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!



Hermitian Symmetric Subspaces of Siegel Space
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 $U(p,q)/U(p)\times U(q)$.


