v) = l(u O v)(u* O v*) = wrl(u)u* O i(v)v*

If p = q and r = s, it follows that

i(u)u* = <u*, u> and i(v)v* = <v*, v>,

while if p q, it follows that either p > q or r > s, so that both sides of (5.51)are again zero.

5.19. The Diagonal Mapping

Consider now the case F = E and let the diagonal mapping A: E - E Q Ebe defined by

A=i1+i2.Then the product u* n v* of two elements u* and v* of n E* can be writtenin the form

u* n v* = A ^ (u* Q v*). (5.52)

In fact, if j 1 and j2 denote the canonical injections of E* into E* Q E*,Formula (5.42) yields

u* O v* = (J1) ^ u* n (J2) A v* = (T1) ^ u* n (Tt2) ^ v*.

Applying A ^ we obtain

A^(u* p v*) = A^(ic1)^u* A A^(ic2)^v*

=(itloA)^u* A(i20A)^v*= n l^v*

= u* n V.

Formula (5.52) shows that A" is the structure map of the algebra n E*(see Section 2.1).

Exterior Algebra Over a Direct Sum

5.20. Direct Sums of Several Vector Spaces

Now consider the direct sum of r vector spacesr

E= Q E,.p=1

Then an r-linear mapping

/i: AEl x . x AEr - nEis defined by

..., Ur) = (l) A U1 A ... A (r) A ur up e Ep, i : Ep - E.

The r-linear mapping l'/ induces a linear map

f: AEl Q ... Q AEr - nEsuch that

125

f (u 1 ®" O ur) = (i 1) A U 1 n ... n (ir) A Ur (5.53)

The same argument as in the case r = 2 shows that f is a homomorphismand, in fact, an isomorphism of the graded algebra n E 1 O O A Eronto n E. Formula (5.53) shows that f is homogeneous of degree zero.Hence we may write

n (E1 +0 ... 3 Er) n E 1 0 ... j n Erand

U 1 O .. OX Ur = (l 1) n U 1 A A (lr) n Ur.

Comparing the homogeneous subspaces of degree p in the relation weobtain

n (E 1 ®... 0 Er)]p~ A P 1 E l 0 ... O A PrEr

P1++Pr=P

5.21. Exterior Algebra Over a Graded Vector Space

Let E = 7. 1 E. be a graded vector space, where the subspaces E. are homo-geneous of degree k.. Then there exists precisely one gradation in the algebran E such that the injection i : E - n E is homogeneous of degree zero.

The uniqueness follows immediately from the fact that the algebra n Eis generated by the vectors of E and the unit element 1 (which is necessarilyof degree zero). To prove the existence of such a gradation consider firstthe algebra n E,. By assigning the degree pk1 to the subspace n PEA we maken E, into a graded algebra. Now writing

AE= AElp...p AEr,


we recall that the gradations of the n E, induce a gradation in the algebran E. Clearly the injection i : E -+ n E is homogeneous of degree zero and sothe proof is complete.

The algebra n E together with the above gradation is called the gradedtensor algebra over the graded vector space E. The subspace of homo-geneous elements of degree k is given by

(n E)k = n p1E1 O O ' ` prEr(p)

where the sum is extended over all r-tuples (Pi' ... , pr) subject tor

= k.i=1

Suppose now that the vector space E has finite dimension and that thegradation is positive. Then the Poincare polynomial of the graded spacen E, is given by

Pi(t) = (1 + t")"i n = dim E1, i = 1, ..., r.

Since the space n E is a tensor product of the spaces n E. we obtain for thePoincare polynomial P(t) of A E (in view of Section 2.6) the expression

P(t) = (1 + tk 1)" 1 ... (1 + tkr)"r.

PROBLEMS

1. Let E be a vector space and A E be an exterior algebra over E. Show that

uAv=ltA(u®x v) u,veAE

where n and are the canonical projections of E Q+ E onto E and it = n + ?t2.

2. Let E = E 1 + E2 be a decomposition of E and set E 12 = E 1 n E2.(a) Establish a natural isomorphism

/i:E1/E12 O+ E2/E12 E/E12

(b) Consider the canonical projections

p1: E1 -+ E1/E12 p2: E2 - E2/E12 p: E -+ E/E12

and let p : E 1 Q+ E2 - E be the linear map given by

p(x1, x2) = x1 + x2

Show that the diagram

x1eE1,x2eE2.

AE1 Q AE2 AE

cpl)

(E1/E12) Ox n (E2/E12) = n (E/E12)

is commutative.

Ideals in AE

Ideals in AE5.22. Graded Ideals

Suppose that

1 => P, IP=1n APE,P

127

is a graded left ideal in the algebra A E. Then we have, for every p-vectoru e A PE and any element v= L vq E 1,

v A U=> vq A U= (- 1)PgU A vq E Iq q

and so 1 is a two-sided ideal. The same argument shows that every gradedright ideal is two-sided.

Now let a e A PE be an arbitrary homogeneous element, and considerthe graded subspace Ia of A E consisting of the elements u A a, u E A E.Clearly Ia is a graded left ideal in A E, and hence it is a two-sided ideal.Since a e Ia, it follows that Ia is the smallest (graded) ideal in A E containinga, i.e.,

IQ = fl1.

aEl

Ia is called the ideal generated by a. A homogeneous element a 0 is calleda divisor of an element u E A E if there exists an element v e A E such thatu= a A v or equivalently, if u E Ia.

More generally, every homogeneous subspace A c A PE generates agraded ideal I A defined by

1 A = ui n a; ul E A E, a E A . (5.54)

1 A is the intersection of all graded ideals containing A. If B is a subspaceof A it is clear that IB c IA. Now consider two homogeneous subspacesA A PE and B A qE. It will be shown that 1 A = 1 B if and only if A = B(and hence p = q). Clearly A = B implies that 1 A = 'B Conversely, assumethat 1 A = 'B Then every element b e B can be written in the form

b= U1 n a1 U1 E A q- PE, a E A

and hence it follows that q >_ p. The same argument shows that p q,whence p = q. Consequently, the U1 are scalars, and so b e A, i.e., B c A.Similarly we obtain that A c B, whence A = B.

As a special case of this result we have

Ia = Ib a e A PE, b e A qE

if and only if a = ,%b, ,% 0.


The ideal 'APE, p >_ 0, will be denoted by I". It follows from (5.54) that

1p = ME.j1 p

The ideals Ip form a filtration of the algebra A E, i.e.,

The ideal I' = I E is often denoted by A + E,

AE= A'E.j>o

If E is of dimension n we have

1n = A 'E

and

I p = O if p > n.


Let

E=E1QE2be a direct decomposition of E. Then we have (considering IE1 as an idealin AE)

I E1 = A + E1 Q A E2 . (5.55)

In fact, since A+ E, c 1E 1, it follows that A+ E 1 Q A E2 c 'E1 Conversely,let y A v (y e E1, v e A E) be a linear generator of IE1. Writing

v= a1® b, a1 e A E1, b,e A E2,

we obtain

yAveA+E1pAE2

and so

I E1 c A +E1 O A E2 .

Writing A E in the form

AE= AE1 Q AE2 = (r Q AE2) Q (A +E1 Q AE2)

= A E2 Q (n +E1 O A E2),

Ideals in AE

we obtain, in view of (5.55), the relation

AE = 'E1 +Q A E2 .

5.24. Linear Maps

129

(5.56)

Let p: E - F be a linear map and consider the induced homomorphismco A: n E - n F. Generalizing the result of Section 5.8, we shall prove that

ker q = 1 ker q . (5.57)

Let E' c E be a subspace such that

E = ker p Q E'.

Then we can write co = 0 Q cp', where gyp' denotes the restriction of co toE' c E. Since gyp' is injective so is (q' ) A and hence (see Section 5.16)

ker(p ^) = ker 0 A p n E' = n + ker co p n E'.

In view of (5.55), we have

n + ker p Q AE' = 1ker p .

Combining these relations, we obtain (5.57).

5.25. Invertible Elements, Maximum and Minimum Ideals

Proposition 5.25.1. An element z = zi, z E A `E, of A E is invertible if andonly if zo 0. z is nilpotent if and only if zo = 0.

PROOF. Since nilpotency and invertibility are mutually exclusive properties,it is sufficient to show that z is nilpotent (respectively invertible) if z e A + E(respectively z n + E).

If z e A + E, then z e A + F, where F is a finite-dimensional subspace of E.It follows that zm = 0 for m > dim F and so z is nilpotent. If z n + E, then,for some 0,

,z = 1 - a aE n+E.

Now consider the identity

(1 -a) n (1 1 -(k+ 1)!ak+i

Since a is nilpotent, it follows from this relation that 1 - a has an inverseand hence z has an inverse as well.


Corollary I. If z e A E is invertible, then z' is a polynomial in z.

Corollary II. Every proper ideal in A E is contained in A + E and so A Ehas a maximum ideal, namely A +E.

PROOF. If 1 A + E is an ideal in A E, then 1 contains invertible elements.Hence, 1 E 1 and so 1= A E.

Proposition 5.252. Let E be an n-dimensional vector space and let e be abasis vector for n E. Then for every element u 0 of n E, there exists anelement v e A E such that

uAv=e and vAu=±e.

PROOF. Let u = u, e A `E and assume that ur 0 and u = 0 for i < r.Choose a basis {ev} (v = 1, ... , n) of E such that e = el A A e". Then

ur - v 1..... vre A A eyr.v 1

Without loss of generality we may assume that l. r 0. Multiplying u byv = r)_ ler+l A ... A en,

we obtain

u A v=ur A v=el A A e"=e

and

vAu=vAur=er+l A...Aer=(-1)r("-r)e,which proves the proposition.

Corollary. If E has finite dimension, then every (two-sided) nontrivial ideal 1in A E contains 1" = A" E and so A E has a minimum ideal, namely A "E.Conversely, f E is a vector space such that A E has a minimum ideal, then E has

finite dimension.

PROOF. Let 1 0 be an ideal in A E and u 0 be an arbitrary element in I.Then by the above proposition there is an element v e A E such that u A v = ewhence I" I. To prove the second part consider the ideals Iq = p,q A pE,q >_ 0. If E has infinite dimension, it follows that flq Iq = 0 and so A Ehas no minimum ideal.

5.26. The Annihilator

Let u e A E be a homogeneous element. Then a graded ideal N(u) in thealgebra A E is determined by

N(u) = ker µ(u). (5.58)

Ideals in AE

N(u) is called the annihilator of u. It follows from the definition that

N(O)= AE, N(1)=O

and that

131

N(u) N(v) whenever u divides v.

More generally, if U E A E is a homogeneous subspace of n E, the space

N(U) = n N(u) (5.59)UEU

is called the annihilator of U. As an intersection of graded ideals N(U) isitself a graded ideal. For U = E we obtain that

N(E) _" ' '

0 if dim E _En E i = n.f dim

It follows from the definition that N(V) N(U) whenever V c U.Now consider the special case U = n "F, where F is a subspace of E.

N( A "F) consists of the elements u e A E satisfying

yi yIEF.

It follows from the definition of N( A "F) that

N(F) N( A 2F) c ... c N( n pF) .... (5.60)

Proposition 5.26.1. Let F be an m-dimensional subspace of E. Then theannihilator N( n m - p + 'F) coincides with the ideal generated by n "F,

N(nm-p+1F)=1APF 0< p<m+ 1. (5.61)

PROOF. If z1 n n zp, zi e F, is a generator of 1 A nF and y E F(j = p, ... , m)are arbitrary vectors, then

µ(z i n ... A z p) (yp n ... A ym) = z 1 A ... A Zp n yp n ... A ym = 0,

whence 1 A pF c N(n m - p + 'F). To prove that

N(nm-p+iF) 1APF

we show first that every element u e 1 APF can be written as

u = u' + ai Q b. U' E 1 A P+'F, a E A "F, b. E H, (5.62)

where H is a complementary subspace to F in E. Without loss of generalitywe may assume that u is of the form

u=anv aen"F,veAE.


Since E = F Q H, we have

veF,b;e AH,

u= a n = (a n v;)pb;.

Let u' be the sum of all terms in this equation for which v; has positive degree.Then we have

a E A "F, b; E A H, u' E I n p+ 1 F,

which proves (5.62).Now we prove (5.61) by induction on p. For p = 0 we have

N( n m+ 'F) = N(0) = n E and I A of = n E

and so (5.61) is correct. Now assume that (5.61) holds for the integer p andlet u e N( n m- "F) be an arbitrary element. In view of (5.60) we haveu e N(n - P + 'F) and hence by the induction hypothesis, u e 1 A pF . Inview of (5.62), we can write

u = u' + a p b, u' E 1 n p+ 1F, a1 E A "F, b. E A H. (5.63)

Now let {eµ} (µ = 1, ... , m) be a basis for F. Then (5.63) can be written inthe form

u = u' + eµ1 n ... A eµp O cµ1, ..., µp cµ1, ..., µp E / \ H. (5.64)

Choose a fixed p-tuple (µ 1, ... , µ p) and let (µp + ,, ..., µm) be the comple-mentary (m - p)-tuple. Multiplying (5.64) by eµp +1 n n eµm and observ-ing that u e N(n - pF) and u' E 1 A p + 1 F c N(n - "F), we obtain

0 = e p cµ1....°µp eE AmF,e 0.

It follows that cµ 1. µp = 0 and so u = u' E 1 A p + 1F This completes the proof.

Applying Proposition 5.26.1 for the special cases p = 1 and p = m, weobtain immediately the following

Corollary. Let u = xl n n xm be a nonzero decomposable m-vectorand X c E be the subspace generated by the vectors x, (i = 1, ..., m). Then

N(u)=1X (p= 1)and

N(X)=I (P=m).

In particular, u is divisible by a vector y 0 if and only f y n u = 0.

Ideals in A E

5.27.

As an application of the results of Section 5.26 we prove

133

Proposition 5.27.1. Let E be an n-dimensional vector space and cP : E -+ n Ea linear map. Then

x n (px = 0 x E E (5.65)

if and only if

( p x = x n v x E E (5.66)

for some fixed element v e A E. The element v is uniquely determined mod n "E.

PROOF. It is trivial that (5.66) implies (5.65). Conversely, suppose (5.65)holds. Then it follows that

x n (py + y n (px = 0 x, y E E.

Now we proceed by induction with respect to n. For n = 0 the propositionis trivial. Assume now that it is correct for dim E = n - 1. Choose anarbitrary vector a 0, a e E. Since a n cpa = 0, it follows from (5.61) withm = p = 1 that there is an element c e A E such that

spa = a n c.

Now define a mapping ti: E -+ n E by

6x=cpx-x n c. (5.67)

Then we have

6a=(pa-a n c=0, (5.68)

x n 6x = x n cpx = 0, (5.69)

and

an6x=ancpx-anxnc=-xncpa+xnanc=0. (5.70)Now consider the 1-dimensional subspace E 1 generated by a and writeE= E 1 Q F. Since

AE= AE1QAF,

it follows that

6x=1®i0x+a®Qi1x xeE,

where i and 61 are well-defined linear maps. Multiplying by a we obtainin view of (5.70)

0=a® i0x xeE,

134

whence t70 x = 0 and so

5 Exterior Algebra

6x = a Qx 61 x x E E. (5.71)

Now let y e F be an arbitrary vector. Then relations (5.71) and (5.69)yield

- (1 A a) OO (y A 61 y) _ (1 OO y) A (a OO 61 y) = y A 6y = 0

whence

y A 61y=0 yeF.

Now it follows from the induction hypothesis that

61 y= y A v1 y E F, (5.72)

where v1 E A F is a fixed element. Since every vector x e E can be writtenin the form x = ,a + y, . E IT, y e F, we obtain, in view of (5.68), (5.71)and (5.72),

6x=6a+6y=6y=a®x 61y=a®x (y A v1)

= -y A (a A v1) = -X A (a A v1) (5.73)

Combining (5.67) and (5.73) we find that

(px=6x+x A c=x A (c-a A v1)

and so the induction is closed.To prove the uniqueness part of the proposition assume that u1 E A E

and u2 E A E are two elements such that

cpx=XAU1 and cpx=XAU2.

Then

XA(U2-U1)=0 xeE

and so u2 - u1 E N(E) = A'SE.

Corollary. Let p: E -p A k + 'E be a linear map such that

xA4x=O xeE.

Then there exists an element u e A kE such that

cpx=xAU xeE.

If o 0, then u is uniquely determined.

PROOF. In view of the above proposition, we can write

cpx = x A v,

Ideals in AE 135

where v e A E. Applying the projection Pk +1: n E - n k + 'E to this relation,we obtain

(Ox-Pk+1(x n v)=x n pkv=x n u u= pkv

and so the first part of the corollary follows. Suppose now that

(px = x n u 1 and cpx = x n u2 u 1, u2 E n IE.

Then u2 - u 1 E ME. On the other hand we have that u2 - u 1 E n kE whenceu2 - u 1 E ME E n n kE. But, if p 0, it follows that k + 1 < n and so weobtain u2 - u 1 = 0.

If co is a homogeneous linear map, Proposition 5.27.1 can be extendedto spaces of infinite dimension.

Proposition 5.27.2. Let (p: E - n E be a homogeneous mapping of degree ksuch that x n cpx = 0. Then there exists an element u e n kE such thatcpx = x n u. If o 0 the element u is uniquely determined by co.

PROOF. We prove first the uniqueness of u. Assume that

cpx=x n u1 and (px=x n u2.Then

and so

xn(u2-u1)=0 xeE

u2 - u 1 E N(E).

If the dimension of E is infinite we have N(E) = 0, whence u2 = u 1. Ifdim E = n, we have N(E) = A" E. If o 0, it follows that k n and sou2 -u1 =0.

To show the existence of u we state first the following

Lemma. Let p : E -p n k + 'E be a linear map such that for a subspace F offinite dimension

y n (py = 0 yeF.

Then there exists an element v e n kE such that

(py = y n v y E F.

The above lemma is proved by induction on dim F in the same way asProposition 5.27.1.

Now consider a linear map p: E - n k + 'E satisfying x n cpx = 0. Wemay assume that o 0. Choose a E E such that cpa 0. Let H be a sub-space of dimension k + 1 such that a e H. Then, by the above lemma, thereexists an element u e n kE such that

coy = y n u y E H. (5.74)


Now we show that

cpx = x n u for every x e E.

Let x e E be an arbitrary vector and consider the subspace H 1 c E gener-ated by x and H. In view of the above lemma there exists an elementv e n kE such that

cpz = z n v zeH1.

From (5.74) and (5.75) we obtain

y n (u-v)=0 yeH,

(5.75)

whence u - v e N(H). In view of (5.61) we have N(H) = 1 n k+ 1H. On theother hand u - v is of degree k and so u - v = 0.

Proposition 5.27.2 permits us to give explicitly the forms of a derivationof odd degree and an antiderivation of even degree in n E.

Let 8 be a homogeneous derivation of degree k, where k is odd. Thenwe have 8 = p) where co denotes the restriction of 8 to E (see Section 5.11).In view of (5.32), we have x n cpx = 0 and hence, by the Corollary toProposition 5.27.1, there exists an element a e n kE such that

(px =anx x e E.


Bu =JaAU u e n "E, p odd,

0 u e n "E, p even.

If 8 is an antiderivation of even degree in n E, we have again x n cpx = 0,where co denotes the restriction of 8 to E. Hence co can be written in the form

(px=anx aen"E.Now (5.34) gives

Hu =

PROBLEMS

anu uen"E,poddo un pEE , p even.

1. Let 1 be a right ideal in AE. Assume that a n b e l and b n E. Prove that a e l.

2. Show that the generator of a graded principal ideal is homogeneous. Conclude thatthere exist nongraded principal ideals in n E.

3. Construct a left ideal in n E which is not a right ideal.

4. Given a subspace F E prove that the algebra n (E/F) is isomorphic to the algebran E/1 F .

5. Let E = E 1 + E2 be a decomposition of E and consider the homomorphism

(P,,: nEl Q AE2 -+ AE,

Ideals and Duality 137

which is induced by the linear map

rp: E1 Q E2 - E1 + E2.

Determine the kernel of p .

6. Let (p be linear transformation of a finite-dimensional vector space E and denotethe Fitting null- and 1-components of (p by F0((p) and F1((p). Prove that

F0(q) = IF,.() and AF1((p).

Hint : See Problem 5, Section 2, Chapter XIII of Linear Algebra.

Ideals and Duality

5.28.

In this paragraph E, E* will denote a pair of dual finite-dimensional vectorspaces. Let 1 be a graded ideal in A E. Then the orthogonal complement Ilis stable under the operations i(h), h e E. In fact, if u* a Il is an arbitraryelement we have, for every u e 1,

<i(h)u*, u> = <u*, h n u> =0,

and so i(h)u* e I'. Conversely, if 1 is a graded subspace of A E such that Ilis stable under every i(h), then 1 is an ideal. Let u e 1 and h E be arbitrary.Then we have, for every u* a I',

<u*, h Au> = <i(h)u*, u> =0

whence

h n u e (Il)l = I.

Now let F be a subspace of E and Fl be the orthogonal complement.Then the subalgebra A (Fl) and the ideal 1 F are orthogonal complementswith respect to the scalar product in A E and A E*,

A (Fl) = (I F)'.

In fact, consider the canonical projection

m : E --> E/F

and the injection

(5.76)

J: E* F'.In view of Section 2.23 of Linear Algebra, the spaces E/F and Fl are dual withrespect to the scalar product defined by

<Y*, mx> = <JY*, x> Y* e Fl, x e E. (5.77)


This shows that the mappings it and j are dual. Consequently, the inducedhomomorphisms

Th A: n E - n (E/F)

and

j A . n E* E- n (Fl)

are dual as well. This implies that

Im j A = (ker It A )l.

Since ker it = F we have, in view of (5.57)

ker it A = 1 F .

On the other hand, it follows from Section 5.8 and the definition off that

Im j = n Im j = n (F').Combining the above relations we obtain (5.76).

As an immediate consequence of (5.76) we have the formula

(AF)' = IFl, (5.78)

which is obtained by applying (5.76) to the subspace Fl c E* and takingorthogonal complements on both sides.

Proposition 5.28.1. Let F c E, F* c E* be two dual subspaces. Then theideals IF and IF* are dual as well.

PROOF. Let F 1 = (F*)' and F i = Fl. Then we have the direct decom-positions

E=F+QF1

and

E* = F* Q F i.

This yields, in view of (5.56)

nE = IF Q AF1

and

AE* = 1F* o AFT.

Since A Fi = n (Fl) = (IF)' and A F1 = n (F*1) = IF*, it follows that theideals 1 F and 1 F* are dual. E

Proposition 5.28.2. Let F c E be a subspace. Then the subalgebra n F isstable under the operations i(h*), h* E E*. Conversely, if A is a subalgebraof n E stable under i(h*), h* E E*, then there exists a subspace F c E suchthat A = AF.

Ideals and Duality 139

PROOF. Let Y1 n n yp, y E F, be any decomposable element of n F.Then, for every h* E E*,

i(h*)(Y1 n ... n yp) = (-1y+l<h*, Y;>Y1 n ... n y; n ... n yp E AF.

Hence A F is stable under i(h*). Conversely, assume that A is a subalgebraof n E which is stable under every i(h*). Then A is stable under every operatori(u*), u* E A E. If A = 0 the proposition is trivial and so we may assumethat A 0. We show first that 1 E A. Let u 0 be an element of A. We writeu in the form

ru= u p u p E A "E, ur 0.p=0

Since ur 0, there exists an r-vector u* E n rE* such that <u*, urn = 1.Applying i(u*), we obtain

i(u*)u = <u*, ur> = 1.

Since i(u*) u e A and <u*, ur> = 1, it follows that 1 E A.Now consider the subspace F = A n E. It will be shown that A = n F.

Clearly, A F c A. Letn

u = up up e n pE, n = dim E,p=0

be an arbitrary element of A and assume by induction that

uE n F for v > q.

Since n n + 'E = 0, this relation is correct for q = n. Define v by

n

v=u-v=q+ 1

(5.79)

Since u e A and uE A F c A, for v >- q + 1, we have v e A. Now letu* E n q- 1E* be an arbitrary element. Applying i(u*) to (5.79) we obtain

l(u*)v = l(u*)uq _ 1 + l(u*)uq

<u*, uq- 1> + l(u*)uq (5.80)

Since A is stable under i(u*), we have i(u*)v e A. Since 1 E A, it follows from(5.80) that i(u*)uq e A. On the other hand i(u*)uq is of degree 1 and hence

i(u*)uq E A n E = F.

Now let y* E Fl be arbitrary. Then

Y* n u*, uq> = + <Y*, i(u*)uq> = 0.

Hence, uq is orthogonal to the ideal 'F'. Now it follows from (5.78) thatuq E A F and so the induction is closed.

MO 5 Exterior Algebra

PROBLEMS

1. Let 1 be a subspace of n E. Prove that 1 is a left ideal if and only if the orthogonalcomplement 1 is stable under i(h) for every h e E.

2. Given a subspace F E prove that

AF = n ker i(u*).U* E IF1

3. Let h e E be a fixed vector. Define operators a : AE -+ AE and b : AE* + AE* by

au = µ(h)u u e A E

and

bu* = i(h)u* u* e AE*.

(a) Show that (n E, a) is a graded differential space and (n E*, (5) is a graded dif-ferential algebra.

(b) Prove that

H( A E) = 0 and H( A E*) = 0.

4. Let E, E* be a pair of dual vector spaces and consider a graded subspace U A E.

Show that U is an ideal in n E if and only if U1 is stable under the operations i(h),he E.

The Algebra of Skew-Symmetric Functions

5.29. Skew-Symmetric Functions

Let E be an n-dimensional vector space and consider the space T "(E) ofp-linear functions in E (see Section 3.18). Then, if Fe T "(E) and 6 E S p,an element 6F E T "(E) is defined by

(o) (x i , ... , xp) _ F(xa(1) , ... , xy(p)).

The function F is called skew-symmetric, if

o F = EQ .

The skew-symmetric functions form a subspace of T "(E) which will be de-noted by A"(E).

Every p-linear function F in E determines a skew-symmetric p-linearfunction A'F given by

A = 1 E 6 .P Q

AF is called the skew-symmetric part of F. In particular, if F is skew-sym-metric, then A'F _ 'F.

The Algebra of Skew-Symmetric Functions 141

Thus the operator A : T "(E) -+ T "(E) is a projection operator. It is calledthe antisymmetry operator.

Now let F e TP(E) and Y e T(E) and consider the (p + q)-linear function1F Y (see Section 3.18). A simple calculation shows that

A(D kY) = A(AID kY) = A(D AtY). (5.81)

This formula implies that

A(F kY) = A(A F AtY).

Thus the skew-symmetric part of a product depends only on the skew-symmetric parts of the factors.

5.30. The Algebra A (E)

The Grassmann product of two skew-symmetric functions F e AP(E) andY E A(E) is the (p + q)-linear skew-symmetric function F A Y given by

b n tt ()Explicitly,

A x ... x = 1 E x ... x 'Y x ... x

Proposition 5.30.1. The Grassmann product has the following properties :

F A Y=(-1)P9' A F F e AP(E), Y e A(E), (A)

('F n kY) n X= F n (kY n X) F, '-Y, X e A(E). (B)

PROOF. (A) Observe that

where 6 is the permutation

(1,...,p,p + 1,..., p + q)-+(P + 1,..., p +q, 1,...,p)Applying A to this equation and observing that E, = (-1)' we obtain (A).

(B) Let F e AP(E), Y e A(E), and X e Ar(E). Then

A n X= p+ q+ r)!. A X

=(p +q+r)!AA(D ) XJp!q!r!

-(p+q+r)!t r

A(D Y X ).Pq

142

Similarly,

=(p+q+r)!tq. trt ( )p..

and so (B) follows.The A -product makes the direct sum

n

A(E) = AP(E)p=o

5 Exterior Algebra

into a graded associative algebra called the Grassmann algebra over E.

5.31. Homomorphisms and Derivations

A linear map p : E -+ F induces a homomorphism c0A : A(E) E- A(F) givenby

((PAkJ')(xi, ..., xp) = 'P E AP(F).

Next, let p be a linear transformation of E. Then p determines a derivationin A(E) given by

P

1 ... , xp) = (x 1, ..., (pxy , ..., xp).v=1

To show that this is indeed a derivation consider the derivation oT (cp)induced by p in the algebra T(E) (see Section 3.18). It follows from thedefinitions that

= eT (co)d E A(E).

Moreover, the operator oT (cp) satisfies the relations

eT o cr = 6 o eT (p)

whence

eA(co) ° A = A o oT (co)

It follows that

Op)(F n i =(p+g)!OA

n Y + D n

OA(co)l A Y + D A OA(co)%Y.


5.32. The Operator iA(h)

Recall from Section 3.19 the definition of the linear operator iA(h) : T(E) -T'(E). It follows that if F E A(E), then

iA(h) = i1(h)b.

Thus,

(iA(h) ) (x 1, : , x p - 1) = I (h, x 1, ..., x_ 1) F E A( E).

iA(h) is called the substitution operator in the algebra A(E).

Lemma. Let A denote the antisymmetry operator. Then

iA(h) o A = A o iA(h).

PROOF. Let F E T"(E). It is sufficient to consider the case

D = f1 ... fp f1 E E*.

Then we have

A = 1 E .fQ Q(1) Q(p)pand thus,

1ACh) (4 ) = i 1(h) (Ai:F)

1

= i 1(h) EQ fa(1) ' ... fQ(p)p-

1

= EQ fa(1)(h)fa(2) ... fa(p)p Q

'p= EQ fµ(h)fQ(2) ' ... fQ(p)

p µ= 1 Q(1)=µ

'p_ fµ(h)Aµ

p µ=1

where

Aµ = EQ fa(2) ' ... fQ(p)Q(i)=µ

(5.82)

Now we show that

Aµ = (-1)"-1(p - 1)!A(fl O ... OO fµ ... fp).

In fact, fix µ and let 6µ E Sp be the permutation defined by

6µ(1)=2,..., 6µ0u- 1)=µ, 6µ0u)= 1, 6µ0u+ 1, ...6µ(p) = p.


Then every permutation 6 E S, determines a permutation TQ E S, given by

iQ = 6 o 6µ .

Since

E(6µ) = (-1)µ-1,

E(#EQ) = (-1)µ - 1 E(6).

Thus we obtainµ _ 1 ,.

Aµ = (-1) E(i)ft(1) ' ... ft(µ) ' ... ft(p)t(µ) = µ

= (-1)µ-1(p - 1)!A(f1'... fp).

Equations (5.82) and (5.83) yield

i h A= 1)! -1 µ-1fµ(h)A(fi ...

..A( ) ( ) ) .. fµ fp)p µ

_1A ( 1)µ - lfµ(h)fl ... ' fµ ' ... ' f p

p µ

= 1 A µ- li(- ) µ(h) (fl ... f p)p µ

= A1A(h)1F

and so the lemma is established.

(5.83)

0

Proposition 5.32.1. The operator iA(h) is an antiderivation in the algebra A'(E),

lA(h) ( A 'F) = 1A(h)I) A 'F + (- 1)(F A LA(h)'F (F E A(E), 'F E A(E).

PROOF. Since

we have to show that

1A(h)A( `F) =p A(iA(h) F 'F) + (_ 1)p q A((F iA(h)`F).p+q p+q

By the lemma,

1 p+qiA(h)A( 'F) = ( -1)µ -1 Aiµ(h) (lF. 'F).

p+qµ=1


Now,

h = iµ(h)i

µ P"()( ) 1(F' i_ (h)tY µ > - p + 1µ p

(see Formula 3.18). Thus we obtain

P

i(h)A(F ) = (-1)µ-1A(iµ(h)iF kY)p+qµ=1

1 p+q+ (-1)µ - 1 A( iµ _ p(h)i)

p + q µ=p+

1P

= A (-1)µ - 1 lµ(h)D ijp+q µ=1

1 p+q+ A (F. (-1)µ -1lµ- p(h)i

p+q µ=p+1

P A(iA(h)l kY) + (-1)P q A(F' iA(h)`Y),p+q p+qwhich completes the proof.

5.33. The Isomorphism A E* 4 T' (E)

We show that the Grassmann algebra over E is isomorphic to the exterioralgebra over E*. In fact, consider the isomorphism a: ®PE* 3 T(E) (seeSection 3.20). It is easily checked that the diagram

TP(E)

jA

commutes, where icA denotes the alternator (see Section 4.2). Since

Im icA = XP(E*) and Im A = AP(E),

it follows that a restricts to a linear isomorphism

a : X P(E*) AP(E).

Next observe that, in view of the commutative diagram above,

oc(u n v) = ajA(u Q v) = Aoc(u Q v)

=Aau av = +q au A av ueXPE* veXgE*.

146

Thus the map

defined by

f3: X(E*) - A(E)

5 Exterior Algebra

f3(u) = p !a(u) u e X p(E*)

is an algebra isomorphism.Composing this isomorphism with the algebra isomorphism

ri: AE* X (E*),

obtained in Section 5.3, we obtain an algebra isomorphism

A E* X(E*) A(E).

Under this isomorphism,

(1) The homomorphism cP ^ : A E* - A F* corresponds to the homo-morphism p": A(E) - A(F) (see Sections 5.9 and 5.31).

(2) The derivation 8 A ((p) corresponds to the derivation (see Sections5.10 and 5.31).

(3) The operator i(h) corresponds to the substitution operator iA(h)(see Sections 5.14 and 5.32).

5.34. The Algebra A,(E)

Denote by A (E) (p > 1) the space of skew-symmetric p-linear mappings inthe dual space E* and set A0(E) = r. Then we have the n -multiplicationbetween A (E) and A9(E) (see Section 5.30). It makes the direct sum

n

A , (E) = A p(E)p=0

into an associative algebra which is isomorphic to the algebra A E.Now we show that the scalar product between T "(E) and T(E) defined

in Section 3.22 restricts to a scalar product between A"(E) and A (E).In fact, it is easy to check the relation

<D, Q`P> = <o D, 'P> D E A(E), 'P e A(E).

It follows that

Now suppose that D 1 E A"(E) is an element such that ' 1 > = 0 forevery ' 1 E A (E). Then we have for every 'P ETp(E)

= <AD1, '> = <D1, A'> = 0

whence = 0. In the same way it follows that if '1 E A (E) is an elementsuch that <D 1, ' 1 > = 0 for every 1 E A"(E), then 'P 1 = 0.


Thus a scalar product <, >A between the spaces AP(E) and A (E) is definedby

'' = 1 qi HEAP E 'P EA(E).P

x1 n ... n D(x1, ..., xp) xE E

<fl A" . A fps 'P>A - `I'(fl, ... , fp) feE*.

1

<, x 1 ^ ... A x p>A = x 1 A ... A xp>P

= 2A(x1 OO ... OO xp)> = <A D, x1 OO ... OO xp>

= <(D, x 1 ® .. OX xp> = D(x 1, ..., xp).

The second formula is obtained in the same way.Finally we show that

<fi n ... A fp, x1 n ... A xp>A = det(f (x;)) ff E E*, xi e E.

In fact,

1

<fl n ... ^ fps xl n ... n xP>A = <fl n ... ^ fps xl n ... n xp>P

= <f1 n ... n fp,A(xl p...Qxp)>

EQ f1(x) ... f(x)Q

= det(f, (x;)).

Mixed Exterior Algebra

Throughout this chapter E*, E will denote a pair of dual vector spaces over a field of characteristiczero.

The Algebra n (E*, E)

6.1. Skew-Symmetric Maps of Type (p, q)

A skew-symmetric map of type (p, q) from E*, E into a vector space H is a(p + q)-linear mapping

/i:E* x x E* x E x x E Hp 9

which satisfies

YI (x (1), ... , x (p)' Xt(1),.. . , Y (x 1 ... , x; X1,. . . , X9)

for all permutations o E Sp and i e S9.

Proposition 6.1.1. Every skew-symmetric map cui of type (p, q) determines aunique linear map

f:APE*®such that

... ^ xp) 0 (x 1 n ... n X9)] = Y' (aC i , ... , x; X 1, ..., X9).

PROOF. The uniqueness follows from the fact that the products

(x i A ... n xp) O (x1 A ... n x9)

span the space (A PE*) Q (A 9E).

148

The Algebra A (E*, E)

To prove existence define a linear map

g: (®PE*) ® (®'E) - H

by

g(x 1 ®... ® xp ® x 1 ®... ® xq) = Y (x 1 ..., xp ; x 1, ..., xq)

(see Section 1.20) and consider the bilinear mapping

f3:(®PE*) x (®9E) H

given by

149

f3(u, u) = g(u ® u)

The skew-symmetry of c implies that f3(u, u) depends only on the vectors7r1u and ire u where

7r1: ®PE* + APE* and 2®E+ + A "E

are the canonical projections (see Section 5.3). Thus a bilinear mapping

y: A PE* x A qE -+H

is defined by

y(it1 u, 2 u) = f3(u, u)

This map, in turn, induces a linear map

f:(ApE*)®(AlE)+H.

It follows that

.f [(x 1 n ... n xp) ® (x 1 A A xq)]

y [ir1(x 1 ® ... ® xp ), 7r 2(x 1 A ... A xq)]

_ f3(xi 0."®xp,x1 ®."®xq)

= g(xi 0... ®xp ®x1 ®... ®xq)

= i/i(x i , ... , xp ; x 1, ... , xq)

and so the proof is complete.

6.2. The Algebra A (E*, E)

The mixed exterior algebra over the pair E*, E, denoted by A (E*, E) isdefined to be the canonical tensor product of the algebras A E* and A E(see Section 2.2),

A (E*, E) = A E* ® AE.

150 6 Mixed Exterior Algebra

The multiplication in this algebra will be denoted by. and is determined bythe equation

(u* Q u) (v* p u) = (u* n u*) Q (u n u) u*, u* E A E*, u; u e A E.

Thus A (E*, E) is a graded associative algebra with unit element 1 Q 1.It is generated by the elements 1 Q 1, x* Q 1 and 1 Q x with x* E E* andxeE.

If w e A (E*, E), we shall define wk (k >_ 0) by

wk= ww.....k.

andk

k>_1

w° = 1.

Now consider the bigradation of A (E*, E) by the subspaces

A 9(E*, E) = A PE* Q A 9E.

Clearly,

w1 , w2 = (_ 1)P1P2+9192w2 . w1

Since

w 1 E A 9i (E*, E), w2 E A 92 (E*, E). (6.1)

P1P2 + g1g2 = (Pi + g1)p2 + (p2 + g2)g1 (mod 2),

it follows that p 1 p2 + q 1 q2 is even whenever Pi + q1 and p2 + q2 are andthat

w 1 w2 = w2 w 1 if p1 + q1 and p2 + q2 are even.

In this case we have the binomial formula

(w1 + w2)k = w1 w2.i+ j=k

The scalar product between A E* and A E defined by

<x 1 n ... n xp, x 1 ^ ... ^ xq > _Jo ifpq,ldet(<x, x3>) if p = q,

(see Section 5.6) induces an inner product in A (E*, E) via

<u* Q u, u* Q u> = <u*, vXu*, u> u*, u* E A E*, u, u e A E. (6.2)

(The symmetry and the nondegeneracy are easily checked.)Now fix an element z e A (E*, E) and, denote by µ(z) the left multiplication

by z,

µ(z)w = z w we A (E*, E);

let i(z) be the dual operator,

<i(z)w1, w2 > = <w1, µ(z)w2 > w 1, w2 E A (E*, E).

1

The Algebra A (E*, E)

Then, if z e A 9(E*, E) and w e A s(E*, E),

i(z)w e A s- p(E*, E) if r> p and s >_ q

and

151

i(z)w = 0 otherwise.

It follows from the definition that

µ(u* Q u) = µ(u*) Q µ(u) u* E A E*, u e A E,

where the operators on the right-hand side are the left multiplications inA E* and A E respectively. In particular,

µ(1 Q u) = 1 p µ(u) u e A E

and

µ(u* O 1) = µ(u*) O l u* e AE*.

Dualizing the relation

µCzl z2) = µCzl) ° µ(z2) z1, z2 E A (E*, E),

we obtain

i(zl . z2) = i(z2) o i(Z1). (6.3)

Note that if z e A 9(E*, E) and w e A p(E*, E), then i(z)w is the elementin A g(E*, E) = r given by

i(z)w = <z, w>.

Next, consider the flip operators

QE: A (E*, E) -+ A (E, E*) and QE* : A (E, E*) -+ n (E*, E)

given by

QE(u* Q u) = u Q u* u* E A E*, u e A E

and

QE*(u O u*) = u* p u u*E AE*,uE AE.

They are algebra isomorphisms as well as isometries with respect to theinner products in A (E*, E) and in A (E, E*).

The subspace

of = Op E, where AE = A p(E*, E),p>_0

of A (E*, E) is obviously a subalgebra. It is called the diagonal subalgebra.Formula (6.1) implies that the diagonal subalgebra is commutative. More-over, of is stable under the operators i(z) if z e AE. Finally, the restriction ofthe inner product in A (E*, E) to of is nondegenerate.


Next, let F, F* be a second pair of dual vector spaces and let

(p*:E*E-F*

and

,i:EE-F,

be a pair of dual maps. Then we have the induced algebra homomorphisms

/jA O (P : n (E*, E) n (F*, F)and

P ^ p /, A : A (E*, E) F-- A (F*, F).

It is easy to check that these maps are dual.Since l/i ^ Q (p is an algebra homomorphism, we have the relation

(/jA O 4, )(z w) _ (i/i O 4 n )z (/1 O 4 /, )W

or equivalently,

z, w E A (E*, E)

(/j ®4,)0/1(Z) = µ[/ O 4 )z] ° (I ®4A).

Dualizing, we obtain

i(z) ° ((p A O 'I' A) _ (`p A O 'I' A) ° i[(/i A O `I' )z]. (6.4)

Finally, note that a pair of dual isomorphisms P : E 3 F, 4* : E* + F*induces an algebra isomorphism

cx, : A (E*, E) A (F*, F)

given by

a = l OO P

Next, consider the linear map

TE : A (E*, E) -+ L( A E; A E)

defined by

TE(a* p b)u = <a*, u>b u e A E

(see Section 1.26). Since

= 0 if a* E A pE*, u E A 9E, p q,

TE restricts to linear maps

A 9(E*, E) - L( A pE; A 9E).

The Algebra A (E*, E) 153

The dual of the linear transformation

TE(a* ®b) : n E -, AE

is the linear transformation TE(b ® a*). Finally, recall from Section 1.26 thatTE is a linear isomorphism if E has finite dimension.

6.3. The Box Product of Linear Transformations

Let (P1 (i = 1, ... , p) be linear transformations of E. Then a linear transfor-mation

n"E

is given by

((p 1...

(P pX x 1 n ... A xP) = 7P17(1) A ... A (ppQ

It is called the box product of the (o. In particular,

((pl co2Xx1 ^ x2) = (plx1 ^ (p2 x2 - (p1x2 ^ (p2x1.

The box product formula can be written in the form

((p 1... (p p)(x 1 A ... ^ x p) = q) (1) x 1 n ... n xP

Q

This show that the box product is symmetric,In fact, let a SP be any permutation. Then we have

((pt(1) ... 41(P)Xx 1 A ... A xP) = (p(7 (1)x 1 A ... A (pQt(P) xPQ

Setting 6i = p, we obtain

(q, ... 4t(P))(x 1 A ... A x p) = (Pp(1) x 1 A ... A P(P) xPP

= (4 1 ... (p p)(x 1

It follows from the definition of the box product that

np(p.p

P

1

n ... n xp).

Proposition 6.3.1. The operator TE satisfies the relation

TE(Z 1 ... zP) = TE(z l) ... TE(z p) ; e E* ® E.


PROOF. It is sufficient to consider the case z = y* ®y e E (i = 1, ..., p). Then we have, for x, e E (i = 1, ... , p),

TE(z 1 ... Zp) (x1

n ... n xp)

= TE(Y* n ... A yp ® Y t n ... n yP) (x 1 n ... n xp)

=<Yi n...Ayp,x1A."Axp>(Y1A."Ayp)

= y p, xP)> (y1 n ... n yp)Q

= c <y, X(l)>Y1 n ... A <Yp, xe,(P)>Yp

= c T (z 1)x (1) ^ ... A TQ

(TE(z 1) ... TE(z p)) (x 1 A ... ^ x p).

Corollary. Let z e E* ® E. Then

TE(z") = n P(TE z).

The composition Product

6.4

We now define a second multiplication in the space n E* ® n E which willbe denoted by o and called the composition product. Given u*, v* e A E* andu,veAE,set

(u* ® u) o (v* ® v) = <u*, v>v* ® u. (6.7)

(cf. Section 1.26).Then we have the relation

?'E(w 1 o w2) = TE(w 1) o TE(w2) w 1, w2 e A (E*, E), (6.8)

where TE is the operator defined by (6.5) and the right-hand side is the com-position of the linear transformations TE(w 1) and TE(w2). Note that ifdim E < oo, then TE is a linear isomorphism (cf. Section 1.26) and so thecomposition product is determined by relation (6.8) in this case.

It follows easily from the definition that the composition algebra is as-sociative. In particular,

(ui ®u1)o(u2 ®u2)o...o(up ®up)

- <u1, u2/ <u2, u3/ ... <u p_ 1, up/up ®u1.

The Composition Product 155

The kth power of an element w in the composition algebra will be denotedby w®,

k

Note that the composition algebra has no unit element unless dim E < oo(see Section 1.26).


A 9(E*, E) o A s(E*, E) = 0 if p s (6.10)

and

A 9(E*, E) A;(E*, E) c A 9(E*, E). (6.11)

In particular,

A p(E*, E) o A p(E*, E) A p - p(E*, E)

and so the subspaces AE = A p(E*, E) are subalgebras. Moreover, TErestricts to homomorphisms

T, : 0 p(E) -+L(A "E ; ARE) for each p.

Formula (6.11) shows that the composition product is not homogeneouswith respect to the usual gradation in A (E*, E). However, if we introduce anew gradation (called the cross-gradation) by setting

Deg w= p- q w e A 9(E*, E),

then we have, for any two homogeneous elements w 1 and w2 for whichw 1 o w2 0,

Deg(w 1 o w2) = Deg w1 + Deg W 2.

Thus, the composition product preserves the cross-gradation.

6.5

Lemma I. Let w 1, W 2 E A (E*, E). Then

w1 o w2 = [i O TE(w1)]w2 = [TE(w2)* O 1]w1

PROOF. We may assume that w1 = u* © u and W2 = u* Q u. Then formula(6.7) yields

[i O TE(wl)]w2 = u* O TE(wl)V = <u*, U>(U* O u) = W1 o W2.

The second relation is established in the same way.

156

Lemma II. Let w e A p(E*, E) and z e E* Q E. Then

(TE(z)^ Q a)w = w o zP

and

6 Mixed Exterior Algebra

(i O 7E(z) A )w = z P o w.

PROOF. Applying Lemma I with w1 = w and w2 = zP, we obtain

w o ZP = [TE(zP)* 0 ljw.

By the corollary to Proposition 6.3.1, TE(z") = A P(TEz). Thus we obtain,since w e A p(E*, E),

w o zP = [(A P(TE z))* Q i] w = [ TE(z) ^ Q i] w.

The second formula follows by a similar argument.

The following proposition states a relation between the multiplications inthe mixed exterior algebra and the composition algebra.

Proposition 6.5.1. Let zv e E* Q E (v = 1, ... , p), p > 2 and z e E* Q E.Then,

l(z)(z 1 ... zP) - , <z, Z )Z 1 ... Zv ... ZP .

v

- (ZµoZoZy + ZvoZoZµ).Z1v<µ

PROOF. We may assume that zv = y* Q yv, where y* E E* and yv e E and thatz = x* ® x. Then we have (see Corollary II to Proposition 5.14.1)

i(z) (z1 ... zP) = [i(x) OO i(x*)J (Yi n ... A yp OX y1 n ... n YP)= (-1)``-v<Y*, x><x*, Yv>Yi A ... A yµ A ... A yp

O Y1 n ... n Yv n ... n yP.

Now write this sum in three parts with v = µ, v <u and v> to obtain

i(z) (z 1 ... zP)

n=<Y*,x><x*,Yv>Yin...n y*A...Aypoyl A...ApvA...A yPV

- <Yux><x*, Yv>(Y* OX Y, )[Yi n . v µ Pv<µ

- <Yu, x> <x* Yv> (Y* O Yµ) [Yi n ... n yµ n ... A y* n ... n Yp]µ<v

0 y1 n...A5 A...Ayvn...A yr].

Poincare Duality

Since, in view of (6.9)

<Ya, x><x*, Yv>(Y* ® Yµ) = zµ ° Z ° zv

it follows that

i(Z)(Z 1 ... Zr,) = <z, Z>Z 1 ... Zv ... ZPV

- (zµ°Z°Zv).Z1 ... 2 ...Zµ ...z,V<µ

- (Zµ°Z°Zv).Z1 ...Zµ...ZV...Zp.

µ<V

157

Interchanging the roles of µ and v in the last term and combining it with thepreceding term now completes the proof.

Corollary. Let z e E* ® E and w e E* ® E. Then

i(z)w" = <z, w>wp-1 - (w o z o w). wp- 2 p > 2,

Poincare Duality

In Sections 6.6-6.16 E denotes an n-dimensional vector space over a field ofcharacteristic zero.

6.6. The Isomorphism TE

Consider the linear isomorphism TE from A (E*, E) to L( A E; A E) givenby

TE(a* ® b)u = <a*, u>b u e A E

(see Section 6.5). It restricts for each pair (p, q) to an isomorphism

TE : A 9 (E*, E) 3 L(A "E ; ME).

This isomorphism satisfies the relation

tr(TEw1 ° TEw2) = <w1, w2> w1 e T9(E*, E), w2 e T (E*, E). (6.12)

In fact, let w 1 = a* ®b, a* e APE*, be ME, w2 a b* ®a, b* e anda e A "E. Then

(w1, w2> = <a*, a><b*, b>.

On the other hand, Formulas (6.7) and (6.8) yield

tr(TE(a* ® b) ° TE(b* ® a)) = <a*, aXb*, b>

and so Formula (6.12) is established.

158

6.7. The Unit Tensors

Define tensors 1E e A (E*, E) and & e A (E*, E) by

DE = TE 1''AE

and


p=TE1(l,1) (P=0,...,n).

Then E is the unit element of the composition algebra whilep is the unitelement of the subalgebra A(E). The tensorp will be called the unit tensor ofdegree p. We shall denote1 simply by . Thus,

= TE 1(iE).

The tensor " can be written in the form

" = e Q e,

where a*, a is a pair of dual basis vectors of A "E* and A "E. Clearly,

"E = gyp. (6.13)

p=0

The unit tensor & coincides with the pth power of in the algebra A (E*, E),

p = fp (p = 0, ..., n). (6.14)

In fact, the corollary to Proposition 6.3.1 yields

TE(MP) = A "TE(S) = A plE = 'APE (6.15)

and so (6.14) follows.Next observe that Formula (6.12) applied with w 1 = p and w2 e A p(E*, E)

yields

Thus,

In particular,

<p , w> = tr TE(w).

< p, w> = tr TE(w) w e A p(E*, E). (6.16)

<, u* Q u> = <u*, u> u* e A PE*, u e A "E. (6.17)

Poincare Duality

Proposition 6.7.1. The unit tensors satisfy the relation

159

q)p+q p q (6.18)P 9 (p+q\

and

i( ) =n p + q

q6.19)p-q p >_ q ()

PROOF. The first part follows directly from Formula (6.14). To prove thesecond part consider first the case q = 1. Then the corollary of Proposition6.5.1 (applied with z = w = ) yields

l( P= <,4 P-1 -( a o ) P-2.

Since, by (6.16), <, > = n and o = , we obtain

i(P = nP - 1 - (p - 1)t3' -1

=(n-p+1)tP-1.Thus (6.19) is correct for q = 1. Now the general formula follows via induc-tion on q.

. ni. <pp, p) _ (p = 0,. . . , n)

P

II. i( q) n = n - q (q = 0, ..., n).

6.8. The Poincare Isomorphism

Choose a basis vector e of A "E and let e* be the unique basis vector of A "E*such that <e*, e> = 1. Then, as noted previously,

e*©e=Now define linear maps

and

by setting

and

De: AE -- AE*

De:AE*--,AE

De u = i(u)e* u E A E

Deu* = i(u*)e u* e AE*.


In particular, De(1) = e*, De(e) = 1 and De(1) = e, De(e*) = 1. Note that,for any nonzero scalar ,,

De = a 'De and D =As immediate consequences of the definitions we have

De(u n v) = i(v)De(u) u, v E A E

and

De(u* n v*) = i(v*)De(u*) u*, v* E A E*.

Moreover, the operators De and De restrict to linear maps

Dp:APE-_^E* p=0,...,n

and

D": APE* -- ^ " PE p = 0, ..., n.

Theorem 6.8.1

1. De and De preserve the scalar products

<Deu, Deu*> _ <u*, u>

2. The duals Dp and (DP)* are given by

Dp = (-1)1 P p = 0, ..., n

and

(DP)* _ (-1)Pt" - PAD" - P p=0,...,n.3. DPoD"_p=(-1"-P)l p=0,...,n

and

DP o D" - P = (-1)P(" - P)l p = 0, ... , n.

In particular, the maps De, De, D, and DP are all linear isomorphisms.

PROOF

1. It is sufficient to consider the case u e A PE, u* E A PE*. Then we have,in view of (6.17),

<De u,Deu*> _ <i(u)e*, l(u*)e>

<in- p, i(u)e* ® l(u*)e> = /in- p, i(u* ® u) >.

Since the diagonal subalgebra is commutative,

<in- p, i(u* ® u) n> = <u* ® u, lC n-p nThus we obtain, in view of the Corollary to Proposition 6.7.1,

<De u,Deu*> _ <u* ® u, ip> _ <u*, u>.

Poincare Duality

2. Let u e A PE and v e A" - E. Then

161

Dell, v> = <i(u)e*, v> = i(v)i(u)e*= (- 1)I' - P)i(u)i(v)e* = (- 1)' - P)l D" _ P v, u>.

This proves the first relation. The second relation is obtained in the sameway.

3. Let v E Then, by (2) and (1),

(DP ° D" -,,)u, v> = (- 1)"" - P)<D" _ P u, D" - Pv> = (- 1)Pc" - P)<u, v>

which gives the first part of (3). The second part is established in the sameway

6.9. Naturality

Let (p : E F be a linear isomorphism from E to a second vector space F.Choose a basis vector e of n "E and set f = (p A e. Note that the vectorf * = (qp ^) - le* satisfies < f '*, f> = 1.

We show that the diagrams

AE " ' AF

ae "' DI

AE* car AF*

1+"' DI

AE* c^) AF* AE - AF

commute.In fact, Formula 5.40 yields

((p A ° D f ° (p A )U = i(u)gp ^ f * = De u

and so the first diagram commutes. The commutativity of the second diagramfollows in the same way.

6.10. The Isomorphism DE

In this section we introduce a canonical linear isomorphism

DE : n (E*, E) A (E*, E)

(not depending on the choice of a basis vector of n "E). In fact, let

DEW = i(w)w e A (E*, E).

If a*, a is a pair of dual basis vectors of ME and n "E*, we have,

DE(u* ® u) = i(u* ® u)(e* ® e) = De u ® Deu*. (6.20)

162

Thus, the operators De, De, and DE are related by

DE = (De e De) ° QED

where QE denotes the flip map defined in Section 6.2.The corollary to Proposition 6.7.1 shows that

DE 1P =n - P (p = 0, ... , n).

In particular,

DE(1) = and DE(1) = 1.

Moreover, it is immediate from (6.3) that

0

(6.21)

DE(wl w2) = i(w2)DE wI w1, w2 e A (E*, E). (6.22)

To state the analogue of Theorem 6.8.1 we introduce the involution E ofA (E*, E) given by

E(Z) = (- 1)P(n - P) + 9(n - 9)Z z e A 9(E*, E).

Then we have

Theorem 6.10.1

1. DE is an isometry.2. The dual operator is given by

DE = SZE ° DE .

3. DE-czE.

PROOF

1. By Theorem 6.8.1, Part (1) and Formula (6.20) we have

(DE(u* +® u), DE(v* +® v)> = <1)eu ® IYu*, De v ® Dev*>= <De u, Dev*>(De v, DeU*>

= <u* ® u, v* ® v>.

2. This follows from Theorem 6.8.1, Part (2) and (6.20x).3. This is a consequence -of (1) and (2). CJ

Finally, observe that the isomorphism DE restricts to a linear auto-marphism DA of the space AE.

Theorem &10.1 implies that

DA = Dit

and


Da = i.

Poincare Duality 163

6.11. Naturality

Let q : E - F be a linear isomorphism and consider the induced isomorphism; : A (E*, E) A (F*, F) (see Section 6.2).

Then the diagram

A (E*, E) ;) A (F*, F)

DE DF

A (E*, E) A (F*, F)

commutes. This follows from the naturality of De and De, Formula (6.21) andthe relation

QFo; =where aq = QP A Q (q, ^) -1.

6.12. The Intersection Product

We introduce a second product structure, the intersection product, in n E bysetting

u rn v= De[(De) -1 u n (De) - 1 v] u, v e A E.

Thus if u E A PE and v E A 9E, then u r v E A P+ 9-"E. In particular, u ( v= 0ifp+q>2norp+q<n.

It is immediate from the definition that

u r P)(" - 9)v rn u u E APE, v E A 9E.

Moreover, if e e A "E is the element used to define De, then

u r a=er u=u u e A E.

The intersection product makes A E into an algebra called the intersectionalgebra.

Now we show that De is an isomorphism from the intersection algebra tothe exterior algebra,

De(u r v) = De u n De v u, v e A E.

In fact, let u e A PE and v e A 9E. Then Theorem 6.8.1, Part (3) gives

De(u rn v) = (- 1)P(P)+( 9" - 9)(D" - P) - 1 u ^ (D" - 9) - 1 v

= DP u A D9 v = De u n De v.


Now let u e A PE and v e Then u r v is a scalar. It will be denoted byJ(u, v). Theorem 6.8.1, Part (3) implies that

J(u, v) _ (u r v)<e*, e> _ <De(U r v), e>

_ <De U A De v, e> _ <De v, i(De U)e>

_ <De v, DeDe U> _ (-1)P(" - P)<De v, U

_ (-1)P(" P'<e*, v n u> _ <e*, U n v>.

Thus,

J(u, v) _ <e*, u n v>.

To obtain a geometric interpretation of J(u, v) let E be a real vector space.Orient E via the determinant function 0 given by

0(x1, ... , x") _ <e*, xi n ... n x"> x E E.

Now consider two oriented subspaces E 1 and E2 of dimensions p andn - p. Choose positive bases

a 1, ..., aP and aP + 1, ... , a"

of E 1 and E2 respectively and set

U=a1 v=aP+1

Then

J(u, v) = 0(a 1, ..., a").

This shows that

i. J(u, v) 0 if and only if E 1 + E2 = E ; that is, if and only if E = E 1 p E2.

ii. Suppose that E = E1 p E2 holds. Then the orientation of E coincideswith the orientations induced by E1 and E2 (see Section 4.29 of LinearAlgebra) if and only if J(u, v) > 0.

6.13. The Duals of the Basis Elements

Now let {e}, {e*v} be a pair of dual bases. Then the dual images of the basisvectors eel n n (v1 < < vP) of A PE are given by

De (eel ^ ^ 1)Ep=1(Vz - Oe*vp + 1 ^ ... A e*V I, (6.23)

where (v+ 1, ..., v") is the complementary (n - p)-tuple.To prove (6.23) we write e* in the form

e* = E, a*v1 n ... n e*Vn,

where o denotes the permutation (1, ..., n) - (v1, ... , v").

Poincare Duality

Then we have

De(evl ^ ... A evp) = E i(ev1 A ... A evpxe*v1 A ... A e*v")

= E,ri(evp) ... i(ev1xe*v1 ^ ... n e*' )

= E a*vp + 1 n ... n e*v".

The relations v 1 < < vp and vp + 1 < < vn imply that

E = (-1)''_°and so Formula (6.23) is proved. In the same way it follows that

De(e*v1 ^ ... ^ e*') = (- evn.

In view of Theorem 6.8.1, Part (3) and (6.23) we obtain

De(e*p + 1 ^ ... n e*vn) = DeDe(_ 1)Ep=1(v1-i)ev l

^ ... A evp

= (_ 1)P(n-P)(_ n ... A evp

165

i.e.,

De(e*Vp + 1 ^ ... n 1)"(- n ... A evp , (6.24)

(6.23) yields, in the -case p = n - 1,

De(ei A ... A ei n ... A en) = (-1)n- `e*`. (6.23)

6.14. The External Product

Consider the (n - 1)-linear skew-symmetric mapping of E x x E into

E* defined byn-1

[x 1, ... , xn _ 1] = De(x 1 A ... ^ xn - 1).

The vector [xi, ... , x_ 1] is called the external product of the vectorsx, (i = 1, ..., n - 1). Clearly the definition of the external product dependson the choice of the basis vector e* e A nE*. It follows that

<[x1, ..., x-], xi>=<e*, x1 n...nx- nx1>=0 (i=1, ..., n-1)n1 n1showing that the external product is orthogonal to all factors. Now considerthe external product of n- 1 vectors x*v e E*,

[x*1, ..., x*n-1] = De(x*1 n ... n x*n-1).

From Theorem 6.8.1, Part (1) we obtain

<[x1,..., xn1], [x*1, ..., x*"- 1]> = {x*l A ... ^ x*n- 1, x A ... A x

= det(<x*`, x;>).


This yields the Lagrange identity

... , x,,_], [x*l, ... , x*"-1]> = det(<x*`, x;>) 0 < i, j < n - 1.

For the external product of (n - 1) basis vectors e we obtain, from (6.25)

This formula gives, in the case n = 3,

[e1, e2] = e*3, [e2, e3] = e* 1

i= , ..., n.

[e3, e1] = e#2

6.15. Euclidean Spaces

Suppose now that E is an n-dimensional oriented Euclidean space. Then allthe spaces A "E (1 <p <- n) are Euclidean and there exists precisely one unitvector e e A "E which represents the given orientation. Since E is self-dual wemay set E* = E. Then the Poincare isomorphisms coincide and are given by

De u = i(u)e u e AE.

De maps A "E onto A "- pE (0 < p < n). Theorem 6.8.1, Part (1) impliesthat

(Deu, DeV) _ (u, u) u, u e A E

and so De is an isometric mapping. From Theorem 6.8.1, Part (2) we obtain

Dp = (6.26)

where Dp denotes the adjoint of D.Assume now that n is even, n = 2m. Then the operator Dm defines a linear

transformation of AmE,

Dm : A mE A mE.

The above formula yields

Dm = (-1)mDm.

It follows that the transformation Dm is selfadjoint or skew depending onwhether m is even or odd.

PROBLEMS

1. Define the operation of GL(E) on A (E*, E) by

az = [(a ^ )-1 p «^ ]z x e GL(E).

Prove the following relations :

a. a(z 1 + z2) = az 1 + az2b.c. (ad)z = a(flz)d. iz = z (i being the identity map).

Poincare Duality

Show that

where is the unit tensor.

2. Verify the relations

app=gyp (Ocp<n),

i(1Qu)'=u uEApE

i(u* Q 1)p = u* Q 1 u* E A PE*

n+q,p n w =n 9 <q, w w E A (E*, E)

P

3. Verify the formulas

De i(v*)u = (-1)q(n _ )v* n Deu U E ARE, V* E /\*

D%(v)u* v n Deu* y e A I E, u* E qE*

DE i(z)w = (-1)P(S - n)+ qc' - DE w z e A (E*, E), we A (E*, E).

4. Prove that

Deµ(z) = i(z)DA z e 0(E*, E).

167

5. Let E and F be vector spaces of dimension n and let a : E - F be a linear isomorphism.

a. Denote by DQ and Df the Poincare isomorphisms

Deu* = i(u*)e, Df v* = i(v*) f,

where a is a basis vector of A' E and f = a e. Show that

Df o (a ") - i= a A 0D.

b. If F = E prove that

DE(ocz) = aDE(z) z e A (E*, E)

(for the definition of az see Problem 1).

6. Let E be an oriented Euclidean plane. Show that the linear map

De : E 4 E

is a rotation with rotation angle + ir/2.

7. Let E be an oriented 3-dimensional Euclidean space. Prove that

x x y= De(x n y) x, y c E

(see Section 7.16 of Linear Algebra). Use the above relation to prove the formulas

(x1 X YX2 X Y2) = (x1, x2)(Y1, Y2) - (x1, Y2xx2, Yi)

and

(x x y) x z = y(x, z) - x(y, z).


8. Let {e}, {e*"} be a pair of dual bases of E and E*. Prove that

DL, (x 1 n ... n xp)

= p!(- 1 Z =1(vi - i) /e*V1, x1\ ... <e*vn, x >e*vP + 1 ^ ... A a*v".

v1<...<VP

In the above sum, (vp+ 1, ... , vn) denotes the ordered complement (v1, ..., vp) inthe natural order.

9. Show that the restriction of the isomorphism DE to the subspaces A (E*, E) andA °(E*, E) is the identity.

10. Show that the components of DE Z are given by (see Problem 8)

(DEZ)p+ 1..... Vn = (- 1)' 1(v1 - 1

1 1 < ... < Vn,

where (v1, ... , vp) is the complementary p-tuple of (vp+ 1, ... , vn) and (/2i,. . . , pq)is the complementary q-tuple of (pq + 1, ... , pn) in the natural order.

11. Let E* and E be two dual 3-dimensional spaces and a e E, a & 0 and b* E E* be twogiven vectors. Prove that there exists a vector x e E such that [a, x] = b* if and onlyif <b*, a> = 0, and prove that, if xo is a particular solution of [a, x] = b*, then thegeneral solution is given by x = xo + Aa, e f.

12. Consider an (n - 1)-linear skew-symmetric map q of E into a vector space F. Provethat there exists exactly one linear map x : E* - F such that

(p(x1, ... , x,,- 1) = x[xl,... , xn_ 1] x,, E E (V = 1, ... , n - 1).

13. If a is a linear automorphism of E, prove that

[wc1, ... , det a((x-1)*[x1, ... , x,,_1] x,, E E (V = 1, ... , n - 1).

14. Let e,, (v = 1, ..., n) be a basis of E such that e1 n n e n = e. Given n - 1vectors

xi = 4 c eV (i = 1, ... , n - 1),v

show that the vector y* = [x1, ..., xn ] has componentsA

5 11 1

,Iv = (1)_V1 Av nn-1 ' n-1 " n-1

15. Using the formula in Problem 14 derive the relation

n H

V1 v=1

n nv 1 v n-1n - 1 nv ' '

V1 V1

n

=V= 1

(v= 1, ... , n).

A

nl n ... n1 zAv n

yn-1 sn-1

from the Lagrange identity.

Applications to Linear Transformations 169

16. Find the minimum polynomial, the characteristic polynomial, the trace, and thedeterminant of D. Hint: In case dim E = 2m consider the restriction of Ds to thesubspace n m(E*, E).

17. Suppose E is a Euclidean space of dimension n = 2m and set E* = E. Prove that

n m(E, E) = X 2(A mE) 8 Y2(A mE)

is a decomposition of n m(E, E) into orthogonal spaces stable under Ds (see Sections4.2 and 4.10 with E replaced by n mE).

Applications to Linear Transformations

In Section 6.16 E and E* denote a dual pair of n-dimensional vector spaces.If p : E - E is a linear transformation, then the induced homomorphism

P A : AE - AE will also be denoted by A (P and the restriction of (p to AEwill be denoted by n p(p.

6.16. The Isomorphism T

Let t e E* ® E be the unit tensor for E and E* (see Section 1.26). Then tdetermines a linear map

T:L(E;E)_E*®E

defined by

T (q,) = (i q e L(E ; E).

(Notice that this isomorphism T is the inverse of the T in Section 1.26.) Inparticular, we have

T(i) = t.

Now let a e E and b* a E* be arbitrary vectors. Then it follows that

<T(co), b* ® a> = ®:)t, b* ® a> = <t, P*b* ® a>

= <(p*b*, a> = <b*, cpa>,

whence

<T(cp), b* ® a> = <b*, cpa> (p e L(E; E), a e E, b* e E*. (6.27)

Using this relation, we obtain

<T (q,), t> = <e, tr (p,

T(cp), t> = tr p p e L(E; E). (6.28)

170

Similarly, a linear map

T : L(E*; E*) _ E* ® E

is defined by


?(x) = (x ® c )t x e L(E* ; E*).

An argument similar to that given above shows that

< T (x), b* O a = <xb*, a> = <b*, x*a) x e L(E* ; E*), a e E, b* E E*.

Comparing this relation with (6.27) we find that

T(qp) = T(rp*) (p e L(E; E). (6.29)

Now let l/i e L(E; E) be a second linear transformation. Then we have

T (Y' p) = (l ® = (:® Y') [(l ® p)t] = (l ® Y') T ((p),

whence

T(fi ° q,) = (i/i ® l)T(co). (6.30)

With the aid of (6.28) and (6.30) we shall prove that T and T are linearisomorphisms. In fact, for any p e L(E; E), i/i* e L(E*; E*) we have

T(/,*)) = l)t) = <(l ® t)= <T(i/i ° qi), t> = tr(i/i ° qi),

< T (q,), T (,fr*)> = tr(/i ° q,) rp, l/i e L(E; E).

Since the bilinear function (i//, rp) - tr(i/i o q,) is nondegenerate, it follows thatT and T are linear isomorphisms.

The Skew Tensor Product of n E* and A E

6.17. The Algebra A E* p n E

Recall that so far we have defined two multiplications in the space A E* ®A E, the canonical tensor product of the algebras A E* and A E (see Section6.1) and the composition product (see Sections 6.4 and 6.5). We shall nowintroduce a third algebra structure in this space, namely the skew tensorproduct of the graded algebras A E* and A E. The corresponding multi-plication will be denoted by A.

The Skew Tensor Product of A E* and A E 171

Thus,

(u* ® u) n (U* ® U) = (-1)9r(u* n U*) ® (u n U)

U* E A E*, u E A 9E, U* E A rE, U E A E.

A E* © A E is an associative algebra with unit element 1 ® 1. It is generatedby the elements 1 ® 1, x* ® 1 and 1 ® x where x* a E* and x e E. Theformula above implies that

W1 A W (- 1)(P+ 9)(r+s)W2 A W1 2 - 1

w 1 E A PE* ® A 9E, W 2 E A rE* ® A E. (6.31)

as is easily verified. In particular, if p + q or r + s is even, then

W1AW2=W2AW1.

For w e A E* ® A E we shall set

1wk= k>1

k.k

and

Then (6.32) yields the binomial formula

(6.32)

(w1 + w2)k = wi n ww1, w2 e A PE* ® ^ 9E, pq even.i+j=k

Note that the products in the algebras A E* Q A E and A E* ® A E areconnected by the relation

(u* ® u) A (U* ® U) = (-1)9r(u* ® u) (U* ® U) u E A 9E, U* e A rE*.

In particular,

z1 n ... A zk = (_ 1)k(k- 1)/2z1 ... zk zi e E* ® E, (6.33)

and so1)k(k - 1)/ 2 zk

6.18. The Inner Product in E* © E

zeE* ® E.

Consider the bilinear function (, >> in the space A E* ® A E defined by

* * ((-1y <u*, U)<U*, u> if p = s and r = q,®U,V ®U>> =

0 otherwise,

where u* e A PE*, u e A 9E, U* e A rE*, v e A E. It is easily checked that thisbilinear function is an inner product.


Next, recall from Section 5.15 the isomorphism f : A E* Q A E -+A (E* Q E) given by f(u* ® v) = 'A u * n JA v where i : E* -+ E* Q E andj : E - E* Q E denote the inclusion maps.

Proposition 6.18.1. Define an inner product in E* Q E by

(x* © x, y* © Y) = <x*, y> + <Y*, x>.

Then f is an isometr y,

«.f (u* ® u), f(v* ® v)) = (u* ® u, v* ® v>>.

PROOF. We may assume that

u=xln...nxqand

v*=y*A...A y*, v= y1 A... Ay.

Then we have

f (u* ® u) = (ix t A ... A ix) A (jx 1 A ... ^ Jxq)

f(v* ® v) = (iy* n ... n iy*) ^ (jy 1 n ... A JYj

whence

(f (u * ® u), .f (v* ® v)) = (<(ix* n ... n i4) n (jx 1 n ... A Jxq),

(iy n ... A iy*) n (j y1 n ... A

Thus,

(f(u* ® u), f(v* ® v)) = 0 ifp+r q+s.On the other hand, if p + r = q + s, then

(f(u* u v* v = det «iy*>> «ixQ,JYo>(1 x p , iy ) (1 x , jya)

But, since

(ix, iy) = 0 a = 1, ... , P,

(ixa , JYa) = (x, Ya>> P = 1, ... , q,

(Jxp, iy*) = «Y*, xp>> Y = 1, ..., r,

(Jxp, JYa>> = 0, b = 1, ... , s,

it follows that

( f (u* ® u), f (v* ® v)) = 0 unless 5 = p and r = q.

The Skew Tensor Product of A E* and A E 173

Ifs = p and r = q, we obtain

«.f (u* O u),.f (U* O U)) = (-1)P9 det(<x« , ys>)det(<yy, , xp>)- ( 1)' <u*, U><U*, u>

= ru* ® , ®

which completes the proof. aFinally note that the inner product defined above and the inner product

defined in Section 6.2 are obtained from each other by the formula

(W1, wz) = (-1)' <w1, w2> w1 E A 9(E*, E), w2 E A p(E*, E). (6.34)

In particular,

«Z1 A ... ^ ZP, W1 ^ ... n WP>> = (- 1)P<Z1 n ... ^ ZP, W1 ^ ... n WP>

zi EE* Q E, wi EE* Q E. (6.35)

7 Applications to LinearTransformations

In this chapter E continues to denote an n-dimensional vector space over a field of characteristiczero.

All the problems concerning this chapter are collected at the end of the chapter.

The Isomorphism DL7.1. Definition

Recall from Section 6.2 the isomorphism

TE: A (E*, E) -> L(n E; A E).

Let

DL:L(AE; AE) L(AE; AE)

denote the isomorphism defined by the commutative diagram

n (E*, E) - L( A E ; A E)

A (E*, E) L( A E; A E)

where DE is the Poincare map. If u* E A E* and u e A E, the linear trans-formation DL ° TE(u* ® u) is explicitly given by

[DL o TE(u* ® u)]u = <Deu, u>Deu*.

From the results of Section 6.10 and Formula (6.12) we obtain the follow-ing relations:

DL = SZL ° DL (7.1)

DL = SZL (7.2)

DL ('APE) = lnn-PE

tr(DL X Dj) = tr(x ° f3) a, fi e L( A E; A E),

where cZL denotes the involution in L( A E; A E) given by cZL = TE ° E °

174

The Isomorphism DL

The last relation implies that

tr(DL a) = tr a a e L(A E; AE).

175

Proposition 7.1.1. Let e be a basis vector of n E. Then we have for a, fJEL(nE; AE)

DL a = D° a* ° (De}-1

DL(x ° fi) = DL(f3) ° DL(a)

PROOF. Let a = TE(u* ® u). Then we have, by Theorem 6.8.1,

CDe ° TE(u* ® u)* ° De J(U) = De(<De u, u>u*) _ <Deu, V>Deu*

= DL TE(u* ® u)(U)

and so (1) follows. (2) is an immediate consequence of (1). 0

Next observe that the image of the diagonal subalgebra 0(E) ander TE isthe space L( A "E; A "E). Thus DL restricts to a linear automorphismDL of p= o L( A "E; A "E). Moreover, since cZL reduces to the identity in thisspace, Formulas (7.1) and (7.2) show that DL is a selfadjoint involution.

Let (P be a linear transformation of E and consider the induced transforma-tion i ® cP A of A (E*, E). To simplify notation we shall set

Then

Now Formula (6.4) yields the relations

i(Z) o 4j* = * o i(4Z)

i(z) o = o i(4" z).

Moreover, if = TE(z), Lemma II of Section 6.3 implies that

w = cD(w) w e A (E*, E)

(7.3)

w° p= 4* w we /1 E*(7.4)

Setting w = gyp, we obtain

zp = gy(p) (P = 4, ... , n). (7.5)

7.2. The Determinant

Recall from Section 4.4 of Linear Algebra that the determinant of a lineartransformation ( is defined by the equation

i px 1, ... , det P A(x 1, ..., x).

176 7 Applications to Linear Transformations


(A ncp*)e* = det p e*

where e* is a basis vector of A "E*. Thus,

det (p = tr( A nip*) = tr( A "gyp).

Similarly,

( ^ e* = (A det p e,

where a is a basis vector of A E.Since " = e* ® e, where a*, a is a pair of dual basis vectors of n "E* and

A nE, the relations above imply that

fi(n) = det (p -" .

Now Formula (7.5) shows that

z" = det ( & z e E* ® E.

Proposition 7.2.1. Let z e E* ® E and set TE(z) = (p. Then for 0 < p < n,

l(DE zP) = det ( "-P

1. D zP = det _( E ) n p

and

2.

Zp o DE(z" - P) = det (p ' Lp

DE(z" - P) o zP = det (p .P .

PROOF. Using (7.5), the second Formula (7.3) and Part (ii) of the Corollary toProposition 6.7.1 we obtain

z'(DE 9) = n

= (1( p) }tn = det pi(p) n = det

which proves the first Relation (1). The second Relation (1) is obtained in thesame way.

(2) In fact, the first Relation (7.4) and Part (1) of the proposition yield

ZP DE Z" P = I'(DE Z" P) = det (p. P .

The second Formula (2) is established in a similar way.

Corollary (Laplace formula). Let p be a linear transformation of E. Then

1. Pip ° DL( ^ "- P(p) = det (p 1^PE

and

2. DL(^ Pip) ° A n P( = det (p i pL .

The Isomorphism DL 177

PROOF. Apply TE to Formula (2) in the proposition and observe that TE(zp)= A pq,

Proposition 7.2.2. Let z e E* ® E and set T (z) =gyp. Then

1. i(D Zp)Zq=(2n p - q

En p

and

det , `p+q-n

2. DEZp DEZq =2n p pq det q, . D zp+q-"E .n-

PROOF. In fact, applying (7.5) the second Relation (7.3) and Part (1) ofProposition 7.2.1 we obtain

i(DE ZP)z = 1(DE Zp)D( q)

= I i(D*DE det q cDi(4n _

2n-p-q)detP _ `gyp + q - n)n-p

(2n - p - 91cpz)det

n-p(2) Since the restriction of DE to the diagonal algebra A(E) is an involu-

tion, we have, in view of (6.22),

DEL1(DE ZP)Zq] - DEL1CDE Zp)DE(DE zq)

= DE[DE(DE Zq DE ?)] = DE(zq) ' DE(zp).

Now (2) follows from (1), since AE is commutative.

7.3. The Adjoint Tensor

Consider the symmetric (n - 1)-linear mapping

Ad:(E*®E) x ... x (E*®E)_.,E*®E

given by

Ad(Z 1, ..., Zn _ 1) = DE(Z l ... Zn _ ) ; e E* ® E,and set

1

ad(z) = Ad(z, ... , z).(n-i).


Thus,

ad(z) = DE(Zn - 1) ;

ad(z) is called the adjoins tensor of z. Observe that ad(z) does not dependlinearly on z except in the case n = 1.

Since DE is an involution in the diagonal subalgebra AE, we have therelation

DE ad(z) = zn -1 z e E* Q E.

Moreover, Proposition 7.2.1, (2) implies that

ad(z) o z = z o ad(z) = det TE(z)&

Next, let z e E* Q E and define tensors B p(z) E E* Q E (p = 0, ..., n - 1)by the expansion

p- 1ad(z + _ B p(Z)a," p - 1.

p=0

Proposition 7.3.1. The tensors B p(z) are given by

B p(z) = 1 (P = 0, ..., n - 1).

PROOF. Since, by the binomial formula (see Section 6.2)n-1

we have

(Z + ), )n - 1 = 1- pp n - 1 p

p=0

n-1ad(z + A) _ DE(zp n _ pP," - 1-p.

p=0

But, since A(E) is commutative,

and so we obtain

It follows that

DE(Z"' n - p - 1) = 1(Zp 4n - p - 1) n = 1(Zp) p + 1

n- 1ad(z + ) = 1(Zp), + 1

an p - 1.p=0

B p(z) = 1 (P = 0, ..., n - 1).

Next we prove the Jacobi identity

DE(ad z)" = (det TE(z))p -1 z" p 1 <p < n. (7.6)

Proposition 7.2.2, (1) yields

i(ad z)z9 = (n + 1 - q)det TE(z)z4-1

The Isomorphism DL

whence, by induction,

i ad z p z9 = n+ p q det T Z pz9 - p.P

In particular,

i(ad Z)p - 1)Z" - 1 = p(det TE Z)p - 1z " - P 1 < p < n.

On the other hand,

i [(ad Z)p - 1] Z" - 1 = i [(ad Z)p - 1 ] DE ad z

and thus, in view of Formula (6.22)

i((ad Z)p - 1)Z" - 1 = DE [(ad z). (ad z)" - 1]

= pDE(ad z)p.

Combining these relations we obtain the Jacobi identity.Setting p = n - 1 in this identity yields the formula

ad ad z = (det TE(z))"- 2z z e E* ® E, n > 2.

7.4. The Classical Adjoint Transformation

179

Let 4' -1 be linear transformations of E. Fix a nonzero determinantfunction 0 in E. Then, for every n-tuple of vectors (x 1, ..., x"), a linear trans-formation i (x 1, ..., x") of E is defined by

SZe(x 1, ... , x")h = EQ 0(qp 1 4n - 1 XQ(n - 1) '

It is easy to check that SZs(x 1, ... , x") is skew-symmetric in the x. Thus thecorrespondence

(x1, ..., x") H 1, ..., xn)

defines a skew-symmetric n-linear mapping

E x x E -+L(E ; E).n

Hence, there is a unique element in L(E; E), denoted by Ad(p1, ..., c°?such that

SZs(x 1, ... , x") = 0(x 1, ..., xn)Ad(q,1, ..., c_ 1).

Now the above definition reads

Ad(cp1, ... , c_ 1)h 0(x 1, ..., x") = E, 0((p 1 X(1), ... , 4_ 1 x_ 1) ' h) x7(n) .Q


In particular, if {e1, ..., en} is a basis of E such that 0(e 1, ..., en) = 1, thenwe have, for h e E,

Ad((p1, ... , q_ 1)h = c, 0((p1 eQ(1), ... , (pn - 1e(_ _ 1),Q

We shall show that the function Ad is symmetric. In fact, let be a permuta-tion of (1, ..., n - 1) and set

t(aC1, ... , xn)h = c (1) (pt(n - 1)X7(_ 1), h)x0(n)Q

It has to be shown that SZt = SZo . Extend i to a permutation of (1, ..., n) bysetting i(n) = n and let a = o r'. Then we have

t(aC 1, ... , xn)h = Ea 0((p 1xa(1), ... , (pn - 1 Xa(n - 1), h)xa(n)a

= SZA(x 1, ..., xn)h,

whence SZt = SZo .

For a single linear transformation (p we set

ad(q,) =1

Ad((P> ... > (p).(n _ 1)n-1

Since

n

0(cox 1, ... , h, ... , (pxn)xQ(n) _ (n - 1) ! o(rpx 1, ..., h, ... , (pxn)xi ,Q

Q(n)i = 1

it follows that ad((p) coincides with the classical adjoint of (p as defined inSection 4.6 of Linear Algebra.

Proposition 7.4.1. The operators DL and Ad are connected by the relation

1. DL(q l D ... D (Pn - 1) = Ad((p 1, ... , (pn - 1)

(p e L(E; E), v = 1, ... , n - 1.

In particular

2. DL(^n-1(p) = ad(p) P e L(E; E).

PROOF. Without loss of generality we may assume that the q are of the form

(p= TE(a* Q a) a* a E*, aa E.

Fix a basis {e1, ..., en} of E such that 0(e 1, ..., en) = 1 and set

e 1 A ... n en = e e* 1 n ... n e*n = e*.

The Isomorphism DL

Then we have for h e E

= E, 0(<a i , 1, ... , <an - 1, eQ(n - 1)>an - 1, h)e (n)Q

181

= i , <a_1, 1)> . 0(a 1, ...,

It follows that for h* E E*

<h*, Ad((p1,... , (Pn-1)h>

= <an- 1, eQ(n- 1)><h*, 0(a1, ... , an- 1, h)

= <ai A A an_1 n h*, e1 A A

<ai n n aq_1 n h*,e> <e*,a1 n n n h>

= <h*, a><a*, h>,

where

a = i(a i n ... n a_ 1)e and a* = i(a 1 n ... n a_ 1)e*.

Thus,

<h*, Ad((p1, ..., <h*, a><a*, h>. (7.7)

On the other hand,

DL((p1 ... p, - 1) = DL TE[(a i ® a 1) ... (a_ 1 ® a_ 1)]

= D L TE [(a 1 ^ ... A a_ 1) ® (a1 A ... A a_ 1)]

TEDE[(al ^ ... ^ an_ 1) ® (a1 A ... A an- 1)]

= TE[i(a 1 n ... ^ a_ 1)e* ® i(a i n ... n a_ 1)e]

= TE(a* ® a).

It follows that

<h*, DL((P1 ... (Pn- 1)h> = <h*, TE(a* ® a)h>= <h*, a><a*, h>.

Now the proposition follows from Relations (7.7) and (7.8).

Corollary I. Let p be a linear transformation of E. Then

iad(p) = det (p

and

ad(p) o p = det p i

In particular, ad(p) is a linear isomorphism if and only if p is.


PROOF. In fact, Proposition 7.4.1 and the Laplace formula (see Section 7.2)yield

ad((p) ° (p = (p o ad((p) = (p ° DL( ^ "- 1(p) = det (p 1.

Corollary II. If p and cui are linear transformations, then

ad(/i o (p) = ad((p) o ad(/i).

PROOF. In view of Proposition 7.4.1 and Proposition 7.1.1, (2),

ad(/i ° (p) = DL ^ "- 1(Y, ° (p) = DL(^ "- 1,1/ ° ^ "-1(p)

= DL( A "- 1(p) o DL( A1 ,/i) = ad(/i) o ad((p).

Corollary III. Let (p be a linear transformation of E. Then

DL(A p ad (p) = (det (p)p - 1 ^ " p(p (P = 1, ..., n).

In particular,

ad ad((p) = (det (p)"-2(p n > 2.

PROOF. Apply the proposition and the Jacobi identity (see Section 7.3).

Corollary IV. Let a* a E*, av e E (v = 1, ..., n - 1) and set

(Pv = TE(a* ® a).

Assume that either {a*} or {av} is linearly dependent. Then

Ad((p1, ..., (p"-1) = O.

Characteristic Coefficients

7.5. Definition

Consider for each p > 1 the p-linear function C, in L(E ; E) given by

Cp((p 1, ..., p1,) = tr((p 1 ... p1,)

and set

Co = 1.

Since the box product is commutative, the functions Cp are symmetric. Notethat

C 1((p) = tr (p (p e L(E ; E).

Characteristic Coefficients 183

The function C will be denoted by Det,

Det(p1, ..., cn) = tr((p 1 ... cn)Now consider the tensors z e E* a E given by

TE1(4)) v= 1,...,p.Then Proposition 6.3.1 and Formula (6.16) imply that

Cp((p 1, ..., p p} = tr(TE(z 1) ... TE(zp)) = <z 1 ... (7.9)

In particular,

Cp=O ifp>n.For a single linear transformation P we set

C = 1 C ... _P

Thus, in view of (6.6),

Cp((p) = tr( A Pop) (p = 0, ..., n).

Clearly,

C(AP) = APCp(q,) A e F.

Relation (7.9) implies that

Cp(4) = <(T E 1(p)P, p> (P = 0, ..., n). (7.10)

Recall from Section 7.2 that

det cP = tr( A p).

Thus

Cn(4) = det gyp.

Equivalently,

1-! Det(cp,... , 4)) = det (p.n

Replacing cp by Q + /i in this relation we obtain the formula

1

det(gp +) = Det(gp, ..., (p, ..., i/).n.p+q=n

P 9

Proposition 7.5.1. The CP(cp) satisfy the relation

n

det(gp + )i) = C_ A e F.p=0

Thus p(cp) is the pth characteristic coefficient of cp.


PROOF. Let z e E* ® E be the unique tensor such that TE(z) = gyp. Then we have

(p+)1= TE(z+'U)

whence

det(gp + ti) = <(z + )", i">.

Expanding by the binomial formula we obtain"

det(gp + t1) _ <p z" -

p p

p = C"- p(co) (= 0, ..., n).

It follows that"

det(gp + i) = C"- p(co) V'.p=o

Corollary. If q, and i/i are linear transformations, then the characteristicpolynomials of i/i o q, and q, o ,/i coincide.

PROOF. In fact, since the trace of a composition does not depend on theorder,

Cp(11/ ° (p) = tr n p(/, ° q,) = tr( n p,/, ° n pep)

= tr( n pip o n " 1i) = Cp(co ° i/i) (p = 0, ..., n).

7.6. The Linear Transformations Ap(p)

Let q be a linear transformation and writen-1

ad(gp + 1) -p=o

where A p(ip) E L(E ; E).In particular,

A_ (q,)(q,) = ad(gp).

Proposition 7.6.1. The transformation A p(co) is given by

P

(-1)vCp- 0, ..., n - 1).=0


PROOF. From the formula (see Corollary I to Proposition 7.4.1)

((p + i) o ad((p + i) = det((p + ti).

we obtainn n- 1 n

L, A,_ 1((P)An P + A p(co)1P = C p((p)),n P

1.

p=1 p=0 p=0

Comparing the coefficients of on both sides of this equation yields therecursion formulas

(P ° Ap- i(P) + A(p) = Cp((P) l (p = 0, ..., n),where A_ ((P) = 0 and An((p) = 0. From these relations we obtain

P

A p((P) = (-1)VCp - (P = 0, ..., n). (7.11)v=0

In particular, for p = n - 1,n- 1

ad((p) = (- 1)VCn - v - 1((P)(Pvv=0

Cayley-Hamilton theorem. Every linear transformation cp satisfies its char-acteristic equation; that is,

n

L, (- 1)VCn - v((p)(pv = 0.v=0

PROOF. Apply (7.11) for p = n observing that An((p) = 0.

Proposition 7.62. The trace of the linear transformation A p((p) is given by

tr Ap((P) _ (n - p)Cp((P) (p = 0, ..., n).

PROOF. Let z e E* Q E be the unique tensor such that TE(z) _ (p. Consider thetensors B p(z) determined by the expansion

n- 1ad(z + i) _ B p(Z)a,n- p- 1

p=0

(see Section 7.3). Since

TEBP(z) = A(p) (P = 0, ..., n - 1),

tr A (p) _ <, B p(z)> (p = 0, ..., n - 1). (7.12)

Next observe that, by Proposition 7.3.1,

Bp(z) = i(zP}p + (P = 0, ..., n - 1).


It follows from (7.10) that

<, B ,,(z)> = l( )B p(Z) = l( 1

= i(z")i()ip+ 1 = (n - p)<z", p> = (n - p)Cp(co). (7.13)

Relations (7.12) and (7.13) yield

tr A p((p) = (n - p)Cp(co) (p = 0, ... , n).

Finally, we show that the characteristic coefficients of the classical adjointtransformation are given by

Cp(ad (p) = (det (p)P 1C_(p) i(p = 1, ..., n). (7.14)

In particular,

det ad((p) = (det (p)" -1 and tr ad((p) = C" _ 1((p).

In fact, taking the trace on both sides of the Jacobi identity (7.6) andobserving that for a E L( A E; A E)tr DL X = tr a (see Section 7.1) we obtain

tr A" ad((p) = (det (p)P- 1 tr n ""p (p = 1, ... , n)

whence (7.14).

7.7. The Trace Coefficients

Let Trp denote the symmetric p-linear function in L(E ; E) given by

Tr ... = 1 tr > 1p

and set

Tro = n.

The pth trace coefficient of a linear transformation p is defined by

Trp((p, ... , gyp) p >_ 1

and

Tro((p) = n.

Thus

Tr p((p) = tr (pn p > 1.

In particular,

Trp(i) = n.

Note that, in contrast to the characteristic coefficients, the trace coefficientsdo not vanish in general for p> n.


Proposition 7.7.1. The trace coefficients and the characteristic coefficients areconnected by the relation

1 p-1 _Cp(co) = - (- 1)P v

- 1 Cv((p)Trp- (p) p ? 1.P v=0

-

PROOF. Taking the trace in the formula in Proposition 7.6.1 and using Propo-sition 7.6.2 we obtain

P

(n - (-1)VCp-v=0

P

= nCp(co) + (- 1)VCp - v((p)tr (pvv=i

It follows that

P

Cp(p) = - - (-1)vCp - v((P)tr pvP v=1

1_(- 1)P q9 v.

P v=o

7.8. Application to the Elementary Symmetric Functions

Fix a basis {e1, ..., en} of E and consider the linear transformation cP given by

cpev = ,vev (v = 1, ..., n).

A simple calculation shows that

C(p) = ..., n)while

where Qp and sp denote the symmetric polynomials given by

... , ,n) = ,v1 ...P P > 1

v1<...cvp

and

P

sp(a,1, ..., = viNow Proposition 7.7.1 yields the classical recursion formulas for the Qp interms of the sp,

1

op= - (- 1)P v-

1 ovsp - v (p= 1, ... , n).

P v=0


7.9. Complex Vector Spaces

Let E be an n-dimensional complex vector space and let E denote the 2n-dimensional real vector space. Let DE be a nonzero determinant function inE. Regard DE as a C-valued n-linear function in E and set

0 = (- l)n0E n DE ,

where DE is defined by

DE(x 1, ... , xn) = DE(X 1, ... , xn) xv a E.

Then 0 is linear over and skew-symmetric. To show that 0 is real-valuedand nonzero (and hence a determinant function in choose a basis{al, ... , an} of E. Since 0(zl, ... , zn) = 0 whenever the vectors {z} arelinearly dependent (over C) it follows that

0(al, ... , an, ial, . lan)

(- i)n

. , a7(n))DE laid l), ... ,

Thus,

(n!)2

= DE(a 1, ... , an)AE(a 1, ... , an)

= I DE(a 1, ... , an) 12.

and so 0 is a nonzero determinant function in E. If DE is replaced by ),DE,where ), is a nonzero complex number, then 0 changes into I , U 20. Thisshows that the orientation of E determined by DE is independent of DEand so E carries a natural orientation.

Proposition 7.9.1. Let cp be a linear transformation of E and let cps denote thecorresponding linear transformation of E. Let f and f denote the character-istic polynomials of p and Then

li(t) _ .f(t) f(t).PROOF. We show first that

det (p = det (p det (p. (7.15)

In fact, let DE be a nonzero determinant function in E and set 0 = inAE nDE . Then

q A = (- i)n(p*DE A (p*DE

But

(p*DE = det (p DE, (p*DE = det cP DE


and so we obtain

0 = (-1)" det (p det P 0E 0E _ det (p 20

whence (7.15).Replacing by co - ti in this relation yields

det(q, - ti) = det(gp - ti)det(cp - ti)

whence

Je(t) =.f(t) I(t).

Corollary. The characteristic coefficients of p are given by

Cr((P) = (r = 0, ..., n).p+q=r

PROBLEMS

In the problems below T is the map defined in Section 6.16.

1. Find all linear transformations p of E such that T(cp) is decomposable.

2. Let

r

T(p) _ > a O a*` a E E, a*` E E*

be a representation of the tensor T(p) such that the vectorsa1 and a*` (i = 1, ... , r)are linearly independent. Show that r =

3. Let E, E* be a pair of finite-dimensional dual vector spaces and consider the linearmap

f:E*QE_ L(E;E)defined by f (b* O a)x = <b*, x>a. Prove that

a. f= T*b. 1= T'.

4. Verify the relation

<zp, (ad z)p> = n()(det T -'(z))" 1 < p < n, z e E* O E.p

5. Show that

det A nip det A "- nip = (det 'p)0 < p < n.

6. Show that the coefficient of )J - n in the characteristic polynomial of an n x n-matrix is (-1)" _ p times the sum of all principal minors of order p.


7. Let E be a Euclidean space and set E* = E. Write

L(E; E) = S(E; E) Q+ A(E; E)

where S(E; E) denotes the space of selfadjoint transformations and A(E; E) denotesthe space of skew transformations (see Linear Algebra Problem 3, Chapter IX,Section 2). Prove that

T(S(E; E)) = Y2(E)

and

T(A(E; E)) = X 2(E).

(see Sections 4.2 and 4.10).

8. Let E be a real vector space and consider the bilinear function b in 0(E, E*) definedby

I(u, v) = <Dsu, v> u, v e 0(E, E*).

a. Prove that D is symmetric and has signature zero.Hint: Make E into a Euclidean space and set E* = E; then consider first A m(E, E)

in the case dim E = 2m. See problems 7 above, Problems 3 and 4 after Section 6.14,Problem 4, Section 5, Chapter IX of Linear Algebra, and Problem 3, Section2, Chapter IX of Linear Algebra.

b. Given two linear transformations 'p, t// e H( A E; A E) (see problem 13) prove that

D(T(co), T(/i)) =

and conclude that the signature of the bilinear function

F(cp, /,) = b(T(p), T(/i))

is zero.

c. If dim E = 2m and gyp, cli are any two linear transformations of E such that p isregular, prove that

T(i)m) = det p tr A "(p' ° cli)

and

tr A m = det p tr A m -1.

Conclude that

= det p

where am and &m are the mth characteristic coefficients of 'p and P -1 respectively.

d. If n = 2 show that,

T(cli)) = tr p tr cl' - tr(' ° cli).


9. Use one of the formulas (2) in Proposition 7.2.2 to derive the classical Laplaceexpansion formula for a determinant:

Let A = (a;) be an n x p matrix and fix a p-tuple (2i, ... , 2p) such that 21 << 2p. Let (2p+ 1, ... , 2n) denote the complementary (n - p)-tuple in increasing

order. Then

det A = (-1)E =1det detV1<...<Vp

where (vp+ 1, ... , vn) is the complementary (n - p)-tuple of (vl, ... , vn). Heredenotes the q x q matrix consisting of the rows 111..... /39 and the columns

OC 1, ... , OC9 .

10. Show that the rank of the adjoint transformation is given by

r(ad gyp) = n if r(p) = n

r(ad gyp) = 1 if r(cp) = n - 1

r(ad gyp) = 0 if r(cp) < n - 2.

Use these relations to prove the formula

ad ad p = (det p)n - 2(p.

11. Given an n x n matrix A = define the matrix by

(v, u = 1, ... , n).

1 µ n«n ... an ...

Using the Jacobi identity in Section 7.3 prove the classical Jacobian identities:

«µP + 1 «µP + 1Vp + l Vn

«µn . .. «µnvr + 1 vn

(det A)'1

(v1 <...<vp,µl <...<µp)

where (µp+ 1, ... , µn) and (vp+ 1, ... , vn) are the complementary (n - p)-tuples of(µl, ... , l p) and (vl, ... , vp) respectively.

12. Use the formula

<De u, v> = <e*, u n v> u e A "E, v E A n- pE

to derive the classical Laplace expansion formula (see Problem 9).Hint: Given an n x n matrix define the vectors x by

x1 = >Ja4 e i = 1, ... , n.v

Then apply the above formula with

u = x1 n n xp and v = xp +1 n n x.


13. Let E be a Euclidean space and consider the space H( A E; A E) of homogeneouslinear transformations A E -+ A E of degree zero. Then the Poincare isomorphisminduces a linear automorphism DH of H( A E; A E). Given an isometry cp : E -+ E,prove that A p is an eigenvector of DH with eigenvalue + 1.

14. Consider the isomorphism

*:H(AE; AE) + H(AE*; AE*)

defined by Prove that * commutes with DL,

DL (DLcO)* P E H(A E; AE).

Skew and Skew-HermitianTransformations

Throughout this chapter E denotes an n-dimensional vector space over a field of characteristiczero.

The Characteristic Coefficients of a SkewLinear Transformation

8.1. Definition

Let E be an n-dimensional inner product space see Section 1.25 and denotethe inner product by (, ). It determines a linear isomorphism o

Q:E-=>E*

via

<0x, y> = (x, y) x,yEE.

Thus E can be regarded as self-dual. Hence every subspace F E determinesa second subspace Fl (the orthogonal complement of F). Its dimensionis given by

dim Fl = n - dim F.

In particular, if the restriction of the inner product to F is nondegenerate,then F r Fl = 0 and so we have the direct decomposition

E=FQ+F'.

A basis {e1, ..., of E is called orthogonal, if

eµ) = 0 v µ.

Note that (e', ev) 0 (v = 1, ..., n), since the inner product is nondegenerate.

193

194 8 Skew and Skew-Hermitian Transformations

Every inner product space admits an orthogonal basis. In fact, considerthe function

Q(x) = (x, x) x E E .

Since

(x, y) = i (Q(x + y) - Q(x) - Q(y)),

it follows that Q 0. Thus there is a vector e1 such that Q(e 1) 0. LetE 1 denote the subspace spanned by e 1. Then

E=E1Q+Ei

and the restriction of the inner product in E to E i is again nondegenerate.Now assume by induction that Ei admits an orthogonal basis {e2, ...,

Then {e1, e2, ..., is an orthogonal basis of E.Every linear transformation ( of E determines an adjoins transformation

P by the equation

((Px, y) = (x, 5y) x, y e E.

If P = - cp, then cP is called a skew linear transformation. The skew lineartransformations form a subspace of L(E; E) which will be denoted by Sk(E).Its dimension is given by

dim Sk(E) _ (21.l8.2. The Isomorphisms 'E and lI'E

Consider the linear operator D : A 2E -+ Sk(E) given by

E(a n b)x = (a, x)b - (b, x)a x e E.

We show that 'DE is an isomorphism. Since

dim Sk(E) _ (2) =dim n ZE,

it is sufficient to prove that 'E is injective.Choose an orthogonal basis {e1, ..., en} of E and set

E(e n e;) = (i <J).

Then

Bpi,{x) = (e,, x)e; - (e;, x)ei x e E.

The Characteristic Coefficients of a Skew Linear Transformation 195

Thus the relation

0i <j

implies that p1!' = 0 and so J?E is injective.Thus

E: A 2E - Sk(E)

is a linear isomorphism. The inverse isomorphism will be denoted by 'FE.

Remark. In Section 8.3 we shall derive a formula which expresses thecharacteristic coefficients of a skew transformation p in terms of the tensor"E(P) (see Theorem 8.3.1).

Next consider the projections

7LE:E Q E -+ A 2E, E*:E* O E* -+ ^ 2E*

and

7C: LE -+ Sk(E),

where

and

2E(a p b) = a n b, E*(a* O b*) = a* n b*

p) = P q 1P E LE.

We shall show that the isomorphisms 'DE and TE : E* Q E - LE (see Section(6.2) are connected by the following commutative diagrams :

E* Q E 3 LE E* Q E ' LE

W

EQE

nEIrL*

W

A 2E - Sk(E) A 2E* ` - A 2E E

In fact, let a* E E* and b e E. Then

O l)(a* O b) = T la* n b

and so

n E* Q E* n

O l)(a* O b)x = (a- la*, x)b - (b, x) r la*


On the other hand, set TE(a* Q b) = (p. Then

cpx = <a*, x>b = (Q- la*, x)b.

Hence

px = (b, x)Q la*

and so

((p - P)x = (Q- la*, x)b - (b, x)Q- la*.

It follows that

7Lo TE = E° O l)

and so the first diagram commutes. The commutativity of the second diagramfollows from i o E (Q -1 Q i) = 7LE* o (i Q a).

8.3. The Isomorphism i

Let E* be the dual space of E. Then the bilinear function (, ) defined by

Kx* © x, y* © y>> = <x*, y> + <y*, x>

(cf. Section 6.18) determines an inner product in the direct sum E* p E, as iseasily checked. Observe that this inner product does not depend on the innerproduct in E.

Now consider the isomorphism Q : E - E* induced by the inner productin E (see Section 8.1) and define a linear transformation i of E* Q E bysetting

(x* p x) = (x* + ax) +O (_ox*1x* + x).

A simple calculation shows that

2(x* p x) = 2(ox p (-o-1x*))

and so i is a linear automorphism of E* Q E.Moreover, i is selfadjoint with respect to the inner product (, ). In fact,

i(x* +O x), y* +O y» = <x*, y> + <Qx, y> - <y*, Q-1x*> + <y*, x>

= <x*, y> + <y*, x> + (x, y) - (y*, x*)

= (x* +O x, i(y* © y)>>.

Now extend i to an algebra automorphism

i: A (E* Q E) - A (E* Q E).

Then i A is selfadjoint with respect to the induced inner product in A (E* Q E).


The automorphism i A determines via the canonical isomorphism

f:AE*Q A E 4 A(E*QE)(see Section 5.15) an algebra automorphism of n E* 2 n E which will alsobe denoted by i A A straightforward computation shows that

i A (x* p 1) = x* Q 1- 1 Q 7 1x* x* E E*

and

tA(1Qx)= JXQ1+1®x xeE.

These relations yield

i A (x* p x) = (x* n Qx) O 1- 1 O (Q- lx* n x)

+ x* Q x+ ox O o 1x* x* E E*, x e E. (8.2)

Let QE be the linear automorphism of E* Q E given by

QE(x* O x) = Qx 0 Q- ix*.

Then Formula (8.2) yields

X A QE(x* Q x) = (Qx A x*) Q 1- 1 Q (x A cr x*)

+ ox ® Q- 1x* + x* O x. (8.3)

Adding (8.2) and (8.3) we obtain

n [x* 0 x + QE(x* OX x)] = 2(x* OO x + QE(x* OX x)) x* E E*, x e E.

Thus,

X n (U + QE U) = 2(u + QE U) u E E* Q E.

In particular, if QEu = u, then

i A u - 2u. (8.4)

On the other hand, subtracting (8.3) from (8.2) yields

iA [x* O x - QE(x* Q x)] = 2[(x* n ox) O 1 - 1 O (o - tx* n x)]

= 2[7LE*(x* p ox) 0 1- 1 0 E(° 1x* 0 x)],

where

7LE : E Q E + A 2E and 7LE* o E* Q E* -+ A 2E*

are the projections defined in Section 8.2. Thus we have the relation

'L (u - QEu) = 2[7CE*(l O Q)u O 1 - 1 O E E* p E. (8.5)

In particular, if QEu = - u, this reduces to

X n u- 7LE*(l Q Q)u Q 1- 1 Q 7LE(° t QX l)u. (8.6)


Next, consider the linear isomorphism TE: E* Q E -+ L(E; E) (seeSection 6.2). It is easily checked that the adjoint transformation of TE(u)is given by

TE(u) = TE(QE u) u e E* Q E.

Thus TE(u) is skew if and only if u satisfies

QE(u) = -u.

Lemma. Let (P be a skew linear transformation and set

u = TE 1C(P), Z = E((P)

Then

i A u- 2(Q A Z Q 1- 1®z).

PROOF. Since (P is skew, u satisfies

QEU = -U.

Thus Formula (8.6) yields

t, u= 7LE*(l Q Q)u Q 1- 1 Q 7LE(Q- 1 Q l)u.

In view of the second diagram (8.1)

7LE*(l 0 Q)u = 7LE*(l 0 Q) TE 1(4))

= Q n 'E 7L((P)

Since (P is skew, 7r((p) = 2(p, and so we obtain

E*(l Q Q)u = 20 `I'E((P) = 20 z. (8.7)

On the other hand, the first diagram (8.1) yields

7LE(Q- 1 0 l)u = 7LE(Q- 1 0 l)TE 1((p) = PE 1ir((P) = 2'PE 1((p) = 2Z. (8.8)

Relations (8.7), and (8.8) yield

i u= 2(QA z Qx 1- 1 ®z)

and so the lemma is proved.

Theorem 8.3.1. The characteristic coefficients of a skew linear transformationare given by

C2 p(4)) = ((' E (P)p, ('PE 4)))

and

C2p+1C4)=0 p>_0.

PROOF. Set (p = TE(u), u e E* Q E. Then, according to (7.10),

Ck((P) = <Uk, k> = <uk, k> (k = 0, ... , n).


Next, observe that (see Section 6.17)uk = (- 1)k(k - 1)/2uk

and

k = (- 1)k(k- 1)/2tk

(see Formula (6.33) of Section 6.17). Using Formula (6.34) in Section 6.18we obtain

Ck(p) = <uk, tk> = ( 1)k<<uk, tk>>. (8.9)

Now consider the automorphism i. Since p is skew, the lemma showsthat

TA u= 2(z* Q 1- 1 Q z), (8.10)

where

z = `PE cp and

On the other hand, since QED _ , Formula (8.4) yields

= 2&

Thus, since TA preserves products in the algebra n E* Q A E,

TA() = 2kk.

Inserting this into (8.9) and observing that TA is selfadjoint with respect tothe inner product (, >> we obtain

Ck(co) = ( 1)k2 -k(uk i n (k)»

( 1)k2 - k«i (uk)k>>

(- 1)k2 - k«(i u)k,k». (8.11)

Raising (8.10) to the kth power (in the algebra n E* Q A E) and usingthe binomial formula yields

n u)k = 2k(z* QX 1 - 1 QX z)k = 2k (- 1)'zQ (8.12)i+j=k

From (8.11) and (8.12) we obtain

Ck(co) = ( 1)k (1)J

\\z*i O z,, k>>. (8.13)i+j=k

Now observe that

E n 2jE* O n 2jE

while

k E n kE* QX n kE.


Thus Formula (8.13) shows that

Ck((o) = 0 if k is odd

(see the definition of the inner product (,) in Section 6.18).

On the other hand, if k = 2p, then, (8.13) yields

C2 ((p) _ (_ 1)P«z#n OXP

Since

P=(_1)Pc2P-1)2P=(_1)P2P

(see (6.33), Section 6.18) it follows that

C2p((P) = <<z#P ® zP, 4'>>

and so, in view of (6.33)

C2p((p) _ <z#P ® z, 2P>.

Finally, applying (6.17), we obtain

C2 p((p) _ <z#P, zP> = <(C A z)", z"> = (z", z") = (('11E (P)PS ('PE (P)P)

This completes the proof of the theorem.

The Pfaffian of a Skew Linear Transformation

8.4. The Pfaffian

Let E be an inner product space of dimension n = 2m and let p,, , p,be skew linear transformations. Every cp determines an element

E A ZE.

Thus,

"E((P 1) A ... A "E((Pm) E A "E.

Now fix a basis vector a of A "E and set

Pfa((P 1, ... , (Pm) _ (`I E((P 1) n ... A E((Pm), a)

Equivalently,

E(o1) n...n E(cm) = a1aPfa(o1> ..., co) a.m

The scalar Pfa((P 1, ... , Corn) is called the Pfaffian of (P1, ..., co. Since everytwo elements q'E(rpµ) commute, it follows that Pfa is a symmetric m-linearfunction in Sk(E).

The Pfaffian of a Skew Linear Transformation 201

The Pfaffian of a single skew transformation (p is defined by

f= 1 Pf... .a((P) m a ((P, > (P)


m

Pfa((p) = (Ì'E((P)m, a)

Clearly,

mPfa((p) E V

Proposition 8.4.1. If i is an isometr y of E, then

Pfa(t ° (p j ° 'L - 1, ... , 'L ° (p m ° - 1) = det -r Pfa((P 1 , ... , (pm)

PROOF. Since i preserves the inner product we have

E(ix n x y) = i° E(x n y) o f -1 x, y e E

whence

i n `1'E(p) = ° 'L -1) (P E Sk(E).

It follows that

E(t° (1 ° - 1) n ... n LIJE( ° (m ° - 1)

= 'L A E((p 1) ^ ... ^ 'L A E((Pm) = n (Ì'E((p 1) ^ ... A qJ ((pm))

whence

Pfa(t°p1or 1,...,'L°(p m°'L-1)

(t n (Ì'E((p1) ^ ... A E(Pm)), a) = (1 E((p1) ^ ... A E((pm), x n 1 a)

But since i is an isometry

i l a= det i- 1 a= det i a

and so we obtain

Pfa(t ° (p1 ° 'L - 1, ... , 'L ° (p m ° 'L - 1) = det i(LI'E((p 1) ^ ... A "E((pm), a)

= det Pfa((p 1, ... , (pm)

Proposition 8.4.2. Let a and b be basis vectors of A E. Then

Pfa((P) Pfh((p) = (a, b)det (p.

In particular,

Pfa((p)2 = (a, a)det (p.


PROOF. In fact, since

' m=1 Pf((P)a= 1 PfbbE P) (a, a) a (b, b)( ) (, )

it follows that

Pfa((p)Pfb((p)1 Pf Pf

((q'E P)"`,( 'E P)"`) =a s b b (a, b) = a b

b = Aa, t e r). Now applying Theorem 8.3.1 with p = m and observingthat det cp, we obtain

det =1 PfP) Pf .a((a, b)(, )

8.5. Direct Sums

Let E and F be inner product spaces. Then an inner product is induced inthe direct sum E p F by

(x1 +O Yi, x2 +O Y2) = (x1, x2) + (Y1, Y2) x1, x2 E E, Yi, Y2 E F.

Let

iE:E-E +QF, iF:FEE E)Fand

denote the natural inclusions and projections. Then the isomorphismst'E, and t'E®F (see Section 8.2) are connected by the relations

E®F(1E) n (x n Y) = 'E° E(x A Y) ° E x e E, y e E

and

E®F(1F) n (x A Y) = 1F ° DF(x A Y) ° F

as is easily checked. The relations above imply that

xeF, yeF,

(iE) n "E(SP) = '11E®F('E ° ° E) P E Sk(E)

and

(F) A '11F(k) = '11E©F('F ° Y ° F) 'I' E Sk(F).

Since

(P E 3i// = lE0(P07LE + lF0l//07LF,

we obtain

(E) A "E(SP) + (iF) n F(Y') = '11E©F('P 4') E Sk(E), c e Sk(F). (8.14)

The Pfaflian of a Skew Linear Transformation 203

Proposition 8.5.1. Let E and F be inner product spaces of dimensions 2p and 2qrespectively. Choose basis vectors a and b of A 2E and A 2"F. Then

PfQ®b((P Q i) = PfQ(q,) Pfh((p) cP e Sk(E), ,,Ii e Sk(F).

PROOF. Set "E('P) = u, LI'F(l//) = v and `I'E©F((P O+ i//) = w, where

w = (lE) n u + (1F) n v.

It follows that

wp + q = ((lE) n u)k A ((1F) nv)1 = uk 0 v'.

k+1= p+q k+1= p+q

Since u k = 0 for k > p and v 1 = 0 for 1> q, this formula reduces to

Wp+q = up OX Uq.

It follows that (see Section 5.15)

PfQ®b((P Q i//) = (wp+q, a O b)

= (up Q uq, a Q b) = (up, a) . (vq, b)

= Pf p)

8.6. Euclidean Spaces

Let E be an oriented Euclidean space of dimension n = 2m and let a be theunique unit vector in A "E which represents the orientation. Then we set

PfE(co1.... , om) = ("E('P 1) n ... A E((Pm), e) pv e Sk(E)

and

PfE = 1 P e Sk(E).(SP) m

Observe that if the orientation of E is reversed then PfE is changed to -PfE .

Proposition 8.4.1 shows that the Pfaffian is invariant under properrotations. Proposition 8.4.2 implies that

PfE((P)2 = det cP cP e Sk(E).

Finally, by Proposition 8.5.1,

PfE©F( +O i/i) = PfE((p) PfE('') cP, cli e Sk(E).

EXAMPLE 1. Choose an orthonormal basis {e1, ..., of E and define cP by

coe2µ _ 1 =)e2 µ )µ E r

cpe2µ = - i%µ e2µ - 1 (µ = 1, ..., m).


Then, when n = 2, ' E(cp) = ,1(el n e2), and so

PfE((p) =

as follows from Proposition 8.5.1.

EXAMPLE 2. Let E be an n-dimensional complex vector space with a Hermitianinner product (, )H and let E denote the underlying real vector space. GiveE the natural orientation (see Section 7.9) and the positive definite innerproduct

(x, y) = Re(x, y)H x, y e E.

Let i be the skew transformation of E given by

Then

1

as follows from Example 1.

Skew-Hermitian Transformations

In Section 8.7 we shall derive an analogue of Theorem 8.3.1 for skew-Hermitian maps.

8.7. The Isomorphisms 6 and i

Let E be an n-dimensional complex vector space and let E* be the complexdual space. Define an inner product in E* Q E by setting

(x* Q x, y* O+ y) = <x*, y> + <y*, x>. (8.15)

Extend this inner product to the (complex) exterior algebra A (E* E).

This inner product determines via the canonical isomorphism

A (E* Q E) AE* Q AE

(see Section 5.15) an inner product in A E* Q A E. A simple calculationshows that (see Formula (6.34))

<<u* OO u, v* OO v>> = (-1)p<u*, v> <v*, u> u*, v* e A PE*, u, v e ARE.

Next, assume that E is equipped with a Hermitian inner product (, )H .Then an inner product is defined in the underlying real vector space E by

(x, y) = Re(x, y)H x, y e E.

Skew-Hermitian Transformations 205

The multiplication by i in E determines a skew linear transformation inE which will be denoted by i (see Example 2, Section 8.6)

i(x) = i x.

Recall from Section 8.2 the inverse isomorphisms

IE : n 2E - SkE and L: n 2E

and set

J =E(lpg)

Then J E n 2 E.Next define a (real) linear isomorphism Q : E E* by

Qx, y> = (x, y) - i(x, i Y) X E E, y e E.

Since i is skew and an isometry it follows that

QiR(x) = - iQ(x) x e E.

The map o- determines a (real) linear map i : ER E* Q E by

i(x) = kr(x) Q x x E E.

Formula (8.15) implies that

(ix, iy) = <iQx, y> + <icJY, x>= i(x, y) + (x, i y) + i(y, x) + (y, ix)= 2i(x, y) + (x, i y) + (y, ix) = 2i(x, y).

Thus i satisfies

(8.16)

«ix, iy) = 2i(x, y) x, y e E. (8.17)

Next observe that ix n ix = 0, x E E, where the symbol n c indicatesthat the multiplication is taken in the complex algebra n (E* Q E), and so iextends to a homomorphism (see Section 5.5)

i A: n E + n (E* Q E).

Relation (8.17) implies that

<<X A u, i u) = (2i)"(u, u) u, y e pE . (8.18)

Lemma 1. Consider the direct decomposition

n 2(E* +E E) = [n 2E* O 1] +O [E* O E] +E [1 O n 2E]

and write

iAU=iLUQ1+XMU+10XRu uEn2E,


where iL u E A 2E*, iM u E E* O E, i R u E A 2E. Then the following relationshold:

XL "E[l08 4] _ - 2it L '11E(4) (p e Sk(E)

XR 1-E[l , (p] = 2lxR '11E(4) (p E

Here [, ] denotes the commutator and q'E : Sk(E) - A 2E.

PROOF. It follows from the definition of i that

XL(x n y) = krx n c iO y = - Qx A Qy.

This equation yields, in view of (8.16),

t L(l 08 x n y+ x n i y) _ -o i (x) n c 0Y - Qx n c Qi y= iQx n Qy + Qx n c 10y= 2krx n Qy = - 2ixL(x n y). (8.19)

Next observe that the isomorphism t'E satisfies

[(p E(x n y)] _ IE((px n y + x n (py) (p E

Applying this relation with (p = i we obtain

[ l 08 , E(x A y)] = E(' 08 x A ) + x A l y)

whence

'E[i08 E(x n y)] = i y.

Formulas (8.19) and (8.20) imply that

XL '1'E[, E(x A y)] _ - 2ixL(x A y)

whence

(8.20)

XL'E [lpg, "E u] _ - 2lxL(u) u e

Since 'DE: A 2E - Sk(E) is an isomorphism, the first formula in the lemmafollows. The second formula is proved in the same way.

Corollary. If cp is a skew-Hermitian transformation of E, then

iLY'E(4') = 0 (8.21)

and

XR '11E(4') = 0. (8.22)

PROOF. Apply the lemma observing that (p] = 0.

Lemma II. Let (p be a skew linear transformation of E and consider the complexlinear transformation i ° (p + (p ° i . Then

XM `I'E(4) = T E 1(i ° (p + (p ° i08)

Skew-Hermitian Transformations 207

PROOF. It follows from the definition of i that

XM(x A y) = icrx ®c y - icry ® x.

This yields, in view of (8.16),

TE XM(x n y)(z) _ - <icry, z>x + <icrx, z>y = - i<Qy, z>x + i<Qx, z>y_ -i{(y, z) - i(y, ijz)}x + i{(x, z) - i(x,

i{(x,z)y - (y, z) x} + {(x,iE(x n y)z + DE(x n y)i z.

Thus we have the relation

TE XM(x n y) = i FE(x n y) + E(x n y) ° i

whence

TE XM(u) _X08 cDE(u) + E(U) o lpg u e n 2Epg .

Setting DE(u) _ cP we obtain

TE iM'1' p) = i$8 ° cP + cP o i cP Sk(ER)

and so the lemma is proved.

Corollary. A skew-Hermitian transformation (p of E satisfies the relation

TE XM `PE((p) = 2up. (8.23)

Lemma III. A skew-Hermitian transformation of E satisfies

A''E cP = 2l TE 1(co)

In particular,

X A J = -2k.

PROOF. Apply formulas (8.21)-(8.23) observing that

xAu=xLu®l+xMu+1®iRu

Theorem 8.7.1. The characteristic coefficients of a skew-H ermitian transforma-tion are given by

Cp(p) = i"(CI E p)p, JP) (p = 0, ..., n).

Thus, if p is even, then Cp(cp) is real and if p is odd, then C,((p) is imaginary.

PROOF. Set T E' (gyp) = u. Then we have, in view of formulas (7.10), (6.33),and (6.34)

<U", i"> = <ub, fb> = ( 1)p«up, »


Now set LI'E (P = z. Then, by Lemma III,

X A Z = 2iu

and

X A J = -2-t.

It follows that

1 1

Cp((P) = P2 (2i) n Zp, 'L A JpNow using relation (8.18) and (6.33), we obtain

C(p) =1p 1

p(2l)2p(zp, JP) = i (zp, JP) = i"((tPE (P)p, JP).

2(2,)

Corollary. Let E have the natural orientation (see Section 7.9). Then thePfaffian of p is given by

Pfq = 1 det .

PROOF. Applying Theorem 8.7.1 with p = n we obtain

det cP = i"(J", (`I'E P)")

Now write

J" _ , e,

where a is the positive unit vector in A 2nEpg . Since J = it followsthat

= 1

(see Example 2, Section 8.6). Thus,

det cP = l"(e, (LE q)") = j"Pf((p)

Symmetric Tensor Algebra

In this chapter E denotes a vector space over a field of characteristic zero.


The results of this chapter are, in most cases, isomorphic to analogousresults in Chapter 5. Consequently, most proofs are omitted or presentedin a highly abbreviated form. Modulo occasional changes in terminologythe reader should be able to read the proofs in Chapter 5 as substitutes forthe proofs omitted here.

9.1. Symmetric Mappings

Let E and F be vector spaces and let

P

be a p-linear mapping. Then cP is called symmetric if cP = Qcp for every permu-tation o (see Section 5.1). Since every permutation is a product of trans-positions, it follows, that a mapping p is symmetric if and only if p = icpfor every transposition i. Every p-linear mapping p determines a symmetricp-linear mapping Scp by

Scp is called the symmetric part of cP and S is called the symmetry operator.If p is symmetric, we have Scp = cp.

209

210 9 Symmetric Tensor Algebra

Proposition 9.1.1. Let ( be a p-linear mapping of E x x E into F andf : ®"E -+ F be the induced linear map. Then (p is symmetric if and only ifM(E) c ker f (cf. Section 4.9).

PROOF. If (p is symmetric, we have, for each transposition i,

f(x1 ...O x, - i-1(x1 ®... O x,))

_ (p(x 1, ..., xp) - i(p(x 1, ... , xp) = 0.

Since the products x1 Q ® x, generate ®-"E, it follows that f reducesto zero in M"(E). Conversely, if M"(E) c ker f, it follows that

((p - x p) (x 1, ..., xp) = f[(x1 ®... O x p) - (r - 1(x1 ©... O x,)] = 0

for every transposition i and hence (p is symmetric.


Let

P

be a symmetric p-linear mapping from E into a vector space S. We shall saythat V p has the universal property (with respect to symmetric maps) if itsatisfies the following conditions:

v 1: The vectors V p(x 1, ... , xp), x a E, generate S.v2: If

P

is any symmetric p-linear mapping, then there exists a linear mapf:S - H such that the diagram

H

V 1'

S

commutes.

Conditions v 1 and v 2 are equivalent to the following condition

v : If

cp:E x x E-->HP

is a symmetric p-linear mapping, then there is a unique linear mapf:S H which makes the diagram above commute.


Suppose now that

W:E x x E-+S and Vp:E x x E-+Sp p

211

are symmetric maps with the universal property. Then there is a linearisomorphism f : S S such that

fo Vp= Vp.

This is shown in exactly the same way as in the skew-symmetric case (cf.Section 5.3).

To establish existence, set

V pE = ®E/M(E)

(where M"(E), the subspace of ®"E, is defined as in Section 4.9) and defineVpby

V "(x1, ... , xp) = ?L(x 1 ®... ® xp),

where ?r denotes the projection.

Definition. The pth symmetric power of E is a pair (S, V p), where

p

is a symmetric p-linear mapping with the universal property. The space S isalso called the pth symmetric power of E and is denoted by V "E.

9.3. Symmetric Algebra


VE= Q VpEp=0

(where V °E = f and V 'E = E) and identify each V pE with its image underthe canonical injection ip : V pE -+ V E. We thus obtain

V E = V pE.p=0

Assigning to the elements of V pE the degree p, we make V E into a gradedvector space.

As in Section 5.4 on exterior algebra we construct the homogeneouslinear isomorphism

f:®E/M(E)-+ V E


such that

fit(x1 0 ... O xp) = x1 v ... V xp

and use it to induce a multiplication in V E. Then V E becomes an associativecommutative graded algebra with the scalar 1 as unit element. It followsfrom the definition of the product that

(x1 V ... V xp)(xp+1 V ... V xp+9) = x1 V ... V xp+9.

Hence we shall denote the product of two elements u and v by u v v. Thenthe above formula reads

(x1 V ... V xp) V (xp+1 V ... V xp+9) = x1 V ... V xp+9. (9.1)

The graded algebra V E is called the symmetric algebra over E. It is clear thatthe vectors of E together with the scalar 1 form a system of generators forthe algebra V E.

As in Section 5.4 on exterior algebra, we define the kth power of an elementue VE by

1uk= u V V u k> 1

k.k

u° = 1.

Then we have the binomial formula

(u + v)k = u` V vj u, v e V E. (9.3)i+j=k

The algebra V E has the following universal symmetric algebra property:Let A be an associative algebra with unit element a and cP : E - A be alinear map such that

(px (p y = (p y px.

Then there exists a unique homomorphism h : V E -+ A such that h(1) = eand h o i = cp, where i denotes the injection E -+ V E.

Moreover, if U is any associative algebra with unit element and E : E -+ Uis a linear map such that the pair (U, E) satisfies the universal propertyabove, then U is the symmetric algebra over E.

9.4. Symmetric Algebras Over Dual Spaces

Let V E, V E* be symmetric algebras over a pair of dual spaces E, E*,and consider the induced isomorphisms

f : QE/M(E) - V E, g: QE*/M(E*) -= V E*.

Symmetric Tensor Algebra 213

It follows from Section 4.16 that f and g induce a scalar product <, > in V Eand V E* such that

/x*1 v ... v x*P, x1 v ... v xp> = perm(<x*i, xj>)

= ,1µ<V PE*, V IE> = 0 if p q.

p>_1

From v1 we obtain that <, > is uniquely determined by (9.4). It followsfrom (9.4) that the restriction of <, > to the pair V PE*, V pE is nondegeneratefor each p, and so induces duality between these spaces. In particular, therestriction of <, > to V 1E* = E* and V 'E = E is just the original scalarproduct.

9.5. Homomorphisms and Derivations

Suppose that cP : E -+ F is a linear map. Then cP can be extended in a uniqueway to a homomorphism cP : V E -+ V F such that cP (1) = 1. The homo-morphism cP is given by

q (x1 v...vcpxp x,EE

and is homogeneous of degree zero. If l/i : F -+ G is a linear map into a thirdvector space G, then we have

and the identity map of E induces the identity in V E,

'V = 1.

It follows from (9.5) and (9.6) that if cP is injective (surjective) then cP is alsoinjective (surjective). The fact that p preserves products can be expressedby the relation (see Section 9.6 for µ(a))

P V ° µ(a) = a) ° cP , a e V E. (9.7)

Suppose now that p* : E* -- F* is a linear map dual to cp. Then theinduced homomorphism (gyp*) : V E* * V F* is dual to cP

('P*) _ ('P v)*.

The homomorphism (gyp*) will be denoted by (p V .

If lfi* : F* - G* is a linear map dual to cli we have the composition formula

(f,°q,)v=(Pv°JV.


Now let p be a linear map of E into itself. Then p extends in a uniqueway to a derivation B (cp) in the algebra V E. The derivation 8 (') isgiven by

P

e v (x 1 V ... V xP) = x 1 V ... V (px j V ... V XPj=1

and is homogeneous of degree zero. The derivation property of O ((p)can be expressed by the formula

8 V µ(a) = µ(e (4)a) + u(a) ° 8 (q) a e V E. (9.8)

If tji : E - E is a second linear map, we have the composition formula

O , 'I' - /i °i') = 0 (p)0O OI') - e AcIi) ° e

If (p* : E* * E* is dual to cp, then the induced derivation 0 (p*) of thealgebra V E* will be denoted by 8 " (gyp). The linear maps 8 (gyp) and 0 V ((p)

are dual,

ev(e) =

If /j* : E* - E* is dual to i//, we have the composition formula

(9 V (p 0-,/J 4') = e V(/,) 0 O"() - e v (`I,) o O v (cI,)

9.6. The Operator i(a)

Fix a E V E and consider the linear map µ(a): V E -+ V E given by

µ(a)u = a v u u E V E.

Clearly,

µ(a v b) = µ(a) o µ(b) a, b E y E. (9.9)

Now let

i(a) : V E* 4 V E*

be the dual map. It is determined by the equation

<i(a)u*, v> = <u*, a A U> u* E V E*, u e V E.

If a E V PE, then i(a) is homogeneous of degree p. In particular,

i(a)u* = <u*, a> u* E V E

and

i(a)u * = 0 u * E V rE, r <p.


Dualizing Formulas (9.9) and (9.7), we obtain

i(a v b) = i(b) o i(u) a, b e V E,

and

i(a)°cp" _ cP" oi((p,a) a e V E, cpe L(E; F).

Finally, if (p : E - E, (p* : E* - E* is a pair of dual linear transformations,then, dualizing (9.8), we have

i(a) o 0 V (gyp) = i(V V((p)a) + 0 V (p) ° i(a) a e V E.

Now consider the operator i(h), where h e E. Observe that i(h) is homo-geneous of degree -1. As in Proporition 5.14.1 one may show that i(h) is aderivation in the algebra V E*,

i(h)(u* v v*) = i(h)u* v u* + u* v i(h)u*.

This yields the Leibniz formula

i(h)r(u* v u*) _(r)

i(h)Pu* v i(h) v*.p+q=r p

Finally, observe that if an element u* E V PE (p - 1) satisfies the equationi(h)u* = 0, for every h e E, then u* = 0 (cf. Proposition 5.14.2).

9.7. Zero Divisors

In this section it will be shown that the algebra V E* has no zero divisors(of course, the same holds for V E).

Consider, for each p > 0, the subspace IP c V PE* given by

IP = V µE*.Ii-P

Clearly, Ip is an ideal in the algebra V E* and we have the sequence

VE*=JoD11D'2D....The ideal I 1 is also denoted by V +E* . Every two ideals

J, = V µE* and F' = V µEL>p µ>_P

are dual. If a e V PE, then i(a) restricts to an operator from Iq to Iq _ P forq<p.

Lemma. Let u* a IP be an element such that

i(h)Pu* = 0 for every h e E.

Then u* = 0.


PROOF. If p = 1, the lemma follows from the remark at the end of Section9.6. Now assume that the lemma is true for p - 1 (p > 2) and let u* E IP,be an element satisfying

i(h)Pu* = 0 h E E.

Replacing h by h + k (. E I,) we obtainP p i(h)i(k)P - u * = 0.

a=o µSince . is arbitrary, this yields

i(h)µi(k)P - µu* = 0 (µ = 0, ... , p).

In particular,

Thus, by induction,

i(h)P-1 i(k)u* = 0 h, k E E.

i(k)u* = 0 k E E.

Now applying the lemma for p = 1 we obtain u* = 0 and so the inductionis closed.

Proposition 9.7.1. The algebra V E* has no zero divisors.

PROOF. Let u* and U* be two nonzero elements in V E*. Assume first thatu* and U* are homogeneous of degree p and q respectively and that p > 1and q > 1. In view of the lemma, there exists h e E such that i(h)Pu* 0.

Now consider the elements i(h)Pu* (µ = 0, 1, ...). Since i(h)°U = U 0and i(h)4+ 1 U* = 0, there is an integer r > U such that

i(h)rv* 0 while i(h)r + 1 U* = 0.

Now the Leibniz formula yields

ihP+ru* V U* = P+r ihPu* V ihrv*.P

Since i(h)Pu* is a nonzero scalar and i(h)rv* 0, it follows thati(h)P+r(U* V U*) 0

whence u* V U* 0.In the general case

and

P

u* _ u u E V aE*, up 0a.=o

9

U* _ Un UK E VK E*, U9 0.K=0


Then

u* v U* _ ua*. v UK + uP v U9.

217

Since up v u9 0, it follows that u* v u* 0. This completes the proofof the proposition.

9.8. Symmetric Algebra Over a Direct Sum

Consider two vector spaces E and F and the direct sum E Q F. In this sectionwe shall establish an isomorphism between V (E Q F) and the canonicaltensor product V E O V F. Define a linear map

f:VE® VF-+ V(EQF)

by

f(u OO u) _ (i1) u v (i2) u,

where it and i2 are the inclusions. A straightforward calculation shows thatf is an algebra homomorphism (cf. Section 5.15).

To show that f is an isomorphism, consider the linear map

ri:EQF-+ VEQ VF

given by

ri(z) = 7riz Q 1+ 1 O 2 z z e E Q F,

where 7r1 and r2 denote the projections. Since

1(z1) . rI(z2) _ (Z2) ' j(z) z1, z2 E E Q F,

ri extends to an algebra homomorphism

ri:V(EQF)-+ VEQ VF

(see Section 9.3).It is easy to check that

j f (x O 1) = x 0 1, ?1.f (1 O y) = 1 0 y

and

flj(x®y)=x®y x e E, y e F.


and fotj=i.Thus f is an isomorphism.

Next, let E* and F* be spaces dual to E and F respectively. Define ascalar product between E* Q F* and E Q F in the usual way and consider


the induced scalar product between V (E* p F*) and V (E p F). Then wehave the relation (cf. Formula (5.51))

<f (u* ® v*), f(u ® v)> = <u* ® v*, u ® v> = <u*, uXv*, v>.

Finally, assume that F = E and let A: E -+ E p E be the diagonal map.Then we have the relation (cf. Formula (5.52))

0" f(u* ® v*) = u* v v* u*, v* e V E*.

9.9. Symmetric Tensor Algebras Over a Graded Space

Let E = . i E1 be a graded vector space and let the vectors of E, behomogeneous of degree k,. Then there exists precisely one gradation in thealgebra V E such that the injection i : E - V E is homogeneous of degreezero. V E together with this gradation is called the graded symmetric algebraover the graded vector space E. The subspace of homogeneous elements ofdegree k is given by

(V E)k = (V p' E 1) ®... ® (V E,),(P)

where the sum is extended over all r-tuples (pi, ... , pr) subject tor

p1k1 = k.

9.10. Symmetric Algebra Over a Vector Space of FiniteDimension

Suppose now that E is a vector space of dimension n and that ea (a = 1, ..., n)is a basis of E. Then the products

ea 1 v ... v ea a 1 < a2 < ... < ap (9.10)P

form a basis of V "E. In fact, it follows immediately from v 1 and the com-mutativity of V E that the products (9.10) generate V E. To prove linearindependence let E* be a dual space of E and a*(a = 1, ..., n) be a dualbasis. Then Formula (9.4) yields

<e* v ... v e*RP, ea1 v ... v eaP> = perm(<e*f;, e21>) = perm(5 )

and thus the products (9.10) are linearly independent.The above result shows in particular that

dimVPE=(n+p-11 p>0.(9.11)

n 1


The basis vectors of V PE can be written in the formn n

JJ k! e1 V ... V en" kv>=p.

i= 1 v= 1

9.11. Poincare Series

For the Poincare series of the graded algebra V E we obtain, from (9.11)

(t) t Y.( 1) tP=o P p=o p (1

whence

P(t) = (1 - t) - n.

(Here E has dimension n.)Now suppose that E = Ei is a positively graded vector space of finite

dimension and that the vectors of E. are homogeneous of degree ki . Thenthe Poincare series of V Ei is given by

Pi(t) = (1 - t"9 - ni ni = dim E1.

Hence, the Poincare series of E is

P(t) = (1 - tkl)-nl .. . (1 - tkr) - nr.

9.12. Homogeneous Functions

A homogeneous function of degree p in E is a map h : E -- F which satisfies

h(i%x) = il!h(x) ,% E f.

The homogeneous functions of degree p form a vector space HP(E). Theproduct of two homogeneous functions h and k of degree p and q respectivelyis the homogeneous function h k of degree p + q given by

(h k) (x) = h(x)k(x) x e E.

This multiplication makes the direct sum

H(E) _ Hp(E)P

into a commutative associative algebra. Its unit element is the homogeneousfunction ho of degree zero given by

h0(x)=1 x E E.


Now let D be a symmetric p-linear function in E. Then a homogeneousfunction h, of degree p is given by

he(x) = ... , x).

In terms of the substitution operator we can write

h,(x) =

It follows from the definition (see Section 9.15 for S"(E)) that

= 1,h + µh,, D, 'P E S"(E), , µ E r

and that

h , = h h, 1D, 'I' E S"(E).

Thus, the correspondence D - h determines an algebra homomorphism

i : S(E) -+ H(E).

The map i is injective. In fact, assume that iD = 0 for some D E S"(E). Then

i(x)M = 0

for every x e E and so the lemma in Section 9.7 implies that D = 0.On the other hand, i is not in general surjective. As an example let E be a

Euclidean space and define h e H 1(E) by

h(x) = (x) (x, x)112,

where E is a function in E satisfying

E(x) . > 0,

Then h is homogeneous of degree 1 but it is not additive and hence not alinear function. Thus it is not contained in Im i.

PROBLEMS

1. Consider the problems of Chapter 5, and carry them over to symmetric algebrawhenever possible.

2. Let F c E be a subspace, and define IF to be the ideal in V E generated by the vectorsof F. If F1 is a complementary subspace, prove that

VE=IFp VF1

and

IF =V F® V F1.

3. If p : E - F is a linear map, prove that

Im ep = V Im p and ker (p = Iker p.

Polynomial Algebras

4. If

P

221

is a simultaneously symmetric and skew- symmetric p-linear mapping (p >_ 2),prove that p = 0.

Polynomial Algebras

9.13. Polynomial Algebras

A monomial of degree p in n variables in a field f is a function

P

which satisfies

P(t 1, ... , tn) = P(1, ... , 1)til ... tnn

where k 1 + + kn = p.Thus every monomial of degree p can be written in the form

pP(t1,... , tn) = atil ... rn" k= p.

In particular, a monomial of degree zero is an element of r.The monomial of degree 1 given by

Pi(t 1, ... , tn) = ti (i= 1, ... , n)

will be denoted by t.The monomials of degree p generate a vector space with respect to the

usual operators. It will be denoted by rn.

The product of a monomial P of degree p and a monomial Q of degree qis the monomial of degree p + q defined by

(P ' Q) (t 1, ... , tn) = P(t 1, ... , tn)Q(t 1, ... , tn).

This multiplication makes the direct sum

Prn - rnP

into a commutative associative algebra called the polynomial algebra in nvariables over r with the scalar 1 as unit element. It is generated by 1 and themonomials ti (i = 1, ..., n). The elements of rn are called homogeneouspolynomials of degree p.

222

9.14.

9 Symmetric Tensor Algebra

Now we shall establish an isomorphism between V E (dim E = n) and thepolynomial algebra in n variables. In fact, fix a basis {e1, ..., of E. Thenevery element u e V "E can be uniquely written in the form

= 1 ...VPu = en" vi p, c e f(V) ii

(cf. Section 9.10). Thus it determines a homogeneous polynomial Pu ofdegree p given by

Pu(t 1, ..., tn) - t:" Vi = p

Since,

= I,Pu + 4uP, A, p e f u, v E V "E,

this defines a linear map

q:V"E-urn.

Conversely, every polynomial

P(t1, ..., tIt) = cvt,(v)

homogeneous of degree p, determines an element Up e V "E given by

and so we have a linear map

/i : f - V E.


l/i o p = i, (p01/1=1

and so q and i/i are inverse isomorphisms. Finally, observe that

u V V = Pu Pv U, v e V E

and thus q is an algebra isomorphism,

(p : V E F~n .

In particular, we have

(p(e)=t i (1= 1, ... , n).

The Algebra of Symmetric Functions 223

It follows that the polynomial- algebra has no zero divisors since V E hasno zero divisors (see Proposition 9.7.1).

PROBLEMS

1. Prove that V E is a principal ideal domain if and only if dim E = 1. Hint: See ChapterXII of Linear Algebra.

2. Let E be a pseudo-Euclidean space of dimension n (see Section 9.17 of Linear Algebrafor the definition) and index r. Consider the symmetric tensor algebra V E and choosefor each p >_ I a subspace T" V "E of maximal dimension such that the restrictionof the scalar product to T" is negative definite. Prove that the Poincare polynomialof the graded space T = T" is given byiii 1

PT(t)=2(lt)[(lt)s_(l+t)s] s=n-r.

Prove an analogous formula for the exterior algebra :

PT(t) = i (I - tf [(I + t)5 - (1 - to s = n - r.

The Algebra of Symmetric Functions

9.15. Symmetric Functions

A p-linear function I in E is called symmetric, if

Q E Sp.

The symmetric functions form a subspace of TP(E) (cf. Section 3.18) denotedby S(E).

Every p-linear function fi determines a symmetric function Sit' by

P acalled the symmetric part of I.

If t e S(E), then ScI = ct and so S is a projection operator. Moreover, Ssatisfies (cf. Formula (5.81))

S(I a '1') = S(S o '1') = S(I o S'V) , P E T(E)

whence

S(cb Q '1') = S(S Q S'1').

The symmetric product of e S' (E) and P E 5(E) is defined by

Vq'=(p+q)!SI xlP.pq!

224

Explicitly,

9 Symmetric Tensor Algebra

1( v ) (x1, ..., xp + q) _ (1)' ... , 1) ' ... , q)).

The symmetric product satisfies the relations

and

(bv'1')v X= bv('l'vX).This multiplication makes the direct sum

S(E) = >2SP(E)p=0

into a commutative associative algebra, called the algebra of symmetricfunctions in E.

A linear map gyp: E - F induces a homomorphism p* : S(E) - S(F)given by

((p* qI) (x 1, ..., xp) = '4'((px 1, ... , (px p) 'I' E SP(F)

and a linear transformation q of E determines a derivation 0S(p) in thealgebra S(E). It is defined by

P

(())(x1, ... , xp) = (x1, ..., rpx,, , ... , xp).v=1

This is shown in the same way as for the algebra A(E) in Section 5.31.

9.16. The Operator is(h)

Let is(h) : S'(E) - S'(E) also denote the restriction of the operator is(h) definedin Section 3.19 to S'(E). Thus,

(is(h)(b) (x 1, ... , x,_) = b(h, x 1, ... , x,_) b e S(E).

is(h) is called the substitution operator in the algebra S'(E).In exactly the same way as in Section 5.32 it is shown that

is(h) o S = 50 is(h)

(see the lemma in Section 5.32) and consequently,

is(h) (st' v 'F) = is(h)1 v 'F + 'b v is(h)'F

(see Proposition 5.32.1). Thus, is(h) is a derivation in the algebra S(E).Note that the operator is(h) is dual to the multiplication operator ,us(h)

in S(E).

The Algebra of Symmetric Functions

Finally, we have the commutative diagram (see Section 3.20)

OpE* ) T"(E)

ns

1°

O pE * ) T "(E)

225

where its denotes the symmetrizer (see Section 4.15). It implies that a restrictsto a linear isomorphism V PE* = S"(E) for each p and in fact to an algebraisomorphism Y(E*) 3 S(E).

Since V PE* Y(E) (see Section 4.15), it follows that V PE* S"(E).In particular,

dimSpE = n+p- 1 (p=0,1,...).p

9.17. The Algebra S,(E)

In exactly the same way we obtain the algebra

S,(E) = S(E),p=0

where S(E) denotes the space of symmetric p-linear functions in E*. Thescalar product between T"(E) and Tp(E) determines a scalar product betweenS"(E) and S(E) given by

tb, W = 1 tb'b ESpE q' eS E.p

It follows from the definition that (cf. Section 5.34)

and

In particular,

<tb, x1 V ... V b(x1, ... , xp) xE E

<f1 V ... v fps'P>s - 'P(f1, ... , fp) fEE*.

<fi V V fp, x1 V V xp> = perm(f (x;)).


9.18. Homogeneous Functions and HomogeneousPolynomials

Let

pP(t 1, ... , tn) = Ck l , ..., k t !' ... tnn k= p(k) v

be a polynomial of degree p. It determines a homogeneous function hp ofdegree p by

h C { 1)k ... ( n)kn

(k)

(here the kare exponents!), where x = L vev . This yields a homomorphism

6: rn - H(E).

On the other hand, the polynomial P determines an element Up E S"(E)given by (cf. Section 9.14)

Up = C 1,...,kf(f i i ... (f1) kv = p(k) v

where {f 1, ..., f'} is the dual basis of {e1, ... , end (the kv are again exponents).Moreover, the correspondence P H u defines an isomorphism

i: rn - S'(E)

(see Section 9.16).We shall show that the diagram (see Section 9.12 for z).

IT,, S(E)

(9.12)

H(E)

commutes. Since all maps are algebra homomorphisms, it is sufficient toshow that

= 6(ti) (i = 1, ..., n).

But, since l/Jti = f', we have ii/i(ti) = f On the other hand, 6f i = ` = f `whence (9.12).

Clifford Algebras

10In this chapter E denotes a vector space over a field with characteristic zero and ( , ) denotes a

(possibly degenerate) symmetric bilinear function in E.

Basic Properties


Let A be an associative algebra with unit element eA. A Clifford map from Eto A is a linear map p which satisfies

(gpx)2 = (x, x)eA x e E

or equivalently,

(P(x)(P(y) + (P(y)(P(x) = 2(x, y)eA x, y e E.

A Clifford algebra over E is an associative algebra CE with unit element etogether with a Clifford map iE : E -- * CE subject to the following conditions:

Cl : The subspace Im iE generates the algebra CE.C2 : To every Clifford map gyp : E -- * A there exists a homomorphismf : CE -- A which makes the diagram

E A

iE

CE

(10.1)

commutative.

Conditions Cl and C2 are equivalent to the following condition

C : To every Clifford map p : E -- * A there exists a unique homomorphismf : CE - A such that Diagram (10.1) commutes.

227

228 10 Clifford Algebras

In fact, it is easy to check that Ci and C2 imply C. Conversely, assume thatiE satisfies C. Then C2 follows immediately. To establish Ci denote by Athe subalgebra of CE generated by Im iE and by iE the map iE considered as amap into A. Then, clearly, iE is a Clifford map. Hence there is a uniquehomomorphism f : CE - A such that

fo 'E = lE.

On the other hand, if j : A - CE is the inclusion map, we have

J°lE = lENow consider the map j of: CE - CE. Then the relations above imply that

r,(J ° f) ° lE = J ° (f ° 1E) = J ° lE = lE .

On the other hand,

t ° lE = lE .

Thus the uniqueness part of Condition C implies that

j°f= 1.Hence j is surjective and so Ci follows.

10.2. Examples

(1) Let be the real axis with the negative definite inner product given by

(x, y) _ - xy x, y E .

We show that C is a Clifford algebra over . Let iE : - C be the linear mapgiven by

iE(x) - i x x E {l.

Then iE satisfies C1. To establish C2, let

be any Clifford map. Since p is linear,

q(x) = xE[Q

Set p(1) = a. Since p is a Clifford map, it follows that a2 = - eA. Thus qP isof the form

p(x) = x a, a2 = - eA .

Now define f : C - A by setting

f(x+iy)=xeA+ya x,ye.

Basic Properties 229

Then f is an algebra homomorphism as is easily checked. Moreover,

f lE(Y) = f (iy) = y' a= (p(y) y e

and so iE satisfies C2.(2) Let 2 be the plane with a negative definite inner product. We show that

the algebra of quaternions (see Linear Algebra Section 7.23) is a Cliffordalgebra over 2.

Choose an orthonormal basis {x 1, x2} of 2 and let {e, a 1, e2, e3} be anorthonormal basis of 0-0. Define a linear map iE : 2 - 0-{1 by setting

iE(xl) = e1, jE(x2) = e2.

Then

(iE(x))2 = - (x, x). e = (x, x)- e x e

and so iE is a Clifford map. It is easily checked that iE satisfies C 1. To establishC2, let p: 2 -+ A be a Clifford map and set q (x1) = a1, q (x2) = a2. Then(p is of the form

(px = 21a1 + 22a2,

where

x = 21x1 + 22x2

and the vectors ai (i = 1, 2) satisfy the relations

a= a=a2 = -eala2 + a2 a1 = 0.

Now define f : D-LI - A by

f(2e+21e1 +22e2 +23e3)=2eA+21a1 +22a2 +23a1a2.

Then f is a homomorphism and satisfies

fiE(x) = f'E(21x1 + 22x2) = 21a1 + 22a2 = cpx x E 2

fo'E=(p.

10.3. Uniqueness and Existence

In this section we shall show that there is a Clifford algebra over every innerproduct space, unique up to an isomorphism.

First suppose that CE, CE are Clifford algebras over E and let iE : E - CEand iE : E -+ CE denote the corresponding Clifford maps. Then, by C2, thereare homomorphisms f : CE -p CE and g : CE - CE such that

lE = f o lE an lE = g o lE .


It follows that

iE = (f ° g) ° iE and lE = `g ° f) ° iE .

Now Condition C 1 implies that

fog=i and g ° = i.

Thus f and g are inverse isomorphisms. This shows that a Clifford algebraover E, if it exists, is uniquely determined up to an isomorphism.

To prove existence consider the tensor algebra ®E (see Section 3.2) andlet J denote the two-sided ideal in ®E generated by the elements

xeEwhere 1 is the 1-element of Q °E = r. Define CE to be the quotient algebra

CE = Q E/J

and let

7L : O E - CE

be the canonical projection. Let iE : E - CE be the linear map

lE = 7L ° JE ,

where JE : E --* ®E denotes the inclusion map. We shall show that (CE, 'E)is a Clifford algebra over E.

First observe that, for x e E,

(iE x)2 = (7tJE x)2 = lt(JEx)2 = 7L(x QX x) = (x, x). 1

and so iE is a Clifford map. Since the algebra ®E is generated by E and sinceit is surjective, it follows that iE satisfies C1. To establish C2, let p : E - A beany Clifford map. In view of Section 3.3, qP extends to a homomorphism

h:®E-+A.

This homomorphism satisfies

h(x O x - (x, x). 1) = (q x)2 - (x, x)eA = 0.

It follows that J c ker h and so h factors over it to induce a homomorphism

f:CE - A.

Clearly,

flE(x) - firjE(x) =,r1C(JC) _ h(x) = lp(JC) JC E E

and so C2 follows.Thus to every inner product space E there exists a unique Clifford algebra

CE (up to an isomorphism). It is called the Clifford algebra over E.


EXAMPLE. If ( , ) = 0, then J is generated by the products x Q x and soCE = n E. Thus the Clifford algebra is a generalization of the exterioralgebra.

10.4. The Injectivity of iE

Proposition 10.4.1. The map iE:E - CE is injective.

PROOF. If ( , ) = 0, then CE = A E and the injectivity of iE follows fromSection 5.8. Now assume that ( , ) is nondegenerate. Then, if xo E ker iE,we have for y e E

2(x0, y)e = 1E(x0) ' IE.Y + 1EV 1E(x0) = 0

whence x0 = 0.In general write E = E0 Q E1, where E0 is the nullspace of ( , ) and E1

is a subspace such that the restriction ( , ) of( , ) to E 1 is nondegenerate.Then the Clifford map i, : E1 - CEI is injective.

Let it0 : E -+ E0 and n, : E -+ E1 be the projections and consider the map

(pl E E1 CEI.

Then

((p1x)2 = (llitlx)2 = (itlx, it1x)

where e1 denotes the unit element of CEI. Now,

(?t, x, ?t, x) = (x, x) - 2(x, 7t0 x) + (7t0 x, it0 x) = (x, x) x e E

and so we obtain

((p, x)2 = (x, x)e, x e E. (10.2)

Next consider the linear map p: E - n E0 Q CEI given by

(px = tax a e, + 1 Q p1 x X E E.

Then we have, in view of (10.2),

((px)2 = (7c0x)2

O e 1 + ic0 x OO (p 1 x - 7C0 X O (p 1 x + 1 O ((p 1x)2

= (x,x).(1 Oe1)

and so (p is a Clifford map. Thus, by C2, there is a homomorphism f : CE -pn E0 Q CE 1 such that (p = f o 'E. Since (p is injective, it follows that iE isinvective.

Henceforth we shall identify E with its image under iE . Then E becomes asubspace of CE and we have the relation

x,yeE.

232

In particular,

10 Clifford Algebras

x2 = xeE.

Now properties C1 and C2 can be rephrased as follows:

C1: The algebra CE is generated by the subspace E.C2 : Every Clifford map p: E - A extends to a homomorphism f : CE - A.

10.5. Homomorphisms

Let F be a second vector space and let ( , ) be a symmetric bilinear functionin F. A linear map p: E - F is called an isometry, if

(qx, coy) = (x, y) x, y e E.

Let p: E - F be an isometry. Then there is a unique homomorphismcPc : CE -+ CF which makes the diagram

E 'P )F

(10.3)

CE Cp F

commutative. In fact, consider the linear map

Then

(cPF x)2 = (lF (px)2 = ((ox, (px)eF = (x, x)eF x e E,

and so cPF is a Clifford map. Thus it extends to a homomorphism cPc : CE - CF.To prove uniqueness, suppose that CPC : CE - CF is a second homomor-

phism making Diagram (10.3) commute. Then

Pc 1E(x) = cc 1E(x) x e E

and so C 1 implies that Pc = cPcIf i/i : F - H is a second isometry, then clearly,

( ° P)C - C ° CPC

Moreover, the identity map of E induces the identity in CE,

These properties imply that if p is an isometric isomorphism, then cPc is analgebra isomorphism.

Basic Properties

10.6. The 712-Gradation of CE

Consider the linear automorphism w of E given by

w(x) = - x x E E.

Since w is an isometry, it induces a homomorphism

WE : CE -k CE.

Moreover, since w2 = i, it follows that

wE = 1.

233

Thus WE is an involution of the algebra CE.Next, consider the subspaces CE and CE of CE consisting of the elements

CE = ker(WE - 1)

and

CE = ker(WE + 1).

Since WE is an involution, it follows that

CE = CE $ C.

Moreover,

CE CE c CE

C. CE c CE, C. CE c CE

C. CcCE c C.

In particular, CE is a subalgebra of CE.The elements of CE (respectively CE) are called homogeneous of even

(respectively odd) degree. This defines a 12-gradation of CE. The map WE iscalled the degree involution of CE.

It follows from the definitions that the subspaces CE and CE are linearlygenerated by the products

C: x 1 xp x1 E E, p even

and

C: x 1 xp x1 E E, p odd.

A subspace U of CE which is stable under the degree involution is agraded subspace. To prove this, set

Uo =Urn CE, U1 = UnCE.

Then

V = U0 E1 U 1.

234

In fact, let u e U and set

u0 = 2(u + (J)E u), u 1 = 2 u (J)E u)

Then u0 e U0, u1 e U1, and u = u0 + u1.



Let E and F be inner product spaces. Define an inner product in the directsum E Q F by setting

(x1 0 Y1' x2 0 Y2) = (x1, x2) + (Y1' Y2) x1, x2 E E, Y1' Y2 E F.

Now consider the Clifford algebras CE, CF, and CE®F .

Theorem 10.7.1. The 12-graded algebra CE®F is isomorphic to the skew tensorproduct of the 712-graded algebras CE and CF,

CE®F" CE QX CF

PROOF. To simplify notation we write E Q F = H. Let i : E - H and j : F - Hdenote the inclusion maps. Since these maps are isometries they inducehomomorphisms

IC: CE - CH and Jc : CF -+ CH.

Now define a linear map f : CE Q CF -+ CH by

.f (a O b) = ic(a) 'jc(b) a E CE, b E CF .

We show that f is an algebra homomorphism. Since iC and jc are homo-morphisms it is sufficient to show that

ic(a) ' jc(b) = (-1)jc(b) ' ic(a) a = deg a, f = deg b.

Moreover, in view of C 1, we may assume that

a = x1 xp x1EE

and

b=Y1 ... yjeF.Yq

Then

ic(a) ' jc(b) = ic(x 1) ... iC(x p) ' ic(Y 1) .. ic(Yq)

Since every two vectors iC(xl) and JC(yq) are orthogonal with respect to theinner product in H, it follows that

jc(YJ) + jc(Yj) ' 0.


Thus we obtain

iC(a) JC(b) _ (- 1)'j(y1) ... Jc(yq)lc(x 1) ... ic(x p)

_ (-1)' jc(b) ic(a) = (- 1)j(b) ' ic(a).To show that f is an isomorphism we construct an inverse homomorphism.Consider the linear map

given by

(x O+ y) = x Q eF + eE Q y x E E, y E F.

It satisfies

(rl(xEJ y))2=x2QeF+x®y-x®y+eEQy2

= x 2 Q eF + eE Q y2 = [(x, x) + (y, y)] eE O eF

_ (x O+ y, x O+ y)eH

Thus ri is a Clifford map and so it extends to a homomorphism

g : CH - CE Q CF.

It follows from the definitions off and g that

gf (x O eF) = gic(x) = g(x O+ 0) = ri(x Q 0) = x Q eF x e E.

Similarly,

gf (eE O+ y) = eE Q y y e F.

Finally,

gf (eE O eF) = eE O eF

Since the algebra CE Q CF is generated by the elements x Q eF, eE Q y, andeE Q eF ; this implies that

gof=i.On the other hand,

fg(x O+ y) = f(x O+ y) = f(x O eF) + f(eE O y)

= i(x) + j(y) = x Q y x e E, y e F

and

fg(eH) = eH.


fig=i.Thus f is an isomorphism.


EXAMPLE. Suppose that the inner product in E is degenerate and let Eo denotethe null-space. Choose a second subspace F of E such that

E=E0O+F.

Then we have, in view of Theorem 10.7.1 and the example in Section 10.3,CE A Eo ©CF .

Proposition 10.7.2. Let p : E - F be an isometry. Then

1. If p is injective, so is Pc;2. If co is surjective, so is Pc.

PROOF OF 1. Set Im p = F 1. Then p determines an isometric isomorphismQP i :E 4 F1. Now write F = F 1 $ F2 where F2 is orthogonal to F 1. Thenthe diagram

CE ( CF 1

CF

commutes as is easily checked (i 1 denotes the obvious inclusion map). Thediagram shows that c°c is injective.

PROOF OF 2. Set ker p = E1 and write E = E1 $ E2 where E2 is orthogonalto E1. Then q induces an isometric isomorphism P2: E2 -- F and the diagram

CE f CE 1 QX CE2 2 CE2

CF

commutes, where it2 is the obvious projection. Since it2 is surjective, it followsthat c°c is surjective.

10.8. The Involution SE

Given a Clifford algebra CE consider the opposite algebra CEP (cf. LinearAlgebra Section 5.1) and let denote the multiplication in CE P. Then the in-clusion

C oPPJ --* E

is a Clifford map and so it extends to a homomorphism

SE : CE - CE P

For x1 E E we have

SE(x 1 ... xp) = x i x2 ... xp = xp ... x 1.

Clifford Algebras Over a Finite-Dimensional Space 237

Clearly,

S2E - t

and so SE is an involution. Moreover,

SEX=x xeE.

Finally, SE commutes with the degree involution,

SE°0E _ 0E°SE.

Next, let u e CE and set

u - SE WE(u).

In particular,

x = -x xeE.

The correspondence u H u defines a linear involution of CE. It satisfies

u ' 3 = SE WE(u U) - SE(W E(u) ' WE(V ))

- SE(wE(U)) ' SE(WE(u)) = U u.

PROBLEM

Show that the map a - a (see Section 10.8) in the cases CE = C and CE = H coincideswith the usual conjugation.

Clifford Algebras Over aFinite-Dimensional Space

In this paragraph E denotes an n-dimensional vector space.

10.9

Proposition 10.9.1. Let dim E = n. Then

dim CE = 2.

PROOF. First consider the case n = 1. Fix a nonzero vector a in E and let Adenote the vector space generated by a and a. Then

e a = a e = a and a2 = (a, a)e.


Thus A is an algebra. It is easy to check that the inclusion map E - Aextends to an isomorphism CE 3 A. Thus, dim A = 2.

In the general case choose an orthogonal basis {ei} (i = 1, ... , n) of Eand denote by Ei the 1-dimensional subspace of E generated bye, (i = 1 , , n).Then we have the orthogonal decomposition

E=E1Q+...D+En.

Thus, by Theorem 10.7.1 (cf. Section 5.20)

CE = CE1 O O CEn

and so

dim CE = 2.

Corollary. If {xi} (i = 1, ... , n) is any basis of E, then the 2n vectors

<J),...,xl ...xn

form a basis of CE.

PROOF. It follows from the relation xix; + x;xi = 2(x1, x;)e that the vectorsabove generate the space CE. Since, by the proposition, dim CE = 2, theymust form a basis of CE.

10.10. The Canonical Element eo

Consider the linear map

E:AE-4CE

given by

1E(x 1 A ... A xp) = E x(11 ... xp) (1 <- P <- n).P

We show that E is a linear isomorphism. In fact, choose an orthogonal basis{e1, ... , en} of E. T en ei e; = - e; ei (i an so

E(eil A ... A eip) = ei1 ... eip (11 < i2 < ... < ip).

Thus E takes a basis of A E into a basis of CE and so it is a linear isomorphism.Now choose a nonzero determinant function A in E. Then E determines

an element ee e CE by the equation

E(x 1 A A xn) = 0(x 1, ... , xn) ee xv e E. (10.4)

eo is called the canonical element in CE (with respect to the determinantfunction A).


Now choose a basis {e1, ..., en} of E such that (e1, e;) = 0 (i j) and0(e 1, ..., en) = 1. Then

ee = e1 en. (10.5)

Next observe that the determinant function 0 determines a scalar 2e suchthat the Lagrange identity

det((xi, yj)) = 2e 0(x1, ... , xn)0(y1, ... , yn) xv e E, yv e E, (10.6)

holds. Setting xv = yv = ey we obtain

(e1, e 1) ... (en, en) = 2e .


ee = (_ 1)ncn - 112e2 .. en

(10.7)

= (_ 1)n(n - 112(e1, e 1) ... (en, en) ' e= (_ 1)n(n - 1)/22 e e.

Thus the square of the canonical element is given by

ee = (_ 1)ncn - 1)/22e e. (10.8)

Since 2e 0 if and only if the inner product in E is nondegenerate it followsthat

1. If the inner product is nondegenerate, then ee is invertible in CE,2. If the inner product is degenerate, then ee = 0.

Proposition 10.10.1. The canonical element ee satisfies the relation

ee x = xeE.

Thus,

uECE.

In particular, fn is odd,

uECE

and if n is even,

ee u = COE(u) ee uECE.

PROOF. Choose an orthogonal basis {e1, ..., en} of E and write

ee = A. e 1 ... en A E r.

(see Formula (10.5)). Then we have

ee ei = 2e 1 ... e e, = 1)n `2(e1, e.)e 1 ... e1... E

and=2ei .e1 ...e =(-1)`-12(ei., e.)e1 ...e....ei n i i n

240

whence

eo ei = (-1)n -' et eo .

Thus, by linearity,

xeE.

10.11. Center and Anticenter


The center of a Clifford algebra CE, denoted by ZE, consists of the elements awhich satisfy

a subalgebra of CE. Since CE is generated by E, it followsthat an element a e CE is in the center if and only if

xeE.

Next observe that ZE is stable under the degree involution. In fact, ifa e ZE, then

WE(a) ' x = - WE(a) ' WE(x) _ - WE(a ' x) = - WE(x ' a)= - WE(x) ' wE(a) = x ' wE(a)

and so WE(a) E ZE. Thus ZE is a graded subspace of CE and hence a gradedsubalgebra,

ZE = ZE O+ ZE, ZE = ZE r C, ZE = CE r ZE.

The anticenter of CE, denoted by AZE, consists of the elements a whichsatisfy

uECE.

Since CE is generated by E, an element a e CE is in AZE if and only if

xeE.

As above it follows that the anticenter is stable under the degree involutionand hence it is a graded subspace of CE.

Proposition 10.10.1 shows that

1. If n is odd, then eo E ZE,2. If n is even, then eo E AZE .

Next, assume that the inner product in E is nondegenerate and considerthe linear map'PE CE - CE given by

coE(u) = eo u u E CE .

Since eo is invertible, APE is a linear isomorphism.


Now let u e ZE. Then we have, in view of Proposition 10.10.1,

coE(u)x = eoux = eoxu = (-1)"-'xeou= (-1)" 'x cPE(u) x e E.

Similarly, if a e AZE, then

(PE(a) x = (-1)"x (PE(a) x e E.

These relations show that

1. If n is odd, then APE restricts to linear automorphisms of ZE and AZE,2. If n is even, then cPE interchanges ZE and AZE.

Lemma I. Assume the inner product in E is nondegenerate. Then (AZE)' = 0.

PROOF. Let a E (AZE)'. Then

ax=-xa xeE

and so

aeo=(-1)"eoa.On the other hand, Proposition 10.10.1 yields, since a e CE,

eo a = wr' (a) eo = (-1)" - 'a eo .


aeo=0.Since the inner product is nondegenerate, eo is invertible and we obtaina=0.

Lemma II. Assume that the inner product in E is nondegenerate. Then ZE = (e).

PROOF. The lemma is trivial for n = 1. Assume that it holds for n - 1 and letE be an n-dimensional inner product space. Choose a vector a e E such that(a, a) 0 and write

E = (a) $ F,

where F denotes the orthogonal complement of a. Thus we can write

u = 1 ®v + a ®w ueCE veCF, weCF

and

x= 1®y+a®eF xeE,yeF.These relations imply that

ux=a®v-a2®w+ 1®vy+a®wy=a®v-(a,a)1®w+ 1®vy+a®wy.

242

Similarly,


1Qyv-a®x yw.

Now assume that u is in the center of E. Then u x = x u and we obtain

a 0 (wy + yw) = 0 y e F (10.9)

and

1 Q (vy - yv) - 2(a, a)1 Q w = 0. (10.10)

Thus, w e AZE and so w e (AZE)'. Hence Lemma I implies that w = 0. NowFormula (10.10) yields

vy=yv yeF.

This shows that v e ZF and so v e ZF. Hence, by induction, v = 2. eF. Itfollows that

u=

and so the induction is closed.

Proposition 10.11.1. Assume that the inner product in E is nondegenerate. Then

1. If n is odd, ZE = (e) + (ee), AZE = 0,2. If n is even, ZE = (e), AZE _ (eo).

PROOF OF 1. Let n be odd. Assume that a e AZE. Then ax = - xa, x e E,and so

It follows from Proposition 10.10.1 that a ee = 0 whence a = 0.Next observe that, by Lemma II, ZE = (e). Since n is odd, the map APE

restricts to an isomorphism ZE 4 Z. Since (pE(e) = ee we obtain ZE = (ee)whence ZE = (e) + (ee).

PROOF OF 2. Let n be even. Then we have, for a e ZE, ax = xa, x e E, and so

On the other hand, by Proposition 10.10.1,

a ee = WE(a) ee.

It follows that WE(a) = a whence a e Z. Now Lemma II implies thata = 2. e whence ZE = (e).

Since the map c°E for even n interchanges center and anticenter, it followsthat AZE = (ee). This completes the proof of the proposition.


10.12. The Algebra C _ E

Given an inner product space E denote by - E the space E with the innerproduct

(x, y) _ _ - (x, y) x, y E E

and let C _ E be the Clifford algebra over - E. Denote the multiplication inC_E by Thus,

xeE.

Now fix a nonzero determinant function 0 in E. Then it is easy to check that

' A - (-1)"/LA

(cf. Formula (10.6)).Now define a linear map p: E -+ C _ E by

(10.11)

px = ee ° x x E E,

where ee denotes the canonical element of C_ E Then we have, in view ofProposition 10.10.1,

cpx°cpx = ee °x°ee °x = (-1)"-1ee gee °x°x.

Since, by (10.8),

ee o ee = 122e = (- 1)" 1)12( -1)"AA e

and

x°x = (x,x)_ e=

it follows that

px o cpx = (-1)" 1)12(x, x)AA e.

Now assume that 0 can be chosen such that

A = (-1)" - 1)12

Then the relation above reads

cpx o cpx = (x, x) e x e E.

Thus p extends to a homomorphism

(p: CE -+ CE.

We show that

(10.12)

p(eA) _ (-1)" - '2(e)e n+ 1. (10.13)


In fact, choose an orthogonal basis {e1, ..., en} of E such that A(e1, ..., en)= 1. Then eA = e 1 en and thus

(P(eA) _ (p(e 1) o ... o (p(en)

(eA oe1)o...o(eA oen)

= (_ 1)n(n - 1'2(eA )n

+ 1

Similarly, consider the linear map il/ : - E - CE given by

xE-E.Suppose that A can be chosen such that

A = (_ 1)n(n + 1)/ 2

Then

e

and hence, by the result above (applied to - E), il/ extends to a homomorphism

- CE.

Proposition 10.12.1. Assume that n is even, n = 2m, and that A can be chosensuch that 2A = (-1)m. Then the algebras CE and C_E are isomorphic.

PRooF. Since n = 2m and 2e = (-1)m,A = (_ 1)n(n - 1)/2 _ (_ 1)n(n + 1)/2

Thus we have the homomorphisms

(p: CE - C_E and Ii: C_E - CE.

To show that (p and il/ are isomorphisms we establish the relations

il/ o q = w and (poll/ _ (,gym E .

In fact, let x e E. Then

i/i(ee o x) = i/i(ee) il/x

_ (_ 1)mee+ 1 = (_ 1)menA+ 1 , eA x

_ (_ 2 . x = (- 1)m(eo)m+1 . x.

Since eo = e, (by (10.8)), it follows that

i/np(x) = (- 1)mx = w(x) x e E.

But (p and i// are homomorphisms and so the equation above implies that

il/ (p = w.

In the same way it is shown that (p o i// = o/ E .


10.13. The Canonical Tensor Product of Clifford Algebras

If E is an inner product space and E = + 1 we shall set E = E if E = 1 and

E be an even-dimensional inner product space whichadmits a determinant function 0 such that ee = E e (E = + 1). Then, for anyinner product space F,

CE®eF ti CE ® CF

PROOF. Write E Q EF = H and let i : E - H and j : F - H denote the inclusionmaps. Since E has even dimension, Proposition 10.10.1 implies that

xeE. (10.14)

Moreover, if is : CE - CH and Jc : CAF - CH are the induced homomorphisms,then (cf. the proof of Theorem 10.7.1)

ic(eA) ' Jc(b) = Jc(b). ic(eA) b e CAF . (10.15)

Now let p : F - CH be the linear map given by

Sp(y) = ic(eA) ' Jc(Y) Y E F

Then, in view of (10.15),

P(Y)2 = ic(ee)Jc(YZ) = E(Y, Y)E ic(eE) ' Jc(eF)

= E2(Y, y). eH = (Y, y) ' eH.

Thus, qP extends to a homomorphism

p: CF - CH.

Next note that, in view of (10.14),

ic(x) ' SP(Y) = ic(x)ic(eA)Jc(Y) = - ic(eA)ic(x)Jc(Y)

= sp(y).c(x) x e E, y e F

whence

Next, define

by setting

ic(a) Wp(b) = Wp(b) ic(a) a e CE, be CF.

D : CE O CF -+ CH

(a O b) = ic(a) Wp(b) a e CE, b e CF.

Then b is a homomorphism. To show that it is an isomorphism we constructan inverse map.


First define a linear map i/i : H - CE Q CF by

(xO+Y)=xOeF+C Qy.Then, in view of 10.14,

(ifr(x O+ Y))2 = x2 Q eF + E[(xeA + eA x) Q y] + E2(ee O y2)

[(x, x) + E(Y, Y)]eE O eF = (x O+ Y, X O+ Y)H eE Q eF.

Thus i/i extends to a homomorphism

kY:CH- CEOCF

Note that

'T'ic(a) = a Q eF a e CE.

It follows that

('1' ° 1) (x O eF + eE O y) _ '[i(x) + ic(eA)J(Y)]

= x Q eF + E(eA O eF)(eA O Y)

= x Q eF + E2(eE O Y)

=x®eF+eEQy xeE,yeF.This shows that q'0 i _ t. On the other hand,

(I ° ') (x O Y) _ l [x O eF + E(eA O Y)]= i(x) + Eic(eA)ic(eA)Jc(Y)

= i(x) + E2ic(eE)J(Y)

= i(x) + j(y) = x Q y x e E, y e F

and so I o ' = i. It follows that I is an isomorphism.

10.14. The Direct Sum of Dual Spaces

Let E 1 and E2 be dual n-dimensional spaces and consider the direct sum

E=E1QE2.Then a nondegenerate inner product is defined in E by

(x1 0 x2, Y1 0 Y2) = 2[<x1, Y2> + <Y1, x2>] x1, Y1 E E1, x2, Y2 E E2

(note that this is not the usual inner product in the direct sum!). In particular,the restriction of twice the inner product in E to E1 x E2 coincides with thescalar product between E 1 and E2.

Proposition 10.14.1. There is an isomorphism CE 4 L( A E 1).


PROOF. First recall the definition of the multiplication and the substitutionoperators in A E 1. Identifying E 1 with E2 we have the relations

µ(x1)2 = 0 x1 EE1

i (x2)2 = 0 x2 E E2

and (see Corollary I to Proposition 5.14.1)

1(x2) °µ(x1) + µ(x1) ° 1(x2) = <x1, x2>1 x1 E E1, x2 E E2.

Now define a linear map p: E - L( A E 1) by setting

(p(x) = µ(x1) + 1(x2) x E E,

where x = x 1 $ x2, x 1 E E 1, x2 E E2. Then the relations above yield for allxeE

p(x)2 = µ(x 1) ° µ(x 1) + µ(x 1) ° 1(x2) + 1(x2) ° µ(x 1) + 1(x2) ° 1(x2)

=Thus p extends to a homomorphism

q . CH - L(A E 1).

We show that p is an isomorphism. Since

dim L( A E1) = (2n)2 = 22n = dim CE,

it is sufficient to show that p is surjective. This is a consequence of thefollowing.

Lemma. Let E*, E be a pair of finite-dimensional dual spaces. Then the algebraL( A E) is generated by the operators µ(x) and i(x*), x e E, x* E E*.

PROOF. Recall from Section 6.2 the linear isomorphism

TE: APE*Q AE4L(A'E; AE)given by

TE(a* Q b)u = <a*, u>b.


TE(a* Q b) = µ(b) ° i(a*).

Since every vector a* E A PE* is generated by products x; A A xp,x* E E* and every vector b e A qE is generated by products Y1 A A yq,y E E, and since

µ(Y 1 A ... A Yq) = µ(Y 1) ° ... ° µ(Yq)and

1(x i A ... A x) = i(x) ° ... ° i(x 1)

the lemma follows because L( A E) = Qp L( A pE; A E).


Proposition 10.14.2. Let E be a 2n-dimensional vector space with a non-degenerate inner product. Assume that there exists an involution w of E suchthat th = - w. Then the Clifford algebra CE is isomorphic to the algebra oflinear transformations of A E 1, where E 1 = ker(w - i).

PROOF. Consider the subspaces E 1 and E2 consisting of the vectors x whichsatisfy, respectively, wx = x and wx = -x. Then E = E1 $ E2. In fact, letx e E and set

x l - 2(x + wx)

x2 - 2(x - wx).

For x1 E E1, yl E E1, we have

(x1, yl) = (wx1, wy1) = -(w2x1, yl) = -(x1, yl)

whence (x1, yl) = 0. Similarly,

(x2 , y2) = 0 x2 , y2 E E2 .

Thus the restrictions of the inner product to E1 x E1 and E2 x E2 are zero.On the other hand, the restriction of the inner product to E1 x E2 is non-degenerate. In fact, fix x1 E E 1 and assume that (x 1, y2) = 0 for every y2 E E2.Then we have for y e E

(x1, y) - (x1, yl) + (x1, y2) = 0

whence x1 = 0. Thus a scalar product between E 1 and E2 is defined by

<x1, x2> = 2(x1, x2) x1 E E1, x2 E E2.

It satisfies the relation

(x1 O+ x2, yl e y2) = (x1, y2) + (yl, x2)= 2[<xl, y2> + <y1, x2>]

Thus Proposition 10.14.1 implies that CE ^L( A E 1).

PROBLEMS

1. Let CE be the Clifford algebra over an n-dimensional inner product space and denotethe left multiplication by an element a e CE by µ(a). Show that

det µ(x) = (x, x)2" -1

x E E

and

tr µ(a) = 0 if a E CE .

2. The isomorphisms bE and 11E. Let E be an n-dimensional vector space with a non-degenerate inner product. Identify E with the dual space and denote by i(x) (x e E) thesubstitution operator in n E.

Clifford Algebras Over a Complex Vector Space 249

i. Show that the isomorphism bE defined in Section 10.10 satisfies the relations forall x e E and all a e A "E

b E(x A a) = x b E(a) - b E 1(x)a

and

bE(a A x) = ,(a) x + (-1 )'Eb1(x)a.

ii. Let 11E denote the inverse isomorphism. Show that for all x e E and all u e CE

'1E( x ' u) = x A ii ,(u) + I(x)rl E(U)

and

11E(U ' x) _ 11E(U) A x -

iii. Let 7rE. CE - r be the linear map given by

nE u = no r1 E(u) u E CE

where no : A E -+ r is the obvious projection. Prove the following relations for allxeEandallu,veCE:

iE(x ' u) = n0 I(X)E(U)

iE(U ' x) _ - n0 E(wE u)

nE ° WE = nE

iE(U v) = 7rE(v. U).

3. Use the linear map nE (see Problem 2 (iii),) to define a bilinear function QE in CE bysetting

QE(u, v) = nE(u v).

i. Show that QE is symmetric and nondegenerate.ii. Show that QE(x, y) _ (x, y), x, y e E.iii. Prove the relation

QE(U ' W, v ' W) = QE(W U, w v) U, V, W E CE.

Clifford Algebras over a Complex Vector Space

10.15. Clifford Algebras Over Complex Vector Spaces

Let C" be an n-dimensional complex vector space and let ( , ) be a non-degenerate symmetric bilinear function in C". Note that if ( , )1 and ( , )2

are two such bilinear functions, then there is a linear isomorphism p of C"such that

(qx, (Py) i = (x, y)2 x, y e C".

Thus the corresponding Clifford algebras are isomorphic. We shall denotethe Clifford algebra over C" by C,,.


EXAMPLE : n = 1. The Clifford algebra C 1 is isomorphic to the algebraC Q C. In fact, choose a vector e 1 E C such that (e 1, a 1) = 1. Then the vectors{e, e 1 } form a basis of C 1. Now define a linear map sb: C 1 - C Q C bysetting

b(ae + f e 1) = (a + fl, a - f3) a, f e C.

Then 'b is an algebra isomorphism as is easily checked. Thus,

C1 CQC.

The Element ee. A normed determinant function in C" is a determinantfunction A which satisfies

0(x1, ... , xn) . 0(y1, ... , yn) = det((xi, yj)) xi, yj e C".

It is easy to see that a normed determinant function always exists and that itis determined up to sign. Thus the canonical element ee corresponding to anormed determinant function satisfies (see (10.8))

Hence Theorem 10.13.1 yields an isomorphism

C2m+k ti C2m ® Ck. (10.16)

Proposition 10.15.11. Let n be even, n = 2m, and assume that the inner productis nondegenerate. Then

C2m L( A Cm). (10.17)

PROOF. Write C" as an orthogonal sum

C"= AQBwhere dim A = m and dim B = m. Choose orthonormal bases {aµ} and{bµ}, µ = 1, ..., m in A and B respectively and define an involution w ofC" by setting

w(aµ) = ibµ u= 1, ... , m

iaµ µ = 1, ... , m.

Then for µ, v = 1, ..., m we have the relations

(waµ , 0, (wbµ , 0

and

(waµ , i(bµ , (aµ,

w is a skew transformation. Now Proposition 10.14.2 shows that

C" L(A E1),

where E 1 is the subspace of E determined by the equation wx = x.

Clifford Algebras Over a Real Vector Space 251

Combining Formulas (10.16) and (10.17) we obtain for all m > 1 therelations

C2m L( gym)

and (using the example above)

C2m+ 1 L( A Cm) Q+ L( A Cm).

10.16. Complexification of Real Clifford Algebras

Let F be a real inner product space (not necessarily of finite dimension) andconsider the complexification E _ C Q F. Define an inner product in E by

(A O x, µ O Y)E _ 2µ(x, Y) 2, u E C, x, y E F.

Then the inclusion map j : F - E is an isometry and so it extends to a (real)homomorphism jc : CF -p CE. Now consider the complex linear map

(p.CQCF - SCE

given by

(p(A O a) _ 2 ' Jc(a) 2 E C, a E CF.

To show that (p is an isomorphism we construct an inverse homomor-phism. Consider the linear map i/i : E - C Q CF given by

i/4x+iy)= 1Qx x+i®y.

Then

(li(x + iy))2 = 1 O x2 - 1 O Y2 + i(x y + y x)_ [(x, x) - (Y, Y)](1 O e) + 2i(x, Y)(1 O e)

e) x,yeF.Thus i/i extends to a homomorphism i/i : CE -+ C Q CF It follows from thedefinitions that ll/ o co = l and co o = i. Thus p is an isomorphism,

p:COCF4CE

Clifford Algebras Over a Real Vector Space

10.17

Let E be a real n-dimensional vector space with a nondegenerate innerproduct. Recall that E can be decomposed in the form

E=E+QE-,


where the restriction of the inner product to E + (respectively E) is positive(respectively negative) definite. If dim E + = p and dim E - = q, we shall saythat the inner product is of type (p, q). Clearly, p + q = n. The differences = p - q is called the signature of the inner product.

The Clifford algebra over an inner product space of type (p, q) will bedenoted by C(p, q). We shall write

C(n, 0) = C(+)

and

C(0, n) = Cn( - ).

Thus, by Examples 1 and 2 of Section 10.2,

C1(-) ^C and C2(-) ^Q-i.

Next recall from Section 9.19 of Linear Algebra that a normed determinantfunction in E is a determinant function A which satisfies

det((x1, y;)) = 1, ... , xn) 0(y 1, ... , yn) xv e E, yv e E.

Every inner product space admits a normed determinant function A and Ais uniquely determined up to sign. Thus the scalar 2n (see Section 10.10)corresponding to a normed determinant function is given by

o =This implies that the corresponding canonical element ee of the Cliffordalgebra C(p, q) satisfies

eo = (-1)(1/2)n(n-1)+'?e.

In particular,

1. If ( , ) is positive definite, eo = (-1)(1 /2)n(n - 1)e

2. If ( , ) is negative definite, eo = (-1)(h/2)(t+ 1)e

3. Ifn=2mandp=q=m,eo=e.

Theorem 10.17.1. There are algebra isomor phisms

C(p,q)©C2(+)=C(q+2, p) p,q?0and

C(p, q) © C2(-) C(q, p + 2) p, q ? 0.

In particular, for n > 0,

Cn(-) © C2(+) ~ Cn+2(+)

and

C(+)® C2(-) ti Cn+2(-).


PROOF. Observe that the canonical elements of C2(+) and C2(-) satisfyee = -e and apply Theorem 10.13.1.

Theorem 10.17.2. Assume that s - 0 (mod 4). Then

C(p, q) C(q, p)In particular,

if n - 0 (mod 4).

PROOF. Set s = 4k. Then

n=2q+s=2q+4k=2m, wherem=q+2kand

m-q=2k.Thus, if A is the normed determinant function, then

=(-1)q=(-1)m.Now apply Proposition 10.12.1.

10.18. Inner Product Spaces With Signature Zero

Suppose that E is an inner product space with signature zero. Then p = q andso dim E = 2p. We shall show that

CE L(n E 1),

where dim E 1 = p.Choose an orthogonal decomposition E = E + Q E - such that the re-

striction of the inner product to E + (respectively E) is positive (respectivelynegative) definite. Thus dim E+ = dim E- = p. Let and (v =1, ... , p) be orthonormal bases of E + and E - respectively,

(ar, aµ) = 5vµ V, U = 1, ... , p

(br, bµ) = v, µ = 1, ... , p.

Define an involution w of E by setting

wa= b and wb - av = 1, ... , p.Then w is a skew transformation as is easily checked. Thus by Proposition10.14.2,

CE L(n E ),where E 1 is the kernel of w - i.

EXAMPLE. Consider the algebra C(2, 2). Then, by the result above,

C(2, 2) ^' L(4).


On the other hand, the second formula in Theorem 10.17.1 applied withp = 0 and q = 2 yields

C(2, 2) C2(-) O C2(-) "' fNl Q u-fl.

Thus,

fl--fl Q fl--fl

The following proposition establishes an explicit isomorphism between thesealgebras.

Proposition 10.18.1. There is a canonical algebra isomorphism

fb: fl--l Q u--fl -

PROOF. Identify with 0-fl and define a linear map

f : fl--l Q fl--fl -

by setting

fb(pOq)x =

where q denotes the conjugate of q. It is easily checked that fb is an algebrahomomorphism. To show that it is an isomorphism define positive definiteinner products in 0-fl ® 0-fl and L(4) by

(p O q, p' O q') = (p, p')(q, q')

and

(q,, /,) = 4 tr(rp ° /,) P, /i e

respectively. Observe that the adjoint transformation of fb(p Q q) is given by

(p O q) = DO O q)

Thus we have

(fb(p O q), fb(p' O q')) = 4 tr(fb(p O q) ° fb(p' O q'))= 4 tr fb(pp' O qq')

Now it is easy to check that for a e 0-fl and b e 0--fl

tr fb(a Q b) = 4(a, e)(b, e).

It follows that

(fb(p O q), fb(p' O q')) = (pp', e)(qq', e) = (p', p)(q', q)=(p®q,p'Qq').

Thus fb is an isometry and hence a linear isomorphism.


10.19. Clifford Algebras of Low Dimensions

In this and the following section we shall determine the structure ofand for 1 < n < 8.

(1) n = 1. We show that

C 1(+)

In fact, let e 1 be a unit vector in . Then the vectors e, e 1 form a basis ofC 1( +). Now define a linear map t': C 1( +) -+ [ Q [ by setting

J(ae+fe1)=(a+f3,a-f3) a,fe.Then 1 is a homomorphism as is easily checked. Moreover, 1 is a linearisomorphism and so an isomorphism of algebras. This proves that C 1( +)

On the other hand, it has been shown in Example 1, Section 10.2, that

C1(-) C.

(2) n = 2. We show that

C2(+)

Fix an orthonormal basis {e1, e2 } of 2 and define a linear map p: 2 -L(2) by setting

gp(x)e 1 = ae 1 + f3e2

gp(x)e2 = fe 1 - ae2 ,

where

x=ael+le2.Then it is easy to check that

p(x)2 = (a2 + f32)1 = (x, x). 1.

Thus q extends to a homomorphism t : C2(+) -+ To show that 1 isan isomorphism note that the transformations t(x), x e E, are selfadjoint andhave trace zero. On the other hand, t(e) = 1. Thus, if A denotes the subspaceof CE spanned by e, e 1, and e2, 1 determines an isomorphism from A to thespace of selfadjoint transformations of 2 Finally, it is easy to check that(e2 e1) is the transformation e1 -+ e2, e2 -+ -e1 and so it is skew. Sinceevery transformation of 2 is the sum of a selfadjoint and a skew transforma-tion, it follows that 1 is an isomorphism.

10.20

Using the results of Section 10.19 and Theorem 10.17.1 we shall now deter-mine the Clifford algebras Ck( +) and Ck( -) for k < 8 explicitly. Recall thatthe direct sum of two algebras A and B is the algebra A Q B with multi-plication defined by

(a1 Q b1) (a2 Q b2) = (a1a2 Q b1b2).


We have the following isomorphisms :

n = 3: C3(+) C1(-) ®C2(+) C O L(I2)C3(-)C1(+)®C2(-)(l

$l)®D' I1-OeI1-O

n =4: C4(±) C2(-) O C2(+) U-0 O L(11R2)

n = 5: C5(+) C3(-) O C2(+) (U-0 e U-0) O L(l 2)

I1-0 O L(l 2) e Il-0 O L(l 2 )C5(-) C3(+) O C2(-) C O L([1R2) p U-O

n =6: C6(+) C4(-) O C2(+) U-0 O L(l 2) O L(l 2) U-0 ® L(l 4)

C6(-) ti C4(+) ®C2(-) = DO®L(E2) ®I1-O ~ D-O ®I1-O ®L(E2)

L(l 4) Ox L(l 2) L(l 8)

n = 7: C7(+) C5(-) OO C2(+) C Ox L(1R2) O U-0 O L(1R2)

C O U-0 O L({R4)

C7(-) C5(+) Q C2(-) [U-0 O L(2) $ U-0 O L(2)] O U-0D-0 O D--0 O L(R 2) e D--0 O D--0 O L(2 )

O O+ O O+

n = 8: C8(+) C4(-) O C4(+) U-0 O O U-0 O

O O L(W6).

These results are combined in the following table:

C(+) C(-)

CL(B 2) 0-U

C O U-U Q+ U-U

0-U® L(B2) U-() O L(B2)

(0-il O L(B2) O (0-ll O L(B2)) C® L(B2) O H0-U ®L(B4) L(B 8)

C® 0-U ®L(B4) L(B 8) O L(B 8)L(B16) L(B16)

10.21. The Algebras and

Since the canonical element of C8( +) ^C8( -) satisfies eo = 1, Theorem10.13.1 yields isomorphisms

Cn+8(+) ti Cn(+) ® L(W6)

and

Cn + 8(-) ti Cn(-) ® L(W6).


These isomorphisms, together with the isomorphisms in the table, determinethe structure of the Clifford algebras and for all n.

10.22. The Algebras C(p, q)

Finally, we determine the algebra C(p, q) for an indefinite inner product oftype (p, q)

Case 1: p = q. Then it was shown in Section 10.18 that

C(p, p) ^L( A DU'). (10.18)

Case 2. p > q. Write p = q + s. Then we have an orthogonal decomposi-tion

n = = (f 0 f) = + 0 (fSince has even dimension and since the canonical element ofC(q, q) satisfies ee = e, Theorem 10.13.1 gives

C(p, q) C(q, q) O CS( +).

Now using Formula (10.18) we obtain

C(p,q)L(A )QCS(+) p>q,s= p - q.Case 3: p <q. Set q - p = r. Then we obtain

C(p, q) = C(p, p + r) C(p, p) O Cr(- )

Thus

C(p, q) L( A fiP) © Cr(-) p <q, r = q - pNow the results of Section 10.21 give the structure of C(p, q) for all p, q.

EXAMPLE. The Clifford algebra over the Minkowski space (p = 3, q = 1) isgiven by

C(3, 1) = C2(+) Q L(A W) L(E2) O L(4).

PROBLEMS

1. Establish explicit isomorphisms

C3(+) L(C2)

C3(-) H $ HC7(+) L(Cs)

C7(-) "' L(Ps) O+ L(B8).


2. Let C2 be a complex vector space of dimension 2 with a positive definite Hermitianinner product

i. Consider the linear transformations (p which satisfy

q) +rp=trq).i.

Show that these transformations form an algebra (over B) and that this algebra isisomorphic to 0-fl.

ii. Use this algebra to establish a canonical isomorphism

C O H L(C2).

3. Let E be an n-dimensional Euclidean space and consider the symmetric bilinearfunction QE in CE defined in Problem 3 after Section 10.14.

i. Show that the signature s of QE is given by

nn nncos

4+sin

4n> 1.

ii. Conclude that

5n+4 = - 5n+8 = 16s n > 1.

iii. Show that QE is positive definite only if n = 1.

4 Derivations. Let E be a vector space over a field F.

i. Let 9 be a derivation in CE which restricts to a linear transformation ( of E. Showthat (p is skew.

ii. Show that every skew transformation 'p of E extends uniquely to a derivation in CE.

5. Antiderivations. Recall that an antiderivation in CE is a linear transformation Qwhich satisfies

Sl(ab) = Sl(a)b + WE(a)b a, b E CE

(cf. Section 5.8 of Linear Algebra).

i. Show that the map Q = i - wE is an antiderivation in CE.ii. Assume that the inner product in E is nondegenerate. Show that every antideriva-

tion Q in CE which restricts to a linear transformation of E is of the form

Q = , (1- wE) , E r.

6. Clifford algebras over even dimensional spaces. Let CE be the Clifford algebra over aneven-dimensional real vector space with a nondegenerate inner product. Show that CEis simple (that is, the only two-sided ideals in CE are CE and (0)). Conclude that ahomomorphism p from CE to any associative algebra A which satisfies ap(e) = eA isinjective. Hint: Consider the complexification of CE and apply Proposition 10.15.1.

7. Clifford algebras over spaces of odd dimension. Let CE be the Clifford algebra over anodd-dimensional space and assume that the canonical element satisfies eo = e.Consider the linear transformations

spa = 2(e + eo)a a E CE


and

via = 2(e - ee)a a E CE.

Set J" + = Im p and % - = Im /i. Show that and % - are two-sided ideals in CEand that

CE=TApply the result to complex and real Clifford algebras.

8. Show that the center of a Clifford algebra over an infinite-dimensional vector spacewith a nondegenerate inner product is the subspace spanned by e.

11Representations ofClifford Algebras

Basic Concepts

11.1. Representations of an Algebra

Let A be an associative algebra with unit element e. A representation of Ain a vector space V is a homomorphism R :A - L(V) where L(V) denotesthe algebra of linear transformations of V such that R(e) = i. A representa-tion is called faithful, if the map R is injective.

A subspace W c V is called stable under R, if it is stable under everytransformation R(a), a E A. A representation is called irreducible, if the onlystable subspaces are W = V and W = 0. In particular, if R is surjective, thenthe representation is irreducible. In fact, assume that W is a stable subspace.Then R is stable under every linear transformation of V. This is only possibleifW=0orW=V.

Two representations R 1 and R2 of A in V1 and V2 respectively are calledequivalent, if there is a linear isomorphism b: V1 V2 such that

o R 1(a) = R 2 (a) o a E A.

In this case we shall write R1 ' R2.Next let P and Q be representations of A in U and V respectively. Then a

representation of A in U $ V, denoted by P $ Q, is given by

(P O+ Q) (a) = P(a) $ Q(a) a E A.

It is called the direct sum of P and Q. Similarly, the tensor product of P andQ, denoted by P O Q, is the representation in U O V defined by

(P O Q) (a) = P(a) O Q(a) a E A.

260

Basic Concepts 261

Clearly, if P, P2 and Q1 Q2, then

P, e Q, --P2 e Q2 and P, Q Q, "'P2 Q Q2.

11.2. Representations of a Clifford Algebra

Let CE be the Clifford algebra over an inner product space E and let Rbe a representation of CE in an n-dimensional vector space V. Then Rrestricts to a linear map RE : E - L(V). We show that this map is injectiveif the inner product in V is nondegenerate. In fact, assume that RE(xo) = 0for some xo E E. Then

R(xo y + yxo) = R(xo) ° R(y) + R(y) ° R(xo) = 0 for every y e E.

Since

xo y + yxo = 2(x0, y)e,

we obtain

(xo, y) = 0 y E E

whence xo = 0. Thus RE is injective.

11.3. Orthogonal Representations

A representation of a Clifford algebra CE in a Euclidean space V is calledorthogonal, if

(R(x)u, R(x)v) = E(x, x) (u, v) x e E, u, v e V,

where c = + 1. It is called positive orthogonal, if c _ + 1 and negativeorthogonal, if c = -1.

Thus, if R is positive orthogonal, then

(R(x)u, R(x)v) = (x, x). (u, v).

It follows from this equation that i(x) o R(x) = (x, x). ,,x e E.On the other hand, R(x) o R(x) = R(x2) = (x, x). i, x e E. These relations

imply that i(x) = R(x). Similarly, if R is negative orthogonal, then the trans-formations R(x) are skew.

Proposition 11.3.1. Assume that the inner product in E is positive (respectivelynegative) definite. Then every representation of CE in a Euclidean space isequivalent to a positive (respectively negative) orthogonal representation.

PROOF. Let dim E = k and choose a basis {e,, ..., ek} of E such that

a;) = E b1; (i, j = 1, ... , k).

262 11 Representations of Clifford Algebras

Then

e1 e; + e; e1 = 2c51. e (i, j = 1, ..., k).

In particular, e? = E e, and so the elements e1 are invertible. Thus, if CEdenotes the multiplicative group of invertible elements in CE, then e1 E C.Let G denote the subgroup of CE generated by the elements e (i = 1, ..., k)and e. The relations above imply that G is a finite group.

Now introduce a new definite inner product in V by setting

(u, v)0 = (R(a)u, R(a)v).aeG

Then we have for g e G

(R(g)u, R(g)v)0 = (R(a)R(g)u, R(a)R(g)v)aeG

= (R(ag)u, R(ag)v) = (R(a)u, R(a)v)aeG aeG

= (u, v)0 u, v E V.

Thus,

(R (g)u, R(g)v)0 = (u, v)0 g e G. (11.1)

Next observe that, since both inner products of V are definite of the sametype, there is a linear automorphism'b of V such that

(b(u), b(v)) = (u, v)0 u, v e V.

Now set

P(a) = o R(a) o b -' a e CE.

Then P is a representation of CE equivalent to R. Moreover, Relation (11.1)yields

(P(g)u, P(g)v) = (bR(g) b 1(u), bR(g) b 1(v))

= (R(g) F 1(u), R(g) F 1(v))o

= (F 1(u), F-'(v))0 = (u, v) g e G, u, v e V.

Thus,

(P(g)u, P(g)v) = (u, v)

In particular, if we set P(e1) = o (i = 1, ..., k), then

(6i u, 6i v) = (u, v) u, v E V,

and so

6 o a. = i (i = 1, ..., k).

The Twisted Adjoint Representation 263

On the other hand, we have

6i o 6i = 6i = P(e) _ (e,, e.) l = c. l (1 = 1, ... , k).

These relations yield

61 = E a1 (1= 1, ... , k)

and so, by linearity,

P(x) = E P(x) x E E.

It follows that

(P(x)u, P(x)v) _ (P(x)P(x)u, v) = E (P(x)2u, v)

= E (P(x2)u, v) = c. (x, x) (u, v) x e E.

This relation shows that P is an orthogonal representation. In particular, ifthe inner product in E is positive (respectively negative) definite, P is apositive (respectively negative) orthogonal representation.

The Twisted Adjoint Representation

11.4

Definition. Let E be an n-dimensional vector space with a nondegenerateinner product. Denote by CE the multiplicative group of invertible elementsin CE. Then a representation of the group CE in CE is defined by

ad(a)u = WE(a)ua -' a e CE, u e CE,

where WE denotes the degree involution, ad is called the twisted adjointrepresentation of C.


ad WE(a) _ WE o ad(a) o (11.2)

We show that the kernel K of ad consists of the elements e, t 0. Clearly,

ad() e)u = = u u e CE

and so ) e e K. Conversely, assume that ad(a) = i. Then

WE(a)u = ua u E CE . (11.3)

Setting u = e we obtain WE(a) = a. Now Equation (11.3) yields

au = ua u E CE

whence a e ZE and so a e Z. Now Lemma II, Section 10.11, implies thata = , e, t e T. Since a E CE, it follows that t 0.


11.5. The Clifford Group

Let rE denote the subgroup of CE consisting of those elements a for which Eis stable under ad(a). rE is called the Cli/ ord group of E. Every elementh e E which satisfies (h, h) 0 is contained in FE. In fact, since for x e E

ad(h)x = -hxh-1 = x - 2(h'x)hh

it follows that ad(h)x e E, x e E, and so h e E

Proposition 11.5.1. The Cli/ ord group is stable under the degree involutionand under the antiautomorphism SE (see Section 10.8).

PROOF. (1) Let a E E Then Formula (11.2) yields

ad(wE a)x = WE ad(a)wE 1(x) _ - WE ad(a)x = ad(a)x e E x E E.

Thus, WE(a) E E(2) Let a E E Then a' E rE and so we have

WE(a -1)xa E E x E E.

Applying SE yields

SE(a)x SE WE(a -1) E E x e E

whence, since WE commutes with SE

WE(SE(a))x(SE(a)) -1 E E x e E.

Thus, SE(a) E rE .

11.6. The Map 2E

Recall from Section 10.8 the antiautomorphism a - a defined by a =SE WE(a). Proposition 11.5.1 implies that the Clifford group is stable underthis map.

Now consider the (nonlinear) map o: CE - CE given by

o(a) = as a E CE .

Proposition 11.6.1. If a E rE, then

o(a) _ Aa e Aa E r*

Moreover, the map 2E rE - r* given by AE(a) _ 2a is a homomorphism fromrE to r* (the multiplicative group of r).


PROOF. Since rE is stable under conjugation, a e fE. Now let x e E and sety = wE(a)xa-1. Then y e E and so SE(y) = y. It follows that

SE(a) - 1 xSE(wE a) = WE(a)xa - 1

whence

xSE(wE a)a - SE(a)W E(a)x x e E.

Since

SE(wE a) = a and SE(a) = W E(a),

we obtain

x(aa) = WE(aa)x x e E.

Thus, setting as = b, we have

xb = WE(b)x x e E.

Now write b = bo + b 1 with bo C CE and b 1 C C. Then the equation aboveyields

xbo = box and xb 1 = - b 1 x x C E.

Thus, bo C ZE and b1 C (AZE)1. Now Lemmas I and II, Section 10.11, showthat bo = Aa a and b 1 = 0 whence b = ,a e and so 'bE(a) = e. Finally,since a is invertible, it follows that Aa 0 and so C r*.

Now consider the map AE : rE - r* defined by

E(a) = AE(a)e a C rE .

To show that 2E is a homomorphism, let a e rE and b e E Then

E(ab) = ab ab = abba.

By the first part of the proposition,

bb = AE(b) e.

It follows that

E(a b) = AE(b)aa = AE(b)AE(a)e = AE(a)AE(b)e

whence

AE(ab) = AE(a)AE(b)

Corollary I. The map 9 satisfies

8(WE(a)) = 9(a) a e 1E.

PROOF. In fact,

e(W E(a)) = WE(a)WE(a) = WE(a)WE(a) = WE(aa) = ,E(a) - e = 0(a).


Corollary II. The homomorphism 2E satisfies the following two relations:

1. AE(WE(a)) _ AE(a) aErE.

2. AE(ad(b)a) _ AE(a) a, b e E.

PROOF. (1) follows from Corollary I.

(2) Let a E rE and b e E Then, since 2E is a homomorphism, the firstrelation yields

AE(ad(b)a) _ AE(w(b)ab -1) _ AE(w(b))AE(a)AE(b) - 1

_ AE(b)AE(a)AE(b) - 1 = AE(a)

Proposition 11.6.2. Fix a E rE and let to denote the restriction of ad(a) to E.Then to is an isometry.

PROOF. Since for x e E

O(x)= - (x, x) e,

it follows that

AE(x) _ - (x, x) x e E.

Now Part (2) of Corollary II to Proposition 11.6.1 yields

(ad(a)x, ad(a)x) _ - AE(ad(a)x) _ - AE(x) _ (x, x)

whence

(tax, tax) _ (x, x) x E E.

In particular, consider a vector h e E with (h, h) 0 (recall that thenh e fE). Since

hx + xh = 2(x, h)e,

we obtain

(h, x)ih(x) = x- 2 h h x x E E.

This equation shows that

ih(h) _ - h,

while

ih(Y) = Y if (h, y) = 0.

Thus th is the reflection in the plane orthogonal to h.


11.7. The Homomorphism E: rE -- O(E)

Let O(E) denote the group of isometries of E. Then a homomorphism

E : rE - O(E)

is defined by

(FE(a) = to a E rE .

Proposition 11.7.1. The homomorphism (FE is surjective.

PROOF. Let h e E be a vector such that (h, h) 0 and let ph denote the reflec-'tion in the plane perpendicular to h. Then (FE(h) = ph.

By Lemma I below, every isometry of E is generated by reflections andso the proposition follows.

Lemma I. Let E be an n-dimensional vector space with a nondegenerate innerproduct. Then every isometry t of E is the product of at most n + 1 reflections.

PROOF. We show first that if a and b are any two vectors such that (a, a) =(b, b) 0, then there is a reflection p such that p(a) = ± b.

In fact, since

(a + b, a + b) + (a - b, a - b) = 4(a, a) 0,

it follows that (a + b, a + b) 0 or (a - b, a - b) 0. Replacing b by -bif necessary, we may assume that (a - b, a - b) 0. Now set h = a - band consider the reflection

P() (h, h)( )Then p(a) = a - (a - b) = b.

Now we prove the lemma by induction on n. If n = 1, then ix = E x,E = ±1, and so t is a product of at most two reflections. Suppose now thatthe lemma holds for a space with dimension n - 1 and let t be an isometry ofE (dim E = n). Choose a vector a e E such that (a, a) 0 and set b = i(a).Then (b, b) = (a, a) and so, by the remark above, there is a reflection p suchthat p(a) = ± b. Now set

ti =p-lot.Then t i(a) = + a and so t i restricts to an isometry of the orthogonal com-plement, E1, of a. Thus, by induction, t i is the product of at most n reflectionsof E1. Every such reflection extends to a reflection of E. Thus, t is the productof at most n + 1 reflections and the induction is closed.

Corollary I. The Clifford group rE is generated by the elements h e E whichsatisfy (h, h) 0.


PROOF. Let a E rE and set (FE(a) = t. By Lemma I, t is of the form

= Phi°".°Phr h1EE

where (h1, h.) 0. Since (FE(h1) = ph;, it follows that

FE(a - t h l ... hr) = t 't = 1.

Thus,

a-1h1 ...hr = 2*Ef*and so

a h, _ (2 tht)h2 ... h,..

Corollary II. The homomorphism (FE satisfies

det (FE(a) a = WE(a) a E rE .

PROOF. Consider the homomorphism p: rE - I~E given by

(p(a) = det (FE(a) a.

Then we have, for x e E, (x, x) 0,

(p(x) _ - x = WE(x)

and so (11.4) holds in this case. Now apply Corollary I.

11.8. The Group Pin

(11.4)

From now on E will be a real vector space with a (nondegenerate) innerproduct of type (p, q). We shall write CE = C(p, q) and rE = r(p, q). Recallfrom Section 11.6, the homomorphism AE : r(p, q) - The group Pin(p, q)is the subgroup of r(p, q) consisting of the elements a which satisfy

AE(a) = ±1.

Since, for x e E, AE(x) _ - (x, x), it follows that all the vectors x e Efor which (x, x) = ±1 are contained in Pin(p, q). In particular, - e e Pin(p, q).Moreover, Corollary I to Proposition 11.7.1 shows that Pin(p, q) is generatedby these vectors.

Next, denote by O(p, q) the group of isometries of U r and consider thehomomorphism (FE: r(p, q) - O(p, q) defined in Section 11.7. We shall showthat the restriction (F of E to Pin(p, q) is still surjective. In fact, let a e O(p, q).Then, by Proposition 11.7.1, there is an element b e r(p, q) such that(FE(b) = a. Now set

ba =

12E(b) Itr2 .


Then

, a = 2E(b) = + 1E()b

-E( )I

and so a e Pin(p, q). Moreover, tE(a) = 'E(b) = a. Thus tE restricts to asurjective homomorphism

: Pin(p, q) - O(p, q)

Now we show that ker I = S°, where S° is the subgroup of Pin(p, q)consisting of the elements a and - e. In fact, if a e ker I, then a e ker ad,and so a = 2. e (see Section 11.4). It follows that AE(a) = 22. On the otherhand, since a e Pin(p, q), ( 2E(a) ( = 1. These relations yield 22 = 1 whence2= + 1 and so a = ±e.

In view of the results above we have the exact sequence of groups

1 - S° ` Pin(p, q) O(p, q) - 1

where i denotes the inclusion map.

11.9. The Group Spin

Let Spin(p, q) denote the subgroup of Pin(p, q) consisting of the elementswhich satisfy WE(a) = a. Thus,

Spin(p, q) = Pin(p, q) n C°(p, q).

Formula (11.4) shows that an element a e Pin(p, q) is contained in Spin(p, q)if and only if det t(a) = + 1; that is, if and only if P(a) is a proper isometry.Thus the homomorphism I restricts to a surjective homomorphism kfrom Spin(p, q) to the group SO(p, q) of proper isometries. SinceS° c Spin(p, q) it follows that the kernel of this homomorphism is again S°.Thus we have the exact sequence

1 - S° ` Spin(p, q) Z SO(p, q) - 1.

PROBLEMS

1. Show that the homomorphism E and the bilinear function QE (see Section 10.14,Problem 3) are connected by the relation

)E(a) = QE(a, a) a E FE.

2. Let B" be an n-dimensional Euclidean space and write Pin(n, 0) = Pin(n) andSpin(n, 0) = Spin(n). Determine the groups Pin(n) and Spin(n) (n = 1, 2, 3) explicitly.In particular, show that the group Spin(3) is isomorphic to the group of unitquaternions.


The Spin Representation

11.10

Let F be a 2n-dimensional Euclidean space. Recall that a complex structurein F is a linear transformation J which satisfies

J 2 = - l and (Jx, J y) = (x, y) x, y e E.


j= -Jand so J is a skew transformation.

Next, consider the 2n-dimensional complex vector space E = C Q Fand define an inner product in E by

(2 O x, µ O y) = 2µ(x, y) 2, µ E C, x, y e F.

Let w denote the linear transformation of E given by

w(2 Q x) = i2 Q Jx 2 E C, x e F.

Then we have

w2(2Qx) _ (-2)Q(-x) _ 2®xwhence

cc,2=l.

Thus w is an involution. Moreover,

(w(2 O x), 2 Q x) = (i2 Q Jx, 2 Q x) = i22(Jx, x) = 0 2 E C, x e E

and so w is skew.Let E1 and E2 denote the following subspaces of E:

E1 = {x (wx = x} and E2 = {x (wx = -x}.

Then Proposition 10.14.2 shows that

CE L(A E1).

Moreover, the isomorphism CE L( A E 1) is obtained as follows: Let(p: E - L( A E 1) be the linear map given by

gp(x)u = x 1 A u+ i(x2)u x e E

x = x1 Q x2 x1 E E1, x2 E E2

and extend it to a homomorphism RE : CE - L( A E 1). Then this homo-morphism is an isomorphism,

RE : CE 3 L(A E 1).

In particular, the representation RE is irreducible.

The Spin Representation

11.11. The Spin Representation

271

Recall that the inclusion map j : F - E induces a homomorphism jc : CF -p CE.Thus the representation RE determines a representation RF of CF by complexlinear transformations of n E 1 given by

RF(a) = RE(1 Q a) a e CF.

RF is called the spin representation of CF. A representation of a real algebrain a complex vector space V is called irreducible if the only stable (complex)subspaces are W = V and W = (0). We show that the spin representation isirreducible. In fact, assume that W is a stable subspace of n E 1. Let b e CE andwrite

b = 2Qa AEC,aECF

(see Section 10.16). Then we have for w e W

RE(b)w = RE(2 O a)w = ARE(1 Q a)w = ),RF(a).

Since W is a complex subspace of n E 1, it follows that RE(b)w E Wand so W isstable under RE. But RE is irreducible and so it follows that W = n E 1, orW = (0).

11.12. The Hermitian Inner Product in A E 1

Since E = C Q F, we have the complex conjugation z H z in E given by2 Q x H Z Q x. Now introduce a positive definite Hermitian inner productin E by setting

(z 1, Z2)H = (z 1, z2) z 1, z2 E E.

Then we have an induced Hermitian inner product in n E. It is given by

(z 1 n ... n Z, W 1 n ... n Wp)H

= (z1 n n Z ,w1 n n w,)

Proposition 11.12.1. Let x e F. Then the transformation RF(x) is Hermitiansymmetric.

PRooF. Let iH(z) denote the substitution operator in n E 1 corresponding tothe Hermitian inner product. We show that

iH(z) = 1(2) z E E. (11.5)


In fact, let z, z2, ... , z p E E 1 and w1,..., w, E E 1. Then

(iH(Z)(w 1 n ... n Wp), Z2 n ... A Zp)H

= (W 1 A "A WP, Z A Z2 A ... A Zp)H

= (W 1 n ... A W p, Z A Z 2 A ... A Z p)

= (i(Z)(W 1 n ... n w y), z2 A ... A Zp)

= (i(Z) (W 1 n ... A Wp), Z2 A ... A Zp)H,

and so Formula (11.5) follows.Next, let x e F and set

x1 =2(x+wx)=2(1ax+i®Jx)and

x2=2(x-wx)=2(1Qx-IQJx).These relations show that x2 = xl.

Now consider the linear transformation RF(x) of A E 1. Then, sincel(x2) = i(x 1) = lH(x 1)

(RF(x)u, V)H = (x 1 A U, V)H + (l(x2)u, V)H

= (u, 1H(x 1)V)H + (IH(x 1)u, V)H

= (u, l(x2)v)H + (u, x 1 A V)H = (u, RF(x)v)H

and so the proposition is proved.

Corollary. If x e F, then

(RF(x)u, RF(x)v)H = (x, x)(u, V)H u, v e A E 1.

In particular, f x is a unit vector, then RF(x) is a unitary transformation.

PROOF. In fact, by the proposition,

(RF(x)u, RF(x)v)H = (RF(x)2u, v)H = (x, x `u, v)H

11.13. The Half Spin Representations

The spin representation restricts to a representation RF of the subalgebraCF in A E 1. Now write

(A E1)+ = ME1 and (A E1)- = ME1.p even p odd

Then the spaces (A E 1) + and (A E 1) - are stable under the transformationsRF(a), a E CF °. Thus we have induced representations

RF : CF - L(A E 1) +

The Wedderburn Theorems 273

and

RF :C F - L(n E1)-

called the half spin representations.

Proposition 11.13.1. The homomorphisms RF and RF are isomorphisms. Inparticular, the half spin representations are irreducible.

PROOF. First observe that the maps RF and RF are injective. SinceE 1) + = 2" -1, we have dims L(n E 1) + = 2" - 2 and so dimR L(n E 1) +

= 22n -1. On the other hand, dim CF = 22" - 1. Thus RF is an isomorphism.The same argument applies to R.

The Wedderburn Theorems

In this paragraph we establish an algebraic structure theorem which will be used laterto study the representations of C8( - ).

11.14. Invariant Linear Maps

Let E and F be finite-dimensional vector spaces and let R be a representationof the algebra L(E) in F. Thus R is a homomorphism

R : L(E) - L(F).

A linear map a : E - F will be called R-invariant, if it satisfies

a o P = R(qp) o a qP E L(E). (11.6)

The R-invariant linear maps form a subspace of L(E; F) denoted by LR(E; F).A linear transformation ll/ of F is called R-invariant, if

R(ip) ° il/ = i/i ° R(ip) cP E L(E). (11.7)

These transformations form a subspace of L(F) denoted by LR(F).

11.15. The Isomorphism R

Let

R:LR(E;F)OE-+F

be the linear map given by

R(a Q x) = a(x) a e LR(E; F), x e E.


We show that the diagram

LR(E;F) Q E R F

LR(E;F)QER

F

commutes. In fact, let p e L(E) and a E LR(E;F). Then we have, in view of(11.6),

(FRL(1 O P)(a O x)] = R(co)FR(a O x)

Thus,

(Z O 4,) = R(co) ° (FR (p E L(E). (11.8)

11.16

Theorem 11.16.1 (Wedderburn). R is a linear isomorphism,

(FR : LR(E ; F) O E 4 F.

PROOF. We shall construct an inverse map

- LR(E; F) O E.

Choose a pair of dual bases {e*v}, {ev} (v = 1, ..., n) of E* and E and set

Pv = TE(e*µ O ev),

where TE : E* Q E 4 L(E) is the canonical isomorphism defined by (6.5).Observe that TE satisfies

(p0 TE(x* Q x) = TE(x* Q cpx) (p e L(E). (11.9)

Next, define linear maps Tµ:F - L(E; F) by

Tµ(y)x = <e*v, x>R(cpv)y x e E, y e F. (11.10)v

Lemma. The maps Tµ have the following properties:

1. Tµ(y) E LR(E; F), y E F.2. Tµ(ax) = <e*µ, x>a, a E LR(E; F).3. T '(y)eµ = y, y e F.

µ

PROOF. (1) It has to be shown that

T µ(y) 0 ( = R(4,) 0 T µ(y) i/i e L(E).


Fix y E F. Then, by (11.10)

R(p)Tµ(y)x = <e*v, x>R(p °

On the other hand,

T µ(y)cox = <e*v, )yv

Hence we have to establish the relation

275

<e*v, x>R(co ° <e*v, (11.11)v v

Formula (11.9) yields, for x* = e*µ and x = ev,

P ° = TE(e*µ O coev)

It follows that

<e*v, x>cP ° _ <e*v, x> TE(e*µ O Peti,)

v v

= TE(e*µ O px) = TE (e*tL Q <e*v, t x>evv

_ <e*v, (px> q .

v

Thus,

<e*v, <e*v, (Px>qvv v

Applying R to this formula we obtain (11.11).(2) Let a E E. Then

Tµ(ocx)a = <e*v, a>R(q )(ax)v

_ <e*v, a>a(q,vx)v

Since

it follows that

(pox = <e*µ, x>ev,

Tµ(ocx)a = <e*v, a><e*µ, x>a(ev)

_ <e*µ, x> ' a(a)

whence

Tµ(ax) _ <e*µ, x>a.


(3) In fact, since

TE (e*% ®eV) = 1,

it follows that

T L(y)e, = Y Y E F.L

This completes the proof of the lemma.

Now let

be the linear map given by

11'(Y) _ TL(y) © eµ .L

Then we have for a e LR(E; F) and x e E

kIIbR(a Q x) = '(ax) _ TL(ax) Q eµL

_ <e*L, x>a Q eL = a Q xL

whence

On the other hand, for y e F,

R '41(Y) = R TL(y) O eL = T L(Y)(eL) = Y

whence FR qi = i. It follows that 'b is an isomorphism.

Corollary.

dim F = dim LR(E; F) dim E.

Thus dim E divides dim F.

11.17. The Isomorphism BR

Observe that the composition map

L(F) x L(E ; F) - L(E ; F)

restricts to a bilinear mapping

LR(F) x LR(E; F) - LR(E; F).


To simplify notation we set LR(E; F) = U. Then a linear map

8R : LR(F) - L(U)

is defined by

0R(Ii) = ° a / E LR(F), a e U.

Clearly,

OR(i/i2 ° 'i'1) = OR(i/i2) ° OR(i/i1)

and so 9 is an algebra homomorphism.

11.18

Theorem 11.18.1. (Wedderburn). eR is an isomorphism.

PROOF. We construct an inverse map

L(U) - LR(F)

Let y e L(U) and set

277

R(Y) = R ° (Y 0 1)0 1 1.

Then SPR(Y) E L(F). Since, by (11.8), R(p) ° cZR(Y) =CR(Y) ° R(ip), p e L(E),it follows that SZR(Y) E LR(F).

Now we show that

eR ° R = l and R ° eR = 1. (11.12)

In fact, let y e L(UR) and a e U. Then

[(OR ° R)(Y)]a = [OR(tR ° (Y © l) ° I ' )] a

Now fix x e E. Then; by definition of tR,

It follows that

[OR i)(a ® x) = tR(Y(a) © x) = y(a)x.

Thus,

(eR ° Y Y E L(U)

and so the first relation (11.12) follows.To establish the second relation, let i/i e LR(F). Then

= R ° (0R(fr) ©l) ° R ' .


From the proof of Theorem 11.16.1 we have for y e F

R 1(Y) = TL(y) O eµµ

whence

(OR(l//) OX l)tR 1(Y) = L, OR(lfr)T µ(Y) O eµ .µ

It follows that

OO l) I 1(Y) = L, µ(Y)eµ] = L, T(y)e] = (Y)µ µ

(see the lemma in Section 11.16). We thus obtain

R ° (OR(S) ® 1)0 R 1

i.e.,

0R) = l/J i/i e LR(F)

This completes the proof of Theorem 11.18.1.

11.19

Theorem 11.19.1. Let A be an associative algebra with unit element a and letR be a representation of the algebra A Q L(E) in a vector space F. Then thereexists both a representation R of A in a vector space U and an isomorphismI: U Q E 3 F which makes the diagrams

UQE R (a)®q UQE

1i a E A, cP E L(E) (11.13)

F

commute. Thus R is equivalent to R Q i where i denotes the standard repre-sentation of L(E) in E.

PROOF. Define representations P and Q of A and L(E) in F by setting

P(a) = R(a Q i) a E A

and

Q((P) = R(e O q,) (P E L(E).

The Wedderburn Theorems 279

Let LQ(F) denote the subspace of F which is invariant under Q. We show that

P(a) E LQ(F) a E A.

In fact, let p e L(E). Then

P(a) ° Q(q,) = R(a O l) ° R(e O P)

= R(a O q,) = R[(e O q,) ° (a O l)]

= R(e O q,) ° R(a O l)

= Q(q,) ° P(a).

Now set U = LR(E; F). Then, by Theorem 11.16.1, there is an isomorphism

UxE +Fsuch that

1 ° (l O co) = e L(E). (11.14)

By Theorem 11.18.1 there is an algebra isomorphism

Q : L(U) 3 LQ(F).

It is defined by

Q(y) = ° (y O l) ° -' y e L(U). (11.15)

Thus a representation Rv of A in U is given by

Ru(a) = SZ - 'P(a) a E A.


P(a) = QRu(a) = ° (Ru(a) Q t) o I 1

Thus,

° (Rv(a) Q l) = P(a) ° a e A. (11.17)

Relations (11.14) and (11.17) yield

° (Ru(a) O 4') = ° (Ru(a) O l) ° (l O 4:))

= P(a)°t 6(i O co)

= P(a) ° Q((p) ° Y

= R(a O 4,) ° (1'.

Thus Diagram (11.13) commutes and the proof of Theorem 11.19.1 is com-plete.


Representations of Ck(- )

11.20. The Radon-Hurwitz Number

In this section we shall denote Ck( -) simply by Ck. Let R be a representationof Ck in a Euclidean n-space IJn. We show that then k - n - 1. In fact, byProposition 11.3.1 we may assume that R is a (negative) orthogonal repre-sentation. Let {e1, ..., ek} be an orthonormal basis of fik and set R(e1) =(i = 1, ..., k). Then we have the relations

t6io6 + 6;o6 = - 25,in particular, 6? = - i (i = 1, ..., k). Moreover, the remark precedingProposition 11.3.1 shows that o is skew (i = 1, ... , k). Now fix a unit vectora e n and set a,(a) = a,(i = 1, ... , k). Then

(a, a3= (a, o 1 a) = 0 (i= 1, ... , k)

and

(at, a) = (a' a, 6; a) = (6; 6 a, o a).

It follows that

2(ai , a) = - ((6; o + o o )a, a) = a) = 2

whence

(a1, a) = &, (i, J = 1, ..., k).

Thus the vectors a, a1, ..., ak form an orthonormal (k + 1)-frame in Rn.This implies that k + 1 - n. Thus to every n > 1 there is a largest k > 0such that Ck represents in pn. This number is called the Radon-Hurwitznumber of n and will be denoted by K(n). It follows from the above that

K(n) - n - 1. (11.18)

Proposition 11.20.1. The Radon-Hurwitz number satisfies the functionalequation

K(16n) = K(n) + 8 n > 1.

PROOF. In view of Theorem 10.17.1, Section 10.21, and the table in Section10.20, we have an isomorphism

'V: C,,8 Ck O L(U 16).

Thus, if P is a representation of Ck in lin, then

Q=(PO1)0k

is a representation of Ck + 8 in [jn a [ 16 16n It follows that

K(16n) > K(n) + 8.

Representations of Ck( -) 281

Conversely, let Q be a representation of C,, + 8 in g 16n and set R = Q o I11.Then R is a representation of C,, Q C8 in 16n By Theorem 11.19.1 (appliedwith A = C,, and E = p 16) there is a representation of C,, in a vector spaceU, where

U 0 g16n

It follows from this relation that dim U = n. Thus,

K(n) > K(16n) - 8.

Theorem 11.20.2. Write

n= a>0,0-b-3,godd.Then the Random-Hurwitz number of Din is given by

K(n)=8a+26- 1 n> 1.

In particular, fn is odd, then K(n) = 0.

PROOF. In view of Proposition 11.20.1 we have only to show that

K(2". q) = 2b - 1 0 - b < 3, q odd.

This will be proved in Lemma II of the next section.

11.21

Lemma I. Let q be odd and 0 - b < 3. Then

K(26 . q) <7

PROOF. Assume that C,, represents in n and k > 8. Write k = l + 8, l > 0.Then C, ^C1 Q C8 ^C1 Q L('6) represents in Rn. Thus, by Theorem11.19.1, 16 divides n and so n cannot be of the form 2b q(0 - b - 3, q odd).

Lemma II. Let q be odd and 0 - b - 3. Then

K(26 q) = 2 b - 1.

PROOF. It has to be shown that, for odd q,

1. K(q) = 0.2. K(2q) = 1.3. K(4q) = 3.4. K(8q) = 7.

(1) K(q) = 0: By Lemma I, K(q) < 7. Thus it has to be shown that if C,,represents in D and 0 - k < 7, then k = 0. By the table in Section 10.20


every Ck (1 <_ k < 7) contains C as a subalgebra. Thus a representationof Ck (1 < k < 7) in determines a representation of C in . This isimpossible, since q is odd. It follows that q = 0.

(2) K(2q) = 1: We show that

K(2q) <_ 1. (11.19)

Lemma I implies that K(2q) <_ 7. Now the isomorphisms in Section 10.20show that Ck contains 0-0 as a subalgebra if 2 < k < 7 and so a representationof Ck (2 <_ k < 7) in 2j induces a representation of 0-0 in 2j This is im-possible since 2q is not divisible by 4 (see Lemma III below). Thus, k = 1and so (11.19) is proved.

On the other hand, C1 C represents in 2j and so

K(2q) >_ 1.

It follows that K(2q) = 1.

(3) K(4q) = 3: We show first that

K(4q) _< 3. (11.20)

By Lemma I, K(4q) <_ 7. Thus we have to show that for 4 < k < 7 the algebraCk does not represent in By the isomorphisms in Section 10.20 everysuch algebra is of the form

Ck = Bk O

where Bk contains 0-0 as a subalgebra. In fact,

B4 = 0-0, B5 = 0-0 Q C,

B6 = 0-0 Q 0-0, B7 = 0-0 Q 0-0 Q 0-0 Q 0-0.

Now let R be a representation of Ck in Then Theorem 11.19.1 (appliedwith A = Bk and E = 2) shows that there is a representation R of Bk ina vector space U where

X44 U O 2.

Since 0-0 c Bk, R determines a representation of 0-0 in U. Thus dim U isdivisible by 4 (see Lemma III below) and hence dim is divisible by 8.This is impossible since q is odd, and so (11.20) follows.

On the other hand,

K(4q) _> 3. (11.21)

In fact, write

C3 = 0-0 Q 0-0

(see Section 10.20). Then a representation R of C3 in (0-0) is given by

R(p Q+ q)x = xE 0-0.

Representations of Ck( -)

Thus

RQ...QR4

is a representation of C3 in D" and (11.21) follows.(4) K(8q) = 7: By Lemma I,

K(8q) < 7.

To show that

K(8q) > 7,

we construct a representation of C-, in 8q. Write

C-, = Q

(see Section 10.20) and set

R(a Q 1) = a a,1 E L(8).

Then R is a representation of C-, in and so

RQ...QR4

283

is a representation of C-, in 8q. Thus, K(8q) > 7.This completes the proof of Lemma II and hence the proof of the theorem.

EXAMPLES

K(2)=21-1=1, K(4)=22-1=3K(6)=21-1=1, K(8)=23-1=7

K(10)=2'-1=1, K(16)=8+2°-1=8.

Lemma III. Let R be a representation of 0-II an n-dimensional vector space E.Then n is a multiple of 4.

PROOF. Write

R(p) (x) = p. x p e 0-0, x e E.

We shall say that a family of vectors x,, ..., xk E E generates E over 0-[l, ifevery x e E can be written in the form

k

x = pi.xi piE 0-0.i=1


Let m be the least number such that E is generated by m vectors and letX1 , ... , xm be such a family. Then it is easy to show that the relation

m

i= 1

implies that pt = 0 (i = 1, ... , m). Now choose a basis {e, a 1, e2, e3} of 0-fl.Then it follows that the 4m vectors

xi , e 1 xi , e2 xi , e3 xi (l = 1, . m )

form a basis of E over B. Thus, n = 4m.

11.22. Orthogonal Multiplications

Let E and F be Euclidean spaces. An orthogonal multiplication between Eand F is a bilinear mapping E x F - F, denoted by (x, y) - x y with thefollowing properties :

i. Ix'YI = Ixllyl, xeE, yeF.ii. There is an element e E E such that e y = y, y E F.

Property (i) implies, in view of the symmetry of the inner products, that

(x 1 ' Y x2 .y) _ (x 1, x2) I Y I2 x1, x2 EE, yeF

and

(x'Y1, x'Y2) = IxI2(Y1, Y2) xeE,Y1, Y2 EF.

As an example consider the algebra of quaternions (E = F ^0-[l) (see Section7.23 of Linear Algebra).

Given an orthogonal multiplication denote the orthogonal complementof e by E 1. Every vector a e E 1 determine a linear transformation 6a of Fgiven by

6a(Y) = a ' Y Y E F.

Relations (i) and (ii) imply that this transformation satisfies

(6a Y, 6a Y) =1 a l 2 ' I Y 12

and

(6a Y' Y) = 0 a E E1, y e F.

In particular, if I aI = 1, then 6a is a rotation of F.

Representations of Ck( -)

11.23. Orthogonal Systems of Skew Transformations

285

Let F be an n-dimensional Euclidean space and let {o, ..., 6k} be a familyof skew linear transformations of F. The family {o, ..., 6k} will be calledorthogonal, if

(ay, 6; Y) = (Y, y) y e F.

This relation is equivalent to the relation

0,0cr + 6jo6 =

In fact, (11.22) implies that

(0 Y 6 j z) + (6i z, 6; Y) = 2(y, z) 5, y, z e F.

Since the Q are skew, it follows that

(6 j 6i Y, z) + (z, 6i 6; Y) = - 2(Y, z)

whence

(6jai -+ 6iaj)y - -2(Sijy y e F.

Conversely, assume that (11.23) holds. Then we have

(6i Y 6 j Y) _ - (6; 6i Y, y) = 2i5r,(Y, y) + (a16; Y, y)

= 2511(y, y) - (6; Y, o 1 Y) y e F.

It follows that

(11.22)

(11.23)

(61Y, 6; Y) = 51,(Y, y).

Every orthogonal family of k skew transformations of F determines anorthogonal multiplication between +1 and F. In fact, choose an ortho-normal basis {e, a 1, ..., ek } in k+ 1 and set

x ' Y = AY + A'a1(Y) x e 1, y e F,

where

Then, clearly, e y = y, y e F. Moreover,

x y 2 = 221y12 + 22 21(Y, o 1 Y) + A`A'(a. Y, ay)

= 221Y12 + aAiVtyJ2

= (22 + 2121). FyI2 =

and so we have an orthogonal multiplication k +1 x F - F.


Conversely, let k +1 x F - F be an orthogonal multiplication. Denotethe orthogonal complement of e by k and choose an orthonormal basis{el, ... , ek} of R. Define 6 by

a.(y) = e, y y E F.

Then the 6 form an orthogonal system of skew transformations.

11.24. Orthogonal Multiplications and Representations of Ck

Let E x F - F be an orthogonal multiplication and denote the orthogonalcomplement of e by E 1. Consider the linear map P : E 1 - L(F) given by

P(a)y = a y a E E, y E F.

Then

(y, P(a)y) _ (e, a) (y, y) = 0

and so the transformations P(a) are skew. This implies that

(P(a)2y, z) _ - (P(a)y, P(a)z) _ - (a ' y, a ' z) _ - (a, a) (y, z) y, z E F,

whence

P(a)2 = -(a, a). i.

Now introduce a negative definite inner product in E 1 by setting

(a, b) - _ -(a, b) a, b e E 1.

Then the equation above reads

P(a)2 = (a, a) - l a E E 1 .

Thus P extends to a homomorphism from the Clifford algebra Ck,(dim E 1 = k) into L(F) and so it determines a representation of Ck in F.

Conversely, let R be a representation of Ck in F. By Proposition 11.3.1R is equivalent to a (negative) orthogonal representation P of Ck in F. Psatisfies the relations

P(a)2 = P(a2) _ (a, a) - l = -(a, a). i a e E 1

and

(P(a)y, P(a)y) _ -(a, a) ' (y, y) _ (a, a) ' (y, y) a e E 1, y e F.

In particular, the transformations P(a) are skew.Now set E _ (e) Q k and define a bilinear mapping E x F - F by

setting

x'y=Ay+P(a)y,where x e E, y e F, x= Ae + a, a e k.


Then, clearly, e y = y. Moreover,

I x y 12 = 221 y 12 + 22(y, P(a)y) + I P(a)y 12 = 22 I y 12 + (a, a) ' I y 12

=IxI21yI2 xeE,yeF,and so this bilinear mapping is an orthogonal multiplication between Eand F.

Thus there is a one-to-one correspondence between orthogonal multi-plications E x F -* F and representations of Ck in F (dim E = k + 1).

Theorem 11.24.1. Assume that there exists an orthogonal multiplication" x " -* Rj". Then n = 1, 2, 4, or 8.

PRo0F The orthogonal multiplication in " determines a representation ofC,_1 in . Thus, n - 1 <_ K(n). On the other hand, in view of Relation(11.18) K(n) < n - 1. Thus,

K(n) = n - 1. (11.24)

Now write

a>0,0<b<-3,goddThen, by Theorem 11.20.2,

K(n) = 8a + 2b - 1

and so Equation (11.24) implies that

8a + 2b = 16a 2b q;

that is,

1) 0<b<-3.It follows that

(11.25)

8a>16a-1.Since 8x < 16X - 1 for x> 1, x e , we obtain a = 0. Now Formula (11.25)shows that q = 1. Thus, n = 2b (0 -< b < 3); i.e., n = 1, 2, 4, 8.

Remark. If n = 1, 2, 4, or 8, there are indeed orthogonal multiplicationsin p". For n = 1 and n = 2 we have the ordinary multiplication of real(respectively complex) numbers, for n = 4 the algebra of quaternions andfor n = 8 the algebra of Cayley numbers (see Problem 5).

11.25. Orthonormal k-Frames on S"-1

Let S"' be the unit sphere in p". A (continuous) orthonormal k -frame onS' is a system of k continuous maps X1: S" -1 -p satisfying

(x, X.(x)) = 0 and (X 1(x), X;(x)) = x e S'.


Now let 61, ..., 6k be an orthonormal system of k skew transformationsof " and set

X.(x) = 61(x) x e S" - 1, (i = 1, ..., k).

Then we have for x e

(x, X.(x)) _ (x, ajj(x)) = 0

and

(X 1(x), X,{x)) =

Thus every orthogonal system of k skew transformations of " determinesan orthonormal k-frame on S" -' . Now, combining the results of Sections11.23 and 11.24 and Theorem 11.20.1, we see that the (n - 1)-sphere admitsan orthogonal k-frame with k = K(n).

1-frame on S 1,

3-frame on S3,1-frame on Ss,

Hence there is an orthonormal 7-frame on S',1-frame on S9,

8-frame on S 15,

9-frame on 531

(see the examples in Section 11.21).

Remark. F. Adams has shown that this is in fact the best possible result;that is, there are no orthonormal k-frames on S" - ' if k> K(n), (cf. Ann. ofMath. 75 (1962) p. 603.).

PROBLEMS

1. Suppose that an orthogonal multiplication is defined in a Euclidean space E. Denotethe orthogonal complement of e by E1. Show that the following conditions areequivalent:

1. -(x, y) x,yEE1.2. x2 = xEE1.

2. Cross-products in Euclidean spaces. A cross-product in a Euclidean space F is abilinear mapping F x F -+ F, denoted by x, which satisfies the conditions:

1. (x, x x y) = 0 and (y, x x y) = 0 for all x, y E F.2. lx x yl2 = (x,y)2 forallx,yEF.i. Show that a cross-product is skew-symmetric.

ii. Let E be a Euclidean space with an orthogonal multiplication which satisfiesthe relation

(x y, e) _ (x, y) x, y E E.


Let F be the orthogonal complement of a and denote by n the projection nx =x- (x, e)e, x e E. Define a bilinear mapping F x F-+ F by

x x y= n(x y) x, y e F.

Show that this map is a cross-product in F.

3. The complex cross-product. Let C3 be a 3-dimensional complex vector space with apositive Hermitian inner product (, ). Choose a normed determinant function D.Then the complex cross-product of two vectors a and b is defined by the equation

(a x b, x) = D(a, b, x) x e C 3.

Show that the complex cross-product has the following properties :

1. ( l a l + 2a2) x b = .Z1(a 1 x b) + Z2(a2 x b) for all ) 2 E C .2. (a x b, a) = 0 and (a x b, b) = 0.3. a x b = -bxa.4. (al x b1, a2 x b2) _ (al, a2). (b1, b2) - (a1, b2)(b1 , a,).

5. la x bl2

= 1a121b12 - I(a b)12.

6. a x (b x c) _ ()b - (a, b)c.4. Orthogonal multi plications in C4. Let C4 be a complex 4-dimensional vector space with

a positive definite Hermitian inner product. Choose a normed determinant function0 and a unit vector e. Let C3 denote the orthogonal complement of e. Then a normeddeterminant function D is defined in C3 by the equation

D(y1, .y2, .y3) = O(e, .y1, .y2, .y3) .yi E C3.

Define a multiplication in C4 by the equations

xl .yl = -(x1, y1)e + x1 x yl/1,e .y 1 - /2y 1 x 1 (/1,e) _ ) x 1 ) E C

()).(e)= )µ e # u E C.

i. Prove the formula

Ix yl2 = Ixl2. lyl2 x, y E C4.

ii. The conjugate of an element x e C4 is defined by x = a,e - x1 , where x = Ae + x 1,E C, x 1 E C3. Show that

=x = xeC4.

iii. Verify the formulas x y2 = (x y) y and x2 y = x (x y).

5. The algebra of Cayley numbers. Let E be an 8-dimensional Euclidean space. Choosea complex structure J in E (that is, a skew transformation J which satisfies J2 = - i)and use it to make E into a 4-dimensional complex space H. Define a Hermitianinner product in H by setting

(x, y)H = (x, y) + i(x, J y) x, y E H.

Choose a normed determinant function 0 in H.

i. Show that the multiplication defined in Problem 4 makes E into a real (non-associative) division algebra with a as unit element. It is called the algebra of Cayleynumbers.

290

ii. Verify the relations

and

11 Representations of Clifford Algebras

(ax, ay) = a 2(x, y) x, y E E

(xa, ya) = (x, y) I a f 2.

6. The cross-product in B. With the notation and hypotheses of Problem 5, let Fdenote the (7-dimensional) orthogonal complement of a in E.

i. Prove the relation

(xy, e) = - (x, y) x, y E F.

Conclude that the Cayley multiplication in E determines a cross-product in F.

ii. Let x E E and write x = ae + y where a E R, y E F. Show that the conjugateelement of x (see Problem 4, (ii)) is given by x = ae - y. Conclude that if x is anon-zero Cayley number, then

xx

--

(x, x)

iii. Show that the relation x x y = 0 holds if and only if y = fix, ,. E fly.

7. Recall from Section 10.20 that C6(-) L(B ). Construct an explicit isomorphismb : C6(-) 3 L(B ) in the following way: Regard 6 as the underlying real vectorspace of C 3 and define a negative inner product in B6 by setting (x, y) _ = -(x, y),where (x, y) = Re(x, y)H. Show that the map gyp: L(B 8) given by coa (x) = ax(Cayley multiplication) a E 6, x E is a Clifford map and that the homomorphismb: C6 - extending p is an isomorphism.

Consider the linear transformations of B8 given by

(x, e)H a WA(x) = (e, x)H e,

w(x) = (x, e)e and x H x

and describe the corresponding elements in C.

8. Use Problem 7 to construct explicit isomorphisms

C7(-) 3 L(P8) O L(B8)

and

C8(-) L(B ' 6).

Index

A

adjoint tensor 178A(E) 142A.(E) 146algebras

A(E) 142A.(E) 146C_ E 243Cn(+) 256C(-) 256C (P, q) 257S(E) 224S.(E) 225T(E) 82T(E) 78T.(E) 81®E/M(E) 94®E/N(E) 89

alternator 85annihilators 130, 131anticommutative flip operator 46anticommutative tensor products of

graded algebras 46, 120, 170antiderivations 43, 112, 144, 258

a-antiderivations 115of graded algebras 48

antisymmetry operator 98, 141

B

bilinear mappings 1, 18box product 153

C

canonical element eo 238Cayley numbers 289C_ E 243characteristic coefficients 182classical adjoint transformation 179classical Jacobian identities 191Clifford algebras 227f f

anticenter of 240center of 240complexif ication of real Clifford

algebras 251existence of 229uniqueness of 229

Clifford group 264Clifford map 227C(+) 256Cn(-) 256composition algebra 33composition product 154contraction operator 72

C (P, q) 257cross-products 288

D

DE 161

degree involution 233derivations 66, 70, 110, 123, 142, 258diagonal mapping 124diagonal subalgebra 151divisors 127

291

292

DL 174dual spaces 31, 71, 92, 106, 123

dual differential spaces 54dual G-graded spaces 46dual graded differential spaces 56

E

eo 238rlE 248exterior algebras 103

over a direct sum 120over a graded vector space 125over dual spaces 106over inner product spaces 107

exterior powerof an element 103of a vector space 101

external product 165

F

filtrations 128flip-operator 42

G

graded differential spaces 55graded ideals 127Grassmann algebra 142Grassmann products 141

Hhalf spin representations 272homogeneous functions 219

I

i (a) 117, 214iA (h) 143i (h) 118i(h) 79intersection algebra 163intersection product 163invariant linear maps 273

Index

is (h) 224isomorphisms

DE 161DL 174

rlE 248E 194R 273'T'E 194a 204T 36, 169TE 157T 196, 204T®09R

SE

J

76

276

248

Jacobi identity 178

K

Kiinneth formula for graded differentialspaces 55

Kunneth theorem 53

L

Lagrange identity 108, 166Laplace formula 176

AE 264

MM(E) 93mixed exterior algebras 149mixed tensors 71MP (E) 91multilinear mappings 3

N

N (E) 88Nolting algebras 108N( E) 84

Index

0operators

1(a) 117,214i (h) 118iA (h) 143

(h) 79is (h) 224

opposite algebra 236orthogonal multiplications 284, 286,

289orthogonal systems of skew transforma-

tions 285orthogonal k-frames on S' 287

P

E 194, 267R 273Pin 268Poincare duality 157Poincare isomorphism 159Poincare series 44, 219polynomial algebras 221

'T'E 194

R

Randon-Hurwitz number 280representations 260

equivalent 260faithful 260irreducible 260orthogonal 261

S

S(E) 224S.(E) 225SE 236a 204skew-Hermitian transformations 204skew linear transformations 194

Pfaffian of 200skew-symmetric functions 140skew-symmetric mappings 96, 148skew tensor products (see anticommuta-

tive tensor products)Spin 269

293

Spin representation 270structure map 41

of the homology algebra 56substitution operators 79, 143, 224symmetric algebras 212

graded symmetric algebra over agraded vector space 218

over a direct sum 217over dual spaces 212

symmetric functions 223symmetric power 211symmetric product 223symmetrizer 92

T

P 169T 196, 2047'E 1577'(E) 827''(E) 787'.(E) 81tensor algebras 62

graded tensor algebra over a gradedvector space 126

mixed 72over a G-graded vector space 67over an inner product space 75

tensor products 10, 26existence of 9intersections of 19of algebra homomorphisms 42of algebras 41of basis vectors 17of Clifford algebras 245of differential algebras 57of differential spaces 50of direct sums 14of dual differential spaces 54of factor spaces 13

of G-graded vector spaces 44of inner product spaces 33of linear maps 21of representations 260of subspaces 13

uniqueness of 8tensors 60

adjoint 178decomposable 60, 71invariant 75metric 76skew-symmetric 85, 89

294 Index

symmetric 91 symmetric algebras 212®E/M(E) 94 symmetric p-linear mappings 210®E/N(E) 89 tensor algebras 639R 276 ®E 62trace coefficients 186trace form 37twisted adjoint representation 263

wU Wedderburn theorems 273

universal property forbilinear mappings 5, 6exterior algebras 105 xmultilinear mappings. 5, 6skew-symmetric maps 99 SE 248

9 780387

multilinear algebra greub

Documents

chapter b

old chapter

chapter proceeds

chapters b

springerverlag new york

new material

end of chapters

prefacethe book