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Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeOct 15th 2009

started Lie algebra cohomology,

(for the moment mainly to record that reference on super Lie algebra cocycles)

• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeSep 21st 2010

polished and expanded Lie algebra cohomology: added an Idea-section, collected the different definitions together, added explanations to the definition via oo-Lie algebra morphisms, expanded the section on Extension, started an Examples, section

• CommentRowNumber3.
• CommentAuthorzskoda
• CommentTimeSep 21st 2010

The words "infinitesimal gauge transformation" in one entry point to gauge transformation while in gauge transformation to infinitesimal object. At both places allusion is just half-clear so far. Could you have exact statement ? Infinitesimal gauge transformations are infinitesimal object in which category/setup ? Can this explanation be more than allusive playing with words ?

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeSep 21st 2010
• (edited Sep 21st 2010)
• CommentRowNumber5.
• CommentAuthorUrs
• CommentTimeSep 21st 2010

• CommentRowNumber6.
• CommentAuthorUrs
• CommentTimeFeb 2nd 2011
• (edited Feb 2nd 2011)

I was being asked, and so I added a textbook reference to Chevalley-Eilenberg algebra, to Lie algebra cohomology and and a pointer to an article to nonabelian Lie algebra cohomology

• CommentRowNumber7.
• CommentAuthorUrs
• CommentTimeFeb 26th 2018
• (edited Feb 26th 2018)

I have recorded the following fact (here) form Solleveld 02, theorem 2.28:

Let

1. $(\mathfrak{g}, [-,-])$ be a Lie algebra of finite dimension;

2. $(V, \rho)$ a $\mathfrak{g}$-Lie algebra module of finite dimension, which is reducible;

3. $\mathfrak{h} \hookrightarrow \mathfrak{g}$ a sub-Lie algebra which is reductive in $\mathfrak{g}$ in that its adjoint representation on $\mathfrak{g}$ is reducible

4. such that

$\mathfrak{g} = \mathfrak{h} \ltimes \mathfrak{a}$

is a semidirect product Lie algebra (hence $\mathfrak{a}$ a Lie ideal).

Then the invariants in Lie algebra cohomology of $\mathfrak{a}$ (equivalently with respect to $\mathfrak{h}$ or all of $\mathfrak{g}$) coincide with the relative Lie algebra cohomology (using the invariant subcomplex!):

$H^\bullet(\mathfrak{a}; V)^{\mathfrak{h}} \;\simeq\; H^\bullet(\mathfrak{g}, \mathfrak{h}; V) \,.$