Lagrangian subspaces and complements

Question

Let $G(n,2n)$ denotes the Grassmannian of $n$-dimensional subspaces of $\mathbb C^{2n}$. Fix a symplectic form on $\mathbb C^{2n}$. Then $G(n,2n)$ has a natural involution $\sigma: V \mapsto V^{\perp}$ where $V^{\perp}$ is the orthogonal complement with respect to the symplectic form. Now my question is does this involution induced from an involution of $\mathbb C^{2n}$ ?

Answer 1

We can generalize your question with the following claim:

Given the standard symplectic manifold $(\mathbb{R}^{2n}, \omega_{st})$, we obtain the involution $\sigma : G(n, 2n) \to G(n, 2n) : V \mapsto V^{\omega}$. There exists a (possibly nonlinear and non-involutive) map $\psi : \mathbb{R}^{2n} \to \mathbb{R}^{2n}$ which induces $\sigma$ if and only if $n=1$.

For $n=1$, $G(1,2)$ is the set of lines through the origin, so that $\sigma = id$ and thus $\psi = id$ works just fine.

Now consider $n > 1$. Suppose that $\psi$ exists and aim for a contradiction. A crucial but general observation is that for any two subsets $A, B \subset \mathbb{R}^{2n}$, we have $\psi(A \cap B) = \psi(A) \cap \psi(B)$.

Step 1: We show that for every $p \in \mathbb{R}^{2n}$, $\psi(p)$ and $p$ are colinear. (This step is true also for $n=1$.)

If $p=0$, that is if $\{p\} = \cap_{V \in G(n, 2n)} V$, then $\{\psi(p)\} = \cap_{V \in G(n, 2n)} \psi(V)= \cap_{V \in G(n, 2n)} \sigma(V) = \{0\}$ so that $\psi(0) = 0$ and the claim holds.

If $p \neq 0$, then there exists a symplectic basis $\{p_1 = p, q_1, \dots, p_n, q_n \}$ of $(\mathbb{R}^{2n}, \omega_{st})$. Given $\Theta = (\theta_2, \dots, \theta_n) \in [0, 2\pi)^{n-1}$, let $L_{\Theta}$ be the Lagrangian subspace generated by $p_1$ and the vectors $p_i \cos \theta_i + q_i \sin \theta_i$ (where $i = 2, \dots, n$). Observe that $Span(p_1) = \cap_{\Theta \in [0, 2\pi)^{n-1}} L_{\Theta}$, so that

$$ Span(\psi(p_1)) \subseteq \psi(Span(p_1)) = \cap_{\Theta \in [0, 2\pi)^{n-1}} \psi(L_{\Theta}) = \cap_{\Theta \in [0, 2\pi)^{n-1}} \sigma(L_{\Theta}) = \cap_{\Theta \in [0, 2\pi)^{n-1}} L_{\Theta} $$

where the last equality follows from the fact that each $L_{\Theta}$ is Lagrangian. Hence $p$ and $\psi(p)$ are colinear.

Step 2: We show that for every $p \in \mathbb{R}^{2n} \setminus \{0\}$, $\psi(p)$ and $p$ are not colinear. (This step does not work for $n=1$.)

This step follows from the fact that when $n > 1$, we can find a linear subspace $V \in G(n, 2n)$ which contains a symplectic sub-subspace which constains any prescribed vector $p \neq 0$.

Given $p \neq 0$, consider a symplectic basis as above. For $\Theta = (\theta_2, \dots, \theta_{n-1}) \in [0, 2\pi)^{n-2}$ (mind the difference), let $V_{\Theta} \in G(n, 2n)$ be generated by $p_1, q_1$ and the vectors $p_i \cos \theta_i + q_i \sin \theta_i$ (where $i = 2, \dots, n-1$). Let also $W_{\Theta} \in G(n, 2n)$ be the subspace generated by $p_n, q_n$ and the same vectors $p_i \cos \theta_i + q_i \sin \theta_i$ (where $i = 2, \dots, n-1$). Observe that $Span(p_1, q_1) = \cap_{\Theta \in [0, 2\pi)^{n-2}} V_{\Theta}$ and that $\psi(V_{\Theta}) = \sigma(V_{\Theta}) = V_{\Theta}^{\omega} = W_{\Theta}$. Hence

$$ \psi(p_1) \subset \psi(Span(p_1, q_1)) = \cap_{\Theta \in [0, 2\pi)^{n-2}} \psi(V_{\Theta}) = \cap_{\Theta \in [0, 2\pi)^{n-2}} W_{\Theta} = Span(p_n, q_n). $$

Since $p_1 \not \in Span(p_n, q_n)$, the claim follows.

Both steps together yield the sought-after contradiction.

Remark: In your situation, as you are considering $n$ even, for any $p \neq 0$ you can find a symplectic $V \in G(n, 2n)$ which constains $p$. As $\psi(V) = \sigma(V)$ and $V$ are complementary, the second step is readily proved.