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.



Subvarieties of Lagrangian Grassmannians
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.


