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.