EDIT: two years later, I considered instead this at the level of the Tannaka–Krein categories. If \mathcal{NC}<\mathcal{C}<\mathcal{P} at parameter N, then \langle\mathcal{C},e_N\rangle=\mathcal{P} if and only if N\leq 4. This may still be true, but I tried to relax to \mathcal{NC}_2<\mathcal{C}<\mathcal{P} and this doesn’t work for \mathcal{C}_{O_N^*}, the half-liberated orthogonal group. For every N, because u_{ij}u_{NN}u_{k\ell}=u_{k\ell}u_{NN}u_{ij} in C(O_N^*), quotienting to u_{NN}=1 forces the isotropy to be commutative. This is related to “uniformity” in the sense of Banica.

The following is an approach to the maximality conjecture for S_N\subset S_N^+ which asks what happens to a counterexample S_N\subsetneq \mathbb{G}_N\subsetneq S_N^+ when you quotient C(\mathbb{G}_N)\to C(\mathbb{G}_N)/\langle 1-u_{NN}\rangle. If C(\mathbb{G}_N)/\langle 1-u_{NN}\rangle is noncommutative, you generate another counterexample S_{N-1}\subsetneq \mathbb{G}_{N-1}\subsetneq S_{N-1}^+.}

Most of my attempts at using this approach were doomed to fail as I explain below.

Let \mathbb{G}\subseteq S_N^+ be a quantum permutation group with (universal) algebra of continuous functions C(\mathbb{G}) generated by a fundamental magic representation u\in M_N(C(\mathbb{G})). Say that \mathbb{G} is classical when C(\mathbb{G}) is commutative and genuinely quantum when C(\mathbb{G}) is noncommutative.


Definition 1 (Commutator and Isotropy Ideals)

Given \mathbb{G}\subseteq S_N^+, the commutator ideal J_N\subset C(\mathbb{G}) is given by:


\displaystyle J_N=\langle [u_{ij},u_{kl}]\,\mid\, 1\leq i,j,k,l\leq N\rangle.

The isotropy ideal I_N\subset C(\mathbb{G}) is given by:

\displaystyle I_N=\langle 1-u_{NN}\rangle.

Lemma 1

The commutator ideal J_N\subset C(\mathbb{G}) is equal to the ideal

\displaystyle K_N:=\langle u_{i_1j_1}u_{i_2j_2}u_{i_1j_3},u_{i_1j_1}u_{i_2j_2}u_{i_3j_1}\,\mid \, 1\leq i_k,j_k\leq N,\,j_1\neq j_3,i_1\neq i_3\rangle.

Proof:

\begin{aligned} [u_{i_2j_2},u_{i_3j_1}] & =u_{i_2j_2}u_{i_3j_1}-u_{i_3j_1}u_{i_2j_2} \\ \implies u_{i_1j_1}[u_{i_2j_2},u_{i_3j_1}] & = u_{i_1j_1}u_{i_2j_2}u_{i_3j_1}\qquad (i_1\neq i_3)\\ \implies u_{i_1j_1}u_{i_2j_2}u_{i_3j_1} & \in J_N, \end{aligned}
with a similar statement for u_{i_1j_1}u_{i_2j_2}u_{i_1j_3}.

On the other hand:
\begin{aligned} [u_{i_1j_1},u_{i_2j_2}] & =u_{i_1j_1}u_{i_2j_2}-u_{i_2j_2}u_{i_1j_1} \\ & =u_{i_1j_1}u_{i_2j_2}\sum_{k=1}^Nu_{i_1k}-\sum_{l=1}^{N}u_{i_1l}u_{i_2j_2}u_{i_1j_1} \\ & = u_{i_1j_1}u_{i_2j_2}u_{i_1j_1}+u_{i_1j_1}u_{i_2j_2}\sum_{k\neq j_1}u_{i_1j_1}u_{i_2j_2}u_{i_1k}-\\&u_{i_1j_1}u_{i_2j_2}u_{i_1j_1}\sum_{l\neq j_1}u_{i_1l}u_{i_2j_2}u_{i_1j_1}\\ \implies [u_{i_1j_1},u_{i_2j_2}]& \in K_N \end{aligned}

Proposition 1

The commutator and isotropy ideals are Hopf*-ideals. The quotient C(\mathbb{G})\to C(\mathbb{G})/J_N gives a classical permutation group G\subset S_N, the classical version G\subseteq \mathbb{G}, and the quotient C(\mathbb{G})\to C(\mathbb{G})/I_N giving an isotropy quantum subgroup. If \mathbb{G} is classical, this quotient is the isotropy subgroup of N for the action \mathbb{G}\curvearrowright {1,2,\dots,N}.

Via C(S_N^+)\to C(S_N^+)/J_N, the classical permutation group is a quantum subgroup S_N\subset S_N^+. It is conjectured that for all N it is a maximal subgroup.


Theorem 1 (Wang/Banica/Bichon)

S_N\subseteq S_N^+ is maximal for 1\leq N\leq 5.

By the universal property, the isotropy quantum subgroup of S_N^+ is S_{N-1}^+\subset S_N^+. The isotropy subgroup S_{N-1}^+\subset S_N^+ is genuinely quantum except at N=4 (due to S_3^+=S_3). That the isotropy quantum subgroup of S_N^+ is genuinely quantum if and only if N\geq 5 is the same as saying that for C(S_N^+), J_N\subset I_N if and only if N\leq 4.

Definition 2

A quantum permutation group \mathbb{G}\subseteq S_N^+ with fundamental magic representation u\in M_N(C(\mathbb{G})) has free three-orbitals if a product of three generators u_{i_1j_1}u_{i_2j_2}u_{i_3j_3} is zero for trivial reasons only. That is, if two adjacent factors are distinct entries from the same row or column. For such a quantum permutation group the generators are non-zero, the commutators [u_{ij},u_{kl}] are non-zero (except for entries along the same row or column — therefore \mathbb{G} is genuinely quantum), and all of the generators of K_N are non-zero.

Call a strictly intermediate quantum permutation group S_N\subsetneq \mathbb{G}_N\subsetneq S_N^+ exotic. In [McC], Th. 3.31 it was shown that if S_N\subsetneq \mathbb{G}\subsetneq S_N^+ then \mathbb{G} has free three-orbitals.

Conjecture 1

Let \mathbb{G}\subseteq S_N^+ be a quantum permutation group with free three-orbitals. Then


\displaystyle J_N\subset I_N\implies N\leq4.

Remark


Conjecture 1 can be weakened to J_N\subset I_N\implies N\leq 5 but I expect that if this weaker version is true, then in fact the stronger version is true. A basic relevant calculation towards Conjecture 1 is:
\begin{aligned} u_{11}u_{21} & =0 \\ \implies u_{11}\left(\sum_{k=1}^N u_{k3}\right)u_{21} & =0 \\ \implies u_{11}\left(\sum_{k=3}^N u_{k3}\right)u_{21} & =0 \\ \implies u_{11}\left(\sum_{k=3}^{N-1}u_{k3}\right)u_{21}&=-u_{11}u_{N3}u_{21} \\ \implies u_{11}\left(\sum_{k=3}^{N-1}u_{k3}\right)u_{21} & \in I_N \end{aligned}
Via K_3=J_3, this calculation at N=3 implies S_3=S_3^+. At N=4 it gives J_4\subset I_4 (and also, via a universal property, S_3=S_3^+). At, say, N=5 we have:


\displaystyle u_{11}(u_{33}+u_{34})u_{21}\in I_5.


We are stuck however, because we want to show that e.g. u_{11}u_{33}u_{21}\not \in I_5.

We also get the following very strange non-classical relation (more here): for S\subset{1,2,\dots,N}, define u_{S,j}=\sum_{i\in S}u_{ij}. Let R_N:={3,4,\dots,N} Then, where |f|=f^*f, for any subset S\subseteq R_N:


\displaystyle|u_{11}u_{S,3}u_{21}|^2=|u_{11}u_{(R_N\cap \overline{S}),3}u_{21}|^2.



So, for example at N=7:


\displaystyle |u_{11}u_{33}u_{21}|^2=|u_{11}(u_{43}+u_{53}+u_{63}+u_{73})u_{21}|^2.

Conditional Proof of Maximality Conjecture: Suppose that S_N\subsetneq \mathbb{G}_N\subsetneq S_N^+ with N>5. By taking a sequence of quotients, by I_N, I_{N-1},\cdots, \,I_{6}, we generate a sequence of quantum subgroups \mathbb{G}_M\subseteq S_M^+ for M=5,\cdots,N-1. The inclusions S_M\subset \mathbb{G}_M hold because (using \mathbb{G}_M\subset S_{M+1} with a 1_{C(\mathbb{G}_M)} in the (M,M) entry):

\begin{aligned} C(\mathbb{G}_{M})/{J_{M+1}}&=(C(\mathbb{G}_{M+1})/I_{M+1})/J_{M+1}\\ &=C(\mathbb{G}_{M+1})/( I_{M+1}+J_{M+1})\\&=\left(C(\mathbb{G}_{M+1})/ J_{M+1}\right)/I_{M+1}\\&=\left(C(\mathbb{G}_{M+1})/ J_{M+1}\right)/I_{M+1}\\&C(S_{M+1})/{I_{M+1}}=C(S_{M}).\end{aligned}

By Conjecture 1, these are genuinely quantum subgroups.
In particular, we have that S_{5}\subsetneq \mathbb{G}_{5}\subseteq S_{5}^+. However by Theorem 1, we must have \mathbb{G}_5=S_5^+.

Now consider S_6\subsetneq \mathbb{G}_6\subseteq S_6^+. Via C(\mathbb{G}_6)/J_6=C(S_6), and C(\mathbb{G}_6)/I_6=C(S_5^+) we have that S_5^+\subset \mathbb{G}_6 and S_6\subset \mathbb{G}_6. But by Brannan, Chirvasitu and Freslon ([BCF], Th. 3.3) these topologically generate S_6^+, and so \mathbb{G}_6=S_6^+. Inductively \mathbb{G}_N=S_N^+, and so the maximality conjecture holds.

An approach that does not work

I have been banging at this approach quite a bit, with very little success. One thing I was trying to do was maybe show that Conjecture 1 holds more generally. This basically boiled down to looking at ideals rather than Hopf ideals. This generalised Conjecture 1 would look something like:

Suppose that a \mathrm{C}^*-algebra \mathcal{A} is generated by the entries of a magic unitary u\in M_N(\mathcal{A}) with free three orbitals and abelianisation C(S_N). Then if N>5, it is never the case that J_N\subset I_N.

So in fact this is not the case, by results from [McC]. For each N\geq4 there exists a magic unitary v with entries in M_N(\mathbb{C}) with free three orbitals (indeed free orbitals of every order). Now, embed this in C(S_N)\oplus M_N(\mathbb{C}), and where \mathbf{1}_{j\to i} are the standard generators of C(S_N). Consider the magic unitary u\in C(S_N)\oplus M_N(\mathbb{C}) given by:

u_{ij}=\mathbf{1}_{j\to i}+v_{ij}

Now this magic unitary generates a \mathrm{C}^*-algebra with free three orbitals and abelianisation C(S_N). The commutator ideal is J_N=M_N(\mathbb{C}). Each v_{ij} is a rank one operator on M_N(\mathbb{C}) and so, I_N generated by 1-u_{NN}, contains something in M_N(\mathbb{C}) and so M_N(\mathbb{C})\subset I_N, and so J_N\subset I_N at every N. Failure.


[BCF] M. Brannan, A. Chirvasitu and A. Freslon, Topological generation and matrix models for quantum reflection groups, Adv Math, 363, 1-26 (2020)


[McC] J.P. McCarthy, Tracing the orbitals of the quantum permutation group, Arch. Math., 121, 211-224 (2023)

  1. I found a much easier proof: assume all u{i_1j_1}u_{i_2j_2}u_{i_1j_3} are zero. Sum up j_1=1,\dots,N except for j_1=j_3. Out pops (1-u_{i_1j_3})u_{i_2j_2}u_{i_1j_3}=0\implies u_{i_2j_2}u_{i_1j_3}=u_{i_1j_3}u_{i_2j_2}u_{i_1j_3}. This implies that [u_{i_1j_3},u_{i_2j_2}]=0 but exotic quantum permutation groups have no non-trivial commuting generators. Then the labels can be permuted using characters coming from the inclusion of S_N} ↩︎