Killing Form of Symplectic Lie Algebra

Theorem
Let $\mathbb K\in\{\C,\R\}$.

Let $n$ be a positive integer.

Let $\mathfrak{sp}_{2n}(\mathbb K)$ be the Lie algebra of the symplectic group $\operatorname{Sp}(2n, \mathbb K)$.

Then its Killing form is $B : (X,Y) \mapsto (2n+2)\operatorname{tr}(XY)$.

Lemma
Let $R$ be a ring with unity.

Let $n$ be a positive integer.

Let $E_{ij}$ denote the matrix with only zeroes except a $1$ at the $(i,j)$th position.

Let $X,Y\in R^{2n\times 2n}$.

Let $X= \begin{pmatrix}X_{11} & X_{12} \\ X_{21} & X_{22}\end{pmatrix}$ and $Y= \begin{pmatrix}Y_{11} & Y_{12} \\ Y_{21} & Y_{22}\end{pmatrix}$

Then
 * $\displaystyle \sum_{i,j\mathop =1}^{2n}\operatorname{tr}\left( (X E_{ij} Y)^t E_{ij} \right) = \operatorname{tr}(Y)\operatorname{tr}(X)$
 * $\displaystyle \sum_{i,j\mathop =1}^n\operatorname{tr}\left( (X E_{ij} Y)^t E_{j+n,i+n} \right) = \operatorname{tr}(Y_{12}^t X_{21})$

Proof
Use Trace of Alternating Product of Matrices and Almost Zero Matrices.

Use Definition:Frobenius Inner Product and Trace in Terms of Orthonormal Basis and the fact that the $(E_{ij})_{i\leq n,j\geq n+1}, (E_{ij})_{i\geq n+1,j\leq n}, (E_{ij}-E_{j+n,i+n})/\sqrt2$ are an orthonormal basis of $\mathfrak{sp}_{2n}$.

Also see

 * Killing Form of Orthogonal Lie Algebra