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!