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 $B: \tuple {X, Y} \mapsto \paren {2 n + 2} \map \tr {X Y}$.
Proof
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 $\mathfrak {sp}_{2 n}$.
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