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](http://katlas.org/wiki/Main_Page) or the [OEIS](http://www.research.att.com/~njas/sequences/). Does such a thing exist?



Where can I easily look up / calculate (abelian) group cohomology?
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 $K\_2(-)$).


