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 $A_n$.