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Finally cracked this egg.

Preprint here.

I thought I had a bit of a breakthrough. So, consider the algebra of a functions on the dual (quantum) group $\widehat{S_3}$. Consider the projection:

$\displaystyle p_0=\frac12\delta^e+\frac12\delta^{(12)}\in F(\widehat{S_3})$.

Define $u\in M_p(\widehat{S_3})$ by:

$u(\delta^\sigma)=\langle\text{sign}(\sigma)1,1\rangle=\text{sign}(\sigma)$.

Note

$\displaystyle T_u(p_0)=\frac12\delta^e-\frac12 \delta^{(12)}:=p_1$.

Note $p_1=\mathbf{1}_{\widehat{S_3}}-p_0=\delta^0-p_0$ so $\{p_0,p_1\}$ is a partition of unity.

I know that $p_0$ corresponds to a quasi-subgroup but not a quantum subgroup because $\{e,(12)\}$ is not normal.

This was supposed to say that the result I proved a few days ago that (in context), that $p_0$ corresponded to a quasi-subgroup, was as far as we could go.

For $H\leq G$, note

$\displaystyle p_H=\frac{1}{|H|}\sum_{h\in H}\delta^h$,

is a projection, in fact a group like projection, in $F(\widehat{G})$.

Alas note:

$\displaystyle T_u(p_{\langle(123)\rangle})=p_{\langle (123)\rangle}$

That is the group like projection associated to $\langle (123)\rangle$ is subharmonic. This should imply that nearby there exists a projection $q$ such that $u^{\star k}(q)=0$ for all $k\in\mathbb{N}$… also $q_{\langle (123)\rangle}:=\mathbf{1}_{\widehat{S_3}}-p_{\langle(123)\rangle}$ is subharmonic.

This really should be enough and I should be looking perhaps at the standard representation, or the permutation representation, or $S_3\leq S_4$… but I want to find the projection…

Indeed $u(q_{(123)})=0$…and $u^{\star 2k}(q_{\langle (123)\rangle})=0$.

The punchline… the result of Fagnola and Pellicer holds when the random walk is is irreducible. This walk is not… back to the drawing board.

I have constructed the following example. The question will be does it have periodicity.

Where $\rho:S_n\rightarrow \text{GL}(\mathbb{C}^3)$ is the permutation representation, $\rho(\sigma)e_i=e_{\sigma_i}$, and $\xi=(1/\sqrt{2},-1/\sqrt{2},0)$, $u\in M_p(G)$ is given by:

$u(\sigma)=\langle\rho(\sigma)\xi,\xi\rangle$.

This has $u(\delta^e)=1$ (duh), $u(\delta^{(12)})=-1$, and otherwise $u(\sigma)=-\frac12 \text{sign}(\sigma)$.

The $p_0,\,p_1$ above is still a cyclic partition of unity… but is the walk irreducible?

The easiest way might be to look for a subharmonic $p$. This is way easier… with $\alpha_\sigma=1$ it is easy to construct non-trivial subharmonics… not with this $u$. It is straightforward to show there are no non-trivial subharmonics and so $u$ is irreducible, periodic, but $p_0$ is not a quantum subgroup.

It also means, in conjunction with work I’ve done already, that I have my result:

Definition Let $G$ be a finite quantum group. A state $\nu\in M_p(G)$ is concentrated on a cyclic coset of a proper quasi-subgroup if there exists a pair of projections, $p_0\neq p_1$, such that $\nu(p_1)=1$, $p_0$ is a group-like projection, $T_\nu(p_1)=p_0$ and there exists $d\in\mathbb{N}$ ($d>1$) such that $T_\nu^d(p_1)=p_1$.

## (Finally) The Ergodic Theorem for Random Walks on Finite Quantum Groups

A random walk on a finite quantum group is ergodic if and only if the driving probability is not concentrated on a proper quasi-subgroup, nor on a cyclic coset of a proper quasi-subgroup.

The end of the previous Research Log suggested a way towards showing that $p_0$ can be associated to an idempotent state $\int_S$. Over night I thought of another way.

Using the Pierce decomposition with respect to $p_0$ (where $q_0:=\mathbf{1}_G-p_0$),

$F(G)=p_0F(G)p_0+p_0F(G)q_0+q_0F(G)p_0+q_0F(G)q_0$.

The corner $p_0F(G)p_0$ is a hereditary $\mathrm{C}^*$-subalgebra of $F(G)$. This implies that if $0\leq b\in p_0F(G)p_0$ and for $a\in F(G)$, $0\leq a\leq b\Rightarrow a\in p_0F(G)p_0$.

Let $\rho:=\nu^{\star d}$. We know from Fagnola and Pellicer that $T_\rho(p_0)=p_0$ and $T_\rho(p_0F(G)p_0)=p_0F(G)p_0$.

By assumption in the background here we have an irreducible and periodic random walk driven by $\nu\in M_p(G)$. This means that for all projections $q\in 2^G$, there exists $k_q\in\mathbb{N}$ such that $\nu^{\star k_q}(q)>0$.

Define:

$\displaystyle \rho_n=\frac{1}{n}\sum_{k=1}^n\rho^{\star k}$.

Define:

$\displaystyle n_0:=\max_{\text{projections, }q\in p_0F(G)p_0}\left\{k_q\,:\,\nu^{\star k_q}(q)> 0\right\}$.

The claim is that the support of $\rho_{n_0}$, $p_{\rho_{n_0}}$ is equal to $p_0$.

We probably need to write down that:

$\varepsilon T_\nu^k=\nu^{\star k}$.

Consider $\rho^{\star k}(p_0)$ for any $k\in\mathbb{N}$. Note

\begin{aligned}\rho^{\star k}(p_0)&=\varepsilon T_{\rho^{\star k}}(p_0)=\varepsilon T^k_\rho(p_0)\\&=\varepsilon T^k_{\nu^{\star d}}(p_0)=\varepsilon T_\nu^{kd}(p_0)\\&=\varepsilon(p_0)=1\end{aligned}

that is each $\rho^{\star k}$ is supported on $p_0$. This means furthermore that $\rho_{n_0}(p_0)=1$.

Suppose that the support $p_{\rho_{n_0}}. A question arises… is $p_{\rho_{n_0}}\in p_0F(G)p_0$? This follows from the fact that $p_0\in p_0F(G)p_0$ and $p_0F(G)p_0$ is hereditary.

Consider a projection $r:=p_0-p_{\rho_{n_0}}\in p_0F(G)p_0$. We know that there exists a $k_r\leq n_0$ such that

$\nu^{\star k_r}(p_0-p_{\rho_{n_0}})>0\Rightarrow \nu^{\star k_r}(p_0)>\nu^{\star k_r}(p_{\rho_{n_0}})$.

This implies that $\nu^{\star k_r}(p_0)>0\Rightarrow k_r\equiv 0\mod d$, say $k_r=\ell_r\cdot d$ (note $\ell_r\leq n_0$):

\begin{aligned}\nu^{\star \ell_r\cdot d}(p_0)&>\nu^{\star \ell_r\cdot d}(p_{\rho_{n_0}})\\\Rightarrow (\nu^{\star d})^{\star \ell_r}(p_0)&>(\nu^{\star d})^{\star \ell_r}(p_{\rho_{n_0}})\\ \Rightarrow \rho^{\star \ell_r}(p_0)&>\rho^{\star \ell_r}(p_{\rho_{n_0}})\\ \Rightarrow 1&>\rho^{\star \ell_r}(p_{\rho_{n_0}})\end{aligned}

By assumption $\rho_{n_0}(p_{\rho_{n_0}})=1$. Consider

$\displaystyle \rho_{n_0}(p_{\rho_{n_0}})=\frac{1}{n_0} \sum_{k=1}^{n_0}\rho^{\star k}(p_{\rho_{n_0}})$.

For this to equal one every $\rho^{\star k}(p_{\rho_{n_0}})$ must equal one but $\rho^{\star \ell_r}(p_{\rho_{n_0}})<1$.

Therefore $p_0$ is the support of $\rho_{n_0}$.

Let $\rho_\infty=\lim \rho_n$. We have shown above that $\rho^{\star k}(p_0)=1$ for all $k\in\mathbb{N}$. This is an idempotent state such that $p_0$ is its support (a similar argument to above shows this). Therefore $p_0$ is a group like projection and so we denote it by $\mathbf{1}_S$ and $\int_S=d\mathcal{F}(\mathbf{1}_S)$!

Today, for finite quantum groups, I want to explore some properties of the relationship between a state $\nu\in M_p(G)$, its density $a_\nu$ ($\nu(b)=\int_G ba_\nu$), and the support of $\nu$, $p_{\nu}$.

I also want to learn about the interaction between these object, the stochastic operator

$\displaystyle T_\nu=(\nu\otimes I)\circ \Delta$,

and the result

$T_\nu(a)=S(a_\nu)\overline{\star}a$,

where $\overline{\star}$ is defined as (where $\mathcal{F}:F(G)\rightarrow \mathbb{C}G$ by $a\mapsto (b\mapsto \int_Gba)$).

$\displaystyle a\overline{\star}b=\mathcal{F}^{-1}\left(\mathcal{F}(a)\star\mathcal{F}(b)\right)$.

An obvious thing to note is that

$\nu(a_\nu)=\|a_\nu\|_2^2$.

Also, because

\begin{aligned}\nu(a_\nu p_\nu)&=\int_Ga_\nu p_\nu a_\nu=\int_G(a_\nu^\ast p_\nu^\ast p_\nu a_\nu)\\&=\int_G(p_\nu a_\nu)^\ast p_\nu a_\nu\\&=\int_G|p_\nu a_\nu|^2\\&=\|p_\nu a_\nu\|_2^2=\|a_\nu\|^2\end{aligned}

That doesn’t say much. We are possibly hoping to say that $a_\nu p_\nu=a_\nu$.

## Quasi-Subgroups that are not Subgroups

Let $G$ be a finite quantum group. We associate to an idempotent state $\int_S$quasi-subgroup $S$. This nomenclature must be included in the manuscript under preparation.

As is well known from the GNS representation, positive linear functionals can be associated to closed left ideals:

$\displaystyle N_{\rho}:=\left\{ f\in F(G):\rho(|f|^2)=0\right\}$.

In the case of a quasi-subgroup, $S\subset G$, my understanding is that by looking at $N_S:=N_{\int_S}$ we can tell if $S$ is actually a subgroup or not. Franz & Skalski show that:

Let $S\subset G$ be a quasi-subgroup. TFAE

• $S\leq G$ is a subgroup
• $N_{\int_S}$ is a two-sided or self-adjoint or $S$ invariant ideal of $F(G)$
• $\mathbf{1}_Sa=a\mathbf{1}_S$

I want to look again at the Kac & Paljutkin quantum group $\mathfrak{G}_0$ and see how the Pal null-spaces $N_{\rho_6}$ and $N_{\rho_7}$ fail these tests. Both Franz & Gohm and Baraquin should have the necessary left ideals.

### The Pal Null-Space $N_{\rho_6}$

The following is an idempotent probability on the Kac-Paljutkin quantum group:

$\displaystyle \rho_6(f)=2\int_{\mathfrak{G}_0}f\cdot (e_1+e_4+E_{11})$.

Hence:

$N_{\rho_6}=\langle e_1,e_3,E_{12},E_{22}\rangle$.

If $N_{\rho_6}$ were two-sided, $N_{\rho_6}F(\mathfrak{G}_0)\subset N_{\rho_6}$. Consider $E_{21}\in F(\mathfrak{G}_0)$ and

$E_{12}E_{21}=E_{11}\not\in N_{\rho_6}$.

We see problems also with $E_{12}$ when it comes to the adjoint $E_{12}^{\ast}=E_{21}\not\in N_{\rho_6}$ and also $S(E_{12})=E_{21}\not\in N_{\rho_6}$. It is not surprise that the adjoint AND the antipode are involved as they are related via:

$S(S(f^\ast)^\ast)=f$.

In fact, for finite or even Kac quantum groups, $S(f^\ast)=S(f)^\ast$.

Can we identity the support $p$? I think we can, it is (from Baraquin)

$p_{\rho_6}=e_1+e_4+E_{11}$.

This does not commute with $F(G)$:

$E_{21}p_{\rho_6}=E_{21}\neq 0=p_{\rho_6}E_{21}$.

The other case is similar.

Back before Christmas I felt I was within a week of proving the following:

Ergodic Theorem for Random Walks on Finite Quantum Groups

A random walk on a finite quantum group $G$ is ergodic if and only if $\nu$ is not concentrated on a proper quasi-subgroup, nor the coset of a ?normal ?-subgroup.

The first part of this conjecture says that if $\nu$ is concentrated on a quasi-subgroup, then it stays concentrated there. Furthermore, we can show that if the random walk is reducible that the Césaro limit gives a quasi-subgroup on which $\nu$ is concentrated.

The other side of the ergodicity coin is periodicity. In the classical case, it is easy to show that if the driving probability is concentrated on the coset of a proper normal subgroup $N\lhd G$, that the convolution powers jump around a cyclic subgroup of $G/N$.

One would imagine that in the quantum case this might be easy to show but alas this is not proving so easy.

I am however pushing hard against the other side. Namely, that if the random walk is periodic and irreducible, that the driving probability in concentrated on some quasi-normal quasi-subgroup!

The progress I have made depends on work of Fagnola and Pellicer. They show that if the random walk is irreducible and periodic that there exists a partition of unity $\{p_0,p_1,\dots,p_{d-1}\}$ such that $\nu^{\star k}$ is concentrated on $p_{k\mod d}$.

This cyclic nature suggests that $p_0$ might be equal to $\mathbf{1}_N$ for some $N\lhd G$ and perhaps:

$\Delta(p_i)=\sum_{j=0}^{d-1}p_{i-j}\otimes p_j,$

and perhaps there is an isomorphism $G/N\cong C_d$. Unfortunately I have been unable to progress this.

What is clear is that the ‘supports’ of the $p_i$ behave very much like the cosets of proper normal subgroup $N\lhd G$.

As the random walk is assumed irreducible, we know that for any projection $q\in 2^G$, there exists a $k_q\in \mathbb{N}$ such that $\nu^{\star k_q}(q)\neq 0$.

Playing this game with the Haar element, $\eta\in 2^G$, note there exists a $k_\eta\in\mathbb{N}$ such that $\nu^{k_\eta}(\eta)>0$.

Let $\overline{\nu}=\nu^{\star k_\eta}$. I have proven that if $\mu(\eta)>0$, then the convolution powers of $\mu\in M_p(G)$ converge. Convergence is to an idempotent. This means that $\overline{\nu}^{\star k}$ converges to an idempotent $\overline{\nu}_\infty$, and so we have a quasi-subgroup corresponding to it, say $\overline{p}$.

The question is… does $\overline{p}$ coincide with $p_0$?

If yes, is there any quotient structure by a quasi-subgroup? Is there a normal quasi-subgroup that allows such a structure?

Is $\overline{p}$ a subgroup? Could it be a normal subgroup?

As nice as it was to invoke the result that if $e$ is in the support of $\nu$, then the convolution powers of $\nu$ converge, by looking at those papers which cite Fagnola and Pellicer we see a paper that gives the same result without this neat little lemma.

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.

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.

Slides of a talk given at Munster Groups 2019, WIT.

Abstract: It is a folklore theorem that necessary and sufficient conditions for a random walk on a finite group to converge in distribution to the uniform distribution – “to random” – are that the driving probability is not concentrated on a proper subgroup nor the coset of a proper normal subgroup. This is the Ergodic Theorem for Random Walks on Finite Groups. In this talk we will outline the rarely written down proof, and explain why, for example, adjacent transpositions can never mix up a deck of cards. From here we will, in a very leisurely and natural fashion, introduce (and motivate the definition of) finite quantum groups, and random walks on them. We will see how the group algebra of a finite group is the algebra of functions on a finite quantum group. Freslon has very recently proved the Ergodic Theorem in this setting, and we present ongoing work towards an Ergodic Theorem in the more general finite quantum group setting; a result that would generalise both the folklore and group algebra Ergodic Theorems.

### Introduction

Every finite quantum group has finite dimensional algebra of functions:

$\displaystyle F(G)=\bigoplus_{j=1}^m M_{n_j}(\mathbb{C})$.

At least one of the factors must be one-dimensional to account for the counit $\varepsilon:F(G)\rightarrow \mathbb{C}$, and if this factor is denoted $\mathbb{C}e_1$, the counit is given by the dual element $e^1$. There may be more and so reorder the index $j\mapsto i$ so that $n_i=1$ for $i=1,\dots,m_1$, and $n_i>1$ for $i>m_1$:

$\displaystyle F(G)=\left(\bigoplus_{i=1}^{m_1} \mathbb{C}e_{i}\right)\oplus \bigoplus_{i=m_1+1}^m M_{n_i}(\mathbb{C})=:A_1\oplus B$,

Denote by $M_p(G)$ the states of $F(G)$. The pure states of $F(G)$ arise as pure states on single factors.

In the case of the Kac-Paljutkin and Sekine quantum groups, the convolution powers of pure states exhibit a periodicity of sorts. Recall for these quantum groups that $B$ consists of a single matrix factor.

In these cases, for pure states of the form $e^i$, that is supported on $A_1$ (and we can say a little more than is necessary), the convolution remains supported on $A_1$ because

$\Delta(A_1)\subset A_1\otimes A_1+B\otimes B$.

If we have a pure state $\nu$ supported on $B=M_{\sqrt{\dim B}}(\mathbb{C})$, then because

$\Delta(B)\subset A_1\otimes B+B\otimes A_1$,

then $\nu\star\nu$ must be supported on, because of $\Delta(A_1)\subset A_1\otimes A_1+B\otimes B$, $A_1$.

Inductively all of the $\nu^{\star 2k}$ are supported on $A_1$ and the $\nu^{\star 2k+1}$ are supported on $B$. This means that the convolutions powers of a pure state, in these cases, cannot converge to the Haar state.

The question is, do the results above about the image of $A_1$ and $B$ under the coproduct hold more generally? I believe that the paper of Kac and Paljutkin shows that this is the case whenever $B$ consists of a single factor… but does it hold more generally?

To find out we go back and do some sandboxing with the paper of Kac and Paljutkin. Which is a pleasure because that paper is beautiful. The blue stuff is my own scribbling.

## Finite Ring Groups

Let $G$ be a finite quantum group with notation on the algebra of functions as above. Note that $A_1$ is commutative. Let

$p=\sum_{i=1}^{m_1}e_i$,

which is a central idempotent.

### Lemma 8.1

$S(p)=p$.

Proof: If $S(\mathbf{1}_G-p)p\neq 0$, then for some $i>m_1$, and $f\in M_{n_i}(\mathbb{C})$, the mapping $f\mapsto S(f)p$ is a non-zero homomorphism from $M_{n_i}(\mathbb{C})$ into commutative $A_1$ which is impossible.

If $S(\mathbf{1}_G-p)p=g\neq 0$, then one of the $S(I_{n_i})\in A_1\oplus B$, with ‘something’ in $A_1$. Using the centrality and projectionality of $p$, we can show that the given map is indeed a homomorphism.

It follows that $S(p)p=p\Rightarrow S(S(p)p)=S(p)=S(p)p=S(p)$, and so $p=S(p)$ $\bullet$

### Lemma 8.2

$(p\otimes p)\Delta(p)=p\otimes p$

Proof: Suppose that $(p\otimes p)\Delta(f)=b$ for some non-commutative $f\in M_{n_i}(\mathbb{C})$. This means that there exists an index $k$ such that $f_{(1)_k}\otimes f_{(2)_k}\in A_1\otimes A_1$. Then for that factor,

$f\mapsto \Delta(f)(p\otimes p)$

is a non-null homomorphism from the non-commutative into the commutative.

We see that $(p\otimes p)\Delta(f)=0$ for all $f\in B$. Putting $a=\mathbf{1}-p$ we get the result $\bullet$

The following says that $p$ is a group-like projection. We know from previous work that if a state is supported on a group-like projection that it will remain supported on it. In particular, any state supported on $A_1$ will remain there.

### Lemma 8.3

$(p\otimes \mathbf{1}_G)\Delta(p)=p\otimes p=(\mathbf{1}_G\otimes p)\Delta(p)$.

Proof: Since $\Delta$ is a homomorphism, $\Delta(p)$ is an idempotent in $F(G)\otimes F(G)$I  do not understand nor require the rest of the proof.

### Lemma 8.4

$A_1=F(G_1)$ is the algebra of functions on finite group with elements $i=1,\dots,m_1$, and we write $e_i=\delta_i$. The coproduct is given by $(p\otimes p)\Delta$.

We have:

$(p\otimes p)\Delta(e_i)=\sum_{t\in G_1}\delta_{it^{-1}}\otimes \delta_t$,

$S(\delta_i)=\delta_{i^{-1}}$,

$\varepsilon(e_i)=\delta_{i,1}$,

as $e_1=\delta_e$.

The element $\Delta(\delta_i)$ is a sum of four terms, lying in the subalgebras:

$A_1\otimes A_1,\,B\otimes B,\,A_1\otimes B,\,B\otimes A_1$.

We already know what is going on with the first summand. Denote the second by $P_i$. From the group-like-projection property, the last two summands are zero, so that

$\Delta(\delta_i)=\sum_{t\in G_1}\delta_{t}\otimes \delta_{t^{-1}i}+P_i$.

Since the $\delta_i$ are symmetric ($\delta_i^*=\delta_i$) mutually orthogonal idempotents, $P_i$ has similar properties:

$P_i^*=P_i,\,P_i^2=P_i,\,P_iP_j=0$

for $i\neq j$.

At this point Kac and Paljutkin restrict to $B=M_{n_{i+1}}(\mathbb{C})$, that is there is only one summand. Here we try to keep arbitrarily (finitely) many summands in $B$.

Let the summand $M_{n_i}(\mathbb{C})$ have matrix units $E_{mn}^i$, where $m,n=1,\dots,n_i$Kac and Paljutkin now do something which I think is a little dodgy, but basically that the integral over $G$ is equal on each of the $\delta_i$, equal on each of the $E_{mm}^i$, and then zero off the diagonal.

It does follow from above that each $P_i\in B\otimes B$ is a projection.

Now I am stuck!