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: Torsion product Tor^R_1(,)

[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.

[new update on May 12, 2013]

I would like to address Mariano's inquiry below: The Cohomology I mean here is Borel Group Cohomology. See for example, http://arxiv.org/pdf/1106.4772v6.pdf, p.26, Appendix D: Group cohomology and p.44, Appendix J: Calculations of group cohomology, Sec 4. Some useful tools in group cohomology.

See also: http://arxiv.org/pdf/1110.3304.pdf. For Segal and Mitchison's $\mathcal{H}^d_{SM}(G; Z)$ These two Ref show: \begin{equation} \mathcal{H}^d_{SM}(G; Z)=H^d(BG; Z),\;\; \mathcal{H}^d_{SM}(G; U(1))=H^{d+1}(BG; Z) \end{equation}

Also this question: The relationship between group cohomology and topological cohomology theories

If it helps, I can change all my $H^d[G,U(1)]$ above to $\mathcal{H}^d[G,U(1)]$