### Introduction

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

$p=\sum_{i=1}^{m_1}e_i$,

which is a central idempotent.

### Lemma 8.1

$S(p)=p$.

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:

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

$S(\delta_i)=\delta_{i^{-1}}$,

$\varepsilon(e_i)=\delta_{i,1}$,

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:

$P_i^*=P_i,\,P_i^2=P_i,\,P_iP_j=0$

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 $m,n=1,\dots,n_i$Kac 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!