Consider the Lie Group $\text{Sp}(2n,\mathbb{C})=\{g\in\text{Mat}_{2n}\mid\ J=g^TJg\}$ where $J=\begin{pmatrix} 0 & 1_n \\ -1_n & 0 \end{pmatrix} $. The corresponding Lie Algebra is $\text{sp}(\text{2n},\mathbb{C})=\{g\in\text{Mat}_{2n}\mid\ g^TJ+Jg=0\}$.
How do I determine a basis for the Lie-Algebra $\text{sp}(\text{2n},\mathbb{C})$?
Thanks in advance!
First of all, write down the general matrix $S=\begin{pmatrix} A & B \\ C & D \end{pmatrix}$ then, look at the equation $S^TJ+JS=0$, ie $$\begin{pmatrix} A & B \\ C & D \end{pmatrix}\begin{pmatrix} 0 & 1_n \\ -1_n & 0 \end{pmatrix}+\begin{pmatrix} 0 & 1_n \\ -1_n & 0 \end{pmatrix}\begin{pmatrix} A & B \\ C & D \end{pmatrix}=0$$ Solving it blockwise would then give you $S=\begin{pmatrix} A & B \\ C & -A^T \end{pmatrix}$, where $B$ and $C$ are symmetric, ie $B=B^T,C=C^T$. So your basis is a free choice on $A$, symmetric $B$ and $C$.
By the way, a great book on Lie Algebras is Erdmann-Wildon's Lie Algebras.