Killing Form of Symplectic Lie Algebra

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

Let $n$ be a positive integer.

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

Then its Killing form is $B: \tuple {X, Y} \mapsto \paren {2 n + 2} \map \tr {X Y}$.

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 $\tuple {i, j}$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:
 * $\ds \sum_{i, j \mathop = 1}^{2 n} \map \tr {\paren {X E_{i j} Y}^t E_{i j} } = \map \tr Y \map \tr X$


 * $\ds \sum_{i, j \mathop = 1}^n \map \tr {\paren {X E_{ij} Y}^t E_{j + n, i + n} } = \map \tr {Y_{1 2}^t X_{2 1} }$

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 $\sequence {E_{i j} }_{i \mathop \le n, j \mathop \ge n + 1}, \sequence {E_{i j} }_{i \mathop \ge n + 1, j \mathop \le n}, \sequence {E_{i j} - E_{j + n, i + n} } / \sqrt 2$ are an orthonormal basis of $\mathfrak {sp}_{2 n}$.

Also see

 * Killing Form of Orthogonal Lie Algebra