# $W_1,…,W_k\subset V$ lagrangian subspaces $\Rightarrow \exists\, L\subset V$ lagrangian with $L\cap W_i=\{0\}$ for all $i$

Let $(V,\omega)$ be a symplectic vector space and $W_1,...,W_k\subset V$ Lagrangian subspaces. Prove there is a Lagrangian subspace $L\subset V$ such that $L\cap W_i=\{0\}$ for all $i$.

I've noticed the following: Take the special case where $k=2$ and $W_1\cap W_2=\{0\}$. If $\{e_{i1},...,e_{in}\}$ is a basis for $W_i$ (where $2n=\dim V$), we define $L:=\text{span}\{v_1,...,v_n\}$, where:

$$v_i:=e_{1i}+e_{2i}$$

It's is easy to check that $\omega(v_i,v_j)=0$ for all $i,j$, and the fact that $W_1\cap W_2=\{0\}$ allows us to conclude that $\dim L=n$ and that $L\cap W_i=\{0\}$, so we are done.

I'm trying to extend this idea somehow to the general case, but I can't see how.

Any suggestions?

Consider the Lagrangian Grassmannian $\mathcal{L}(V)$ consisting of all Lagrangian subspaces of $V$; this set is known to be a smooth manifold of dimension $n(n+1)/2$. For each $1\leq i \leq k$ and each $0\leq m\leq n$ we have the subset $$\Sigma_m(W_i) = \{ L\in\mathcal{L}(V)|\dim(L\cap W_i) = m\}.$$ (Allowing $m$ to vary would give us the Maslov cycle associated to $W_i$.) Our goal is to show that the set $$\Sigma_0(W_1)\cap\cdots\cap\Sigma_0(W_k) = \mathcal{L}(V) - \left(\bigcup_{m\geq 1,i}\Sigma_m(W_i)\right)$$ is nonempty, and we can do this by showing that $\Sigma_m(W_i)$ has codimension at least 1 in $\mathcal{L}(V)$ for each $m\geq 1$.
How do we establish this last claim? For convenience, choose a complex structure $J\colon V\to V$ that is compatible with $\omega$, so that $\langle\cdot,\cdot\rangle:=\omega(\cdot,J\cdot)$ is an inner product. Suppose $L\in\mathcal{L}(V)$ is transverse to the Lagrangian subspace $JW_i$. Then we can choose a linear transformation $T\colon W_i\to W_i$ for which $$L = \{w+JTw|w\in W_i\},$$ so that we can think of $L$ as the graph of $T$. If we fix an orthonormal basis for $W_i$ and let $A$ denote the matrix representation of $T$ with respect to this basis, we can check that such a graph is Lagrangian precisely when $A$ is symmetric. That is, choosing $L\in\mathcal{L}(V)$ transverse to $JW_i$ amounts to choosing a symmetric $n\times n$ matrix, giving us $n(n+1)/2$ degrees of freedom. So $\{L\in\mathcal{L}(V)|L\pitchfork JW_i\}$ is open in $\mathcal{L}(V)$.
Now suppose we want $L\in\Sigma_m(W_i)$ transverse to $JW_i$. Let's rechoose our basis $v_1,\ldots,v_n$ for $W_i$ so that $L\cap W_i=span(v_1,\ldots,v_m)$. Then $Tv_i=0$ for $1\leq i\leq m$, so the first $m\times m$ block of our matrix $A$ is the zero matrix, and we've lost $m(m+1)/2$ of our degrees of freedom in choosing $A$. (We lose an additional $m(n-m)$ degrees of freedom, since the bottom-left and top-right blocks must be zero, but the choice of $m$-dimensional subspace of $n$ returns this freedom to us.) So $\Sigma_m(W_i)$ is a codimension-$m(m+1)/2$ subset of $\mathcal{L}(V)$.
Finally, we obtain $\Sigma_0(W_1)\cap\cdots\Sigma_0(W_k)$ by removing a finite collection of subspaces from $\mathcal{L}(V)$, each of codimension at least 1. Thus this remaining set of subspaces is nonempty.