Centre of $\mathfrak{sp}(2n,k)$

Question

How can I show that $\mathfrak{sp}(2n,K)=\{A \in M_{2n}(K)\,|\, A^TJ_{sp}+J_{sp}A=0\}$ has trivial centre? ($J_{sp}=\begin{pmatrix} 0 & Id_n\\ -Id_n & 0\end{pmatrix}$)

Answer 1

It is convenient to use the fact that the elements of $\mathfrak{sp}(2n,K)$ are the matrices $\left(\begin{smallmatrix}A&B\\C&-A^T\end{smallmatrix}\right)$ such that $A,B,C\in M_n(K)$ and that both $B$ and $C$ are symmetric.

So if such a a matriz belongs to the center of $\mathfrak{sp}(2n,K)$, how to prove that it must be the null matrix? It is clear that $A$ must commute with every matrix $M\in M_n(K)$. Therefore, $A$ is a diagonal matrix, that is $A=\lambda\operatorname{Id}$. But you know that$$\left[\begin{pmatrix}\lambda\operatorname{Id}&0\\0&-\lambda\operatorname{Id}\end{pmatrix},\begin{pmatrix}0&\operatorname{Id}\\0&0\end{pmatrix}\right]=0$$In other words $\left(\begin{smallmatrix}0&2\lambda\operatorname{Id}\\0&0\end{smallmatrix}\right)=0$. So, $\lambda=0$.

What about an element of the center of the type $\left(\begin{smallmatrix}0&B\\C&0\end{smallmatrix}\right)$? It turns out that$$0=\left[\begin{pmatrix}0&B\\C&0\end{pmatrix},\begin{pmatrix}0&0\\\operatorname{Id}&0\end{pmatrix}\right]=\begin{pmatrix}B&0\\0&-B\end{pmatrix}.$$So, $B=0$. A simillar argument shows that $C=0$.

Answer 2

The Lie algebra $\mathfrak{sp}(2n,K)$ over a field $K$ of characteristic zero has non-degenerate Killing form, see here. Hence its center, which is an abelian ideal, is zero.