LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-09-23 (gpt-6-sol)

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The quadratic Casimir element is independent of the choice of dual bases

Statement

Let $g$ be a complex semisimple Lie algebra with Killing form $B$ . For any pair of $B$ -dual bases $(x_{i}), (x^{i})$ , the tensor $\sum_{i} x_{i} \otimes x^{i} \in g \otimes g$ , and hence its image $\sum_{i} x_{i} x^{i} \in U (g)$ under multiplication, is independent of the chosen pair.

Facts & Assumptions

Given: Two pairs of dual bases of a complex semisimple Lie algebra with respect to its nondegenerate Killing form (The Killing form is invariant and nondegenerate on a complex semisimple Lie algebra).

Proof

technique · direct

1.1given

Nondegeneracy identifies $g$ with $g^{*}$ , and under that identification the tensor $\sum_{i} x_{i} \otimes x^{i}$ is the image of the identity map on $g$ . Therefore it depends only on the form, not on the chosen dual bases.

2.1step 1.1∎

Multiplication $g \otimes g \to U (g)$ (The universal enveloping algebra as a tensor quotient) sends that basis-independent tensor to $\sum_{i} x_{i} x^{i}$ . Its image is therefore basis independent as well.

Depends on

Used by

Arakawa-Suzuki Casimir operators on tensor functors Definition
The quadratic Casimir element Definition
The quadratic Casimir element is central Proposition

Dependency tree · two levels

4 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)