Taken from Random Walks on Finite Quantum Groups by Franz & Gohm

In this section we want to represent the algebras on Hilbert spaces and obtain spatial implementations for the random walk. On a finite quantum group A we can introduce an inner product

\langle a,b\rangle=\eta(a^*b),

where a,\,b\in A and \eta and \eta is the Haar state. Because the Haar state is faithful we can think of A as a finite dimensional Hilbert space. Further we denote by \|\cdot\| the norm associated to this inner product. We consider the linear operator

W:A\otimes A\rightarrow A\otimes Ab\otimes a\mapsto \Delta(b)(1_A\otimes a).

It turns out that this operator contains all information about the quantum group and thus it is called its fundamental operator. We discuss some of its properties.


W is unitary.

Proof : Using (\eta\otimes I_A)\circ\Delta=\eta(\cdot)1_A it follows that

\displaystyle \|W(b\otimes a)\|^2=\|\Delta(b)(1_A\otimes a)\|^2=(\eta\otimes \eta)\left((1_A\otimes a^*)\Delta(b^*b)(1_A\otimes a)\right)

\displaystyle=\eta\left(a^*\left[(\eta\otimes I_A)\Delta(b^*b)\right]a\right)=\eta(a^*\eta(b^*b)a)=\eta(b^*b)\eta(a^*a)

=(\eta\otimes\eta)(b^*b\otimes a^*a)=\|b\otimes a\|^2.

A similar computation works for \sum_i b_i\otimes a_i instead of b\otimes a. Thus W is isometric and, because A is finite dimensional, also unitary \bullet

I am confused here — I can’t see the second, third nor the last two equalities.

It can easily be checked using Sweedler’s notation that with the antipode S the inverse W^{-1}=W^* can be written explicitly as

W^{-1}(b\otimes a)=[(I_A\otimes S)\Delta b](1_A\otimes a).


W satisfies the Pentagon Equation


This is an equation on A\otimes A\otimes A and we have used the leg notation W_{12}=W\otimes 1_AW_{23}=1_A\otimes WW_{13}=(1_A\otimes \tau)(W\otimes 1_A)(1_A\otimes \tau).

Proof : Just a straight, fun, calculation \bullet


The pentagon equation expresses the coassociativity of the comultiplication \Delta. Unitaries satisfying the pentagon equation have been called multiplicative unitaries.

The operator L_c of left multiplication by c\in A:

L_a:A\rightarrow Aa\mapsto ca

will often be written as c in the following. It is always clear whether c\in A or c:A\rightarrow A is meant. We can also look at left multiplications as a faithful representation L of the C*-algebra A acting on itself. In this sense we have


\Delta (a)=W(a\otimes 1_A)W^* for all a\in A.

Proof : Here \Delta (a) and a\otimes 1_A are left multiplication operators on A\otimes A. The formula can via another calculation \bullet

By left multiplication we can also represent a random walk on a finite quantum group A. Then j_n(a) becomes an operator on a (n+1)-fold tensor product of A. To get used to it let us show how the pentagon equation is related to Proposition 3.1.

Theorem 5.2

j_n(a)=W_{01}W_{02}\cdots W_{0n}(a\otimes 1_A\otimes\cdots\otimes 1_A)W^*_{0n}\cdots W^*_{02}W^*{01}.

{W_{01}W_{02}\cdots W_{0n}}_{|A}={W_{n-1,n}W_{n-2,n-1}\cdots W_{01}}_{|A},

where {\,}_{|A} means restriction to A\otimes 1_A\otimes\cdots\otimes1_A and this left position gets the number zero.

I can’t make head nor tail of this notation. The only thing that makes sense to me is something like

\underbrace{A}_{0}\otimes \underbrace{A}_{1}\otimes \cdots\otimes \underbrace{A}_{n}

and W_{oi}:A^{\otimes (n+1)}\rightarrow A^{\otimes (n+1)} acts something like:

a\otimes \cdots\otimes\underbrace{b}_{i}\otimes\cdots \mapsto \Delta(a)(1_A\otimes 1_A\otimes\cdots\otimes \underbrace{b}\otimes\cdots).

Not much point doing the proof when I can’t understand the notation (the proof doesn’t really help with the notation either).

It is an immediate but remarkable consequence of this representation that we have a canonical way of extending our random walk to B(A), the C*-algebra of all (bounded) linear operators on A. Namely, we can for n\geq 0 define the random variables

\displaystyle J_n:B(A)\rightarrow B\left(\bigotimes_{0}^nA\right)\cong \bigotimes_0^nB(A),

T\mapsto W_{01}W_{02}\cdots W_{0n}(T\otimes I_A\otimes\cdots I_A)W^*_{0n}\cdots W^*_{02}W^*_{01}

There is not much point going any further until I can clear up the notation of Theorem 5.2.

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