Timeline for The relation between group cohomology and the cohomology of the classifying space
Current License: CC BY-SA 4.0
13 events
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| Sep 30, 2021 at 21:56 | history | edited | Kevin Walker | CC BY-SA 4.0 |
added 65 characters in body
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| Sep 30, 2021 at 21:53 | history | rollback | Kevin Walker |
Rollback to Revision 1
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| Sep 30, 2021 at 21:53 | history | edited | Kevin Walker | CC BY-SA 4.0 |
URLs were incorrectly incorporating right-bracket
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| Jun 1, 2018 at 21:51 | vote | accept | Xiao-Gang Wen | ||
| Nov 13, 2014 at 12:02 | comment | added | Xiao-Gang Wen | Christoph: Thank you very much. $H^n(BG,Z_m)\cong H^n_B(G,Z_m)\cong H^n_t(G,Z_m)$ are the relations I am asking. We know that $H^n(BG,Z)\cong H^n_B(G,Z)\cong H^n_t(G,Z)$, but we do not have $H^n(BG,R/Z)\cong H^n_B(G,R/Z)\cong H^n_t(G,R/Z)$. So I do not know if $H^n(BG,Z_m)\cong H^n_B(G,Z_m)\cong H^n_t(G,Z_m)$ is valid for continuous group $G$. You seems suggest that it is valid. Is there any ref? Also what is "topological group cohomology"? | |
| Nov 13, 2014 at 7:45 | comment | added | Christoph Wockel | Sorry, there are too many $H^n$ flying around. Let us call $H^n_{t}(G,A)$ the topological group cohomology and $H^n(BG,A)$ the cohomology of the classifying space with discrete coefficients $A$. What we have are the isomorphisms $H^n(BG,Z_m)\cong H^n_B(G,Z_m)\cong H^{n+1}_B(G,A)$ for $A$ as described (and $G$ locally compact and finite-dimensional), and also $H^n(BG,Z_m)\cong H^n_t(G,Z_m)\cong H^{n+1}_t(G,A)$. Could you now repeat your question? | |
| Nov 11, 2014 at 14:07 | comment | added | Xiao-Gang Wen | Or I wonder if we can show $H_B^*(G,Z_n)=H^*(BG,Z_n)$ for continuous group. If not, how can we modify the relation to make it valid. | |
| Nov 11, 2014 at 11:53 | comment | added | Xiao-Gang Wen | Christoph: thank you very much for explaining. Using your above construction, now we have $H^n_B(G,A)\simeq H^{n+1}_B(G,Z_n)$ for measurable group cohomology, or $H^n(BG,A)\simeq H^{n+1}(BG,Z_n)$ for topological cohomology, which is very very interesting. My question here is can we use a similar construction to find the relation between group cohomology $H^*_B(G,Z_n)$ and topological cohomology $H^*(BG,M)$ with some properly chosen $M$ (where $G$ can be continuous). | |
| Nov 11, 2014 at 8:45 | comment | added | Christoph Wockel | @Xiao-Gang: Yes, this runs under the name "dimension shifting". Take $E Z_n$ the group of (equivalence classes of) measurable $Z_n$-valued functions on the unit interval. With the distance in measure this should be a contractible polish group. Then take $C(G,E Z_n)$, which is still contractible and polish. Moreover, it is soft, so the topological group cohomology (and under the above assumption on $G$) of it vanishes (see the appendix in our paper). Since $Z_n$ is finite, it acts properly discontinuously on $C(G,E Z_n)$ and so $E:=C(G,E Z_n)\to C(G,E Z_n)/Z_n=:A$ has a continuous section. | |
| Nov 8, 2014 at 12:57 | comment | added | Xiao-Gang Wen | Christoph: Do you have an example of a short exact sequence $Z_n\to E\to A$ such that that $H^n_B(G,E)$ vanishes? | |
| Nov 7, 2014 at 7:51 | comment | added | Christoph Wockel | Our results work also for finite-dimensional locally compact groups. The problems with $Z_n$-coefficients would be to find a short exact sequence $Z_n \to E \to A$ where you see that $H^n_B(G,E)$ vanishes. (To this sequence you would like to apply the long exact sequence to in oder to identify $H^n_B(G,Z_n)$ with $H^n_B(G,A)$ in case $H^n_B(G,E)$ vanishes. For $Z$ you can take $Z \to R \to U(1)$.) | |
| Nov 7, 2014 at 4:33 | comment | added | Xiao-Gang Wen | Christoph Wockel: Thank you very much. Do your results apply to Borel group cohomology $H^d_B(G,Z_n)$ when $G$ is continuous? | |
| Nov 6, 2014 at 21:56 | history | answered | Christoph Wockel | CC BY-SA 3.0 |