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 2
n
dimensional symplectic vector space
V
with a symplectic structure 𝜔, the set of all 2
k
dimensional symplectic subspaces
S
⊂
V
is called the
symplectic Grassmannian
Gr
^{Sp}
(2
k
, 2
n
). This is an open subset of the real Grassmannian
Gr
(2
k
, 2
n
) 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
Gr
^{Sp}
(2
k
, 2
n
) is introduced and identified as
and we discussed natural inclusions
The inclusion
Gr
^{Sp}
(2
k
, 2
n
) ⊂
Gr
(2
k
, 2
n
) is the complement of a hyperplane in 𝜦
^{2k}
V
, and the inclusion
is the intersection with
, and topologically it is a deformation retract. In particular, the deformation retract of
Gr
^{Sp}
(2
k
, 2
n
) 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
Gr
^{Sp}
(2
k
, 2
n
) 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
2
k

dimensional symplectic subspace
S
in
a
2
n

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
Gr
^{Sp}
(2
k
, 2
n
) (Theorem 9)
In
[2]
, each element in
Gr
^{Sp}
(2
k
, 2
n
) is considered as an element of 𝜦
^{2k}
V
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 2
n
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) Thesymplectic complementofPisP𝜔:= {u∊V｜ 𝜔 (u,v) = 0 for allv∊P}

(2) ThenullspaceofPisN(P) :=P∩P𝜔and its dimension is called the nullityn(P) := dimN(P)

(3) Thesymplectic rank,or simply rank, ofPisr(P) :=

(4) Any two vectorsuandvwith 𝜔 (u,v) = 0 (resp.g(u,v) = 0) are called 𝜔orthogonal(resp.gorthogonal)
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+Q𝜔and (iii)P⊂QiffP𝜔⊃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
⊂
P
^{𝜔}
(resp.
P
^{𝜔}
⊂
P
), and this condition is equivalent to 𝜔｜
_{P}
= 0; namely
P
=
N
(
P
) (resp.
P
^{𝜔}
=
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
⊕
P
^{𝜔}
.
It is wellknown 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 iffP⊥is symplectic.

(3)P is coisotropic iffP⊥is isotropic.
Symplectic Grassmannians
Definition 2.
Given any
k
≤
n
, the set of all 2
k
dimensional symplectic linear subspaces in a 2
n
dimensional symplectic vector space
V
is called the
symplectic Grassmannian
, and it is denoted as
Gr
^{Sp}
(2
k
,
V
), or simply
Gr
^{Sp}
(2
k
, 2
n
).
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
(2
k
, 2
n
) of all 2
k
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
(2
k
, 2
n
) with a homogeneous space of
O
(2
n
) with the isotropy subgroup the product of orthogonal groups of
P
and
P
^{⊥}
. Therefore, we have
Gr
(2
k
, 2
n
) ≃
O
(2
n
) /
O
(2
k
)
O
(2
n
– 2
k
). Moreover, for the same reason, there is also a canonical identification between
Gr
^{Sp}
(2
k
, 2
n
) 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
Gr
^{Sp}
(2
k
, 2
n
) and
Gr
^{Sp}
(2
n
– 2
k
, 2
n
) 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 2
k
≤
n
.
In
[2]
,
acts on
, it also acts on
Gr
(2
k
, 2
n
) which is the space of linear subspaces of
V
. we have a disjoint union decomposition
where 𝓞
_{r}
is the orbit of rank
k
subspaces. 𝓞
_{k}
=
Gr
^{Sp}
(2
k
, 2
n
) is the unique open orbit in
Gr
(2
k
, 2
n
). In particular the complement of
Gr
^{Sp}
(2
k
, 2
n
) is a hypersurface in
Gr
(2
k
, 2
n
).
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 𝜔
^{k}
(𝜍) denotes
where
e
_{1}
,
e
_{2}
, ...,
e
_{2k}
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
Gr
^{Sp}
(2
k
, 2
n
).
𝜔
basis and
𝜔
decomposition of symplectic subspaces
Lemma 3.
Let
(
V
, 𝜔,
J
,
g
)
be a
2
n

dimensional Hermitian vector space
.
For any
2
k

dimensional symplectic subspace S in V
,
there is an ordered orthonormal basis
{
u
_{1}
,
v
_{1}
, ...,
u
_{k}
,
v
_{k}
}
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
And there is a maximizing oriented orthonormal pair of vectors
u
_{1}
and
v
_{1}
in
S
satisfying
. Here,
g
(
J
u
_{1}
,
v
_{1}
) is nonzero because
S
is symplectic.
For each unit vector
w
in
S
∩
span
{
u
_{1}
,
v
_{1}
}
^{⊥}
, we observe that the function

f(𝜽) :=g(Ju1,cos𝜽v1+ sin 𝜽w)
has a maximum value at 𝜽 = 0 and hence the derivative of
f
at 𝜽 = 0 vanishes, 0 =
f
′ (0) =
g
(
Ju
_{1}
,
w
), i.e.
w
⊥
Ju
_{1}
. Similarly, we also obtain
w
⊥
Jv
_{1}
. Therefore any
g
orthogonal vector
w
to span {
u
_{1}
,
v
_{1}
} is also 𝜔orthogonal, and a maximizing pair (
u
_{1}
,
v
_{1}
) is isolated.
By Lemma 1,
S
∩
span
{
u
_{1}
,
v
_{1}
}
^{⊥}
is symplectic. Thus we can repeat the above process for
S
∩
span
{
u
_{1}
,
v
_{1}
}
^{⊥}
and obtain an ordered orthonormal basis {
u
_{1}
,
v
_{1}
, ...,
u
_{k}
,
v
_{k}
} 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 𝜔orthogonal except (
u
_{1}
,
v
_{1}
)
i
= 1, ...,
k
.
Because the value of
g
(
Ju
_{i}
,
v
_{i}
) 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 2
n
dimensional Hermitian vector space. For any 2
k
dimensional symplectic subspace
S
in
V
, an ordered orthonormal basis
on
S
such that

1 ≥ λ1 > λ2 > ... > λm > 0
where
,
i
= 1, ...,
m
, and 𝜔orthogonal for other pairs in the basis is called a 𝜔
basis
. Moreover, we call each
in
S
a λ
_{i}
component of
S
and
S
=
S
_{λ1}
⊕ ... ⊕
S
_{λm}
a 𝜔decomposition (or symplectic decomposition) of
S
.
Remark.
The referee comments that the 𝜔decomposition is simply the decomposition of the eigenspaces of
A
^{2}
, where
A
is the nondegenerate skewsymmetric matrix associated to the symplectic form 𝜔.
It is natural to ask the uniqueness of 𝜔decomposition of a symplectic subspace
S
in (
V
, 𝜔,
J
,
g
). At first, we consider the following Lemma.
Lemma 5.
Let S be a
2
k

dimensional symplectic subspace in a
2
n

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
) = λ
_{1}
,
then span
{
u
,
v
} ⊂
S
_{λ1}
.
Proof
. By applying the 𝜔decomposition of
S
given by the basis
B
,
u
can be written as
u
=
a
_{0}
u
_{0}
+
a
_{+}
u
_{+}
where
u
_{0}
and
v
_{+}
are orthonormal vectors in
S
_{λ1}
and
, respectively. Since
u
is a unit vector, the coeffcients
a
_{0}
and
a
_{+}
satisfy
. Similarly,
v
can be written as
v
=
b
_{0}
v
_{0}
+
b
_{+}
v
_{+}
, and we have
.
Since
S
_{λ1}
and
are also 𝜔orthogonal, we get 𝜔 (
u
_{0}
,
v
_{+}
) = 𝜔 (
v
_{0}
,
u
_{+}
) = 0, and we obtain
Here we use the fact that the value of 𝜔 on each
S
_{𝜆i}
is in [𝜆
_{i}
, 𝜆
_{i}
] so that we have 𝜔
^{2}
(
u
_{0}
,
v
_{0}
) ≤ 𝜆
_{1}
and 𝜔
^{2}
(
u
_{+}
,
v
_{+}
) ≤ 𝜆
_{2}
. To get the equality in the above, we need
a
_{+}
=
b
_{+}
= 0 so that
u
and
v
are in
S
_{𝜆1}
. This gives the Lemma 5.
By applying Lemma 5 inductively, we conclude the following Theorem.
Theorem 6.
For each
2
k

dimensional symplectic subspace S in a
2
n

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 1paramameter family of symplectic subspaces which in fact gives a path from [
S
] in
Gr
^{Sp}
(2
k
, 2
n
) to an element in
.
Theorem 7.
For each
2
k

dimensional symplectic subspace S in a
2
n

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 𝜔basis {
u
_{1}
,
v
_{1}
, ...,
u
_{k}
,
v
_{k}
} of
S
such that

g(Ju1,v1) ≥g(Ju2,v2) ≥ ... ≥g(Juk,vk) > 0,
and
S
=
span
{
u
_{1}
,
v
_{1}
} ⊕... ⊕
span
{
u
_{k}
,
v
_{k}
}. 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 𝜔decomposition of
S
, we need to show that the construction of symplectic subspace
S
(
t
) is independent of the choice of the 𝜔basis on
S
.
First, we observe {
U
_{1}
(
t
) ,
V
_{1}
(
t
) , ...,
U
_{k}
(
t
) ,
V
_{k}
(
t
)} is a 𝜔basis of
S
(
t
).
By applying
one can check that

(i) {U1(t) ,V1(t) , ...,Uk(t) ,Vk(t)} is an orthonormal basis ofS(t),

(ii) 𝜔 (U1(t) ,V1(t)) ≥ 𝜔 (U2(t) ,V2(t)) ≥ ... ≥ 𝜔 (Uk(t) ,Vk(t)) > 0, and

(iii) 𝜔orthogonal for other pairs in {U1(t) ,V1(t) , ...,Uk(t) ,Vk(t)}.
Thus {
U
_{1}
(
t
) ,
V
_{1}
(
t
) , ...,
U
_{k}
(
t
) ,
V
_{k}
(
t
)} is a 𝜔basis of
S
(
t
), indeed.
It is useful to note that if 𝜔 (
u
_{i}
,
v
_{i}
) = 𝜔 (
u
_{j}
,
v
_{j}
), then 𝜔 (
U
_{i}
(
t
) ,
V
_{i}
(
t
)) = 𝜔 (
U
_{j}
(
t
) ,
V
_{j}
(
t
)). Therefore if
span
{
u
_{i}
,
v
_{i}
} and
span
{
u
_{j}
,
v
_{j}
} are in the same component of 𝜔decomposition of
S
, span {
U
_{i}
(
t
) ,
V
_{i}
(
t
)} and span {
U
_{j}
(
t
) ,
V
_{j}
(
t
)} must be in the same component of 𝜔decomposition of
S
(
t
) for each
t
.
Second, we want to show that all 𝜔basis on
S
produces the same
S
(
t
) at each
t
having the same 𝜔decomposition of
S
(
t
).
Suppose {
u
_{1}
,
v
_{1}
, ...,
u
_{m}
,
v
_{m}
} in the given basis {
u
_{1}
,
v
_{1}
, ...,
u
_{k}
,
v
_{k}
} forms the λ
_{1}
component of
S
, i.e.

λ1= 𝜔 (u1,v1) = ... = 𝜔 (um,vm)
and

Sλ1=span{u1,v1, ...,um,vm} .
For another 𝜔basis
B'
on
S
, by Lemma 5
B'
has an ordered orthonormal pair
u
,
v
contained in
S
_{λ1}
with 𝜔 (
u
,
v
) = λ
_{1}
. As the above, we define
U
(
t
) and
V
(
t
)
and we want to show that
U
(
t
) and
V
(
t
) are in
span
{
U
_{1}
(
t
) ,
V
_{1}
(
t
) , ...,
U
_{m}
(
t
) ,
V
_{m }
(
t
)} for each
t
.
Since
u
and
v
are in
S
_{λ1}
, 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
) = λ
_{1}
, we get
Here we use the fact that {
u
_{1}
,
v
_{1}
, ...,
u
_{m}
,
v
_{m}
} is a subset of the 𝜔basis producing λ
_{1}
component in
S
. By combining above three equations in (
A
2) and (
A
3), we obtain
which implies
x
_{i}
=
w
_{i}
and
y
_{i}
= –
z
_{i}
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
{
U
_{1}
(
t
) ,
V
_{1}
(
t
) , ...,
U
_{m}
(
t
) ,
V
_{m}
(
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 𝜔basis on
S
produces the same
S
(
t
) at each
t
having the same 𝜔decomposition 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 𝜔bases on a 2
k
dimensional symplectic subspace
S
is acted by block diagonal unitary matrices in
U
(
k
) ⊂
O
(2
k
). Thus we have the following corollary.
Corollary 8.
Let S be a
2
k

dimensional symplectic subspace in a
2
n

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
(
a
_{1}
,
a
_{2}
, ...,
a
_{m}
)
which is the set of block diagonal unitary matrices in
whose blocks have sizes
(2
a
_{1}
, 2
a
_{2}
, ..., 2
a
_{m}
)
as real matrices
.
Remark.
From (
A
4) and (
A
5) in Theorem 7, we conclude that the action of block diagonal unitary matrices
U
(
a
_{1}
,
a
_{2}
, ...,
a
_{m}
) on the set of all the 𝜔bases on
S
and the set of all the 𝜔bases 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
Gr
^{Sp}
(2
k
, 2
n
), indeed.
Theorem 9.
Let
(
V
, 𝜔,
J
,
g
)
be a
2
n

dimensional Hermitian vector space. Then the complex Grassmannian
is a strong deformation retract of the symplectic Grassmannian
Gr
^{Sp}
(2
k
, 2
n
).
Proof
. First, we take a 𝜔basis on
S
for each [
S
] in
Gr
^{Sp}
(2
k
, 2
n
). By applying Theorem 7 for
S
and the 𝜔basis, we obtain an 1parameter family
S
(
t
) of symplectic subspaces which is uniquely determined by 𝜔decomposition of
S
. As in Theorem 7, the construction of symplectic subspace
S
(
t
) is independent of the choice of the 𝜔basis on
S
. Therefore we have the following well defined map
which can be easily seen to be a strong deformation retract from
Gr
^{Sp}
(2
k
, 2
n
) to
.
Remark.
1. If we vary the complex structure
J
on
V
, then the corresponding
will move inside
Gr
^{Sp}
(2
k
, 2
n
) and cover the whole symplectic Grassmannian (see
[2]
).

2. ThusandGrSp(2k, 2n) are homotopically equivalent to each other.

3. The dimensions of λispacesSλiin a 𝜔decomposition ofSis preserved under deformation.
Acknowledgements
The author would like to thank the referees for their valuable comments which helped to improve the manuscript. This research is supported by Basic Science Research Program through the National Research Foundation of Korea(NRF) funded by the Ministry of Education, Science and Technology(No.2014027205).