I have been working for a number of months on a paper/essay based on this talk (edit: big mad long draft here). After the talk the seminar host Teo Banica suggested a number of things that the approach could be used to look at, and one of these was orbits and orbitals. These are nice, intuitive ideas introduced by Lupini, Mancinska (missing an accent), and Roberson introduced orbitals.

I went to Teo’s quantum permutations tome, and Chapter 13 (p.297) orbits and orbitals are introduced, and it is remarked that, where we are studying quantum permutation groups $\mathbb{G}< S_N^+$, a certain relation on $N\times N\times N$ is believed not to be transitive. This belief is expressed also in the fantastic nonlocal games and quantum permutations paper, as well as by Teo here.

One of the things that the paper has had me doing is using CAS to write up the magic unitaries for a number of group duals, and I said, hey, why don’t I try and see is there any counterexamples there. My study of the $\widehat{S_3} led me to believe there would be no counterexamples there. The next two to check would be $\widehat{S_4} and the dual of the quaternion group $\widehat{Q}. I didn’t get called JPQ by Professor Des MacHale for nothing… I had to look there. OK, time to explain what the hell I am talking about.

I guess ye will have to wait for the never-ending paper to see exactly how I think about quantum permutation groups… so for the moment I am going to assume that you know what compact matrix quantum groups… but maybe I can put in some of the new approach, which can be gleaned from the above talk, in bold italics. A quantum permutation group $\mathbb{G}\leq S_N^+$ is a compact matrix quantum group whose fundamental representation $u^{\mathbb{G}}\in M_N(C(\mathbb{G}))$ is a magic unitary. The relation that was believed not to be transitive is:

$(j_3,j_2,j_1)\sim_3 (i_3,i_2,i_1)\Leftrightarrow u_{j_3i_3}u_{j_2i_2}u_{j_1i_1}\neq 0$,

that is the indices are related when there is a quantum permutation $\varsigma$ that has a non-zero probability of mapping:

$(\varsigma(j_3)=i_3)\succ(\varsigma(j_2)=i_2)\succ (\varsigma(j_1)=i_1)$. (*)

This relation is reflexive and symmetric. If we work with the universal (or algebraic) level, then $e\in\mathbb{G}$ will fix all indices giving reflexivity, if a quantum permutation $\varsigma\in\mathbb{G}$ can map as per (*), it’s reverse $\varsigma^{-1}:=\varsigma\circ S$ will map, with equal probability of $\varsigma$ doing (*):

$(\varsigma^{-1}(i_3)=j_3)\succ(\varsigma^{-1}(i_2)=j_2)\succ (\varsigma^{-1}(i_1)=j_1)$,

so that $\sim_3$ is symmetric.

Now, to transitivity. We’re going to work with the algebra of functions on the dual of the quaternions, $F(\widehat{Q}):=\mathbb{C}Q$. Working here is absolutely fraught what with coefficients $i$ and $-1$ and elements of $Q$ of the same symbol. Therefore we will use the $\delta^g$ notation. Consider the following vector in $F(\widehat{Q})$:

$\displaystyle (u^{\widehat{\langle j\rangle}})_{,1}:=\frac{1}{4}\left[\begin{array}{c}\delta^1+\delta^j+\delta^{-1}+\delta^{ij}\\ \delta^1+i\delta^{j}-\delta^{-1}-i\delta^{-j} \\ \delta^1-\delta^{j}+\delta^{-1}-\delta^{-j} \\ \delta^{1}-i\delta^{j}-\delta^{-1}+i\delta^{-j}\end{array}\right]$.

This vector is the first column of a magic unitary $u^{\widehat{\langle j\rangle}}$ for $\widehat{\langle j\rangle}\cong \widehat{\mathbb{Z}_4}\cong \mathbb{Z}_4$, and the rest of the magic unitary is made by making a circulant matrix from this. Do the same with $k\in\widehat{Q}$, another magic unitary $u^{\widehat{\langle k\rangle}}$, and so we have $\widehat{Q} via:

$u^{\widehat{Q}}=\left(\begin{array}{cc}u^{\widehat{\langle j\rangle}} & 0 \\0 & u^{\widehat{\langle k\rangle}} \end{array}\right)$.

Now for the counterexample: $u^{\widehat{Q}}_{67}u^{\widehat{Q}}_{41}u^{\widehat{Q}}_{87}\neq 0$ so  $(6,4,8)\sim_3 (7,1,7)$ and $u_{78}^{\widehat{Q}}u_{14}^{\widehat{Q}}u_{78}^{\widehat{Q}}\neq0$ so $(7,1,7)\sim_3(8,4,8)$, but $u^{\widehat{Q}}_{68}u^{\widehat{Q}}_{44}u^{\widehat{Q}}_{88}=0$ so $(6,4,8)$ is not related to $(8,4,8)$ and so $\sim_3$ is not transitive.

That $u^{\widehat{Q}}_{68}u^{\widehat{Q}}_{44}u^{\widehat{Q}}_{88}=0$ is a bit of algebra, and I guess the others are too… but instead we can exhibit states $\varsigma_2,\,\varsigma_1\in S(F(\widehat{Q}))$ such that $\varsigma_2(|u^{\widehat{Q}}_{67}u^{\widehat{Q}}_{41}u^{\widehat{Q}}_{87}|^2)>0$ and $\varsigma_1(|u_{78}^{\widehat{Q}}u_{14}^{\widehat{Q}}u_{78}^{\widehat{Q}}|^2)>0$ instead. The algebra structure of $F(\widehat{Q})$ is:

$\displaystyle F(\widehat{Q})=\mathbb{C}\oplus\mathbb{C}\oplus\mathbb{C}\oplus\mathbb{C}\oplus M_2(\mathbb{C})\subset B(\mathbb{C}^6)$.

Define $\varsigma_2$ to be the vector state associated with $\xi_2:=(0,0,0,0,1/\sqrt{2},1/\sqrt{2})$. Then:

$\|u^{\widehat{Q}}_{67}u^{\widehat{Q}}_{41}u^{\widehat{Q}}_{87}(\xi_2)\|^2=\frac14$.

$\varsigma_2\in\widehat{Q}$ is a quantum permutation such that:

$\mathbb{P}[(\varsigma_2(7)=6)\succ (\varsigma_2(1)=4)\succ (\varsigma_2(7)=8)]=\frac{1}{4}$.

Similarly the vector state $\varsigma_1$ given by $\xi_1:=(0,0,0,0,0,1)$ has

$\|u^{\widehat{Q}}_{78}u^{\widehat{Q}}_{14}u^{\widehat{Q}}_{78}(\xi_1)\|^2=:\mathbb{P}[(\varsigma_1(8)=7)\succ (\varsigma_1(4)=1)\succ (\varsigma_1(8)=7)]>0$.

Now, classically we might expect that $\varsigma_2\star \varsigma_1$ (convolution) might have the property that:

$(\varsigma_2\star\varsigma_1)(|u^{\widehat{Q}}_{68}u^{\widehat{Q}}_{44}u^{\widehat{Q}}_{88}|^2)>0$,

but as we have seen the product in question is zero.

Edit: The reason this phenomenon happens is that $u_{11}^{\widehat{Q_8}}$ and $u_{55}^{\widehat{Q_8}}$ are projections to random/classical permutations! I have also found a counterexample in the Kac-Paljutkin quantum group for similar reasons.

In the paper under preparation I think I should be able to produce nice, constructive, proofs of the transitivity of $\sim_1$ and $\sim_2$, constructive in the sense that in both cases I think I can exhibit states on $C(\mathbb{G})$ that are non-zero on suitable products of $u_{ij}$, using I think the conditioning of states:

$\displaystyle\varsigma\mapsto \frac{\varsigma(u_{ij}\cdot u_{ij})}{\varsigma(u_{ij})}$.

There is also something here to say about the maximality of $S_N. All must wait for this paper though (no I don’t have a proof of this)!