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Let $G$ be a finite group. A categorical Schur detector for $G$ is a set $\mathcal{S}$ of proper subgroups $S \subsetneq G$ such that the total restriction map $$ \mathrm{rest}_{\mathcal{S}} : \mathrm{H}^3(G; \mathrm{U}(1)) \to \prod_{S \in \mathcal{S}} \mathrm{H}^3(S; \mathrm{U}(1)) $$ is an injection. The name is a riff on the suggestion of Epa and Ganter to call $\mathrm{H}_3(G;\mathbb{Z})$ the categorical Schur multiplier of $G$ (Platonic and alternating 2-groups, Higher Structures 1(1):122–146, 2017).

I know of two classes of groups which do not admit categorical Schur detectors: the cyclic and the binary dihedral (aka dicyclic) groups of prime power order. (Of course, a binary dihedral group of prime power order is necessarily of order $2^n$.) This follows from the fact that for any finite subgroup $G \subset \mathrm{SU}(2)$, there is an isomorphism $\mathrm{H}^3(G;\mathrm{U}(1)) \cong \mathbb{Z}/\lvert G\rvert$.

Question: Are there any other finite groups which do not admit categorical Schur detectors?

Of course, one could ask the same question about cohomology of other degrees, or with other coefficients. $\mathrm{H}^3(-;\mathrm{U}(1))$ is of particular interest in "moonshine", because it is home to global symmetry anomalies of 2D QFTs.

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    $\begingroup$ I record the trivial remark that a group of non-prime-power order always admits a categorical Schur detector, namely the set of its Sylow subgroups. Note that my definition of "categorical Schur detector" requires that each subgroup in the set be proper. $\endgroup$ Dec 5 '20 at 16:25
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    $\begingroup$ A variant of your question has been studied in the literature: the case when the coefficients are $\mathbb Z/p$, and the common kernel of restrictions to all proper subgroups is called essential cohomology. Adem and Karagueuzian show that a p-group has essential cohomology if all its elements of order p are central. (I gave a different proof, giving a degree in which this happens.) $\endgroup$ Dec 5 '20 at 17:42
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    $\begingroup$ @NicholasKuhn Thank you. I did not know the term "essential cohomology". Is the paper of you mention arxiv.org/abs/math/0612133? $\endgroup$ Dec 5 '20 at 20:18
  • $\begingroup$ Yes. And it appeared in Advances in 2007. $\endgroup$ Dec 6 '20 at 2:52
  • $\begingroup$ @NicholasKuhn Thanks! $\endgroup$ Dec 7 '20 at 3:18
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Inspired by YCor's remark I realized that every finite abelian $p$-group $G$ of rank three will have essential elements in $H^3(G;U(1))$. I think that this works for $p=2$ as well, but here's an argument for $p$ odd. Let $y_1,y_2,y_3$ be linearly independent elements of $H^1(G;\mathbb{F}_p)\cong \mathrm{Hom}(G,\mathbb{F}_p)$. If $S$ is any proper subgroup of $G$, then the image of $H^1(G;\mathbb{F}_p)\rightarrow H^1(S;\mathbb{F}_p)$ has rank two. Thus the images of $y_1,y_2,y_3$ in $H^1(S;\mathbb{F}_p)$ are no longer linearly independent, and so the product $y_1y_2y_3$ is an element of $H^3(G;\mathbb{F}_p)$ that is essential.

It remains to show that $y_1y_2y_3$ has non-zero image under the map $H^3(G;\mathbb{F}_p)\rightarrow H^3(G;U(1))$ induced by the inclusion of additive groups $\mathbb{F}_p\rightarrow U(1)$. For this, look at the long exact sequence in cohomology coming from the short exact coefficient sequence $0\rightarrow \mathbb{F}_p\rightarrow U(1)\rightarrow U(1)\rightarrow 0$. It suffices to show that the image of $H^2(G;U(1))\rightarrow H^3(G;\mathbb{F}_p)$ does not contain $y_1y_2y_3$. I struggle with $U(1)$ coefficients, so I would instead show the equivalent fact that $y_1y_2y_3$ is not in the image of $H^3(G;\mathbb{Z})\rightarrow H^3(G;\mathbb{F}_p)$. This is because $H^3(G;\mathbb{Z})$ is detected by proper subgroups. Roughly speaking this is because for a finite cyclic group $C$, $H^1(C;\mathbb{Z})=0$. In terms of the Kunneth theorem, the only way to get elements in degree three for a product of cyclic groups is as a `Tor' term coming from a pair of elements of $H^2(C;\mathbb{Z})$ and $H^2(C';\mathbb{Z})$ for cyclic subgroups $C$, $C'$. But such an element is detected on the subgroup $C\times C'$.

This should generalize: if $G$ is a finite abelian $p$-group then every element of $H^{k+1}(G;\mathbb{Z})$ will be detected on subgroups of rank $k$. If so, then for $G$ an abelian $p$-group $H^3(G;U(1))$ will contain essential elements if and only if $G$ has rank 1 or 3.

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