Where can I easily look up / calculate (abelian) group cohomology?

Question

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?

Answer 1

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 $K_2(-)$).