# The Symplectic group is connected

Let $K = \mathbb{R}, \mathbb{C}$ be a field and consider the skew-symmetric matrix
$$J = \left( \begin{matrix} 0 & I_n \\ -I_n & 0 \end{matrix} \right)$$
where $I_n$ is the unit matrix of order $n$. Then I define the symplectic group:
$$\mathbf{Sp}_{2n}(K) = \left\{ A \in \mathbf{SL}_{2n}(K) \mid A^t J A = J \right\}$$
This is a Lie group. (see: https://en.wikipedia.org/wiki/Symplectic_group )

I want to prove the following:

The Lie group $\mathbf{Sp}_{2n}(K)$ is connected.

My strategy is as follows:

1. I proved the following Theorem: If a Lie group $G$ acts transitivelly on a connected differentiable manifold $X$, and there is a point $x \in X$ such that the stabilizer $G_x$ is connected, then so is $G$.
2. I use induction on $2n$.
3. For $n=1$ we have $\mathbf{Sp}_{2}(K) = \mathbf{SL}_{2}(K)$ which I proved it is connected.
4. Let $n > 1$. Then $\mathbf{Sp}_{2n}(K)$ acts naturally on $X = K^{2n}-\{ 0 \}$. I need to prove this action is transitive.
5. I want to prove that the stabilizer of $e_1 = (1,0,…,0)^t \in X$ is diffeomorphic to $\mathbf{Sp}_{2n-2}(K) \times K^m$ where $m = \mathrm{function}(n)$ is some integer depending on $n$.
6. Then $K^m$ is connected, and $\mathbf{Sp}_{2n-2}(K)$ is connected by induction. $\Rightarrow$ Their product is connect. The Theorem then implies that $\mathbf{Sp}_{2n}(K)$ is connected.

$\square$

I have here two gaps I need to complete:

1. The action defined above is transitive.
2. The stabilizer of $e_1 = (1,0,…,0)^t \in X$ is diffeomorphic to $\mathbf{Sp}_{2n-2}(K) \times K^m$. Surely the stabilizer is a block matrix of the form $$\left( \begin{matrix} 1 & u^t \\ 0 & A \end{matrix} \right)$$ but I need to find restrictions on $A$ that will give me a submatrix which is symplectic, hopefully a block which is in $\mathbf{Sp}_{2n-2}(K)$ or in $\mathbf{Sp}_{n-2}(K)$.