Killing Form of Symplectic Lie Algebra

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

Let $n$ be a positive integer.

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

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

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 $\left({i, j}\right)$th position.

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

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

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 $\left({E_{i j} }\right)_{i \le n, j \ge n + 1}, \left({E_{i j} }\right)_{i \ge n + 1, j \le n}, \left({E_{i j} - E_{j + n, i + n} }\right) / \sqrt2$ are an orthonormal basis of $\mathfrak{sp}_{2 n}$.

Also see

 * Killing Form of Orthogonal Lie Algebra