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In the case of a finite classical group G, we can show that if we have i.i.d. random variables \zeta_i\sim\nu\in M_p(G), that if \text{supp }\nu\subset Ng, for Ng a coset of a proper normal subgroup N\rhd G, that the random walk on G driven by \nu, the random variables:

\xi_k=\zeta_k\cdots \zeta_1,

exhibits a periodicity because

\xi_k\in Ng^{k}.

This shows that a necessary condition for ergodicity of a random walk on a finite classical group G driven by \nu\in M_p(G) is that the support of \nu not be concentrated on the coset of a proper normal subgroup.

I had hoped that something similar might hold for the case of random walks on finite quantum groups but alas I think I have found a barrier.

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Slides of a talk given at Munster Groups 2019, WIT.

Abstract: It is a folklore theorem that necessary and sufficient conditions for a random walk on a finite group to converge in distribution to the uniform distribution – “to random” – are that the driving probability is not concentrated on a proper subgroup nor the coset of a proper normal subgroup. This is the Ergodic Theorem for Random Walks on Finite Groups. In this talk we will outline the rarely written down proof, and explain why, for example, adjacent transpositions can never mix up a deck of cards. From here we will, in a very leisurely and natural fashion, introduce (and motivate the definition of) finite quantum groups, and random walks on them. We will see how the group algebra of a finite group is the algebra of functions on a finite quantum group. Freslon has very recently proved the Ergodic Theorem in this setting, and we present ongoing work towards an Ergodic Theorem in the more general finite quantum group setting; a result that would generalise both the folklore and group algebra Ergodic Theorems.

Introduction

Every finite quantum group has finite dimensional algebra of functions:

\displaystyle F(G)=\bigoplus_{j=1}^m M_{n_j}(\mathbb{C}).

At least one of the factors must be one-dimensional to account for the counit \varepsilon:F(G)\rightarrow \mathbb{C}, and if this factor is denoted \mathbb{C}e_1, the counit is given by the dual element e^1. There may be more and so reorder the index j\mapsto i so that n_i=1 for i=1,\dots,m_1, and n_i>1 for i>m_1:

\displaystyle F(G)=\left(\bigoplus_{i=1}^{m_1} \mathbb{C}e_{i}\right)\oplus \bigoplus_{i=m_1+1}^m M_{n_i}(\mathbb{C})=:A_1\oplus B,

Denote by M_p(G) the states of F(G). The pure states of F(G) arise as pure states on single factors.

In the case of the Kac-Paljutkin and Sekine quantum groups, the convolution powers of pure states exhibit a periodicity of sorts. Recall for these quantum groups that B consists of a single matrix factor.

In these cases, for pure states of the form e^i, that is supported on A_1 (and we can say a little more than is necessary), the convolution remains supported on A_1 because

\Delta(A_1)\subset A_1\otimes A_1+B\otimes B.

If we have a pure state \nu supported on B=M_{\sqrt{\dim B}}(\mathbb{C}), then because

\Delta(B)\subset A_1\otimes B+B\otimes A_1,

then \nu\star\nu must be supported on, because of \Delta(A_1)\subset A_1\otimes A_1+B\otimes B, A_1.

Inductively all of the \nu^{\star 2k} are supported on A_1 and the \nu^{\star 2k+1} are supported on B. This means that the convolutions powers of a pure state, in these cases, cannot converge to the Haar state.

The question is, do the results above about the image of A_1 and B under the coproduct hold more generally? I believe that the paper of Kac and Paljutkin shows that this is the case whenever B consists of a single factor… but does it hold more generally?

To find out we go back and do some sandboxing with the paper of Kac and Paljutkin. Which is a pleasure because that paper is beautiful. The blue stuff is my own scribbling.

Finite Ring Groups

Let G be a finite quantum group with notation on the algebra of functions as above. Note that A_1 is commutative. Let

p=\sum_{i=1}^{m_1}e_i,

which is a central idempotent.

Lemma 8.1

S(p)=p.

Proof: If S(\mathbf{1}_G-p)p\neq 0, then for some i>m_1, and f\in M_{n_i}(\mathbb{C}), the mapping f\mapsto S(f)p is a non-zero homomorphism from M_{n_i}(\mathbb{C}) into commutative A_1 which is impossible.

If S(\mathbf{1}_G-p)p=g\neq 0, then one of the S(I_{n_i})\in A_1\oplus B, with ‘something’ in A_1. Using the centrality and projectionality of p, we can show that the given map is indeed a homomorphism. 

It follows that S(p)p=p\Rightarrow S(S(p)p)=S(p)=S(p)p=S(p), and so p=S(p) \bullet

Lemma 8.2

(p\otimes p)\Delta(p)=p\otimes p

Proof: Suppose that (p\otimes p)\Delta(f)=b for some non-commutative f\in M_{n_i}(\mathbb{C}). This means that there exists an index k such that f_{(1)_k}\otimes f_{(2)_k}\in A_1\otimes A_1. Then for that factor, 

f\mapsto \Delta(f)(p\otimes p)

is a non-null homomorphism from the non-commutative into the commutative.

We see that (p\otimes p)\Delta(f)=0 for all f\in B. Putting a=\mathbf{1}-p we get the result \bullet

The following says that p is a group-like projection. We know from previous work that if a state is supported on a group-like projection that it will remain supported on it. In particular, any state supported on A_1 will remain there.

Lemma 8.3

(p\otimes \mathbf{1}_G)\Delta(p)=p\otimes p=(\mathbf{1}_G\otimes p)\Delta(p).

Proof: Since \Delta is a homomorphism, \Delta(p) is an idempotent in F(G)\otimes F(G)I  do not understand nor require the rest of the proof.

Lemma 8.4

A_1=F(G_1) is the algebra of functions on finite group with elements i=1,\dots,m_1, and we write e_i=\delta_i. The coproduct is given by (p\otimes p)\Delta.

We have:

(p\otimes p)\Delta(e_i)=\sum_{t\in G_1}\delta_{it^{-1}}\otimes \delta_t,

S(\delta_i)=\delta_{i^{-1}},

\varepsilon(e_i)=\delta_{i,1},

as e_1=\delta_e.

The element \Delta(\delta_i) is a sum of four terms, lying in the subalgebras:

A_1\otimes A_1,\,B\otimes B,\,A_1\otimes B,\,B\otimes A_1.

We already know what is going on with the first summand. Denote the second by P_i. From the group-like-projection property, the last two summands are zero, so that

\Delta(\delta_i=\sum_{t\in G_1}\delta_{t}\otimes \delta_{t^{-1}i}+P_i$.

Since the \delta_i are symmetric (\delta_i^*=\delta_i) mutually orthogonal idempotents, P_i has similar properties:

P_i^*=P_i,\,P_i^2=P_i,\,P_iP_j=0

for i\neq j.

At this point Kac and Paljutkin restrict to B=M_{n_{i+1}}(\mathbb{C}), that is there is only one summand. Here we try to keep arbitrarily (finitely) many summands in B.

Let the summand M_{n_i}(\mathbb{C}) have matrix units E_{mn}^i, where m,n=1,\dots,n_iKac and Paljutkin now do something which I think is a little dodgy, but basically that the integral over G is equal on each of the \delta_i, equal on each of the E_{mm}^i, and then zero off the diagonal. 

It does follow from above that each P_i\in B\otimes B is a projection.

Now I am stuck!

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…

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Just some notes on section 1 of this paperFlags and notes are added but mistakes are mine alone.

Definition

Let C(G) be the algebra of continuous functions on a compact matrix quantum group. Such an object is given by a matrix u=\{u_{ij}\}_{i,j=1}^N which generates C(G) as a C*-algebra. Furthermore, there exists a C*-algebra homomorphism \Delta:C(G)\rightarrow C(G)\otimes C(G) such that

\displaystyle \Delta(u_{ij})=\sum_{k=1}^N u_{ik}\otimes u_{kj},

and both u and u^T are invertible in M_N(C(G)).

Any subgroup G\subset \text{GL}(N,\mathbb{C}) is such an object, with the u_{ij}\in C(G) given by u_{ij}(g)=g_{ij}\in\mathbb{C}. Furthermore

\mathrm{C}_{\text{comm}}\langle u_{ij}\rangle \cong C(G).

We say that \rho=(\rho_{ij})_{i,j=1}^{d_\rho}\in M_{d_{\rho}}(C(G)) is a representation if it is invertible and

\displaystyle \Delta(\rho_{ij})=\sum_{k=1}^{d_\rho}\rho_{ik}\otimes\rho{kj}.

The transpose \rho^T=(\rho_{ji})_{i,j=1}^N\in M_{d_{\rho}}(C(G)) is also invertible and so we have:

Proposition

The C*algebra generated by the \rho_{ij} is also the algebra of continuous functions on a compact matrix quantum group.

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Background

I am trying to prove an Ergodic Theorem for Random Walks on Finite Quantum Groups. A random walk on a quantum group driven by \nu\in M_p(G) is ergodic if the convolution powers (\nu^{\star k})_{k\geq 0} converge to the Haar state \int_G.

The classical theorem for finite groups:

Ergodic Theorem for Random Walks on Finite Groups

A random walk on a finite group G driven by a probability \nu\in M_p(G) is ergodic if and only if \nu is not concentrated on a proper subgroup nor the coset of a proper normal subgroup.

Not concentrated on a proper subgroup gives irreducibility. A random walk is irreducible if for all g\in G, there exists k\in\mathbb{N} such that \nu^{\star k}(\{g\})>0.

Not concentrated on the coset of a proper normal subgroup gives aperiodicity. Something which should be equivalent to aperiodicity is if

p:=\gcd\{k>0:\nu^{\star k}(e)>0\}

is equal to one (perhaps via invariance \mathbb{P}[\xi_{i+1}=t|\xi_{i}=s]=\mathbb{P}[\xi_{i+1}=th|\xi_{i}=sh]).

If \nu is concentrated on the coset a proper normal subgroup N\rhd G, specifically on Ng\neq Ne, then we have periodicity (Ng\rightarrow Ng^2\rightarrow \cdots\rightarrow Ng^{o(g)}=Ne\rightarrow Ng\rightarrow \cdots), and p=o(g), the order of g.

In Markov chain theory, ergodicity is equivalent to irreduciblity and aperiodicity.

The theorem in the quantum case should look like:

Ergodic Theorem for Random Walks on Finite Quantum Groups

A random walk on a finite quantum group G driven by a state \nu\in M_p(G) is ergodic if and only if “X”.

Irreducibility

At the moment I have some part of X;the irreducibility bit. As is well known since Pal (1996), it is possible to have a probability not concentrated on a quantum subgroup be reducible. This led Franz & Skalski to generalise quantum subgroups to group-like-projections, which I will say correspond to quasi-subgroups following Kasprzak & Sołtan.

I have shown that if \nu is concentrated on a proper quasi-subgroup S, in the sense that \nu(P_S)=1 for a group-like-projection P_S, that so are the \nu^{\star k}. The analogue of irreducible is that for all q projections in F(G), there exists k\in\mathbb{N} such that \nu^{\star k}(q)>0. If \nu is concentrated on a quasi-subgroup S, then for all k, \nu^{\star k}(Q_S)=0, where Q_S=\mathbf{1}_G -P_S.

I have also shown on the other hand that if the random walk is reducible that it must be concentrated on a proper quasi-subgroup. Franz & Skalski show that group-like-projections also have a correspondence with idempotent states. The Césaro Means

\displaystyle \nu_n:=\frac{1}{n}\sum_{k=1}^n\nu^{\star k},

converge to an idempotent state \nu_\infty. If \nu^{\star k}(q)=0 for all k then the \nu_{\infty}(q)=0 also, so that \nu_\infty\neq \int_G (as the Haar state is faithful). I was able to prove that \nu is supported on the quasi-subgroup given by the idempotent \nu_\infty.

I believe this result — irreducible if and only if not concentrated on a quasi-subgroup — holds more generally than just in finite quantum groups.

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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.

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Slides of a talk given to the Functional Analysis seminar in Besancon.

Some of these problems have since been solved.

“e in support” implies convergence

Consider a \nu\in M_p(G) on a finite quantum group such that where

M_p(G)\subset \mathbb{C}\varepsilon \oplus (\ker \varepsilon)^*,

\nu=\nu(e)\varepsilon+\psi with \nu(e)>0. This has a positive density of trace one (with respect to the Haar state \int_G\in M_p(G)), say

\displaystyle a_\nu=\nu(e)\eta+b_\psi\in \mathbb{C}\eta\oplus \ker \varepsilon,

where \eta is the Haar element. 

An element in a direct sum is positive if and only if both elements are positive. The Haar element is positive and so b_\psi\geq 0. Assume that b_\psi\neq 0 (if b_\psi=0, then \psi=0\Rightarrow \nu=\varepsilon\Rightarrow \nu^{\star k}=\varepsilon for all k and we have trivial convergence)

Therefore let

\displaystyle a_{\tilde{\psi}}:=\frac{b_\psi}{\int_G b_\psi}

be the density of \tilde{\psi}\in M_p(G).

Now we can explicitly write

\displaystyle \nu=\nu(e)\varepsilon+(1-\nu(e))\tilde{\psi}.

This has stochastic operator

P_\nu=\nu(e)I_{F(G)}+(1-\nu(e))P_{\tilde{\psi}}.

Let \lambda be an eigenvalue of P_\nu of eigenvector a. This yields

\nu(e)a+(1-\nu(e))P_{\tilde{\psi}}(a)=\lambda a

and thus

\displaystyle P_{\tilde{\psi}}a=\frac{\lambda-\nu(e)}{1-\nu(e)}a.

Therefore, as a is also an eigenvector for P_{\tilde{\psi}}, and P_{\tilde{\psi}} is a stochastic operator (if a is an eigenvector of eigenvalue |\lambda|>1, then \|P_\nu a\|_1=|\lambda|\|a\|_1\leq \|a\|_1, contradiction), we have

\displaystyle \left|\frac{\lambda-\nu(e)}{1-\nu(e)}\right|\leq 1

\Rightarrow |\lambda-\nu(e)|\leq 1-\nu(e).

This means that the eigenvalues of P_\nu lie in the ball B_{1-\nu(e)}(\nu(e)) and thus the only eigenvalue of magnitude one is \lambda=1, which has (left)-eigenvector the stationary distribution of P_\nu, say \nu_\infty.

If \nu is symmetric/reversible in the sense that \nu=\nu\circ S, then P_\nu is self-adjoint and has a basis of (left)-eigenvectors \{\nu_\infty=:u_1,u_2,\dots,u_{|G|}\}\subset \mathbb{C}G and we have, if we write \nu=\sum_{t=1}^{|G|}a_tu_t,

\displaystyle \nu^{\star k}=\sum_{t=1}^{|G|}a_t\lambda_t^ku_t,

which converges to a_1\nu_\infty (so that a_1=1).

If \nu is not reversible, it is a standard argument to show that when put in Jordan normal form, that the powers P_{\nu}^k converge and thus so do the \nu^{\star k} \bullet

Total Variation Decrasing

Uses Simeng Wang’s \|a\star_Ab\|_1\leq \|a\|_1\|b\|_1. Result holds for compact Kac if the state has a density.

Periodic e^2 is concentrated on a coset of a proper normal subgroup of \mathfrak{G}_0

e_2+e_4 is a minimal projection (coset) in the quotient space of the normal subgroup (to be double checked) given by \langle e_1,e_3\rangle

Supported on Subgroup implies Reducible

I have a proof that reducible is equivalent to supported on a pre-subgroup.

Diaconis–Shahshahani Upper Bound Lemma for Finite Quantum GroupsJournal of Fourier Analysis and Applications, doi: 10.1007/s00041-019-09670-4 (earlier preprint available here)

Abstract

A central tool in the study of ergodic random walks on finite groups is the Upper Bound Lemma of Diaconis and Shahshahani. The Upper Bound Lemma uses Fourier analysis on the group to generate upper bounds for the distance to random and thus can be used to determine convergence rates for ergodic walks. The Fourier analysis of quantum groups is remarkably similar to that of classical groups. This allows for a generalisation of the Upper Bound Lemma to an Upper Bound Lemma for finite quantum groups. The Upper Bound Lemma is used to study the convergence of ergodic random walks on the dual group \widehat{S_n} as well as on the truly quantum groups of Sekine.

In a recent preprint, Haonan Zhang shows that if \nu\in M_p(Y_n) (where Y_n is a Sekine Finite Quantum Group), then the convolution powers, \nu^{\star k}, converges if

\nu(e_{(0,0)})>0.

The algebra of functions F(Y_n) is a multimatrix algebra:

F(Y_n)=\left(\bigoplus_{i,j\in\mathbb{Z}_n}\mathbb{C}e_{(i,j)}\right)\oplus M_n(\mathbb{C}).

As it happens, where a=\sum_{i,j\in\mathbb{Z}_n}x_{(i,j)}e_{(i,j)}\oplus A, the counit on F(Y_n) is given by \varepsilon(a)=x_{(0,0)}, that is \varepsilon=e^{(0,0)}, dual to e_{(0,0)}.

To help with intuition, making the incorrect assumption that Y_n is a classical group (so that F(Y_n) is commutative — it’s not), because \varepsilon=e^{(0,0)}, the statement \nu(e_{(0,0)})>0, implies that for a real coefficient x^{(0,0)}>0,

\nu=x^{(0,0)}\varepsilon+\cdots= x^{(0,0)}\delta^e+\cdots,

as for classical groups \varepsilon=\delta^e.

That is the condition \nu(e_{(0,0)})>0 is a quantum analogue of e\in\text{supp}(\nu).

Consider a random walk on a classical (the algebra of functions on G is commutative) finite group G driven by a \nu\in M_p(G).

The following is a very non-algebra-of-functions-y proof that e\in \text{supp}(\nu) implies that the convolution powers of \nu converge.

Proof: Let H be the smallest subgroup of G on which \nu is supported:

\displaystyle H=\bigcap_{\underset{\nu(S_i)=1}{S_i\subset G}}S_i.

We claim that the random walk on H driven by \nu is ergordic (see Theorem 1.3.2).

The driving probability \nu\in M_p(G) is not supported on any proper subgroup of H, by the definition of H.

If \nu is supported on a coset of proper normal subgroup N, say Nx, then because e\in \text{supp}(\nu), this coset must be Ne\cong N, but this also contradicts the definition of H.

Therefore, \nu^{\star k} converges to the uniform distribution on H \bullet

Apart from the big reason — that this proof talks about points galore — this kind of proof is not available in the quantum case because there exist \nu\in M_p(G) that converge, but not to the Haar state on any quantum subgroup. A quick look at the paper of Zhang shows that some such states have the quantum analogue of e\in\text{supp}(\nu).

So we have some questions:

  • Is there a proof of the classical result (above) in the language of the algebra of functions on G, that necessarily bypasses talk of points and of subgroups?
  • And can this proof be adapted to the quantum case?
  • Is the claim perhaps true for all finite quantum groups but not all compact quantum groups?

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