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Question: Given a generic finite abelian group $G=\mathbb{Z}_{N^{(1)}} \times \cdots \times \mathbb{Z}_{N^{(k)}}$.

(1) What is the explicit forms of its cohomology group (see my definition) in a generic $n$: $$ H^n(G,R/\mathbb{Z})=H^n(G,U(1)) =? $$

(2) what are their explicit $n$-cocycles? $$ \omega_{}^{}(A_1,A_2,\dots, A_n): (G)^n \to U(1) $$

Please be very explicit in $n=4$ for priority (and the secondary are $n=5$ and $n=2$) of interests. But the generic formulas for any $n$ are the best. Thank you. References are very welcome.

ps. below I provide the answer I know for $n=3$. Please feel free providing partial answers for other $n$.

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    $\begingroup$ Using the Long exact sequence for the short exact sequence $O\to Z\to R\to U(1)\to 0$ and the fact that G is finite and R divisible, youncan reduce to computing cohomology with values in Z. Then you can use the Künneth formula, you get reduced to computing the cohomology of cyclic groups, and that is explained in every textbook. If you want explicit cocycles, that's more work, but the Künneth formula is actually explicit, and gives you explicit cocycles built out of cup products. $\endgroup$ – Mariano Suárez-Álvarez Jan 17 '14 at 23:21
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    $\begingroup$ Thanks Mariano, I know how to compute cohomology group. But I wonder whether there are explicit formulas for them (i.e. people already computed in the literature? such as my answer below)? n-cocycles need to solve cocycle conditions, so are there known results in any Ref? $\endgroup$ – wonderich Jan 17 '14 at 23:26
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    $\begingroup$ The computation is of course known and, by now, part of folklore: it is probably easier to carry out the sketch in my comment above than go looking for references. (I can't make a lot of sense of your answer, by the way) Moreover, everything I wrote above is as explicit as things come. $\endgroup$ – Mariano Suárez-Álvarez Jan 17 '14 at 23:28
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    $\begingroup$ For example, if $k=1$, the formula you wrote for $H^3$ gives zero. Yet $H^3(G,\mathbb R/\mathbb Z)$ is isomorphic to $H^4(G,\mathbb Z)$, which is not zero because G is cyclic. Apart from that, most of the notation in your description of cocycles is not explained at all, and what is explained I don't understand (what do you mean by «labels», the brackets plainly do more than group terms, A, B and C only appear on the left hand sides of your equations , &c), so I cannot really tell if the formula is correct or not, because it does not mean anything to me! $\endgroup$ – Mariano Suárez-Álvarez Jan 20 '14 at 0:35
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    $\begingroup$ those are Dijkgraff-Witten theory and 2+1D Chern-Simons theory; and TQFT, topological orders to me. $p$ "labels" $Z_N$ means $p \in Z_N$. $\endgroup$ – wonderich Jan 20 '14 at 1:23
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3 years later.

My recent paper (https://arxiv.org/abs/1703.03266) answers this question.

More precisely, let $\mathbb{k}$ be an algebraically closed field of characteristic zero. By $\mathbb{k}^*$ we denote the multiplicative group $\mathbb{k}-\{0\}$. Let $G=\mathbb{Z}_{m_{1}}\times\cdots \times\mathbb{Z}_{m_{n}}$ where $m_i|m_{i+1}$ for $1\le i\le n-1$ and $(B_{\bullet},\partial_{\bullet})$ be its normalized bar resolution. Applying $\text{Hom}_{\mathbb{Z}G}(-,\mathbb{k}^*)$ one gets a complex $(B^*_{\bullet},\partial^*_{\bullet})$.

Let $\alpha:=(\alpha_{11},\dots,\alpha_{1n},\dots,\alpha_{k1},\dots,\alpha_{kn})$ where $0\leq\alpha_{ij}<m_{j}$ for $1\leq i\leq k$. For $1\leq r\leq n$, $[a,b]\subseteq[1,k]$, $a,b\in\mathbb{N}$ and $\alpha$, denote

$$ \eta_{r,[a,b]}^{\alpha}:=\left\{ \begin{array}{ll}[\frac{\alpha_{br}+\alpha_{b-1,r}}{m_{r}}]\cdots[\frac{\alpha_{a+1,r}+\alpha_{ar}}{m_{r}}]&\;\;\;a-b\text{ odd}\\ [\frac{\alpha_{br}+\alpha_{b-1,r}}{m_{r}}]\cdots[\frac{\alpha_{a+2,r}+\alpha_{a+1,r}}{m_{r}}]\alpha_{ar}&\;\;\;a-b\text{ even} \end{array}\right. $$ The following $\omega\in \text{Hom}_{\mathbb{Z}G}(B_k,\mathbb{k}^*)$ \begin{eqnarray} &&\omega([g_1^{\alpha_{11}}\cdots g_n^{\alpha_{1n}},\dots,g_1^{\alpha_{k1}}\cdots g_n^{\alpha_{kn}}])\\\notag &=&\prod_{l=1}^{k}\prod_{\begin{array}{ccc}1\leq r_{1}<\cdots<r_{l}\leq n\\\lambda_1+\cdots+\lambda_l=k,\lambda_1\text{ odd}\\\lambda_i\ge1\text{ for }1\le i\le l\end{array}}\zeta_{m_{r_1}}^{(-1)^{\sum_{1\leq i<j\leq l}\lambda_{i}\lambda_{j}}\eta_{r_{1},[a_{1},b_{1}]}^{\alpha}\cdots\eta_{r_{l},[a_{l},b_{l}]}^{\alpha}a_{r_1^{\lambda_1}\cdots r_l^{\lambda_l}}} \end{eqnarray} where $a_{l}=1,b_{l}=\lambda_{l},\dots,a_{1}=\lambda_{2}+\cdots+\lambda_{l}+1,b_{1}=\lambda_{1}+\cdots+\lambda_{l}=k$ and $0\leq a_{r_1^{\lambda_1}\cdots r_l^{\lambda_l}}<m_{r_1}$ for $1\leq r_1<\cdots<r_l\leq n$ makes a complete set of representatives of $k$-cocycles of the complex $(B_{\bullet}^*,\partial_{\bullet}^*)$. Moreover, $$\text{H}^k(G,\mathbb{k}^*)=\prod_{r=1}^n\mathbb{Z}_{m_r}^{\sum_{j=1}^k(-1)^{k+j}\binom{n-r+j-1}{j-1}}.$$

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  • $\begingroup$ Nice answer and congrats on your paper! $\endgroup$ – Linus Rastegar Nov 15 '17 at 3:04
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    $\begingroup$ Very excellent paper. See page 10-12 of this paper: arxiv.org/abs/1405.7689 (in Phys Rev Lett). They derive essential the similar formulas from quantum fields. And also arxiv.org/abs/1708.04264. Both results are consistent and deal with the similar issues you have. $\endgroup$ – annie heart Nov 15 '17 at 16:59
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An algorithmic way to describe the standard n-cocycle (cocycles respect to the bar resolution) for abelian groups is given in Lyndon's paper The cohomology theory of group extensions (It is no more than the LHS spectral sequences in a very particular case).As Mariano said, first you need to describe the n-cocycles for cyclic groups (constructing a chain homomorphism between the cyclic resolution and the Bar resolution). After that you only need to follow the ideas of Lyndon's paper in order to construct explicitly the standard n-cocycles.

By the way, in Lyndon's paper also there is an explicit description of $H^{n}(A,U(1))\cong H^{n+1}(A,\mathbb{Z})$ for $A$ a finite abelian group.

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(Not a complete answer.) Let me say what I have tried in $n=3$ (this is only a partial answer I already know).

Given a generic finite abelian group $G=\mathbb{Z}_{N^{(1)}} \times \cdots \times \mathbb{Z}_{N^{(k)}}$. I compute that 3rd-cohomology group is $$ H^3(G,R/\mathbb{Z}) \simeq \! \bigoplus_{1 \leq i < j < l \leq k} \! Z_{N^{(i)}} \oplus Z_{{\gcd} (N^{(i)},N^{(j)})} \oplus Z_{{\gcd} (N^{(i)},N^{(j)},N^{(l)})} \, . $$ ($R/\mathbb{Z}=U(1)$)

I also compute the corresponding 3-cocycles are: $$ \omega_{{I}}^{(i)}(A,B,C) = \exp \left( \frac{2 \pi i p^{(i)}_{{I}}}{N^{(i)\;2}} \; a^{(i)}(b^{(i)} +c^{(i)} -[b^{(i)}+c^{(i)}]) \right) \\ \omega_{{II}}^{(ij)}(A,B,C) = \exp \left( \frac{2 \pi i p_{{II}}^{(ij)}}{N^{(i)}N^{(j)}} \; a^{(i)}(b^{(j)} +c^{(j)} - [b^{(j)}+c^{(j)}]) \right) \\ \omega_{{III}}^{(ijl)} (A,B,C) = \exp \left( \frac{2 \pi i p_{{III}}^{(ijl)}}{{\gcd}(N^{(i)}, N^{(j)},N^{(l)})} \; a^{(i)}b^{(j)}c^{(l)} \right) $$ with $p^{(i)}_{{I}}$ labels the group element $Z_{N^{(i)}}$, $p_{{II}}^{(ij)}$ labels the group element in $Z_{{\gcd} (N^{(i)},N^{(j)})}$ and $p_{{III}}^{(ijl)}$ labels the group element in $Z_{{\gcd} (N^{(i)},N^{(j)},N^{(l)})}$.

I am still interested in knowing other $n$, especially $n=2,4,5$. Please reply in the specific forms in $n=3$ as I did for $H^3(G,R/\mathbb{Z})$ and $\omega$. Thank you.

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    $\begingroup$ ps. this won't be the final answer. Just a partial result on $n=3$. Please help on generic $n$ or $n=4$. Many thanks. $\endgroup$ – wonderich Jan 18 '14 at 6:19
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    $\begingroup$ Here $A=(a^{(1)},a^{(2)},\dots,a^{(k)})$, and so does $B,C$ define the same way. Here $[b^{(i)}+c^{(i)}] \equiv (b^{(i)}+c^{(i)}) \text{mod} Z_N^{(i)}$. $\endgroup$ – wonderich Jan 20 '14 at 2:32

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