Timeline for Reference request for cohomology of coverings
Current License: CC BY-SA 3.0
23 events
| when toggle format | what | by | license | comment | |
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| Nov 17, 2014 at 23:21 | vote | accept | Mohammad Farajzadeh-Tehrani | ||
| Nov 17, 2014 at 20:23 | answer | added | Ben Wieland | timeline score: 6 | |
| Nov 17, 2014 at 16:02 | comment | added | Danny Ruberman | @Mohammad: you get an exact sequence $\ldots H^k(X) \overset{p^*}{\to} H^k(X') \overset{t^*-1}{\to} H^k(X') \to H^{k+1}(X) \to \cdots$. This identifies the image of $p^*$ with the kernel of $t^*-1$, or in other words the subgroup left invariant by $t^*$. | |
| Nov 17, 2014 at 16:01 | comment | added | Danny Ruberman | @MatthiasWendt: Yes, of course you're right that this can be done via a spectral sequence. My point was that there's a more elementary approach where you see what's going on at the chain level. Plus, I really like that Milnor paper! | |
| Nov 17, 2014 at 10:43 | comment | added | Matthias Wendt | @DannyRuberman: the long exact sequence associated to the short exact sequence of complexes is exactly the one arising from the $E_2$-degeneration of the Cartan-Leray spectral sequence (which follows from cohomological dimension 1 of $\mathbb{Z}$). I guess my explanation with spectral sequences obscured the simplicity of the situation... | |
| Nov 17, 2014 at 1:35 | comment | added | Mohammad Farajzadeh-Tehrani | @ Ruberman: Could you please add few more lines on how you conclude the result from the long exact sequence . | |
| Nov 17, 2014 at 1:12 | comment | added | Danny Ruberman | For the special case of an Z covering, there is an easy approach due to Milnor (in his beautiful paper called Infinite Cyclic Coverings). For $X'\to X$ a Z-covering with covering group generated by t, there is a short exact sequence of chain groups $0 \to C_*(X') \to C_*(X') \to C_*(X) \to 0$ where the first map is $t_* -1$ and the second is $p_*$. A quick look at the associated long exact sequence gives the requested result. Perhaps this can be iterated (to get $Z^n$ coverings) and combined with easy results about finite coverings to get the general case. | |
| Nov 17, 2014 at 1:04 | history | edited | Danny Ruberman | CC BY-SA 3.0 |
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| Nov 16, 2014 at 23:54 | answer | added | Aleksey | timeline score: 1 | |
| Nov 15, 2014 at 22:29 | history | edited | Mohammad Farajzadeh-Tehrani | CC BY-SA 3.0 |
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| Nov 15, 2014 at 21:10 | comment | added | Qiaochu Yuan | @Mohammad: probably you also want to specify the coefficients; you said $\mathbb{Q}$ was okay? | |
| Nov 15, 2014 at 20:05 | comment | added | Mohammad Farajzadeh-Tehrani | @ Matthias: Now It is very explicit as you wanted. | |
| Nov 15, 2014 at 20:04 | history | edited | Mohammad Farajzadeh-Tehrani | CC BY-SA 3.0 |
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| Nov 15, 2014 at 19:24 | comment | added | Matthias Wendt | Could you please formulate your situation precisely? I do not really understand the formulation with the quotients of maximal abelian covering... Generally, the problem will be that if $X$ and $G$ have sufficiently complicated rational cohomology, it will be difficult to see why the differentials $H^0(G,H^p(X,M))\to H^2(G,H^{p-1}(X,M))$ (and then the higher ones...) should be trivial. | |
| Nov 15, 2014 at 17:49 | history | edited | Mohammad Farajzadeh-Tehrani | CC BY-SA 3.0 |
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| Nov 15, 2014 at 17:48 | comment | added | Mohammad Farajzadeh-Tehrani | I am not sure, I hope so ;) I am looking at certain quotients of maximal abelian covering and I just need it for this case. | |
| Nov 15, 2014 at 17:37 | comment | added | Francesco Polizzi | But why are you sure that this is true in such a generality? | |
| Nov 15, 2014 at 17:33 | comment | added | Matthias Wendt | If I got the indices right, the differentials on the page $E^{p,q}_2=H^p(G,H^q(X,M))$ are $E^{p,q}_2\to E^{p+2,q-1}$ - so that some differentials may be non-trivial on $H^0(G,H^q(X,M))$. Again, if I get the indices right, the filtration of the $E_\infty$-term gives a factorization $H^i(X/G)\twoheadrightarrow K_i\to H^i(X)$ of the pullback, where $K_i$ is the kernel of all differentials starting at $H^0(G,H^i(X,M))$. This seems to say that not every $G$-invariant cohomology class is a pullback. | |
| Nov 15, 2014 at 17:10 | comment | added | Matthias Wendt | In general, there is the Cartan-Leray spectral sequence $H^p(G,H^q(X,M))\Rightarrow H^{p+q}(X/G,M)$, see Theorem 7.9, Section VII of Brown's book on cohomology groups. $H^0(G,H^q(X,M))$ appears in the $E^2$-term, but there may be differentials interfering... | |
| Nov 15, 2014 at 17:03 | comment | added | Mohammad Farajzadeh-Tehrani | Cohomology over $\mathbb{Q}$ is fine for me, but I need a reference for the infinite covering mainly. | |
| Nov 15, 2014 at 16:59 | comment | added | Matthias Wendt | For a finite group $G$ and coefficients prime to the order of $G$, this is explained in the answers to MO-question mathoverflow.net/questions/57071 | |
| Nov 15, 2014 at 16:55 | history | edited | Mohammad Farajzadeh-Tehrani |
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| Nov 15, 2014 at 16:40 | history | asked | Mohammad Farajzadeh-Tehrani | CC BY-SA 3.0 |