*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 and . Then we have a *random walk on a finite group . *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 we see that the transition operator satisfies the formula ()

.

Suppose now that consists of functions on a finite group and is the comultiplication which encodes the group multiplication, i.e.

.

We also have

,

where is the stochastic matrix.

This calculation makes perfect sense when and since we’ve said nothing about what should be this makes perfect sense — and with linearity we get all the properties of the stochastic operator we could possibly want.

Now calculate

,

.

Now reindex the second sum here and at the same time map and we may then conclude that for all . 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 . On the one hand, one may choose (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 . In other words, we get a convolution semigroup of states.

Inasmuch as I can tell, we have

.

On the other hand, stationarity of the random walk can be obtained if is chosen such that . In particular, we may choose the unique Haar state of the finite quantum group .

## Proposition 4.1

*The random walks on a finite quantum group are stationary for all choices of if and only if .*

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

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