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For an example I'm trying to understand, I need to calculate some cohomology group of some $\mathbb Z$-module with coefficients in some other $\mathbb Z$-module (with no interesting actions). (In particular, letting $\mathbb C^\times$ be the multiplicative group of the complex numbers with the discrete topology, I would like to know ${\rm H}^2(\mathbb C^\times,\mathbb C^\times)$, which is probably trivial but I have no idea how to show it is.) Having never really learned much cohomology theory, I know some basic definitions and I know how to turn an extension of groups into an exact sequence in cohomology. But this doesn't help me do particular computations unless I at least know the necessary cohomologies for pieces of the group. (For example, I may know that $\mathbb C^\times = S^1 \times \mathbb R$ as discrete groups, but I don't know any $\mathbb H^\bullet(S^1,\mathbb C^\times)$ or $\mathbb H^\bullet(\mathbb R,\mathbb C^\times)$.)

What would be best is some reference book or, as I won't be near a good library for at least two days, a good website that lists sufficiently many cohomology groups and pedagogically explains some ways to do explicit computations to extend their list. Something like (but presumably easier than) the Knot Atlas or the OEIS. Does such a thing exist?

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The nice thing about the Knot Atlas is that all the invariants it lists are algorithmically computable. Group cohomology isn't. Even for abelian groups (provided the abelian group isn't finitely presentable). So it's not clear to what extent you can generate tables -- certainly there isn't an efficient way. My understanding is people like Antonio Montalban have strong results of the form that $H^3$ of torsion free abelian groups is highly non-computable in general. –  Ryan Budney Oct 17 '10 at 4:49
    
I don't know the answer, but I wonder why you think it's trivial? The H^1 analogue is enormous. –  Minhyong Kim Oct 17 '10 at 5:30
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Where in real life do you find $H^2(\mathbb C^\times,\mathbb C^\times)$? –  Mariano Suárez-Alvarez Oct 17 '10 at 6:57
    
@Mariano S-A and @Minhyon Kim: There is a result by Deligne and Milne that says, roughly, for a group G there is a unique-up-to-monoidal-isomorphism symmetric monoidal functor Rep(G) \to Vect, at least if G is affine algebraic. I'm reading through the paper that I think this result is in, but thought I'd test on the easiest non-affine-algebraic group, namely Z. Then, I think, the monoidal (I haven't restricted to symmetric) functors Rep(Z) \to Vect are parameterized by H^2(C^\times,C^\times). So I don't really expect this to vanish, but someone told me the result holds for all groups. –  Theo Johnson-Freyd Oct 17 '10 at 17:10
    
The symmetric functors should be elements of H^2 represented by symmetric 2-cocycles. So really I'd like to compute that. But I often realize that I'm not very good at computing cohomologies, hence the question. –  Theo Johnson-Freyd Oct 17 '10 at 17:11
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1 Answer

up vote 15 down vote accepted

This group is best understood in terms of the universal coefficient formula, i.e., in terms of the homology of the involved group. Hence, if $A$ is any abelian group we have $H_1(A)=A$ and the addition map $A\times A\rightarrow A$ induces a Pontryagin product on homology making $H_\ast(A)$ a (graded) commutative algebra. We also have that the square of an element of $H_1(A)$ is zero (in $H_2(A)$). This does not directly follow in the presence of $2$-torsion but it follows by functoriality; given $a\in H_1(A)=A$ we have a group homomorphism $\mathbb Z\rightarrow A$ taking $1$ to $a$ and we are reduced to showing that $1\cdot1=0\in H_2(\mathbb Z)$ but $H_2(\mathbb Z)=0$.

Hence we get an algebra map $\Lambda^\ast A\rightarrow H_\ast(A)$. This is an isomorphism in degrees $1$ and $2$ and an isomorphism in all degrees if $A$ is torsion free. This is proved easily by noting that both sides commutes with filtered direct limits so that one is reduced to the case when $A$ is finitely generated and then using the Künneth formula to reduce to the case when $A$ is cyclic in which case $H_2(A)=0$ and $H_i(A)=0$ for $i\geq2$ if $A=\mathbb Z$.

Now, using the universal coefficient formula and the fact that the coefficient group $\mathbb C^\ast$ is injective we get $H^2(\mathbb C^\ast,\mathbb C^\ast)=\mathrm{Hom}(\Lambda^2\mathbb C^\ast,\mathbb C^\ast)$. Of course as abelian group $\Lambda^2\mathbb C^\ast$ is humongous as is the coefficient group $\mathbb C^\ast$ but they appear naturally in some cases. For instance $H^2(\mathbb C^\ast,\mathbb C^\ast)$ is the group of characteristic classes of degree $2$ for complex line bundles with integrable connection (in the smooth case, otherwise think local systems). Also the related group $\mathbb R\bigotimes S^1$ appears in the solution of Hilbert's third problem. We also have that $K_2(\mathbb C)$ is a quotient of $\Lambda^2\mathbb C^\ast$ and similar but worse groups (such as $S^2\Lambda^2\mathbb C^\ast$) appear in the relation with generalisations of Hilbert's third problem (see for instance J. Dupont: Scissors congruences, group homology and characteristic classes, Nankai Tracts in Mathematics).

I may misrepresent the feelings of a lot of people if I say that algebraic topologists are resigned to the appearance of such large abstract groups whereas algebraic geometers want to believe that they have more structure (such as $K_1(-)$ being the multiplicative group scheme) but haven't really been able to realise this (there are some tantalising results such as Bloch et al's results on the deformation theory of $K_2(-)$).

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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? –  Theo Johnson-Freyd Oct 17 '10 at 18:14
    
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. –  Torsten Ekedahl Oct 17 '10 at 18:51
    
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 $=2$. –  Theo Johnson-Freyd Oct 22 '10 at 4:07
    
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"? –  Theo Johnson-Freyd Oct 22 '10 at 4:24
    
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. –  Torsten Ekedahl Oct 22 '10 at 4:41
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