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After completing a piece of work, I like to record some things that I would like to work on next. The previous time was a little over 11 months ago, and was heavily geared towards random walks on quantum groups. The need at that time was to start properly learning some compact quantum groups. I made a start on this: my plan was to write up some notes on compact quantum groups… however these notes [these are real rough], were abandoned like the Marie Celeste in May 2020. What happened was that I had a technical question about the construction of the reduced algebra $C_r(\mathbb{G})$ of continuous functions on a compact quantum group $\mathbb{G}$, and the expert who helped me suggested some intuition that I could use for quantum permutations… this kind of set off a quest to find a better interpretation for quantum permutations that started with this talk, then led to this monstrosity, and finally to this paper that I am proud of (but not sure if journals will concur).

So anyway some problems and brief thoughts. I tried and failed to resist the urge to use the non-standard notation and interpretation used in this new paper… I guess this post is for me… if you want to understand the weird “$\varsigma$ is a quantum permutation” and “$\varsigma\in\mathbb{G}$“, etc you will have to read the paper.

### No Quantum Alternating Group

So the big long crazy draft of has some stuff about why there is no quantum cyclic or alternating groups but these are arguments rather than proofs. A no-go theorem here looks as follows:

A finite group $G has no quantum version if whenever $\mathbb{G} is a quantum permutation group with group of characters of $C_u(\mathbb{G})$ equal to $G$, then $\mathbb{G}=G$.

I know the question for $A_N$ is open… is it formally settled that there is no quantum $\mathbb{Z}_N$? Proving that would be a start. A possible strategy would be to construct from $G<\mathbb{G}$ and a quantum permutation like $\varsigma_{e_5}\in\mathfrak{G}_0$ a character on $C_u(\mathbb{G})$ that is in the complement of $G$ in $S_N$. The conclusion being that there can be no quantum permutation in $C_u(\mathbb{G})$ that is $C_u(\mathbb{G})$ is commutative. There might be stuff here coming from alternating (pun not intended) projections theory that could help.

### Ergodic Theorem for Random Walks on Compact Quantum Groups

From April 2020:

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.

What I would be interested in doing here is seeing can I maybe use the language from the Ergodic Theorem to prove some partial results in this direction. The analysis is possibly a little harder than I am used to. What I might want to show is that if $\phi\in C(\mathbb{G})$ is an idempotent state (I am not sure are there group-like projections $\mathbf{1}_{\phi}$ lying around, I think there are maybe here), and $p_{\phi}$ its support projection in $\ell^{\infty}(\mathbb{G})$, that the convolution of quantum permutations $\varsigma_1,\varsigma_2$ such that $p_\phi(\varsigma_i)=1$, that $p_\phi(\varsigma_2\star \varsigma_1)=1$. This would almost certainly require the use of a group-like projection. Possibly restricting to to quantum permutation groups we would then have a good understanding of non-Haar idempotent “quasi-subgroups”:

A quasi-subgroup of a compact (permutation?) quantum group is a subset of $\mathbb{G}$ that (for $C_u(\mathbb{G})$) contains the identity, is closed under reversal, and closed under the quantum group law. A quasi-subgroup is a quantum subgroup precisely when it is closed under wave function collapse.

The other thing that could be done here would be to refine two aspects of the finite theory. One, from the direction of cyclic shifts, in order to make the definition of a cyclic coset more intrinsic (it is currently defined with respect to a state), and two to further study the idea of an idempotent commuting with something like a “finite order” deterministic state (see p.27).

### Maximality Conjecture

For $N\leq 5$, there is no intermediate quantum subgroup $S_N<\mathbb{G}. That is there is no comultiplication intertwining surjective *-homomorphism $\pi:C_u(S_N^+)\twoheadrightarrow C_u(\mathbb{G})$ to noncommutative $C_u(\mathbb{G})$ such that there is in addition there is another such map $\pi_C:C_u(\mathbb{G})\twoheadrightarrow F(S_N)$.

This well-known conjecture is that there is no intermediate quantum subgroup $S_N<\mathbb{G} at any $N$.

Let $\mathfrak{G}_0$ be the Kac–Paljutkin quantum group of order eight and consider the quantum permutation $\varsigma_0:= (\varepsilon+ \varsigma_{e_5})/2\in \mathfrak{G}_0$, the state space of the algebra of functions on $\mathfrak{G}_0$, $F(\mathfrak{G}_0)$. Where $u^{\mathfrak{G}_0 and $\pi_{\mathfrak{G}_0}:C(S_N^+)\twoheadrightarrow F(\mathfrak{G}_0)$, $\varsigma\circ \pi_{\mathfrak{G}_0}\in S_N^+$. Where $\pi_C:C(S_N^+)\rightarrow F(S_N)$ and (by abuse of notation) $h_{S_N}:=h_{S_N}\circ\pi_C\in S_N^+$ the Haar state of $S_N$ in $S_N^{+}$. Where $\varsigma_1:=h_{S_N}\star \varsigma_0$, consider the idempotent state on $C(S_N^+)$:

$\displaystyle \varsigma:=w^{*}-\lim_{n\rightarrow \infty}\frac{1}{n}\sum_{k=1}^{n}\varsigma_1^{\star k}$.

There are three possibilities and all three are interesting:

$\varsigma=h_{S_N^+}$ — this is what I expect to be true. If this could be proven, the approach would be to hope that it might be possible to construct a state like $\varsigma_{e_5}$ on any compact quantum group. The state $\varsigma_{e_5}$ has a nice “constraint” property: I can only map 1 or 2 to 3 or 4 and vice versa. A starting idea in the construction might be to take the unital $\mathrm{C}^*$ algebra generated by a non-commuting pair $u^{\mathbb{G}}_{i_1j_1},u^{\mathbb{G}}_{i_2j_2}$. Using Theorem 4.6, there is a *-representation $\pi_2:\mathrm{C}^*(u^{\mathbb{G}}_{i_1j_1},u^{\mathbb{G}}_{i_2j_2})\rightarrow B(\mathbb{C}^2)$ such that for some $t\in (0,1)$ we have a representation and from this representation we have a nice vector state that is something like $\varsigma_{e_5}$. Well it doesn’t have the “constraint” property… one idea which I haven’t thought through is to condition this lovely quantum permutation in a clever enough way… we could condition it to only map $j_1$ to $i_1$ or some $i_3$… but that seems to only be the start of it.

$\varphi=h_{\mathbb{G}}$ for $S_N<\mathbb{G} — this would obviously be a counterexample to the maximality conjecture.

$\varphi$ is a non-Haar idempotent — this is a possibility that I don’t think many have thought of. It wouldn’t disprove the conjecture but would be an interesting example. This might be something non-zero on e.g. $|u_{31}u_{22}u_{11}|^2$ but zero on some strictly positive $|f|^2\in C(S_N^+)$.

### Doing something with abelian quantum permutation groups

I can’t really describe this so instead I quote from the paper:

It could be speculated that the dual of a discrete group $\Gamma=\langle\gamma_1,\dots,\gamma_k\rangle$ could model a $k$ particle “entangled” quantum system, where the $p$-th particle, corresponding to the block $B_p$, has $|\gamma_p|$ states, labelled $1,\dots,|\gamma_p|$. Full information about the state of all particles is in general impossible, but measurement with $x(B_p)$ will see collapse of the $p$th particle to a definite state. Only the deterministic permutations in $\widehat{\Gamma}$ would correspond to classical states.

### Quantum Automorphism Groups of Graphs

So I want to read Schmidt’s PhD thesis and maybe answer the question of whether or not there is a graph whose quantum automorphism group is the Kac–Paljutkin quantum group. Also see can anything be done in the intersection of random walks and quantum automorphism groups.

### Write locally compact to compact dictionary

I want to able to extract results on locally compact quantum groups to compact quantum groups.

### Use of Stopping Times and Classical Probabilistic Methods for Random Walks on Quantum Permutation Groups

This is a bit mad… bottom of p.35 to p.36.

Maybe take $x_0:=1\in\{1,\dots,N\}$ and define a Markov chain on $\{1,\dots,N\}$ using a quantum permutation. So for example you measure $\varsigma$ with $x(1)$ to get $x_1$. Then measure $x(x_1)$ and iterate. The time taken to reach $N$ or some other $k\in\{1,\dots,N\}$ or maybe hit all of $\{1,\dots,N\}$… this is a random time $T$. Is there any relationship between the expectation of $T$ and the distance of the convolution powers of $\varsigma$ to the Haar state. Lots of things to think about here.

### Alternating Measurement

More mad stuff. See Section 8.5.

### Infinite Discrete Duals with no Finite Order Generators

If $\Gamma$ has a finite set of finite-order generators it is a quantum permutation group. What about if there are only infinite order generators? I guess this isn’t a quantum permutation group (subgroup of some $S_N^+$) but maybe using Goswami and Skalski such a dual can be given the structure of a quantum permutation group on infinite many symbols. At the other end of the scale… are the Sekine quantum groups quantum permutation groups?

### Other random walk questions

From April 2020:

1. Look at random walks on quantum homogeneous spaces, possibly using Gelfand Pair theory. Start in finite and move into Kac? — No interest in this by me at the moment
2. Following Urban, study convolution factorisations of the Haar state. — ditto.
3. Examples of non-central random walks on compact quantum groups — Freslon and coauthors have cornered the market on interesting examples of random walks on compact quantum groups… I don’t think I will be spending time on this.

Note that Simeng Wang has sorted: extending the Upper Bound Lemma to the non-Kac case. There are a handful of other problems here, here, and here that I am no longer interested in.

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.

I am currently (slowly) working on an essay/paper where I expand upon the ideas in this talk. In this post I will try and explain in this framework why there is no quantum cyclic group, no quantum $S_3$, and ask why there is no quantum alternating group.

### Quantum Permutations Basics

Let $A$ be a unital $\mathrm{C}^*$-algebra. We say that a matrix $u\in M_N(A)$ is a magic unitary if each entry is a projection $u_{ij}=u_{ij}^2=u_{ij}^*$, and each row and column of $u$ is a partition of unity, that is:

$\displaystyle \sum_ku_{ik}=\sum_k u_{kj}=1_A$.

It is necessarily the case (but not for *-algebras) that elements along the same row or column are orthogonal:

$u_{ij}u_{ik}=\delta_{j,k}u_{ij}$ and $u_{ij}u_{k j}=\delta_{i,k}u_{ij}$.

Shuzou Wang defined the algebra of continuous functions on the quantum permutation group on $N$ symbols to be the universal $\mathrm{C}^*$-algebra $C(S_N^+)$ generated by an $N\times N$ magic unitary $u$. Together with (leaning heavily on the universal property) the *-homomorphism:

$\displaystyle \Delta:C(S_N^+)\rightarrow C(S_N^+)\underset{\min}{\otimes}C(S_N^+), u_{ij}\mapsto \sum_{k=1}^N u_{ik}\otimes u_{kj}$,

and the fact that $u$ and $(u)^t$ are invertible ($u^{-1}=u^t)$), the quantum permutation group $S_N^+$ is a compact matrix quantum group.

Any compact matrix quantum group generated by a magic unitary is a quantum permutation group in that it is a quantum subgroup of the quantum permutation group. There are finite quantum groups (finite dimensional algebra of functions) which are not quantum permutation groups and so Cayley’s Theorem does not hold for quantum groups. I think this is because we can have quantum groups which act on algebras such as $M_N(\mathbb{C})$ rather than $\mathbb{C}^N$ — the algebra of functions equivalent of the finite set $\{1,2,\dots,N\}$.

This is all basic for quantum group theorists and probably unmotivated for everyone else. There are traditional motivations as to why such objects should be considered algebras of functions on quantum groups:

• find a presentation of an algebra of continuous functions on a group, $C(G)$, as a commutative universal $\mathrm{C}^*$-algebra. Study the the same object liberated by dropping commutativity. Call this the quantum or free version of $G$, $G^+$.
• quotient $C(S_N^+)$ by the commutator ideal, that is we look at the commutative $\mathrm{C}^*-$algebra generated by an $N\times N$ magic unitary. It is isomorphic to $F(S_N)$, the algebra of functions on (classical) $S_N$.
• every commutative algebra of continuous functions on a compact matrix quantum group is the algebra of functions on a (classical) compact matrix group, etc.

Here I want to take a very different direction which while motivationally rich might be mathematically poor.

### Weaver Philosophy

Take a quantum permutation group $\mathbb{G}$ and represent the algebra of functions as bounded operators on a Hilbert space $\mathsf{H}$. Consider a norm-one element $\varsigma\in P(\mathsf{H})$ as a quantum permutation. We study the properties of the quantum permutation by making a series of measurements using self-adjoint elements of $C(\mathbb{G})$.

Suppose we have a finite-spectrum, self-adjoint measurement $f\in C(\mathbb{G})\subset B(\mathsf{H})$. It’s spectral decomposition gives a partition of unity $(p^{f_i})_{i=1}^{|\sigma(f)|}$. The measurement of $\varsigma$ with $f$ gives the value $f_i$ with probability:

$\displaystyle \mathbb{P}[f=f_i\,|\,\varsigma]=\langle\varsigma,p^{f_i}\varsigma\rangle=\|p^{f_i}\varsigma\|^2$,

and we have the expectation:

$\displaystyle \mathbb{E}[f\,|\,\varsigma]=\langle\varsigma,f\varsigma\rangle$.

What happens if the measurement of $\varsigma$ with $f$ yields $f=f_i$ (which can only happen if $p^{f_i}\varsigma\neq 0$)? Then we have some wavefunction collapse of

$\displaystyle \varsigma\mapsto p^{f_i}\varsigma\equiv \frac{p^{f_i}\varsigma}{\|p^{f_i}\varsigma\|}\in P(\mathsf{H})$.

Now we can keep playing the game by taking further measurements. Notationally it is easier to describe what is happening if we work with projections (but straightforward to see what happens with finite-spectrum measurements). At this point let me quote from the essay/paper under preparation:

Suppose that the “event” $p=\theta_1$ has been observed so that the state is now $p^{\theta_1}(\psi)\in P(\mathsf{H})$. Note this is only possible if $p=\theta_1$ is non-null in the sense that

$\displaystyle \mathbb{P}[p=\theta_1\,|\,\psi]=\langle\psi,p^\theta(\psi)\rangle\neq 0.$

The probability that measurement produces $q=\theta_2$, and $p^{\theta_1}(\psi)\mapsto q^{\theta_2}p^{\theta_1}(\psi)\in P(\mathsf{H})$, is:

$\displaystyle \mathbb{P}\left[q=\theta_2\,|\,p^{\theta_1}(\psi)\right]:=\left\langle \frac{p^{\theta_1}(\psi)}{\|p^{\theta_1}(\psi)\|},q^{\theta_2}\left(\frac{p^{\theta_1}(\psi)}{\|p^{\theta_1}(\psi)\|}\right)\right\rangle=\frac{\langle p^{\theta_1}(\psi),q^{\theta_2}(p^{\theta_1}(\psi))\rangle}{\|p^{\theta^1}(\psi)\|^2}.$

Define now the event $\left((q=\theta_2)\succ (p=\theta_1)\,|\,\psi\right)$, said “given the state $\psi$, $q$ is measured to be $\theta_2$ after $p$ is measured to be $\theta_1$“. Assuming that $p=\theta_1$ is non-null, using the expression above a probability can be ascribed to this event:

$\displaystyle \mathbb{P}\left[(q=\theta_2)\succ (p=\theta_1)\,|\,\psi\right]:=\mathbb{P}[p=\theta_1\,|,\psi]\cdot \mathbb{P}[q=\theta_2\,|\,p^{\theta_1}(\psi)]$
$\displaystyle =\langle\psi,p^{\theta_1}(\psi)\rangle\frac{\langle p^{\theta_1}(\psi),q^{\theta_2}(p^{\theta_1}(\psi))\rangle}{\|p^{\theta^1}(\psi)\|^2}$
$=\|q^{\theta_2}p^{\theta_1}\psi\|^2.$

Inductively, for a finite number of projections $\{p_i\}_{i=1}^n$, and $\theta_i\in{0,1}$:

$\displaystyle \mathbb{P}\left[(p_n=\theta_n)\succ\cdots \succ(p_1=\theta_1)\,|\,\psi\right]=\|p_n^{\theta_n}\cdots p_1^{\theta_1}\psi\|^2.$

In general, $pq\neq qp$ and so

$\displaystyle \mathbb{P}\left[(q=\theta_2)\succ (p=\theta_1)\,|\,\psi\right]\neq \mathbb{P}\left[(p=\theta_1)\succ (q=\theta_1)\,|\,\psi\right],$

and this helps interpret that $q$ and $p$ are not simultaneously observable. However the sequential projection measurement $q\succ p$ is “observable” in the sense that it resembles random variables with values in $\{0,1\}^2$. Inductively the sequential projection measurement $p_n\succ \cdots\succ p_1$ resembles a $\{0,1\}^n$-valued random variable, and

$\displaystyle \mathbb{P}[p_n\succ \cdots\succ p_1=(\theta_n,\dots,\theta_1)\,|\,\psi]=\|p_n\cdots p_1(\psi)\|^2.$

If $p$ and $q$ do commute, they share an orthonormal eigenbasis, and it can be interpreted that they can “agree” on what they “see” when they “look” at $\mathsf{H}$, and can thus be determined simultaneously. Alternatively, if they commute then the distributions of $q\succ p$ and $p\succ q$ are equal in the sense that

$\displaystyle \mathbb{P}\left[(q=\theta_2)\succ (p=\theta_1)\,|\,\psi\right]= \mathbb{P}\left[(p=\theta_1)\succ (q=\theta_1)\,|\,\psi\right],$

it doesn’t matter what order they are measured in, the outputs of the measurements can be multiplied together, and this observable can be called $pq=qp$.

Consider the (classical) permutation group $S_N$ or moreover its algebra of functions $F(S_N)$. The elements of $F(S_N)$ can be represented as bounded operators on $\ell^2(S_N)$, and the algebra is generated by a magic unitary $u^{S_N}\in M_N(B(\ell^2(S_N)))$ where:

$u_{ij}^{S_N}(e_\sigma)=\mathbf{1}_{j\rightarrow i}(e_\sigma)e_{\sigma}$.

Here $\mathbf{1}_{j\rightarrow i}\in F(S_N)$ (‘unrepresented’) that asks of $\sigma$… do you send $j\rightarrow i$? One for yes, zero for no.

Recall that the product of commuting projections is a projection, and so as $F(S_N)$ is commutative, products such as:

$\displaystyle p_\sigma:=\prod_{j=1}^Nu_{\sigma(j)j}^{S_N}$,

There are, of, course, $N!$ such projections, they form a partition of unity themselves, and thus we can build a measurement that will identify a random permutation $\varsigma\in P(\ell^2(S_N))$ and leave it equal to some $e_\sigma$ after measurement. This is the essence of classical… all we have to do is enumerate $n:S_N\rightarrow \{1,\dots,N!\}$ and measure using:

$\displaystyle f=\sum_{\sigma\in S_N}n(\sigma)p_{\sigma}$.

A quantum permutation meanwhile is impossible to pin down in such a way. As an example, consider the Kac-Paljutkin quantum group of order eight which can be represented as $F(\mathfrak{G}_0)\subset B(\mathbb{C}^6)$. Take $\varsigma=e_5\in \mathbb{C}^6$. Then

$\displaystyle\mathbb{P}[(\varsigma(1)=4)\succ(\varsigma(3)=1)\succ(\varsigma(1)=3)]=\frac{1}{8}$.

If you think for a moment this cannot happen classically, and the issue is that we cannot know simultaneously if $\varsigma(1)=3$ and $\varsigma(3)=1$… and if we cannot know this simultaneously we cannot pin down $\varsigma$ to a single element of $S_N$.

### No Quantum Cyclic Group

Suppose that $\varsigma\in \mathsf{H}$ is a quantum permutation (in $S_N^+$). We can measure where the quantum permutation sends, say, one to. We simply form the self-adjoint element:

$\displaystyle x(1)=\sum_{k=1}^Nku_{k1}$.

The measurement will produce some $k\in \{1,\dots,N\}$… but if $\varsigma$ is supposed to represent some “quantum cyclic permutation” then we already know the values of $\varsigma(2),\dots,\varsigma(N)$ from $\varsigma(1)=k$, and so, after measurement,

$u_{k1}\varsigma \in \bigcap_{m=1}^N \text{ran}(u_{m+k-1,m})$, $u_{k1}\varsigma\equiv k-1\in\mathbb{Z}_N$.

The significance of the intersection is that whatever representation of $C(S_N^+)$ we have, we find these subspaces to be $C(S_N^+)$-invariant, and can be taken to be one-dimensional.

I believe this explains why there is no quantum cyclic group.

#### Question 1

Can we use a similar argument to show that there is no quantum version of any abelian group? Perhaps using $F(G\times H)=F(G)\otimes F(H)$ together with the structure theorem for finite abelian groups?

### No Quantum $S_3$

Let $C(S_3^+)$ be represented as bounded operators on a Hilbert space $\mathsf{H}$. Let $\varsigma\in P(\mathsf{H})$. Consider the random variable

$x(1)=u_{11}+2u_{21}+3u_{31}$.

Assume without loss of generality that $u_{31}\varsigma\neq0$ then measuring $\varsigma$ with $x(1)$ gives $x(1)\varsigma=3$ with probability $\langle\varsigma,u_{31}\varsigma\rangle$, and the quantum permutation projects to:

$\displaystyle \frac{u_{31}\varsigma}{\|u_{31}\varsigma\|}\in P(\mathsf{H})$.

Now consider (for any $\varsigma\in P(\mathsf{H})$, using the fact that $u_{21}u_{31}=0=u_{32}u_{31}$ and the rows and columns of $u$ are partitions of unity:

$u_{31}\varsigma=(u_{12}+u_{22}+u_{32})u_{31}\varsigma=(u_{21}+u_{22}+u_{23})u_{31}\varsigma$

$\Rightarrow u_{12}u_{31}\varsigma=u_{23}u_{31}\varsigma$ (*)

Now suppose, again without loss of generality, that measurement of $u_{31}\varsigma\in P(\mathsf{H})$ with $x(2)=u_{12}+2u_{22}+3u_{33}$ produces $x(2)u_{31}\varsigma=1$, then we have projection to $u_{12}u_{31}\varsigma\in P(\mathsf{H})$. Now let us find the Birkhoff slice of this. First of all, as $x(2)=1$ has just been observed it looks like:

$\Phi(u_{12}u_{31}\varsigma)=\left[\begin{array}{ccc}0 & 1 & 0 \\ \ast & 0 & \ast \\ \ast & 0 & \ast \end{array}\right]$

In light of (*), let us find $\Phi(u_{12}u_{31}\varsigma)_{23}$. First let us normalise correctly to

$\displaystyle \frac{u_{12}u_{31}\varsigma}{\|u_{12}u_{31}\varsigma\|}$

So

$\displaystyle\Phi(u_{12}u_{31}\varsigma)_{23}=\left\langle\frac{u_{12}u_{31}\varsigma}{\|u_{12}u_{31}\varsigma\|},u_{23}\frac{u_{12}u_{31}\varsigma}{\|u_{12}u_{31}\varsigma\|}\right\rangle$

Now use (*):

$\displaystyle\Phi(u_{12}u_{31}\varsigma)_{23}=\left\langle\frac{u_{23}u_{31}\varsigma}{\|u_{23}u_{31}\varsigma\|},u_{23}\frac{u_{23}u_{31}\varsigma}{\|u_{23}u_{31}\varsigma\|}\right\rangle=1$

$\displaystyle \Rightarrow \Phi(u_{12}u_{31}\varsigma)=\left[\begin{array}{ccc}0 & 1 & 0 \\ 0 & 0 & 1 \\ \Phi(u_{12}u_{31}\varsigma)_{31} & 0 & 0 \end{array}\right]$,

and as $\Phi$ maps to doubly stochastic matrices we find that $\Phi(u_{12}u_{31}\varsigma)$ is equal to the permutation matrix $(132)$.

Not convincing? Fair enough, here is proper proof inspired by the above:

Let us show $u_{11}u_{22}=u_{22}u_{11}$. Fix a Hilbert space representation $C(S_3^+)\subset B(\mathsf{H})$ and let $\varsigma\in\mathsf{H}$.

The basic idea of the proof is, as above, to realise that once a quantum permutation $\varsigma$ is observed sending, say, $3\rightarrow 2$, the fates of $2$ and $1$ are entangled: if you see $2\rightarrow 3$ you know that $1\rightarrow 1$.

This is the conceptional side of the proof.

Consider $u_{23}\varsigma$ which is equal to both:

$(u_{11}+u_{21}+u_{31})u_{23}\varsigma=(u_{31}+u_{32}+u_{33})u_{23}\varsigma\Rightarrow u_{11}u_{23}\varsigma=u_{32}u_{23}\varsigma$.

This is the manifestation of, if you know $3\rightarrow 2$, then two and one are entangled. Similarly we can show that $u_{22}u_{13}\varsigma=u_{31}u_{13}\varsigma$ and $u_{22}u_{33}=u_{11}u_{33}$.

Now write

$\varsigma=u_{13}\varsigma+u_{23}\varsigma+u_{33}\varsigma$

$\Rightarrow u_{11}\varsigma=u_{11}u_{23}\varsigma+u_{11}u_{33}\varsigma=u_{32}u_{23}\varsigma+u_{22}u_{33}\varsigma$

$\Rightarrow u_{22}u_{11}\varsigma=u_{22}u_{33}\varsigma$.

Similarly,

$u_{22}\varsigma=u_{22}u_{13}\varsigma+u_{22}u_{33}\varsigma=u_{31}u_{13}\varsigma+u_{22}u_{33}\varsigma$

$\Rightarrow u_{11}u_{22}\varsigma=u_{11}u_{22}u_{33}\varsigma=u_{11}u_{11}u_{33}\varsigma=u_{11}u_{33}\varsigma=u_{22}u_{33}\varsigma$

Which is equal to $u_{22}u_{11}x$, that is $u_{11}$ and $u_{22}$ commute.

### Question 2

Is it true that if every quantum permutation in a $\mathsf{H}$ can be fully described using some combination of $u_{ij}$-measurements, then the quantum permutation group is classical? I believe this to be true.

### Quantum Alternating Group

Freslon, Teyssier, and Wang state that there is no quantum alternating group. Can we use the ideas from above to explain why this is so? Perhaps for $A_4$.

A possible plan of attack is to use the number of fixed points, $\text{tr}(u)$, and perhaps show that $\text{tr}(u)$ commutes with $x(1)$. If you know these two simultaneously you nearly know the permutation. Just for completeness let us do this with $(\text{tr}(u),x(1))$:

The problem is that we cannot assume that that the spectrum of $\text{tr}(u)$ is $\{0,1,4\}$, and, euh, the obvious fact that it doesn’t actually work.

What is more promising is

However while the spectrums of x(1) and x(2) are cool (both in $\{1,2,3,4\}$), they do not commute.

### Question 3

Are there some measurements that can identify an element of $A_4$ and via a positive answer to Question 3 explain why there is no quantum $A_4$? Can this be generalised to $A_n$.

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.