Disconnecting the Lagrangian Grassmannian

Question

Let $(V, \omega)$ be a symplectic vector space of dimension 2n. This has a Lagrangian Grassmannian $\Lambda(V)$ of Lagrangian subspaces of $V$. Now consider the following subvariety: Fix a half-dimensional vector space $K \subseteq V$ and define $S_{K} = \{ U \in \Lambda(V) \ | \ U \cap K \neq 0 \}$, the space of all Lagrangians which intersect $K$ non-trivially. Presumably this has codimension 1.

Question: Is the space $\Lambda(V) \setminus S_{K}$ connected? In other words, given two Lagrangians $L_{1}, L_{2}$ which are transverse to $K$, is it possible to deform one into the other while preserving the property of being transverse to $K$?

Update: There are two extreme cases that work:

i) K is Lagrangian: Choose a reference Lagrangian L transverse to K. This allows us to put the symplectic vector space into standard form $L \oplus L^{\ast}$. Then the Lagrangians which are transverse to $K$ are given by graphs of symmetric matrices, which form a vector space. Hence $\Lambda(V) \setminus S_{K}$ is connected.

ii) K is symplectic. Then since $K \cap K^{\omega} = 0$, the symplectic vector space can be written as the direct sum $(V, \omega) = (K, \omega|_{K}) \oplus (K^{\omega}, \omega|_{K^{\omega}})$. Then Lagrangians which are transverse to $K$ are automatically transverse to $K^{\omega}$ and therefore can be expressed as the graph of an isomorphism $A: K \to K^{\omega}$. The property of being Lagrangian is then the condition that $A^{\ast}(\omega|_{K^{\omega}}) = - \omega|_{K}$. Hence $\Lambda(V) \setminus S_{K}$ is isomorphic to a symplectic group, which is connected.

Answer 1

It looks as though $M_{K} := \Lambda(V) \setminus S_{K}$ is connected. Here is the way I can see it: First of all, we know that when $K$ is Lagrangian $M_{K}$ can be identified with $Sym^{2}(L^{\ast})$, where $L \in M_{K}$ is a reference Lagrangian, and when $K$ is symplectic $M_{K}$ is a torsor for $Sp(K, \omega|_{K})$. Both of these spaces are connected. The idea for a general subspace $K \subseteq V$ is to see $M_{K}$ as a fibration involving both of the above cases.

So let $(V, \omega)$ be a $2n$-dimensional symplectic vector space, $K \subseteq V$ an arbitrary $n$-dimensional subspace. Then $U = K \cap K^{\omega}$ is isotropic of dimension $p$, and $U^{\omega} = K + K^{\omega}$ is coisotropic of dimension $2n - p$.

Define $P_{K} = \{ W \in Gr(U^{\omega}, n-p) \ | \ W \text{ isotropic}, W \cap K = 0 \}$, the space of all $n-p$ dimensional isotropic subspaces of $U^{\omega}$ which are transverse to $K$. We have a natural map $G : M_{K} \to P_{K}$ sending a Lagrangian $L$ to the isotropic subspace $L \cap U^{\omega}$. Given an isotropic $W \in P_{K}$, the fibre over this point is the set $G^{-1}(W) = \{ L \in \Lambda(V) \ | \ W \subseteq L, L \cap U = 0 \}$, and this is isomorphic to $M_{(U+W)/W} = \{\bar{L} \in \Lambda(W^{\omega}/W) \ | \ \bar{L} \cap (U + W)/W = 0 \}$. But $(U + W)/W$ is a Lagrangian subspace of $W^{\omega}/W$, and so we are reduced to the above case: $M_{K}$ is fibred over $P_{K}$ with fibre isomorphic to the space of $p \times p$ symmetric matrices.

Lets now look at $P_{K}$ more closely. Since $K \subseteq U^{\omega}$ and $K \cap K^{\omega} = U$, the subspace $K/U \subseteq U^{\omega}/U$ is symplectic. Furthermore, given any isotropic $W \in P_{K}$, the quotient $(W+U)/U$ is a Lagrangian subspace transverse to $K/U$. Hence we have a natural map $F : P_{K} \to M_{K/U}$ given by sending an isotropic $W$ to the Lagrangian $(W+U)/U \subseteq U^{\omega}/U$. Note that since $K/U$ is symplectic we know from above that $M_{K/U}$ is a torsor for $Sp(K/U)$. Note also that the composition $F \circ G : M_{K} \to M_{K/U}$ is doing the symplectic reduction of Lagrangians. Given a Lagrangian $\bar{L} \in M_{K/U}$, the fibre $F^{-1}(L)$ over this point is given by the splittings of the following short exact sequence

$0 \to U \to \pi^{-1}(\bar{L}) \to \bar{L} \to 0$,

where $\pi : U^{\omega} \to U^{\omega}/U$ is the projection map. Hence $F^{-1}(L)$ is an affine space for $Hom(\bar{L}, U)$. In fact $F: P_{K} \to M_{K/U}$ is an affine bundle for $Hom(\tau, U)$, where $\tau$ is the tautologial bundle over $M_{K/U} \subseteq \Lambda(U^{\omega}/U)$.

Therefore we have two fibrations $M_{K} \to P_{K} \to M_{K/U} \cong Sp(K/U)$, with fibres isomorphic to $Sym_{p \times p}$ and $Mat_{(n-p) \times p}$. In particular, $M_{K}$ is connected.