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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 $ker(i_g^2)=im(i_g^3)$?

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  • $\begingroup$ What is an anti-unitary operation? And what do you mean by other center? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Apr 4, 2017 at 20:58
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    $\begingroup$ @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. $\endgroup$
    – Theo Johnson-Freyd
    Commented Apr 4, 2017 at 23:56
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    $\begingroup$ @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 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 $g$. $\endgroup$
    – Theo Johnson-Freyd
    Commented Apr 5, 2017 at 14:19
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    $\begingroup$ 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 $Z(M) = \int_M P^*\alpha \in U(1)$. $\endgroup$
    – Theo Johnson-Freyd
    Commented Apr 5, 2017 at 14:23
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    $\begingroup$ 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 $i_g\alpha$. $\endgroup$
    – Theo Johnson-Freyd
    Commented Apr 5, 2017 at 14:34

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