The end of the previous Research Log suggested a way towards showing that $p_0$ can be associated to an idempotent state $\int_S$. Over night I thought of another way.

Using the Pierce decomposition with respect to $p_0$ (where $q_0:=\mathbf{1}_G-p_0$),

$F(G)=p_0F(G)p_0+p_0F(G)q_0+q_0F(G)p_0+q_0F(G)q_0$.

The corner $p_0F(G)p_0$ is a hereditary $\mathrm{C}^*$-subalgebra of $F(G)$. This implies that if $0\leq b\in p_0F(G)p_0$ and for $a\in F(G)$, $0\leq a\leq b\Rightarrow a\in p_0F(G)p_0$.

Let $\rho:=\nu^{\star d}$. We know from Fagnola and Pellicer that $T_\rho(p_0)=p_0$ and $T_\rho(p_0F(G)p_0)=p_0F(G)p_0$.

By assumption in the background here we have an irreducible and periodic random walk driven by $\nu\in M_p(G)$. This means that for all projections $q\in 2^G$, there exists $k_q\in\mathbb{N}$ such that $\nu^{\star k_q}(q)>0$.

Define:

$\displaystyle \rho_n=\frac{1}{n}\sum_{k=1}^n\rho^{\star k}$.

Define:

$\displaystyle n_0:=\max_{\text{projections, }q\in p_0F(G)p_0}\left\{k_q\,:\,\nu^{\star k_q}(q)> 0\right\}$.

The claim is that the support of $\rho_{n_0}$, $p_{\rho_{n_0}}$ is equal to $p_0$.

We probably need to write down that:

$\varepsilon T_\nu^k=\nu^{\star k}$.

Consider $\rho^{\star k}(p_0)$ for any $k\in\mathbb{N}$. Note

\begin{aligned}\rho^{\star k}(p_0)&=\varepsilon T_{\rho^{\star k}}(p_0)=\varepsilon T^k_\rho(p_0)\\&=\varepsilon T^k_{\nu^{\star d}}(p_0)=\varepsilon T_\nu^{kd}(p_0)\\&=\varepsilon(p_0)=1\end{aligned}

that is each $\rho^{\star k}$ is supported on $p_0$. This means furthermore that $\rho_{n_0}(p_0)=1$.

Suppose that the support $p_{\rho_{n_0}}. A question arises… is $p_{\rho_{n_0}}\in p_0F(G)p_0$? This follows from the fact that $p_0\in p_0F(G)p_0$ and $p_0F(G)p_0$ is hereditary.

Consider a projection $r:=p_0-p_{\rho_{n_0}}\in p_0F(G)p_0$. We know that there exists a $k_r\leq n_0$ such that

$\nu^{\star k_r}(p_0-p_{\rho_{n_0}})>0\Rightarrow \nu^{\star k_r}(p_0)>\nu^{\star k_r}(p_{\rho_{n_0}})$.

This implies that $\nu^{\star k_r}(p_0)>0\Rightarrow k_r\equiv 0\mod d$, say $k_r=\ell_r\cdot d$ (note $\ell_r\leq n_0$):

\begin{aligned}\nu^{\star \ell_r\cdot d}(p_0)&>\nu^{\star \ell_r\cdot d}(p_{\rho_{n_0}})\\\Rightarrow (\nu^{\star d})^{\star \ell_r}(p_0)&>(\nu^{\star d})^{\star \ell_r}(p_{\rho_{n_0}})\\ \Rightarrow \rho^{\star \ell_r}(p_0)&>\rho^{\star \ell_r}(p_{\rho_{n_0}})\\ \Rightarrow 1&>\rho^{\star \ell_r}(p_{\rho_{n_0}})\end{aligned}

By assumption $\rho_{n_0}(p_{\rho_{n_0}})=1$. Consider

$\displaystyle \rho_{n_0}(p_{\rho_{n_0}})=\frac{1}{n_0} \sum_{k=1}^{n_0}\rho^{\star k}(p_{\rho_{n_0}})$.

For this to equal one every $\rho^{\star k}(p_{\rho_{n_0}})$ must equal one but $\rho^{\star \ell_r}(p_{\rho_{n_0}})<1$.

Therefore $p_0$ is the support of $\rho_{n_0}$.

Let $\rho_\infty=\lim \rho_n$. We have shown above that $\rho^{\star k}(p_0)=1$ for all $k\in\mathbb{N}$. This is an idempotent state such that $p_0$ is its support (a similar argument to above shows this). Therefore $p_0$ is a group like projection and so we denote it by $\mathbf{1}_S$ and $\int_S=d\mathcal{F}(\mathbf{1}_S)$!

## But is it a subgroup? But is it a normal subgroup? Can we construct a normal quasi-subgroup?

My feeling at the moment is that $p_0$ is probably not a subgroup. One idea I had was to look at an element $f\in F(G)$ such that $f\mathbf{1}_S-\mathbf{1}_Sf\neq 0$, and see can I get anything from that.

A good idea would be to look for a counterexample. Perhaps dual groups might be a good place to look? After all, we can construct non-Haar idempotents for any non-abelian finite $G$ using $\mathbb{C}G$. Let $H\leq G$ be non-normal and consider $\mathbf{1}_H$. If it is a state, it is an idempotent state. We need to find a unitary representation $\rho$ of $G$ and a unit vector $\xi$ such that:

$\mathbf{1}_G(s)=\langle \rho(s)\xi,\xi\rangle$

Bekka, de la Harpe, and Vallette say this is possible (Exercise C. 6. 7).

Now… what does periodicity look like in $\mathbb{C}$. Is it even possible?

Suppose that $p_0,p_1,\dots p_{d-1}$ are cyclic for $T_u$. Let

$p_i=\sum_{g\in G}\alpha_g^i\delta^g$.

It is straightforward to show that for all $k\in\mathbb{N}$:

$T^k_u(p_i)=\sum_{g\in G}\alpha^i_g u(\delta_g)^k\delta^g$.

Letting $k=d$ shows that all $u(\delta^g)$ are $d$-th roots of unity:

$u(\delta^g)=e^{2\pi i k_g/d}$.

This means that $u^{\star d}$ is a characteristic function.

What we need to invoke now is irreducibility. Every subgroup $H\leq G$ induces a projection:

$p_H=\frac{1}{|H|}\sum_{h\in H}\delta^h$.

Every element $g\in G$ is in some subgroup of $G$ (say, $\langle g\rangle$).

We are assuming that the random walk given by $u\in M_p(\widehat{G})$ is irreducible. Therefore there exists a $k_g\in\mathbb{N}$ such that

$u^{\star k}(p_{\langle g\rangle})=u(p_{\langle g\rangle})^k>0$

this seems promising but does not go anywhere.

My feeling is that periodicity is not compatible with irreducibility for random walks on $\widehat{G}$. We will see can we show this on St Patrick’s Day! Perhaps assuming irreducibility will be faster (and showing it implies aperiodicity?)