# H^d[U(1)^n,U(1)] of the Borel cohomology and Chern-Simons theory

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: $\begin{cases} H^3(Z_n,U(1))=Z_n \newline H^3(Z_n\times Z_m,U(1))=Z_n \times Z_m \times Z_{gcd(n,m)} \newline H^3(Z_n \times Z_m\times Z_o,U(1))=Z_n \times Z_m \times Z_o \times Z_{gcd(n,m)}\times Z_{gcd(n,o)} \times Z_{gcd(m,o)} \times Z_{gcd(n,m,o)} \end{cases}$

$\begin{cases} H^3[U(1),U(1)]=Z, \newline H^3[U(1)\times U(1),U(1)]=Z\times Z\times Z=(Z)^3 \text{(to be checked)} \newline H^3[U(1)\times U(1)\times U(1),U(1)]=Z\times Z \times Z \times Z \times Z \times Z \times Z=(Z)^7 \text{(to be checked)} \newline \end{cases}$

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 $$H^3[U(1)^n,U(1)] (\text{guessed})\to H^3[\mathbb{Z}_p^n,U(1)]$$ as

$$\mathbb{Z}^{n+\frac{1}{2}n(n-1)+\frac{1}{3!}n(n-1)(n-2)} (\text{guessed})\to \mathbb{Z}_p^{n+\frac{1}{2}n(n-1)+\frac{1}{3!}n(n-1)(n-2)}$$

On the other hand, we know the fact that however $$H^4(B(U(1)^n),\mathbb{Z}) \to H^3[\mathbb{Z}_p^n,U(1)]$$

broken down from $$\mathbb{Z}^{n+\frac{1}{2}n(n-1)} \to \mathbb{Z}_p^{n+\frac{1}{2}n(n-1)+\frac{1}{3!}n(n-1)(n-2)}$$

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: $$\mathcal{H}^d_{SM}(G; Z)=H^d(BG; Z),\;\; \mathcal{H}^d_{SM}(G; U(1))=H^{d+1}(BG; Z)$$

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

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So you are making guesses that you say you can prove are wrong (by invoking the universal coefficient theorem) and then asking whether your guesses are right? – Steven Landsburg Apr 22 '13 at 1:19
Thank you Steven for the comment. The inconsistency also happens to the known fact $H^3[U(1),U(1)]=Z$. So what I was driving to is "whether there is some extra constraint on universal coefficient theorem" applying to continuous group $U(1)$? Or something else was wrong. Although my (Q1) is just a guessed answer, I somehow feel it still makes some sense. The understanding of my guess is based on K matrix multiplet Chern-Simons theory K_{ij} a_i \wedge d a_j, where the classification should lead to the same answer as (Q1). For people who are interested in PHYSICS motivation for my problem(contd) – Idear Apr 22 '13 at 2:04
For people who are interested in PHYSICS motivation for my problem, please take a look on: "Symmetry protected topological orders and the group cohomology of their symmetry group" arxiv.org/abs/1106.4772v6 or the Science paper: Science 338, 1604 (2012) sciencemag.org/content/338/6114/1604.abstract The formulas I used are listed in 1106.4772v6 paper. Again, my intuition for my guessed answer of (Q1) is from: Chern-Simons theory. For example, $N \times N$ K matrix multiplet Chern-Simons theory $K_{ij} a_i \wedge d a_j$ with $U(1)^N$ gauge group. – Idear Apr 22 '13 at 2:10
Can you say more precisely which cohomology you consider? You say "Moore cohomology". Do you mean Borel-Moore cohomology, which is ordinary cohomology for compact spaces as we have here? Or do you mean the equivariant Moore cohomology, in which case you have to specify an action of a group on your spaces? Or do you mean a version of group cohomology? If so, which one (algebraic group cohomology, continuous group cohomology, smooth group cohomology)? – Konrad Waldorf Apr 23 '13 at 12:10
@Idear: why do you keep capitalizing PHYSICS? – YangMills Apr 27 '13 at 23:47

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 $CS(A)$:

$$e^{2\pi i CS(A)}=\exp\Bigl[ 2\pi i k \int_{X_3} A \wedge dA\Bigr]$$

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]

$$k\in H^4(BG,\mathbb{Z})$$

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

$$H^4(B(U(1)),\mathbb{Z}) \cong \mathbb{Z}$$

, 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:

$$e^{2\pi i CS(A)}=\exp\Bigl[ 2\pi i K_{ij} \int_{X_3} A_{i} \wedge dA_{j}\Bigr]$$

For $G=U(1)^n$, CS actions $CS(A)$ is classified by

$$H^4(B(U(1)^n),\mathbb{Z}) \cong \mathbb{Z}^{n+\frac{1}{2}n(n-1)}$$

Let's shortly sketch the proof of $H^4(B(U(1)^n),\mathbb{Z}) \cong \mathbb{Z}^{n+\frac{1}{2}n(n-1)}$ $$H^d(BU(1),\mathbb{Z}) \simeq \begin{cases} \mathbb{Z} & \text{if d\in even} \newline 0 & \text{if d\in odd} \end{cases}$$

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:

$$\begin{cases} H^3[\mathbb{Z}_p,U(1)]=\mathbb{Z}_p, \text{(confirmed)}\newline H^3[\mathbb{Z}_p^2,U(1)]=(\mathbb{Z}_p)^3 \text{(confirmed)} \newline H^3[\mathbb{Z}_p^3,U(1)]=(\mathbb{Z_p})^7 \text{(confirmed)} \newline H^3[\mathbb{Z}_p^n,U(1)]=\mathbb{Z_p}^{n+\frac{1}{2}n(n-1)+\frac{1}{3!}n(n-1)(n-2)} \text{(confirmed)} \newline \end{cases}$$

Since the group $\mathbb{Z}$ is torsion free, however, the terms due to torsion products $Tor$ vanish in this case (thanks to the discussion: Torsion product Tor^R_1(,)).

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:



$${\sum_{i=1}^n {K_{(ii)}} \epsilon^{\kappa\sigma\rho} {A_\kappa^{(i)}} \partial_{\sigma} A^{(i)}_{\rho} }$$

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: $${\sum_{i\neq j}^n {K_{(ij)}} \epsilon^{\kappa\sigma\rho} {A_\kappa^{(i)}} \partial_{\sigma} A^{(j)}_{\rho} }$$

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 $H$. $$H^d(B{H},{\mathbb{Z}}) \simeq H^d({H},{\mathbb{Z}}) ,$$

Moreover, the universal coefficient theorem(UCThm) shows the isomorphism $$H^{d} (H, \mathbb{Z}) \simeq H^{d-1} ({ H}, U(1)) \qquad \forall n>1.$$

We have then: $$H^4 ({BH}, \mathbb{Z}) \simeq H^3 ({ H}, U(1))$$

It is this relation: $$$H^4 ({BH}, \mathbb{Z}) \simeq H^3 ({ H}, U(1))$$$

allure me from this finite gauge groups $H$ result, $$\begin{cases} H^3[\mathbb{Z}_p,U(1)]=\mathbb{Z}_p, \text{(confirmed)}\newline H^3[\mathbb{Z}_p^2,U(1)]=(\mathbb{Z}_p)^3 \text{(confirmed)} \newline H^3[\mathbb{Z}_p^3,U(1)]=(\mathbb{Z_p})^7 \text{(confirmed)} \newline H^3[\mathbb{Z}_p^n,U(1)]=\mathbb{Z_p}^{n+\frac{1}{2}n(n-1)+\frac{1}{3!}n(n-1)(n-2)} \text{(confirmed)} \newline \end{cases}$$

to prompt the guessed result of: $$\begin{cases} H^3[U(1),U(1)]=\mathbb{Z}, \newline H^3[U(1)^2,U(1)]=(\mathbb{Z})^3 \text{(to be checked)} \newline H^3[U(1)^3,U(1)]=(\mathbb{Z})^7 \text{(to be checked)} \newline H^3[U(1)^n,U(1)]=\mathbb{Z}^{n+\frac{1}{2}n(n-1)+\frac{1}{3!}n(n-1)(n-2)} \text{(to be checked)} \newline \end{cases}$$

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 $$\begin{cases} H^4(B(U(1)),\mathbb{Z})=\mathbb{Z}, \newline H^4(B(U(1)^2),\mathbb{Z}) \cong \mathbb{Z}^{3} \newline H^4(B(U(1)^3),\mathbb{Z}) \cong \mathbb{Z}^{6} \newline H^4(B(U(1)^n),\mathbb{Z}) \cong \mathbb{Z}^{n+\frac{1}{2}n(n-1)} \newline \end{cases}$$ This is what I should ask in (Q1).

(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 $$H^3[U(1)^n,U(1)] (\text{guessed})\to H^3[\mathbb{Z}_p^n,U(1)]$$ as

$$\mathbb{Z}^{n+\frac{1}{2}n(n-1)+\frac{1}{3!}n(n-1)(n-2)} (\text{guessed})\to \mathbb{Z}_p^{n+\frac{1}{2}n(n-1)+\frac{1}{3!}n(n-1)(n-2)}$$

We know that however $$H^4(B(U(1)^n),\mathbb{Z}) \to H^3[\mathbb{Z}_p^n,U(1)]$$

broken down from $$\mathbb{Z}^{n+\frac{1}{2}n(n-1)} \to \mathbb{Z}_p^{n+\frac{1}{2}n(n-1)+\frac{1}{3!}n(n-1)(n-2)}$$

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: $$\begin{cases} H^3[U(1),U(1)]=\mathbb{Z}, \newline H^3[U(1)^2,U(1)]=(\mathbb{Z})^3 \text{(to be checked)} \newline H^3[U(1)^3,U(1)]=(\mathbb{Z})^7 \text{(to be checked)} \newline H^3[U(1)^n,U(1)]=\mathbb{Z}^{n+\frac{1}{2}n(n-1)+\frac{1}{3!}n(n-1)(n-2)} \text{(to be checked)} \newline \end{cases}$$

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

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If what you said is: $U(1) \cong \mathbb R \setminus \mathbb Z \cong \S^{1}$ Why not try to calculate the $H^3[U(1)\times U(1),U(1)] \cong H^d[S^{1} \times S^{1},S^{1}]$ in that way as a product? In that case, you can simply use the cellular cohomology where the cohomology groups are directly obtained via applying the $Hom( ,Z)$ onto the corresponding homology chain complex $H_{k}[U(1)\times U(1),U(1)]$.

Step1: Observe or calculate the $H_{k}[U(1)\times U(1),U(1)]$ via Device Lemma

Lemma 2.34 in A.Hatcher's Algebraic Topology If X is a CW complex, then: (a) $H_{k}(Xn,Xn−1)$ is zero for k!=n and is free abelian for k=n, with a basis in one-to-one correspondence with the n-cells of X. (b) $H_{k}(X^{n})=0$ for k > n. In particular, if X is finite-dimensional then $H_{k}(X)=0$ for k > $dimX$. (c) The inclusion i :Xn->X induces an isomorphism I: $H_{k}(X^{n})->H_{k}(X)$ if k < n.

Step2: Calculate the cohomology via applying the $Hom( ,Z)$

Ans1:I do not think the results are correct for Q1.

I don't know if there's any restriction by doing this through cellular cohomology in PHYSIC background.

For further ref. you can search by the keyword"Eilenberg-MacLane space" to deal with the infinite dimensional case.

Again,I strongly suggest that you can just switch the subscripts to suit the cohomology... You're just using the UCThm for homology in the notations for UCThm of cohomology, which makes it confusing...Moreover, I'm not sure whether you're talking about external product.

Ans2:So it's meaningless to talk about 'inconsistency' unless you correctly present the Q2 arguments.

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Thank you. Let me please provide one more PHYSICS motivation for the problem, it concerns: Dijkgraaf-Witten theory: Robbert Dijkgraaf, Edward Witten, Topological Gauge Theories and Group Cohomology, Commun. Math. Phys. 129 (1990), 393 See also: ncatlab.org/nlab/show/Dijkgraaf-Witten+theory Hopefully it drives more attention to this seemly simple calculation. $\begin{cases} H^3[U(1),U(1)]=Z,\newline H^3[U(1)\times U(1),U(1)]=?(\text{my guess}(Z)^3) \newline H^3[U(1)\times U(1)\times U(1),U(1)]=?(\text{my guess}(Z)^7) \end{cases}$ I like to compare it to Chern-Simons theory. – Idear Apr 22 '13 at 23:44
Update as an answer. – Henry.L Apr 26 '13 at 9:34
Thank you Henr, let me think a bit more. – Idear Apr 27 '13 at 16:18