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The Casimir tensor is invariant under the diagonal adjoint action

Statement

Let $g$ be a finite-dimensional complex semisimple Lie algebra, let $B$ be its Killing form, and let $(x_{a})$ and $(x^{a})$ be $B$ -dual bases. The tensor $Ω : = \sum_{a} x_{a} \otimes x^{a} \in g \otimes g$ corresponds to the identity map under $Φ_{B} (u \otimes v) (y) : = u B (v, y)$ , and is invariant under the diagonal adjoint action:

$([x, \cdot] \otimes 1 + 1 \otimes [x, \cdot]) Ω = \sum_{a} [x, x_{a}] \otimes x^{a} + x_{a} \otimes [x, x^{a}] = 0 (x \in g) .$

Facts & Assumptions

Given: A finite-dimensional complex semisimple Lie algebra $g$ , its Killing form $B$ , and a pair of $B$ -dual bases $(x_{a}), (x^{a})$ .

[F1]

The Killing form is invariant and nondegenerate on a complex semisimple Lie algebra: $B$ is symmetric, nondegenerate, and invariant, so $B ([u, v], w) = B (u, [v, w])$ .

[F2]

The quadratic Casimir element is independent of the choice of dual bases: $\sum_{a} x_{a} \otimes x^{a}$ is independent of the chosen pair of $B$ -dual bases.

Proof

technique · identify $\mathfrak g\otimes\mathfrak g$ with $\operatorname{End}(\mathfrak g)$ using the nondegenerate form, then compute the two adjoint actions

1.1F1givenalgebra

Nondegeneracy [F1] makes $β_{B} : v \mapsto B (v, \cdot)$ an isomorphism $g \to g^{*}$ . Since $g$ is finite-dimensional, $u \otimes λ \mapsto (y \mapsto λ (y) u)$ is an isomorphism $g \otimes g^{*} \to End (g)$ : in any basis and its dual, it sends the tensor basis to the matrix-unit basis. Their composite is $Φ_{B}$ , hence $Φ_{B}$ is an isomorphism. Duality gives $\sum_{a} x_{a} B (x^{a}, y) = y$ for every $y$ , so $Φ_{B} (Ω) = I_{g}$ .

2.1F1step 1.1algebra

Invariance and symmetry [F1] imply $B ([x, u], v) = - B (u, [x, v])$ : indeed $B (u, [x, v]) = B ([u, x], v) = - B ([x, u], v)$ . Thus $Φ_{B} (([x, \cdot] \otimes 1) Ω) (y) = [x, y]$ , while $Φ_{B} ((1 \otimes [x, \cdot]) Ω) (y) = - \sum_{a} x_{a} B (x^{a}, [x, y]) = - [x, y]$ , using step 1.1's dual-basis identity.

3.1F2step 1.1step 2.1algebra∎

The two images in step 2.1 sum to zero for every $y$ . Since $Φ_{B}$ is injective by step 1.1, the sum of tensors is zero. The tensor itself is independent of the dual bases by [F2], so this is the diagonal adjoint invariance of the canonical Casimir tensor.

The Casimir tensor is invariant under the diagonal adjoint action

Statement

Facts & Assumptions

Proof

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