# Hermitian Symmetric Subspaces of Siegel Space

Let $\mathbb{H}_g$ denote Siegel space, and $M$ denote an order 4 element of the unitary subgroup $U(n)(\mathbb{R})$with $p$ eigenvalues equal to $i$, and $q$ eigenvalues equal to $-i$, $p+q=g$. Consider $M$ as an element of the symplectic group $SP_{2g}(\mathbb{R})$ by identifying $U(n)(\mathbb{R})$ with a stabilizer of a point.If we consider the fixed point $S$ set of $M$ acting on $\mathbb{H}_g$, we get a complex, totally geodesic subvariety, which must therefore be a hermitian symmetric subspace. By computing at the tangent space, we can check that $S$ has complex dimension $pq$. Going through the classification table, it is natural enough to guess that $S$ is in the AIII family, isomorphic to the non-compact dual of the Grassmanian $G(p+q,q)$.

Question: Is $S$ indeed this hermitian symmetric space?

I would be interested in how one goes about determining what $S$ is through group theoretic methods.

Thanks!

Use the bounded model (Harish-Chandra realization) of the Siegel upper half-space $\mathfrak H_{p+q}$. Then $U(n)$ is "diagonalized" into blocks, so it's easy to see its action.

EDIT: as @jacob commented, probably my initial reaction was off by a sign, ... so, in greater detail: let $$M \;=\; \pmatrix{ i1_p & 0 & 0 & 0 \\ 0 & -i1_q & 0 & 0 \\ 0 & 0 & -i1_p & 0 \\ 0 & 0 & 0 & i1_q }$$ where the bottom right block is the transpose-inverse of the upper left block. The action in cooresponding coordinates on the bounded model is (as @jacob speculated) $$\pmatrix{z & u \\ u^\top & t} \longrightarrow \pmatrix{i & 0 \\ 0 & -i} \pmatrix{z & u \\ u^\top & t} \pmatrix{i & 0 \\ 0 & -i} \;=\; \pmatrix{ -z & u \\ u^\top & -t}$$

The fixed points of the element are the bounded-model $p\times q$ complex matrices, which are the bounded model for $U(p,q)/U(p)\times U(q)$.

• Are you certain? It seems to me that it acts via, if $K$ is an element of $U(n), Z--> KZK^T$. In which case the fixed points are the off-diagonals. (I mean $M^T$ to denote the transpose of $M$) Jul 3, 2014 at 18:13
• @jacob, you may be right. Let me write it out in a bit more detail... Jul 3, 2014 at 18:40
• Thanks! Can you please explain how to be sure that this is indeed the bounded model for $U(p,q)$? I know that the bounded $p\times q$ matrices give such a mode (through looking at your notes) but is it obvious that this particular embedding has an action of $U(p,q)$? Jul 3, 2014 at 18:54
• @jacob, there is a natural $SU(p,q)\to Sp_{p+q}(\mathbb R)$ by taking the imaginary part of the hermitian form whose isometry group is the unitary group. The center of the maximal compact of $SU(p,q)$ is a circle (motivating the $SU$ instead of $U$, to get rid of the extra circle in the center of the whole $U(p,q)$), and maps to the circle in the center of the maximal compact of the symplectic group, so the map is "hermitian", and will give a holomorphic equivariant map of the symmetric spaces. Taking Harish-Chandra realizations of both should produce a version of the in-coordinates thing. Jul 3, 2014 at 19:23
• Also, that copy of $U(p,q)$ is exactly the commutator of your special element $M$, which can be seen in various ways. Jul 3, 2014 at 19:25

$S$ cannot be isomorphic to a complex Grassmanian, or any other compact complex manifold, for that matter. The reason is that $\mathbb{H}_g$ is an open subset of a complex vector space, so a holomorphic embedding $X\hookrightarrow \mathbb{H}_g$ of a compact complex manifold would violate the maximum principle.

• Oops! Thanks Kevin. I should have said the non-compact dual of $G(P+q,q)$. I've edited the question to say that. Jul 3, 2014 at 17:18