Center of the symplectic group

Question

I am trying to figure out what the center of $Sp(n)$ is. I know that $Sp(n) = U(n) \cap Sp(n,\mathbb{C})$, where $Sp(n,\mathbb{C})$ are all $2n \times 2n$ complex matrices $A$ such that $A^TJA = J$, where $$ J = \begin{pmatrix} 0 & -I_n\\ I_n & 0 \end{pmatrix}. $$ I was able to find the center of most of the other classical matrix groups using the Spectral theorem and permutation matrices, but for this one I'm having a hard time. Maybe after I use the spectral theorem on an element of the center, I could decompose the unitary matrix that diagonalizes it by an element in $Sp(n)$ and some other matrix?

Answer 1

Let $A$ be in the centre of $Sp(n)$.

Then $A$ commutes with $S=D\oplus\overline{D}$, where $D$ is any diagonal matrix whose diagonal elements are distinct unimodular complex numbers with positive imaginary parts. Since $S$ is a diagonal matrix with distinct diagonal entries, $A$ must be a diagonal matrix. Write $A=D_1\oplus D_2$ where $D_1$ and $D_2$ are $n\times n$ diagonal matrices.

$A$ also commutes with all matrices of the form $U\oplus\overline{U}$ where $U$ is unitary. In turn, $D_1$ and $D_2$ commute with all $n\times n$ complex matrices, because $M_n(\mathbb C)$ is the linear span of $U(n)$. Hence they are scalar matrices and $A=aI_n\oplus bI_n$ for some scalars $a$ and $b$.

Now, from $AJ=JA$ and $A^TJA=J$, we get $a=b=\pm1$. Hence $A=\pm I_{2n}$. Such an $A$ obviously does commute with all matrices in $Sp(n)$. Therefore the centre of $Sp(n)$ is $\{I_{2n},-I_{2n}\}$.

Answer 2

Here's the answer via Cayley Transform. There's an exercise from Artin's Algebra lurking in the background (a "miscellaneous exercise" in the Linear Groups chapter, either edition).

Suppose $Z$ is in the center of the symplectic group, $G$
(i.) since $J \in G\implies ZJ = JZ$
(ii.) focus on $U$, the set of all $2n \times 2n$ matrices such that $A \in U$ when $\det\big(A+I\big)\neq 0$
(iii.) define $A'$ as the Cayley Transform of $A$. Note that said transform is an involution. It is immediate that $ZA-AZ = \mathbf 0 \implies ZA'-A'Z = \mathbf 0$
(iv.) (from the exercise), $A\in U\cap G$ iff $J^TA'$ is symmetric

So consider arbitrary symmetric matrix $S$, then $Z$ commutes with $S$.
proof:
$ZS-SZ =\mathbf 0 \iff \delta \cdot ZJS -(\delta\cdot JS)Z=\mathbf 0$
for some $\delta \gt 0$. Well $(\delta\cdot JS)$ does not have an eigenvalue of $-1$ by selecting $\delta$ small enough, hence it has a Cayley Transform $(\delta\cdot JS)\mapsto T$, so applying the Cayley Transform again $T\mapsto (\delta\cdot JS)$ and $J^T (\delta\cdot JS)$ is symmetric, hence $T\in G$ by (iv.), so $Z$ commutes with $T$ and by (iii.) $Z$ commutes with $(\delta\cdot JS)$.

Thus $Z$ commutes with arbitrary symmetric matrices $\implies Z \propto I$, using e.g. If a symmetric matrix commutes with all symmetric matrices, is it then a multiple of the identity?