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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… :-)
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…
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 $G^{\times_k}$ a cofibrant replacement $C(\{U^k_i\}) \stackrel{\simeq}{\to} G^{\times_n}$ and then setting
$H^n_{diffr}(G,A) := sPSh(\int^{[k]} \Delta[k] \otimes C(\{U^k_i\}) , \mathbf{B}^n A)$But even though each $C(\{U^k_i\})$ is cofibrant, the above totalization is, while related by a zig-zag of weak equivalences to $\mathbf{B}G$, not cofibrant. What is a cofibrant replacement for $\mathbf{B}G$ is
$\int^{[k]} \mathbf{\Delta}[k] \otimes C(\{U^k_i\})$where the fat Delta is the Bousfield-Kan resolution $\mathbf{\Delta}[k] = N([k]/\Delta[op]^{op})$ of the point in $[\Delta^{op}, sSet]_{proj}$. So the “true” intrinsic cohomology $H^n_{true}(G,A)$ of $\mathbf{B}G$ is obtained by mapping out of this bigger guy. By pullback along the Bousfield-Kan map we hence have a natural morphism
$H^n_{naive}(G,A) \to H^n_{diffr}(G,A) \to H^n_{true}(G,A) \,.$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.
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