Today, for finite quantum groups, I want to explore some properties of the relationship between a state , its density (), and the support of , .

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

,

and the result

,

where is defined as (where by ).

.

An obvious thing to note is that

.

Also, because

That doesn’t say much. We are possibly hoping to say that .

Actually this is not difficult. Note that . Therefore

,

and note that as the product of positive elements, is positive, and as is faithful, we have . Using traciality of we have:

.

Note now that . Suppose now that is an idempotent. We want to compare now both and , and also consider the fact that is an idempotent in the ‘convolution algebra’ .

First off… perhaps in this case ? Or rather where is normalised .

Firstly, if is idempotent then it can be thought of as the indicator function on a quasi-subgroup, and its support . It is trivial that is idempotent in the convolution algebra.

We are thinking about for a -cyclic partition of unity 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 but this is true in this -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 correspond in a bijective way with idempotent states as left-eigenvectors). Just consider the support of an idempotent state , , and its density .

The proposal is that:

.

This is about idempotent states only. It is not true in general for states. For example, for classical , if , the density is:

,

but the support is which is not equal to .

We know that , etc. Just to make the notation clear,

.

This might already be in a paper of Franz and Skalski… let me check. Corollary 4.2 implies that if is an idempotent state, there exists a group-like projection such that .

So we have that the density of an idempotent corresponds with .

The question remains, is ? 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 be an idempotent state. The support of coincides with the group-like projection, , associated to , and furthermore the density of is, up to a scaling factor, equal to this support.

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

Suppose that is irreducible and periodic. Then there exists a -cyclic partition of unity , and where ,

.

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

My proposal is that somehow for , this cannot be done, but for , this can be done, that can be associated to an idempotent state , and perhaps ?

One possibility would be to show that for only. It might possibly be a good idea to show that the state also shows cyclic behaviour.

There are other possibilities. Show that . We know at least as is in the support of that these powers converge.

Actually no need for these tildes, in this context.

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

### The Random Walk given by

To ease notation let .

We want to show that:

Question: is . Answer: Yes. Because and by definition is the smallest such projection. Thus is in the null-space and we can show that:

.

Actually I can rule out any being a group-like projection and so idempotent. This is because . Therefore they cannot associate with idempotent states.

### Show me the idempotent!

So we will look back at and note that does not have this barrier.

We know from the second last paragraph that just because , it does not follow that we can form an idempotent state. The proposal now is that the invariant corresponds with an idempotent state and that .

Let us state for the moment that the right-(positive)-eigenvector has a left-eigenvector such that:

.

For ease of notation here let us write .

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

Recall that that for :

.

Therefore, an idempotent state is a state such that:

,

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.

.

So forget about and start thinking about, where , . Could we have:

?

It might be the case that we *can *associate to the fixed point of a stochastic operator a fixed state , but this gives:

,

but of course this merely asserts and this does not an idempotent make!

So we are looking at this . 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 converges. Carbone and Pautrat say that if we decompose:

,

that the restriction of to is ergodic… but this doesn’t quite cut the mustard because it doesn’t differentiate between and . In particular, this restriction is most natural when we have the antecedent…

Of course convergence implies convergence . Perhaps ? Perhaps this limit argument is not what I need?

Let us try and calculate directly… PERHAPS it might be possible to show that there is no ‘part’ of

.

If this is the case, for ,

,

and because , it must be the case that is supported by … 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 . If I can prove:

- (fairly sure this is true), and
- (we’ll see), then

,

and is an idempotent! A route to progress!

## Leave a comment

Comments feed for this article