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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 $\mathbf{Sp}_{n-2}(K)$.

Please help me to fill these two gaps.

LinAlgMan
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  • When you say "let $K$ be a field" I assume you mean "let $K = \mathbb{R}$ or $\mathbb{C}$"? Otherwise you just get some group with no obvious Lie structure. – Qiaochu Yuan May 12 '13 at 20:16
  • Is the following argument correct? Transitivity: let $x,y \in X$ be nonzero vectors, then each can be complete to a symplectic basis $B_x$ and $B_y$ respectively. Then the action of the group is transitive on the bases and so there is a symplectic matrix sending $x$ to $y$. Hence, Sp(2n,K) is acting transitively on $X = K^{2n}-{0}$. – LinAlgMan May 13 '13 at 20:21
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    One way I know of showing that a compact Lie group, $G$, is path-connected is to find a standard maximal torus, $T$, and then prove that $\cup_{g \in G} gTg^{-1} = G$. This is sufficient to establish that $\exp \colon \mathfrak{g} \to G$ is surjective (c.f. Chapter 9 of Kristopher Tapp's Matrix Groups for Undergraduates for a proof of this), which immediately implies path-connectedness. – Doug Sep 10 '13 at 06:31
  • I found a proof in https://www.jnu.ac.in/Faculty/vedgupta/matrix-gps-gupta-mishra.pdf. Here he use the technology called symplectic transvections – Wuming Liu Aug 18 '23 at 08:33

1 Answers1

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Here is my attempt to the two gaps.

For the first question, given a vector $\alpha=(\alpha_1,\alpha_2)^t\in K^{2n}-\{0\},$ where $\alpha_i\in K^n$. Consider the following symplectic matrices, $\begin{bmatrix} A & 0\\ 0 & (A^t)^{-1}\end{bmatrix}, A$ is invertible, $\begin{bmatrix} I & B\\ 0&I\end{bmatrix}, B$ is symmetric, every block is of order $n$. If $\alpha_2=0\in K^n,$ find an invertible matrix $A$ of order $n$, such that $A\alpha_1^t=(1,0,\cdots,0)^t\in K^n$, then $\begin{bmatrix} A & 0\\ 0 & (A^t)^{-1}\end{bmatrix}$ sends $\alpha$ to $e_1=(1,0,\cdots,0)^t\in K^{2n}.$ If $\alpha_2\neq 0,$ find a symmetric matrix $B$ such that $\alpha_1^t+B\alpha_2^t=0$, then $\begin{bmatrix} 0&I\\-I&0\end{bmatrix}\begin{bmatrix} I&B\\0&I\end{bmatrix}\alpha=(\alpha_2,0)^t, $ which falls into the formal case.

For the second one, the stabilizer of $e_1$, denoted $G_{e_1},$ is homeomorphic to $Sp_{2n-2}(K)\times K^{2n-1}.$ To show this, consider the standard symplectic basis $e_1,\cdots,e_n,e_{n+1},\cdots,e_{2n},$ a matrix $P\in G_{e_1}$ sends $e_{n+1}$ to $\tilde{e}_{n+1}\in e_{n+1}+span\{e_1,\cdots,e_n,\hat{e}_{n+1},\cdots,e_{2n}\}\cong K^{2n-1}$. Once $\tilde{e}_{n+1}$ is known, we can form a new standard symplectic basis $e_1,w_2,\cdots, w_n, \tilde{e}_{n+1},w_{n+2},\cdots,w_{2n},$ where $w_i=e_i-<e_i,\tilde{e}_{n+1}>e_1.$ This decides a homeomorphism $G_{e_1}\cong K^{2n-1}\times Sp_{2n-2}(K)$ by assigning $P\in G_{e_1}$ the coordinate coefficients(transformation matrix) with respect to the above chosen basis.

Tsoshamry
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