symplectic geometry 2

8/6/2019 Symplectic Geometry 2

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Outline Review of the basics. Using the polar decomposition. A coordinate description of the symplectic group. Eigenvalues of a

Lecture 2The symplectic group.

Shlomo Sternberg

July 9, 2010

Shlomo Sternberg

Lecture 2 The symplectic group.

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The purpose of today’s lecture is to assemble various facts aboutthe symplectic group. Many of these facts are special cases of general facts about Lie groups, but we will prove them directly in

the symplectic case.

We begin with a review of some definitions and facts abutsymplectic vector spaces that we proved last time.

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1 Review of the basics.

2 Using the polar decomposition.The group Sp (V ) is connected.The dimension of Sp (V ).

3 A coordinate description of the symplectic group.

4 Eigenvalues of a symplectic matrix.

5 The Lie algebra of Sp (V ).

6 Polar decomposition of elements of Sp (V ).

7 The Cartan decomposition of sp (V ).8 Compact subgroups of Sp (V ).

9 Gaussian generators of Sp (V ).

Shlomo Sternberg


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Symplectic vector spaces.

Let V be a finite dimensional vector space over the real numbers.

A symplectic structure on V consists of an antisymmetric bilinearformω : V × V → R

which is non-degenerate. A vector space equipped with asymplectic structure is called a symplectic vector space.

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The simplest example.

A basic example is R2 with

ωR2

ab

,

c d

:= det

a b c d

= ad − bc .

We will call this the standard symplectic structure on R2.So if u , v ∈ R

2 then ωR2 (u , v ) is the oriented area of the

parallelogram spanned by u and v .

Shlomo Sternberg


O

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Special kinds of subspaces.

If W is a subspace of symplectic vector space V then W ⊥ denotesthe symplectic orthocomplement of W :

W ⊥

:= {v ∈ V | ω(v , w ) = 0, ∀w ∈ W }.

A subspace is called

1 symplectic if W ∩ W ⊥ = {0},

2 isotropic if W ⊂ W ⊥,

3 coisotropic if W ⊥ ⊂ W , and

4 Lagrangian if W = W ⊥.

Shlomo Sternberg


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Symplectic subspaces.

Since (W ⊥)⊥ = W by the non-degeneracy of ω, it follows that W

is symplectic if and only if W

⊥

is. Also, the restriction of ω to anysymplectic subspace W is non-degenerate, making W into asymplectic vector space.

Conversely, to say that the restriction of ω to W is non-degeneratemeans precisely that W ∩ W ⊥ = {0}.

Shlomo Sternberg


O tli R i f th b i U i th l d iti A di t d i ti f th l ti Ei l f

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Normal forms.

Let V be a symplectic vector space. For any non-zero e ∈ V wecan find an f ∈ V such that ω(e , f ) = 1 and so the subspace W

spanned by e and f is a two dimensional symplectic subspace.Furthermore the map

e →

10

, f →

01

gives a symplectic isomorphism of W with R2 with its standardsymplectic structure. We can apply this same construction to W ⊥

if W ⊥

= 0. Hence, by induction, we can decompose anysymplectic vector space into a direct sum of two dimensionalsymplectic subspaces:

V = W 1 ⊕ · · · W d .

Shlomo Sternberg


Outline Review of the basics Using the polar decomposition A coordinate description of the symplectic group Eigenvalues of a

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V = W 1 ⊕ · · · W d

where dim V = 2d proving that

every symplectic vector space is even dimensional

and where the W i are pairwise (symplectically) orthogonal andwhere each W i is spanned by e i , f i with ω(e i , f i ) = 1. In particularthis shows that

all 2d dimensional symplectic vector spaces are isomorphic, and

isomorphic to a direct sum of d copies of R2 with its standard

symplectic structure.

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Existence of Lagrangian subspaces.

Let us collect the e 1, . . . , e d in the above construction and let L bethe subspace they span. It is clearly isotropic. Also,e 1, . . . , e d , f 1, . . . , f d form a basis of V . If v ∈ V has the expansion

v = a1e 1 + · · · ad e d + b 1f 1 + · · · + b d f d

in terms of this basis, then ω(e i , v ) = b i . So v ∈ L⊥ ⇒ v ∈ L.Thus L is Lagrangian. So is the subspace M spanned by the f ’s.

Conversely, if L is a Lagrangian subspace of V and if M is acomplementary Lagrangian subspace, then ω induces anon-degenerate linear pairing of L with M and hence any basise 1, · · · e d picks out a dual basis f 1, · · · .f d of M giving a basis of V

of the above form.

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Consistent Hermitian structures.

In terms of the basis e 1, . . . , e d , f 1, . . . , f d introduced above,consider the linear map

J : e i → −f i , f i → e i .

It satisfies

J 2 = −I , (1)

ω(Ju , Jv ) = ω(u , v ), and (2)

ω(Ju , v ) = ω(Jv , u ). (3)

Any J which satisfies two of the three conditions aboveautomatically satisfies the third. (1) says that J makes V into ad -dimensional complex vector space. (2) says that J is asymplectic transformation, i.e acts so as to preserve the symplecticform ω. (3) says that ω(Ju , v ) is a real symmetric bilinear f orm.

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All three conditions (really any two out of the three) say that( , ) = ( , )ω,J defined by

(u , v ) = ω(Ju , v ) + i ω(u , v )

is a semi-Hermitian form whose imaginary part is ω. For the J

chosen above this form is actually Hermitian, that is, the real partof ( , ) is positive definite.

For the proof of various facts about the symplectic group it will be

convenient to focus on the positive definite case.

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g p p p y p g p g

Recall that a complex structure on a real vector space V is an

automorphism J : V → V such that J 2

= −Id .Definition

A complex structure J on a symplectic vector space V withsymplectic form ω is called compatible if the bilinear form g

defined byg (u , v ) := ω(Ju , v )

is a positive definite inner product on V .

A compatible complex structure makes V into a Hermitian vector

space that is, into a complex inner product space with Hermitianmetric

h(v , w ) = g (v , w ) + i ω(v , w ).

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Starting with such a compatible complex structure, let e 1, . . . , e d be an orthonormal basis of V thought of as a d -dimensionalcomplex vector space with Hermitian form h. Let f j := −Je j . Thenω(e j , e k ) = ω(f j , f k ) = 0 for all j and k and

ω(e i , f j ) = Imh(e i , −Je j ) = δij .

So the e , f form a symplectic basis of V when we think of V as areal symplectic vector space.

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Review of some facts from linear algebra.

We recall the following facts from linear algebra. Let V be a real

vector space with a positive definite scalar product, g . If A : V → V is linear transformation its adjoint is defined byg (Au , v ) = g (u , A†v ) for all u , v ∈ V . If A = A† we say that A isself-adjoint, and we say that a linear transformation O isorthogonal if OO † = Id .

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Square roots of non-negative self-adjoint matrices.

It is a theorem that every self-adjoint matrix C can be diagonalized- more precisely, that there is a decomposition of V into a direct

sum of mutually orthogonal subspaces such that the restriction of C to each subspace is multiplication by a real number. If C = B 2

where B is self-adjoint (so that C is also), then the decompositionfor B works for C . In particular, if C is non-negative g (Cu , u ) ≥ 0

for all u ) then C has a unique non-negative square root.

Shlomo SternbergLecture 2 The symplectic group.


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Polar decomposition.

Proposition

Let A : V → V be an invertible linear transformation on a real

vector space with a positive definite scalar product, g. Then we

can write

A = PO

where P is self-adjoint and positive definite and O is orthogonal

and this decomposition is unique.

Proof.AA† is self-adjoint and positive. So it has a unique positive squareroot. If A = PO then AA† = P 2 showing that P and hence O isunique. If P is the square root of AA† then(P −1A) · (P −1A)† = P −1AA†P −1 so O := P −1A is orthogonal.



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Using the polar decomposition in symplectic geometry.

For any real vector space V let Riem(V ) denote the convex opensubset of the space S 2(V ∗) consisting of all positive definitesymmetric bilinear forms on V .

Now let (V , ω) be a symplectic vector space, and let J (V , ω)denote the space of all compatible positive definite complexstructures on V . We have associated to each J ∈ J (V , ω) anelement g of Riem(V ). So we have defined a map

G : J (V , ω) → Riem(V ), J → g .



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On the other hand, for every k ∈ Riem(V ), there is an invertiblelinear transformation A : V → V which is uniquely determined by

k (u , v ) = ω(Au , v ).

Since ω is anti-symmetric it follows that A is skew-adjoint(A = −A†) with respect to k . Let us apply the polar

decomposition A = PO . We have

A = −A† = −O †P = (O †PO )(−O −1).

We conclude from the uniqueness of the polar decomposition that

OP = PO and O = −O †.

So O 2 = −Id . In other words O is a complex structure which weshall now denote by J . So A = PJ = JP and J 2 = −I d .



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We claim that J is ω-compatible. Indeed

ω(Ju , v ) = ω(AP −1u , v ) = k (P −1u , v ) = k (P −12 u , P −

12 v ).

So g (u , v ) = k (P −12 u , P −

12 v ) is positive definite and J is

ω-compatible. We have proved

Theorem

k → J as defined above is a map

F : Riem(V ) → J (V , ω)

and

F ◦ G = Id .



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The space J (V , ω) has a topology (inherited from the topology of the space of all linear transformations of V ) as does the spaceRiem(V ) (as an open convex set in S 2(V ∗)). The maps G and F

are clearly continuous. Since Riem(V ) is convex, and hencecontractible we conclude that

Proposition

J (V , ω) is contractible.



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The group Sp (V ) is connected.

The group Sp (V ) is connected.

Fix a compatible complex structure J so that the correspondingunitary group (which we may denote by U (V , J )) is the subgroupconsisting of elements which commute with J . Let e 1, . . . e d be anorthonormal basis for J and (e , f ) = (e 1, . . . , e d , −Je 1, . . . , −Je d )

the corresponding symplectic basis of V (as a real vector space). If J is another compatible complex structure with a correspondingsymplectic basis (e , f ), there is a symplectic transformationcarrying (e , f ) to (e , f ). So Sp (V ) acts transitively on J (V , ω)and

J (V , ω) = Sp (V )/U (V , J ).

Since J (V , ω) is contractible and U (d ) is connected we see thatSp (V ) is connected.



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The dimension of Sp (V ).


All symplectic structures on V are equivalent. This means thatGl (V ) acts transitively on the open subset of ∧2(V ∗) consisting of symplectic structures. The dimension of ∧2(V ∗) is 2d (2d − 1)/2

and the dimension of GL(V ) is (2d )2

. SoProposition

The dimension of Sp (V ) is

2d (2d + 1)2 .



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The dimension of the space of Lagrangian subspaces.

Fix J ∈ J (V , ω) and a Lagrangian subspace L of V . Choose abasis e 1, ....., e d of L which is orthonormal relative to the metricg = G (J ), so e 1, ..., e d is an orthonormal basis of V relative to theassociated Hermitian form (and complex structure). If L isanother Lagrangian subspace and e

1, . . . , e

d an orthonormal basis

of it, then there will be a unitary map U such that Ue j = e j for all j . So the group U (V , J , h) which is a subgroup of Sp (V ) actstransitively on the space of Lagrangian subspaces. The stabilizergroup of L consists of those unitary transformations which are real.

So if dim V = 2d we can say that the space of all Lagrangiansubspaces of V is diffeomorphic to U (d )/O (d ). In particular, it isa manifold of dimension

d 2 −d (d − 1)

2=

d (d + 1)

2.



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Fix a J ∈ J (V , ω) whose associated symmetric form is g = G (J ).We let M † denote the adjoint of M ∈ Gl (V ) relative to g .M ∈ Sp (V ) if and only if ω(Mu , Mv ) = ω(u , v ) for all u , v ∈ V .Since ω(u , v ) = g (u , Jv ) this says that g (Mu , JMv ) = g (u , Jv ), i.ethat

M †JM = J .

In terms of a standard symplectic basis our choice of J will have

the block decomposition J =

0 −I

I 0

where I is the d × d

identity matrix. Suppose we use the corresponding blockdecompositions for M :

M =

A B

C D

.



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M =

A B

C D

so M † =

At C t

B t D t

where At denotes the transpose of A as a d × d matrix etc. Then

M †JM =

C t A − At C C t B − At D D t A − B t C D t B − B t D

.

So the condition M †JM = J becomes

At C = C t A, B t D = D t B , and At D − C t B = I .



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We can rewrite the condition M †JM = J as

M † = JM −1J −1

so det M = det M † = det M −1 and hence det M =±

1. SinceSp (V ) is connected, this implies that

det M = 1

for any M ∈ Sp (V ).



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The eigenvalues of a symplectic matrix.

If λ is an eigenvalue of a real matrix so is λ. So the non-realeigenvalues of any real matrix occur in complex conjugate pairs.

The eigenvalues of M † are the same as the eigenvalues of M . If M is symplectic so that M † = JM −1J −1, the eigenvalues of M † arethe same as the eigenvalues of M −1. So if λ is an eigenvalue of asymplectic matrix, so is λ−1.

In more detail:



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TheoremThe eigenvalues of M ∈ Sp (V ) occur as either

real pairs λ and λ−1, λ = ±1

complex pairs λ and λ−1 = λ λ = ±1

complex quadruples, λ not real, |λ| = 1, λ , λ−1, λ , λ−1,

λ = −1 or

λ = 1.

The multiplicity of −1 and of 1 is even.

All that remains to be proved is the last assertion.



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Proof.

We have already verified that the eigenvalues which are not = ±1are of the first three types. The product of all the eigenvalues of

these types is 1, and det M = 1, so the multiplicity of −1 is even.There are an even number of eigenvalues of the first four types andthe size of M is even, so the multiplicity of 1 must also beeven.



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The Lie algebra of Sp (V ).

By definition, this consists of all linear transformations ξ of V suchthat the corresponding one parameter group

exp t ξ = 1 + t ξ +1

2t 2ξ2 +

1

3!t 3ξ3 + · · ·

is a subgroup of Sp (V ). Differentiating the equation

ω(exp t ξu , exp t ξv ) ≡ ω(u , v )

with respect to t and setting t = 0 gives the condition

ω(ξu , v ) + ω(u , ξv ) = 0, ∀u , v ∈ V

as the condition for ξ to belong to the Lie algebra of Sp (V ). Wedenote this Lie algebra by sp (V ). (The above condition is alsosufficient as can be seen by solving a linear diff erential equation.)



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If we set M = exp t ξ and differentiate the equation M †JM ≡ J

with respect to t and set t = 0 we obtain the condition

ξ†J + J ξ = 0.

If we use a block decomposition ξ =a b

c d

we obtain theconditions

b = b†, c = c

†, d = −a†.

Notice that the dimension of the possible b or c is d (d +1)2 while the

dimension of the possible a is d 2 so once again we verify that thedimension of Sp (V ) is d (2d + 1).



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Polar decomposition of elements of Sp (V ).

Fix a compatible complex structure J and hence a correspondingpositive definite scalar product g on V . Every invertible lineartransformation the has a polar decomposition relative to g . LetM ∈ Sp (V ) and M = PO its polar decomposition. I wish toshow that P and O both belong to Sp (V ). Then since O preservesboth ω and g it belongs to U (V , J ). As J is fixed, I will denoteU (V , J ) by U (V ). I will denote the set of positive definite matricesbelonging to Sp (V ) by P. I will show that the map

P × U (V ) → Sp (V ), (P , O ) → PO

is a diffeomorphism and that the space P is contractible. Thisimplies that Sp (V ) is homotopically equivalent to U (V ). Since thefundamental group of U (V ) is Z we conclude that thefundamental group of Sp (V ) is also Z.



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Polar decomposition of elements of Sp (V ), continued.

Suppose that M ∈ Sp (V ). We know thatM † = JM −1J −1 ∈ Sp (V ) and hence that MM † ∈ P. Since the P

in the polar decomposition of M is obtained by taking the positivedefinite square root of MM † we will have proved our result if weshow that the positive definite square root of an element of P

belongs to P. For this we begin with a lemma:



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Lemma

Let M ∈ Sp (V ) be symmetric, i.e. M = M † so that all the

eigenvalues of M are real. Let V λ denote the eigenspace

corresponding to the eigenvalue λ. Then

V ⊥λ =

µ|µλ=1

V µ

where the orthogonal complement is taken with respect to ω.



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If u ∈ V λ and v ∈ V µ then

ω(u , v ) = ω(Mu , Mv ) = λµω(u , v )

so if λµ = 1 then V µ ⊂ V ⊥λ . So the direct sum of the V µ with

µλ = 1 is contained in V ⊥

λ . We will have proved the lemma if weshow that in all cases the sum of the dimensions of theV µ, µλ = 1 is dim V − dim V λ, since dim V ⊥λ = dim V − dim V λ.

We will prove this dimensional equality by examining the variouspossibilities for eigenvalues as provided by the theorem. Noticethat since M is symmetric, all eigenvalues are real.



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Proof.

If λ = 1, then µλ = 1 means that µ = 1 and the sum of thedimensions of the V µ in question is the sum of the dimensions of the eigenspaces corresponding to eigenvalues = 1. So we are ok.Notice that in this case V 1 is a symplectic subspace of V . Similarlyfor λ = −1. If λ = ±1 then the µ in question consist of all

µ = λ−1

. But, by the theorem, the multiplicity of λ−1

is the sameas the multiplicity of λ. So

dim V − dim V λ = dim V − dim V λ−1

and since M is diagonalizable, this is the sum of the dimensions of the V µ, µ = λ−1, so we are also ok in this case. Notice that in thecase λ = ±1, the space V λis isotropic but V λ ⊕ V λ−1 is symplectic.



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Back to our claims about the polar decomposition of

elements of Sp (V ).

Suppose that T ∈ P so that all its eigenvalues are positive. We

will show that there is a one parameter group s → T s of elementsof P with T 1 = T . In particular, taking s = 1

2 will show that thepositive square root of T belongs to Sp (V ) which is what we wantto prove.

We proceed by examining the eigenspaces of T :



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T acts as the identity on V 1 so on this subspace we may take

T s

≡ id .If λ = 1 then λ > 0 and V λ ⊕ V λ−1 is a symplectic subspace onwhich T has the block decomposition

T =λI 0

0 λ−1I

, λ > 0.

So if we set σ = log λ then the entire one parameter group

T

s

:=e s σI 0

0 e −σI

belongs to Sp (V λ ⊕ V λ−1 ). We have proved all our claims.



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Let p ⊂ sp (V ) denote the set of symmetric elements, i.e. thosesatisfying ξ† = ξ. Our proof shows that the exponential map exprestricts to a diffeomorphism exp : p → P. Let t = u (V ),the Lie algebra of U (V ), and let g = sp (V ). From the above polardecomposition we see that

g = t ⊕ p

as vector spaces. Since U (V ) = Sp (V ) ∩ O (V ) it follows that p isstable under conjugation by elements of U (V ) and hence that

[t , p] ⊂ p.

Since the commutator of two symmetric matrices is antisymmetric,we have [p, p] ⊂ t . Since t is a Lie subalgebra, we have [t , t ] ⊂ t .



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The Cartan decomposition of sp (V ).

We have shown that

g = t ⊕ p

[t , t ] ⊂ t

[t , p] ⊂ p

[p, p] ⊂ t .

This is an example of what is known as a Cartan decomposition of a real semi-simple Lie algebra.

Shlomo Sternberg



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Compact subgroups of Sp (V ).

Let H ⊂ Sp (V ) be a compact subgroup. By averaging over H wecan find a k ∈ Riem(V ) which is invariant under all elements of H , and hence so is the corresponding J H . So H is a subgroup of

U (V , J H ). Since Sp (V ) acts transitively on J (V , ω) all suchU (V , J ) are conjugate. So we have proved

Proposition

Any compact subgroup of Sp (V ) is a subgroup of some U (V , J )

for some compatible J and all U (V , J ) are conjugate under Sp (V ).

Shlomo Sternberg



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Linear Optics.

I wish to show that matrices of the formI dI

0 I

d ∈ R and

I 0S I

S = S t

generate the symplectic group. This will be a somewhat tediouscomputation. But a consequence of will be the that the groupSp (4) corresponds to linear optics just as Sp (2) corresponds to

Gaussian optics. Indeed the 4 × 4 matrix

I dI

0 I

corresponds to

straight line propagation in a medium of constant index of

refraction while the matrix

I 0S I

S = S t corresponds to a

linearized version of Snell’s law.

Shlomo Sternberg



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Let G be the subgroup of Sp (V ) generated by matrices of the form

I dI

0 I

d ∈ R and

I 0S I

S = S t .

We wish to show that G is all of Sp (V ). To start, we have

I I

0 I

I 0

−I I

I I

0 I

=

0 I

−I 0

so

0 I

−I 0

∈ G .

Shlomo Sternberg



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Next

0 I −I 03

=

0 −I I 0

so

0 −I I 0

∈ G . Next

0 −I

I 0

I 0

−S I

0 I

−I 0

=

I S

0 I

so all matrices of the formI S

0 I

, S = S t

belong to G .

Shlomo Sternberg



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If S = S t and is invertible, we have

I S −1

0 I I 0

−S I I S −1

0 I

= 0 S −1

−S 0

so all matrices of the form

0 S −1

−S 0

with S symmetric and invertible belong to G . Now

0 −I

I 0

0 S −1

−S

0 =

S 0

0S −1 .

So all matrices of the form

S 00 S −1

with S symmetric and

invertible belong to G .Shlomo Sternberg



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I claim that the following lemma holds:

Lemma

Every non-singular n × n matrix can be written as the product of

three non-singular symmetric matrices.

Assuming the lemma, we find that any matrix of the form

A 00 (A−1)t

(with A nonsingular) belongs to G . Indeed, write

A = S 1S 2S 3 so (A−1)t = S −11 S −1

2 S −13 and each of the matrices

S i 0

0 S −1

i

belongs to G .

I temporarily postpone the proof of the lemma.

Shlomo Sternberg



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I next will show that any A B

C D ∈ Sp (V ) with A non-singular

belongs to G : Consider the equation

I 0

−E I

A B

C D

=

A B

C − EA D − EB

.

If we choose E = CA−1 we get C − EA = 0. We know from thecondition of belonging to Sp (V ) that At C = C t A which tells usthat E = CA−1 is symmetric. So for this choice of E we have

A B

C D

= I 0

E I A B

0 D − EB

.

Shlomo Sternberg



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A B

C D

= I 0

E I A B

0 D − EB

.

We know that the first factor belongs to G . In the second factorwe know that D − EB = (At )−1 because the full matrix belongs toSp (V ). Now

A B

0 (A−1)t

=

A 00 (A−1)t

I A−1B

0 I

and both factors on the right belong to G since A−1B is

symmetric. So (modulo the lemma) we have shown that any

matrix

A B

C D

∈ Sp (V ) with A non-singular belongs to G .

Shlomo Sternberg



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Now we consider the case where A is singular, say of rank r < d .Row and column reduction says that we can find non-singular n × n

matrices Q and R such that RAQ =

I r 00 0

where I r is the r × r

identity matrix. By pre and post multiplying byR 0

0 (R t )−1

andQ 00 (Q t )−1

which belong to G we are left with showing that

matrices A B

C D ∈ Sp (V ) with A =

I r 0

0 0 belong to G .

Shlomo Sternberg



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Write down the block decomposition of C corresponding to theblock decomposition of A, i.e.

C =

C 1 C 2C 3 C 4

so that

At C = C 1 C 2

0 0 .

The condition that At C be symmetric implies that C 2 = 0 andthat C 1 is symmetric. Let

E :=0 0

0 I d −r

so

I E

0 I

∈ G .

Shlomo Sternberg



I E

A B

A + EC B + ED

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Then

I E

0 I

A B

C D

=

A + EC B + ED

C D

and

A + EC =

I r 0C 3 C 4

.

I claim that this matrix is non-singular which would then completethe proof that G = Sp (V ) (up to the proof of the lemma). Indeed,

if this matrix were singular, it would mean that C 4 is singular,which would mean that there is a non-zero vector v whose first r

components vanish, and is sent into 0 by C (since C 2 = 0). Butthen Av = 0 and Cv = 0 contradicting the conditionD t A − B t C = I . So we are left with proving the

Lemma

Every non-singular n × n matrix can be written as the product of

three non-singular symmetric matrices.

Shlomo Sternberg



http://find/

http://goback/



By polar decomposition A = PO where P is positive definite (inparticular symmetric) and O is orthogonal. So we are reduced toproving

LemmaEvery orthogonal matrix O can be written as the product of two

symmetric matrices.

Shlomo Sternberg



http://find/

http://goback/



Proof.

We can block diagonalize O , that is write O = RBR −1 where R isorthogonal and B consists of a block of matrices along thediagonal where each block is either a two by two block of arotation matrix or is a one by one block of ±1. If B = S 1S 2 thenRBR −1 = (RS 1R −1) · (RS 2R −1) ) and the conjugate of a

symmetric matrix by an orthogonal matrix is symmetric. So we arereduced to the block diagonal case, which means that we arereduced to the one or two dimensional case. A one by one matrixis always symmetric while

cos θ sin θ

− sin θ cos θ

=

1 00 −1

cos θ sin θsin θ − cos θ

Shlomo Sternberg


http://find/

http://goback/

symplectic geometry 2

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