In the case of a finite classical group $G$, we can show that if we have i.i.d. random variables $\zeta_i\sim\nu\in M_p(G)$, that if $\text{supp }\nu\subset Ng$, for $Ng$ a coset of a proper normal subgroup $N\rhd G$, that the random walk on $G$ driven by $\nu$, the random variables:

$\xi_k=\zeta_k\cdots \zeta_1$,

exhibits a periodicity because

$\xi_k\in Ng^{k}$.

This shows that a necessary condition for ergodicity of a random walk on a finite classical group $G$ driven by $\nu\in M_p(G)$ is that the support of $\nu$ not be concentrated on the coset of a proper normal subgroup.

I had hoped that something similar might hold for the case of random walks on finite quantum groups but alas I think I have found a barrier.

Let $N\rhd G$ be a proper normal subgroup so that $G/N$ is a finite group. Consider the pure state $\delta^{Ng}\in M_p(G/N)$. In the classical case, the convolution of pure states remains pure. This is because for a finite group $H$, the pure states are precisely the delta measures and the convolution of delta measures is a delta measure again:

$\delta^{g_1}\star \delta^{g_2}=\delta^{g_1g_2}$.

In particular convolution powers of pure states remain pure.

The algebra $F(G/N)$ is a subalgebra of $F(G)$ consisting of functions constant on cosets. Suppose $\nu\in M_p(G)$ is concentrated on a coset $Ng$. Then its support projection $P_{\nu}$ is less than the support projection of $\delta^{Ng}$:

$P_\nu\leq P_{\delta^{Ng}}$.

In the classical case, we have that:

$P_{\nu^{\star k}}\leq P_{\delta^{Ng^k}}$.

Presumably this is a special case of, for states $\nu,\,\mu\in M_p(G)$,

$P_{\nu}\leq P_{\mu}\Rightarrow P_{\nu\star \nu}\leq P_{\mu\star \mu}$.

What we are looking at here is $\nu$ which is not pure being comparable to $\delta^{Ng}$ which is pure. The state $\nu^{\star k}$ isn’t pure but it’s support is less than the support of a pure state on $F(G/N)$ and so not ergodic.

An obvious candidate for a probability on a finite quantum group $G$ to be concentrated on the coset of a proper normal quantum subgroup $N$… well what should be the analogue of $\delta^{Ng}$ for a quantum quotient group?

Something that comes to mind is that a pure state on $F(G/N)$ would correspond to $\delta^{Ng}$. Alternatively $\delta_{Ng}$ might be a minimal projection in the algebra of functions.

At any rate, we would want the convolution of pure states to remain pure. However this is not true in general in the quantum case.

Take the algebra of functions on a quantum group given by:

$F(G)=\mathbb{C}S_3\otimes F(S_3)$.

This has algebra:

$F(G)\cong \mathbb{C} S_3\otimes F(S_3)\cong (\mathbb{C}^2\oplus M_2(\mathbb{C}))\otimes \mathbb{C}^6\cong \mathbb{C}^{12}\oplus \bigoplus_{\sigma\in S_3}M_{2,\sigma}(\mathbb{C})$.

Consider the pure state $\rho$ concentrated on the $M_{2,(12)}(\mathbb{C})$ factor, say as simple as $\frac12(E^{11}+E^{22})$.

Then the pure state $\rho\star \rho$ is no longer concentrated on a single summand but rather across the factors $M_{2,e}(\mathbb{C})$, $\mathbb{C}_{1,e}$, $\mathbb{C}_{2,e}$.

This means that $\rho\star \rho$ is not pure.

While it may be the case that for any state $\nu\in M_p(G)$ such that

$P_\nu\leq P_{\rho}$ and $P_{\nu^{\star k}}\leq P_{\rho^{\star k}}$,

we can no longer say that $\rho^{\star k}$ is pure and so cannot know that $P_{\rho^{\star k}}< P_{\pi}= \mathbf{1}_G$.

Presumably it is possible to find a quantum group $H$ such that $G$ is a quotient of $H$ by a normal quantum subgroup $N$ so that

$F(G)\cong F(H/N)$,

and $F(H/N)$ a subspace of $F(H)$.

Now take an element of $\nu\in M_p(H)$ such that

$P_{\nu}\leq P_{\rho}$.

Because $\rho$ is pure, classically we know that $\nu$ is not ergodic. We cannot be sure of this now in the quantum case.

We cannot argue that $P_{\rho^{\star k}}$ cannot be the whole of $G$.