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An expository piece, the watered down version of the madness here… should be on the Arxiv Thursday.

What started about ten months ago as a technical question to an expert, led to a talk, and led to me producing this weird production here.

Now that it is complete, although I really like all its contents (well except for the note to reader and introduction I spilled out very hastily), I can see on reflection it represents rather than a cogent piece of mathematics, almost a log of all the things I have learnt in the process of writing it. It also includes far too much speculation and conjecture. So I am going to post it here and get to work on editing it down to something a little more useful and cogent.

EDIT: Edited down version here.

Giving a talk 17:00, September 1 2020:

See here for more.

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 $a_\nu p_\nu=a_\nu$.

In my pursuit of an Ergodic Theorem for Random Walks on (probably finite) Quantum Groups, I have been looking at analogues of Irreducible and Periodic. I have, more or less, got a handle on irreducibility, but I am better at periodicity than aperiodicity.

The question of how to generalise these notions from the (finite) classical to noncommutative world has already been considered in a paper (whose title is the title of this post) of Fagnola and Pellicer. I can use their definition of periodic, and show that the definition of irreducible that I use is equivalent. This post is based largely on that paper.

## Introduction

Consider a random walk on a finite group $G$ driven by $\nu\in M_p(G)$. The state of the random walk after $k$ steps is given by $\nu^{\star k}$, defined inductively (on the algebra of functions level) by the associative

$\nu\star \nu=(\nu\otimes\nu)\circ \Delta$.

The convolution is also implemented by right multiplication by the stochastic operator:

$\nu\star \nu=\nu P$,

where $P\in L(F(G))$ has entries, with respect to a basis $(\delta_{g_i})_{i\geq 1}$ $P_{ij}=\nu(g_jg_{i^{-1}})$. Furthermore, therefore

$\nu^{\star k}=\varepsilon P^k$,

and so the stochastic operator $P$ describes the random walk just as well as the driving probabilty $\nu$.

The random walk driven by $\nu$ is said to be irreducible if for all $g_\ell\in G$, there exists $k\in\mathbb{N}$ such that (if $g_1=e$) $[P^k]_{1\ell}>0$.

The period of the random walk is defined by:

$\displaystyle \gcd\left(d\in\mathbb{N}:[P^d]_{11}>0\right)$.

The random walk is said to be aperiodic if the period of the random walk is one.

These statements have counterparts on the set level.

If $P$ is not irreducible, there exists a proper subset of $G$, say $S\subsetneq G$, such that the set of functions supported on $S$ are $P$-invariant.  It turns out that $S$ is a proper subgroup of $G$.

Moreover, when $P$ is irreducible, the period is the greatest common divisor of all the natural numbers $d$ such that there exists a partition $S_0, S_1, \dots, S_{d-1}$ of $G$ such that the subalgebras $A_k$ of functions supported in $S_k$ satisfy:

$P(A_k)=A_{k-1}$ and $P(A_{0})=A_{d-1}$ (slight typo in the paper here).

In fact, in this case it is necessarily the case that $\nu$ is concentrated on a coset of a proper normal subgroup $N\rhd G$, say $gN$. Then $S_k=g^kN$.

Suppose that $f$ is supported on $g^kN$We want to show that for $Pf\in A_{k-1}$Recall that

$\nu^{\star k-1}P(f)=\nu^{\star k}(f)$.

This shows how the stochastic operator reduces the index $P(A_k)=A_{k-1}$.

A central component of Fagnola and Pellicer’s paper are results about how the decomposition of a stochastic operator:

$P(f)=\sum_{\ell}L_\ell^*fL_{\ell}$,

specifically the maps $L_\ell$ can speak to the irreducibility and periodicity of the random walk given by $P$. I am not convinced that I need these results (even though I show how they are applicable).

## Stochastic Operators and Operator Algebras

Let $F(X)$ be a $\mathrm{C}^*$-algebra (so that $X$ is in general a  virtual object). A $\mathrm{C}^*$-subalgebra $F(Y)$ is hereditary if whenever $f\in F(X)^+$ and $h\in F(Y)^+$, and $f\leq h$, then $f\in F(Y)^+$.

It can be shown that if $F(Y)$ is a hereditary subalgebra of $F(X)$ that there exists a projection $\mathbf{1}_Y\in F(X)$ such that:

$F(Y)=\mathbf{1}_YF(X)\mathbf{1}_Y$.

All hereditary subalgebras are of this form.

This sandbox is going to take from a variety of sources, mostly Shuzhou Wang.

## C*-Ideals

Let $J\subset C(X)$ be a closed (two-sided) ideal in a non-commutative unital $C^*$-algebra $C(X)$. Such an ideal is self-adjoint and so a non-commutative $C^*$-algebra $J=C(S)$. The quotient map is given by $\pi:C(X)\rightarrow C(X)/C(S)$, $f\mapsto f+J$, where $f+J$ is the equivalence class of $f$ under the equivalence relation:

$f\sim_{J} g\Rightarrow g-f\in C(S)$.

Where we have the product

$(f+J)(g+J)=fg+J$,

and the norm is given by:

$\displaystyle\|f+J\|=\sup_{j\in C(S)}\|f+j\|$,

the quotient $C(X)/ C(S)$ is a $C^*$-algebra.

Consider now elements $j_1,\,j_2\in C(S)$ and $f_1,\, f_2\in C(X)$. Consider

$j_1\otimes f_1+f_2\otimes j_2\in C(S)\otimes C(X)+C(X)\otimes C(S)$.

The tensor product $\pi\otimes \pi:C(X)\otimes C(X)\rightarrow (C(X)/C(S))\otimes (C(X)/ C(S))$. Now note that

$(\pi\otimes\pi)(j_1\otimes f_1+f_2\otimes j_2)=(0+J)\otimes(f_1+J)+$

$(f_2+J)\otimes(0+J)=0$,

by the nature of the Tensor Product ($0\otimes a=0$). Therefore $C(X)\otimes C(S)+C(S)\otimes C(X)\subset \text{ker}(\pi\otimes\pi)$.

### Definition

A WC*-ideal (W for Woronowicz) is a C*-ideal $J=C(S)$ such that $\Delta(J)\subset \text{ker}(\pi\otimes\pi)$, where $\pi$ is the quotient map $C(G)\rightarrow C(G)/C(S)$.

Let $F(G)$ be the algebra of functions on a classical group $G$. Let $H\subset G$. Let $J$ be the set of functions which vanish on $H$: this is a C*-ideal. The kernal of $\pi:F(G)\rightarrow F(G)/J$ is $J$.

Let $\delta_s\in J$ so that $s\not\in H$. Note that

$\displaystyle\Delta(\delta_s)=\sum_{t\in G}\delta_{st^{-1}}\otimes\delta_t$

and so

$\displaystyle(\pi\otimes \pi)\Delta(\delta_s)=\sum_{t\in G}\pi(\delta_{st^{-1}})\otimes \pi(\delta_t)$.

Note that $\pi(\delta_t)=0+J$ if $t\not\in H$. It is not possible that both $st^{-1}$ and $t$ are in $H$: if they were $st^{-1}\cdot t\in H$, but $st^{-1}\cdot t=s$, which is not in $H$ by assumption. Therefore one of $\pi(\delta_{st^{-1}})$ or $\pi(\delta_t)$ is equal to zero and so:

$(\pi\otimes\pi)\Delta(\delta_s)=0$,

and so by linearity, if $f$ vanishes on a subgroup $H$,

$\Delta(f)\subset \text{ker}(\pi\otimes\pi)$.

In this way, WC*-ideals generalise functions which vanish on distinguished subgroups. In fact, without checking all the details, I imagine that first isomorphism theorem can show that $F(G)/ J=F(H)$. Let $\pi_H:F(G)\rightarrow F(H)$ be the ring homomorphism

$\displaystyle\pi_H\left(\sum_{t\in G}a_t\delta_t\right)=\sum_{t\in H}a_t\delta_t$.

Then $\text{ker}\,\pi_H=J$, $\text{im}\,\pi_H=F(H)$, and so we have the isomorphism of rings, which presumably carries forward to the algebras of functions level…

Just some notes on the pre-print. I am looking at this paper to better understand this pre-print. In particular I am hoping to learn more about the support of a probability on a quantum group. Flags and notes are added but mistakes are mine alone.

#### Abstract

From this paper I will look at:

• lattice operations on $\mathcal{I}(G)$, for $G$ a LCQG (analogues of intersection and generation)

## 1. Introduction

Idempotent states on quantum groups correspond with “subgroup-like” objects. In this work, on LCQG, the correspondence is with quasi-subgroups (the work of Franz & Skalski the correspondence was with pre-subgroups and group-like projections).

Let us show the kind of thing I am trying to understand better.

Let $F(G)$ be the algebra of function on a finite quantum group. Let $\nu,\,\mu\in M_p(G)$ be concentrated on a pre-subgroup $S$. We can associate to $S$ a group like projection $p_S$.

Let, and this is another thing I am trying to understand better, this support, the support of $\nu$ be ‘the smallest’ (?) projection $p\in F(G)$ such that $\nu(p)=1$. Denote this projection by $p_\nu$. Define $p_\mu$ similarly. That $\mu,\,\nu$ are concentrated on $S$ is to say that $p_\nu\leq p_S$ and $p_\mu\leq p_S$.

Define a map $T_\nu:F(G)\rightarrow F(G)$ by

$a\mapsto p_\nu a$ (or should this be $ap_\nu$ or $p_\nu a p_\nu$?)

We can decompose, in the finite case, $F(G)\cong \text{Im}(T_\nu)\oplus \ker(T_\nu)$

Claim: If $\nu$ is concentrated on $S$$\nu(ap_S)=\nu(a)$I don’t have a proof but it should fall out of something like $p_\nu\leq p_S\Rightarrow \ker p_\nu\subseteq \ker p_S$ together with the decomposition of $F(G)$ above. It may also require that $\int_G$ is a trace, I don’t know. Something very similar in the preprint.

From here we can do the following. That $p_S$ is a group-like projection means that:

$\Delta (p_s)(\mathbf{1}_G\otimes p_S)=p_S\otimes p_S$

$\Rightarrow \sum p_{S(1)}\otimes (p_{S(2)}p_S)=p_S\otimes p_S$

Hit both sides with $\nu\times \mu$ to get:

$\sum \nu(p_{S(1)})\mu(p_{S(2)}p_S)=\nu(p_S)\mu(p_S)$.

By the fact that $\nu,\,\mu$ are supported on $S$, the right-hand side equals one, and by the as-yet-unproven claim, we have

$\sum \nu(p_{S(1)})\mu(p_{S(2)})=1$.

However this is the same as

$(\nu\otimes\mu)\Delta(p_S)=1\Rightarrow (\nu\star \mu)(p_S)=1$,

in other words $p_{\nu\star \mu}\leq p_S$, that is $\nu\star \mu$ remains supported on $S$. As a corollary, a random walk driven by a probability concentrated on a pre-subgroup $S\subset G$ remains concentrated on $S$.