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.

Advertisements