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This post is basically a quantum extension of Is every finite group a group of “symmetries”?
Here finite quantum group means finite dimensional Hopf ${\rm C}^{\star}$-algebra.

Frucht's theorem states that every finite group is the automorphism group of a finite graph. Wang defined here a notion of quantum automorphism group. The application to a finite space of $n$ elements is called "the quantum permutation group of $n$ symbols", and its quantum subgroups are called quantum permutation groups of degree $n$. Bichon introduced here the quantum automorphism groups of finite graphs, these are quantum permutation groups. See also this survey of Banica-Bichon-Collins.

Kojima proved here that every finite group is realized as the full isometry group of some compact hyperbolic $3$-manifold. This book of Goswami-Bhowmick introduces the notion of quantum isometry group.

General question: Is every finite quantum group a quantum symmetry group?

Sub-question 1: Is a finite quantum permutation group a twisted finite group?

Sub-question 2: Is every finite quantum group a quantum permutation group?
Answer: no, the smallest counter-example has dim. $24$ (see this paper of Banica-Bichon-Natale).

Sub-question 3: Is every finite quantum permutation group of dimension $n$ a quantum permutation group of degree $n$?

Sub-question 4: Is every finite quantum permutation group a quantum automorphism group of a finite graph?

Sub-question 5: Is every finite quantum group $\mathbb{G}$ a quantum subgroup of the quantum automorphism group of a finite dimensional ${\rm C}^{\star}$-algebra $\mathcal{A}$? Ok for $\mathcal{A} = C(\mathbb{G})$?
Answer (Bhowmick): yes, it follows trivially from the definition, using Haar state.

Sub-question 6: Is every finite quantum group a quantum isometry group?

Sebastien Palcoux
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  • Small point: Kojima points out in his paper that his argument generalizes Greenberg's result on groups realizable as isometry groups of hyperbolic 2-manifolds. – Ryan Budney May 24 '19 at 02:42
  • Is any non trivial finite quantum permutation group a quantum automorphism group of a graph? – JP McCarthy Jan 29 '21 at 23:12
  • For question three there is no counter examples in duals of finite abelian groups as they all have two generators. If one has order $n$ it is cyclic otherwise both less than or equal $n/2$ and you can construct two Fourier-type magic unitaries and do block diagonal of those. Presumably no issue to push that to all group duals. – JP McCarthy Jan 29 '21 at 23:28
  • *finite SIMPLE groups not abelian groups. – JP McCarthy Jan 30 '21 at 09:12
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    No counterexamples to Q3 in group duals https://math.stackexchange.com/a/4007044/19352 – JP McCarthy Jan 31 '21 at 16:44
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    @JPMcCarthy: Thanks for your investigation! – Sebastien Palcoux Jan 31 '21 at 17:29
  • Is any non trivial finite quantum permutation group a quantum automorphism group of a graph? Yes: there is a dimension 256 example https://arxiv.org/abs/2111.12362 – JP McCarthy Dec 12 '21 at 17:27
  • @JPMcCarthy: Good! I guess the question it answers is not "Is any..." but "Is there any...", right? – Sebastien Palcoux Dec 13 '21 at 02:56
  • @Sebastien Palcoux yes, this is what I meant: "is there any?" rather than "Are all?" The answer below answers the second question. – JP McCarthy Dec 13 '21 at 05:00

1 Answers1

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The answer to sub-question 4 is no.

See here, The Frucht property in the quantum group setting.

To expand slightly, this paper gives four explicit finite quantum permutation groups which are not the quantum automorphism groups of a finite graph: the duals of $S_3$, $A_4$, and $A_5$, and the Kac--Paljutkin quantum group of order eight, denoted $G_0$.

If the dual of $S_3$ is the quantum automorphism group of a graph, then, taking abelianisations, the (classical) automorphism group of the graph must be $\mathbb{Z}_2$. The paper above shows that if the dual of $S_3$ acts on a graph then a (classical) abelian group larger than $\mathbb{Z}_2$ acts on the graph too. Therefore the dual of $S_3$ is not the quantum automorphism group of a graph.

In fact it is shown that if the dual of $S_3$ acts on a graph then the dual of a(n infinite) free product acts on the graph too.

A similar story holds for the duals of $A_4$ and $A_5$.

If $G_0$ is the quantum automorphism group of a graph, then the graph has classical automorphism group $\mathbb{Z}_2\times\mathbb{Z}_2$. The above shows that if $G_0$ acts on a graph then so does the dihedral group of order eight. Therefore $G_0$ is not the quantum automorphism group of a graph.

JP McCarthy
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