I thought I had a bit of a breakthrough. So, consider the algebra of a functions on the dual (quantum) group $\widehat{S_3}$. Consider the projection:

$\displaystyle p_0=\frac12\delta^e+\frac12\delta^{(12)}\in F(\widehat{S_3})$.

Define $u\in M_p(\widehat{S_3})$ by:

$u(\delta^\sigma)=\langle\text{sign}(\sigma)1,1\rangle=\text{sign}(\sigma)$.

Note

$\displaystyle T_u(p_0)=\frac12\delta^e-\frac12 \delta^{(12)}:=p_1$.

Note $p_1=\mathbf{1}_{\widehat{S_3}}-p_0=\delta^0-p_0$ so $\{p_0,p_1\}$ is a partition of unity.

I know that $p_0$ corresponds to a quasi-subgroup but not a quantum subgroup because $\{e,(12)\}$ is not normal.

This was supposed to say that the result I proved a few days ago that (in context), that $p_0$ corresponded to a quasi-subgroup, was as far as we could go.

For $H\leq G$, note

$\displaystyle p_H=\frac{1}{|H|}\sum_{h\in H}\delta^h$,

is a projection, in fact a group like projection, in $F(\widehat{G})$.

Alas note:

$\displaystyle T_u(p_{\langle(123)\rangle})=p_{\langle (123)\rangle}$

That is the group like projection associated to $\langle (123)\rangle$ is subharmonic. This should imply that nearby there exists a projection $q$ such that $u^{\star k}(q)=0$ for all $k\in\mathbb{N}$… also $q_{\langle (123)\rangle}:=\mathbf{1}_{\widehat{S_3}}-p_{\langle(123)\rangle}$ is subharmonic.

This really should be enough and I should be looking perhaps at the standard representation, or the permutation representation, or $S_3\leq S_4$… but I want to find the projection…

Indeed $u(q_{(123)})=0$…and $u^{\star 2k}(q_{\langle (123)\rangle})=0$.

The punchline… the result of Fagnola and Pellicer holds when the random walk is is irreducible. This walk is not… back to the drawing board.

I have constructed the following example. The question will be does it have periodicity.

Where $\rho:S_n\rightarrow \text{GL}(\mathbb{C}^3)$ is the permutation representation, $\rho(\sigma)e_i=e_{\sigma_i}$, and $\xi=(1/\sqrt{2},-1/\sqrt{2},0)$, $u\in M_p(G)$ is given by:

$u(\sigma)=\langle\rho(\sigma)\xi,\xi\rangle$.

This has $u(\delta^e)=1$ (duh), $u(\delta^{(12)})=-1$, and otherwise $u(\sigma)=-\frac12 \text{sign}(\sigma)$.

The $p_0,\,p_1$ above is still a cyclic partition of unity… but is the walk irreducible?

The easiest way might be to look for a subharmonic $p$. This is way easier… with $\alpha_\sigma=1$ it is easy to construct non-trivial subharmonics… not with this $u$. It is straightforward to show there are no non-trivial subharmonics and so $u$ is irreducible, periodic, but $p_0$ is not a quantum subgroup.

It also means, in conjunction with work I’ve done already, that I have my result:

Definition Let $G$ be a finite quantum group. A state $\nu\in M_p(G)$ is concentrated on a cyclic coset of a proper quasi-subgroup if there exists a pair of projections, $p_0\neq p_1$, such that $\nu(p_1)=1$, $p_0$ is a group-like projection, $T_\nu(p_1)=p_0$ and there exists $d\in\mathbb{N}$ ($d>1$) such that $T_\nu^d(p_1)=p_1$.

## (Finally) The Ergodic Theorem for Random Walks on Finite Quantum Groups

A random walk on a finite quantum group is ergodic if and only if the driving probability is not concentrated on a proper quasi-subgroup, nor on a cyclic coset of a proper quasi-subgroup.