PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-05

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The quadratic Casimir element is central

Statement

For every $x \in g$ , the quadratic Casimir element $C$ from The quadratic Casimir element satisfies

$[x, C] = 0.$

Hence $C \in Z (U (g))$ .

Facts & Assumptions

Given: A complex semisimple Lie algebra $g$ , its Casimir element $C$ , and $x \in g$ .

Proof

technique · direct

1.1

Choose dual bases $x_{i}, x^{i}$ for the Killing form. Then $[x, C] = \sum_{i} [x, x_{i}] x^{i} + x_{i} [x, x^{i}]$ .

givenalgebra

2.1

The basis-independent tensor $Ω : = \sum_{i} x_{i} \otimes x^{i}$ corresponds, through the Killing-form identification $g ≅ g^{*}$ , to the identity endomorphism of $g$ . Invariance of the form says that this identity tensor is fixed by the diagonal adjoint action. Hence $\sum_{i} [x, x_{i}] \otimes x^{i} + x_{i} \otimes [x, x^{i}] = 0.$ Multiplying the tensor factors and using step 1.1 gives $[x, C] = 0$ . The quadratic Casimir element is independent of the choice of dual bases

step 1.1algebra

3.1

Since $[x, C] = 0$ for every $x \in g$ , the element $C$ commutes with the generators of $U (g)$ and is therefore central.

step 2.1∎

Depends on

Used by

A noninvariant quadratic PBW element need not be central Counterexample
The leading PBW symbol of a central element is invariant Lemma
The quadratic Casimir eigenvalue on a highest-weight module is (λ,λ+2ρ) Proposition

Dependency tree · two levels

5 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Pavel Etingof, Representations of Lie Groups (standard reference, not scraped)
Alexander Kleshchev, Lectures on Infinite Dimensional Lie Algebras (standard reference, not scraped)