4
$\begingroup$

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!

$\endgroup$
2
$\begingroup$

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 $U(p,q)/U(p)\times U(q)$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 $M$) $\endgroup$ – jacob Jul 3 '14 at 18:13
  • $\begingroup$ @jacob, you may be right. Let me write it out in a bit more detail... $\endgroup$ – paul garrett Jul 3 '14 at 18:40
  • $\begingroup$ 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 $U(p,q)$? $\endgroup$ – jacob Jul 3 '14 at 18:54
  • $\begingroup$ @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. $\endgroup$ – paul garrett Jul 3 '14 at 19:23
  • $\begingroup$ Also, that copy of $U(p,q)$ is exactly the commutator of your special element $M$, which can be seen in various ways. $\endgroup$ – paul garrett Jul 3 '14 at 19:25
2
$\begingroup$

$S$ 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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Oops! Thanks Kevin. I should have said the non-compact dual of $G(P+q,q)$. I've edited the question to say that. $\endgroup$ – jacob Jul 3 '14 at 17:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.