This paper referenced in the title (preprint here) tries to establish some very basic properties of a so-called exotic quantum permutation group, one intermediate to the classical and quantum permutation groups:

\displaystyle S_N\subsetneq \mathbb{G}\subsetneq S_N^+.

If a compact matrix quantum subgroup \mathbb{H}\subseteq \mathbb{G} (here \mathbb{G} is generic, not exotic) given by a quotient \pi:C(\mathbb{G})\to C(\mathbb{H}), is such that for all monomials f in the generators u_{ij}\in C(\mathbb{G}):

\displaystyle h_{C(\mathbb{H})}(\pi(f))=h_{C(\mathbb{G})}(f),

then in fact \mathbb{H}=\mathbb{G}. In the paper above I explored calculating the Haar state on C(S_N^+) using invariances related to the inclusion S_N\subseteq S_N^+… the point being that the Haar measure on exotic \mathbb{G} shares these invariances too… and perhaps, very speculatively, we could show that these Haar states coincide on such monomials… and a famous open problem in the theory would be solved.

The paper above fails at generators of length four though. The invariances do not determine the Haar measure on S_N^+ and we have to include a representation theory result around the law of the main character to give the explicit formulae. If we leave out the representation theory input, we have the Haar state up to a parameter.

However, a very basic algebraic result, that a product of three generators in exotic C(\mathbb{G}) can be zero for trivial reasons only, allows us to put bounds on the Haar state. If these bounds could force the fourth moment of the main character (with respect to the Haar state) to equal 14, then the Haar state on length four monomials would be equal for both. Alas… they only bound them less than or equal to 15. At this point the paper gives up… but this moment is a whole number, either 14 or 15… so I added the relation that the fourth moment is 15… and what comes out is that the parameter above is zero… but when this parameter is zero we find, in exotic \mathbb{G}:

\displaystyle h(|u_{22}u_{11}u_{23}|^2)=0,

but this is not the case, as u_{22}u_{11}u_{23} is not zero for trivial reasons, and the Haar state is faithful… the fourth moment equal to 15 gives classical S_N\subseteq S_N^+. Therefore the fourth moment for exotic \mathbb{G}\subset S_N^+ is forced equal to 14, giving the same explicit formula for length four generators as for S_N^+. Further work needed here.