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Kostant cohomology in degrees zero and top

Statement

Assume the Axiom of Choice. In the setting of Kostant's nilradical cohomology theorem: $H^{0} (n^{+}, V) = C_{λ}$ is the highest-weight line, i.e. the $n^{+}$ -invariants; $H^{∣ Φ^{+} ∣} (n^{+}, V) = C_{w_{0} \cdot λ}$ for the longest element $w_{0} \in W$ (Weyl length equals inversion number); and $H^{k} (n^{+}, V) = 0$ for $k > ∣ Φ^{+} ∣$ . When $Φ^{+} = \emptyset$ , $n^{+} = 0$ and $H^{0} (n^{+}, k) = k$ with all higher cohomology zero.

Facts & Assumptions

Given: The setting of Kostant's nilradical cohomology theorem.

[L1]

$H^{k} (n^{+}, V) ≅ ⨁_{ℓ (w) = k} C_{w \cdot λ}$ as $h$ -modules for every $k \geq 0$ (Kostant's nilradical cohomology theorem).

[L2]

The length function satisfies $ℓ (w) = ∣ Φ_{w} ∣ \leq ∣ Φ^{+} ∣$ , there is a unique longest element $w_{0}$ , characterized by $w_{0} (Φ^{+}) = Φ^{-}$ and $ℓ (w_{0}) = ∣ Φ^{+} ∣$ , and $ℓ (1) = 0$ ; the unit is the only element of length $0$ (Weyl length equals inversion number, Length and longest Weyl-group element).

[L3]

The invariants in $V$ are the vectors killed by $n^{+}$ ; for the irreducible highest-weight module $V = L (λ)$ the invariants are the highest-weight line $V_{λ} = C v_{λ}$ (Degree-zero Lie algebra cohomology is the invariant subspace, Weight and weight space, Integral, dominant, and strictly dominant weights, The Weyl vector rho for a chosen positive system).

Proof

technique · specialize the length grading in Kostant's theorem at the two endpoints

1.1L1L2L3

Degree zero: by [L2] the only element of length $0$ is $1$ , and $1 \cdot λ = 1 (λ + ρ) - ρ = λ$ ; by [L1] this gives $H^{0} (n^{+}, V) = C_{λ}$ , the highest-weight line, which is independently the space of $n^{+}$ -invariants by [L3].

1.2L1L2

Top degree: by [L2] the only element of length $∣ Φ^{+} ∣$ is $w_{0}$ , so [L1] gives $H^{∣ Φ^{+} ∣} (n^{+}, V) = C_{w_{0} \cdot λ}$ .

2.1L1L2L3∎

Degrees above the top: by [L2] no $w$ has $ℓ (w) > ∣ Φ^{+} ∣$ , so the direct sum in [L1] is empty and $H^{k} (n^{+}, V) = 0$ for $k > ∣ Φ^{+} ∣$ . If $Φ^{+} = \emptyset$ then $n^{+} = 0$ , $W = {1}$ , $∣ Φ^{+} ∣ = 0$ and $w_{0} = 1$ , so the theorem reduces to $H^{0} (0, V) = V$ for the trivial coefficient module $k$ and vanishing in all positive degrees; this is the rank-zero case of the same formulas.

Depends on

Used by

Kostant cohomology for sl2 Example

Dependency tree · two levels

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Sources

Faisal Al-Faisal, On the Representation Theory of Semisimple Lie Groups (University of Waterloo MMath thesis, 2010), §3.4 printed pp.73–77 (standard reference, not scraped)
Peter Woit, Lie Algebra Cohomology and the Borel–Weil–Bott Theorem (Math G4344, Spring 2012), printed pp.1–7 (standard reference, not scraped)