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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

1.1

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

givenalgebra

2.1

For ${s l}_{2}$ , the Weyl vector satisfies $ρ (h) = 1$ , so the scalar from The quadratic Casimir eigenvalue on a highest-weight module is $(λ, λ + 2 ρ)$ is $(λ, λ + 2 ρ) = λ (λ + 2) / 8$ . Therefore $C v_{λ} = λ (λ + 2) v_{λ} / 8$ .

step 1.1algebra∎

Depends on

The quadratic Casimir eigenvalue on a highest-weight module is $(\lambda,\lambda+2\rho)$

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

10 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)

sl2sl2​ Casimir and highest-weight eigenvalue