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)?



Maximal isotropic subspaces of $V\oplus V^*$
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 $L$. 


