LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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The quadratic Casimir element is independent of the choice of dual bases

Statement

The tensor $\sum_{i} x_{i} \otimes x^{i} \in g \otimes g$ , and hence the element $C = \sum_{i} x_{i} x^{i} \in U (g)$ from The quadratic Casimir element, is independent of the chosen pair of dual bases.

Facts & Assumptions

Given: Two pairs of dual bases of a complex semisimple Lie algebra with respect to its Killing form.

Proof

technique · direct

1.1

The Killing form 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.

given

2.1

Multiplication $g \otimes g \to U (g)$ sends that basis-independent tensor to the element $\sum_{i} x_{i} x^{i}$ from The quadratic Casimir element, so the Casimir element is basis independent as well.

step 1.1∎

Depends on

The quadratic Casimir element

Used by

The quadratic Casimir element is central Proposition

Cited to discharge well-definedness by The quadratic Casimir element.

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)