Subvarieties of Lagrangian Grassmannians

Question

Let $LG(n,2n)$ be the Lagrangian Grassmannian parametrizing Lagrangian subspaces (so of dimension $n$) of $\mathbb{C}^{2n}$. Then $LG(n,2n)\subset G(n,2n)$, where $G(n,2n)$ is the Grassmannian of subspaces of dimension $n$ of $\mathbb{C}^{2n}$.

Fix a subspace $H\subset \mathbb{C}^{2n}$ of dimension $n+2$, and denote by $G(n,H)\subset G(n,2n)$ the Grassmannian of subspaces of dimension $n$ that are contained in $H$.

The intersection $X_n := LG(n,2n)\cap G(n,H)\subset G(n,2n)$ parametrizes Lagrangian subspaces of $\mathbb{C}^{2n}$ that are contained in $H$.

For instance, for $n = 2$ we have $G(2,H) = G(2,4)$ and hence $X_2 = LG(2,4)$.

In general, is the variety $X_n$ smooth and irreducible? By any chance is $X_n$ a well-known variety appearing under some name in the literature?

Addition: Is the subvariety $Y_n\subset G(n+2,2n)$, parametrizing $(n+2)$-dimensional subspaces $H\subset \mathbb{C}^{2n}$ that are co-isotropic, homogeneous as well? Is there a formula for the dimension of $Y_n$?

Thank you very much.

Answer 1

Note first that, if $L$ is a Lagrangian contained in $H$ then $L^\perp = L$ contains $H^\perp$. So $X_n$ is non-empty only when $H$ is co-isotropic for your symplectic form.

When $H$ is co-isotropic, the symplectic form $\omega$ induces a symplectic form on $H/H^\perp$ (which has dimension $4$) and Lagrangians of $\mathbb C^{2n}$ contained in $H$ are in bijection with Lagrangian subspaces of $H/H^\perp$ (since they all contain $H^\perp$). Thus $X_n$ is a copy of $LG(2,4)$ inside $LG(n,2n)$. It is homogeneous under the parabolic subgroup of $Sp(2n,\mathbb C)$ preserving $H$, so it is smooth and irreducible.

Answer 2

Co-isotropic subspaces of dimension $n+k$ are in bijection with isotropic subspaces of dimension $n-k$.

The variety of isotropic subspaces of dimension $n-k$ of a symplectic vector space of dimension $2n$ is the symplectic Grassmannian $SG(n-k,2n)$. Its dimension is given by

$$\dim(SG(n-k,2n)) = 2n(n-k)-\frac{3(n-k)^2-(n-k)}{2}.$$

In particular, when $k = 0$ you get the Lagrangian Grassmannian $LG(n,2n)$ parameterizing maximal isotropic subspaces, and its dimension is given by $$\dim(LG(n,2n)) = \frac{n(n+1)}{2}.$$