*Cross-Posted from Math Stackexchange.*

Ergodic TheoremA random walk on a finite group $G$ driven by a probability $\nu\in M_p(G)$ isergodicif $\operatorname{supp}(\nu)$ is not concentrated on a proper subgroup $S\subset G$ nor the coset of a normal subgroup $N\triangleleft G$.

In this case the convolution powers of $\nu$ converge to the uniform distribution $\pi$ on $G$:

$$\nu^{\star k}\rightarrow \pi.$$

Where $\|\cdot \|=\frac12\|\cdot\|_{\ell_1}$, $$(\nu\star \nu)(g)=\sum_{t\in G}\nu(gt^{-1})\nu(t),$$ $d_\alpha$ is the dimension of a representation $\rho_\alpha:G\rightarrow \operatorname{GL}(V)$, $$\hat{\nu}(\rho)=\sum_{t\in G}\nu(t)\rho(t),$$ and $T^*$ denotes the conjugate transpose of $T$ in $\operatorname{GL}(V)$, Diaconis & Shahshahani proved the following:

Upper Bound LemmaWhere $\operatorname{Irr}(G)\backslash \tau$ is the set of non-trivial unitary irreducible representations on $G$: $$\|\nu^{\star k}-\pi\|^2\leq \frac{1}{4}\sum_{\rho_\alpha\in \operatorname{Irr}(G)\backslash \tau}d_\alpha \operatorname{Tr}[\widehat{\nu}(\rho_\alpha)^k(\widehat{\nu}(\rho_\alpha)^*)^k].$$

The Upper Bound Lemma still holds if the random walk driven by $\nu$ is not ergodic.

Note that the sum over the non-trivial irreducible representations is *equal* (up to a constant) to $\|\nu^{\star k}-\pi\|_{\ell_2}^2$ and so can detect convergence.

**Question: Can the Upper Bound Lemma be used to prove the Ergodic Theorem?**

Can the Upper Bound Lemma show that for $\nu^{\star k}$ to converge to $\pi$ it is necessary that $\nu$ is not supported on a subgroup (irreducibility)? I suspect aperiodicity (not concentrated on the coset of normal subgroup) might be harder.

My own MSc thesis should be a good reference for some of this.

**Background:** It is possible to prove an Upper Bound Lemma for finite *quantum* groups, however finding necessary and sufficient conditions for convergence to uniform (convergence to the Haar state) is an open problem. If the Upper Bound Lemma can yield the necessary and sufficient conditions for convergence in the classical case, perhaps something similar might be possible in the quantum case.

upperbound. But I don't see why you want to use it? It is trivial that irreducibility and aperiodicity are necessary. The hard part of the ergodic theorem is the sufficiency. $\endgroup$ – Nate Eldredge Oct 20 '17 at 13:08