Every finite quantum group has finite dimensional algebra of functions:

\displaystyle F(G)=\bigoplus_{j=1}^m M_{n_j}(\mathbb{C}).

At least one of the factors must be one-dimensional to account for the counit \varepsilon:F(G)\rightarrow \mathbb{C}, and if this factor is denoted \mathbb{C}e_1, the counit is given by the dual element e^1. There may be more and so reorder the index j\mapsto i so that n_i=1 for i=1,\dots,m_1, and n_i>1 for i>m_1:

\displaystyle F(G)=\left(\bigoplus_{i=1}^{m_1} \mathbb{C}e_{i}\right)\oplus \bigoplus_{i=m_1+1}^m M_{n_i}(\mathbb{C})=:A_1\oplus B,

Denote by M_p(G) the states of F(G). The pure states of F(G) 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 B consists of a single matrix factor.

In these cases, for pure states of the form e^i, that is supported on A_1 (and we can say a little more than is necessary), the convolution remains supported on A_1 because

\Delta(A_1)\subset A_1\otimes A_1+B\otimes B.

If we have a pure state \nu supported on B=M_{\sqrt{\dim B}}(\mathbb{C}), then because

\Delta(B)\subset A_1\otimes B+B\otimes A_1,

then \nu\star\nu must be supported on, because of \Delta(A_1)\subset A_1\otimes A_1+B\otimes B, A_1.

Inductively all of the \nu^{\star 2k} are supported on A_1 and the \nu^{\star 2k+1} are supported on B. 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 A_1 and B under the coproduct hold more generally? I believe that the paper of Kac and Paljutkin shows that this is the case whenever B 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 G be a finite quantum group with notation on the algebra of functions as above. Note that A_1 is commutative. Let


which is a central idempotent.

Lemma 8.1


Proof: If S(\mathbf{1}_G-p)p\neq 0, then for some i>m_1, and f\in M_{n_i}(\mathbb{C}), the mapping f\mapsto S(f)p is a non-zero homomorphism from M_{n_i}(\mathbb{C}) into commutative A_1 which is impossible.

If S(\mathbf{1}_G-p)p=g\neq 0, then one of the S(I_{n_i})\in A_1\oplus B, with ‘something’ in A_1. Using the centrality and projectionality of p, we can show that the given map is indeed a homomorphism. 

It follows that S(p)p=p\Rightarrow S(S(p)p)=S(p)=S(p)p=S(p), and so p=S(p) \bullet

Lemma 8.2

(p\otimes p)\Delta(p)=p\otimes p

Proof: Suppose that (p\otimes p)\Delta(f)=b for some non-commutative f\in M_{n_i}(\mathbb{C}). This means that there exists an index k such that f_{(1)_k}\otimes f_{(2)_k}\in A_1\otimes A_1. Then for that factor, 

f\mapsto \Delta(f)(p\otimes p)

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

We see that (p\otimes p)\Delta(f)=0 for all f\in B. Putting a=\mathbf{1}-p we get the result \bullet

The following says that p 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 A_1 will remain there.

Lemma 8.3

(p\otimes \mathbf{1}_G)\Delta(p)=p\otimes p=(\mathbf{1}_G\otimes p)\Delta(p).

Proof: Since \Delta is a homomorphism, \Delta(p) is an idempotent in F(G)\otimes F(G)I  do not understand nor require the rest of the proof.

Lemma 8.4

A_1=F(G_1) is the algebra of functions on finite group with elements i=1,\dots,m_1, and we write e_i=\delta_i. The coproduct is given by (p\otimes p)\Delta.

We have:

(p\otimes p)\Delta(e_i)=\sum_{t\in G_1}\delta_{it^{-1}}\otimes \delta_t,



as e_1=\delta_e.

The element \Delta(\delta_i) is a sum of four terms, lying in the subalgebras:

A_1\otimes A_1,\,B\otimes B,\,A_1\otimes B,\,B\otimes A_1.

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

\Delta(\delta_i)=\sum_{t\in G_1}\delta_{t}\otimes \delta_{t^{-1}i}+P_i.

Since the \delta_i are symmetric (\delta_i^*=\delta_i) mutually orthogonal idempotents, P_i has similar properties:


for i\neq j.

At this point Kac and Paljutkin restrict to B=M_{n_{i+1}}(\mathbb{C}), that is there is only one summand. Here we try to keep arbitrarily (finitely) many summands in B.

Let the summand M_{n_i}(\mathbb{C}) have matrix units E_{mn}^i, where m,n=1,\dots,n_iKac and Paljutkin now do something which I think is a little dodgy, but basically that the integral over G is equal on each of the \delta_i, equal on each of the E_{mm}^i, and then zero off the diagonal. 

It does follow from above that each P_i\in B\otimes B is a projection.

Now I am stuck!