## Quasi-Subgroups that are not Subgroups

Let be a finite quantum group. We associate to an idempotent state a *quasi-*subgroup . This nomenclature must be included in the manuscript under preparation.

As is well known from the GNS representation, positive linear functionals can be associated to closed left ideals:

.

In the case of a quasi-subgroup, , my understanding is that by looking at we can tell if is actually a subgroup or not. Franz & Skalski show that:

Let be a quasi-subgroup. TFAE

- is a subgroup
- is a two-sided or self-adjoint or invariant ideal of

I want to look again at the Kac & Paljutkin quantum group and see how the Pal null-spaces and fail these tests. Both Franz & Gohm and Baraquin should have the necessary left ideals.

### The Pal Null-Space

The following is an idempotent probability on the Kac-Paljutkin quantum group:

.

Hence:

.

If were two-sided, . Consider and

.

We see problems also with when it comes to the adjoint and also . It is not surprise that the adjoint AND the antipode are involved as they are related via:

.

In fact, for finite or even Kac quantum groups, .

Can we identity the support ? I think we can, it is (from Baraquin)

.

This does not commute with :

.

The other case is similar.

This tells us something. Unfortunately that something is rather ‘finite-specific’, but I guess we have thrown our eggs in that basket at this stage. The algebra of functions on a finite quantum group is a ‘full’ matrix algebra:

,

and the only central projections are of the form:

.

This probably does not give us much *unless* we can attack from a representation-theoretic-angle.

## More Notes

One thing I have not explicitly done is show that in the case of a periodic random walk is that the projection is a group-like projection. This is pretty vital to anything that I am doing.

There are two ways to come at this problem. The first is to show that satisfies:

.

It is possibly easier to come at this from the other side: show that is the support of an invariant state. Franz & Skalski show that if is idempotent, there exists a group-like projection such that:

.

This includes a normalisation factor. It might be an idea to prove that this is indeed the support of .

First of all, we do have that:

.

Note

.

Let be the support projection of , the smallest projection such that . This implies that so that is a projection. Thus:

,

so that . Let . We can show that and for all . In particular:

.

It is clear that is not enough to show that is the support projection — for example .

We need to use the fact, where , that

,

and that is faithful. Suppose that so that is a projection.

Consider

.

Note

,

so that , and as is faithful, , and so the group-like projection of an idempotent state is also its support.

To finish this off, we must show that the idempotent state associated to is indeed idempotent… it should be by . I have a little confusion here… we have invariant states and we have idempotent states. These are not precisely the same thing… I am fairly sure that this “invariant state” is , which is not a state. We might have something like:

.

The problem is this right eigenvector… does it necessarily give a left eigenvector that is a state… This is a problem. Recall that . Can we maybe get from an (actual) invariant state to an idempotent?

Perhaps using the convolution in ,

?

This is now next week’s problem.

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March 13, 2020 at 12:02 pm

Research Note: 13 March 2020 | J.P. McCarthy: Math Page[…] question remains, is ? Ah, we showed in the last note that the group-like projection associated to an idempotent states coincides with its […]