Timeline for Explicit 2-cocycle from a 2nd cohomology group $H^2[Q_8 \times \mathbb{Z}/2\mathbb{Z}, U(1)]$
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 9, 2016 at 17:47 | vote | accept | wonderich | ||
| Sep 9, 2016 at 10:00 | comment | added | Ehud Meir | As far as I see it the results agree: Let us choose $n=lcm(M,N)$. Then as a map $f:\mathbb{Z}_M\to \mathbb{Z}_n$, $f$ will have the form $g\mapsto gN/gcd(M,N)$ or a multiple thereof. a similar statement holds for $r$. But now the multiplication is going to be $$((g_1,h_1),(g_2,h_2))\mapsto h_1g_2MN/gcd^2 = h_1g_2lcm/gcd$$. And translating this back again to $exp$, we get the expression you wrote. | |
| Sep 9, 2016 at 5:54 | comment | added | wonderich | But then your answer does not match my result if you go through this. Your map $f$ and $g$ should not map to the same $Z_n$. | |
| Sep 8, 2016 at 20:30 | comment | added | Ehud Meir | in this case we will take lcm(M,N) and not gcd(M,N). | |
| Sep 8, 2016 at 15:14 | comment | added | wonderich | The above, $α(g_1,h_1,g_2,h_2)=\exp[(2πi/\text{gcd}(M,N))g_1h_2]$, we have $g_1,g_2 \in G=Z_M$ and $h_1,h_2 \in H=Z_N$. The group homomorphism is not the same as you suggested as from $Z_M,Z_N$ to $Z_n=Z_{\text{gcd}(M,N)}$. It is $Z_M\to Z_M$ and $Z_N \to Z_N$ instead. | |
| Sep 8, 2016 at 15:09 | comment | added | wonderich | In this case, the n prescribed in your answer should be $n=\text{gcd}(M,N)$, but then $f: G=Z_M→Z_n=Z_\text{gcd}(M,N)$ and $r: H=Z_N→Z_n=Z_\text{gcd}(M,N)$ do not hold. Instead, we have $f: G=Z_M→Z_M$ and $r: H=Z_N→Z_N$. Thus, something can be wrong in your statement -- could you crosscheck? Thanks Ehud. | |
| Sep 8, 2016 at 15:04 | comment | added | wonderich | I don't know whether this is consitent with what you said: I know the case that $f: G=Z_M→Z_n$ and $r: H=Z_N→Z_n$, the 1-cocycle for $Z_M$ is $\alpha(g)=\exp[(2\pi i/M) g]$, the 1-cocycle for $Z_N$ is $\alpha(h)=\exp[(2\pi i/N) h]$, the2-cocycle for $Z_M \oplus Z_N$ is $\alpha(g_1,g_2,h_1,h_2)=\exp[(2\pi i/\text{gcd}(M,N)) g_1 h_2]$. | |
| Sep 8, 2016 at 0:09 | comment | added | Ehud Meir | you can determine explicitly the two cocycles by the formula written above. In the case of $Q_8$ and $\mathbb{Z}_2$ it is going to be the following: $alpha((g_1,h_1),(g_2,h_2)) = 1$ of $h_1=0$, and $\phi(g_2)$ if $h_1=1$ (we are talking here about residues modulo 2), where $\phi$ is a homomorphism from $Q_8$ to $\mathbb{Z}_2$. Since there are four possible homomorphisms, you will get four possible non-equivalent cocycles. | |
| Sep 7, 2016 at 21:20 | comment | added | wonderich | I am puzzled by part of your answer. Suppose we take $G=\mathbb{Z}_M$ and $H=\mathbb{Z}_N$ as an example. Then what will be the group homomorphism be? $f: G=\mathbb{Z}_M \to \mathbb{Z}_n$ and $r: G=\mathbb{Z}_N \to \mathbb{Z}_n$? What is $n$ here gcd$(M,N)$ or lcm$(M,N)$? And should both homomorphisms $f$ and $g$ give the same $n$? | |
| Sep 7, 2016 at 16:21 | comment | added | wonderich | I also know that $H^2[Q_8 \times Z_2, U(1)]=H^1[Q_8, Z_2]=\text{Tor}^Z[Z_2, H^2(Q_8,Z)]=\text{Tor}^Z[Z_2, H^1(Q_8,U(1))]=\text{Tor}^Z[Z_2, Z_2 \times Z_2]=Z_2 \times Z_2= Z_2 \oplus Z_2$ | |
| Sep 7, 2016 at 16:19 | comment | added | wonderich | Thanks Ehud. Say $G=Q_8$, and $H=Z_2$. Do you imply that if we know the explicit 1-cocycle of $H^1[G,U(1)]$ and the explicit 1-cocycle of $H^1[H,U(1)]$, then I am able to determine the explicit 2-cocycle of $H^2[G \times H,U(1)]$? | |
| Sep 7, 2016 at 14:48 | history | answered | Ehud Meir | CC BY-SA 3.0 |