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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 $A_n$.

Giving a talk 17:00, September 1 2020:

See here for more.

This post follows on from this one. The purpose of posts in this category is for me to learn more about the research being done in quantum groups. This post looks at this paper of Schmidt.

# Preliminaries

## Compact Matrix Quantum Groups

The author gives the definition and gives the definition of a (left, quantum) group action.

### Definition 1.2

Let $G$ be a compact matrix quantum group and let $C(X)$ be a $\mathrm{C}^*-algebra$. An (left) action of $G$ on $X$ is a unital *-homomorphism $\alpha: C(X)\rightarrow C(X)\otimes C(G)$ that satisfies the analogue of $g_2(g_1x)=(g_2g_1)x$, and the Podlés density condition:

$\displaystyle \overline{\text{span}(\alpha(C(X)))(\mathbf{1}_X\otimes C(G))}=C(X)\otimes C(G)$.

## Quantum Automorphism Groups of Finite Graphs

Schmidt in this earlier paper gives a slightly different presentation of $\text{QAut }\Gamma$. The definition given here I understand:

### Definition 1.3

The quantum automorphism group of a finite graph $\Gamma=(V,E)$ with adjacency matrix $A$ is given by the universal $\mathrm{C}^*$-algebra $C(\text{QAut }\Gamma)$ generated by $u\in M_n(C(\text{QAut }\Gamma))$ such that the rows and columns of $u$ are partitions of unity and:

$uA=Au$.

_______________________________________

The difference between this definition and the one given in the subsequent paper is that in the subsequent paper the quantum automorphism group is given as a quotient of $C(S_n^+)$ by the ideal given by $\mathcal{I}=\langle Au=uA\rangle$… ah but this is more or less the definition of universal $\mathrm{C}^*$-algebras given by generators $E$ and relations $R$:

$\displaystyle \mathrm{C}^*(E,R)=\mathrm{C}^*( E)/\langle \mathcal{R}\rangle$

$\displaystyle \Rightarrow \mathrm{C}^*(E,R)/\langle I\rangle=\left(\mathrm{C}^*(E)/R\right)/\langle I\rangle=\mathrm{C}*(E)/(\langle R\rangle\cap\langle I\rangle)=\mathrm{C}^*(E,R\cap I)$

where presumably $\langle R\rangle \cap \langle I \rangle=\langle R\cap I\rangle$ all works out OK, and it can be shown that $I$ is a suitable ideal, a Hopf ideal. I don’t know how it took me so long to figure that out… Presumably the point of quotienting by (a presumably Hopf) ideal is so that the quotient gives a subgroup, in this case $\text{QAut }\Gamma\leq S_{|V|}^+$ via the surjective *-homomorphism:

$C(S_n^+)\rightarrow C(S_n^+)/\langle uA=Au\rangle=C(\text{QAut }\Gamma)$.

_______________________________________

## Compact Matrix Quantum Groups acting on Graphs

### Definition 1.6

Let $\Gamma$ be a finite graph and $G$ a compact matrix quantum group. An action of $G$ on $\Gamma$ is an action of $G$ on $V$ (coaction of $C(G)$ on $C(V)$) such that the associated magic unitary $v=(v_{ij})_{i,j=1,\dots,|V|}$, given by:

$\displaystyle \alpha(\delta_j)=\sum_{i=1}^{|V|} \delta_i\otimes v_{ij}$,

commutes with the adjacency matrix, $uA=Au$.

By the universal property, we have $G\leq \text{QAut }\Gamma$ via the surjective *-homomorphism:

$C(\text{QAut }\Gamma)\rightarrow C(G)$, $u\mapsto v$.

### Theorem 1.8 (Banica)

Let $X_n=\{1,\dots,n\}$, and $\alpha:F(X_n)\rightarrow F(X_n)\otimes C(G)$, $\alpha(\delta_j)=\sum_i\delta_i\otimes v_{ij}$ be an action, and let $F(K)$ be a linear subspace given by a subset $K\subset X_n$. The matrix $v$ commutes with the projection onto $F(K)$ if and only if $\alpha(F(K))\subseteq F(K)\otimes C(G)$

### Corollary 1.9

The action $\alpha: F(V)\rightarrow F(V)\otimes C(\text{QAut }\Gamma)$ preserves the eigenspaces of $A$:

$\alpha(E_\lambda)\subseteq E_\lambda\otimes C(\text{QAut }\Gamma)$

Proof: Spectral decomposition yields that each $E_\lambda$, or rather the projection $P_\lambda$ onto it, satisfies a polynomial in $A$:

$\displaystyle P_\lambda=\sum_{i}c_iA^i$

$\displaystyle \Rightarrow P_\lambda A=\left(\sum_i c_i A^i\right)A=A P_\lambda$,

as $A$ commutes with powers of $A$ $\qquad \bullet$

# A Criterion for a Graph to have Quantum Symmetry

### Definition 2.1

Let $V=\{1,\dots,|V|\}$. Permutations $\sigma,\,\tau: V\rightarrow V$ are disjoint if $\sigma(i)\neq i\Rightarrow \tau(i)=i$, and vice versa, for all $i\in V$.

In other words, we don’t have $\sigma$ and $\tau$ permuting any vertex.

### Theorem 2.2

Let $\Gamma$ be a finite graph. If there exists two non-trivial, disjoint automorphisms $\sigma,\tau\in\text{Aut }\Gamma$, such that $o(\sigma)=n$ and $o(\tau)=m$, then we get a surjective *-homomorphism $C(\text{QAut }\Gamma)\rightarrow C^*(\mathbb{Z}_n\ast \mathbb{Z}_m)$. In this case, we have the quantum group $\widehat{\mathbb{Z}_n\ast \mathbb{Z}_m}\leq \text{QAut }\Gamma$, and so $\Gamma$ has quantum symmetry.

Warning: This is written by a non-expert (I know only about finite quantum groups and am beginning to learn my compact quantum groups), and there is no attempt at rigour, or even consistency. Actually the post shows a wanton disregard for reason, and attempts to understand the incomprehensible and intuit the non-intuitive. Speculation would be too weak an adjective.

## Groups

A group is a well-established object in the study of mathematics, and for the purposes of this post we can think of a group $G$ as the set of symmetries on some kind of space, given by a set $X$ together with some additional structure $D(X)$. The elements of $G$  act on $X$ as bijections:

$G \ni g:X\rightarrow X$,

such that $D(X)=D(g(X))$, that is the structure of the space is invariant under $g$.

For example, consider the space $(X_n,|X_n|)$, where the set is $X_n=\{1,2,\dots,n\}$, and the structure is the cardinality. Then the set of all of the bijections $X_n\rightarrow X_n$ is a group called $S_n$.

A set of symmetries $G$, a group, comes with some structure of its own. The identity map $e:X\rightarrow X$, $x\mapsto x$ is a symmetry. By transitivity, symmetries $g,h\in G$ can be composed to form a new symmetry $gh:=g\circ h\in G$. Finally, as bijections, symmetries have inverses $g^{-1}$, $g(x)\mapsto x$.

Note that:

$gg^{-1}=g^{-1}g=e\Rightarrow (g^{-1})^{-1}=g$.

A group can carry additional structure, for example, compact groups carry a topology in which the composition $G\times G\rightarrow G$ and inverse ${}^{-1}:G\rightarrow G$ are continuous.

## Algebra of Functions

Given a group $G$ together with its structure, one can define an algebra $A(G)$ of complex valued functions on $G$, such that the multiplication $A(G)\times A(G)\rightarrow A(G)$ is given by a commutative pointwise multiplication, for $s\in G$:

$(f_1f_2)(s)=f_1(s)f_2(s)=(f_2f_1)(s)$.

Depending on the class of group (e.g. finite, matrix, compact, locally compact, etc.), there may be various choices and considerations for what algebra of functions to consider, but on the whole it is nice if given an algebra of functions $A(G)$ we can reconstruct $G$.

Usually the following transpose maps will be considered in the structure of $A(G)$, for some tensor product $\otimes_\alpha$ such that $A(G\times G)\cong A(G)\otimes_\alpha A(G)$, and $m:G\times G\rightarrow G$, $(g,h)\mapsto gh$ is the group multiplication:

\begin{aligned} \Delta: A(G)\rightarrow A(G)\otimes_{\alpha}A(G)&,\,f\mapsto f\circ m,\,\text{the comultiplication} \\ S: A(G)\rightarrow A(G)&,\, f\mapsto f\circ {}^{-1},\,\text{ the antipode} \\ \varepsilon: A(G)\rightarrow \mathbb{C}&,\, f\mapsto f\circ e,\,\text{ the counit} \end{aligned}

See Section 2.2 to learn more about these maps and the relations between them for the case of the complex valued functions on finite groups.

## Quantum Groups

Quantum groups, famously, do not have a single definition in the same way that groups do. All definitions I know about include a coassociative (see Section 2.2) comultiplication $\Delta: A(G)\rightarrow A(G)\otimes_\alpha A(G)$ for some tensor product $\otimes_\alpha$ (or perhaps only into a multiplier algebra $M(A(G)\otimes_\alpha A(G))$), but in general that structure alone can only give a quantum semigroup.

Here is a non-working (quickly broken?), meta-definition, inspired in the usual way by the famous Gelfand Theorem:

A quantum group $G$ is given by an algebra of functions $A(G)$ satisfying a set of axioms $\Theta$ such that:

• whenever $A(G)$ is noncommutative, $G$ is a virtual object,
• every commutative algebra of functions satisfying $\Theta$ is an algebra of functions on a set-of-points group, and
• whenever commutative algebras of functions $A(G_1)\cong_{\Theta} A(G_2)$, $G_1\cong G_2$ as set-of-points groups.

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.