Firstly I apologize that I am a physicist, with a relatively unrigorous math training. My approach of the problem can be Feynman style. Below $Z$ is the integer $\mathbb{Z}$, and $U(1)$ Abelian group is the same as $\mathbb{R}/\mathbb{Z}$. gcd stands for greatest common divisor. And $Z_1$ is the same as the group $0$.
The question is about $H^d[U(1)^n,U(1)]$ of the Borel cohomology. My questions have two parts (Q1) and (Q2).
My first question (Q1) is whether my guess below is correct:
$
\begin{cases}
H^3[U(1),U(1)]=Z, \newline
H^3[U(1)\times U(1),U(1)]=(Z)^3 \text{(to be checked)} \newline
H^3[U(1)\times U(1)\times U(1),U(1)]=(Z)^7 \text{(to be checked)} \newline
\end{cases}
$
My second question (Q2) is that the above result seems to be inconsistent with the `universal coefficient theorem' and some facts about $H^d[U(1)^n,Z]$ and $H^d[U(1)^n,U(1)]$.
(Q1)
To calculate $H^d[U(1),U(1)]$ directly from the algebraic definition is very tricky to me since $U(1)$ has infinite uncountable many
elements.
Here, I will use a unrigorous physical argument (if you go against it, fine, no problem, but I will say Richard Feynman may do this) to calculate it by first calculating
$H^d[ \Pi_i Z_{n_i},U(1)]$, and then let $n_i\to \infty$. Below I will use the unproven and unrigorous $\lim_{n\to \infty}Z_n=Z$ (which is not true in general).
I start with the known fact:
$
\begin{cases}
H^3(Z_n,U(1))=Z_n \newline
H^3(Z_n\times Z_m,U(1))=Z_n \times Z_m \times Z_{gcd(n,m)} \newline
H^3(Z_n \times Z_m\times Z_o,U(1))=Z_n \times Z_m \times Z_o \times Z_{gcd(n,m)}\times Z_{gcd(n,o)} \times Z_{gcd(m,o)} \times Z_{gcd(n,m,o)}
\end{cases}
$
What I had obtained is:
$
\begin{cases}
H^3[U(1),U(1)]=Z, \newline
H^3[U(1)\times U(1),U(1)]=Z\times Z\times Z=(Z)^3 \text{(to be checked)} \newline
H^3[U(1)\times U(1)\times U(1),U(1)]=Z\times Z \times Z \times Z \times Z \times Z \times Z=(Z)^7 \text{(to be checked)} \newline
\end{cases}
$
My first question is whether my result is correct (Not the method).
(Q2) My second question is that the above result seems to be inconsistent with the
(a)`universal coefficient theorem'
$
\begin{align}
\ \ \ \ H^d(X,M)
\\
\simeq H^d(X,Z)\otimes_{Z} M \oplus
\text{Tor}_1^{Z}(H^{d+1}(X,{Z}),M) ,
\end{align}$
and
(b)the following known facts:
$H^d[U(1),U(1)]=
\begin{cases}
U(1) & \text{ if } d=0, \newline
Z_1 & \text{ if } d=0 \text{ mod } 2,\ \ d>0\newline
Z & \text{ if } d=1 \text{ mod } 2.
\end{cases}$
$H^d[U(1),Z]=
\begin{cases}
Z & \text{ if } d=0 \text{ mod } 2,\newline
Z_1 & \text{ if } d=1 \text{ mod } 2.
\end{cases}
$
This shows that $H^d[U(1),U(1)]=H^{d+1}[U(1),Z]$.
The inconsistency is
\begin{eqnarray}
H^3(U(1),U(1)) &=&[H^3(U(1),Z) \otimes U(1)] \times \text{Tor}^Z_1[H^4(U(1),Z),U(1)] \newline
&=&[Z_1\otimes U(1)] \times \text{Tor}^Z_1[Z,U(1)]=Z_1 \times Z_1 =Z_1
\end{eqnarray}
so it contradicts to $H^3(U(1),U(1))=Z$.
Surprisingly, the inconsistency already happens to the known fact H3[U(1),U(1)]=Z !
The same for $H^3(U(1)\times U(1),U(1))=Z_1$ instead of $(Z)^3$.
The question is why $H^3(U(1),U(1))=Z$ is inconsistent here from the known facts (a)(b). Also the inconsistency at other $H^3(U(1)^n,U(1))$ with $n=2,3,\dots$,etc.
PS. The fact that $\text{Tor}^Z_1[Z,U(1)]=Z_1$ seems to tell me something nontrivial contrary to the naive $\text{Tor}^Z_1[Z_n,U(1)]=Z_n$ at $n \to \infty$. Suppose that $\text{Tor}^Z_1[Z,U(1)] \to Z$ instead, everything seems to match. This (Q2) was the main reason why I had asked this silly question: http://mathoverflow.net/questions/128203/torsion-product-torr-1-closed
[new update on April 27, 2013]
I provide some connections between group cohomology and Chern-Simons theory in the 2nd answer below. And add a question:
(Q3)
whether there is a symmetry breaking picture, such that one can obtain the result of
$H^3[\mathbb{Z}_p^n,U(1)]$ of a discrete $\mathbb{Z}_p^n$ group from a large continuous group, say, from $U(1)^n$ broken down to $\mathbb{Z}_p^n$?
So that, for example, this guessed $H^3[U(1)^n,U(1)]$ broken down to a subgroup picture works
\begin{equation}
H^3[U(1)^n,U(1)] (\text{guessed})\to H^3[\mathbb{Z}_p^n,U(1)]
\end{equation}
as
\begin{equation}
\mathbb{Z}^{n+\frac{1}{2}n(n-1)+\frac{1}{3!}n(n-1)(n-2)} (\text{guessed})\to \mathbb{Z}_p^{n+\frac{1}{2}n(n-1)+\frac{1}{3!}n(n-1)(n-2)}
\end{equation}
On the other hand, we know the fact that however
\begin{equation}
H^4(B(U(1)^n),\mathbb{Z}) \to H^3[\mathbb{Z}_p^n,U(1)]
\end{equation}
broken down from
\begin{equation}
\mathbb{Z}^{n+\frac{1}{2}n(n-1)} \to \mathbb{Z}_p^{n+\frac{1}{2}n(n-1)+\frac{1}{3!}n(n-1)(n-2)}
\end{equation}
does not match when the cohomology group breaks down from $\mathbb{Z}^{n+\frac{1}{2}n(n-1)} \to \mathbb{Z}_p^{n+\frac{1}{2}n(n-1)}$ by the gauge group of $U(1)^n$ Chern-Simons theory breaks down to $\mathbb{Z}^n_p$ Chern-Simons theory.
