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

Theorem

Let \phi be a state on a finite quantum group A. Then the Cesaro mean

\displaystyle \phi_n=\frac{1}{n}\sum_{k=1}^n\phi^{\star n}n\in\mathbb{N}

converges to an idempotent state on A, i.e. to a state \pi such that \pi\star\pi=\pi.

Proof : Let \phi' be an accumulation point of \{\phi_n\}_{n\geq0}, this exists since the states on A form a compact set. We have

\|\phi_n-\phi\star \phi_n\|=\frac{1}{n}\|\phi-\phi^{n+1}\|\leq \frac{2}{n}.

I have no idea where the equality comes from.

Choose sequence \{n_k\}_{k\geq 0} such that \phi_{n_k}\rightarrow \phi', we get \phi\star\phi'=\phi' and similarly \phi'\star \phi=\phi'. By linearity this implies \phi_n\star\phi'=\phi'=\phi'\star \phi_n. If \phi'' is another accumulation point of \{\phi_n\}_{n\geq 0} and \{m_{\ell}\}_{\ell\geq 0} a sequence such that \phi_{m_\ell}\rightarrow\phi'', then we get \phi''\star\phi'=\phi'=\phi'\star\phi'' and thus \phi'=\phi'' by symmetry (??). Therefore the sequence \{\phi_n\}_{n\geq0} has a unique accumulation point, i.e. it converges \bullet

Remark

If \phi is faithful, then the Cesaro limit \pi is the Haar state on A (prove this).

Remark

Due to cyclicity the sequence \{\phi^{\star n}\}_{n\geq 0} does not converge in general. Take, for example, the state \phi=\eta_2 (p.28) on the Kac-Paljutkin quantum group A, then we have

\eta_2^{\star n}=\left\{\begin{array}{ccc}\eta_2&\text{if}& n\text{ is odd}\\ \varepsilon&\text{if}&n\text{ is even}\end{array}\right.,

but

\displaystyle \lim_{n\rightarrow\infty}\frac1{n}\sum_{k=1}^n\eta_2^{\star k}=\frac{\varepsilon+\eta_2}2.

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