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We know that group cohomology $H^2(G,U(1))$ consists of 2-cocycles $\beta(A,B)\in U(1)$ corresponding to elements in the group $H^2(G,U(1))$, where $A\in G,B \in G$. Note that $\beta(A,B)$ satisfies 2-cocycles conditions: $\frac{\beta(A,B)\beta(AB,C)}{\beta(A,BC)\beta(B,C)}=1$, with $A,B,C\in G$.

For example,





(5)$H^2(Z_n \times Z_m,U(1))=Z_{gcd(n,m)}$.

We know (1)$H^2(Z_2,U(1))$ has a 2-cocycle $\beta(A,B)=1$ (up to a 2-coboundary term), this corresponds to the unique element in $Z_1$.

Questions: What are the explicit forms of 2-cocycles $\beta(A,B)$ for the cases of (2)$H^2(Z_2^2,U(1))$,(3)$H^2(Z_2^3,U(1))$?

The answer should look like: For (2), $\beta(A,B)=\beta_1^{n_1}$ with $n_1\in \{ 0,1\}=Z_2$, with $\beta_1$ as a generator of 2-cocycles. For (3), $\beta(A,B)=\beta_1^{n_1}\beta_2^{n_2}\beta_3^{n_3}$ with $n_1,n_2,n_3\in \{ 0,1\}=Z_2$, with $\beta_1,\beta_2,\beta_3$ as generators of 2-cocycles.

Similarly, any answer for explicit 2-cocycles for (4)$H^2(Z_n^k,U(1))$ and (5)$H^2(Z_n \times Z_m,U(1))$?

Any comments, concise/short reference or better understanding will be helpful. I am a physicist, on a modest level trying to absorb this Thank you so much.

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There is a rather nice correspondence between the cohomology of $(\mathbb{Z}/2\mathbb{Z})^n$ (as you described in entries 1-4 of your list) and certain upper-triangular $n \times n$ matrices with 1s on the diagonal. You may want to try to encode these matrices as cocycles. – S. Carnahan Apr 16 '13 at 2:22
up vote 3 down vote accepted

It turns out that by playing around the $U(1)$ form of 2-cocylces, I manage to provide some answers to (2) and (3) and partially (4).

For (2)$H^2(Z_2^2,U(1))=Z_2$, the 2-cocycles are $\beta(b,c)=\beta_1^{n_1}=\exp({i\pi}n_1(b_1 c_2))$, with $b=(b_1,b_2)\in Z_2^2$, $c=(c_1,c_2)\in Z_2^2$. Here $b_1,b_2,c_1,c_2,n_1\in\{0,1\}=Z_2$.

More generally, $H^2(Z_n^2,U(1))=Z_n$, the 2-cocycles are $\beta(b,c)=\beta_1^{n_1}=\exp({i2\pi}\frac{n_1}{n}(b_1 c_2))$, with $b=(b_1,b_2)\in Z_n^2$, $c=(c_1,c_2)\in Z_n^2$. Here $b_1,b_2,c_1,c_2,n_1\in\{0,1,\dots,n-1\}=Z_{n}$.

For (3)$H^2(Z_2^3,U(1))=Z_2^3$, the 2-cocycles are $\beta(b,c)=\beta_1^{n_1}\beta_2^{n_2}\beta_3^{n_3}$. Explicitly, $\beta_1^{n_1}=\exp({i\pi}n_1(b_2 c_3))$, $\beta_2^{n_2}=\exp({i\pi}n_2(b_1 c_3))$, $\beta_3^{n_3}=\exp({i\pi}n_3(b_1 c_2))$, with $b=(b_1,b_2,b_3)\in Z_2^3$, $c=(c_1,c_2,c_3)\in Z_2^3$. Here $b_1,b_2,b_3,c_1,c_2,c_3,n_1,n_2,n_3\in\{0,1\}=Z_2$.

Thus, similarly, we can partially answer (4) for $H^2(Z_n^3,U(1))=Z_n^3$, the 2-cocycles are $\beta(b,c)=\beta_1^{n_1}\beta_2^{n_2}\beta_3^{n_3}$. Explicitly, $\beta_1^{n_1}=\exp({i2\pi}\frac{n_1}{n}(b_2 c_3))$, $\beta_2^{n_2}=\exp({i2\pi}\frac{n_2}{n}(b_1 c_3))$, $\beta_3^{n_3}=\exp({i2\pi}\frac{n_3}{n}(b_1 c_2))$, with $b=(b_1,b_2,b_3)\in Z_n^3$, $c=(c_1,c_2,c_3)\in Z_n^3$. Here $b_1,b_2,b_3,c_1,c_2,c_3,n_1,n_2,n_3\in\{0,1,\dots,n-1\}=Z_{n}$.

The 2-cocycles $\beta(b,c)$ above, have the satisfactory properties: (i) all satisfy 2-cocycles conditions $\frac{\beta(A,B)\beta(AB,C)}{\beta(A,BC)\beta(B,C)}=1$. (2) A generator correspond to an element of $H^2(G,U(1))$ for the given $G$. (3) Any two generators $\beta_i \neq \beta_j$ are not different simply by a 2-coboundary. i.e. $\beta_i \neq \beta_j \frac{\gamma(b)\gamma(c)}{\gamma(bc)}$ for any 1-cochian $\gamma(g)\in U(1)$.

Maybe the above answer implies it will not be too difficult to fully answer (4) and (5).

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