Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $S$ be the Siegel-half plane of dimension $n$, i.e. the set of complex $n \times n$ matrices $Z$ which are symmetric and whose imaginary part is positive-definite. In dimension 1 we can identify $S$ with the Poincaré half-plane.

In dimension 1 there is an action of the symplectic group $Sp_2(\mathbb R)$ on $S$, given by fractional linear transformations. One can extend this action to higher dimensions; we write an element $M$ of the symplectic group $Sp_{2n}(\mathbb R)$ as a block matrix with blocks $A$, $B$, $C$, $D$, and then set

$$ M \cdot Z = (AZ + B)(CZ + D)^{-1}. $$

Now, in dimension 1 one sees that this action is induced by the action of the general linear group on the projective line and the embedding $GL_2(\mathbb R) \hookrightarrow GL_2(\mathbb C)$. However, this does not seem to be true in higher dimensions, as $GL_{2n}(\mathbb C)$ acts on the projective space of dimension $2n$, and not on the space of dimension $n$.

The definition of the action of $Sp_{2n}(\mathbb R)$ on $S$ in higher dimensions thus seems very ad-hoc, which motivates:

Question: How can one find this action naturally in higher dimensions?

[Edit:] Question 2: David's answer is very nice, but I'd like to iterate the question before accepting it. Instead of the Siegel half-plane above, consider the space $U$ of complex $(1,1)$-forms whose imaginary part is positive-definite. In a basis we can identify this with the set of $n \times n$ matrices $Z$ whose antihermitian part is positive-definite. The action of the symplectic group now makes sense on this space as well.

In a basis the Siegel half-plane is a closed subspace of $U$ if $n > 1$. It is easiest to define the action of the symplectic group in a basis, but we can define it without reference to a basis by choosing a hermitian inner product on our vector space. Now, can we find the action of the symplectic group on $U$ in the same fashion as on the Siegel half-plane?

share|improve this question
I don't quite understand your revised question. You seem to be simultaneously saying that you know how to define an action of $Sp_n$ on $U$, and asking how to define such an action. Are you asking for a conceptual interpretation of an action on $U$ that you know how to define in coordinates? –  David Loeffler Oct 16 '11 at 11:25
Well, yes I was wondering about a conceptual interpretation. The action is exactly the same, $M \cdot Z = (AZ + B)(CZ + D)^{-1}$, where $(Z - {}^t \overline Z)/2i$ is positive-definite. It gives a transitive action of $Sp_{2n}$ on the space $U$ of $(1,1)$-forms, and I wonder if there is any way of seeing this action pop up naturally, instead of ad-hoc like this. –  Gunnar Magnusson Oct 16 '11 at 12:01
The HarishChandra-Borel realizations of non-compact type hermitian symmetric spaces inside their compact duals may be of interest. –  paul garrett Oct 16 '11 at 15:51
@Paul: I just found your paper "The classical groups and domains". It looks like my second question is treated in the part on self-adjoint cones; I think the group action on $U$ simply corresponds to the one of $GL_n(\mathbb C)$ restricted to the symplectic group. I had a strong case of not seeing the forest for the trees here - in my head the Siegel half-plane and the space of $(1,1)$-forms are intimitely related, since they both pop up while studying complex tori, while for this question that doesn't matter at all. –  Gunnar Magnusson Oct 16 '11 at 16:53
@Gunnar: the very last page or two of that little essay "does" the Harish-Chandra realization for Sp(n,R) on the last page-or-two. For general simple G with G/K admitting an invariant hermitian structure, the same thing works, suitably abstracted. I think Helgason's "Symmetric spaces..." treats the general case. For classical groups, the computations of course succeed without necessarily couching things in Lie-theoretic or geometric terms. –  paul garrett Oct 16 '11 at 18:19

3 Answers 3

up vote 6 down vote accepted

The Siegel half-plane is just one example of a very general construction: if one lets $G$ be a semisimple Lie group, and $K$ a maximal compact subgroup, then $G$ acts on the quotient $G / K$, and the quotient is known as a "symmetric space". The Siegel upper half-plane is just a symmetric space for $Sp_{2n}(\mathbb{R})$, with its natural action; and from this perspective it's a completely natural thing to consider. It just takes a little while to find explicit coordinates in which you can write down the action, and when you do so, its origins become slightly obscured.

You can already see this for $n = 1$, where $G$ is just $SL_2$. Then you can easily check that the action of $G$ on the upper half-plane $\mathcal{H}$ is transitive and the stabilizer of $i$ is $K = SO_2(\mathbb{R}) \cong S^1$, a maximal compact in $G$; so we can identify $\mathcal{H}$ with $G / K$.

In the higher-dimensional symplectic cases, $G = Sp_{2n}$ has a maximal compact subgroup $K$ isomorphic to the unitary group $U(n)$, and you can check that the quotient $G / K$ is the Siegel upper half-plane.

(The miracle with symplectic groups, which doesn't happen with most real Lie groups, is that the Siegel half-space has a $G$-invariant complex structure. This is very important in the theory of automorphic forms.)

share|improve this answer
Interesting. You wouldn't happen to know if one can obtain a subspace of the space of (1,1)-forms in this way as well? I ask because we can define the same action on the space of forms whose imaginary part is positive-definite. If we then pick a basis and pretend that the transpose of an element makes sense then we find the Siegel half-plane in there. –  Gunnar Magnusson Oct 16 '11 at 8:48
I don't know; it sounds plausible, but I'm not a differential geometer. –  David Loeffler Oct 16 '11 at 8:51
I've just re-read that and spotted a glaring error (I had claimed that the maximal compact subgroup of $Sp_{2n}$ was something that is clearly not compact!). This is what happens when I try and answer MO questions before breakfast. –  David Loeffler Oct 16 '11 at 12:41

A nice way to think about the action of $SP_{2g}(\mathbb{R})$ on $S$ is that $S$ is an $SP_{2g}(\mathbb{R})$-invariant open set in the Lagrangian grassmannian.

Let $V$ be a $2g$-dimensional complex vector space equipped with a symplectic form. Let $L$ be the space of isotropic $g$-dimensional subspaces of $V$. Such a subspace is called a Lagrangian. Clearly, $L$ is a complex manifold equipped with a holomorphic action of $SP_{2g}(\mathbb{C})$.

Choose a splitting of $V$ as $U \oplus U^{\vee}$, where $U$ and $U^{\vee}$ are complementary isotropic subspaces of $V$ and the symplectic pairing between $U$ and $U^{\vee}$ makes $U$ and $U^{\vee}$ into dual spaces. Let $\Omega \subset L$ be the open set consisting of those Lagrangians which are the graph of a map $\phi: U \to U^{\vee}$. The condition that the graph of $\phi$ be Lagrangian is equivalent to the condition that $\phi$ be self adjoint, so $\Omega$ is identified with the space of symmetric $g \times g$ complex matrices. The action of $Sp_{2g}(\mathbb{C})$ on $\Omega$ isn't fully defined, because $\Omega$ isn't $Sp_{2g}$-invariant, but when it is defined you can check that it is given by the formula $Z \mapsto (AZ+B)(CZ+D)^{-1}$.

Choose a real $2g$-dimensional symplectic vector space $V_0$ and identify $V$ with $V_0 \otimes_{\mathbb{R}} \mathbb{C}$. Let $\Omega' \subset L$ be the open set of those Lagrangian's $X$ such that $X \cap V_0 = \{ 0 \}$. Clearly, $\Omega'$ has a holmorphic action of $Sp_{2g}(\mathbb{R})$. If we identify $\Omega$ with $g \times g$ symmetric complex matrices, then $\Omega'$ is those matrices $Z$ for which $\det \Im(Z) \neq 0$.

Obviously, one connected component of $\Omega \cap \Omega'$ is $S$: Symmetric matrices for which $\Im(Z)$ is positive definite. And it turns out that this is actually a connected component of $\Omega'$, so it inherits an action of $Sp_{2g}(\mathbb{R})$. This is the action you asked about.

share|improve this answer

If you can get hold of a copy of Mumford's Lectures on Theta I, you will find a useful discussion of the Siegel upper half plane $H_{g}$ and the action of $Sp_{2g}$ on it in section 4 of chapter II. He gives several coordinate free interpretations. Here is one: Fix the standard symplectic form $\omega$ on $\mathbb{R}^{2g}$. Then $H_{g}$ can be identified with the space of complex structures on $\mathbb{R}^{2g}$ such that $\omega= Im(H)$ for some (uniquely determined) positive definite Hermitean form $H$. It is clear that $Sp_{2g}(\mathbb{R})= Sp(\omega)$ acts the space of such complex structures.

In case you like Hodge structures, let me supplement his Lemma 4.1 with one more (well known) equivalent. (He proves this without quite saying it.) $H_g$ can be identified with the space of pure Hodge structures of type $\lbrace (1,0),(0,1)\rbrace$ polarized by $\omega$. Griffiths generalized this to show that the space of arbitrary polarized pure Hodge structures of fixed type is a homogeneous complex manifold. Unlike the Siegel case, however, it is usually not Hermitean symmetric, so lots of things don't generalize.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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