Let $\mathbb{G}$ be a finite quantum group described by $A=\mathcal{C}(\mathbb{G})$ with an involutive antipode (I know this is true is the commutative or cocommutative case. I am not sure at this point how restrictive it is in general. The compact matrix quantum groups have this property so it isn’t a terrible restriction.) $S^2=I_A$. Under the assumption of finiteness, there is a unique Haar state, $h:A\rightarrow \mathbb{C}$ on $A$.

# Representation Theory

A representation of $\mathbb{G}$ is a linear map $\kappa:V\rightarrow V\otimes A$ that satisfies

$\left(\kappa\otimes I_A\right)\circ\kappa =\left(I_V\otimes \Delta\right)\circ \kappa\text{\qquad and \qquad}\left(I_V\otimes\varepsilon\right)\circ \kappa=I_V.$

The dimension of $\kappa$ is given by $\dim\,V$. If $V$ has basis $\{e_i\}$ then we can define the matrix elements of $\kappa$ by

$\displaystyle\kappa\left(e_j\right)=\sum_i e_i\otimes\rho_{ij}.$

One property of these that we will use it that $\varepsilon\left(\rho_{ij}\right)=\delta_{i,j}$.

Two representations $\kappa_1:V_1\rightarrow V_1\otimes A$ and $\kappa_2:V_2\rightarrow V_2\otimes A$ are said to be equivalent, $\kappa_1\equiv \kappa_2$, if there is an invertible intertwiner between them. An intertwiner between $\kappa_1$ and $\kappa_2$ is a map $T\in L\left(V_1,V_2\right)$ such that

$\displaystyle\kappa_2\circ T=\left(T\otimes I_A\right)\circ \kappa_1.$

We can show that every representation is equivalent to a unitary representation.

Timmermann shows that if $\{\kappa_\alpha\}_{\alpha}$ is a maximal family of pairwise inequivalent irreducible representation that $\{\rho_{ij}^\alpha\}_{\alpha,i,j}$ is a basis of $A$. When we refer to “the matrix elements” we always refer to such a family. We define the span of $\{\rho_{ij}\}$ as $\mathcal{C}\left(\kappa\right)$, the space of matrix elements of $\kappa$.

Given a representation $\kappa$, we define its conjugate, $\overline{\kappa}:\overline{V}\rightarrow\overline{V}\otimes A$, where $\overline{V}$ is the conjugate vector space of $V$, by

$\displaystyle\overline{\kappa}\left(\bar{e_j}\right)=\sum_i \bar{e_i}\otimes\rho_{ij}^*,$

so that the matrix elements of $\overline{\kappa}$ are $\{\rho_{ij}^*\}$.

Timmermann shows that the matrix elements have the following orthogonality relations:

• If $\alpha$ and $\beta$ are inequivalent then $h\left(a^*b\right)=0,$ for all $a\in \mathcal{C}\left(\kappa_\alpha\right)$ and $b\in\mathcal{C}\left(\kappa_\beta\right)$.
• If $\kappa$ is such that the conjugate, $\overline{\kappa}$, is equivalent to a unitary matrix (this is the case in the finite dimensional case), then we have

$\displaystyle h\left(\rho_{ij}^*\rho_{kl}\right)=\frac{\delta_{i,k}\delta_{j,l}}{d_\alpha}.$

This second relation is more complicated without the $S^2=I_A$ assumption and refers to the entries and trace of an intertwiner $F$ from $\kappa$ to the coreprepresention with matrix elements $\{S^2\left(\rho_{ij}\right)\}$. If $S^2=I_A$, then this intertwiner is simply the identity on $V$ and so the the entries $\left[F\right]_{ij}=\delta_{i,j}$ and the trace is $d=\dim V$.

Denote by $\text{Irr}(\mathbb{G})$ the set of unitary equivalence classes of irreducible unitary representations of $\mathbb{G}$. For each $\alpha\in\text{Irr}(\mathbb{G})$, let $\kappa_\alpha:V_{\alpha}\rightarrow V_{\alpha}\otimes A$ be a representative of the class $\alpha$ where $V_\alpha$ is the finite dimensional vector space on which $\kappa_\alpha$ acts.

# Diaconis-Van Daele Fourier Theory

We define a map $\mathcal{F}:A\rightarrow A'$ by

$\displaystyle \mathcal{F}\left(a\right)b=h\left(ba\right).$

We define

$\displaystyle \widehat{A}:=\{\mathcal{F}\left(a\right)\,:\,a\in A\}\subset A'.$

I think in the finite dimensional case that $\widehat{A}=A'$ but I don’t think I necessarily need this.
Van Daele shows that this map allows us to give $\widehat{A}$ the structure of a quantum group with Haar state given by

$\displaystyle \widehat{h}\left(\mathcal{F}\left(a\right)\right)=\varepsilon\left(a\right).$

He also proves a Plancherel theorem which says that

$\displaystyle \widehat{h}\left(\mathcal{F}\left(a\right)^*\mathcal{F}\left(b\right)\right)=h\left(a^*b\right),$

where the star involution in $\widehat{A}$ is given by

$\displaystyle \mathcal{F}\left(a\right)^*\left(b\right)=\overline{\mathcal{F}\left(a\right)S\left(b\right)^*}.$

Given a representation $\kappa_\alpha:V_\alpha\rightarrow V_\alpha\otimes A$ ($\alpha\in\text{Irr}(\mathbb{G})$), following Wang we define the Fourier Transform of $a$ at the representation $\kappa_\alpha$ as the map $\widehat{a}\left(\alpha\right)\in L\left(\overline{V}\right)$ by

$\displaystyle \widehat{a}\left(\kappa\right)=\left(I_{\overline{V}}\otimes \mathcal{F}\left(a\right)\right)\overline{\kappa_{\alpha}}.$

## Diaconis-Van Daele Inversion Theorem

Let $\varepsilon$ be the counit of $A$ and $a\in A$. Then we have

$\displaystyle \varepsilon(a)=\sum_{\alpha\in\text{Irr}(\mathbb{G})}d_{\alpha} \text{Tr}(\hat{a}(\alpha))$

where the sum is over the irreducible representations of $A$.

Proof: Both sides are linear in $a$ so it suffices to check $a=\rho_{kl}^{\beta}$ for $\beta\in\text{Irr}(\mathbb{G})$. The left-hand side reads

$\displaystyle \varepsilon\left(\rho_{kl}^\beta\right)=\delta_{k,l}.$

To calculate the right-hand-side, we calculate for a given representation the trace of $\widehat{a}\left(\alpha\right)$. Let $\overline{e_j}\in \overline{V_\alpha}$ and calculate

$\displaystyle \widehat{\rho_{kl}^\beta}\left(\alpha\right)\overline{e_j}=\left(I_{\overline{V_\alpha}}\otimes\mathcal{F}\left(\rho_{kl}^\beta\right)\right)\sum_{i} \overline{e_i}\otimes\left(\rho_{ij}^\alpha\right)^*$

$\displaystyle =\sum_i \mathcal{F}\left(\rho_{kl}^\beta\right)\left(\rho_{ij}^{\alpha}\right)^* \overline{e_i}$

$\displaystyle =\sum_i h(\left(\rho_{ij}^{\alpha}\right)^*\rho_{kl}^\beta)\overline{e_i}$

This is zero unless $\alpha\equiv \beta$. If $\alpha\equiv\beta$ then we have

$\displaystyle \widehat{\rho_{kl}^\beta}\left(\alpha\right)\overline{e_j}=h\left(\left(\rho_{kj}^\beta\right)^*\rho_{kl}^\beta\right)\overline{e_k}$

$\displaystyle =\frac{1}{d_\beta}\delta_{j,l}\overline{e_k}.$

How much of $\overline{e_j}$ is sent to $\overline{e_j}$:

$\displaystyle \left\langle \overline{e_j},\widehat{\rho_{kl}^\beta}\left(\beta\right)\right\rangle_{\overline{V_{\beta}}}=\frac{1}{d_\beta}\delta_{l,j}\delta_{j,k}$

$\displaystyle \Rightarrow\text{Tr}\left(\widehat{\rho_{kl}^\beta}(\beta)\right)=\sum_j\frac{1}{d_\beta}\delta_{j,l}\delta_{j,k}$

$\displaystyle =\frac{1}{d_\beta}\delta_{k,l}.$

Now $d_\beta$ times this trace is $\delta_{k,l}$ and so we are done $\bullet$
Now there are two convolution theorems at play. The first is Van Daele’s:

## Van Daele’s Convolution Theorem

For all $a,\,b\in A$ we have

$\displaystyle \mathcal{F}\left(a\right)\ast_{\widehat{A}}\mathcal{F}\left(b\right)=\mathcal{F}(a\ast_{A}b),$

where $\ast_{A}$ is a rather messy convolution in $A$ and $\ast_{\widehat{A}}$ is the nice convolution got from $A'$:

$\displaystyle \left(\nu\ast \mu\right)\left(a\right)=\left(\nu\otimes\mu\right)\Delta \left(a\right).$

We also have something we can salvage from Timmermann (third line) which combined with this yields:

## Diaconis-Van Daele Convolution Theorem

For a representation $\kappa_\alpha$ of $\mathbb{G}$ and $a,\,b\in A$ we have

$\displaystyle \widehat{a\ast_{A}b}\left(\alpha\right)=\widehat{a}\left(\alpha\right)\circ\widehat{b}\left(\alpha\right).$

Proof:

$\displaystyle \widehat{\left(a\ast_A b\right)}\left(\alpha\right)=\left(\overline{I_{V_\alpha}}\otimes\mathcal{F}\left(a\ast_A b\right)\right)\overline{\kappa_\alpha}$

$=\displaystyle \left(I_{\overline{V_\alpha}}\otimes \mathcal{F}\left(a\right)\ast_{\widehat{A}}\mathcal{F}\left(b\right)\right)\overline{\kappa_\alpha}$

$= \displaystyle \left(\left(I_{\overline{V_\alpha}}\otimes\mathcal{F}\left(a\right)\right)\overline{\kappa_\alpha}\right)\circ\left(I_{\overline{V_\alpha}}\otimes\mathcal{F}\left(b\right)\right)\overline{\kappa_\alpha}$

$\displaystyle =\widehat{a}\left(\alpha\right)\circ\widehat{b}\left(\alpha\right)\qquad\bullet$

## Van Daele Plancherel Theorem

For all $a,\,b\in A$, we have

$\displaystyle \widehat{h}\left(\mathcal{F}\left(a\right)^*\mathcal{F}\left(b\right)\right)=h\left(a^*b\right).$

As it happens, I should point out that there is a Van Daele Inversion Theorem which says that

$\displaystyle a=\widehat{h}\left(\widehat{S}\left(\cdot\right)\mathcal{F}\left(a\right)\right).$

where $a\in \widehat{\widehat{A}}\cong A$ (and so an element of the dual goes in the $\widehat{S}$) and the antipode on $\widehat{A}$ is given by

$\displaystyle \widehat{S}\left(\mathcal{F}\left(a\right)\right)b=\mathcal{F}\left(a\right)S\left(b\right).$

I haven’t used this… yet.

## Lemma

Where the sum is over irreducible and unitary representations,

$\displaystyle \widehat{h}\left(\mathcal{F}\left(a\right)\ast\mathcal{F}\left(b\right)\right)=\sum_{\alpha\in\text{Irr}(\mathbb{G})}{d_\alpha}\text{ Tr}\left(\widehat{a}\left(\alpha\right)\widehat{b}\left(\alpha\right)\right).$

Proof: The proof uses the convolution theorem of Van Daele and the definition of $\widehat{h}$ to find

$\displaystyle \widehat{h}\left(\mathcal{F}\left(a\right)\ast\mathcal{F}\left(b\right)\right)=\widehat{h}\left(\mathcal{F}\left(a\ast_{A}b\right)\right)=\varepsilon\left(a\ast_A b\right).$

Now use the Diaconis-Van Daele Inversion Theorem

$\displaystyle \varepsilon\left(a\ast_A b\right)=\sum_{\alpha\in\text{Irr}(\mathbb{G})}d_\alpha\text{ Tr}\left(\widehat{a\ast_A b}\left(\alpha\right)\right)$

$\displaystyle =\sum_{\alpha\in\text{Irr}(\mathbb{G})}d_\alpha\text{ Tr}\left(\widehat{a}\left({\alpha}\right)\widehat{b}\left({\alpha}\right)\right)\qquad\bullet$

Two more results that will be used in the final proof.

## Proposition

Suppose that $\mathcal{F}\left(a\right)$ is a state, then where $\kappa_\tau$ is the trivial representation, $\lambda\mapsto \lambda\otimes 1_A$, we have $\widehat{a}\left(\tau\right)=1$.

Proof:

$\widehat{a}\left(\tau\right)\lambda=\left(I_{\mathbb{C}}\otimes\mathcal{F}\left(a\right)\right)\overline{\kappa_{\tau}}\left(\lambda\right)$

$\displaystyle =\left(I_{\mathbb{C}}\otimes \mathcal{F}\left(a\right)\right)\lambda\otimes 1_A^*$

$\displaystyle =\lambda\otimes1=\lambda\qquad\bullet$

## Proposition

Suppose that $\kappa_{\alpha}$ is a non-trivial and irreducible representation, then $\widehat{1_A}\left(\alpha\right)=0$.

Proof: A calculation:

$\displaystyle \widehat{1_A}\left(\alpha\right)\overline{e_j}=\left(I\otimes\mathcal{F}\left(1_A\right)\right)\sum_i\overline{e_i}\otimes\left(\rho_{ij}^\alpha\right)^*$

$\displaystyle =\sum_i\overline{e_i}\mathcal{F}\left(1_A\right)\left(\rho_{ij}^\alpha\right)^*$

$\displaystyle =\sum_{i}h\left(\left(\rho_{ij}^\alpha\right)^*1_A\right)\overline{e_i}.$

Note that $1_A=1_A^*$ is the matrix element of the trivial representation and $\alpha$ is not equivalent to the trivial representation. Therefore, by the first orthogonality relation

$\displaystyle h\left(\left(\rho_{ij}^\alpha\right)^*1_A\right)=0,$

and so $\widehat{1_A}\left(\alpha\right)=0$ as required $\bullet$

Note in particular that $\mathcal{F}\left(1_A\right)=h$. One more result and definition before we prove the upper bound lemma.

## Proposition

If $\kappa_\alpha$ is unitary, then the map, $\widehat{A}\rightarrow L\left(\overline{V_{\alpha}}\right)$, given by $\mathcal{F}\left(a\right)\mapsto \widehat{a}\left(\alpha\right)$ is a $*$-homomorphism.

Proof: Let us call the map $\mathcal{F}(a)\mapsto \hat{a}(\alpha)$ by $\Gamma_\alpha$: we want to show that $\Gamma_{\alpha}(\mathcal{F}(a)^*)=\Gamma_{\alpha}(\mathcal{F}(a))^*$, where the first involution is in $\hat{A}$ and the second is in $L(\overline{V_\alpha})$. First take $\overline{e_i}$. We have

$\Gamma_{\alpha}\left(\mathcal{F}(a)^*\right)\overline{e_i}=\left(I_{\overline{V_\alpha}}\otimes \mathcal{F}(a)^*\right)\overline{\kappa_\alpha}(\overline{e_i})$

$\displaystyle =\left(I_{\overline{V_\alpha}}\otimes \mathcal{F}(a)^*\right)\sum_j \overline{e_j}\otimes\left(\rho_{ij}^\alpha\right)^*$

$\displaystyle =\sum_j\overline{\mathcal{F}(a)S\left(\left(\rho_{ij}^\alpha\right)^*\right)^*}\overline{e_j}$

$\displaystyle =\sum_j\overline{\mathcal{F}(a)\rho_{ji}^\alpha}\overline{e_j}$

$\displaystyle =\sum_j\overline{h(\rho_{ji}^\alpha a)}\overline{e_j}.$

Now looking at the other quantity:

$\displaystyle \Gamma_{\alpha}(\mathcal{F}(a))\overline{e_i}=\hat{a}(\alpha)\overline{e_i}$

$\displaystyle =(I_{\overline{V_\alpha}}\otimes \mathcal{F}(a))\overline{\kappa_\alpha}(\overline{e_i})$

$\displaystyle =(I_{\overline{V_\alpha}}\otimes \mathcal{F}(a))\sum_j\overline{e_j}\left(\rho_{ij}^\alpha\right)^*$

$\displaystyle =\sum_jh\left(\left(\rho_{ij}^\alpha\right)^*a\right)\overline{e_j}.$

Considering that the involution in $L(\overline{V_\alpha})$ is the conjugate-transpose this is enough to show the result $\bullet$

## Definition

Define $\|\cdot\|$ as the norm on $\widehat{A}$ got from the inner product on $\widehat{A}$ given by

$\displaystyle \left\langle\mathcal{F}\left(a\right),\mathcal{F}\left(b\right)\right\rangle=\widehat{h}\left(\mathcal{F}\left(a\right)^*\mathcal{F}\left(b\right)\right).$

i.e. $\|\mathcal{F}\left(a\right)\|=\sqrt{\widehat{h}\left(\mathcal{F}\left(a\right)^*\mathcal{F}\left(a\right)\right)}.$

## Quantum Diaconis-Shahshahani (Upper Bound) Lemma

Suppose that $\mathcal{F}\left(a\right)$ is a state and $N$ a natural number. Then

$\displaystyle\|\mathcal{F}\left(a\right)^{\ast N}-h\|^2=\sum_{\alpha\in\text{Irr}(\mathbb{G})\backslash\{\tau\}}d_\alpha\text{Tr} \left[\left(\widehat{a}\left(\alpha\right)^*\right)^N\left(\widehat{a}\left(\alpha\right)\right)^N\right],$

where the sum is over all non-trivial, irreducible representations.

Proof: Writing

$\displaystyle \|\mathcal{F}\left(a\right)-h\|^2=\widehat{h}\left(\left(\mathcal{F}\left(a\right)-\mathcal{F}\left(1_A\right)\right)^*\left(\mathcal{F}\left(a\right)-\mathcal{F}\left(1_A\right)\right)\right)$

$\displaystyle =\widehat{h}\left(\mathcal{F}\left(a-1_A\right)^*\mathcal{F}\left(a-1_A\right)\right).$

Now using the above lemma, and the fact that the map $\mathcal{F}\left(a\right)\mapsto \widehat{a}\left(\kappa\right)$ is a $*$-homomorphism, we have that this is given by

$\displaystyle \|\mathcal{F}\left(a\right)-h\|^2=\sum_{\alpha}\text{Tr} \left[\widehat{\left(a-1_A\right)}\left(\kappa\right)^*\widehat{\left(a-1_A\right)}\left(\kappa\right)\right],$

where the sum is over all irreducible representations.

Now note that

$\displaystyle \widehat{\left(a-1\right)}\left(\alpha\right)=\widehat{a}\left(\alpha\right)-\widehat{1_A}\left(\alpha\right).$

If $\alpha=\tau$, the trivial representation, then this yields zero as both terms are one. If $\alpha$ is non-trivial, then $\widehat{1_A}\left(\alpha\right)=0$ and we have:

$\displaystyle \|\mathcal{F}\left(a\right)-h\|^2=\sum_{\alpha\in\text{Irr}(\mathbb{G})\backslash\{\tau\}}d_\alpha\text{Tr} \left[\widehat{a}\left(\alpha\right)^*\widehat{a}\left(\alpha\right)\right].$

To get the result, apply the Diaconis-Van Daele Convolution Theorem $N$ times $\bullet$

# What is Next?

The norm that we are measuring $\mathcal{F}(a)^{\ast N}-h$ with is not necessarily the most natural. I want/need to find a norm, $\|\cdot\|'$, on $\hat{A}$ that is perhaps natural in that lower bounds on $\mathcal{F}(a)^{\ast N}-h$ are available. Presumably it will be the case that

$\displaystyle \|\cdot\|'\leq K\|\cdot\|$,

where $K$ is some constant dependent on $A$ or perhaps even $\dim A$.

After this I want to look at this stuff in the cocommutative case…. and the classical case.

Also I heard that there might be a family of quantum groups $A_n$ (of ‘Kac-Paljutkin’-type) on which I could study families of random walks…

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