In the hope of gleaning information for the study of aperiodic random walks on (finite) quantum groups, which I am struggling with here, by couching Freslon’s Proposition 3.2, in the language of Fagnola and Pellicer, and in the language of my (failed) attempts (see here, here, and here) to find the necessary and sufficient conditions for a random walk to be aperiodic. It will be necessary to extract the irreducibility and aperiodicity from Freslon’s rather ‘unilateral’ result.

Well… I have an inkling that because dual groups satisfy what I would call the condition of abelianness (under the ‘quantisation’ functor), all (quantum) subgroups are normal… this is probably an obvious thing to write down (although I must search the literature) to ensure it is indeed known (or is untrue?). Edit: Wang had it already, see the last proposition here.

Let $F(G)$ be a the algebra of functions on a finite classical (as opposed to quantum) group $G$. This has the structure of both an algebra and a coalgebra, with an appropriate relationship between these two structures. By taking the dual, we get the group algebra, $\mathbb{C}G=:F(\widehat{G})$. The dual of the pointwise-multiplication in $F(G)$ is a coproduct for the algebra of functions on the dual group $\widehat{G}$… this is all well known stuff.

Recall that the set of probabilities on a finite quantum group is the set of states $M_p(G):=\mathcal{S}(F(G))$, and this lives in the dual, and the dual of $F(\widehat{G})$ is $F(G)$, and so probabilities on $\widehat{G}$ are functions on $G$. To be positive is to be positive definite, and to be normalised to one is to have $u(\delta^e)=1$.

The ‘simplicity’ of the coproduct,

$\Delta(\delta^g)=\delta^g\otimes\delta^g$,

means that for $u\in M_p(\widehat{G})$,

$(u\star u)(\delta^g)=(u\otimes u)\Delta(\delta^g)=u(\delta^g)^2$,

so that, inductively, $u^{\star k}$ is equal to the (pointwise-multiplication power) $u^k$.

The Haar state on $\widehat{G}$ is equal to:

$\displaystyle \pi:=\int_{\hat{G}}:=\delta_e$,

and therefore necessary and sufficient conditions for the convergence of $u^{\star k}\rightarrow \pi$ is that $u$ is strict. It can be shown that for any $u\in M_p(G)$ that $|u(\delta^g)|\leq u(\delta^e)=1$. Strictness is that this is a strict inequality for $g\neq e$, in which case it is obvious that $u^{\star k}\rightarrow \delta_e$.

Here is a finite version of Freslon’s result which holds for discrete groups.

### Freslon’s Ergodic Theorem for (Finite) Group Algebras

Let $u\in M_p(\widehat{G})$ be a probability on the dual of finite group. The random walk generated by $u$ is ergodic if and only if $u$ is not-concentrated on a character on a non-trivial subgroup $H\subset G$.

Freslon’s proof passes through the following equivalent condition:

The random walk on $\widehat{G}$ driven by $u\in M_p(\widehat{G})$ is not ergodic if $u$ is bimodularwith respect to a non-trivial subgroup $H\subset G$, in the sense that

$\displaystyle u(\underbrace{\delta^g\delta^h}_{=\delta^{gh}})=u(\delta^g)u(\delta^h)=u\left(\underbrace{\delta^h\delta^g}_{=\delta^{hg}}\right)$.

Before looking at the proof proper, we might note what happens when $G$ is abelian, in which case $\widehat{G}$ is a classical group, the set of characters on $G$.

To every positive definite function $u\in M_p(\widehat{G})$, we can associate a probability $\nu_u\in M_p(\widehat{G})$ such that:

$\displaystyle u(\delta^g)=\sum_{\chi\in\hat{G}} \chi(\delta^g) \nu_u(\chi)$.

This is Bochner’s Theorem for finite abelian groups. This implies that positive definite functions on finite abelian groups are exactly convex combinations of characters.

Freslon’s condition says that to be not ergodic, $u$ must be a character on a non-trivial subgroup $H\subset G$. Such characters can be extended in $[G\,:\,H]$ ways.

Therefore, if $u$ is not ergodic, $u_{\left|H\right.}=\eta\in \widehat{H}$.

For $h\in H$, we have

$\displaystyle u(h)=\sum_{\chi\in\widehat{G}}\chi(h)\nu_u(\chi)$,

dividing both sides by $u(h)=\eta(h)\neq 0$ yields:

$\displaystyle\sum_{\chi \in \widehat{G}} (\eta^{-1}\chi)(h)\nu_u (\chi)=1$.

As $\nu_u\in M_p(\widehat{G})$, and $(\eta^{-1}\chi)(h)\in \mathbb{T}$, this implies that $\nu_u$ is supported on characters such that, for all $h\in H$:

$\eta^{-1}(h)\chi(h)=1\Rightarrow \chi=\eta\tilde{\chi}$,

such that $\tilde{\chi}(H)=\{1\}$. The set of such $\tilde{\chi}$ is the annihilator of $H$ in $\widehat{G}$, and it is a subgroup. Therefore $\nu_u$ is concentrated on the coset of a normal subgroup (as all subgroups of an abelian group are normal).

This, via Pontragin duality, is not looking at the ‘support’ of $u$, but rather of $\nu_u$. Although we denote $\mathbb{C}G=:F(\widehat{G})$, and when $G$ is abelian, $\widehat{G}$ is a group (unnaturally, of characters) isomorphic to $G$. Is it the case though that,

$\Delta(\chi)=\chi\otimes\chi$

gives the same object in as

$\displaystyle\Delta(\chi)=\sum_{g\in G}\chi(\delta^g)\Delta(\delta_g)$

$\displaystyle =\sum_{g\in G}\chi(\delta^g)\sum_{t\in G}\delta_{gt^{-1}}\otimes \delta_t$?

Well… of course this is true because $\chi(gh)=\chi(g)\chi(h)$.

We could proceed to look at Fagnola & Pellicer’s work but first let us prove Freslon’s result, hopefully in the finite case the analysis disappears…

Proof: Assume that $u$ is not strict and let

$\Lambda:=|u|^{-1}(\{1\})$.

There exists a unitary representation $\Phi:G\rightarrow B(H)$ and a unit vector $\xi$ such that

$u(g)=\langle \Phi(g)\xi,\xi\rangle$

Cauchy-Schwarz implies that

$|u(g)|\leq \|\Phi(g)\xi\|\|\xi\|=\|\xi\|^2$.

If $h$ is not strict there is an $h$ such that this is an inequality and so $\Phi(h)\xi$ is colinear to $\xi$, it follows that $\Phi(h)\xi=u(h)\xi$.

This implies for $h\in \Lambda$ and $g\in G$:

$|u(gh)|=|\langle \Phi(gh)\xi,\xi\rangle|=|u(h)||\langle \Phi(g)\xi,\xi\rangle|=|u(g)|$,

and so $\Lambda$ is closed under multiplication. Also $u(g^{-1})=\overline{u(g)}$ and so $\Lambda$ and so $\Lambda$ is a subgroup. It follows that $u$ is a character on $\Lambda$, which is not trivial because $u$ is not strict.

I don’t really need to go through the third equivalent condition. If $u$ coincides with a character on a subgroup $\Lambda$, for $h\in \Lambda$

$|u(h)|^2=u(h)\overline{u(h)}=u(h)u(h^{-1})=u(e)=1$,

and so $u$ is not strict $\bullet$

Now let us look at the language of Fagnola and Pellicer. What is a projection in $\mathbb{C}G$? First note the involution in $\mathbb{C}G$ is $(\delta^g)^*=\delta^{g^{-1}}$. The second multiplication is the convolution. This means projections in the algebra are symmetric with respect to the group inverses and they are also idempotents. They are actually equal to Haar states on finite subgroups.

I think periodicity is also wrapped up in Freslon’s result as I think all subgroups of dual groups are normal. Perhaps, oddly, not being concentrated on a subgroup means that the positive function (probability on the dual) is one on that subgroup…

Well… let us start with irreducible. Suppose $u$ fails to be ergodic because it is irreducible. This means there is a projection $p_H=\int_H$ such that that $P_u(p_H)=p_H$ (and support $u$ less than $p_H$?)

Let us look at the first condition:

$P_u(p_H)=(u\otimes I)\Delta(p_H)=\cdots=\frac{1}{|H|}\sum_{h\in H}u(h)\delta^h=p_H\Rightarrow u_{\left|H\right.}=1$.

What now is the support of $u$? Well… some work I have done offline shows that the special projections, the group-like projections, correspond to to $\pi_H$ for $H$ a subgroup of $G$.  If $u$ is reducible, it is concentrated on such a quasi-subgroup, and this means that $u$ coincides with a trivial character on $H$. In terms of Fagnola Pellicer, $P_u(\pi_H)=\pi_H$.

Now let us tackle aperiodicity. It is going to correspond, I think, with being concentrated on a ‘coset’ of a non-trivial character on $H$

Well, we can show that if $u$ is periodic, there is a subset $S\subset G$ such that $u(s)=e^{2\pi i a_s/d}$ for all $s\in S$. We can use Freslon’s proof to show that $S$ is in a subgroup on which $|u|=1$.

Now what I want to do is put this in the language of ‘inclusion’ matrices… but the inclusions for cocommutative quantum groups are trivial so no go…

We can reduce Freslon’s conditions down to irreducible and aperiodic: not coinciding with a trivial character, and not coinciding with a character.