SYMPLECTIC DECOMPOSITION OF SYMPLECTIC SUBSPACES

Abstract

We introduce a decomposition on a symplectic subspace determined by symplectic structure and study its properties. As a consequence, we give an elementary proof of the deformation of the Grassmannians of symplectic subspaces to the complex Grassmannians.

1. INTRODUCTION

On a 2n-dimensional symplectic vector space V with a symplectic structure 𝜔, the set of all 2k-dimensional symplectic subspaces S ⊂ V is called the symplectic Grassmannian GrSp (2k, 2n). This is an open subset of the real Grassmannian Gr (2k, 2n) and contains the complex Grassmannian consisted of those subspaces S which are complex for a linear complex structure on V .

In [2], the symplectic Grassmannian GrSp (2k, 2n) is introduced and identified as

and we discussed natural inclusions

The inclusion GrSp (2k, 2n) ⊂ Gr (2k, 2n) is the complement of a hyperplane in 𝜦2kV , and the inclusion is the intersection with , and topologically it is a deformation retract. In particular, the deformation retract of GrSp (2k, 2n) is as natural as the deformation from to U (n) which is related to the Siegel upper half space representing the space of all compatible complex structures on (V, 𝜔). But the proof in [2] contains tedious issues left to the reader, and it turns out that many readers find diffculties to fill in the detail of arguments. The objective of this paper is to show the deformation retract of GrSp (2k, 2n) via rather elementary methods.

We introduce a notion of 𝜔-basis and 𝜔-decomposition of a symplectic subspace S in (V, 𝜔) determined by the symplectic structure 𝜔 on V (Definition 4), and show that the decomposition is uniquely determined (Lemma 3). Moreover, we obtain the following key Theorem 7,

Theorem. For each 2k-dimensional symplectic subspace S in a 2n-dimensional Hermitian vector space (V, 𝜔, J, g), there is an 1-parameter family of symplectic subspaces S (t), t ∊ [0, 1] in V such that S (0) = S and S (1) is a complex subspace in V . Moreover, it is uniquely determined by the 𝜔-decomposition of S.

As a consequence we show that complex Grassmannian is a strong deformation retract of the symplectic Grassmannian GrSp (2k, 2n) (Theorem 9)

In [2], each element in GrSp (2k, 2n) is considered as an element of 𝜦2kV but in this paper we rather consider it as a linear subspace.

 

2. PRELIMINARIES

In this section, we recall the definition of symplectic Grassmannians and discuss their basic properties and their relationships with real and complex Grassmannians in [2].

In this article, V always denotes a 2n-dimensional symplectic vector space with symplectic form 𝜔. Moreover, we consider a Hermitian structure (𝜔, J, g) on V where J is a complex structure and g is a metric structure on V satisfying

𝜔 (u, v) = g (Ju, v) , g (Ju, Jv) = g (u, v)

for any u, v ∊ V . Note that any two among symplectic, complex and metric structures on V are called compatible with each other if they define the third structure via the relation 𝜔 (u, v) = g (Ju, v) and together they make V into a Hermitian vector space.

Symplectic complement, null space and symplectic rank For any linear subspace P in symplectic vector space (V, 𝜔), we recall the following notions.

(1) The symplectic complement of P is P𝜔 := {u ∊ V ｜ 𝜔 (u, v) = 0 for all v ∊ P} (2) The null space of P is N (P) := P ∩ P𝜔 and its dimension is called the nullity n (P) := dim N (P) (3) The symplectic rank, or simply rank, of P is r (P) := (4) Any two vectors u and v with 𝜔 (u, v) = 0 (resp. g(u, v) = 0) are called 𝜔-orthogonal (resp. g-orthogonal)

The null space N (P) is the largest subset of P where the restriction of 𝜔 vanishes, and the rank r (P) is half of the maximal possible dimension among subspaces in P where 𝜔 is nondegenerate.

Symplectic complements satisfy the following basic properties:

(i) (P𝜔)𝜔 = P, (ii) (P ∩ Q)𝜔 = P𝜔 + Q𝜔 and (iii) P ⊂ Q iff P𝜔 ⊃ Q𝜔.

And we also have dim P = 2 · r (P) + n (P).

Subspaces in symplectic spaces A linear subspace S of V is called symplectic if the restriction of 𝜔 to S defines a symplectic structure on S, equivalently n(S) = 0. Thus, S must be of even dimension. And a subspace P is called isotropic (resp. coisotropic) if P ⊂ P𝜔 (resp. P𝜔 ⊂ P), and this condition is equivalent to 𝜔｜P = 0; namely P = N (P) (resp. P𝜔 = N (P)). When a subspace is both isotropic and coisotropic, we call it Lagrangian.

In particular, one can obtain the following equivalent statements for symplectic subspaces : (1) P is a symplectic subspace in V ; (2) n (P) = 0; (3) dim P = 2 · r (P) and (4) V = P ⊕ P𝜔.

It is well-known that the symplectic complement of a symplectic (resp. isotropic) subspace in (V, 𝜔) is symplectic (resp. coisotropic). As a matter of fact, the similar statements for orthogonal complements are also true (See [2] for detail).

Lemma 1. Let P be any linear subspace in a Hermitian vector space (V, 𝜔, J, g). For the orthogonal complement P⊥ of P in V , we have the following.

(1) JN (P) = N(P⊥) (2) P is symplectic iff P⊥ is symplectic. (3) P is coisotropic iff P⊥ is isotropic.

Symplectic Grassmannians

Definition 2. Given any k ≤ n, the set of all 2k-dimensional symplectic linear subspaces in a 2n-dimensional symplectic vector space V is called the symplectic Grassmannian, and it is denoted as GrSp (2k, V), or simply GrSp (2k, 2n).

Because the symplectic condition on a subspace is an open condition, a natural choice of topology for the symplectic Grassmannian is one given by being an open subset of the real Grassmannian Gr (2k, 2n) of all 2k-dimensional real linear subspaces in V .

We pick a metric g on V so that we obtain orthogonal decomposition V = P⊕P⊥ for each subspace P in V . By considering the orthonormal bases of P and P⊥ which also give one on V , we identify Gr (2k, 2n) with a homogeneous space of O (2n) with the isotropy subgroup the product of orthogonal groups of P and P⊥. Therefore, we have Gr (2k, 2n) ≃ O (2n) /O (2k)O (2n – 2k). Moreover, for the same reason, there is also a canonical identification between GrSp (2k, 2n) and the homogeneous space and if we equip a complex vector space (V, J) with a Hermitian metric g, then the complex Grassmannian can be identified with U (n) /U (k)U (n – k).

Therefrom we have the canonical inclusions

correspond to the following inclusions of homogeneous spaces

 

3. RETRACT OF SYMPLECTIC GRASSMANNIANS

By Lemma 1, the symplectic Grassmannians GrSp (2k, 2n) and GrSp (2n – 2k, 2n) are dual to each other. Thus we could restrict our attention to only those subspaces in V which are at most half dimensional. For the rest of this section, we assume that 2k ≤ n.

In [2], acts on , it also acts on Gr (2k, 2n) which is the space of linear subspaces of V . we have a disjoint union decomposition

where 𝓞r is the orbit of rank k subspaces. 𝓞k = GrSp (2k, 2n) is the unique open orbit in Gr (2k, 2n). In particular the complement of GrSp (2k, 2n) is a hypersurface in Gr (2k, 2n).

On the other hands, we recall is a calibration satisfying the Wirtinger’s inequality which is

and equality sign holds if and only if 𝜍 is complex subspace, i.e. (see [1]). Here 𝜔k (𝜍) denotes

𝜔k (e1 ∧e2... ∧ e2k)

where e1, e2, ..., e2k is an oriented orthonormal basis of 𝜍. Equivalently, is a calibration with contact set . Note we have

In the below, we show that is actually a strong deformation retract of GrSp (2k, 2n).

𝜔-basis and 𝜔-decomposition of symplectic subspaces

Lemma 3. Let (V, 𝜔, J, g) be a 2n-dimensional Hermitian vector space. For any 2k-dimensional symplectic subspace S in V , there is an ordered orthonormal basis {u1, v1, ..., uk, vk} on S such that

𝜔 (u1, v1) ≥ 𝜔 (u2, v2) ≥ ... ≥ 𝜔 (uk, vk) > 0

and 𝜔-orthogonal for other pairs in the basis.

Proof. For each oriented pair of orthonormal vectors u and v in S, we consider

𝜔 (u, v) = g (Ju, v) .

And there is a maximizing oriented orthonormal pair of vectors u1 and v1 in S satisfying . Here, g (Ju1, v1) is nonzero because S is symplectic.

For each unit vector w in S∩ span {u1, v1}⊥, we observe that the function

f (𝜽) := g (Ju1, cos 𝜽v1 + sin 𝜽w)

has a maximum value at 𝜽 = 0 and hence the derivative of f at 𝜽 = 0 vanishes, 0 = f′ (0) = g (Ju1,w), i.e. w ⊥ Ju1. Similarly, we also obtain w⊥Jv1. Therefore any g-orthogonal vector w to span {u1, v1} is also 𝜔-orthogonal, and a maximizing pair (u1, v1) is isolated.

By Lemma 1, S∩ span {u1, v1}⊥ is symplectic. Thus we can repeat the above process for S∩ span {u1, v1}⊥ and obtain an ordered orthonormal basis {u1, v1, ..., uk, vk} of S such that

𝜔 (u1, v1) = g (Ju1, v1) ≥ g (Ju2, v2) ≥ ... ≥ g (Juk, vk) > 0,

and each pair of vectors in the basis is 𝜔-orthogonal except (u1, v1) i = 1, ..., k.

Because the value of g (Jui, vi) can be repeated, we give a refined index to the basis in the above Lemma 3 so as to have the following definition.

Definition 4. Let (V, 𝜔, J, g) be a 2n-dimensional Hermitian vector space. For any 2k-dimensional symplectic subspace S in V , an ordered orthonormal basis

on S such that

1 ≥ λ1 > λ2 > ... > λm > 0

where , i = 1, ..., m, and 𝜔-orthogonal for other pairs in the basis is called a 𝜔-basis. Moreover, we call each

in S a λi-component of S and S = Sλ1 ⊕ ... ⊕Sλm a 𝜔-decomposition (or symplectic decomposition) of S.

Remark. The referee comments that the 𝜔-decomposition is simply the decomposition of the eigenspaces of A2, where A is the non-degenerate skew-symmetric matrix associated to the symplectic form 𝜔.

It is natural to ask the uniqueness of 𝜔-decomposition of a symplectic subspace S in (V, 𝜔, J, g). At first, we consider the following Lemma.

Lemma 5. Let S be a 2k-dimensional symplectic subspace in a 2n-dimensional Hermitian vector space (V, 𝜔, J, g) and let

be a 𝜔-basis on S as in Definition 4. If u, v are ordered orthonormal vectors in S with 𝜔 (u, v) = λ1, then span {u, v} ⊂ Sλ1 .

Proof. By applying the 𝜔-decomposition of S given by the basis B, u can be written as u = a0u0 + a+u+ where u0 and v+ are orthonormal vectors in Sλ1 and , respectively. Since u is a unit vector, the coeffcients a0 and a+ satisfy . Similarly, v can be written as v = b0v0 + b+v+, and we have .

Since Sλ1 and are also 𝜔-orthogonal, we get 𝜔 (u0, v+) = 𝜔 (v0, u+) = 0, and we obtain

Here we use the fact that the value of 𝜔 on each S𝜆i is in [-𝜆i, 𝜆i] so that we have 𝜔2 (u0, v0) ≤ 𝜆1 and 𝜔2 (u+, v+) ≤ 𝜆2. To get the equality in the above, we need a+ = b+ = 0 so that u and v are in S𝜆1 . This gives the Lemma 5.

By applying Lemma 5 inductively, we conclude the following Theorem.

Theorem 6. For each 2k-dimensional symplectic subspace S in a 2n-dimensional Hermitian vector space (V, 𝜔, J, g), the 𝜔-decomposition of S is uniquely determined.

Now, for each symplectic subspace S, we define an 1-paramameter family of symplectic subspaces which in fact gives a path from [S] in GrSp (2k, 2n) to an element in .

Theorem 7. For each 2k-dimensional symplectic subspace S in a 2n-dimensional Hermitian vector space (V, 𝜔, J, g), there is an 1-parameter family of symplectic subspaces S (t), t ∊ [0, 1] in V such that S (0) = S and S (1) is a complex subspace in V . Moreover, it is uniquely determined by the 𝜔-decomposition of S.

Proof. By Lemma 3, we obtain a 𝜔-basis {u1, v1, ..., uk, vk} of S such that

g (Ju1, v1) ≥ g (Ju2, v2) ≥ ... ≥ g (Juk, vk) > 0,

and S = span {u1, v1} ⊕... ⊕ span {uk, vk}. And for each 0 ≤ t ≤ 1 and i = 1, ..., k, we define

and consider

S (t) := span {U1 (t) , V1 (t) , ..., Uk (t) , Vk (t)} .

One can check that S (t) is symplectic for each t, and S (0) = S and S (1) is a complex subspace in V .

To show the family is uniquely determined by the 𝜔-decomposition of S, we need to show that the construction of symplectic subspace S (t) is independent of the choice of the 𝜔-basis on S.

First, we observe {U1 (t) , V1 (t) , ..., Uk (t) , Vk (t)} is a 𝜔-basis of S (t).

By applying

one can check that

(i) {U1 (t) , V1 (t) , ..., Uk (t) , Vk (t)} is an orthonormal basis of S (t), (ii) 𝜔 (U1 (t) , V1 (t)) ≥ 𝜔 (U2 (t) , V2 (t)) ≥ ... ≥ 𝜔 (Uk (t) , Vk (t)) > 0, and (iii) 𝜔-orthogonal for other pairs in {U1 (t) , V1 (t) , ..., Uk (t) , Vk (t)}.

Thus {U1 (t) , V1 (t) , ..., Uk (t) , Vk (t)} is a 𝜔-basis of S (t), indeed.

It is useful to note that if 𝜔 (ui, vi) = 𝜔 (uj , vj ), then 𝜔 (Ui (t) , Vi (t)) = 𝜔 (Uj (t) , Vj (t)). Therefore if span {ui, vi} and span {uj ,vj} are in the same component of 𝜔-decomposition of S, span {Ui (t) , Vi (t)} and span {Uj (t) , Vj (t)} must be in the same component of 𝜔-decomposition of S(t) for each t.

Second, we want to show that all 𝜔-basis on S produces the same S (t) at each t having the same 𝜔-decomposition of S(t).

Suppose {u1, v1, ..., um, vm} in the given basis {u1, v1, ..., uk, vk} forms the λ1- component of S, i.e.

λ1 = 𝜔 (u1, v1) = ... = 𝜔 (um, vm)

and

Sλ1 = span {u1, v1, ..., um, vm} .

For another 𝜔-basis B' on S, by Lemma 5 B' has an ordered orthonormal pair u, v contained in Sλ1 with 𝜔 (u, v) = λ1. As the above, we define U (t) and V (t)

and we want to show that U (t) and V (t) are in span {U1 (t) , V1 (t) , ...,Um (t) , Vm (t)} for each t.

Since u and v are in Sλ1 , we write u and v as

u = x1u1 + y1v1 + ... + xmum + ymvm v = z1u1 + w1v1 + ... + zmum + wmvm,

where the coeffcients are real numbers. Since u and v are unit vectors, we have

and because 𝜔 (u, v) = λ1, we get

Here we use the fact that {u1, v1, ..., um, vm} is a subset of the 𝜔-basis producing λ1-component in S. By combining above three equations in (A-2) and (A-3), we obtain

which implies xi = wi and yi = –zi for each i, and that u and v can be written

Now, U (t) and V (t) can be written

which shows that U (t) and V (t) are in span {U1 (t) , V1 (t) , ..., Um (t) , Vm (t)}. Here we use

g (v + Ju, v + Ju) = g (–u + Jv,–u + Jv) = 2 + 2𝜔 (u, v) = 2 + 2𝜔 (ui, vi) = g (vi + Jui, vi + Jui) = g (–ui + Jvi,–ui + Jvi)

for i = 1, ..., m.

By applying this procedure inductively, we can conclude that all 𝜔-basis on S produces the same S (t) at each t having the same 𝜔-decomposition of S(t).

From the proof of the Theorem 7, we observe that the expression of u and v in (A-2) is related to Therefore, the set of all the 𝜔-bases on a 2k-dimensional symplectic subspace S is acted by block diagonal unitary matrices in U(k) ⊂ O (2k). Thus we have the following corollary.

Corollary 8. Let S be a 2k-dimensional symplectic subspace in a 2n-dimensional Hermitian vector space (V, 𝜔, J, g). Suppose S has a 𝜔-basis as in Definition 4, then the set of all the 𝜔-bases on S is acted by U (a1, a2, ..., am) which is the set of block diagonal unitary matrices in whose blocks have sizes (2a1, 2a2, ..., 2am) as real matrices.

Remark. From (A-4) and (A-5) in Theorem 7, we conclude that the action of block diagonal unitary matrices U (a1, a2, ..., am) on the set of all the 𝜔-bases on S and the set of all the 𝜔-bases on S (t) is compatible via the definition of vectors at t in (A-1).

Now we show that is a strong deformation retract of GrSp (2k, 2n), indeed.

Theorem 9. Let (V, 𝜔, J, g) be a 2n-dimensional Hermitian vector space. Then the complex Grassmannian is a strong deformation retract of the symplectic Grassmannian GrSp (2k, 2n).

Proof. First, we take a 𝜔-basis on S for each [S] in GrSp (2k, 2n). By applying Theorem 7 for S and the 𝜔-basis, we obtain an 1-parameter family S(t) of symplectic subspaces which is uniquely determined by 𝜔-decomposition of S. As in Theorem 7, the construction of symplectic subspace S (t) is independent of the choice of the 𝜔-basis on S. Therefore we have the following well defined map

which can be easily seen to be a strong deformation retract from GrSp (2k, 2n) to .

Remark. 1. If we vary the complex structure J on V , then the corresponding will move inside GrSp (2k, 2n) and cover the whole symplectic Grassmannian (see [2]). 2. Thus and GrSp (2k, 2n) are homotopically equivalent to each other. 3. The dimensions of λi-spaces Sλi in a 𝜔-decomposition of S is preserved under deformation.