3
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

I can't think of a strictly linear-algebraic solution, but here's an approach that should work (despite being a bit unpleasant).

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 \begin{equation} \Sigma_m(W_i) = \{ L\in\mathcal{L}(V)|\dim(L\cap W_i) = m\}. \end{equation} (Allowing $m$ to vary would give us the Maslov cycle associated to $W_i$.) Our goal is to show that the set \begin{equation} \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) \end{equation} 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 \begin{equation} L = \{w+JTw|w\in W_i\}, \end{equation} 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.