I'm trying to prove that the symplectic group $Sp(n)$ acts transitively on the sphere $S^{4n-1}$, and as a consequence $Sp(n)/Sp(n-1)$ is homeomorphic to $S^{4n-1}$. To me $Sp(n)$ is the group of $2n\times 2n$ unitary complex matrices satisfying $AJ=J\bar A$ where J is the matrix

\begin{bmatrix} 0 & -I_n\\ I_n & 0 \end{bmatrix}

It is clear to me that these matrices map the sphere into the sphere. To prove transitivity it is enough to show that any vector in the sphere can be mapped to for example the vector $(1,0,\dots,0)$ via multiplication with a symplectic matrix, and to prove that $Sp(n)/Sp(n-1)$ is homeomorphic to $S^{4n-1}$ it is enough to show that the stabilizer of $(1,0,\dots,0)$ is precisely $Sp(n-1)$. Well, this is the part in where I'm stuck. I don't know how to construct a symplectic matrix mapping a vector in the sphere to $(1,0,\dots,0)$ or to deduce that $Sp(n-1)$ is the stabilizer of that point. I'd appreciate any help with this because everywhere I read these fact are presented as obvious.

[EDIT] Probably it is easier to send the vector $(1,0,\dots,0)$ to any other $x$. In that case I need to construct a symplectic matrix whose first column is $x$.


First, let's figure out a nice description of $Sp(n)$. For $A\in Sp(n)$, write it in the block form $A = \begin{bmatrix} B & C\\ D & E\end{bmatrix}$ where each block is $n\times n$. Then a simple calculation shows that $AJ = J\overline{A}$ iff $A$ has the form $A = \begin{bmatrix} B & -\overline{D}\\ D & \overline{B}\end{bmatrix}$.

Now, given $x = x_1\in \mathbb{C}^{2n}$ of unit length, we extend it to an orthonormal set as follows. Pick $x_2$ to be a unit length vector in the orthogonal complement to $\{x_1, Jx_1\}$, pick $x_3$ to be a unit length vector in the orthogonal complement to $\{x_1, Jx_1, x_2, Jx_2\}$, etc.

If we write $x_i = \begin{bmatrix} y_i \\ z_i\end{bmatrix}$, where both $y_i, z_i \in \mathbb{C}^n$, then the way we choose the $x_i$ guarantees that $\begin{bmatrix} y_i \\ z_i\end{bmatrix}$ is perpendicular to both $\begin{bmatrix} y_j \\ z_j\end{bmatrix}$ (when $j\neq i$) and $\begin{bmatrix} -z_j \\ y_j\end{bmatrix}$ (for any $j$).

Now, we choose $B$ and $D$ so that the block $2n\times n$ matrix $\begin{bmatrix} B\\ D\end{bmatrix} = \begin{bmatrix} x_1 & x_2 & ... & x_n\end{bmatrix}$.

I claim that $A = \begin{bmatrix} B & -\overline{D} \\ D & \overline{B}\end{bmatrix}$ is actually in $U(2n)$. First, since the blocks on the right are just rearrangements and conjugates of things on the left, it's clear that every column of $A$ has unit length. So, it is enough to show that the columns are pairwise perpendicular. This is obvious if both columns come from the left blocks or if both come from the right blocks. The fact that $x_i$ is perpendicular to $Jx_j$ shows it when one column comes from the left block and the other from the right block.

Now, simply note that $A\begin{bmatrix}1 \\ 0 \\ \vdots \\ 0\end{bmatrix} = x$.

$ \ $

Now, let's compute the stabilizer at the point $p = \begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0\end{bmatrix}$. If $Ap = p$, then it follows that the first column of $A$ is $p$, so the first column of $D$ is all $0$s, as is the first column of $B$, except that the top entry in $B$ is $1$.

Because $A\in U(2n)$, if the top left entry is $1$ the rest of the entries in the top row must be $0$.

It follows that $B$ has the form $B = \begin{bmatrix} 1 & 0 \ldots 0 \\ \begin{array} x0 \\ \vdots \\ 0\end{array} & B'\end{bmatrix}$ and the $D$ has the form $D = \begin{bmatrix} 0 & 0\ldots 0\\ \begin{array} x0\\ \vdots \\ 0\end{array} & D'\end{bmatrix}$ where $B'$ and $D'$ are both $(n-1)\times (n-1)$ matrices. If we set $A' = \begin{bmatrix} B' & -\overline{D'}\\ D' & \overline{B'}\end{bmatrix}$, it now follows easily that $A'\in Sp(n-1)\subseteq U(2(n-1))$. This shows that the stabilizer at $p$ is contained in $Sp(n-1)$, embeddeded into $Sp(n)$ as shown.

Conversely, since the first column of any matrix in $Sp(n-1)\subseteq Sp(n)$ is $\begin{bmatrix} 1 \\0 \\ \vdots \\ 0\end{bmatrix}$, $Sp(n-1)$ is a subset of the isotropy group at $p$.


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