Today, for finite quantum groups, I want to explore some properties of the relationship between a state $\nu\in M_p(G)$, its density $a_\nu$ ($\nu(b)=\int_G ba_\nu$), and the support of $\nu$, $p_{\nu}$.

I also want to learn about the interaction between these object, the stochastic operator

$\displaystyle T_\nu=(\nu\otimes I)\circ \Delta$,

and the result

$T_\nu(a)=S(a_\nu)\overline{\star}a$,

where $\overline{\star}$ is defined as (where $\mathcal{F}:F(G)\rightarrow \mathbb{C}G$ by $a\mapsto (b\mapsto \int_Gba)$).

$\displaystyle a\overline{\star}b=\mathcal{F}^{-1}\left(\mathcal{F}(a)\star\mathcal{F}(b)\right)$.

An obvious thing to note is that

$\nu(a_\nu)=\|a_\nu\|_2^2$.

Also, because

\begin{aligned}\nu(a_\nu p_\nu)&=\int_Ga_\nu p_\nu a_\nu=\int_G(a_\nu^\ast p_\nu^\ast p_\nu a_\nu)\\&=\int_G(p_\nu a_\nu)^\ast p_\nu a_\nu\\&=\int_G|p_\nu a_\nu|^2\\&=\|p_\nu a_\nu\|_2^2=\|a_\nu\|^2\end{aligned}

That doesn’t say much. We are possibly hoping to say that $a_\nu p_\nu=a_\nu$.

Actually this is not difficult. Note that $\nu(1-p_\nu)=0$. Therefore

\begin{aligned} \int_G(1-p_\nu)a_\nu&=0\\ \Rightarrow \int_G(a_\nu-p_\nu a_\nu)&=0\end{aligned},

and note that as the product of positive elements, $(1-p_\nu)a_\nu$ is positive, and as $\int_G$ is faithful, we have $a_\nu=p_\nu a_\nu$. Using traciality of $\int_G$ we have:

$a_\nu=a_\nu p_\nu=p_\nu a_\nu=p_\nu a_\nu p_\nu$.

Note now that $a_\nu\overline{\star}a_\nu=a_{\nu^{\star 2}}$. Suppose now that $\nu$ is an idempotent. We want to compare now both $a_\nu$ and $p_\nu$, and also consider the fact that $a_\nu$ is an idempotent in the ‘convolution algebra’ $(F(G),\overline{\star})$.

First off… perhaps in this case $a_\nu=p_\nu$? Or rather $a_\nu=\tilde{p_\nu}$ where $\tilde{p_\nu}$ is normalised $\tilde{p_\nu}=\frac{1}{\int_G p_\nu}p_\nu$.

Firstly, if $\nu$ is idempotent then it can be thought of as the indicator function on a quasi-subgroup, $\nu=\int_S$ and its support $\mathbf{1}_S$. It is trivial that $a_{\int_S}$ is idempotent in the convolution algebra.

We are thinking about $\mathbf{1}_S=p_0$ for a $T_\nu$-cyclic partition of unity $(p_i)$ but I do not believe I have yet shown that this yet corresponds to a idempotent state!

I want to say that the normalisation factor $\frac{1}{\int_Gp_0}=d\in\mathbb{N}$ but this is true in this $T_\nu$-cyclic case but… OK let us think more generally.

So forget about a cyclic partition of unity for the moment (must prove right-eigenvectors of $T_\mu$ correspond in a bijective way with idempotent states as left-eigenvectors). Just consider the support of an idempotent state $\int_S$, $\mathbf{1}_S$, and its density $a_{\int_S}$.

The proposal is that:

$\displaystyle a_{\int_S}=\frac{1}{\int_G\mathbf{1}_S}\mathbf{1}_S$.

This is about idempotent states only. It is not true in general for states. For example, for classical $G$, if $\nu=(\delta^{g_1}/3+2\delta^{g_2}/3)$, the density is:

$a_\nu=|G|(\delta_{g_1}/3+2\delta_{g_2}/3)$,

but the support is $p_\nu=(\delta_{g_1}+\delta_{g_2})$ which is not equal to $\int_G a_\nu \cdot p_\nu$.

We know that $a_{\int_S}=a_{\int_S}\mathbf{1}_S$, etc. Just to make the notation clear,

$\int_Sb=\int_{G}ba_{\int_S}$.

This might already be in a paper of Franz and Skalski… let me check. Corollary 4.2 implies that if $\int_S$ is an idempotent state, there exists a group-like projection $p_S$ such that $a_{\int_S}=\tilde{p_S}$.

So we have that the density of an idempotent corresponds with $\tilde{p_S}$.

The question remains, is $p_S=\mathbf{1}_S$? Ah, we showed in the last note that the group-like projection associated to an idempotent states coincides with its support.

We have a little result.

Let $\int_S\in M_p(G)$ be an idempotent state. The support of $\int_S$ coincides with the group-like projection, $\mathbf{1}_S$, associated to $\int_S$, and furthermore the density of $\int_S$ is, up to a scaling factor, equal to this support.

## From “Invariant States” to Idempotent States (no a scare quotes)

Suppose that $\nu$ is irreducible and periodic. Then there exists a $T_\nu$-cyclic partition of unity $(p_i)$, and where $\overline{\nu}:=\nu^{\star d}$,

$T_{\overline{\nu}}(p_i)=p_i$.

The plan is to associate to these right-eigenvectors $p_i$ (“invariant states”), an idempotent state $\nu_i$.

My proposal is that somehow for $p_{i\neq 0}$, this cannot be done, but for $p_0$, this can be done, that $p_0$ can be associated to an idempotent state $\int_S$, and perhaps $p_0=\mathbf{1}_S$?

One possibility would be to show that $p_i\overline{\star}p_i\sim p_i$ for $i=0$ only. It might possibly be a good idea to show that the state $\mathcal{F}(\tilde{p_1})$ also shows cyclic behaviour.

There are other possibilities. Show that $\mathcal{F}(\tilde{p_0})^{\star k}\rightarrow \mathcal{F}(\tilde{p_0})$. We know at least as $e$ is in the support of $\mathcal{F}(\tilde{p_0})$ that these powers converge.

Actually no need for these tildes, $\tilde{p_i}=d p_i$ in this context.

We are, recall, trying to show that $p_0$ is a group-like projection.

### The Random Walk given by $\mathcal{F}(dp_1)$

To ease notation let $\rho:=\mathcal{F}(dp_1)$.

We want to show that:

$\displaystyle \left(\mathcal{F}(dp_1)\otimes I\right)\circ \Delta(p_i)=p_{i-1}$

Question: is $a_\nu p_1=a_\nu$. Answer: Yes. Because $\nu(p_1)=1$ and by definition $p_\nu$ is the smallest such projection. Thus $p_1-p_\nu$ is in the null-space and we can show that:

$a_\nu p_1=a_\nu(p_\nu+(p_1-p_\nu))=a_\nu p_\nu+\underbrace{a_\nu(p_1-p_\nu)}_{=0}=a_\nu p_\nu=a_\nu$.

Actually I can rule out any $\mathcal{F}(dp_{i\neq 0})$ being a group-like projection and so idempotent. This is because $\varepsilon(p_{i\neq 0})=0$. Therefore they cannot associate with idempotent states.

### Show me the idempotent!

So we will look back at $T_{\overline{\nu}}(p_0)=p_0$ and note that $p_0$ does not have this barrier.

We know from the second last paragraph that just because $T_{\overline{\nu}}(p_i)=p_i$, it does not follow that we can form an idempotent state. The proposal now is that the invariant $T_{\overline{\nu}}(p_0)=p_0$ corresponds with an idempotent state $\varphi$ and that $\varphi=\mathcal{F}(dp_0)$.

Let us state for the moment that the right-(positive)-eigenvector $p_0$ has a left-eigenvector $\varphi_0\in\mathbb{C}G$ such that:

$\varphi_0 T_{\overline{\nu}}=\varphi_0$.

For ease of notation here let us write $T:=T_{\overline{\nu}}$.

In general, we know that just because a right eigenvector is positive, it does not imply that… woah. Slow down. Fixed points of $T$ do not yet correspond with idempotent states! What this means is (just as we defined it!) that this $T$ isn’t what we want.

Recall that that for $\nu,\,\mu\in M_p(G)$:

$\mu T_\nu=\nu\star \mu$.

Therefore, an idempotent state is a state $\int_S$ such that:

$\int_S\star \int_S=\int_S\Leftrightarrow \int_S T_{\int_S}=\int_S$,

that is we are looking for a state that is not only a fixed point (eigenvector with eigenvalue one) of a stochastic operator, but a state that is the fixed point of a stochastic operator such that the stochastic operator is associated to that state, i.e.

$\nu T_\nu=\nu$.

So forget about $T=T_{\nu^{\star d}}$ and start thinking about, where $\varphi=\mathcal{F}(dp_0)$, $T_{\varphi}$. Could we have:

$\varphi T_\varphi=\varphi$?

It might be the case that we can associate to the fixed point of a stochastic operator $T_\nu$ a fixed state $\mu$, but this gives:

$\mu T_\nu=\mu$,

but of course this merely asserts $\nu\star \mu=\mu$ and this does not an idempotent make!

So we are looking at this $\varphi$. Perhaps it might be instructive at the outset to show that positive right eigenvectors of eigenvalue one can give positive left eigenvectors of eigenvalue one (and thus via scaling fixed point states).

Perhaps we need to use that $T^k$ converges. Carbone and Pautrat say that if we decompose:

$F(G)=\sum_{i=0}^{d-1}p_i F(G)p_i+\cdots$,

that the restriction of $T_{\nu}^d$ to $p_iF(G)p_i$ is ergodic… but this doesn’t quite cut the mustard because it doesn’t differentiate between $i=0$ and $i\neq 0$. In particular, this restriction is most natural when we have the antecedent…

Of course convergence $T_{\overline{\nu}^k}\rightarrow T_{\overline{\nu}_\infty}$ implies convergence $\overline{\nu}^{\star k}\rightarrow \overline{\nu}_\infty$. Perhaps $\overline{\nu}_\infty=\varphi$? Perhaps this limit argument is not what I need?

Let us try and calculate $p_0\overline{\star}p_0$ directly… PERHAPS it might be possible to show that there is no ‘part’ of

$\Delta(p_{i\neq 0})\in p_0F(G)p_0\otimes p_0F(G)p_0$.

If this is the case, for $\varphi:=\mathcal{F}(dp_0)$,

$(\varphi\star \varphi)(p_{i\neq 0})=0$,

and because $\sum p_i=\mathbf{1}_G$, it must be the case that $\varphi^{\star 2}$ is supported by $p_0$… We could keep iterating this and get a limit… or something.

That is enough for today.

PS: Just looking over the post, I forgot something that I said I wanted to talk about. Consider these two conditions (possibly too strong). Note $a_\varphi=dp_0$. If I can prove:

• $S(p_0)=p_0$ (fairly sure this is true), and
• $T_\varphi(a_\varphi)=a_{\varphi}$ (we’ll see), then

$a_\varphi=T_\varphi(a_\varphi)=S(a_\varphi)\overline{\star}a_\varphi=a_{\varphi}\overline{\star}a_{\varphi}=a_{\varphi^{\star 2}}$,

and $\varphi$ is an idempotent! A route to progress!