Symplectic Spaces

Published June 22, 2020 (updated: June 23, 2020) in linear-algebra. Tags: linear-algebra, symplectic, vector-spaces.

We come across vector spaces equipped with an inner-product frequently. But instead of an inner product, we could have a skew-symmetric and non-degenerate bilinear form on it. Such spaces are called as symplectic vector spaces. More formally,

Definition: Let $V$ be an $n$ -dimensional real vector space and $\omega: V\times V\to \mathbb{R}$ be a bilinear map such that

1. $\omega(u,v)=-\omega(v,u)$ . (skew-symmetry)
2. $\omega(u,v)=0$ for all $v\in V$ implies that $u=0$ . (non-degeneracy)

Then the pair $(V,\omega)$ is called a symplectic vector space where $\omega$ is called the symplectic form.

The biggest difference when working with a symplectic space as compared to an inner-product space is that the subspaces need not be well-behaved. That is, you can have subspaces $W$ of $V$ which are not symplectic with respect to the restricted form $\omega|_{W\times W}$ . This makes the theory of symplectic spaces much more interesting. Let’s first construct the analog of an orthonormal basis for symplectic spaces.

Let $(V,\omega)$ be a symplectic space. We chose two vectors $u_1$ and $v_1$ from $V$ such that $\omega(u_1,v_1)\neq 0$ . This is possible because of non-degeneracy (condition 2). Normalize one of these vectors so that $\omega(u_1,v_1)=1$ . Now we argue that $V$ can be decomposed as the direct sum of $W$ and $W^{\bot}$ , where $W=\text{span}\{u_1,v_1\}$ and $W^{\bot}=\{v\;|\;\omega(v,w)=0 \;\forall w\in W\}$ .

If $v\in W$ then $v=au_1+bv_1$ . If also $v\in W^{\bot}$ , then this would imply that $a=b=0$ .

If $v\in V$ and we have $\omega(v,u_1)=-b$ , $\omega(v,v_1)=a$ , then we can write $v=(au_1+bv_1)+(v-au_1-bv_1)$ .

So $V=W\oplus W^{\bot}$ . Now we can perform the same set of steps on the space $W^{\bot}$ to get vectors $\{u_2,v_2\}$ . Iterating this many times, we can write $V=W_1\oplus W_2\oplus\hdots W_m$ , where $W_{k}=\text{span}\{u_k,v_k\}$ . Hence the set $\{u_1,u_2,\hdots,u_m,v_1,v_2\hdots,v_m\}$ forms a special basis for our symplectic space $V$ . This basis has the following properties

$\omega(u_i,v_j)=\delta_{ij}$ for $1\leq i,j \leq m$ .
$\omega(u_i,u_j)=0=\omega(v_i,v_j)$ for $1\leq i,j \leq m$ .

This set is called as a symplectic basis for $(V,\omega)$ . This basis is not unique but the number $m$ is invariant, as should be clear from the proof. If we look carefully, we have also shown an important feature of symplectic spaces, that it has to have an even dimension always. In fact given any even number $2n$ , we can easily construct a symplectic space of this dimension: $(\mathbb{R}^{2n},\omega)$ , where we declare the standard basis $\{e_1,\hdots,e_n,e_{n+1},\hdots,e_{2n}\}$ to be the symplectic basis. This fully defines the form $\omega$ because of bilinearity. $(\mathbb{R}^{2n},\omega)$ is the model symplectic space to keep in mind. We call a set of vectors $\{u_1,\hdots,u_k,v_1,\hdots,v_n\}$ to be cross-orthonormal if it is of the above form (need not necessarily form a basis). We will also need to define what an isomorphism is in this category.

An isomorphism between symplectic spaces $(V,\omega_0)\to (W,\omega_1)$ is a linear isomorphism $T:V\to W$ that also preserves the symplectic form, in the following sense
$\omega_0(u,v)=\omega_1(Tu,Tv)$
$T$ is then called a symplectomorphism between $(V,\omega_0)$ and $(W,\omega_1)$ .

Now we investigate the different types of subspaces that a symplectic space could have.

Let $W$ be a vector subspace of $V$ such that $\omega|_{W\times W}$ is non-degenerate. Then we call it a symplectic subspace of $V$ .
Let $W$ be a subspace such that $\omega|_{W\times W}$ is zero, that is $\omega(u,v)=0$ for all $u,v\in W$ . Then we call $W$ an isotropic subspace of $V$ .
If $W^{\bot}$ is an isotropic space, then $W$ is called as a co-isotropic subspace of $V$ .

Given a symplectic basis $\{u_1,u_2,\hdots,u_m,v_1,v_2\hdots,v_m\}$ , its easy to construct examples of such subspaces.

$\text{span}\{u_1,v_1\}$ is symplectic.
$\text{span}\{u_1,u_2,u_3\}$ is isotropic.
$\text{span}\{u_1,u_2,\hdots,u_m,v_1,v_2\}$ is co-isotropic.

We can see from these examples that there is a borderline case that looks like $\text{span}\{u_1,u_2,\hdots,u_m\}$ . This is indeed special and is called as a Lagrangian subspace. Its a subspace that is both isotropic and co-isotropic. We conclude this note with an interesting result involving Lagrangian subspaces.

Let $L$ be a Lagrangian subspace of $V$ and let $L^*$ denote its dual. Then $(V,\omega)$ is naturally symplectomorphic to $L\oplus L^*$ equipped with the following symplectic form
$\omega_1((v,\alpha),(w,\beta))=\beta(v)-\alpha(w)$