### Introduction

Every finite quantum group has finite dimensional algebra of functions:

.

At least one of the factors must be one-dimensional to account for the counit , and if this factor is denoted , the counit is given by the dual element . There may be more and so reorder the index so that for , and for :

,

Denote by the states of . The *pure *states of arise as pure states on single factors.

In the case of the Kac-Paljutkin and Sekine quantum groups, the convolution powers of pure states exhibit a periodicity of sorts. Recall for these quantum groups that consists of a single matrix factor.

In these cases, for pure states of the form , that is supported on (and we can say a little more than is necessary), the convolution remains supported on because

.

If we have a pure state supported on , then because

,

then must be supported on, because of , .

Inductively all of the are supported on and the are supported on . This means that the convolutions powers of a pure state, in these cases, cannot converge to the Haar state.

The question is, do the results above about the image of and under the coproduct hold more generally? I believe that the paper of Kac and Paljutkin shows that this is the case whenever consists of a single factor… but does it hold more generally?

To find out we go back and do some sandboxing with the paper of Kac and Paljutkin. Which is a pleasure because that paper is beautiful. The blue stuff is my own scribbling.

## Finite Ring Groups

Let be a finite quantum group with notation on the algebra of functions as above. Note that is commutative. Let

,

which is a central idempotent.

### Lemma 8.1

.

*Proof: *If , then for some , and , the mapping is a non-zero homomorphism from into commutative which is impossible.

If , then one of the , with ‘something’ in . Using the centrality and projectionality of , we can show that the given map is indeed a homomorphism.

It follows that , and so

### Lemma 8.2

*Proof: *Suppose that for some non-commutative . This means that there exists an index such that . Then for that factor,

is a non-null homomorphism from the non-commutative into the commutative.

We see that for all . Putting we get the result

The following says that is a group-like projection. We know from previous work that if a state is supported on a group-like projection that it will remain supported on it. In particular, any state supported on will remain there.

### Lemma 8.3

.

*Proof: *Since is a homomorphism, is an idempotent in . I do not understand nor require the rest of the proof.

### Lemma 8.4

* is the algebra of functions on finite group with elements , and we write . The coproduct is given by .*

We have:

,

,

,

as .

The element is a sum of four terms, lying in the subalgebras:

.

We already know what is going on with the first summand. Denote the second by . From the group-like-projection property, the last two summands are zero, so that

.

Since the are symmetric () mutually orthogonal idempotents, has similar properties:

for .

At this point Kac and Paljutkin restrict to , that is there is only one summand. Here we *try *to keep arbitrarily (finitely) many summands in .

Let the summand have matrix units , where . Kac and Paljutkin now do something which I think is a little dodgy, but basically that the integral over is equal on each of the , equal on each of the , and then zero off the diagonal.

It does follow from above that each is a projection.

Now I am stuck!

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