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

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.

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

#### (b) $W$ satisfies the Pentagon Equation $W_{12}W_{13}W_{23}=W_{23}W_{12}$.

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

Proof : Just a straight, fun, calculation $\bullet$

#### Remark

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 $c\in A$: $L_a:A\rightarrow A$ $a\mapsto ca$

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

#### (c) $\Delta (a)=W(a\otimes 1_A)W^*$ 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.

$latex$