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In this post here, I outlined some things that I might want to prove. Very bottom of that list was:

Extending the Upper Bound Lemma to the non-Kac case. As I speak, this is beyond what I am capable of. This also requires work on the projection and quantum total variation distances (i.e. show they are equal in this larger category).

After that post, Simeng Wang wrote with the, for me, exciting news that he had proven the Upper Bound Lemma in the non-Kac case, and had an upcoming paper with Amaury Freslon and Lucas Teyssier. At first glance the paper is an intersection of the work of Amaury in random walks on compact quantum groups, and Lucas’ work on limit profiles, a refinement in the understanding of how the random transposition random walk converges to uniform. The paper also works with continuous-time random walks but I am going to restrict attention to what it does with random walks on $S_N^+$.

## Introduction

For the case of a family of Markov chains $(X_N)_{N\geq 1}$ exhibiting the cut-off phenomenon, it will do so in a window of width $w_N$ about a cut-off time $t_N$, in such a way that $w_N/t_N\rightarrow 0$, and, where $d_N(t)$ is the ‘distance to random’ at time $t$, $d_N(t_N+cw_N)$ will be close to one for $c<0$ and close to zero for $c>0$. The cutoff profile of the family of random walks is a continuous function $f:\mathbb{R}\rightarrow[0,1]$ such that as $N\rightarrow \infty$,

$\displaystyle d_N(t_N+cw_N)\rightarrow f(c)$.

I had not previously heard about such a concept, but the paper gives a number of examples in which the analysis had been carried out. Lucas however improved the Diaconis–Shahshahani Upper Bound Lemma and this allowed him to show that the limit profile for the random transpositions random walk is given by:

$\displaystyle d_{\text{TV}}(\text{Poi}[1+e^{-c}],\text{Poi}[1])$

Without looking back on Lucas’ paper, I am not sure exactly how this $d_{\text{TV}}$ works… I will guess it is, where $\xi_1\sim \text{Poi}[1+e^{-c}]$ and $\xi_2\sim \text{Poi}[2]$, and so:

$\displaystyle f_{\text{RT}}(c)=\frac{1}{2}\sum_{k=0}^{\infty}\left|\mathbb{P}[\xi_1=k]-\mathbb{P}[\xi_2=k]\right|$,

and I get $f(0)\approx 0.330$, $f(2)\approx 0.0497$ and $f(-2)\approx 0.949$ on CAS. Looking at Lucas’ paper thankfully this is correct.

The article confirms that Lucas’ work is the inspiration, but the study will take place with infinite compact quantum groups. The representation theory carries over so well from the classical to quantum case, and it is representation theory that is used to prove so many random-walk results, that it might have been and was possible to study limit profiles for random walks on quantum groups.

More importantly, technical issues which arise as soon as $N>4$ disappear if the pure quantum transposition random walk is considered. This is a purely quantum phenomenon because the random walk driven only by transpositions in the classical case is periodic and does not converge to uniform. I hope to show in an upcoming work how something which might be considered a quantum transposition behaves very differently to a classical/deterministic transposition. My understanding at this point, in a certain sense (see here)) is that a quantum transposition has $N-2$ fixed points in the sense that it is an eigenstate (with eigenvalue $N-2$) of the character $\text{fix}=\sum_i u_{ii}$. I am hoping to find a dual with a quantum transposition that for example does not square to the identity (but this is a whole other story). This would imply in a sense that there is no quantum alternating group.

The paper will show that the quantum version of the ordinary random transposition random walk and of this pure random transposition walk asymptotically coincide. They will detect the cutoff at time $\frac12 N\ln N$, and find an explicit limit profile (which I might not be too interested in).

I will skip the stuff on $O_N^+$ but as there are some similarities between the representation theory of quantum orthogonal and quantum permutation groups I may have to come back to these bits.

## Quantum Permutations

### Character Theory

At this point I will move away and look at this Banica tome on quantum permutations for some character theory.

Read the rest of this entry »

I have been working for a number of months on a paper/essay based on this talk (edit: big mad long draft here). After the talk the seminar host Teo Banica suggested a number of things that the approach could be used to look at, and one of these was orbits and orbitals. These are nice, intuitive ideas introduced by Lupini, Mancinska (missing an accent), and Roberson introduced orbitals.

I went to Teo’s quantum permutations tome, and Chapter 13 (p.297) orbits and orbitals are introduced, and it is remarked that, where we are studying quantum permutation groups $\mathbb{G}< S_N^+$, a certain relation on $N\times N\times N$ is believed not to be transitive. This belief is expressed also in the fantastic nonlocal games and quantum permutations paper, as well as by Teo here.

One of the things that the paper has had me doing is using CAS to write up the magic unitaries for a number of group duals, and I said, hey, why don’t I try and see is there any counterexamples there. My study of the $\widehat{S_3} led me to believe there would be no counterexamples there. The next two to check would be $\widehat{S_4} and the dual of the quaternion group $\widehat{Q}. I didn’t get called JPQ by Professor Des MacHale for nothing… I had to look there. OK, time to explain what the hell I am talking about.

I guess ye will have to wait for the never-ending paper to see exactly how I think about quantum permutation groups… so for the moment I am going to assume that you know what compact matrix quantum groups… but maybe I can put in some of the new approach, which can be gleaned from the above talk, in bold italics. A quantum permutation group $\mathbb{G}\leq S_N^+$ is a compact matrix quantum group whose fundamental representation $u^{\mathbb{G}}\in M_N(C(\mathbb{G}))$ is a magic unitary. The relation that was believed not to be transitive is:

$(j_3,j_2,j_1)\sim_3 (i_3,i_2,i_1)\Leftrightarrow u_{j_3i_3}u_{j_2i_2}u_{j_1i_1}\neq 0$,

that is the indices are related when there is a quantum permutation $\varsigma$ that has a non-zero probability of mapping:

$(\varsigma(j_3)=i_3)\succ(\varsigma(j_2)=i_2)\succ (\varsigma(j_1)=i_1)$. (*)

This relation is reflexive and symmetric. If we work with the universal (or algebraic) level, then $e\in\mathbb{G}$ will fix all indices giving reflexivity, if a quantum permutation $\varsigma\in\mathbb{G}$ can map as per (*), it’s reverse $\varsigma^{-1}:=\varsigma\circ S$ will map, with equal probability of $\varsigma$ doing (*):

$(\varsigma^{-1}(i_3)=j_3)\succ(\varsigma^{-1}(i_2)=j_2)\succ (\varsigma^{-1}(i_1)=j_1)$,

so that $\sim_3$ is symmetric.

Now, to transitivity. We’re going to work with the algebra of functions on the dual of the quaternions, $F(\widehat{Q}):=\mathbb{C}Q$. Working here is absolutely fraught what with coefficients $i$ and $-1$ and elements of $Q$ of the same symbol. Therefore we will use the $\delta^g$ notation. Consider the following vector in $F(\widehat{Q})$:

$\displaystyle (u^{\widehat{\langle j\rangle}})_{,1}:=\frac{1}{4}\left[\begin{array}{c}\delta^1+\delta^j+\delta^{-1}+\delta^{ij}\\ \delta^1+i\delta^{j}-\delta^{-1}-i\delta^{-j} \\ \delta^1-\delta^{j}+\delta^{-1}-\delta^{-j} \\ \delta^{1}-i\delta^{j}-\delta^{-1}+i\delta^{-j}\end{array}\right]$.

This vector is the first column of a magic unitary $u^{\widehat{\langle j\rangle}}$ for $\widehat{\langle j\rangle}\cong \widehat{\mathbb{Z}_4}\cong \mathbb{Z}_4$, and the rest of the magic unitary is made by making a circulant matrix from this. Do the same with $k\in\widehat{Q}$, another magic unitary $u^{\widehat{\langle k\rangle}}$, and so we have $\widehat{Q} via:

$u^{\widehat{Q}}=\left(\begin{array}{cc}u^{\widehat{\langle j\rangle}} & 0 \\0 & u^{\widehat{\langle k\rangle}} \end{array}\right)$.

Now for the counterexample: $u^{\widehat{Q}}_{67}u^{\widehat{Q}}_{41}u^{\widehat{Q}}_{87}\neq 0$ so  $(6,4,8)\sim_3 (7,1,7)$ and $u_{78}^{\widehat{Q}}u_{14}^{\widehat{Q}}u_{78}^{\widehat{Q}}\neq0$ so $(7,1,7)\sim_3(8,4,8)$, but $u^{\widehat{Q}}_{68}u^{\widehat{Q}}_{44}u^{\widehat{Q}}_{88}=0$ so $(6,4,8)$ is not related to $(8,4,8)$ and so $\sim_3$ is not transitive.

That $u^{\widehat{Q}}_{68}u^{\widehat{Q}}_{44}u^{\widehat{Q}}_{88}=0$ is a bit of algebra, and I guess the others are too… but instead we can exhibit states $\varsigma_2,\,\varsigma_1\in S(F(\widehat{Q}))$ such that $\varsigma_2(|u^{\widehat{Q}}_{67}u^{\widehat{Q}}_{41}u^{\widehat{Q}}_{87}|^2)>0$ and $\varsigma_1(|u_{78}^{\widehat{Q}}u_{14}^{\widehat{Q}}u_{78}^{\widehat{Q}}|^2)>0$ instead. The algebra structure of $F(\widehat{Q})$ is:

$\displaystyle F(\widehat{Q})=\mathbb{C}\oplus\mathbb{C}\oplus\mathbb{C}\oplus\mathbb{C}\oplus M_2(\mathbb{C})\subset B(\mathbb{C}^6)$.

Define $\varsigma_2$ to be the vector state associated with $\xi_2:=(0,0,0,0,1/\sqrt{2},1/\sqrt{2})$. Then:

$\|u^{\widehat{Q}}_{67}u^{\widehat{Q}}_{41}u^{\widehat{Q}}_{87}(\xi_2)\|^2=\frac14$.

$\varsigma_2\in\widehat{Q}$ is a quantum permutation such that:

$\mathbb{P}[(\varsigma_2(7)=6)\succ (\varsigma_2(1)=4)\succ (\varsigma_2(7)=8)]=\frac{1}{4}$.

Similarly the vector state $\varsigma_1$ given by $\xi_1:=(0,0,0,0,0,1)$ has

$\|u^{\widehat{Q}}_{78}u^{\widehat{Q}}_{14}u^{\widehat{Q}}_{78}(\xi_1)\|^2=:\mathbb{P}[(\varsigma_1(8)=7)\succ (\varsigma_1(4)=1)\succ (\varsigma_1(8)=7)]>0$.

Now, classically we might expect that $\varsigma_2\star \varsigma_1$ (convolution) might have the property that:

$(\varsigma_2\star\varsigma_1)(|u^{\widehat{Q}}_{68}u^{\widehat{Q}}_{44}u^{\widehat{Q}}_{88}|^2)>0$,

but as we have seen the product in question is zero.

Edit: The reason this phenomenon happens is that $u_{11}^{\widehat{Q_8}}$ and $u_{55}^{\widehat{Q_8}}$ are projections to random/classical permutations! I have also found a counterexample in the Kac-Paljutkin quantum group for similar reasons.

In the paper under preparation I think I should be able to produce nice, constructive, proofs of the transitivity of $\sim_1$ and $\sim_2$, constructive in the sense that in both cases I think I can exhibit states on $C(\mathbb{G})$ that are non-zero on suitable products of $u_{ij}$, using I think the conditioning of states:

$\displaystyle\varsigma\mapsto \frac{\varsigma(u_{ij}\cdot u_{ij})}{\varsigma(u_{ij})}$.

There is also something here to say about the maximality of $S_N. All must wait for this paper though (no I don’t have a proof of this)!

In May 2017, shortly after completing my PhD and giving a talk on it at a conference in Seoul, I wrote a post describing the outlook for my research.

I can go through that post paragraph-by-paragraph and thankfully most of the issues have been ironed out. In May 2018 I visited Adam Skalski at IMPAN and on that visit I developed a new example (4.2) of a random walk (with trivial $n$-dependence) on the Sekine quantum groups $Y_n$ with upper and lower bounds sharp enough to prove the non-existence of the cutoff phenomenon. The question of developing a walk on $Y_n$ showing cutoff… I now think this is unlikely considering the study of Isabelle Baraquin and my intuitions about the ‘growth’ of $Y_n$ (perhaps if cutoff doesn’t arise in somewhat ‘natural’ examples best not try and force the issue?). With the help of Amaury Freslon, I was able to improve to presentation of the walk (Ex 4.1) on the dual quantum group $\widehat{S_n}$. With the help of others, it was seen that the quantum total variation distance is equal to the projection distance (Prop. 2.1). Thankfully I have recently proved the Ergodic Theorem for Random Walks on Finite Quantum Groups. This did involve a study of subgroups (and quasi-subgroups) of quantum groups but normal subgroups of quantum groups did not play so much of a role as I expected. Amaury Freslon extended the upper bound lemma to compact Kac algebras. Finally I put the PhD on the arXiv and also wrote a paper based on it.

Many of these questions, other questions in the PhD, as well as other questions that arose around the time I visited Seoul (e.g. what about random transpositions in $S_n^+$?) were answered by Amaury Freslon in this paper. Following an email conversation with Amaury, and some communication with Uwe Franz, I was able to write another post outlining the state of play.

This put some of the problems I had been considering into the categories of Solved, to be Improved, More Questions, and Further Work. Most of these have now been addressed. That February 2018 post gave some direction, led me to visit Adam, and I got my first paper published.

After that paper, my interest turned to the problem of the Ergodic Theorem, and in May I visited Uwe in Besancon, where I gave a talk outlining some problems that I wanted to solve. The main focus was on proving this Ergodic Theorem for Finite Quantum Groups, and thankfully that has been achieved.

What I am currently doing is learning my compact quantum groups. This work is progressing (albeit slowly), and the focus is on delivering a series of classes on the topic to the functional analysts in the UCC School of Mathematical Sciences. The best way to learn, of course, is to teach. This of course isn’t new, so here I list some problems I might look at in short to medium term. Some of the following require me to know my compact quantum groups, and even non-Kac quantum groups, so this study is not at all futile in terms of furthering my own study.

I don’t really know where to start. Perhaps I should focus on learning my compact quantum groups for a number of months before tackling these in this order?

1. My proof of the Ergodic Theorem leans heavily on the finiteness assumption but a lot of the stuff in that paper (and there are many partial results in that paper also) should be true in the compact case too. How much of the proof/results carry into the compact case? A full Ergodic Theorem for Random Walks on Compact Quantum Groups is probably quite far away at this point, but perhaps partial results under assumptions such as (co?)amenability might be possible. OR try and prove ergodic theorems for specific compact quantum groups.
2. Look at random walks on quantum homogeneous spaces, possibly using Gelfand Pair theory. Start in finite and move into Kac?
3. Following Urban, study convolution factorisations of the Haar state.
4. Examples of non-central random walks on compact groups.
5. Extending the Upper Bound Lemma to the non-Kac case. As I speak, this is beyond what I am capable of. This also requires work on the projection and quantum total variation distances (i.e. show they are equal in this larger category)

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.