$\mathrm{Sp}(4, \mathbb{C})$ is simply connected

Question

I'm trying to prove that $\mathrm{Sp}(4, \mathbb{C})$ is simply connected. Note that it is a group of complex $4\times 4$ matrices $A$ satisfying $A^{T}JA = J$, where $$ J = \begin{pmatrix} O & I_{2} \\ -I_{2} & O \end{pmatrix} $$
What I've got so far is the following: consider the standard action of $\mathrm{Sp}(4, \mathbb{C})$ on $\mathbb{C}^{4}\backslash\{0\}$. We can check that the action is transitive (this is not that hard - I can write it down if anyone want), so we have $\mathrm{Sp}(4, \mathbb{C})/\mathrm{Stab}(v)\simeq \mathbb{C}^{4}\backslash\{0\}$. Since $\mathbb{C}^{4}\backslash\{0\}$ is homotopic to $S^{7}$, it is simply connected. Hence it is enough to show that the stabilizer group $\mathrm{Stab}(v)$ is simply connected.

The problem is, I can't figure out what is $\mathrm{Stab}(v)$ is. For example, take $v = e_{1} = (1, 0, 0, 0)^{T}$ and one can check that $A\in \mathrm{Stab}(e_{1})$ has a form of $$ \begin{pmatrix} A&B\\C&D\end{pmatrix} = \begin{pmatrix} 1& * & * & * \\ 0 & * & * & * \\ 0 & 0 & * & * \\ 0 & * & * & * \end{pmatrix} $$ where $A, B, C, D\in \mathrm{M}_{2\times 2}(\mathbb{C})$ satiesfies $$ A^{T}C = C^{T}A, \quad B^{T}D = D^{T}B, \quad A^{T}D-C^{T}B = I_{2}. $$ Actually I hope that the stabilizer group is isomorphic to some other familiar group, but I can't do anything from here. Thanks in advance.

Answer 1

Instead of considering the stabilizer of a single element like $e_1$, it is easier to consider the stabilizer of both $e_1$ and $e_3$. Indeed, any $A\in Sp(4,\mathbb{C})$ which fixes $e_1$ and $e_3$ must also fix their orthogonal complement with respect to the symplectic form, which is just the span of $e_2$ and $e_4$. We then see that this stabilizer group is isomorphic to $Sp(2,\mathbb{C})=SL(2,\mathbb{C})$ which is simply connected.

So, it suffices to show that the space of orbits of the pair $(e_1,e_3)$ is simply connected. This is just the space of pairs of vectors on which the symplectic form evaluates to $1$ (also known as "hyperbolic pairs"). As you say, $e_1$ can map to any point of $\mathbb{C}^4\setminus\{0\}$. Once you have chosen where to map $e_1$, then $e_3$ can map to any element of $\mathbb{C}^4$ whose pairing with the image of $e_1$ under the symplectic form is $1$. So, the space of possible images of $e_3$ for any choice of image of $e_1$ is a codimension $1$ affine subspace of $\mathbb{C}^4$. This means that our orbit space is a fiber bundle over $\mathbb{C}^4\setminus\{0\}$ (the choices for $Ae_1$) with fiber $\mathbb{C}^3$. Since the base and fiber are both simply connected, this implies the orbit space is simply connected.

More generally, for any $n$ we can consider $Sp(2n-2,\mathbb{C})$ as a subgroup of $Sp(2n,\mathbb{C})$, as the subgroup that fixes some particular hyperbolic pair. The quotient space $Sp(2n,\mathbb{C})/Sp(2n-2,\mathbb{C})$ is the space of hyperbolic pairs in $\mathbb{C}^{2n}$, which is a fiber bundle over $\mathbb{C}^{2n}\setminus\{0\}$ with fiber $\mathbb{C}^{2n-1}$ (and so in particular is homotopy equivalent to $S^{4n-1}$). In this way we can show by induction that $Sp(2n,\mathbb{C})$ is simply connected for all $n$.