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…

Advertisements