Taken from Random Walks on Finite Quantum Groups by Franz & Gohm.

The most important special case of the construction from here is obtained when we choose $B=A$ and $\beta=\Delta$. Then we have a random walk on a finite group $A$. Let us first show that this indeed a generalisation of a left-invariant random walk (I must be careful to remember this — I have always worked with right-invariant walks that multiply on the left: these multiply on the right.) Using the coassociativity of $\Delta$ we see that the transition operator $T_\phi=(I_A\otimes\phi)\circ\Delta$ satisfies the formula ( $\checkmark$) $\Delta\circ T_\phi=(I_A\otimes T_\phi)\circ\Delta$.

Suppose now that $B=A$ consists of functions on a finite group $G$ and $\beta=\Delta$ is the comultiplication which encodes the group multiplication, i.e. $\displaystyle\Delta\left(\mathbf{1}_{\{g\}}\right)=\sum_{h\in G}\mathbf{1}_{\{gh^{-1}\}}\otimes\mathbf{1}_{\{h\}}=\sum_{h\in G}\mathbf{1}_{\{h^{-1}\}}\otimes\mathbf{1}_{\{hg\}}$.

We also have $\displaystyle T_\phi\left(\mathbf{1}_{\{g\}}\right)=\sum_{h\in G}p(h,g)\mathbf{1}_{\{h\}}$,

where $[p(h,g)]$ is the stochastic matrix.

This calculation makes perfect sense when $p(h,g)=\phi\left(\mathbf{1}_{h^{-1}g}\right)$ and since we’ve said nothing about what $\phi$ should be this makes perfect sense — and with linearity we get all the properties of the stochastic operator we could possibly want.

Now calculate $\displaystyle (\Delta\circ T_\phi)\mathbf{1}_{\{g\}}=\Delta\left(\sum_{h\in G}p(h,g)\mathbf{1}_{\{h\}}\right)=\sum_{t\in G}\mathbf{1}_{\{t^{-1}\}}\otimes\sum_{h\in G}p(h,g)\mathbf{1}_{\{th\}}$, $\displaystyle[(I_{F(G)}\otimes T_\phi)\circ\Delta]\mathbf{1}_{\{g\}}=(I_{F(G)}\otimes T_\phi)\sum_{t\in G}\mathbf{1}_{\{t^{-1}\}}\otimes\mathbf{1}_{\{tg\}}$ $\displaystyle =\sum_{t\in G}\mathbf{1}_{\{t^{-1}\}}\otimes \sum_{s\in G} p(s,tg)\mathbf{1}_{\{s\}}$.

Now reindex the second sum here $s\rightarrow h$ and at the same time map $s\rightarrow th$ and we may then conclude that $p(h,g)=p(th,tg)$ for all $g,\,h\,t\in G$. This is the left invariance of the random walk.

For random walks on a finite quantum groups there are some natural special choices for the initial distribution $\psi$. On the one hand, one may choose $\psi=\varepsilon$ (the counit) which in the commutative case (i.e. for a group) corresponds to starting at the identity. Then the time evolution of the distributions is given by $\varepsilon\star\phi^{\star n}=\star^{\star n}$. In other words, we get a convolution semigroup of states.

Inasmuch as I can tell, we have $\psi\star \phi=(\psi\otimes\phi)\circ\Delta$.

On the other hand, stationarity of the random walk can be obtained if $\psi$ is chosen such that $\psi\star\phi=\star$. In particular, we may choose the unique Haar state $\eta$ of the finite quantum group $A$.

## Proposition 4.1

The random walks on a finite quantum group are stationary for all choices of $\phi$ if and only if $\psi=\eta$.

Proof : This follows from Proposition 3.2 together with the fact that the Haar state is characterised by its right invariance.