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}}<p_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?)