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First cohomology is derivations modulo inner derivations

Statement

A linear map $δ : g \to M$ is a $1$ -cocycle exactly when

$δ ([x, y]) = x δ (y) - y δ (x) .$

The $1$ -coboundaries are $δ_{m} (x) = x m$ . Hence $H^{1} (g, M) = Der (g, M) / Inn (g, M)$ ; for the adjoint module this is $Der (g) / ad (g)$ .

Facts & Assumptions

Given: A Lie algebra $g$ and a module $M$ .

[L1]

The CE differential has the declared low-degree formula (Chevalley–Eilenberg differential).

[L2]

Cohomology is cocycles modulo coboundaries (Lie algebra cohomology).

[L3]

Ordinary derivations obey the Leibniz rule for the adjoint module (Derivations of Lie algebras).

Proof

technique · compute in degrees zero and one

1.1L1

For a $1$ -cochain $δ$ , [L1] gives $(d δ) (x, y) = x δ (y) - y δ (x) - δ ([x, y])$ . Thus its kernel is precisely the space of module-valued derivations in the displayed sense.

1.2L1L2

For $m \in C^{0} = M$ , [L1] gives $(d m) (x) = x m$ , so the image consists exactly of the inner module-valued derivations. Taking the quotient in [L2] proves the first identification.

2.1L3step 1.11.2∎

If $M = g$ with the adjoint action, step 1.1 becomes $δ ([x, y]) = [δ x, y] + [x, δ y]$ , the derivation law in [L3], while step 1.2 gives $x \mapsto [x, m] = {ad}_{- m} (x)$ ; its span is the same inner-derivation space. Zero algebras and zero modules satisfy the same formulas.

Depends on

Used by

First cohomology with trivial coefficients Example
First Whitehead lemma Theorem

Dependency tree · two levels

9 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

Weibel, Lie Algebra Homology and Cohomology, §7.7 (standard reference, not scraped)