# Maximal isotropic subspaces of $V\oplus V^*$

Let $V$ be a finite dimensional vector space over $\mathbb{R}$ and denote the dual of $V$ by $V^*$. Then for $E$ a subspace of $V$ and $\epsilon\in \Lambda^2E^*$ clearly the space $$L=L(E,\epsilon):=\{X+\xi \in E\oplus V^*: \left.\xi\right|_{E}=\iota_X\epsilon\}$$ is isotropic with respect to the inner product on $V\oplus V^*$ given by $$<X+\xi,Y+\eta> = \frac{1}{2}(\eta(X)+\xi(Y)).$$ My first question is how do we know that $L(E,\epsilon)$ is maximally isotropic, i.e., that the dimension of $L$ is the same as the dimension of $V$?

My second question is regarding the proof that every maximal isotropic is of the form $L(E,\epsilon)$.

The proof I am reading proceeds as follows:

Let $L<V\oplus V^*$ be a maximal isotropic and let $E$ be the projection of $L$ onto $V$. Then it says that since $L$ is a maximal isotropic $L\cap V^*=\text{Ann}(E)$.

But wouldn't this imply that $L(E,\epsilon)=E\oplus \text{Ann}(E)$ no matter the choice of $\epsilon\in \Lambda^2E^*$ (my apologies for the use of $E$ here in two different contexts)?

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First question: The space $\lbrace X+i_X\epsilon: X\in E\rbrace$ is the graph of $\epsilon$ viewed as a mapping $E\to E^*$, so its dimension is the dimension of $E$. The freedom in the choice of $\xi$ is the dimension of $V/E$.
Second question: No, since $E = E\oplus 0$ is not a subset of $L$.