Timeline for Where can I easily look up / calculate (abelian) group cohomology?
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9 events
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| Oct 22, 2010 at 4:48 | comment | added | Torsten Ekedahl | Let me also add that the map $H^2(A,B)\to \mathrm{Hom}(\Lambda^2A,B)$ is given by anti-symmetrisation of a representing cocycle (which can also be described as the commutator map associated to the central extension). | |
| Oct 22, 2010 at 4:41 | comment | added | Torsten Ekedahl | Dear Theo, the exterior algebra imposes the relation $x\land x=0$ so that $\Lambda^2\mathbb Z/2=0$ (this was why I made all this fuss about checking that the square of an element was zero). As for $H^2(\mathbb Z/2,R^\times$ we have that $R^\times=\mathbb Z/2\times R$ and while the second part is injective, the first is not. Hence the universal coefficient formula becomes $H^2(\mathbb Z/2,\mathbb R^\times=\mathrm{Ext}(\mathbb Z/2,\mathbb Z/2)=\mathbb Z/2$. As reference for the universal coefficient formula you can pick (just about) any book on algebraic topology. | |
| Oct 22, 2010 at 4:24 | comment | added | Theo Johnson-Freyd | On the other hand, if I switch out $\mathbb C^\times$ for $\mathbb R^\times$, I can compute directly that ${\rm H}^2(\mathbb Z/2,\mathbb R^\times)=2$, and yet an argument like the one you gives suggests that it should be $\operatorname{Hom}(\wedge^2(\mathbb Z/2),\mathbb R^\times)$? Maybe there's a lot in "using ... fact that the coefficient group $\mathbb C^\times$ is injective" (certainly $\mathbb R^\times$ is not injective). Can you recommend a good reference where I can look up precise statements of things like "the universal coefficient formula"? | |
| Oct 22, 2010 at 4:07 | comment | added | Theo Johnson-Freyd | Hrm. There is something that I'm doing wrong. Let me try an easier example, like ${\rm H}^2(\mathbb Z/2,\mathbb C^\times)$. I can work this out explicitly, and (modulo errors) I get $1$. On the other hand, maybe I don't know what $\wedge^2(\mathbb Z/2)$ is, but I would have guessed that it was $\mathbb Z/2$, as all elements of $(\mathbb Z/2)^{\otimes 2}=\mathbb Z/2$ are antisymmetric? Or is it the quotient of $\otimes^2$ by the symmetric things? If it is the latter, then I do get ${\rm H}^2(\mathbb Z/2,\mathbb C^\times)=\text{Hom}(1,\mathbb C^\times)=1$, but if the former I get $=2$. | |
| Oct 22, 2010 at 3:58 | vote | accept | Theo Johnson-Freyd | ||
| Oct 17, 2010 at 18:51 | comment | added | Torsten Ekedahl | Yes, the switching is given by switching of the input factors. That follows directly from the definition of Pontryagin product. As I said, the only tricky thing is that the square of an element of $H_1(A)$ is zero. Compare, with the cup-product, where switching the factors of $X\times X$ correspond to switching the factors in the cup product but the square of an element of $H^1(X)$ may not be zero when $2$ is not invertible in the coefficient ring. | |
| Oct 17, 2010 at 18:14 | comment | added | Theo Johnson-Freyd | Awesome. You have completely answered the question that I really want to know, and from the discussion in the comments above it looks like the question that for the I asked the answer is "there isn't one". A question: there is an obvious Z/2 action on 2-cocycles, given by switching the two inputs (everything is abelian). Is this the same action as the Z/2 action on \Lambda^2 C^\times, i.e. the sign action? | |
| Oct 17, 2010 at 8:58 | history | edited | Torsten Ekedahl | CC BY-SA 2.5 |
added 1 characters in body
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| Oct 17, 2010 at 8:43 | history | answered | Torsten Ekedahl | CC BY-SA 2.5 |