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    • CommentRowNumber1.
    • CommentAuthorUrs
    • CommentTimeMay 29th 2010

    I moved the proof of the claim that the Segal-Brylinski “differetiable Lie group cohomology” is that computed in the (oo,1)-topos of oo-Lie groupoids from the entry group cohomology to the entry Lie infinity-groupoid and expanded the details of the proof considerably.

    See this new section.

    Towards the end I could expand still a bit more, but I am not allowed to work anymore today… :-)

    • CommentRowNumber2.
    • CommentAuthorUrs
    • CommentTimeMay 29th 2010

    Towards the end I could expand still a bit more,

    Hm, in fact I think I have a gap towards the end. So far it just shows an inclusion of Segal-Brylinski cohomology into the (oo,1)-topos cohomology, not an isomorphisms.

    Hm, darn, I need to think more about it…

    • CommentRowNumber3.
    • CommentAuthorUrs
    • CommentTimeMay 31st 2010
    • (edited May 31st 2010)

    I don’t see the equivalence anymore that I claimed originally, it now seems to me that the intrinsic (oo,1)-topos Lie group cohomology is even finer than the Segal-Brylinski differential cohomology (which refines the “naive” Lie group cohomology).

    I adjusted the entries accordingly, the details are at oo-Lie groupoid – Lie group cohomology.

    But maybe to highlight the sensitive point within the discussion there:

    Brylinski’s differentiable Lie group cohomology is obtained (in paraphrase) by choosing funcorially for each G×k a cofibrant replacement C({Uki})G×n and then setting

    Hndiffr(G,A):=sPSh([k]Δ[k]C({Uki}),BnA)

    But even though each C({Uki}) is cofibrant, the above totalization is, while related by a zig-zag of weak equivalences to BG, not cofibrant. What is a cofibrant replacement for BG is

    [k]Δ[k]C({Uki})

    where the fat Delta is the Bousfield-Kan resolution Δ[k]=N([k]/Δ[op]op) of the point in [Δop,sSet]proj. So the “true” intrinsic cohomology Hntrue(G,A) of BG is obtained by mapping out of this bigger guy. By pullback along the Bousfield-Kan map we hence have a natural morphism

    Hnnaive(G,A)Hndiffr(G,A)Hntrue(G,A).

    First I thought I had shown that the last map is an iso due to the abelianness of A, but I don’t see my argument anymore.