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In a recent preprint (arXiv:2311.04889), my coauthors and I constructed a sequence of graphs with no classical symmetry which nevertheless have quantum symmetry. For graphs this had been an open problem, but we do not know if similar examples exist in other areas of noncommutative geometry.

Question. Are there other known examples of classical (i.e. commutative) spaces which have no classical symmetry, but which do have quantum symmetry?

Here, we mean "commutative space" in the broadest possible sense (for example, a topological space, a probability space, a group, a spin manifold, etc.), with the appropriate notion of quantum symmetry (depending on the context, e.g. quantum automorphisms/isometries/etc.).

(Not so) brief description of our construction as requested by Sebastien Palcoux

See Definition 3.8 in the paper linked above for a formal description.

Consider a linear system $Mx=b$ over $\mathbb{Z}_2$ (the graph will depend on $b$ but the (quantum) automorphism group will not). Create a bipartite graph with vertices of two types:

  1. Variable type vertices: two for each variable corresponding to the two possible values that variable can take in $\mathbb{Z}_2$.
  2. Equation type vertices: for each equation in the system make a vertex for each satisfying assignment to just that equation (so variables not appearing in that equation are not considered), e.g., for the equation $x_1 + x_2 + x_3 = 1$ there would be 4 vertices corresponding to the 4 solutions of this equation.

If an equation type vertex corresponds to the assignment $x_1 \mapsto 0, \ x_2 \mapsto 1$ for instance, make it adjacent to the variable type vertex that corresponds to $x_1$ taking value $0$ and the vertex corresponding to $x_2$ taking value 1. Now color all vertices according to the variable/equation they correspond to. We show that the automorphism group of this vertex-colored graph is isomorphic to the group of solutions to $Mx=0$. For the quantum automorphism group, we consider the solution group of $Mx=0$ (see https://arxiv.org/abs/1606.02278 or Definition 3.1 of our paper linked above) which is defined by taking each $x_i$ as a generator of a finitely presented group where $x_i^2 = \mathrm{id}$ for all $i$ and for each equation, e.g., $x_1 + x_2 + x_3 = b_i$, we include the relation $x_1x_2x_3 = \mathrm{id}$ and the relations requiring that $x_1, x_2$, and $x_3$ pairwise commute. The quantum automorphism group of the graph is the dual of this solution group which is non-classical when the solution group is non-abelian. Note that the automorphism group of the graph is isomorphic to the abelianization of the solution group.

The challenge is then to find a linear system whose solution group is a nontrivial perfect group, i.e., a nontrivial group whose abelianization is trivial. For this we consider the alternating group $A_n$ for $n \ge 7$. We construct a linear system with variables/generators corresponding to the elements of order 2 in $A_n$ and equations/relations corresponding to the triples of such elements that pairwise commute and multiply to identity. For $A_7$, the resulting solution group is the triple cover of $A_7$, which is perfect.

Note that the graph with 560 vertices I mention in the comments below is using a slightly different construction that only has the equation vertices but also uses edge colors (see Definition 3.4 in our paper). The construction above has the advantage that we know how to remove vertex colors in general while preserving the (quantum) automorphism group, but we do not know how to do this for edge colors.

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    $\begingroup$ Another avenue to look at (and again not answering the question - the answer as far as I understand is no, no known examples.): mathoverflow.net/questions/188707/… $\endgroup$ Mar 5 at 8:40
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    $\begingroup$ Fascinating! Could you please include an image of the simplest graph that possesses this property? $\endgroup$ Mar 5 at 8:50
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    $\begingroup$ @SebastienPalcoux I'd love it if such an image could be made that wouldn't be incomprehensible, but even if I am allowed edge and vertex colors, the smallest such graph we construct would have 560 vertices. So I think the image would just end up being a big blob. I can give a brief description if you'd like. $\endgroup$ Mar 5 at 9:10
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    $\begingroup$ @SebastienPalcoux 5880 edges. The only method I can think of for getting such a lower bound is to check all asymmetric graphs on few vertices and show they have no quantum symmetry. There are too many of these so instead you should check asymmetric coherent configurations, which would suffice. There are only 9839 coherent configurations on 15 points (actamath.savbb.sk/pdf/aumb2605.pdf), so this is probably doable. I expect none are asymmetric but with quantum symmetry. Not a great lower bound and I would guess checking for q-symmetry in general is undecidable, so not a great approach. $\endgroup$ Mar 5 at 10:45
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    $\begingroup$ @SebastienPalcoux in addition to the "(not so) brief description" which David added to the question, you can also take a look at these slides, which give a high-level outline of the proof (from slide 10 onwards). Though I guess some parts of the talk, including the description of the graph, might require some verbal explanation, which is not part of the slides. 😅 $\endgroup$ Mar 5 at 15:06

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