TITLE: Finding the action of the symplectic group on the Siegel-half plane
QUESTION [6 upvotes]: 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?

REPLY [5 votes]: 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.