Slides of a talk given to the Functional Analysis seminar in Besancon.

Some of these problems have since been solved.

### “e in support” implies convergence

Consider a $\nu\in M_p(G)$ on a finite quantum group such that where $M_p(G)\subset \mathbb{C}\varepsilon \oplus (\ker \varepsilon)^*$, $\nu=\nu(e)\varepsilon+\psi$ with $\nu(e)>0$. This has a positive density of trace one (with respect to the Haar state $\int_G\in M_p(G)$), say $\displaystyle a_\nu=\nu(e)\eta+b_\psi\in \mathbb{C}\eta\oplus \ker \varepsilon$,

where $\eta$ is the Haar element.

An element in a direct sum is positive if and only if both elements are positive. The Haar element is positive and so $b_\psi\geq 0$. Assume that $b_\psi\neq 0$ (if $b_\psi=0$, then $\psi=0\Rightarrow \nu=\varepsilon\Rightarrow \nu^{\star k}=\varepsilon$ for all $k$ and we have trivial convergence)

Therefore let $\displaystyle a_{\tilde{\psi}}:=\frac{b_\psi}{\int_G b_\psi}$

be the density of $\tilde{\psi}\in M_p(G)$.

Now we can explicitly write $\displaystyle \nu=\nu(e)\varepsilon+(1-\nu(e))\tilde{\psi}$.

This has stochastic operator $P_\nu=\nu(e)I_{F(G)}+(1-\nu(e))P_{\tilde{\psi}}$.

Let $\lambda$ be an eigenvalue of $P_\nu$ of eigenvector $a$. This yields $\nu(e)a+(1-\nu(e))P_{\tilde{\psi}}(a)=\lambda a$

and thus $\displaystyle P_{\tilde{\psi}}a=\frac{\lambda-\nu(e)}{1-\nu(e)}a$.

Therefore, as $a$ is also an eigenvector for $P_{\tilde{\psi}}$, and $P_{\tilde{\psi}}$ is a stochastic operator (if $a$ is an eigenvector of eigenvalue $|\lambda|>1$, then $\|P_\nu a\|_1=|\lambda|\|a\|_1\leq \|a\|_1$, contradiction), we have $\displaystyle \left|\frac{\lambda-\nu(e)}{1-\nu(e)}\right|\leq 1$ $\Rightarrow |\lambda-\nu(e)|\leq 1-\nu(e)$.

This means that the eigenvalues of $P_\nu$ lie in the ball $B_{1-\nu(e)}(\nu(e))$ and thus the only eigenvalue of magnitude one is $\lambda=1$, which has (left)-eigenvector the stationary distribution of $P_\nu$, say $\nu_\infty$.

If $\nu$ is symmetric/reversible in the sense that $\nu=\nu\circ S$, then $P_\nu$ is self-adjoint and has a basis of (left)-eigenvectors $\{\nu_\infty=:u_1,u_2,\dots,u_{|G|}\}\subset \mathbb{C}G$ and we have, if we write $\nu=\sum_{t=1}^{|G|}a_tu_t$, $\displaystyle \nu^{\star k}=\sum_{t=1}^{|G|}a_t\lambda_t^ku_t$,

which converges to $a_1\nu_\infty$ (so that $a_1=1$).

If $\nu$ is not reversible, it is a standard argument to show that when put in Jordan normal form, that the powers $P_{\nu}^k$ converge and thus so do the $\nu^{\star k}$ $\bullet$

### Total Variation Decrasing

Uses Simeng Wang’s $\|a\star_Ab\|_1\leq \|a\|_1\|b\|_1$. Result holds for compact Kac if the state has a density.

### Periodic $e^2$ is concentrated on a coset of a proper normal subgroup of $\mathfrak{G}_0$ $e_2+e_4$ is a minimal projection (coset) in the quotient space of the normal subgroup (to be double checked) given by $\langle e_1,e_3\rangle$

### Supported on Subgroup implies Reducible

I have a proof that reducible is equivalent to supported on a pre-subgroup.