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Recently I have encountered a problem concerning the property of slant product in group cohomology. The problem is as follows: Consider a finite group G (can have anti-unitary operations). And there is a center $Z_N$ of G generated by group element $g$. Then we can construct the following sequence utilizing the slant product: $$H^3(G,U(1))\xrightarrow{i_g^3}H^2(G,U(1))\xrightarrow{i_g^2}H^1(G,U(1)).$$ where slant products are as follows:

for a 3-cocycle $\omega(a,b,c)$, $$i_g^3\omega(a,b)=\omega(g,a,b)\omega(a,b,g)/\omega(a,g,b),$$

for a 2-cocycle $x(a,b)$, $$i_g^2x(a)=x(g,a)/x(a,g)$$.

It is apparent that $im(i_g^3)\subset ker(i_g^2)$. But is it true that $ker(i_g^2)=im(i_g^3)$?

Arun Debray
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Xu Yang
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  • What is an anti-unitary operation? And what do you mean by other center? – მამუკა ჯიბლაძე Apr 04 '17 at 20:58
  • Anti-unitary operation simply means that it will take the complex conjugation of the $U(1)$ element. Then, e.g., the 3-cocycle condition should be written as follows: $\omega(a,b,c)\circ ^a\omega(b,c,d)\circ \omega(a,bc,d)=\omega(ab,c,d)\circ \omega(a,b,cd)$, where if $a$ is anti-unitary, it will take $\omega(b,c,d)$ to $\omega^*$ – Xu Yang Apr 04 '17 at 23:42
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    @მამუკაჯიბლაძე My impression, from the outside, is that this is fairly standard language in more physical contexts (I am guessing OP is a condensed matter theorist; this is confirmed by looking at OP's bio). A typical thing in that world is to work with \bZ_2-graded groups, i.e. groups with a map to \bZ_2, which distinguishes "unitary" from "antiunitary" elements. Then you get an induced action on U(1), and so can take cohomology with twisted coefficients. As for the center, I am guessing OP means simply that g generates the "full center", and not just part of the center. – Theo Johnson-Freyd Apr 04 '17 at 23:56
  • Thank @TheoJohnson-Freyd for the excellent commentary. I am indeed using some languages familiar to physicists. Sorry for the confusion I have made. – Xu Yang Apr 05 '17 at 00:00
  • @XuYang I find myself talking to condensed matter folks a lot these days... – Theo Johnson-Freyd Apr 05 '17 at 00:05
  • My initial reaction was "this can't be true", because it easily fails if you add one to all degrees. (Then e.g. any McKay group --- any finite subgroup of SU(2) --- has H^3(G,U(1)) nontrivial but H^4(G,U(1)) = H^2(G,U(1)) = 0.) But I do not see a counterexample for the case you mention. What examples have you checked? – Theo Johnson-Freyd Apr 05 '17 at 00:08
  • At least this is true for the simplest case where $G=Z_N\times G_1$. In that case we can use Kunneth formula to explicitly show it. I actually don't have any evidence for this stronger conclusion. If you can provide a counter-example then I'm done with it. – Xu Yang Apr 05 '17 at 00:46
  • @Theo, Xu Yang, thanks for the explanations. So $G$ acts on $U(1)$ via complex conjugation, and cohomology is with coefficients in this module? I wonder whether the mysterious construction from Witten's question has something to do with slant products too... – მამუკა ჯიბლაძე Apr 05 '17 at 04:54
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    @მამუკაჯიბლაძე Yes, I think so. Note that what condensed matter theorists call "slant product", the mathematicians I know call "(loop) transgression". Per your comment on http://mathoverflow.net/a/256100/78, a priori it is a map $H^\bullet(BG) \to H^{\bullet-1}(LBG)$. Each central element $g$ in $LBG$ determines a map $BG \to LBG$, and further restricting from $H^{\bullet-1}(LBG)$ to $H^{\bullet-1}(BG)$ gives the slant product with $g$. – Theo Johnson-Freyd Apr 05 '17 at 14:19
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    Slant products appear in condensed matter as follows. An important class of G-protected topological phases are indexed by cohomology classes in $H^\bullet(BG,U(1))$. A "G-protected phase" is something like a TQFT for manifolds equipped with G-bundles. (At least, each G-protected phase determines such a TQFT; I think there is still some question whether a TQFT determines uniquely a G-protected phase.) The action is easy to write down: if $M$ is an oriented $\bullet$-dimensional manifold, $P : M \to BG$ a $G$-bundle, and $\alpha \in H^\bullet(BG,U(1))$, then $Z(M) = \int_M P^*\alpha \in U(1)$. – Theo Johnson-Freyd Apr 05 '17 at 14:23
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    If you have a G-protected $\bullet$-dimensional phase, one thing you can do is to "compactify it on $S^1$" to produce a G-protected $(\bullet-1)$-dimensional phase. The compactification depends on choosing a monodromy around that $S^1$, say $g$. Then, for $N$ a $(\bullet-1)$-dimensional manifold with $G$-bundle, you set $Z^{compactified}(N) = Z(N \times S^1_g)$, where $S^1_g$ is $S^1$ with $g$-monodromy. Of course, to give $N \times S^1_g$ a product $G$-bundle, you had better have that $g$ is central. If you compactify the phase coming from $\alpha \in H^\bullet(BG)$, you get $i_g\alpha$. – Theo Johnson-Freyd Apr 05 '17 at 14:34
  • @XuYang I am stuck on your question. Right now I am inclined to believe it. Here is a strategy for a proof which I was unable to push through. (I will treat just the fully unitary case for now.) Since $g\in Z(G)$, you can consider the quotient group $G_1 = G / \langle g \rangle$. Since $G = \mathbb Z_n . G_1$ is an extension by a cyclic group, $\mathrm H^\bullet(G,U(1))$ is the limit of a spectral sequence whose $E_2$-page is $\mathrm H^\bullet(G_1,\mathrm H^\bullet(\mathbb Z_n,U(1))$, and of course $\mathrm H^\bullet(\mathbb Z_n,U(1))$ is either $0$ or $\mathbb Z_n$ depending on the parity – Theo Johnson-Freyd Apr 05 '17 at 14:51
  • ... of $\bullet$. Then I want to understand how $i_g$ acts on the $E_2$ page. This shouldn't be so bad. See, the $E_0$ page is just the space of all cochains $\mathrm C^\bullet(G_1,\mathrm C^\bullet(\mathbb Z_n,U(1)) = \mathrm C^\bullet(G,U(1))$, so $i_g$ definitely makes sense there, and so presumably makes sense also on the $E_2$ page. Anyway, I expect that on the $E_2$ page it will be clear that $i_g$ is exact at total degree $\bullet = 2$ (and not exact in general). Generally, exactness at some spot is not lost when you run spectral sequences, so if that yoga applies here, you will win. – Theo Johnson-Freyd Apr 05 '17 at 14:56

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