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    • 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

    added Whitehead’s lemma

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

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

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

    4. such that

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

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

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

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