math 747: lie algebras
TRANSCRIPT
Math 747: Lie Algebras
Lecturer:Arun Ram
MathematicsUniversity of Wisconsin
Contents
Contents 1
1 The basics (Wed. Sept 5) 41.1 Homomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 Categories, Functors, and the Universal Enveloping Algebra . . . . . . . . . . . . 41.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2 Forms and Big Examples (Fri. 9/7) 52.1 Favorite Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
3 More Examples (Mon. 9/10) 63.1 Relating the linear groups of symmetries . . . . . . . . . . . . . . . . . . . . . . . 63.2 The free Lie algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73.3 An aside: two dimensional Lie Algebras and Dick Hain (9/12) . . . . . . . . . . . 83.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
4 Complexification (Wed. 9/12) 94.1 Example: su1(C) is a real Lie algebra . . . . . . . . . . . . . . . . . . . . . . . . . 94.2 Example: su2(C) has complexification sl2(C) . . . . . . . . . . . . . . . . . . . . 9
5 Derivations (Fri. Sept 14 & Mon. Sept 17) 115.1 Example: A = C[x] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115.2 Example: A = C[x, x−1] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125.3 Example: A = Uh, where h is abelian . . . . . . . . . . . . . . . . . . . . . . . . 125.4 Example: the Heisenberg Lie algebra . . . . . . . . . . . . . . . . . . . . . . . . . 12
6 Birds and bees of Lie groups and their algebras (Wed. 9/19) 13
7 Birds and bees, part 2 (Fri. 9/21 and Mon. 9.24) 157.1 Example: gln maps to GLn under exponentiation . . . . . . . . . . . . . . . . . . 167.2 Example: On(C) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
1
8 The Virasoro algebra and the beginning of its motivation (Wed. 9/26) 188.1 Semidirect products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198.2 Representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
9 Centralizers and central extensions (Fri. 9/28) 209.1 Central extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
10 Lie algebra cohomology (Mon. 10/1) 21
11 More central extensions (Wed. 10/3) 22
12 Building an Algebra-to-Topology dictionary (Fri. 10/5) 23
13 Deriving the central extension of the Witt algebra (Mon. 10/8) 24
14 Finite dimensional complex semisimple Lie algebras... (Wed. 10/10) 24
15 Bilinear forms and Hopf algebras (Fri. 10/12) 2515.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2615.2 More of what forms can do for us . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
16 (Mon. 10/17) 27
17 (Fri. 10/19) 2917.1 Jordan decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
18 Triangular decomposition (Mon. 10/22) 3118.1 Weights . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
19 The Cartan subalgebra, weights, and root systems (Wed. 10/24) 3219.1 Some people think Lie theory is the same as root systems... . . . . . . . . . . . . 33
20 Finding the sl2’s (Mon. 10/29) 35
21 sl3: the gateway to classification (Wed. 10/31) 3721.1 drawing h∗ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3821.2 Philosophy of classification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
22 Classification (Friday. 11/2) 3922.1 Step 1: How many W , P possibilities are there? . . . . . . . . . . . . . . . . . . . 39
23 More classification: building Dynkin diagrams (Mon. 11/5) 4023.1 But what is a reflection? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4023.2 Dealing with the lattice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4023.3 Classifying the (bigger) general case . . . . . . . . . . . . . . . . . . . . . . . . . 41
24 Coxeter, Cartan, and Kac-Moody (Fri. 11/9) 4224.1 Can we go back? Can we build g from A? . . . . . . . . . . . . . . . . . . . . . . 42
25 More on the Cartan matrix and Kac-Moody Lie algebras (Mon. 11/12) 43
2
26 Affine Lie algebras (Wed. 11/14) 45
27 (Fri. 11/17) 4727.1 Beginning representation theory of KM junk . . . . . . . . . . . . . . . . . . . . . 48
28 Computing the character of M(λ) (Wed. 11/28) 48
A All things aside 50A.1 The Kac-Moody Lie algebra. (9/10) . . . . . . . . . . . . . . . . . . . . . . . . . 50A.2 What is a reductive I hear so much about? (9/12) . . . . . . . . . . . . . . . . . 50A.3 Adeles – the study of everything at once. (9/17) . . . . . . . . . . . . . . . . . . 50A.4 Rob wants to know what this “admissible” junk is (9/17) . . . . . . . . . . . . . 50A.5 Tantalizers (9/19) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51A.6 What is this “affine” I keep hearing? . . . . . . . . . . . . . . . . . . . . . . . . . 51A.7 Homology vs. Cohomology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
B Those other questions 51
3
1 The basics (Wed. Sept 5)
A Lie algebra is a vector space g with a bracket [, ] : g⊗ g→ g (the tensor product implies that[, ] is bilinear) satisfying
(a) (skew symmetry) [x, y] = −[y, x], and
(b) (Jacobi identity) [x, [y, x]] + [y, [z, x]] + [z, [x, y]] = 0,
for all x, y, z ∈ g. Note that a Lie algebra is not an algebra (“Lie” is not an adjective), asalgebras A are vector spaces with a product under which A becomes a (associative) ring withidentity.
1.1 Homomorphisms
An algebra homomorphism is a linear map ϕ : g1 → g2 for which ϕ(xy) = ϕ(x)ϕ(y) (andso ϕ(1g1) = 1g2). A Lie algebra homomorphism is also a linear map ϕ : g1 → g2 for whichϕ([x, y]) = [ϕ(x), ϕ(y)].
1.2 Categories, Functors, and the Universal Enveloping Algebra
A category is a set of objects together with morphisms (functions) between them. Our favoriteexamples are
Alg = (algebras, algebra homomorphisms)
Lie = (Lie algebras, Lie algebra homomorphisms)
A functor is a map between categories
F : C1 → C2
which associates to each object X ∈ C1 an object F (X) ∈ C2, associates to each morphismf : X → Y ∈ C1 a morphism F (f) : F (X)→ F (Y ) ∈ C2, and preserves both identity morphismsand composition of morphisms.
There is a functorL : Alg→ Lie
where if A ∈ Alg, then the underlying vector spaces of A and L(A) are the same, but the producta · b 7→ [a, b] = ab− ba. There is also a functor
U : Lie→ Alg
where if g ∈ Lie, then the underlying vector space of Ug is that of the algebra generated by theelements of g with the relation xy − yx︸ ︷︷ ︸
as in Ug
= [x, y]︸ ︷︷ ︸as in g
. For example, if [x, y] = 0 in g then xy = yx in
Ug. Ug is said to be the universal enveloping algebra of g.So now we have functors
L : Alg→ Lie
andU : Lie→ Alg.
4
It may be tempting to speculate that these two functors are inverses of some kind. However,it is easy to see that Ug is rather large as compared to g (often infinite dimensional), whereasL(A) is no that much smaller than A (especially when A is finite dimensional). However, we dohave the following theorem:
Theorem 1.1. The functor U is left-adjoint to the functor L, i.e.,
HomAlg(Ug, A) ∼= HomLie(g, L(A))
as vector spaces.
1.3 Exercises
1. Classify all low-dimensional lie algebras.
2. Prove theorem 1.1.
3. Is there a right-adjoint to L? i.e., is there some functor F : Lie toAlg for which
HomLie(LA, g) ∼= HomAlg(A,Fg)?
2 Forms and Big Examples (Fri. 9/7)
A representation of g is a Ug-module. A Ug-module is a vector space M with a Ug-action
Ug⊗M →M,
where(u,m) 7→ um
which is bilinear, (i.e., if c1, c2 ∈ C, then
(c1u1 + c2u2)m = c1u1m+ c2u2m
u(c1m1c2m2) = c1um1 + c2um2
for u1, u2 ∈ C,m1,m2 ∈ C) andu1(u2m) = (u1u2)m.
Note: whenever we’re using tensor products, we’re just forcing bilinearity.
2.1 Favorite Examples
Big idea: If A be an algebra, then L(A) is a Lie algebra.
1. For example Mn(C) = {n× n matrices} is a Lie algebra with bracket [a1, a2] = a1a2−a2a2.This is the Lie algebra gln.
This Lie algebra is associated with, but not the same this as, general linear group
GLn(C) = {a ∈Mn(C)|a is invertible }.
Let V be a vector space. Then
gl(V ) = End(V ), with bracket [a1, a2] = a1a2 − a2a2.
5
2. Definesl(V ) = {a ∈ gl(V )|tr(a) = 0}.
This is associated to, but is not the same thing as, SL(V ) = {a ∈ GL(V )| det(a) = 1},the special linear group.
3. Let 〈, 〉 : V × V → C be bilinear form on V . If 〈, 〉 is symmetric, i.e. 〈u, v〉 = 〈v, u〉, then
so(V ) = {a ∈ sl(V )|〈au, v〉+ 〈u, av〉 = 0 for all u, v ∈ V }.
This is related to SO(V ) = {a ∈ SL(V )|〈au, av〉 = 〈u, v〉for all u, v ∈ V }, the specialorthogonal group.
4. If 〈, 〉 is skew symmetric, i.e. 〈u, v〉 = −〈v, u〉, then
sp(V ) = {a ∈ sl(V )|〈au, v〉+ 〈u, av〉 = 0 for all u, v ∈ V }.
This is related to Sp(V ) = {a ∈ SL(V )|〈au, av〉 = 〈u, v〉for all u, v ∈ V }, the symplecticgroup.
5. Let 〈, 〉 : V × V → C be Hermitian, i.e. for vector spaces over C,
a. 〈u, c1v1 + c2v2〉 = c1〈u, v1〉+ c2〈u, v2〉, andb. 〈c1v1 + c2v2, u〉 = c1〈v1, u〉+ c2〈v2, u〉.
Thensu(V ) = {a ∈ sl(V )|〈au, v〉+ 〈u, av〉 = 0 for all u, v ∈ V }.
This is related to SU(V ) = {a ∈ SL(V )|〈au, av〉 = 〈u, v〉for all u, v ∈ V }, the specialunitary group.
Notice that most of these algebras have the same structure, only with different types of forms.
2.2 Exercises
1. Is sp(V ) = {a ∈ gl(V )|〈au, v〉+ 〈u, av〉 = 0 for all u, v ∈ V }?
3 More Examples (Mon. 9/10)
3.1 Relating the linear groups of symmetries
From last time, we learned about gl(V ) = End(V ), sl(V ) = {a ∈ gl(V ) | tr(a) = 0}, and thethree sp(V ), so(V ) and su(V ), all of similar form, but with different forms.
6
The orthogonal group isOn(R) = {A ∈ GLn(R)|AAt = 1}.
The symplectic group. is
Sp2n(C) = {A ∈ GL2n(C)|AJAt = J},
where
J =
1. . .
1−1
. . .−1
.
SP(V ) is “better” though because is is the group of symmetries of V with a skew symmetric form.For example, our favorite set with fixed distance is the sphere. SO(V ) is the group of symmetriesfor a true sphere, SP(V ) is the group of symmetries of a “symplectic” sphere (whatever thatmeans), etc.. (A “level set of 〈, 〉 is a subset of V × V of the form {(u, v) ∈ V × V |〈u, v〉 = c},and SP(V ) preserves this).
What does a skew symmetric form “look like”? Well, what does a vector space look like?First, we choose a basis: V has a basis {e1, . . . , en, e
∗1, . . . e
∗n}. and 〈, 〉 is given by
〈ei, ei〉 = −〈ei, ei〉 = 0, 〈ei, e∗i 〉, 〈ei, ej〉 = 0, 〈e∗i , e∗j 〉 = 0, 〈ei, e∗j 〉 = 0, 〈e∗i , ej〉 = 0.
ThenJ = (〈bi, bj〉)bi,bj∈{e1,...,en,e∗1,...e∗n}
is the matrix of 〈, 〉. So AJAt = J is a translation of 〈Au, v〉 = 〈U,Av〉 by choosing a veryspecific basis. So something we learn by changing to Sp2n by seeing things anew, without reallydoing anything new. (the message: changing bases is one way to do something that is not new,but still very worthwhile)
Aside: Why use the fields we do? There are lots of GLn’s: GLn(C),GLn(R),GLn(Fq), etc.the ones with C,R are “geometric”, but so is Fq in some sense. We change the field as is suits us...
3.2 The free Lie algebra
Motivation: Whenever we learn about a new object, the free version is usually very interesting.
The free Lie algebra on a set {x, y, z} is just the Lie algebra generated by {x, y, z}, i.e. g is thespan of x, y, z, [x, y], [x, z], [y, z], ....
What else is in there? To get a general idea of what other elements appear, start applyingskew symmetry and Jacobi identity relations. For example, g being a Lie algebra gives [y, x] =−[x, y], so we don’t need to include [y, x], [z, x], etc.. We also have the symbol[x, [x, y]] toinclude, but [x, [y, x]] = −[x, [x, y]] is superfluous. Moreover [x, [y, y]] = 0. Also we couldcheck if the Jacobi identity does anything extra for us: 0 = [x, [x, y]] + [y, [x, x]] + [x, [y, x]] =[x, [x, y]] + 0− [x, [x, y]], so no, not in this case.
7
The free Lie algebra is graded by degree (the number of factors bracketed together). Forexample deg([x, [x, y]]) = 3. So
g =⊕n∈Z+
dim(gn) finite
gn, where gn is the span of deg n brackets
3.3 An aside: two dimensional Lie Algebras and Dick Hain (9/12)
Letg = C-span{x, y}, with bracket [x, y] = x.
Let a, b ∈ g . So a = a1x+ a2y and b = b1x+ b2y. Then
[a, b] = [a1x+ a2y, b1x+ b2y]= a1[x, b1x+ b2y] + a2[y, b1x+ b2y]= . . .
= (a1b2 − a2b1)x.
Dick Hain (Duke U) gave a talk Modular forms and derivations of free Lie algebras of rank1 at Sydney U. 10 July 2007 based on the free Lie algebra of rank 2, generated by x and y. Thebackground: Let V = span{x, y}. Then
SL2(C) = {A ∈ GL2(C) | ad− bc = 1}
So SL2(C) acts on V , and thus SL2(C) acts on g. The submodules of g (as SL2-modules)correspond to modular forms.
(a bc d
)[x, y] =
[(a bc d
)x,
(a bc d
)y
]= [ax+ cy, bx+ dy]= (ad− bc)[x, y]
if
x =(
0 10 0
), and y =
(0 01 0
).
3.4 Exercises
1. What is dim(gn)? [For example, dim(g1) = 3.]
((same as number of irreducible polynomials of degree n over some field?? ))
8
4 Complexification (Wed. 9/12)
How can a Lie algebra with complex coefficients not be a complex Lie algebra? And if it’s not,what is its complex counterpart?
Definition. Let g be a real Lie algebra with basis x1, . . . , xn. So
g = R-span{x1, . . . , xn}.
The complexification of g is the complex Lie algebra
gC = C⊗R g = C-span{x1, . . . , xn}
with bracket [xi, xj ] as before.
4.1 Example: su1(C) is a real Lie algebra
Recall su1(C) = {a ∈ sl1(C)|a+ at = 0} with bracket [a, b] = ab− ba. So
s1(C) = {a ∈ C | a+ at = 0}= {x+ iy | a, y ∈ R, x+ iy + (x− iy) = 0}= {x+ iy | a, y ∈ R, 2x = 0}= Ri= R-span{i} with bracket [i, i] = 0.
This is a one-dimensional R-vector space, so has no hope of being a complex Lie algebra! Sosu1(C) is a real Lie algebra, not a complex Lie algebra.
The standard complexification of su1(C) is the Lie algebra
C-span{h} with bracket [h, h] = 0
(we use h since, other wise, we would have two i’s: the elements are αh where α ∈ C, soα = α1 + iα2).
4.2 Example: su2(C) has complexification sl2(C)
Recall
sl2(C) = {A ∈ gl2(C) | tr(A) = 0}
= C-span{(
a bc d
) ∣∣∣ a+ d = 0}
= C-span{x =
(0 10 0
), y =
(0 01 0
), h =
(1 00 −1
)}.
(sometimes people use e = x and f = y)
9
Then
[x, y] = xy − yx
=(
0 10 0
)(0 01 0
)−(
0 01 0
)(0 10 0
)=
(1 00 0
)−(
0 00 1
)=
(1 00 −1
)= h.
The relations[x, y] = h, [h, x] = 2x, and [h, y] = −2y
define sl2 (the rest following similarly as above).Now,
su2(C) ={A ∈ gl2(C)
∣∣∣ tr(A) = 0, A+At = 0}
={(
a bd c
) ∣∣∣ a+ d = 0, a+ a = 0, d+ d = 0, and b+ c = 0}
={(
iα β + iγ−β + iγ −α
) ∣∣∣ α, β, γ ∈ R}
is a real Lie algebra with dimension 3 (it must be real and not complex since it has odddimension). To calculate its complexificaiton, let
σx =(
0 11 0
), σy =
(0 −ii 0
), and σz =
(1 00 −1
).
These are called the Pauli matrices (σ is standard notation from physics), and generate su2(C):
su2(C) = R-span{iσx, iσy, iσz}.
The standard complexification of su2(C) is
C-span {iσx, iσy, iσz} = C-span {σx, σy, σz}
= C-span{
12
(σx + iσy),12
(σx − iσy), σz}
= C-span {x, y, z}= sl2(C).
The moral If you are a “hardcore Lie theorist” you only know sl2(C). If you are a “hardcorephysicist” you only know su2(C). If you are a “hardcore number theorist” you only knowsl2(R) = R-span{x, y, h} with bracket the same. The moral of the story is don’t be so hardcore.
10
5 Derivations (Fri. Sept 14 & Mon. Sept 17)
Rob’s questions: Why are Lie Algebras a natural object?
Attempt 1: Same reason that going to see “March of the Penguins” was a natural thingto do.
Attempt 2: There are connections to Lie groups. We don’t totally understand this con-nection, but they arise in this way.
Attempt 3: A connection to quantum mechanics comes into play... not that QuantumMechanics is such a “natural” thing...
Attempt 4: The work of Dick Hain [3.3] (one of the coolest people in the world these days)uses free Lie algebras to study modular forms.
Attempt 5: Today! Derivations...
Definition. Let A be an algebra. A derivation of A is a linear map d : A→ A (linear means Ionly have to define it on the basis) such that
d(ab) = d(a)b+ ad(b), for a, b ∈ A.
Let Der A be the space of derivations of A.
Theorem 5.1. Der A is a Lie algebra with bracket
[d1,d2] = d1d2 − d2d1.
Proof. Check that the bracket of two d’s satisfies the product rule:
[d1, d2](ab) = (d1d2 − d2d1)(ab)= (d1d2)(ab)− (d2d1)(ab)= d1
(d2(a) · b+ a · d2(b)
)− d2
(d1(a) · b+ a · d1(b)
)=
(d1d2(a) · b+ d2(a) · d1(b) + d1(a) · d2(b) + a · (d1d2)(b)
)−(
(d2d1)(a) · b+ d1(a) · d2(b) + d2(a) · d1(b) + a · (d2d1)(b))
= (d1d2)(a) · b+ a · (d1d2)(b)− (d2d1)(a) · b− a · (d2d1)(b)=
(d1d2 − d2d1
)(a) · b+ a
(d1d2 − d2d1
)(b)
= [d1,d2](a) · b+ a · [d1,d2](b)
5.1 Example: A = C[x]
Let d ∈ Der A. Then
d(x2) = d(x)x+ xd(x) = 2x · d(x).d(x3) = d(x)x2 + xd(x2)
= d(x)x2 + x · 2x · d(x)= 3x2 · d(x).
d(xk) = kxk−1 · d(x).
11
Since C[x] has basis {1, x, x2, . . . }, any derivation d ∈ Der C[x] is determined by your choice ofd(x). But
d : Cx→ C[x],
so d(x) can be any element of C[x]. So
Der C[x] = C[x]-span{∂
∂x
}, where
∂
∂xxk = kxk−1.
For example
d = (x2 + 7x+ 2)(∂
∂x
).
Note: dimC[x]Der C[x] = 1.
5.2 Example: A = C[x, x−1]
Later
5.3 Example: A = Uh, where h is abelian
Let h be a vector space with basis x1, . . . xn and bracket [xi, xj ] = 0. The enveloping algebra ofh is
Uh = C[x1, . . . , xn].
(one way to think about polynomial rings!)
5.4 Example: the Heisenberg Lie algebra
Let C be an arbitrary field. Let h be a vector space with basis x1, . . . , xn and let
h∗ = Hom(h,C)
be the dual (also a vector space) with basis ∂∂x1
, . . . , ∂∂xn
. Note: we’re borrowing notation fromcalculus: instead of writing x∗i , we write ∂
∂xi. It is certainly a dual basis, since ∂
∂xixj = δij . Why
our notation makes sense:(xi
∂
∂xj− ∂
∂xjxi
)f = x1
∂
∂xjf − ∂
∂xjxif
= xi∂
∂xjf −
(δijf + xi
∂
∂xj
)f
= −δijf.
Definition. The Heisenberg Lie algebra g is is
g = h⊕ h∗ ⊕ Cc
(so has basis x1, . . . , xn,∂∂x1
, . . . , ∂∂xn
, c) with bracket
[c, y] = 0, [xi, xj ] = 0,[∂
∂xi,∂
∂xj
]= 0,[
xi,∂
∂xj
]= c
∂
∂xj(xi) = cδij
for all y ∈ g.
12
Note that center of a Lie algebra g is
Z(g) = {z ∈ g | [z, y] = 0 for all y ∈ g}.
So these relations are mostly saying that the Heisenberg Lie algebra is mostly abelian.
The enveloping algebra of g is the Weyl algebra. It has generators x1, . . . , xn,∂∂x1
, . . . , ∂∂xn
, cwith relations
xixj = xjxi,∂
∂xi
∂
∂xj=
∂
∂xj
∂
∂xi,
cxi = xic, c∂
∂xj=
∂
∂xjc,
∂
∂xixj = xj
∂
∂xj+ δijc.
Elements of Weyl algebra are linear combinations of
c`xm11 · · ·x
mnn
(∂
∂x1
)`1· · ·(
∂
∂xn
)`n,
i.e. polynomial coefficient differential operators.
6 Birds and bees of Lie groups and their algebras (Wed. 9/19)
Definition. Let G be a group, and define
µ : G×G→ G via (g1, g2) 7→ g1g2
and
ι : G→ G via g 7→ g−1.
1. G is a topological group if it is also a topological space and µ and ι are continuous maps.
2. G is a Lie group if it is also a manifold and µ and ι are smooth maps.
3. G is a complex Lie group if it is also a complex manifold and µ and ι are holomorphicfunctions.
4. G is a linear algebraic group if it is also an affine variety and µ and ι are regular functions.
5. G is a group scheme if it is also a scheme and µ and ι are morphisms of schemes.
Remark. About these maps:
1. A continuous map is just a morphism of topological spaces .
2. A smooth map is just a morphism of manifolds. Really, “smooth” tells me that I can takederivative. So viewing manifolds with respect to their morphisms, what I care about isnot that they look locally like Rn, but rather that I know what derivatives look like.
13
3. A holomorphic function is just a morphism of complex manifolds. What’s special aboutthe “complex” of complex manifolds? In C, we have complex conjugation, giving it acomplex structure. Holomorphic functions additionally preserve complex conjugation.
4. A regular function is just a morphisms of varieties, rather, things that look like polynomi-als.
The above definitions seem to follow a categorical feel! Another way of talking about thisidea is via “ringed” objects: A ringed space (X,OX) is a topological space X with a space offunctions OX on X. A tangent vector to X at an element x is a linear map
ξx : OX → F, satisfying ξx(f1f2) = ξx(f1) · f2(x) + f1(x) · ξx(f2),
for f1, f2 ∈ OX (so it’s almost a derivation). Now, usually in freshman calculus, we make thismistake:
d
dx(x3 + 2x+ 1)︸ ︷︷ ︸function
= limh→0· · ·︸ ︷︷ ︸
number
,
where we treat the right side as a function instead. A vector field on X is a derivation of OX ,i.e. a linear map d : Ox → Ox such that
d(f1f2) = d(f1) · f2(x) + f1(x) · d(f2),
for f1, f2 ∈ OX .Let (G,OG) be a ringed group (totally unstandard phrase, but it seems to be the right thing
to call it). Then G acts on OG by (left) regular action:
(Lgf)(h) = f(g−1h) for h ∈ G,
and right translation(Rgf)(h) = f(hg) for h ∈ G.
Notice, Rg is actually a left action:
(RgRg′f)(h) = (Rg′f)(hg)= f(hgg′)= (Rgg′f)(h).
A left invariant vector field is a derivation
d : OG → OG, such that LgdLg−1 = d.
(A right invariant vector field is a derivation d : OG → OG such that RgdRg−1 = d.)Suppose d is left invariant, and let dg be the tangent vector at g ∈ G, given by
dgf = (df)(g).
Notice that if d is left invariant, then d is determined by d1:
dgf = (df)(g)= (Lg−1df)(1)= (Lg−1dLgLg−1f)(1)= (dLg−1f)(1)= d1(Lg−1f).
14
Now letT1(G) = {tangent vectors to G at 1}.
So{ left invarient vector fields on G} ↔ T1(G),
and so{ left invarient vector fields on G} ⊂ Der (OG).
Recall that Der (OG) is a Lie algebra under [d, D] = dD −Dd for d, D ∈ Der (OX). Moreover,if d, D ∈ Der (OG) are left-invarient, then [d, D] is also left invariant. So the set of left invarientvector fields on G is a Lie subalgebra of Der (OG). So finally, if (G,OG) is a ringed group, thenthe Lie algebra of the group G is { left invarient vector fields on G} (= T1(G)). (take a tangentvector to the point 1, and move it around by G; since that movement is a morphism, and getout a whole vector field).
For next time: A one parameter subgroup is a morphism γ : R → G. Very fiew of these!Example: find all group homomorphisms γ : R→ GL1(C).
7 Birds and bees, part 2 (Fri. 9/21 and Mon. 9.24)
The Lie algebra of G is g = T1(G) = {tangent vectors to G at 1} with bracket obtained byidentifying T1(G) with the set of left invariant vector fields on G with is a Lie subalgebra ofDer (OG). So we have a map from Lie groups to Lie algebras. THis is a functor!
Let G is a Lie group. A one parameter subgroup of G is a morphism
γ : R→ G
(our favorite Lie group is R under the operation addition).
Example. What are the one parameter subgroups of GL1(C) = C∗:
γ : C→ GL1(C)?
Well, what are the group homomorphisms γ : R → GL1(R), ie.e γ(s + t) = γ(s) + γ(t)? Soγ(t) = a0 + a1t+ a2t
2 + · · · . We want γ(s+ t) = γ(s) + γ(t). The expansion of the left side is
a0
+a1s+ a1t
+a2s2 + 2a2st+ a2t
2
+a3s3 + 3a3s
2t+ 3a3st2 + a3t
3
+a4s4 + 4a4s
3t+ 6a4s2t2 + 4a3st
3 + · · ·
The expansion of the right side isa2
0
+a1a0s+ a1a0t
+a2a0s2 + a2
1st+ a0a2t2
15
+a3a0s3 + a2a1s
2t+ a2a1st2 + a0a3t
3
+a4a0s4 + a3a1s
3t+ a22s
2t2 + a1a3st3 + · · · .
So either 0 = a0 = a1 = a2 = · · · , or
a0 = 1a1 = a1
a2 =12a2
1
a3 =13a2a1 =
13!a3
1
...
So there is a (almost) unique one-parameter subgroup γ : R → GL(R) given by γ(t) = ea1t foreach a1 ∈ R. Really, the one-parameter subgroup is the function, but we like to actually thinkabout it as the image of γ, which is a subgroup of G (which looks locally like R near the identity.
Now suppose γ : R→ G is a one parameter subgroup. Define γ1 as a tangent vector to 1 by
(γ1f) = limt→0
f(γ(t))− f(γ(0))t− 0
(just some convenient derivation. There are many of them, but this is one that we know a lotabout). To Check that this is a tangent vector, we need to check
γ1(f1f2) = f1(1)(γ1f2) + (γ1f1)f2(1).
So now we have a map from one parameter subgroups of G to T1(G) via
γ 7→ γ1.
Is this invertible? Baby example:
{tangent vectors at 1 in GL1(C)} ↔ {one parameter subgroup of GL1(C)}
a1 7→ ea1t.
So left invariant vector fields on G are tangent vectors at 1 to G are one parameter subgroupsof G. We have a functor from Lie groups G to Lie algebras g.
7.1 Example: gln maps to GLn under exponentiation
Recall:
GLn(C) = {g ∈Mn(C) | g−1 exists }, andgln(C) = Mn(C) has basis {Eij | 1 ≤ i, j ≤ n}
16
where Eij has a 1 in the (i, j) entry and 0 elsewhere. so we have a map from gln(C) to theone-parameter subgroups of GLn(C) via Eij 7→ etEij . If i 6= j, then
etEij = 1 + tEij +t2
2!E2ij +
t3
3!E3ij + · · ·
= 1 + tEij + 0 + 0 + · · ·
=i↓
1. . . t
. . .1
← j
= xij .
If, on the other hand, i = j, we have
etEii = 1 + tEii +t2
2!E2ii +
t3
3!E3ii + · · ·
= 1 + tEii +t2
2!Eii +
t3
3!Eii + · · ·
=
1. . .
1et
1. . .
1
= hi(et).
Let G be a Lie group and let g be the Lie algebra. Let {x1, . . . , xn} be a basis of g, andso etx1 , . . . , etxn are one-parameter subgroups in G. The xij(t) and hi(t) are the elementarymatrices for G.
Theorem 7.1 (Linear algebra theorem 1). The elementary matrices generate GLn(C).
7.2 Example: On(C)
Let G = On(C) = {g ∈ GLn(C) | ggt = 1}. There is a Very Important map,
det : GLn(C)→ GL1(C)
with the Very Important property that it is a homomorphism (and therefore a representation).Really, it is a polynomial function:
det(aij) =∑w∈Sn
(−1)`(w)a1w(1) · · · anw(n),
so it is any kind of function that we want it to be (holomorphic, continuous, whatever). Here,
det : On(C)� {±1} since det(g)2 = 1
17
for g ∈ On(C) (certainly, this map is surjective, since if
s1 =
0 11 0
1. . .
1
,
then det s1 = −1).Topologically, SOn(C) and s1SOn(C) are the same, and On is the disjoint union of the
two, with the identity sitting inside of SOn(C). Lie(On) = Lie algebra of On = T1(On(C)) =T1(SOn(C)) (since the identity is in SOn(C)). So we have a functor from Lie groups to Liealgebras via G 7→ T1(G) = g.
In the last example, theorem [7.1] gave us that the exponentiation map took us the otherdirection surjectively. However, in this case, exponentiation won’t get us into the s1SOn(C)component, so this theorem certainly is not true in general!
8 The Virasoro algebra and the beginning of its motivation(Wed. 9/26)
Example: If d ∈ Der (C[x]) the d(xk) = kxk−1d(x), k ∈ Z≥0, where d(x) ∈ C[x]. So Der (C[x]) =C[x]-span{ ∂∂x} and has a C basis xk ∂
∂x .Example: If d ∈ Der (C[x, x−1]) the d(xk) = kxk−1d(x), k ∈ Z, where d(x) ∈ C[x, x−1]. So
Der (C[x, x−1]) = C[x, x−1]-span{ ∂∂x} and has a C basis xk ∂∂x .
Step back and thing about the geometry for a moment:
C[x, x−1] = OX = functions on S1.
since S1 = {eiθ | 0 ≤ θ < 2π} = {z | z ∈ C, |z| = 1}. A functions on S1 is a functions of z. Ifz 7→ z−1 is allowed, then your favorite set of functions includes C[x, x−1. (Another way to thinkof is is functions on C \ 0 is the same as functions on the circle).
The Witt algebra is the Lie alebra with basis . . . , L−2, L−1, L0, L1, L2, . . . and bracket [lm, Ln] =(m− n)Lm+n. Do the computation: If Lm = −xm+1 ∂
∂x , then
[Lm, Ln] = [−xm+1 ∂
∂x,−xn+1 ∂
∂x]
= (−xm+1 ∂
∂x)(−xn+1 ∂
∂x)− (−xn+1 ∂
∂x)(−xm+1 ∂
∂x)
= (xm+1 ∂
∂x)(xn+1 ∂
∂x)− (xn+1 ∂
∂x)(xm+1 ∂
∂x)
= xm+1((n+ 1)xn∂
∂x+ xn+1 ∂
2
∂x2)− xn+1((m+ 1)xm
∂
∂x+ xm+1 ∂
2
∂x2)
= (n+ 1)xm+n+1 ∂
∂x− (m+ 1)xn+m+1 ∂
∂x
= (m− n)(−xm+n+1 ∂
∂x)
= (m− n)Lm+n.
18
(one could have begun by just defining the Witt algebra, but it is good to understand thegeometry hiding in the background).
The Virasoro algebra is the Lie algebra with basis
. . . , L−2, L−1, L0, L1, L2, . . . , and c
and bracket[c, y] = 0, for any y ∈ g,
[Lm, Ln] = (m− n)Lm+n + δm,−n112
(m3 −m)c.
Bam! What’s with the 112?? No one really knows. It’s a very natural and interesting question,
and it comes up in a lot of places, but it’s not clear why it keeps popping up.Where does this come from? We saw c before, in the Heisenberg Lie algebra. It had basis
x1, . . . , xn,∂∂x1
, . . . , ∂∂xn
, cwith bracket
[c, y] = 0, [xi, xj ] = 0,[∂
∂xi,∂
∂xj
]= 0,
[xi,
∂
∂xj
]= c
∂
∂xj(xi) = cδij
for all y ∈ g.
8.1 Semidirect products
Let g be a Lie algebra. Let d be a Lie algebra actions on g by derivations (i.e. d([x, y]) =[dx, y]+[x,dy]). The semidirect product, dog, is g⊕d with bracket [d, x] = d(x) for d ∈ d, x ∈ g.Let R be a ring and G be a group which acts on R by automorphisms. The semidirect productRoG, is the algebra
RoG = {∑g∈G
rgg | rg ∈ R}
with blah given by (r1g1)(r2g2) = r1g1(r2)g1g2. In Math 541, C(G oH) = CG oH. (aut is togrps as der is to lie alg)
8.2 Representations
A representation of G is a homomorphism G→ GL(M). We use M instead of V , since we wantto see it as a module: as soon as we have a map from G in to maps on V , we get an action of Gon V , so V is actually a module. So really, a representation of G “is” a G-module M . A simplemodule is a module with no (stupid) submodules. Many representations (G-modules) have theform
M = M1 ⊗M2 ⊗ · · · .
We’d like the Mi to be simple. The problem is to find the simple modules. THe phiosophy: IfI know the simple R-modules and thie simple G-modules, I should be able to build the simpleR o G-modules. This is true, but takes work. This work is called Clifford Theory (like “rowreduction,” it is an algorithm).
19
9 Centralizers and central extensions (Fri. 9/28)
A representation is a module. A module is a vector space with an action A (a group, and algebra,a Lie algebra, etc). A simple module is a module with no submodules except, of course, 0. Amodule M is decomposable if
M = M1 ⊗M2 as A-modules.
If a ∈ A then the matrix of the action of a, aM on M is block-diagonal in the correct basis.We could also hope for get it into Jordan canonical form, though we have to be careful of ourcontext. Often it is impossible to block-diagonalize , but it is often possible to triangularize aMso it is almost block-diagonal, but stuff all above the diagonal (again, in some good basis of M .)If it is possible for all a ∈ A simultaniously then M has a submodule M1. M is indecomposableif M 6= M1 ⊕M2 asA-modules.
Let M be an A-module. The centralizer algebra of M
EndA(M) = {ϕM →M | ϕaM = aMϕ for all a ∈ A}
which is an algebra under composition.
Lemma 9.1 (Schur). Let A be a C-algebra.If M is simple, then
EndA(M) = C.
Proof. Assume M has no submodules (simple and irreducible are the same word). Let ϕ : M →M ne in EndA(M). Then kerϕ is a submodule of M and im ϕ is a submodule of M . So, sinceM is simple, kerϕ = 0 or kerϕ = M and im ϕ = 0 or im ϕ = M . So either ϕ = 0 or ϕ is anisomorphism. Let c be an eigenvalue of ϕ. Then cidM ∈ EndA(M), so ϕ − cidM ∈ EndA(M).The point of this is that by construction ϕ − cidM is not invertible (since ϕ has an eigenvaluein common with cidM , so ϕ − cidM has an eigenvalue of 0). Therefore ϕ − cidM = 0. Soϕ = cidM .
Something which commutes with the action of A acts on a simple module by a constant.
Aside In physics: charge means exactly this constant: there is some simple module and someoperator which commutes with the symmetries, so there is some constant called the charge.
Where do we find these things which act like constants?
Z(A) = {z ∈ A | za = az for all a ∈ A}
is the center of A.If you have an operator z that commutes with the A action and it’s not in Z(A) then make
A bigger so that is is in the center.
9.1 Central extensions
Let g be a Lie algebra and let c be an abelian Lie algebra (i.e. [c1, c2] = 0 for c1, c2 ∈ c). Acentral extension of g by c is a Lie algebra junk = g⊗ c such that c ⊂ Z(junk) and if x, y ∈ g,then [x, y] = [x, y] + ϕ(x, y) with ϕ ∈ c. So to specify a central extension we must specifyϕ : g× g→ c. There are conditions on ϕ if we want junk to be a Lie algebra.
20
10 Lie algebra cohomology (Mon. 10/1)
First, we need some linear algebra. Let V be a vector space. Any real person thiks about avector space in terms of its vector space, so pick one: let b1, . . . , bn be a basis. The tensor algebraon v is
T (V ) =⊕k∈Z≥0
V ⊗k
where V ⊗k has a basis {bi1 ⊗ · · · ⊗ bik | 1 ≤ i1, . . . , ik ≤ n} and product
(bi1 ⊗ · · · ⊗ bik) · (bj1 ⊗ · · · ⊗ bj`) = bi1 ⊗ · · · ⊗ bik ⊗ bj1 ⊗ · · · ⊗ bj` .
Sometimes we write T k(V ) for V ⊗k, or V ⊗k = V ⊗ · · · ⊗ V . The tensor algebra is the freeassociative algebra generated by b1, . . . , bn. If V has basis b1, . . . , bn and W has basis c1, . . . , cm,then V ⊗W has basis {bi ⊗ cj | 1 ≤ i ≤ n, 1 ≤ j ≤ m}.
Let A be an n×n matrix, i.e. A ∈ End(V ), and let B be an m×m matrix, i.e. B ∈ End(W ).Denote A⊗B matrix of the element of End(V ⊗W ) which is given by (A⊗B)(v⊗w) = Av⊗Bwfor v ∈ V,w ∈ W . A ⊗ B has entries (aijbk`) if A = (aij) and B = (jk`). If A ∈ End(V ) thenA⊗k = A⊗ · · ·⊗A is an elements of End(V ⊗k). So T k maps vector spaces to vector spaces, andis a functos. The exterior algebra is the quotient of T (V ) be the relations v1 ∧ v2 = −v2 ∧ v1.Let ∧
V =⊕k∈Z≥0
k∧V
and∧k V has basis {bi1 ∧ · · · ∧ bik | 1 ≤ i1 < · · · < ik ≤ n}. (everything is bi-linear). So
∧k asa map from vector spaces to vectpr spaces is a functor.
The symmetric algebra is the quotient of T (V ) by the relations v1v2 = v2v1 for v1, v2 ∈ V .Let
S(V ) =⊕k∈Z≥0
Sk(V ).
So Sk maps vector spaces to vector spaces, and is a functor:
S(V ) = C[bi, . . . , bn].
For example,∧n V has basis {bi1 ∧ · · · ∧ bin | 1 ≤ i1 < · · · < in ≤ n}, so
∧n V has basis{b1 ∧ · · · ∧ bn}. Note that if w ∈ Sn then bw(1) ∧ · · · ∧ bw(n) = (−1)`(w) where
Sn → {±1}
w 7→ (−1)`(w)
is the sign homomorphism and A(b1 ∧ · · · ∧ bn) = Ab1 ∧ · · · ∧Abn =∑
1≤ji,...jn≤n aj11bj1 ∧ · · · ∧ajnnbjn =
∑jidistinct
aj11 · · · ajnn(bj1 ∧ · · · ∧ bjn) =∑
j∈Sn(−1)`(w)aj11 · · · ajnn(b1 ∧ · · · ∧ bn) =detA(b1 ∧ · · · ∧ bn). So
∧nA = det(A) if A ∈ End(V ) and dimV = n.Let g be a Lie algebr. Let M be a g-module, i.e. M is a vector space iwht an action of g,
g⊗M →M
(x,m) 7→ xm
such that if x, y ∈ g, m ∈M then (xy − yx)m = [x, y]m, i.e. M is a Ug-module.
21
The cohomology of g wiht coefficients in M is the cohomology of the complex
· · · → Ci(g,M) d→ Ci+1(g,M)→ · · ·
where Ci(g,M) = Hom(∧i g,M) and dφ)(bji ∧ · · · ∧ bji+1 =
∑1≤k≤i+1(−1)k+1bjkϕ(bj1 ∧ · · · ∧
bjk ∧· · ·∧bji+1 +∑
1≤k<`≤i+1(−1)k+`ϕ([bjk , bj` ]∧bj1 ∧· · ·∧ bjk ∧· · ·∧bji+1 where ˆ denotes leavingit out.
Z i(g,M) = ker(diCi(g,M)→ Ci+1(g,M))
Bi(g,M) = im (diCi−1(g,M)→ Ci(g,M))
and
H i(g,M) =Z i(g,M)Bi(g,M)
.
Let g be a Lie algebra, and let L(g) = g ⊗ C[t, t−1] with [x1tk, x2t
`] = [x1, x2]tk+` wherex1, x2 ∈ g, k, ` ∈ Z.
11 More central extensions (Wed. 10/3)
Recall: let g be a Lie algebra and let c be an abelian Lie algebra. A central extension of g by c
is a Lie algebra g⊕ c (as vector spaces) such that
[x, y] = [x, y] + ϕ(x, y)
for x, y,∈ g where ϕ : g× g→ c and c ⊂ Z(g⊕ c) (i.e. [c, x] = 0 for all c ∈ c and x ∈ g⊕ c).How do we find the ϕ? Since [x, y] = −[y, x], we have ϕ(x, y) = −ϕ(y, x). Since [, ] is bilinear,
ϕ is also bilinear. So ϕ : g ∧ g→ c. Then the Jacobi identity,
[x, [y, x]] + [y, [z, x]] + [z, [x, y]] = 0
forcesϕ(x, [y, x]) + ϕ(y, [z, x]) + ϕ(z, [x, y]) = 0.
Since one “clearly recalls” the formula for d in Lie algebra cohomology, one immediately noticesthat this relation is more compactly written as
dϕ = 0
where g acts on c via x · c = 0 for x ∈ g, c ∈ c.Since Z2(g, c) = ker(d : C2(g, c)→ C3(g, c)), this means ϕ is a 2-cocycle.Another way to think about the central extension e = g⊕ c is as
ee� g
x+ c 7→ x
such that ker e ⊂ Z(e). The process of taking elements of g and thinking of them as elementsof e is the same as choosing a section s : g → e of e. THen ϕ : g × g → ker(e) defined by
22
[s(x), s(y)] = s([x, y]) + ϕ(x, y) for x, y ∈ g. If s′ : g → e was a different choice, then the onlychange must have been the choice of the element of c to be added to the element of g:
s′(x) = s(x) + ψ(x), for x ∈ g,
with ψ : g→ c ( i.e. ψ ∈ C1(g, c)). Then s([x, y]) +ϕ(x, y) = [s′(x), s′(y)] = [s(x) +ψ(x), s(y) +ψ(y)] = s′([x, y]) + ϕ′(x, y) = s([x, y]) + ψ([x, y])ϕ′(x, y). So ϕ(x, y) − ϕ(x, y) = ψ([x, y]) =dψ(x, y). SO if ϕ and ϕ′ come from the same e then they differ by an element of B2(g, c) =im ∗ d : C1(g, c) → C2(g, c)). So central extensions of g by c are determined by elements ofH2(g, c) = Z2(g, c)/B2(fg, c).
The universal central extension of g is a surjective Lie algebra homomorphism uu→ g such
that keru ⊂ Z(u) and if ee→ g is another central extension then there is a unique Lie algebra
homomorphism f : u→ e such that e(f(u)) = u(u).First thing to do is think about your favortie examples of universal objects. Examples include
tensor products, free groups, products, localizations, quotients of groups, C[x1, x2, . . . , xn], fildof fractions.... in general, universal objects just build themselves. SO proving existence is just amatter of building the damn thing by following our noses. Hence, we should be able to get theuniversal central extension of the Witt algebra to build itself – this will be the Virasoro.
Last thing: If E → G is a homomorphism of topological groups such that it is a localhomeomorphism, then E is a covering space or covering group. So the central extension is thesame as a cover, and a universal central extension is the same as a universal cover.
12 Building an Algebra-to-Topology dictionary (Fri. 10/5)
Last time, in case you hadn’t noticed, we covered Hatcher, chapter 3.A central extension of a group G is a surjective homomorphism
Ee→ G
such that ker e ⊂ Z(E). A covering group is a surjective homeomorphism of topological groupsE → G such that it is a local homeomorphism, where local homeomorphism means if x ∈ G thereis a neighborhood U of x such that e−1 is a discrete disjoint union of neighborhoods isomorphicto U . If E → G is a covering group, then the kernel of e is a discrete normal subgroup ofE, which forces ker e ⊂ Z(E)! SO it is becoming important to make an Algebra-to-Topologydictionary:
central extension: covering map or covering space.G = DG: G is connected.
universal central extension: universal covercentrally closed simply connected
(where DG = 〈[g1, g2] | g1, g2 ∈ G〉, [g1, g2] = g1g2g−11 g−1
2 )
23
13 Deriving the central extension of the Witt algebra (Mon.10/8)
get notesRecall: The Witt algebra has basis
. . . , L−2, L−1, L0, L1, L2, . . .
with bracket[Lm, Ln] = (m− n)Lm+n.
We want g⊕ c (c is abelian) with c central in g⊕ c and
[Lm, Ln] = (m− n)Lm+n + ϕ(Lm, Ln),
whereϕ : g ∧ g→ h.
In order for g⊕ h to be a Lie algebra,
0 = ϕ(Lm, [Ln, Lp]) + ϕ(Lp, [Lm, Ln]) + ϕ(Ln, [Lp, Lm]).
We can normalize ϕ by a coboundary and use
ϕ′(Lm, Ln) = ϕ(Lm, Ln) + ψ([Lm, Ln])
where ψ : g→ c. Soϕ′(Lm, Ln) = ϕ(Lm, Ln) + (m− n)ψ(Lm+n),
...
14 Finite dimensional complex semisimple Lie algebras... (Wed.10/10)
Ram’s least favorite phrase in grad school was “Let g be a finite dimensional complex semisimpleLie algebra...” However, this is an important phrase, so let’s unwind it. It means:
1. g is a finite dimensional vector space (really, isn’t every vector space?)
2. g is a vector space over C (again, aren’t they all?)
3. g is a Lie algebra (duh - we’re in a Lie algebras course)
4. g is semisimple (ok, maybe there’s something here)
What’s semisimple? It’s a lot like reductive, for one.
Try one: A Lie algebra g is semisimple if it is a direct sum of simple Lie algebras,
g = g1 ⊕ g2 ⊕ · · · ⊕ g`
as Lie algebras. A simple Lie algebra is a Lie algebra with no ideals. An ideal of g is a subspacea such that if x ∈ g, a ∈ a, then [x, a] ∈ a.
24
Try two: A Lie algebra g is semisimple if all of the finite dimensional g-modules are semisim-ple. A g-module M is semisimple if M is a direct sum of simple g-modules:
M ∼= M1 ⊕ · · · ⊕M`
as g-modules. A g-module is simple if it has no submodule.Are these two the same? Well, first, g is a g-module, so one is a special case of two. But
in some sense, all g-modules are controlled by g, i.e. g and {g-modules} are the same data.The reconstruction theorems explain how to “reconstruct” g only from information about thecategory of g-modules.
The dual module is M∗ = Hom(M,C) with actions (xϕ)(m) = ϕ(−xm), with x ∈ g andϕ ∈M∗. Why the minus? Recall: Let G be a group. Let M be a G-modules. The dual of M isthe G-module M∗ = Hom(M,C) with actions (gϕ)(m) = ϕ(g−1m).
If M and N are g-modules, the tensor product is the g-module M⊗N with actions x(m⊗n) =xm⊗ n+m⊗ xn. This is like the thing from group theory where if M and N are G-modules,then M ⊗N is a G-modules with actions g(m⊗ n) = gm⊗ gn.
The trivial g-module is Cv with actions xv = 0. The trivial G-module is Cv with actionsgv = v for g ∈ G. THere are two canonical morphisms
M∗ ⊗M → C
(ϕ,m) 7→ ϕ(m)
andC→M ⊗M∗
1 7→∑bi
bi ⊗ b∗i ,
where the sum is over a basis {bi} of M and {b∗i } is the dual basis in M∗. (see... many times wedo not need an inner product in order to get a dual basis, we just need to calculate Hom(M,C).Notice that the second canonical map looks a lot like the trace... the basis vectors are col vectorswith a 1 in the ith place, and dual basis vector are row vectors with a 1 in the ith place.
homework: With the above definitions, show that these maps are g- and G-module homo-morphisms.
15 Bilinear forms and Hopf algebras (Fri. 10/12)
Last time, we defined (without saying it) a Hopf algebra.
Definition. A Hopf algebra is an algebra U with three maps
∆ : U → U ⊗ U,
ε : U → C,
S : U → U
such that
25
1. If M and N are U -modules, then M ⊗N with action
x(m⊗ n) =∑x
x(1)m⊗ x(2)n
where ∆(x) =∑
x x(1) ⊗ x(2), is a U -notation. [Note: this notation we’re using is calledSweedler notation]
2. The vector space C = vC, with actions xv1 = ε(x)v1 is a U -module.
3. If M is a U -module then M∗ = Hom(M,C) with action
(xϕ)(m) = ϕ(S(x)m)
is a U -module.
4. The canonical maps∪ : M∗ ⊗M → C
and∩ : C→M ⊗M∗
are U -module homomorphisms.
15.1 Examples
1. If G is a group and U = CG = C-span{g ∈ G} with
∆(g) = g ⊗ g the coproduct
ε(g) = 1 the counit
S(g) = g−1 the antipode
is a Hopf algebra
2. If g is a Lie algebra, then U = Ug is a Hopf alegbrea with
∆(x) = x⊗ 1 + 1⊗ x,
ε(x) = 0
S(x) = −x
for x ∈ g
For a very long time, these were the only two examples.
15.2 More of what forms can do for us
Definition. A symmetric bilinear form is
〈, 〉 : M ⊗M → C
such that〈x, y〉 = 〈y, x〉
for x, y ∈M .
26
So notice thatM →M∗
viam 7→ 〈m, ·〉
is a vector space homomorphism.Let M be a U -module. And invarient symmetric bilinear form on M is 〈, 〉 : M ×M → C
such that 〈xm1, ym2〉 = 〈m1, S(x)m2〉 for x ∈ U . If G is a group,
〈gm, gn〉 = 〈m,n〉
i.e. 〈gm, n〉 = 〈m, g−1n〉. The invarient part means is that
M →M∗
viam 7→ 〈m, ·〉
is a U -module homomorphism. On the level of vector spaces, the symmetric form doesn’t knowanything about multiplications in U . But an invariant form does!
Let g be a Lie algebra. Then g is a g-module (under the adjoint representation). An ad-invariant form is
〈, 〉 : g⊗ g→ C
such that〈y, z〉 = 〈z, y〉
〈[x, y], z〉 = 〈y, [x, z]〉.
Definerad〈, 〉 = {x ∈ g | 〈x, y〉 for all y ∈ g}
i.e., the kernel ofg→ g∗
viam 7→ 〈m, ·〉.
〈, 〉 is nondegenerate if rad〈, 〉 = 0. Notice that rad〈, 〉 is an ideal of g, i.e. rad〈, 〉 is asubmodule. So if g is simple rad〈, 〉 is zero of the whole thing.
16 (Mon. 10/17)
We’re still working on what it means for g to be a finite dimensional complex semisimple Liealgebra. Again, “ finite dimensional complex” means finite dimensional vector space over C,and “Lie algebra” should be obvious. So we’re just left with “semisimple”. The definition ofchoice has been that g is semisimple if all of the g modules are semisimple. In particular, g isa semisimple g-module. Reduce to simple g, i.e. it has no g-submodules, i.e. it has no stupidideals.
27
Let 〈, 〉 : g⊗ g→ C be an invariant symmetric bilinear form, i.e.
g→ g∗
x 7→ 〈x, ·〉
is a g-module isomorphism (since g is simple – it’s always a homomorphism). Also, being thatit is simple, the only endomorphisms are constant multiples of the identity. Since g carriesa symmetric bilinear form, if {bi} is a basis of g, then there is a dual basis {b∗i } of g where〈bi, b∗i 〉 = δij .
Homework: Convert section five of T. Halverson and A. Ram’s Partition Algebras to Liealgebras.
So g is semisimple if and only if it has a basis and a dual basis (if it’s semisimple, I get aform on each piece, with which I build a good form over the whole thing, and get my dual basis;if it has a basis and dual basis, I can build a good invariant form with these and show that itmust be semisimple).
Recall that the enveloping algebra of g is the algebra generated by g with the relation[x, y] = xy − yx.
Example. Let g = gln(C) = {n× nmatrices with entries in C} with bracket
[x, y] = xy − yx.
The basis of gln is {E− ij | 1 ≤ i, j,≤ n} where Eij has a 1 in the (i, j) entry and 0’s elsewhere.Let V = C. Then gln acts on C. V has basis {v1, . . . , vn} and
xvi =n∑j=1
xjivj
for x ∈ gln. Then define 〈, 〉 : gln × gln → C by
〈x, y〉 = Tr(xV yV )
where xV is the matric of x acting on V . The dual basis with respect to 〈〉 is E∗ij = Eji. Uglnis generated by the symbols Eij . So EijEij 6= 0 in Ugln.
The Casimir element isκ =
∑bi
bib∗i ∈ Ug
where the sum is over the basis {bi} and the dual basis {b∗i }.
Theorem 16.1. Let κ be the Cassimir element of g.
(a) κ does not depend on the choice of basis.
(b) κ ∈ Z(Ug).
So κ commutes with any action of g.
Proof. (a) is a basic result of what it means to change basis
28
(b) leans hard on our form being ad-invariant
We need to get a picture in our heads. A good way to think about this is to think of g beinga complex semisimple Lie algebra means that is contains lots of sl2 subalgebras. What is sl2?Recall: sl2 is the span of elements x, y, h with bracket [x, y] = h, [h, x] = 2x, [h, y] = −2y.
Suppose M is an sl2-module. So h acts on M . So morally, h is a matrix. So h has at leaston eigenvalue and eigenvector. Let v be an eigenvector for h, i.e.
hv = λv
for some λ ∈ C. So
hxv = (xh+ [hx])v= (xλ+ 2x)v= (λ+ 2)xv.
So sl2 makes more eigenvectors. So
hx2v = (xh+ [hx])xv= (x(λ+ 2) + 2x)v= (λ+ 4)x2v.
And hx3v = (λ+ 6)x3v, and, and, and...If M is finite dimensional, the xkv = 0 for some k ∈ Z+. Let v+ = xk−1v. If M is simple
then v+ generates M . Let hv+ = µv+. Then
hyv+ = (yh+ [h, y])v+
= (yµ− 2y)v+
= (µ− 2)yv+
and hy@v+ = (µ− 4)y2v+, and ...So y` = 0 for some ` ∈ Z+. So {v+, yv+, . . . , y`−1v+} will turn out to be a basis of M .
17 (Fri. 10/19)
Thinking of g a semisimple Lie algebra in terms of embedded sl2 subalgebras, where sl2 is thespan of {h, x, y} with bracket
[x, y] = h, [h, x] = x, [h, y] = −2y.
Let M be an sl2-modules. Let v be an eigenvector for h. So xkv = 0 for some k. Let v+ = xk−1v,where k is minimal. Let µ be such that hv+ = µv+. So
hyv+ = (yh+ [h, y])v+
= (yµ− 2y)v+
= (µ− 2)yv+.
29
Similarly hy2v+ = (µ− 4)y2v+. So
v+, yv+, y2v+, . . . , y`−1v+
are eignevectors for h in M . Similarly, xv+ = 0, xyv+ = µv+, xy2v+ = (2µ − 2)yv+, xy3v+ =(3µ − 6)yv+, . . . . So the span of {v+, yv+, y2v+, . . . , y`−1v+} is a submodule of M . If M issimple, then {v+, yv+, y2v+, . . . , y`−1v+} is a basis of M and the sl2-action is given by:
• h is a diagonal matrix with µ, µ− 2, µ− 4, ... , µ− 2(`− 1) on the diagonal,
• y has ones on the sub-diagonal and zeros elsewhere, and
• x has the weights µ, 2µ− 2, 3µ− 6, ... , (`− 1)µ−(`−12 )2 on the super-diagonal.
But [x, y] = xy − yx, so if we multiply these matrices together, we get the relation that
µ− 2(`− 1) = −(`− 1)µ+ (`− 1)(`− 2)
`µ = 2(`− 1) + (`− 1)(`− 2) = `(`− 1)
so µ = `− 1.
Theorem 17.1. The irreducible finite dimensional sl2 modules are indexed by ` ∈ Z≥0 withbasis {v+, yv+, y2v+, . . . , y`−1v+} and action
hyjv+ = (µ− 2j)yjv+, xyjv+ = (jµ− 2(j
2
))yj−1v+, yyjv+ = yj+1v+.
17.1 Jordan decomposition
Let x be a matrix. Then x = xs + xn with xs semisimple (as a matrix, i.e. it is diagonalizable),xn is nilpotent (i.e. yk = 0 for some k, or Tr(yk) = 0 for all k, or all y’s eigenvalues are zero),and xsxn = xnxs. SO if zx = xz, then zxs = xsz and zxn = xnz.
This decomposition comes from Jordan canonical form, where in an appropriate field, everymatrix can be put into a form which looks like eigenvalues on the diagonal and 1’s in he super-diagonal.
Remark: A matrix A is a C[x]-module:
C[x]→ EndV
x→ A
So A is semisimple exactly when V is a semisimple C[x]-module.
Theorem 17.2 (Jasobson-Morozov). If y is a nilpotent matrix then there exist matrices x andh such that
[x, y] = h, [h, x] = x, [h, y] = −2y.
This choice is relatively unique (with some changes in constants).
30
Example: Let X be a geometric space. Let c be the class in H2(X) corresponding to a linebundle P1 in X. Then multiplications by c is a nilpotent operation on H∗(X) (c pushes up andup and up, and runs out eventually). So sl2 acts on the cohomology ring of X.
18 Triangular decomposition (Mon. 10/22)
How we would like to think of complex semisimple Lie algebras.A triangular decomposition of a Lie algebra is
g = n− ⊕ h⊕ n+
as Lie subalgebras, where
h is abelian
n+ is upper triangular
n− is lower triangular
There is a morphism (Cartan involution)
n+ → h−.
Example: The algebragln = n− ⊕ h⊕ n−,
where h is the set of diagonal matrices, n+ is the set of upper triangular matrices, and n− is theset of lower triangular matrics. Any decent , self-respecting Lie algebra lives inside ofr gln. Infact, g acts on itself by the adjoint representation and so
g→ gln
when dimg = n viax 7→ the matrix of the action of x on g.
g is trying to inherit a triangular decomposition from gln. (From last time, given x ∈ g, thereexist xs, xn, where x = xs + xn, with xs congugate to something diagonal.)
Now, Ug gets a decomposition from g:
Ug is generated by g withgenerators x ∈ g, andrelations xy − yx = [x, y].
Alternatively, we can construct Ug with
generators y ∈ n−, h ∈ h, and x ∈ n+, andrelations g1g2 − g2g1 = [g1, g2].
The relations here allow me to move y’s to the left, x’s to the right, leaving h’s in the middle.So we get
U = U−U0U+ = U− ⊗ U0 ⊗ U+
whereU− = Un−, U0 = Uh, U+ = Un+ = C[h1, . . . , hn],
where {h1, . . . , hn} is a basis of h.
31
18.1 Weights
h is a Lie subalgebra of g. So Uh is a Lie subalgebra of Ug. Let M be a g-modules. Then M isan h-module. So M is an C[h1, . . . , hn]-module. But what are the simple C[h1, . . . , hn]-modules?
Simple modules for a commutative algebra are one-dimensional.
Example: C[h] acts on C[h]/ < f >. M os a C[h]-module determined by the action ofh, a matrix with 1’s below the diagonal and last row −(a0, a1, . . . , ad−1)T , with respect to thebasis {1, h, h2, . . . , hd−1} if f = a0 + a1h + a2h
2 + · · · + ad−1hd−1 (this is rational canonical
form). Inside of M there is an m such that hm = µm. So the span of m is a one-dimensionalsubmodule of M , and is simple. If M is a C[h1, h2, . . . , hn]-module, then there exists m ∈ Mand µ1, . . . , µn ∈ C such that him = µim.
SoMµ = {m ∈M | him = µim}
is a direct sum of simple C[h1, . . . , hn]-modules of “type µ”.
Theorem 18.1. The simple C[h1, . . . , hn]-modules are indexed by elements of Cn. If µ =(µ1, . . . , µn) ∈ Cn then the span of m with actions him = µim is a simple C[h1, . . . , hn]-module.
Let M be a g-module. Let µ ∈ Cn. Then
Cn ∼→ Hom(h,C)
µ→ ϕ
whereϕ : h→ C
byhi 7→ µi.
So write µ ∈ h∗. The µ-weight space of M is
Mµ = {m ∈M | hm = µ(h)m for h ∈ h}.
This is the piece of M which is all simple C[h1, . . . , hn]-modules of type µ on M.In our example, C[h]/ < f > for a polynomial f is an indecomposable C[h]-module.
Note: Nobody knows how to classify the indecomposable C[h1, h2]-modules.
19 The Cartan subalgebra, weights, and root systems (Wed.10/24)
Let g be a Lie algebra. Embed g into gln for some n (i.e. let g act on some module M . ThenEnd(M) = gln for some n and
g→ End(M) = gln
x 7→ xM
32
where xM is the matrix of the action of x on M .So g = n− ⊕ h ⊕ n+ (upper triangular + diagonal + lower triangular). Call h the Cartan
subalgebra. If h has basis h − 1, . . . hn, then Uh = C[h1, . . . , hn]. If M is a g-module, then itis a h-modules. How does M look as an h-module? The irreducible Uh-modules correspond toelements µ ∈ h∗:
µ : h→ C
hi 7→ µi
or µ = (µ1, . . . , µn) ∈ Cn and the corresponding Uh-module is Cm with him = µim (i.e.hm = µ(h)im).
The µ-weight space of M is
Mµ = {m ∈M | hm = µ(h)m∀h ∈ h}.
The generalized weight space is
Mgenµ = {m ∈M | if h ∈ h then (h− µ(h))km = 0 for some k ∈ Z+}.
A Jordan block looks like µ’s on the diagonal and 1’s on the super diagonal, so (h− µ(h))k = 0for some k. Then
M =⊕µ∈h+
Mgenµ
as h-modules. IfM =
⊕µ∈h+
Mµ
then all Jordan blocks of h’s are of size 1 and hence h’s act semisimple. In fact...
Theorem 19.1. If g is semisimple and M is finite dimensional then
M =⊕µ∈h+
Mµ.
A weight vector of weight µ is just an element of Mµ.
19.1 Some people think Lie theory is the same as root systems...
Assume g is reductive. Let g act on g (the adjoint representation). A root is a weight of g.
g =⊕µ∈h+
gµ.
µ keeps track of the h-action. h acts on h by 0 since h is abelian (so [h1, h2] = 0). So h ⊂ g0. Infact h = g0 (since h is the maximal abelian subalgebra [we’re using reductive here]). So
g = g0 ⊕
⊕06=µ∈h+
gµ
= h⊕
(⊕α∈R
gα
)
33
whereR = {µ ∈ h∗ | µ 6= 0, gµ 6= 0}.
R is the root system of g.Let 〈, 〉 : g× g→ C be the Killing form:
〈x, y〉 = Tr(adxady),
where adx is the matrix of x acting on g. For us, semisimple = reductive = form is non-degenerate (If g is a finite dimensional complex simple Lie algebra, then there is a unique, up toconstants, symmetric invariant bilinear form... we’ll just always re-normalize to get the Killingform). Let x ∈ gα and y ∈ gβ. So
α(h)〈x, y〉 = 〈α(h)x, y〉= 〈[h, x], y〉= −〈x, [h, y]〉= −〈x, β(h)y〉= −β(h)〈x, y〉
for all h ∈ h. So if α 6= −β (as functions on h, elements of h∗), then 〈x, y〉 = 0 and so gα ⊥ gβ.So let R+ ⊂ R such that
R = R+ ∪R−
whereR− = {−α | α ∈ R+}.
gα has nonzero pairin only with g−α. So
g = h⊕
⊕α∈R+
(gα ⊕ g−α)
For α, β ∈ R+, (gα ⊕ g−α) ⊥ (gβ ⊕ g−β). I would like a basis that respects this decompositionand acts nicely with respect to our form. Let x ∈ gα and y ∈ gβ. Then
[h, [x, y]] = −[y, [h, x]]− [x, [y, h]]= −[y, α(h)x] + [x, [h, y]]= −α(h)[x, y] + β(h)[x, y]= (α(h) + β(h))[x, y]= (α+ β)(h)[x, y].
So if x ∈ gα and y ∈ gβ, then [x, y] ∈ gα+β. So if x ∈ gα and y ∈ g−α, then [x, y] ∈ h. THe usualchoice of R+ is so that
n+ =⊕α∈R+
gα, n+ =⊕α∈R+
g−α
and g = n− ⊕ h⊕ n+ =(⊕
α∈R+ gα)⊕ h⊕
(⊕α∈R+ gα
).
Let x ∈ gα and y ∈ g−α. We know [x, y] ∈ h. So
〈h, [x, y]〉 = −〈[x, y], h〉= 〈[h, x], y]〉= α(h)〈x, y〉
34
Let Hα ∈ h such that 〈h,Hα〉 = α(h). Hα is well-defined since our form is nondegenerate on h
and if x ∈ gα and y ∈ g−α then [x, y] = αx, y〉Hα.Now choose Xα ∈ gα and Yα ∈ g−α such that [Xα.Yα] = Hα, [Hα, Xα] = 2Xα, and [Hα, Yα] =
−2Yα... hello there Mr. sl2!
20 Finding the sl2’s (Mon. 10/29)
Let g be a complex semisimple (reductive, in characteristic 0) finite dimensional Lie algebra.Then g embeds “faithfully” in gln. So
g = n− ⊕ h⊕ n+.
We saw on Wednesday that we also have
g =
⊕α∈R+
g−α
⊕ h⊕
⊕α∈R+
g−α
where gµ is the µ-weiht space for the action of h, and g0 = h. The roots are
R = {α ∈ h∗ | gα 6= 0, α 6= 0},
and R+ is a choice of half of R such that R = R− ∪R+, where
R− = −R+.
We can and will choose Xα ∈ gα, Yα ∈ g−α, and Hα ∈ h such that [Xα.Yα] = Hα, [Hα, Xα] =2Xα, and [Hα, Yα] = −2Yα.
What’s the point? For each α ∈ R+, there is a copy of sl2 sitting inside of g.
Example: Let g = gln. So n+ is the set of upper triangular matrices, and h is the set ofdiagonal matrices. We often write hi = Eii as a basis of h. Our favorite basis of gln is
{Eij | 1 ≤ i, j ≤ n}
and the action of h on gln is given by
[hi, Ejk] = EiiEjk − EjkEii= δijEik − δikEki= (δij − δik)Ekl= (εj(hi)− εk(hi))Ejk= (εj − εk)(hi)Ejk,
where εj ∈ h∗, which sends hi to δij . So the εj ’s form the dual basis.So Ejk ∈ gεj−εk . Then Ekj ∈ gεk−εj = g−(εj−εk). So we can choose R+ = {εj − εk | 1 ≤ j <
k ≤ n}. Notice that [Ejk, Ekj ] = Ejj − Ekk = Hεj−εk , Xεj−εk = Ejk and Yεj−εk = Ekj
35
We are going to exploit the fact that every g-module is an sl2-module (in several ways).Recall: If M is a simple sl2-module, then M can be realized as a 1-dimensional lattice,
chopped off after a finite number of dots, labeled ykv+, and x moves us up and y moves usdown.
Now, SL2 is the group corresponding to the Lie algebra sl2. Let
X =(
0 10 0
), Y =
(0 01 0
), H =
(1 00 −1
).
So let
xα(c) = ecX = 1 + cX =(
1 c0 1
)x−α(c) = ecY = 1 + cY =
(1 0c 1
)sα = ecXe−c
−1Y ecX = · · · =(c 00 c−1
)(0 −11 0
).
Let sα = sα(1). So
sα =(
0 −11 0
).
Then conjugation by sα is an automorphism of SL2.
sαxα(c)s−1α =
(0 −11 0
)(1 0c 1
)(0 1−1 0
)= x−α(−c).
Any finite dimensional sl2-module is an SL2-module. If G is a group and σ is an automor-phism of G and M is a G-module, then define a new module σ∗(M) whose vector is M and thenew action
gm = σ(g)m
(σ : G→ G).In the case of SL2, us the automorphism σ “conjugation by sα”. In this case σ∗(L(5))
(our highest-weight lattice) is a 6-dimensional simple SL2-module. σ “switches X and Y ” andσ∗(L(5)) is just L(5) flipped over. But the lattice is symmetric, so σ∗(L(5)) ∼= L(5).
Remark: σ is also an automorphism of g.
eadXαe−adYαeadXα ∈ G
and soσα : g→ g.
Note that if y ∈ g, then ecadXy = ecXye−cX .So the Adjoint action of g on g corresponds to conjugation action G on g. SO if g is a
cssLa, M a g-module and α ∈ R+, then σα flips all the strings! So a g-module is symmetricin every direction (α ∈ R+). Recall that the weights of M are µ ∈ h∗ such that Mµ 6= 0. SoM =
⊕µ∈h∗Mµ and σα flips all weights and is a reflection of order 2.
The Weyl group is the group acting on h∗ generated by the reflections sα where sα is thereflection on h∗ induced by the automorphism σα.
36
Remark: Notice that all weights of finite dimensional M are integra, since they are integrafor sl2. If Mµ 6= 0, then µ(Hα) ∈ Z.
21 sl3: the gateway to classification (Wed. 10/31)
Let g = sl3 be the set of 3× 3 matrices with zero traces with bracket
[x, y] = xy − yx.
We want to decompose fg as
g = h⊕
(⊕α∈R
gα
),
where h is the set of traceless diagonal matrices (has dimension 2). Weights µ are elements ofh∗
µ : h∗ → C
viah1 7→ µ(h1)
h2 7→ µ(h2).
Last time we used εi such that
εi
a1 0 00 a2 00 0 a3
= ai.
Note: εi ∈ h∗ and ε1 + ε2 + ε3 = 0.
h∗ = C-span{µ1ε1 + µ2ε2 + µ3ε3 | µ1 + µ2 + µ3 = 0}.
Note that E12, E13, E23 are in sl3 and form a basis of n+. If
h =
a1 0 00 a2 00 0 a3
,
then[h,Eij ] = (εi − εj)(h)Eij .
So Eij ∈ gεi−εj . SoR+ = {ε1 − ε2, ε1 − ε3, ε2 − ε3}
and be have our desired decomposition.
37
21.1 drawing h∗
Each α ∈ R+ gives an embedded sl2 inside g and a reflection sα on h∗.The picture is in three-space, with axes labeled ε1, ε2, and ε3 with a plane going through
(1, 0, 0), (0, 1, 0), (0, 0, 1) shifted back to go through the origin. Off to the side, we drew thehexagonal lattice, with hyperplanes Hε1−ε2 , Hε1−ε3 , and Hε2−ε3 .The copy of sl2 correspondingto α has weights in a one-dimensional lattice (in a line-segment), so in our picture, we shouldhave a copy of this lattice in the α direction (perpendicular to Hα). If λ is a weigh in the αdirectiont, so is a point on the lattice, then
〈λ, α∨
is interpreted as the distance of λ to Hα. So we have a formula for our reflection:
sα = λ− 〈λ, α∨〉α.
The reflection sα fixes points on
Hα = {µ ∈ h∗ | 〈µ, α∨ = 0}
(i.e. all the points on the big lattice in three space which sits on the plane perpendicular to theα direction.
sε1−ε2(ε1 − ε2) = −(ε1 − ε2)
and sε1−ε2 fixes Hε1−ε2 = {µ1ε1 + µ2ε2 + µ3ε3 | µ1 = µ2}So back to our big picture. The hexagonal weight lattice gets embedded into the aforemen-
tioned plane. Each sα corresponds to an automorphism σα of g and induces a flip on weights ofg-modules. If M is a finite dimensional g-module, then
M =⊕µ∈h∗
Mµ
anddim(Msαµ) = dim(Mµ).
Back on our hexagonal pieces, mark x’s on lattice points µ if Mµ 6= 0. The weights of Mare in a Z-lattice and are symmetric under the action of the Weyl group, so one point generatesa hexagon (if it’s not on an H). Now, since we have all these little sl2’s, each string of latticepoints in every α-direction must be filled in (in any direction, M is the union of strings).
So, in general, finite dimensional sl3-modules ”look like” hexagons (with two lengths of sides)with strings filled in perpendicular to the H’s. Any structure must be symmetric. Woo.
21.2 Philosophy of classification
What possible finite collections of weights could correspond to g (g is a g-module)? In order tobuild a g, we need a reflection group acting on a lattice (free Z-module, vector space over Z).But I want g to be finite dimensional, so W must also be finite, of which there are very few!
38
22 Classification (Friday. 11/2)
Let g be a finite dimensional complex semisimple Lie algebra.We can only have one of these if we have a finite reflection group W acting on a lattice P
(it’s true that every such pair gives us a fdcsLa, but it’s not easy to see – this is a big theorem).Review: For every α ∈ R+, we get a symmetry (automorphism) σα of g and a symmetry
(reflection) in h∗.W = 〈sα ∈ R+〉
acts on h∗. The set of weights that actually occue in g-modules, P , is a vector space over Z(inside of h∗).
22.1 Step 1: How many W , P possibilities are there?
Let h∗Z = P and h∗C = h∗. Let h∗R = R⊗Z h∗Z = R-span{ω1, . . . , ωn}, where ω1, . . . , ωn is a Z-basisof h∗Z.
Example: Rank 1 dimh∗ = 1. h∗Z is a line of points, so h∗R is a line, and sαλ = 〈λ, α∨〉α,where 〈λ, α∨〉 is the distance from λ to the hyperplane Hα. The set of points which are fixed bythe reflection sα is
(h∗R)sα = {λ ∈ h∗R | sαλ = λ}.
Since a linear transformation sα fixes 0, and we only have rank 1,
W ∼= Z/2Z.
Up to scaling, there is only one choice of (W,P ) in dimension 1. Since W = 〈sα | α ∈ R+〉, weget
g = g−α ⊕ h⊕ g,
which is three-dimensional... so g must be isomorphic to sl2.
Example: Rank 1 dimh∗ = 2. So dimh∗R = 2. If W = 〈sα, sβ, then if the angle between Hα
and Hβ is not π/2, then they will generate more reflections. This will turn out to be symmetriesof the an n-gon (if W has order 2n). In fact,
W = 〈s1, s1s2 | s21 = 1, (s1s2)4 = 1〉.
Some people play the game with the upper half plane with SL2(Z) acting on it. THis has”fundamental regions” and whatnot. We have ”fundamental chambers” and what not, whichcorrespond to the cone with walls Hα’s.
By thought, we presented by generators s1, s2, and relations
s21 = 1 = s2
2, s1s2s1 · · ·︸ ︷︷ ︸nfactors
= s2s1s2 · · ·︸ ︷︷ ︸nfactors
,
where π/2 = Hα1angleHα2 .
Example: Rank big If dimh∗R is big, then we’ll look for embedded dihedral groups.
39
23 More classification: building Dynkin diagrams (Mon. 11/5)
Our goal is to classify finite dimensional complex semisimple Lie algebras by classifying pairs(W,P ) where h∗Z = P is a Z-lattice, W is a finite subgroup of GL(P ) ∼= GLn(Z) generated byreflections
W = 〈sα | α ∈ R+〉
where R+ us an index set for the reflections in W .
23.1 But what is a reflection?
We can make P into a vector space over Q:
h∗Q = Q⊗Z h∗Z.
A reflections is a (semisimple) element sα ∈ GL(h∗Q) such that
(h∗)α = {λ ∈ h∗Q | sαλ = λ}
has codimension 1 in h∗Q, i.e. if dim(h∗Q) = n, then the number of eigenvalue 1 is n − 1. So insome good basis,
sα =
1
. . .1
ξ
or sα =
1
. . .1 1
1
.
Then there is a α∨ ∈ hQ and α ∈ h∗Q such that
sαµ = µ− 〈µ, α∨〉α.
Note
sαα = α− 〈α, α∨〉α= (1− 〈α, α∨〉)α= ξα.
So 〈α, α∨〉 = 1− ξ. The pair α, α∨ is unique up to a constant. If sα is a reflection in GL(h∗R) inthe ordinary sense, then
s2α = 1, ξ = −1, and 〈α, α∨〉 = 2.
23.2 Dealing with the lattice
Last time, we classified the two cases: in rank 1, W = Z/2Z, and in rank 2, W is a dihedralgroup. In rank two, we just need that Hα1∠Hα2 = π/n (since we need W to be finite). But weforgot! W also has a lattice!
We want W ∈ GL2(Z) for the correct basis. Well, use the basis α1, α2 for h∗R:
s1µ = µ− 〈µ, α∨1 〉α1,
s2µ = µ− 〈µ, α∨2 〉α2,
40
where 〈µ, α∨1 〉, 〈µ, α∨2 〉 ∈ Z. In particular in the basis α1, α2
s1 =(−1 −〈α2, α
∨1 〉
0 1
), s2 =
(1 0
−〈α1, α∨2 〉 −1
).
s1α2 = α2 − 〈α2, α∨1 〉.
Then
s1s2 =(−1 + 〈α1, α
∨2 〉〈α2, α
∨1 〉 〈α2, α
∨1 〉
−〈α1, α∨2 〉 −1
).
so det(s1s2) = 1) and tr(s1s2) = −2 + 〈α1, α∨2 〉〈α2, α
∨1 〉.
So char(s1s2) = t2 − (−2 + 〈α1, α∨2 〉〈α2, α
∨1 〉)t+ 1 and eigenvalues of s1s2 are
−2 + 〈α1, α∨2 〉〈α2, α
∨1 〉 ±
√(−2 + 〈α1, α∨2 〉〈α2, α∨1 〉)2 − 42
. These ought to be e2πi/2 and e−2πi/2 since (s1s2)n = 1. So
−2 + 〈α1, α∨2 〉〈α2, α
∨1 〉
2= cos(2π/n).
So 〈α1, α∨2 〉〈α2, α
∨1 〉 = 2 + 2 cos(2π/n) ∈ Z. THe only possibilities are
n = 1:〈α1, α
∨2 〉〈α2, α
∨1 〉 = 2 + 2
so s1 = s2.n = 2:
〈α1, α∨2 〉〈α2, α
∨1 〉 = 2− 2 = 0
. ...In a Dynkin diagram (a graph), the number of edges between nodes i and j is 〈α1, α
∨2 〉〈α2, α
∨1 〉.
23.3 Classifying the (bigger) general case
Rn ∼= h∗R is a union of closures of connected components Cw of
h∗ \ ∪α∈R+hα
(called the fundamental regions for action of W ). Fix a C1. Let Hα1 , . . . ,Hαn be the walls ofC1. The simple reflections are s1, . . . , sn. THe Cartan matrix of W is(
〈αi, α∨j)
1≤i,j≤n .
41
24 Coxeter, Cartan, and Kac-Moody (Fri. 11/9)
Each α ∈ R+ gives an automorphism σα of g and a reflection sα ∈ GL(h∗) such that
Mµ→Msαµ
is a vector space isomorphism when M is a finite dimensional g-module and µ is a weight(eigenvalue for h action). The Weyl group is
W = 〈sα | α ∈ R+〉.
We get P = h∗Z a lattice, the weights which appear in finite dimensional h-modules.Let h∗R = R⊗Z h∗Z and
C = {λ∨ ∈ h∗R | 〈λ, α∨〉 > 0 for all α ∈ R+}
(points on the positive side of hsα = {λ ∈ h∗R | 〈λ, α∨〉 = 0}). The chamber C is a fundamentalregion for the action of W on h∗R.
Let hsα1 , . . . , hsαn be the walls of C and s1, . . . , s2 be the corresponding reflections.
Theorem 24.1 (Coxeter).
W = 〈s1, . . . , sn | s2i = 1, sisjsi · · ·︸ ︷︷ ︸
mij factors
= sjsisj · · ·︸ ︷︷ ︸mij factors
〉
where π/mij = hsαi∠hsαi .
For our calculations, we noticed that the Cartan matrix,
A =(〈α1, α
∨j 〉)
1≤i,j≤n
is a matrix of integers. Recall tha the formula for a reflection is sαλ = λ − 〈λ, α∨〉α, where〈λ, α∨〉 is the “distance from λ to hsα .
24.1 Can we go back? Can we build g from A?
Let A = (aij)1≤i,j≤n be any matrix. By rearanging rows and columns, we can assume that(aij)1≤i,j≤r is non-singular, where r =rank(A). Let ` =corank(A), so that r + ` = n. Defineh = h′ ⊕ d where h′ has basis h1, . . . , hn and d has basis d1, . . . , d`.
Let α1, . . . , αn ∈ h∗ be given by αi(hj) = aij and αi(dj) = δi,j+r. Let h′ = h′/c where c ={h ∈ h′ | αi(h) = 0 for i = 1, . . . , n}. Let c1, . . . , c` be a basis for c for that h1, . . . , hr, c1, . . . , c`, d1, . . . d`is another basis of h. Define ω1, . . . , ω` ∈ h∗ by
ωi(hj) = 0, ωi(cj) = δij , , ωi(dj) = 0.
(these are the fundamental weights!). Let a be the Lie algebra generated by e1, . . . en, f1, . . . fnand h with relations
[h, h′] = 0, for h, h′ ∈ h,
[ei, fj ] = δijhi, [h, ei] = αi(h)ei, [h, fi] = −αi(h)fi.
42
The Borcher-Kac-Moody Lie algebra is g = a/r, where r is the largest ideal of a such thatr ∩ h = 0. It turns out that this thing is semisimple, though it is rarely finite dimensional. g, a,r are graded by
Q =n∑i=1
Zαi
by setting deg(ei) = αi, deg(fi) = −αi, and deg(h) = 0. So, even though g isn’t finite dimen-sional, we can still write
g = g0 ⊕
(⊕α∈R
gα
)where
R = {α ∈ Q | α 6= o, gα 6= 0}
is the set of roots.Let’s just point out that n+ is the subalgebra generated by the e’s, n− is the subalgebra
generated by the f ’s. So n+ =⊕
α∈R+ g + α where R+ = R ∪Q+ and Q+ =∑
i Z≥0αi. So
g = n− ⊕ h⊕ n+ = g′ o d
g′ = [g, g] = n− ⊕ h′ ⊕ n+
c = Z(g) = Z(g′)
andg′ = g′/c = n− ⊕ h′ ⊕ n+
h = h′ ⊕ d, d acts on g′ by derivationsh′ = h′/c.
25 More on the Cartan matrix and Kac-Moody Lie algebras(Mon. 11/12)
There are a couple of directions we could go from here.A (generalized) Cartan matrix is a matrix A = (aij)1≤i,j≤n such that aij ∈ Z, aii = 2 (makes
stuff sl2’s), aij ≤ 0 if i 6= j and aij 6= 0 if and only if aji 6= 0. WHY?? In this case,
C-span{ei, fi, hi} ∼= sl2
so ei and fi act locally nilpotently on g (adjoint representation) and
si = exp(adei) exp(−adfi) exp(adei).
Here, ad is a code for g acting on itself:
adx(y) = [x, y].
The “exponential map” exp means
exp(S) = 1 + S + S2/2 + S3/3! + · · · .
43
The s1, . . . , sn are automorphisms of g and so g has lots of symmetry (this is the only kind thatwe should care about... stuff with lots of symmetry).
Then define si : h→ h and si : h∗ → h∗ by
sih = h− α(h)hi, and siλ = λ(hi)αi
(by stealing the formula for a reflection). Then indeed, as expected,
sigα = gsiα, and sih = sih
(i.e. on weights and on h, si induce reflections).The Weyl group W ⊂ GL(h∗) is
W = 〈s1, . . . , sn〉
(of course we could have said W ⊂ GL(h), but h and h∗ are dual W -modules, so
(wλ)(h) = λ(w−1h)
for λ ∈ h∗, w ∈ W , and h ∈ h.) This W should look kind of small to us... before, we had areflection for every root. We would like to try to get them al back. However, in the non-finite-dimensional case, we can’t get out an sα for every α such that gα 6= 0! What we do know howto do is get these sα for the real roots:
Rre =n⋃i=1
Wαi.
The imaginary roots areRim = R \Rre.
Recall:
g = h⊕
(⊕α∈R
gα
).
If α ∈ Rre, then α = wαi and
eα = wei, fα = wfi, hα = whi.
This gives us some sl2’s and a sα for every real root!A matrix is symmetrizable if there exists a diagonal matrix
E =
ε1 . . .εn
εi ∈ R≥0 such that AE is symmetric. WHY??
Theorem 25.1. g has a nondegenerate symmetric h-invarient bilinear form
〈, 〉 : g× g→ C.
44
Do a tiny calculation to see that 〈, 〉 is determined by
〈ei, fi〉 = εi, 〈hi, h〉 = αi(h)
for h ∈ h.Most papers on this subject begin with the line
“Let g be a symmetizable Kac-Moody Lie algebra...”
Equivalent would be
“Take some e’s and some f ’s and some h’s, use [ei, fi] = hi, [hi, ej ] = aijej , supposethere are lots os symmetries W , and a symmetric g-invarient bilinear form 〈, 〉...”
We said that if A is a Cartan matrix then ei, fi act on g locally nilpotently . More precisely,we must ask for what k is
(adei)k(ej) = 0?
Well, the first such k is 1− aij ! (if ij is off of the diagonal, i.e. if I’m not acting on ei by ei, soaij ≤ 0, and thus 1− aij is positive... we can find this by experiment) The same is true locallyfor the f ’s:
ad(fi)1−aij (fj) = 0.
Where is this coming from?? Remember, from Coxeter (Theorem 24.1):
W = 〈s1, . . . , sn | s2i = 1, sisjsi · · ·︸ ︷︷ ︸
mij factors
= sjsisj · · ·︸ ︷︷ ︸mij factors
〉
where mij = aijaji. So the fancy version of Serre’s theorem goes...
Theorem 25.2 (Serre). (a) If A is a Cartan matrix then these relations hold in g (i.e. (adei)1−aij (ej) ∈
r, and same with f ’s).
(b) If A is symmetrizable then r is generated by the relations above (call them (S1) and (S2))
26 Affine Lie algebras (Wed. 11/14)
Let g0 be a symmetrizable“?” Kac-Moody Lie algebra with bracket [, ]0 and an invariant sym-metric bilinear form 〈, 〉 (given to me by the symmetrizableness).
Note: Most of the literature takes g0 to be finite dimensional complex semisimple.But something seems to be happening in the community that indicates that this isn’treally what we want. We’re not sure if “symmetrizable” is really what we want, butit’s worth trying out...
The loop Lie algebra isg0[t, t−1] = C[t, t−1]⊗C g0
with bracket[tmx, tny]0 = tm+n[x, y]0.
45
This has a symmetric bilinear form
〈tmx, tny〉 ={〈x, y〉0 m+ n = 0,0 otherwise
The universal central extension of g0[t, t−1] is
g′ = g0[t, t−1]⊕ Cc
with bracket[tmx, tny] = tm+n[x, y]0 + δm,−nm〈x.y〉0c
(c is the “central charge” or something).Recall Der (C[t, t−1]) is the Witt Lie algebra
[di, dj ] = (i− j)di+j
where di = ti+1 ddt . Let d = t d
dt and let
g = g0[t, t−1]⊕ Cc⊕ Cd
where [d, tmx] = (t ddt t
m)x = mtmx, [d, c] = 0, and [c, tmx] = 0. The affine Lie algebra is g (org′ or g′).
With Borchards-Kac-Moody Lie algebra we had
g = n− ⊕ h⊕ n+ = g′ o d
g′ = [g, g] = n− ⊕ h′ ⊕ n+
c = Z(g) = Z(g′)
andg′ = g′/c = n− ⊕ h′ ⊕ n+
h = h′ ⊕ d, d acts on g′ by derivationsh′ = h′/c.
Question: When is the affine Lie algebra Kac-Moody?If g0 is finite dimensional complex semisimple then the answer is yes. In this case the two
realizations play against eachother. g0 has decomplosition as an h0-module
g0 = n−0 ⊕ h0 ⊕ n+0 = h0 ⊕
(⊕α∈R
gα
)
Does the affine Lie algebra have something sumilar?
h = h0 ⊕ CcCd
h′ = h0 ⊕ Cc
h′ = h0
are possible Cartan subalgebras to use.
46
27 (Fri. 11/17)
Our story is that we’re working on Affine Lie algebras. We start with a g0, as symmetrizableK.M. Lie algebra. h0 is a Cartan subalgebra, so
g0 = n−0 ⊕ h0 ⊕ n+0 .
And we take beacket [, ]0 and invariant form 〈, 〉0. Ler g′ = g0[t, t−1] = C[t, t−1] ⊗C g. We taked = t d
dt , so d(tm) = mtm.Note g′ (the one with the central extension) surjects onto g′, the loop Lie algebra. On the
other hand, g′ is contained in g = g′ ⊕ Cd.
Example Let g0 = sl2 = span{x, y, h}. As a K.M. Lie algebra, sl2 should have a Cartanmatrix [???] and g0 = Cy ⊕ Ch ⊕ Cx = g−α ⊕ h0 ⊕ gα. If e1 = x, f1 = y, then h1 = h. So thecartan matrix is A = (2). The sl2[t, t−1] = span{tmx, tmy, tmh | m ∈ Z}.
h′ = h0
h′ = h′0 ⊕ Cc
h = h0 ⊕ Cc⊕ Cd
We want to determine how g looks as an h-module. If x ∈ gα and h+ zc+ wd ∈ h, then
[h+ zc+ wd, tmx] = α(h)tmx+ 0 + wmtmx
= (α(h) + wm)tmx= (α+mδ)(h+ zc+ wd)tmx
where δ ∈ h∗ given byδ(c) = 0, δ(d) = 1
and we’ll extend α to an element of h∗ (it started as an element of h∗0) by
α(c) = 0, α(d) = 0.
Then tmx ∈ gα+mδ. So
g = g0[t, t−1]⊕ Cc⊕ Cd
= h0 ⊕
⊕α ∈ Rm ∈ Z
gα+mδ
⊕(⊕m∈Z
gmδ
)
wheregα+mδ = {tmx | x ∈ gα}
gmδ = {tmh | h ∈ h0}.
Note: [d, tmh] = mtmh = mδ(d)tmh.
47
In g = sl2[t, t−1]⊕ Cc⊕ Cd,
g =
(⊕m∈Z
Ctmx
)⊕
(⊕m∈Z
Ctmy
)⊕ Ch⊕ Cc⊕ Cd⊕
⊕m∈Z 6=0
Ctmh
.
If g0 is finite dimensional complex semisimple then g′ = g0[t, t−1] ⊕ Cc. Clearly, this is nolonger finite dimensional. But it’s KM, so it’s supposed to have a Cartan... 2 × 2. THe newCartan matrix for g′ should be 2 × 2... how do we find it?We need to find e0, e1, f0, f1, h0, h1.What would we like to use? We would like to keep e1 = x, f1 = y, and h1 = h, as before. THeconvention for the rest is e0 = ty, f0 = t−1, and h0 = −h+ c. Did we guess right? To check, wewant these 0-dealies to be an sl2:
[e0, f0] = [ty, t−1x]= t1−1[y, x]0 + 1 · 〈y, x〉c= −h+ tr(yx)c= −h+ c.
Hooray!To find the Cartan matrix:
[h0, e0] = α0(h0)e0 = 2e0, [h1, e0] = α0(h1)e0 = −2e0
[h0, e1] = α1(h0)e1 = −2e1, [h1, e1] = α1(h1)e1 = 2e1
(use the above things we guessed to calculate the brackets) and the Cartan matrix for affine sl2is (
2 −2−2 2
)(can go back and forth with this process).
27.1 Beginning representation theory of KM junk
This means that we want to understand g-modules. Again, we can use h to make weightspaces. But it’s highly unlikely that these modules are finite-dimensional, since g itself is infinitedimensional (most will be infinite dimensional).
28 Computing the character of M(λ) (Wed. 11/28)
Le g = sl2[t, t−1]⊕Cc⊕Cd, where if x, y ∈ sl2 then [tmx, tny] = tm+n[x, y]0+δm+n,0〈x.y〉0c, where[, ]0 is the bracket on sl2 and 〈, 〉) is the invarient form on sl2. c ∈ Z(g) and [d, tmx] = mtmx.sl2 has basis x, y, h with [x, y]0 = h, [h, x] = 2x, and [h, y] = −2y. 〈x1, x2〉0 = tr(x1, x2) wherex1, x2 ∈ sl2 are trace zero matrices.
g is a Kac-Moody Lie algebra with generators
e0 = ty, f0 = t−1x, h0 = −h+ c,
48
ei = x, , f1 = y, h1 = h.
Note: [e0, f0] = [ty, t−1x] = [y, x]0 + 〈y, x〉0c = −h+ c. h has basis {h, c, d}, n+ is generated bye0, e1 and n− is generated by f0, f1.
The adjoint action of h on fg gives
g = h⊕
(⊕α∈R
gα
)
= h⊕
(⊕m∈Z
gα+mδ
)⊕
(⊕m∈Z
g−α+mδ
)⊕
⊕m∈Z6=0
gmδ
where gα+mδ = span{tmx}, g−α+mδ = span{tmy}, gmδ = span{tmh}. This is good notationbecause h∗ has basis ω1, ω0, δ where
ω1(h) = 1, ω1(c) = 0, ω1(d) = 0,
ω0(h) = 0, ω0(c) = 1, ω0(d) = 0,
δ(h) = 0, δ(c) = 0, δ(d) = 1,
and α = 2ω1.If γ ∈ h and p ∈ gα+mδ then
[γ, p] = (α+mδ)(γ)p.
What do we mean?[h, tmx] = tm[h, x] = 2tmx = (α+mδ)(h)tmx
[d, tmx] = mtmx = (α+mδ)(d)tmx.
Bracketing e0, e1, f0, f1 never produces any d’s, since [tmx1, tn, x2]0 + δm+n〈x1, x2〉0c. Some
people would like to work instead with g′ = sl2[t, t−1] ⊕ Cc. The point is that the elementst−mx, t−my, t−my m ∈ Z>0 and y are in n−.
fα−mδ = t−mx
f−α−mδ = t−my
f−mδ = t−mh
Then Un− is generated by these and has basis monomials in these symbols. In other words,wecan take things like
(f−α)3 (f−α−2δ)5 (fα−4δ)
1 ...
49
A All things aside
A.1 The Kac-Moody Lie algebra. (9/10)
In about 1965, Serre wrote down generators and relations for complex semi-simple Lie algebrasg. The data is a matrix of numbers called the Cartan matrix, coming from certain polynomialsand their symmetries. The Kac-Moody very deep idea was that “maybe other matrices are fun”,i.e., they generalized the Cartan algebra to find another Lie algebra with many of the same niceproperties.
A.2 What is a reductive I hear so much about? (9/12)
Something is reductive when it is “almost” semi-simple. A module is semi-simple if it decom-poses as a direct sum of simple modules:
M ∼=⊕i
Mi, Mi is simple.
The idea is a (group / Lie algebra / algebra / ring) is (reductive / semi-simple ) if all of itsmodules are semi-simple. But what kind of modules do we care about? This is where so manydefinitions come in relating to the same idea – they all involve different kinds of modules (finitedimensional, integral, topologically smooth, etc.). Alternately, an object is semi-simple if itsradical is trivial. But there are also many kinds of radicals.
A.3 Adeles – the study of everything at once. (9/17)
The story: Our favorite field is really Q. The next one that we learned was R = Q∞, whichis Q completed at the infinite place. In general, Qp is a completion of Q embedded in R. Thenext one we liked was C[x]... but that’s not a field, so we take C(x). But even in high school,C[x] also wasn’t enough, as not everything we cared about was a polynomial – for some reason,everyone seemed to be excited about trig, and other such junk! So then we move on to C[[x]]and C((x)). Besides, our first encounter with R was to see it as infinite decimal expansionsanyway, so really Q∞ = Q[[ 1
10 ]]. Similarly Qp = [[p]] (it’s hard to tell the difference between pand p−1). So we have all of this number junk... all of which we study important stuff over. Sowhy study all of these things individually if we could study all of them at once?
So that’s what the Adeles are, morally – all of the numbers at once:
Adeles =∏p
Qp,
where p can also be ∞. Our corresponding favorite group would then be GLn(∏p Qp). There
are, of course, finite conditions to help make things more comfortable... all but a finite numberof factors are “trivial”, where trivial usually means sitting inside of Zp = Z[[p]]. So basically, tounderstand the Adeles, you just have to get good at doing everything at once.
A.4 Rob wants to know what this “admissible” junk is (9/17)
An admissible module, as far as Langlands is concerned, is a module V such that (V ∗)∗ ∼= V(via the canonical homomorphism).
50
A.5 Tantalizers (9/19)
Tantalizer is short for “tensor power centralizer.” So, if G is a group and V is a G-module, thenEndG(V ⊗k is a tantalizer, where
EndG(M) = {ϕ ∈ End(M) | ϕgm = gϕm ∀m ∈M, g ∈ G} .
A.6 What is this “affine” I keep hearing?
There are many terms in mathematics which are overused, and affine is one of those terms.It generally means a “flat” space, but the various affines (affine variety, affine BMW-algebra,etc.) sometimes have very little to do with eachother. “Projective” comes as the counterpart inalgebraic geometry, but is not necessarily so in the categorical sense (projective is a categoricalterm, with Hom’s and whatnot).
A.7 Homology vs. Cohomology
Sarah feels that Cohomology wins because it has a ring structure and therefore does more foryou. Ram argues that they are the same thing – V = V ∗ if V is finite dimensional. Evenwhen it’s not finite dimensional, the cohomology ring is graded, so can still be matched up withwith homology groups in some way... so most of the time they are the same thing. This is thesame deal with Atiyah and his K-theory vs. cohomology. Also, same deal with Lie algebras andmusic... see Jacob Laurie.
B Those other questions
• When you go to a talk, how do you manage to get work done and at the same time knowwhat the talk was about?
• How do you read a paper?
• Suppose I go to a professor to ask him a question, and he starts to talk with me aboutthe topic. Then he refers me to a book and/or a paper on the subject. Does this usuallymean that he wants me to read those as opposed to bothering him about it, or is thatan invitation to learn more and then come talk to him again, or something else? [ans.Something about absinth... ]
• Who do we ask for money? When should we start asking people for money? How do you(personally) get money? Does this change as we progress through our careers? How muchmoney do we ask for? Will you buy me a new computer? Is it appropriate to ask myadvisor for a new computer? Lets say I have a friend and their advisor is Professor X andthey want a new computer. What advice would you give them with regard to getting anew computer and asking their advisor for money for it?
• How do you prepare seminar? Do you have some advice how to make it more interesting?How do you decide what to say (and how much)? How do you decide what to write onblackboard?
• Why do mathematicians do math? why not philosophy? why not politics or law? Whynot something “useful”?
51
• Who is the greatest mathematician of all time? How about the past 50 years? past10 year? Who is doing the most exciting math right now? Who is the most influentialmathematician of the past 70 years? past 10 year?
• What are the best journals? How much does it matter which journals your papers appearin?
• Why can’t mathematicians ever figure out how to split a bill at a restaurant?
• How does one grade quickly?
In grading, there are two purposes: one is assessment, and one is feedback. So first, onehas to put priorities on these goals. If you are grading entirely for feedback, you have tofigure out what people are getting wrong and then mark on all of the errors. If it’s forassessment, then you have to figure out what deserves points, and mark all of those. Onceyou’ve made a decision, then it’s important to just go.
The other thing that comes with grading, or any other tedious job, it’s so easy to procras-tinate. Something that one can do is to set up little checkpoints. Like make a bunch ofsmall piles of 5-10, and then make it through a pile before you can go do something.
Now, breaking it up like that can make it so that you are in drastically different moods,compromising fairness. But grading is just not fair. So we have to be honest with ourselvesthat the system is not ideal. But we do want to do the best for our students. And it’sbetter to do the best for our students to learn than to try to be fair.
• Rob has been feeling very conflicted recently about what the goal is in teaching. What ishis job?
Turn it around. When you were a calculus student, what did you want your teacher orTA to do?
• Job apps:
Deadlines come Nov 1, Nov 15, Dec 1, Dec 15... Interviews happen at the Joint meetings,beginning of January.
When do we ask for letter? It takes about 2-4 hours. It takes reading through papers tothat they can comment on them. If we average out 3 hours, with thirty letters to do... amonth is needed.
Make sure you get your application pulled at the universities in which you are very inter-ested. Email someone, give a talk, whatnot.
52