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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeMay 29th 2010
   • PermaLink
   Author: Urs
   Format: MarkdownItexI 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 <a href="http://ncatlab.org/nlab/show/Lie+infinity-groupoid#Cohomology">this new section</a>. Towards the end I could expand still a bit more, but I am not allowed to work anymore today... :-)

   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… :-)

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeMay 29th 2010
   • PermaLink
   Author: Urs
   Format: MarkdownItex> 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...

   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…

3. • CommentRowNumber3.
   • CommentAuthorUrs
   • CommentTimeMay 31st 2010
   • (edited May 31st 2010)
   • PermaLink
   Author: Urs
   Format: MarkdownItexI 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 <a href="http://ncatlab.org/nlab/show/Lie+infinity-groupoid#CohomologyOfLieGroups">oo-Lie groupoid -- Lie group cohomology</a>. 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.

   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.