PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Degree-zero Lie algebra cohomology is the invariant subspace

Statement

Let $a$ be a Lie algebra and $V$ an $a$ -module. In the convention of Chevalley–Eilenberg cochains, evaluation at $1$ identifies $C^{0} (a, V)$ with $V$ , and Chevalley–Eilenberg differential gives $(d^{0} v) (x) = x \cdot v$ for $x \in a$ , $v \in V$ . Hence $H^{0} (a, V) = V^{a} = {v \in V : x \cdot v = 0 for every x \in a},$ so the degree-zero cohomology is the invariant subspace; for the trivial module this reads $H^{0} (a, k) = k$ . This normalizes the degree-zero end of the page and agrees with Zeroth Lie algebra cohomology is invariants.

Facts & Assumptions

Given: A Lie algebra $a$ and an $a$ -module $V$ .

[L1]

Degree-zero cochains are ${Hom}_{k} (Λ^{0} a, V)$ , and evaluation at $1 \in k = Λ^{0} a$ identifies this space with $V$ ; the cochain spaces vanish in negative degrees (Chevalley–Eilenberg cochains).

[L2]

The differential in degree zero is $(d^{0} v) (x) = x \cdot v$ for all $x \in a$ and $v \in V$ (Chevalley–Eilenberg differential), and $d^{1} d^{0} = 0$ (The Chevalley–Eilenberg differential squares to zero).

[L3]

$H^{q} (a, V) = \ker d^{q} / im d^{q - 1}$ (Lie algebra cohomology).

[L4]

The published statement $H^{0} (g, M) = M^{g}$ for the same Chevalley–Eilenberg convention (Zeroth Lie algebra cohomology is invariants); the module identity $[x, y] v = x (y v) - y (x v)$ of Representations of Lie algebras is not needed below, only the action itself.

Proof

technique · compute in degree zero

1.1L1L2

By [L1] an element of $C^{0} (a, V)$ is the same as a vector $v \in V$ , and $d^{0} v = 0$ means exactly that the linear map $x \mapsto x \cdot v$ vanishes, that is $x \cdot v = 0$ for every $x \in a$ . Hence $\ker d^{0} = {v \in V : x \cdot v = 0 for every x \in a}$ .

2.1L1L3step 1.1

The space $C^{- 1} (a, V)$ is zero by [L1], so $im d^{- 1} = 0$ ; substituting into [L3] gives $H^{0} (a, V) = \ker d^{0}$ , and step 1.1 identifies this with the invariant subspace $V^{a}$ .

3.1L4step 1.1∎

For the trivial module the action on $k$ is zero by definition, so every vector is invariant and $H^{0} (a, k) = k$ ; this is the special case $M = k$ of [L4]. If $a = 0$ the condition $x \cdot v = 0$ is vacuous, so $H^{0} (0, V) = V$ ; if $V = 0$ both sides are zero.

Depends on

Used by

Kostant cohomology in degrees zero and top Corollary

Dependency tree · two levels

12 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, Lie Groups and Lie Algebras, MIT 18.755 lecture notes (Spring 2024), §45.2 pp.246–249 and §48.1 pp.259–261 (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)