ExampleConstruction: Literature-sourcedVerification: AI-adaptedverified 2026-09-26 (gpt-6-sol)

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${s l}_{2}$ Casimir and highest-weight eigenvalue

Example

For the Killing form on ${s l}_{2}$ , the quadratic Casimir element is

$C = \frac{1}{8} h^{2} + \frac{1}{4} (e f + f e) .$

On a highest-weight module with highest vector $v_{λ}$ satisfying $h v_{λ} = λ v_{λ}$ , it acts by

$C v_{λ} = \frac{λ (λ + 2)}{8} v_{λ} .$

Facts & Assumptions

Given: The standard basis $e = (\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array})$ , $f = (\begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array})$ , $h = (\begin{array}{cc} 1 & 0 \\ 0 & - 1 \end{array})$ of ${s l}_{2}$ .

Verification

technique · direct

1.1givenalgebra

The Killing form values are $B (h, h) = 8$ and $B (e, f) = B (f, e) = 4$ , so the dual basis is $h / 8, f / 4, e / 4$ . Substituting into the definition of the Casimir gives $C = \frac{1}{8} h^{2} + \frac{1}{4} e f + \frac{1}{4} f e$ .

2.1step 1.1givenalgebra∎

Since $e v_{λ} = 0$ and $[e, f] = h$ , one has $f e v_{λ} = 0$ and $e f v_{λ} = (f e + h) v_{λ} = λ v_{λ}$ . The formula in step 1.1 therefore gives $C v_{λ} = (λ^{2} / 8 + λ / 4) v_{λ} = λ (λ + 2) v_{λ} / 8$ .

Used by

Nothing in the library uses this result yet.

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

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

sl2sl2​ Casimir and highest-weight eigenvalue