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The Borel-Weil-Bott Euler character is a signed dual Weyl character

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $c h$ denote the formal character of a finite-dimensional $g$ -module (The formal character of a finite-dimensional weight module) and let $A$ be the Weyl alternation operator with Weyl denominator $A (ρ) = \prod_{α > 0} (e^{α / 2} - e^{- α / 2})$ (The Weyl alternation operator, The Weyl character formula). For every $λ \in X^{*} (T)$ : if $λ + ρ$ is not regular then $\sum_{i} (- 1)^{i} c h H^{i} (X, L_{λ}) = 0$ ; if $λ + ρ$ is regular with Weyl element $w$ , then $\sum_{i} (- 1)^{i} c h H^{i} (X, L_{λ}) = (- 1)^{ℓ (w)} c h L (w \cdot λ)^{*} = (- 1)^{ℓ (w)} \frac{A (- w_{0} (w \cdot λ) + ρ)}{A (ρ)} .$ Equivalently, with $ν = - w_{0} (w \cdot λ)$ , which is the dominant integral weight of the dual module $L (w \cdot λ)^{*}$ , the last expression is $(- 1)^{ℓ (w)} A (ν + ρ) / A (ρ)$ .

Facts & Assumptions

Given: The Axiom of Choice, the group $G$ , its Borel $B$ , the flag variety $X = G / B$ , the Weyl group $W$ with longest element $w_{0}$ , a weight $λ$ , and its line bundle $L_{λ}$ .

[F1]

Borel-Weil-Bott: if $λ + ρ$ is singular then all $H^{i} (X, L_{λ})$ vanish, and if $λ + ρ$ is regular with Weyl element $w$ then $H^{ℓ (w)} (X, L_{λ}) ≅ L (w \cdot λ)^{*}$ and all other cohomology vanishes; in particular the alternating sum runs over a finite list of finite-dimensional modules (The Borel-Weil-Bott theorem).

[F2]

The formal character is additive: $c h (V \oplus W) = c h V + c h W$ and the zero module has character $0$ , the sum being taken in the completed character ring (Formal characters are additive and multiplicative, The formal character of a finite-dimensional weight module).

[F3]

For a dominant integral weight $ν$ the dual $L (ν)^{*}$ is irreducible of highest weight $- w_{0} ν$ , so $L (ν)^{*} ≅ L (- w_{0} ν)$ by the classification of finite-dimensional irreducibles; the Weyl character formula gives $c h L (μ) = A (μ + ρ) / A (ρ)$ for every dominant integral $μ$ , and the denominator is $A (ρ) = e^{ρ} \prod_{α > 0} (1 - e^{- α}) = \prod_{α > 0} (e^{α / 2} - e^{- α / 2})$ , where the last equality follows from $\sum_{α > 0} α / 2 = ρ$ (Highest weight of the dual representation, Highest-weight classification, The Weyl character formula).

Proof

1.1F1F2given

If $λ + ρ$ is not regular, then by [F1] every $H^{i} (X, L_{λ})$ vanishes, so each character is $0$ and the alternating sum is the finite sum of zero characters, hence $0$ by [F2].

1.2F1F2F3givenalgebra

If $λ + ρ$ is regular with Weyl element $w$ , then by [F1] only $H^{ℓ (w)} (X, L_{λ})$ is nonzero and it is isomorphic to $L (w \cdot λ)^{*}$ ; the alternating sum over the finitely many cohomology groups therefore equals $(- 1)^{ℓ (w)} c h L (w \cdot λ)^{*}$ by additivity [F2]. Since $w \cdot λ$ is dominant integral, [F3] identifies $L (w \cdot λ)^{*} ≅ L (ν)$ with $ν = - w_{0} (w \cdot λ)$ dominant integral, and the Weyl character formula gives $c h L (ν) = A (ν + ρ) / A (ρ) = A (- w_{0} (w \cdot λ) + ρ) / A (ρ)$ .

2.1step 1.1step 1.2∎

Combining the singular case of step 1.1 and the regular case of step 1.2 gives the two asserted evaluations of the alternating sum of characters.

Remarks

The scaffold stated the regular case as $(- 1)^{ℓ (w)} c h L (w \cdot λ) = (- 1)^{ℓ (w)} A (w \cdot λ + ρ) / A (ρ)$ , which is false in rank at least two: for $G = S L_{3}$ and the dominant weight $λ = ω_{1}$ one has $w = 1$ , $H^{0} (X, L_{ω_{1}}) ≅ L (ω_{1})^{*}$ , and $L (ω_{1})^{*}$ has weights $- ω_{1}$ , $ω_{1} - ω_{2}$ , $ω_{2}$ , so its character is $e^{- ω_{1}} + e^{ω_{1} - ω_{2}} + e^{ω_{2}}$ , whereas $c h L (ω_{1}) = e^{ω_{1}} + e^{ω_{2} - ω_{1}} + e^{- ω_{2}}$ ; $W$ -invariance of characters does not identify the two, because it only permutes the weights of a fixed module. The corrected statement above uses $c h L (w \cdot λ)^{*} = A (- w_{0} (w \cdot λ) + ρ) / A (ρ)$ , which coincides with $A (w \cdot λ + ρ) / A (ρ)$ exactly when $- w_{0} (w \cdot λ) = w \cdot λ$ .

Depends on

Used by

An sl3 weight with cohomology in degree one Example

Dependency tree · two levels

49 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

Xiong Rui, Borel-Weil and Borel-Weil-Bott, Lecture 1 (standard reference, not scraped)
Joshua Ng (Hoi Hei Jan Sum), The Borel-Weil-Bott Theorem (Chicago REU 2015) (standard reference, not scraped)