TITLE: H^d[U(1)^n,U(1)] of the Borel cohomology and Chern-Simons theory
QUESTION [9 upvotes]: 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: https://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. 

[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 $\mathcal{H}^d[G,U(1)]$

REPLY [3 votes]: Let me myself comment briefly the calculation of $H^d[U(1)^n,U(1)]$ from another angle, the Chern-Simons theory.
It is known that the relation between Chern-Simons action $CS(A)$:
\begin{equation}
e^{2\pi i CS(A)}=\exp\Bigl[ 2\pi i k   \int_{X_3} A \wedge dA\Bigr]
\end{equation}
as a partition function of $U(1)$ Chern-Simons action $CS(A)$ on a 3-manifold $X_3$.
For a compact gauge group $G$ the Chern-Simons actions are classified by
an element [Ref:D-W] 
\begin{equation}
k\in H^4(BG,\mathbb{Z})
\end{equation}
The  CS actions $CS(A)$ for 
a compact gauge group $G$ are in one-to-one correspondence
with the elements of the cohomology 
group $H^4 (B{ G}, \mathbb{Z})$ of the classifying space
$B{ G}$ with integer coefficients $\mathbb{Z})$. 
For $G=U(1)$, CS actions $CS(A)$ is classified by
\begin{equation}
H^4(B(U(1)),\mathbb{Z}) \cong \mathbb{Z} 
\end{equation}
, and the element $\mathbb{Z}$ is simply the  integer $k$ (level $k$)
appearing in the expression for the $U(1)$ Chern-Simons action on the 3-manifold $X_3$.
For $K_{n\times n}$ Chern-Simons action $CS(A)$ with $U(1)^n$ gauge group:
\begin{equation}
e^{2\pi i CS(A)}=\exp\Bigl[ 2\pi i K_{ij}   \int_{X_3} A_{i} \wedge dA_{j}\Bigr]
\end{equation}
For $G=U(1)^n$, CS actions $CS(A)$ is classified by
\begin{equation}
H^4(B(U(1)^n),\mathbb{Z}) \cong \mathbb{Z}^{n+\frac{1}{2}n(n-1)}
\end{equation}

Let's shortly sketch the proof of $H^4(B(U(1)^n),\mathbb{Z}) \cong \mathbb{Z}^{n+\frac{1}{2}n(n-1)}
$
\begin{equation}
H^d(BU(1),\mathbb{Z}) \simeq  \begin{cases}
                 \mathbb{Z} & \text{if $d\in$ even} \newline
                 0 & \text{if $d\in$ odd}
                 \end{cases}
\end{equation}
For $H^d(B(U(1)^n),\mathbb{Z})$ with $n>1$, we can use  the Kunneth
formula, because the classifying space of the product group 
$U(1)^n$ is the same as the product of the the factorization of classifying spaces. 
That is, $B(U(1)^n) = B(U(1)^{n-1}) \times BU(1)$.
The calculation is analogue to $H^3[\mathbb{Z}_p^n,U(1)]$ case.
It is easier to show that for discrete abelian group:
\begin{equation}
\begin{cases}
H^3[\mathbb{Z}_p,U(1)]=\mathbb{Z}_p,  \text{(confirmed)}\newline
H^3[\mathbb{Z}_p^2,U(1)]=(\mathbb{Z}_p)^3 \text{(confirmed)} \newline
H^3[\mathbb{Z}_p^3,U(1)]=(\mathbb{Z_p})^7 \text{(confirmed)} \newline
H^3[\mathbb{Z}_p^n,U(1)]=\mathbb{Z_p}^{n+\frac{1}{2}n(n-1)+\frac{1}{3!}n(n-1)(n-2)}  \text{(confirmed)} \newline
\end{cases}
\end{equation}
Since the group $\mathbb{Z}$ is torsion free, however, the terms 
due to torsion products $Tor$ vanish in this case (thanks to the discussion: https://mathoverflow.net/questions/128203/torsion-product-torr-1-closed).
To summarize, the contribution of $H^4(B(U(1)^n),\mathbb{Z})$, comes from 
$n$ terms of the form $H^4(BU(1))\simeq \mathbb{Z}$. These
label the different CS actions of diagonal $K_{n\times n}$ matrix type:
\begin{equation}
\end{equation}
\begin{equation}
{\sum_{i=1}^n {K_{(ii)}} \epsilon^{\kappa\sigma\rho} {A_\kappa^{(i)}} \partial_{\sigma} A^{(i)}_{\rho} }
\end{equation}
In addition, there are $\frac{1}{2} n(n-1)$ terms 
of the form $H^2(BU(1)) \simeq \mathbb{Z}$ which label
the CS actions of off-diagonal $K_{n\times n}$ matrix type: 
\begin{equation}
{\sum_{i\neq j}^n {K_{(ij)}} \epsilon^{\kappa\sigma\rho} {A_\kappa^{(i)}} \partial_{\sigma} A^{(j)}_{\rho} }
\end{equation}
So overall, $H^4(B(U(1)^n),\mathbb{Z}) \cong \mathbb{Z}^{n+\frac{1}{2}n(n-1)}
$ (q.e.d.)

Importantly, this classification includes the case of finite gauge groups $H$. 
The isomorphism for a finite gauge groups $H$.
\begin{equation}                           
H^d(B{H},{\mathbb{Z}}) \simeq H^d({H},{\mathbb{Z}})  ,
\end{equation}
Moreover, the universal coefficient theorem(UCThm) shows the isomorphism 
\begin{equation}
H^{d} (H, \mathbb{Z})   \simeq   H^{d-1} ({ H}, U(1))    
\qquad \forall  n>1.
\end{equation}
We have then:
\begin{equation}
H^4 ({BH}, \mathbb{Z})  \simeq  H^3 ({ H}, U(1)) 
\end{equation}

It is this relation: $\begin{equation}
H^4 ({BH}, \mathbb{Z})  \simeq  H^3 ({ H}, U(1)) 
\end{equation}$
allure me from this finite gauge groups $H$ result,
\begin{equation}
\begin{cases}
H^3[\mathbb{Z}_p,U(1)]=\mathbb{Z}_p,  \text{(confirmed)}\newline
H^3[\mathbb{Z}_p^2,U(1)]=(\mathbb{Z}_p)^3 \text{(confirmed)} \newline
H^3[\mathbb{Z}_p^3,U(1)]=(\mathbb{Z_p})^7 \text{(confirmed)} \newline
H^3[\mathbb{Z}_p^n,U(1)]=\mathbb{Z_p}^{n+\frac{1}{2}n(n-1)+\frac{1}{3!}n(n-1)(n-2)}  \text{(confirmed)} \newline
\end{cases}
\end{equation}
to prompt the guessed result of:
\begin{equation}
\begin{cases}
H^3[U(1),U(1)]=\mathbb{Z},  \newline
H^3[U(1)^2,U(1)]=(\mathbb{Z})^3 \text{(to be checked)} \newline
H^3[U(1)^3,U(1)]=(\mathbb{Z})^7 \text{(to be checked)} \newline
H^3[U(1)^n,U(1)]=\mathbb{Z}^{n+\frac{1}{2}n(n-1)+\frac{1}{3!}n(n-1)(n-2)}  \text{(to be checked)} \newline
\end{cases}
\end{equation}
However, the result
$
H^4(B(U(1)^n),\mathbb{Z}) \cong \mathbb{Z}^{n+\frac{1}{2}n(n-1)}
$
already tells me the right way to view this $U(1)^n$ gauge symmetry C-S theory should be classified by $H^4(B(U(1)^n),\mathbb{Z})$
as
\begin{equation}
\begin{cases}
H^4(B(U(1)),\mathbb{Z})=\mathbb{Z},  \newline
H^4(B(U(1)^2),\mathbb{Z}) \cong \mathbb{Z}^{3}  \newline
H^4(B(U(1)^3),\mathbb{Z}) \cong \mathbb{Z}^{6} \newline
H^4(B(U(1)^n),\mathbb{Z}) \cong \mathbb{Z}^{n+\frac{1}{2}n(n-1)} \newline
\end{cases}
\end{equation}
This is what I should ask in (Q1).
So, instead, what I should really ask is, apart from (Q1)(Q2):
(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}
We know 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}
This does not work, where we cannot simply replacing  $\mathbb{Z}$ to $\mathbb{Z}_p$ by symmetry breaking from $U(1)$ to $\mathbb{Z}_p$. It has been known that one may need a symmetry breaking of C-S action with a nonAbelian continuous group broken down to subgroup $Z_p^n$ to produce the all known elements of $H^3[\mathbb{Z}_p^n,U(1)]$. 
So that goes back to my guessed proposal:
\begin{equation}
\begin{cases}
H^3[U(1),U(1)]=\mathbb{Z},  \newline
H^3[U(1)^2,U(1)]=(\mathbb{Z})^3 \text{(to be checked)} \newline
H^3[U(1)^3,U(1)]=(\mathbb{Z})^7 \text{(to be checked)} \newline
H^3[U(1)^n,U(1)]=\mathbb{Z}^{n+\frac{1}{2}n(n-1)+\frac{1}{3!}n(n-1)(n-2)}  \text{(to be checked)} \newline
\end{cases}
\end{equation}
PS. I have to apologize what I had mentioned may be intriguing, the question turns out to overlap different math fields. Instead of directly answering the questions (Q1)(Q2), I now address the questions differently.

Ref:
[Ref:D-W]:Robbert Dijkgraaf, Edward Witten, Topological Gauge Theories and Group Cohomology, Commun. Math. Phys. 129 (1990), 393