# Lagrangian subspaces of symplectic vector spaces

1. Jun 4, 2012

### Kreizhn

1. The problem statement, all variables and given/known data
If $(V,\omega)$ is a symplectic vector space and Y is a linear subspace with $\dim Y = \frac12 \dim V$ show that Y is Lagrangian; that is, show that $Y = Y^\omega$ where $Y^\omega$ is the symplectic complement.

3. The attempt at a solution

This is driving me crazy since I don't think it should be that hard. If we try to go directly, the fact that $\dim Y = \frac12 \dim V$ implies that $\dim Y^\omega = \frac12 \dim V$. At this point, it is actually sufficient to show that Y is either isotropic or co-isotropic, since dimensional arguments will give equality. However, I cannot see why this is the case.

On the other hand, if we assume that $Y \neq Y^\omega$ then without loss of generality, we may assume there exists $w \in Y \setminus Y^\omega$ (otherwise, relabel $Y$ and $Y^\omega$). The space $W = \operatorname{span}(w)$ is isotropic so $W \subseteq W^\omega$ and since $W \subseteq Y$ then $Y^\omega \subseteq W^\omega$. I would like to put this together somehow, but I'm having trouble doing it.

2. Jun 4, 2012

### Kreizhn

Nevermind. I've concluded that the question as phrased was unsolvable. Indeed, consider $\mathbb R^4$ with the standard symplectic basis $\{x_i,y_i\}$. The subspace generated by x1 and y1 is two-dimensional, but is in fact symplectic rather than Lagrangian. Hence the statement cannot be true in general.