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.