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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