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The classical Lie-action BRST differential squares to zero
Statement
For every classical Lie-action BRST input of Classical Lie-action BRST cochain complex and every , Consequently, the quotient is a well-defined vector space, with .
Facts & Assumptions
Given: A characteristic-zero field , a Lie algebra , a unital commutative -algebra , and a Lie algebra representation of on by derivations.
Classical Lie-action BRST cochain complex: the BRST cochains are and the operator is the zero-based two-sum Chevalley–Eilenberg formula for the representation .
The Chevalley–Eilenberg differential squares to zero: for a Lie algebra over a field and a representation on a module, the displayed Chevalley–Eilenberg differential satisfies for every .
Proof
The assumptions in [F1] make a -module because is a Lie algebra homomorphism, so . The cochain spaces and both signed sums in are term-by-term those in [F2].
Therefore [F2] applies to these exact inputs and gives for every .
If , write . Step 2.1 gives , so the image is contained in the kernel and the displayed quotient is well-defined.
Depends on
Used by
Dependency tree · two levels
6 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
- Charles A. Weibel, An Introduction to Homological Algebra, §7.7, Lie Algebra Homology and Cohomology (standard reference, not scraped)