Infinitely divisible states on finite quantum groups

Abstract

In this paper we study the states of Poisson type and infinitely divisible states on compact quantum groups. Each state of Poisson type is infinitely divisible, i.e., it admits n-th root for all $n\ge 1$. The main result is that on finite quantum groups infinitely divisible states must be of Poisson type. This generalizes Böge’s theorem concerning infinitely divisible measures (commutative case) and Parthasarathy’s result on infinitely divisible positive definite functions (cocommutative case). Two proofs are given.

1 Introduction

The space of bounded measures on a compact (semi)group is equipped with a natural convolution operation. The convolution of two probability measures is still a probability measure. Infinitely divisible probability measures are probability measures that admit n-th root for all $n\ge 1$, where the root is also a probability measure. On finite groups such probability measures have been shown to be of Poisson type, see [2] and [18].

A positive definite function on a compact group G is a continuous function $\phi :G \rightarrow {\mathbb {C}}$ such that $[\phi (g^{-1}_ig_j)]_{i,j=1}^{n}$ is a positive semi-definite matrix for all $g_1,\ldots ,g_n\in G$ and for all $n\ge 1$. It is normalized if $\phi (e)=1$, where e is the unit of G. The pointwise product of two normalized positive definite functions on G is again a normalized positive definite function. From this we can define infinitely divisible normalized positive definite functions on a compact group in a natural way. This is thoroughly studied by Parthasarathy [17]. As a special case, he proved that every infinitely divisible normalized positive definite function on a finite group is of Poisson type, although the notion “Poisson type” was not explicitly defined in his paper.

We shall consider the infinite divisibility of states on quantum groups, which provide a more general framework. Our main result is that any infinitely divisible state on a finite quantum group is of Poisson type (in the following also called simply a Poisson state). By taking the finite quantum group to be commutative or cocommutative, we recover the Böge’s result [2] of infinitely divisible probability measures and Parthasarathy’s result [17] on infinitely divisible normalized positive definite functions for finite groups, respectively. We will give two proofs of the main theorem. The first one is based on the ideas of [18] and the second one goes back to [17].

The plan of this paper is as follows. In Sect. 2 we recall the preliminaries on compact quantum groups and introduce the notion of Plancherel triples. In Sect. 3 we explain how to construct a Plancherel triple from an idempotent state on a finite quantum group. Section 4 is devoted to the study of Poisson states on compact quantum groups. Finally in Sect. 5 we prove the main result of this paper, namely that any infinitely divisible state on a finite quantum group is a Poisson state, in two different ways.

2 Preliminaries

2.1 Compact quantum group and its dual

Let us recall some definitions and properties of compact quantum groups. We refer to [21] and [15] for more details.

Definition 2.1

Let A be a unital $C^*$-algebra. If there exists a unital *-homomorphism $\varDelta :A\rightarrow A\otimes A$ such that

• 1. $(\varDelta \otimes \iota )\varDelta =(\iota \otimes \varDelta )\varDelta$;

• 2. $\{\varDelta (a)(1\otimes b):a,b\in A\}$ and $\{\varDelta (a)(b\otimes 1):a,b\in A\}$ are linearly dense in $A\otimes A$;

then $(A,\varDelta )$ is called a compact quantum group and $\varDelta$ is called the comultiplication on A. Here and in the following, $\iota$ always denotes the identity map. We denote ${\mathbb {G}}=(A,\varDelta )$ and $A=C({\mathbb {G}})$. For simplicity, we write $\varDelta ^{(2)}=(\varDelta \otimes \iota )\varDelta$.

Any compact quantum group ${\mathbb {G}}=(A,\varDelta )$ admits a unique Haar state, i.e. a state h on A such that

$$\begin{aligned} (h\otimes \iota )\varDelta (a)=h(a)1=(\iota \otimes h)\varDelta (a),~~a\in A. \end{aligned}$$

Consider an element $u\in A\otimes B(H)$ with $\dim H=n$. By identifying $A\otimes B(H)$ with ${\mathbb {M}}_n(A)$ we can write $u=[u_{ij}]_{i,j=1}^{n}$, where $u_{ij}\in A$. The matrix u is called an n-dimensional representation of ${\mathbb {G}}$ if we have

$$\begin{aligned} \varDelta (u_{ij})=\sum _{k=1}^{n}u_{ik}\otimes u_{kj},~~i,j=1,\dots ,n. \end{aligned}$$

A representation u is called unitary if u is unitary as an element in ${\mathbb {M}}_n(A)$, and irreducible if the only matrices $T\in {\mathbb {M}}_n({\mathbb {C}})$ such that $uT=Tu$ are multiples of identity matrix. Two representations $u,v\in {\mathbb {M}}_n(A)$ are said to be equivalent if there exists an invertible matrix $T\in {\mathbb {M}}_n({\mathbb {C}})$ such that $Tu=vT$. Denote by $\mathrm {Irr}({\mathbb {G}})$ the set of equivalence classes of irreducible unitary representations of ${\mathbb {G}}$. For each $\alpha \in \mathrm {Irr}({\mathbb {G}})$, denote by $u^\alpha \in A\otimes B(H_\alpha )$ a representative of the class $\alpha$, where $H_\alpha$ is the finite dimensional Hilbert space on which $u^\alpha$ acts. In the sequel we write $n_\alpha =\dim H_\alpha$.

Denote $\mathrm {Pol}({\mathbb {G}})=\text {span} \left\{ u^\alpha _{ij}:1\le i,j\le n_\alpha ,\alpha \in \mathrm {Irr}({\mathbb {G}})\right\}$. This is a dense subalgebra of A. On $\mathrm {Pol}({\mathbb {G}})$ the Haar state h is faithful. It is well-known that $(\mathrm {Pol}({\mathbb {G}}),\varDelta )$ is equipped with the Hopf*-algebra structure. That is, there exist a linear antihomormophism S on $\mathrm {Pol}({\mathbb {G}})$, called the antipode, and a unital *-homomorphism $\epsilon :\mathrm {Pol}({\mathbb {G}})\rightarrow {\mathbb {C}}$, called the counit, such that

$$\begin{aligned} (\epsilon \otimes \iota )\varDelta (a)= & {} a=(\iota \otimes \epsilon )\varDelta (a),~~a\in \mathrm {Pol}({\mathbb {G}}), \\ m(S\otimes \iota )\varDelta (a)= & {} \epsilon (a)1=m(\iota \otimes S)\varDelta (a),~~a\in \mathrm {Pol}({\mathbb {G}}). \end{aligned}$$

Here m denotes the multiplication map $m:\mathrm {Pol}({\mathbb {G}})\otimes _{\text {alg}}\mathrm {Pol}({\mathbb {G}})\rightarrow \mathrm {Pol}({\mathbb {G}}),~~a\otimes b\mapsto ab$. Indeed, the antipode and the counit are uniquely determined by

$$\begin{aligned} S(u^\alpha _{ij})= & {} (u^{\alpha }_{ji})^*,~~1\le i,j\le n_\alpha ,~~\alpha \in \mathrm {Irr}({\mathbb {G}}), \\ \epsilon (u^\alpha _{ij})= & {} \delta _{ij},~~1\le i,j\le n_\alpha ,~~\alpha \in \mathrm {Irr}({\mathbb {G}}). \end{aligned}$$

Remark here that $*\circ S\circ *\circ S=\iota$. Also we have for all $a,b\in \mathrm {Pol}({\mathbb {G}})$

$$\begin{aligned} S((\iota \otimes h)(\varDelta (b)(1\otimes a)))= & {} (\iota \otimes h)((1 \otimes b)\varDelta (a)), \end{aligned}$$

(2.1)

$$\begin{aligned} S((h\otimes \iota )((b \otimes 1)\varDelta (a)))= & {} (h\otimes \iota )(\varDelta (b)(a\otimes 1)). \end{aligned}$$

(2.2)

Now we add a remark on the C*-norms on $\mathrm {Pol}({\mathbb {G}})$. We are interested in the following two C*-norms on $\mathrm {Pol}({\mathbb {G}})$:

• 1. the univesal norm:

  $$\begin{aligned} \Vert a\Vert _{{u}}:=\sup \{\Vert \pi (a)\Vert :\pi :\mathrm {Pol}({\mathbb {G}})\rightarrow B({\mathscr {H}}) is a unital *-homomorphism \}, \end{aligned}$$

• 2. the reduced norm:

  $$\begin{aligned} \Vert a\Vert _{ r }:=\Vert \pi _{h}(a)\Vert , \end{aligned}$$

  where $\pi _{h}$ is the GNS representation associated with the Haar state h.

We shall denote by $C_{ u }({\mathbb {G}})$ and $C_{ r }({\mathbb {G}})$ the completions of $\mathrm {Pol}({\mathbb {G}})$ with respect to $\Vert \cdot \Vert _{ u }$ and $\Vert \cdot \Vert _{ r }$, respectively. Then the comultiplication $\varDelta$ and the Haar state h on $\mathrm {Pol}({\mathbb {G}})$ admit extensions to $C_{ u }({\mathbb {G}})$ (resp. $C_{ r }({\mathbb {G}})$), denoted by $\varDelta _{ u }$ and $h_{ u }$ (resp. $\varDelta _{ r }$ and $h_{ r }$), respectively. Both $(C_{ u }({\mathbb {G}}),\varDelta _{ u })$ and $(C_{ r }({\mathbb {G}}),\varDelta _{ r })$ form compact quntum groups.

Note that the counit $\epsilon$ can be always extended to $C_{ u }({\mathbb {G}})$. While, this is not always the case for $C_{ r }({\mathbb {G}})$. If $\epsilon$ can be also extended to $C_{ r }({\mathbb {G}})$, then ${\mathbb {G}}$ is said to be coamenable. An equivalent definition is, ${\mathbb {G}}$ is coamenable iff $\Vert \cdot \Vert _{ r }= \Vert \cdot \Vert _{ u }$. Note that $\Vert \cdot \Vert _{ r }\le \Vert \cdot \Vert _{ u }$ always holds. We refer to [1] for more information. Throughout this paper, we shall always consider compact quantum group ${\mathbb {G}}$ on the universal level, so that the counit can always be extended to the $C({\mathbb {G}})$.

The Peter-Weyl theory for compact groups can be extended to the quantum case. In particular, it is known that for each $\alpha \in \mathrm {Irr}({\mathbb {G}})$ there exists a positive invertible operator $Q_\alpha \in B(H_\alpha )$ such that $\text {Tr}(Q_\alpha )=\text {Tr}(Q^{-1}_\alpha ):=d_\alpha$, which we call quantum dimension of $\alpha$, and the orthogonal relations hold:

$$\begin{aligned} h(u^\alpha _{ij}(u^\beta _{kl})^*)=\frac{\delta _{\alpha \beta }\delta _{ik}(Q_\alpha )_{lj}}{d_\alpha },~~ h((u^\alpha _{ij})^*u^\beta _{kl})=\frac{\delta _{\alpha \beta }\delta _{jl}(Q^{-1}_\alpha )_{ki}}{d_\alpha }, \end{aligned}$$

where $\alpha ,\beta \in \mathrm {Irr}({\mathbb {G}})$, $1\le i,j\le n_\alpha , 1\le k,l\le n_\beta .$

We call ${\mathbb {G}}$ a finite quantum group if the underlying $C^*$-algebra $C({\mathbb {G}})$ is finite dimensional. Note that when ${\mathbb {G}}$ is finite, we have $C({\mathbb {G}})=\mathrm {Pol}({\mathbb {G}})$ and then ${\mathbb {G}}$ is coamenable. In this case each $Q_\alpha$ is identity and h is a trace, i.e. $h(ab)=h(ba)$ for any $a,b\in C({\mathbb {G}})$. Then the orthogonal relation becomes

$$\begin{aligned} h(u^\alpha _{ij}(u^\beta _{kl})^*)=h((u^\alpha _{ij})^*u^\beta _{kl})=\frac{\delta _{\alpha \beta }\delta _{ik}\delta _{jl}}{n_\alpha }, \end{aligned}$$

(2.3)

where $\alpha ,\beta \in \mathrm {Irr}({\mathbb {G}})$, $1\le i,j\le n_\alpha , 1\le k,l\le n_\beta .$ Moreover, the antipode S satisfies $S^2=\iota$. Together with $*\circ S\circ *\circ S=\iota$, one obtains directly that S is *-preserving.

The Pontryagin duality can also be extended to compact quantum groups. We only explain here for finite quantum groups. If ${\mathbb {G}}=(A,\varDelta )$ is a finite quantum group, then we may construct its dual $\hat{{\mathbb {G}}}=({\hat{A}},{\hat{\varDelta }})$ as follows. The underlying finite dimensional C*-algebra ${\hat{A}}$ of $\hat{{\mathbb {G}}}$ is defined as $A'$, the set of all (bounded) linear functionals on A. For $\varphi _1,\varphi _2\in {\hat{A}}$, their convolution product is defined as $\varphi _1\star \varphi _2:=(\varphi _1\otimes \varphi _2)\varDelta$. We may define the involution on ${\hat{A}}$ as: $\varphi ^*:=\overline{\varphi (S(\cdot )^*)}$. Then $({\hat{A}},\star ,*)$ becomes a finite dimensional C*-algebra. For each $\varphi \in {\hat{A}}$, set ${\hat{\varDelta }}(\varphi )(a\otimes b):=\varphi (ab),a,b\in A$. Then $\varDelta (\varphi )\in (A\otimes A)'=A'\otimes A'$ and it is easily seen that ${\hat{\varDelta }}$ defines a comultiplication on ${\hat{A}}$. Hence $\hat{{\mathbb {G}}}=({\hat{A}},{\hat{\varDelta }})$ becomes a finite quantum group. Moreover, it is equipped with a Hopf*-algebra structure with the antipode ${\hat{S}}$ and the counit ${\hat{\epsilon }}$ given by ${\hat{S}}(\varphi ):=\varphi S$ and ${\hat{\epsilon }}(\varphi ):=\varphi (1_A)$, respectively. Starting from the finite quantum group $\hat{{\mathbb {G}}}=({\hat{A}},{\hat{\varDelta }})$, we may also construct its dual. Then in this way we recover the quantum group ${\mathbb {G}}=(A,\varDelta )$.

The Fourier transform${\mathscr {F}}$ on a finite quantum group ${\mathbb {G}}=(A,\varDelta )$ is a map from A to ${\hat{A}}$ defined as ${\mathscr {F}}(a)=h(a\cdot )=h(\cdot a)$, where h is the Haar state on ${\mathbb {G}}$. We shall use ${\hat{a}}$ to denote the its Fourier transform ${\mathscr {F}}(a)$ for simplicity. Then we have the following Parseval’s identity:

$$\begin{aligned} {\hat{h}}({\hat{a}}_1\hat{a}_{2}^*)=c h(a_1a_2^*),~a_1,a_2\in A, \end{aligned}$$

(2.4)

where ${\hat{h}}$ is the Haar state on $\hat{{\mathbb {G}}}$ and $c>0$ is a constant. If we consider the Fourier transform from ${\hat{A}}$ to A, then we obtain a similar equation to (2.4).

2.2 Plancherel triple

In this subsection we introduce the notion of Plancherel triple, which is slightly different from the same notion in [20]. Moreover, one can compare it with the so-called D-pairs in [13]. Let A, B be two finite dimensional C*-algebras. Suppose that

$$\begin{aligned} \langle \cdot ,\cdot \rangle :A\times B\rightarrow {\mathbb {C}}, \end{aligned}$$

is a non-degenerate bilinear form. Then through $\langle ,\rangle$, B can be identified with $A'$. Indeed, the map $B\rightarrow A',b\mapsto \langle \cdot ,b\rangle$ is injective, since $\langle ,\rangle$ is non-degenerate. So B can be viewed as a subspace of $A'$ and then $\dim B\le \dim A'=\dim A$. Similarly, A can be viewed as a subspace of $B'$ and $\dim A\le \dim B'=\dim B$. Hence $\dim A=\dim B$ and $B=A'$ (also $A=B'$).

Definition 2.2

Let the triple $(A,B,\langle ,\rangle )$ be as above. Then it is called a Plancherel triple if

• 1. the comultiplications $\varDelta _A:A\rightarrow A\otimes A$ and $\varDelta _B:B\rightarrow B\otimes B$ are positive, where $\varDelta _A$ is the adjoint of the map given by $B\otimes B\rightarrow B, b_1\otimes b_2\mapsto b_1 b_2$ and $\varDelta _B$ is the adjoint of the map defined through $A\otimes A\rightarrow A, a_1\otimes a_2\mapsto a_1 a_2$;

• 2. the counits $\epsilon _A:=\langle \cdot ,1_B\rangle :A\rightarrow {\mathbb {C}}$ and $\epsilon _B:=\langle 1_A,\cdot \rangle :B\rightarrow {\mathbb {C}}$ are *-homomorphisms;

• 3. the Haar functionals $h_A:=\langle \cdot ,p_B \rangle$ and $h_B:=\langle p_A,\cdot \rangle$ are faithful, positive and tracial, where $p_A\in A$ and $p_B\in B$ are support projections of $\epsilon _A$ and $\epsilon _B$, respectively, i.e.,

  $$\begin{aligned} p_A= & {} p^*_A=p_A^2,~~a p_A= p_A a=\epsilon _A(a) p_A,~~a\in A, \\ p_B= & {} p^*_B=p_B^2,~~b p_B=p_B b=\epsilon _B(b) p_B,~~b\in B. \end{aligned}$$

• 4. The Fourier transforms ${\mathscr {F}}_A:A\rightarrow B,a\mapsto {\hat{a}}$ and ${\mathscr {F}}_B:B\rightarrow A, b\mapsto \check{b}$ satisfy:

  $$\begin{aligned} h_B({\hat{a}}_1{\hat{a}}^*_2)= & {} c h_A(a_1 a^*_2), \end{aligned}$$

  (2.5)
  $$\begin{aligned} h_A(\check{b}_1\check{b}^*_2)= & {} c' h_B(b_1 b^*_2), \end{aligned}$$

  (2.6)

  for all $a_1,a_2\in A, b_1,b_2\in B$ and some $c,c'>0$, where ${\hat{a}}$ and $\check{b}$ are defined through:

  $$\begin{aligned} \langle x,{\hat{a}}\rangle :=h_A(xa),~\langle \check{b},y\rangle :=h_B(by),~~x\in A,~y\in B. \end{aligned}$$

Note that the above definition is self-dual, that is, (B, A, (, )) forms a Plancherel triple if $(A,B,\langle ,\rangle )$ is a Plancherel triple, where $(b,a):=\langle a,b\rangle , a\in A,b\in B$. We have the following properties of a Plancherel triple.

Proposition 2.3

Let $(A,B,\langle ,\rangle )$ be a Plancherel triple. Suppose that $A=\oplus _{\alpha =1}^{m}{\mathbb {M}}_{n_\alpha }({\mathbb {C}})$ with the matrix units $\{e^{\alpha }_{ij}:1\le i,j\le n_{\alpha },1\le \alpha \le m\}$. Let $\{\widetilde{e^{\alpha }_{ij}}:1\le i,j\le n_{\alpha },1\le \alpha \le m\}$ be the dual basis of $\{e^{\alpha }_{ij}:1\le i,j\le n_{\alpha },1\le \alpha \le m\}$.

• (1) We have ${\mathscr {F}}_{A}(e^{\alpha }_{ij})=c_{\alpha }\widetilde{e^{\alpha }_{ji}}$ for some positive number $c_{\alpha }$.

• (2) Suppose that $e^{\alpha _0}=p_A$. Then $\widetilde{e^{\alpha _0}}=\epsilon _A=1_B$. Consequently, $h_B(1_B)=1$ and then $h_B$ is a state.

• (3) We have the orthogonal relation:

  $$\begin{aligned} h_{B}(\widetilde{e^{\alpha }_{ij}}(\widetilde{e^{\beta }_{kl}})^*) =\frac{\delta _{\alpha \beta }\delta _{ik}\delta _{jl}}{c c_\alpha }, \end{aligned}$$

  for all $\alpha ,\beta ,i,j,k,l$, where c is the constant appearing in (2.5).

• (4) We have

  $$\begin{aligned} h_{B}(\widetilde{e^{\alpha }_{ij}}) =\delta _{\alpha _0\alpha }, \end{aligned}$$

  (2.7)

  and

  $$\begin{aligned} \epsilon _{B}(\widetilde{e^{\alpha }_{ij}})=\delta _{ij}, \end{aligned}$$

  (2.8)

  for all $\alpha ,i,j$.

Proof

• (1) For any $a=\sum _{\alpha =1}^{m}\sum _{i,j=1}^{n_{\alpha }}a^{\alpha }_{ij}e^{\alpha }_{ij}\in A$, we have by definition

  $$\begin{aligned} \langle a,{\mathscr {F}}_A(e^{\alpha }_{ij})\rangle =h_A(ae^{\alpha }_{ij}) =a^{\alpha }_{ji}h_A(e^{\alpha }_{jj}) =h_A(e^{\alpha }_{jj})\widetilde{e^{\alpha }_{ji}}(a). \end{aligned}$$

  Since $h_A$ is tracial, $h_A(e^{\alpha }_{jj})$ is independent of j. So

  $$\begin{aligned} \langle a,{\mathscr {F}}_A(e^{\alpha }_{ij})\rangle =c_{\alpha }\widetilde{e^{\alpha }_{ji}}(a),~~a\in A, \end{aligned}$$

  with $c_{\alpha }=h_A(e^{\alpha }_{jj})> 0$. Thus ${\mathscr {F}}_{A}(e^{\alpha }_{ij})=c_{\alpha }\widetilde{e^{\alpha }_{ji}}$, since $\langle ,\rangle$ is non-degenerate.

• (2) For any $a=\sum _{\alpha =1}^{m}\sum _{i,j=1}^{n_{\alpha }}a^{\alpha }_{ij}e^{\alpha }_{ij}\in A$, we have by definition

  $$\begin{aligned} a^{\alpha _0}e^{\alpha _0}=a p_A=\epsilon _A(a)p_A=\langle a,1_B\rangle e^{\alpha _0}. \end{aligned}$$

  Then

  $$\begin{aligned} \widetilde{e^{\alpha _0}}(a)=a^{\alpha _0}=\langle a,1_B\rangle ,~~a\in A. \end{aligned}$$

  Hence $\widetilde{e^{\alpha _0}}=\epsilon _A=1_B$.

• (3) This is a direct consequence of (2.5) and (1).

• (4) By (2) and (3) we have

  $$\begin{aligned} h_{B}(\widetilde{e^{\alpha }_{ij}}) =h_{B}(\widetilde{e^{\alpha }_{ij}}(\widetilde{e^{\alpha _0}})^*) =\frac{\delta _{\alpha _0\alpha }}{c c_{\alpha _0}}. \end{aligned}$$

  In particular, we have

  $$\begin{aligned} 1=h_B(1_B)=h_{B}(\widetilde{e^{\alpha _0}}(\widetilde{e^{\alpha _0}})^*)=\frac{1}{c c_{\alpha _0}} \end{aligned}$$

  This proves (2.7). To show (2.8), recall that by definition,

  $$\begin{aligned} \epsilon _{B}(\widetilde{e^{\alpha }_{ij}})=\langle 1_A,\widetilde{e^{\alpha }_{ij}}\rangle =\widetilde{e^{\alpha }_{ij}}(1_A)=\delta _{ij}. \end{aligned}$$

$\square$

Example 2.4

Let $(A,\varDelta )$ be a finite quantum group with dual $({\hat{A}},{\hat{\varDelta }})$. Then $(A,{\hat{A}},\langle , \rangle )$ forms a Plancherel triple. Here the bilinear form $\langle ,\rangle$ is given by $\langle a,\varphi \rangle :=\varphi (a)$. The comultiplications, counits, Haar states, and Fourier transforms are the usual comultiplications, counits, Haar states, and Fourier transforms on A and ${\hat{A}}$, respectively.

3 Plancherel triple induced from an idempotent state

In this section we will construct a Plancherel triple from an idempotent state on a finite quantum group ${\mathbb {G}}=(A,\varDelta )$. An idempotent state on ${\mathbb {G}}$ is a state $\phi$ such that $\phi \star \phi =\phi$. It is well-known that (see for example [9]) considered as an element in ${\hat{A}}$, $p=\phi$ is a group-like projection in ${\hat{A}}$. By a group-like projection of the finite quantum group $\hat{{\mathbb {G}}}=({\hat{A}},{\hat{\varDelta }})$ we mean a non-zero element $p\in {\hat{A}}$ such that $p=p^*=p^2$ and

$$\begin{aligned} {\hat{\varDelta }}(p)(1\otimes p)=p\otimes p={\hat{\varDelta }}(p)(p\otimes 1),~~{\hat{S}}(p)=p. \end{aligned}$$

(3.1)

Our aim in this section is to show that $(A_\phi ,{\hat{A}}_p,\langle ,\rangle )$ forms a Plancherel triple, where $A_\phi =(\phi \otimes \iota \otimes \phi )\varDelta ^{(2)}(A)$ is a C*-subalgebra of A, ${\hat{A}}_p:=p{\hat{A}}p$ is a C*-subalgebra of ${\hat{A}}$ and the bilinear form is inherited from the one on the pair $(A,{\hat{A}})$. In the following we shall explain the constructions of $A_\phi$ and ${\hat{A}}_p$ in detail.

Before this we remark here that these constructions have already been studied by many people [5, 6, 8, 9]. Many results are well-known and their proofs are omitted here.

3.1 C*-subalgebra $A_{\phi }$

In this subsection we construct and study the C*-subalgebra $A_{\phi }$. Recall that an idempotent state on a compact quantum group ${\mathbb {G}}$ is a state $\phi$ such that $\phi \star \phi =\phi$. Denote by $\text {Idem} ({\mathbb {G}})$ the set of all idempotent states on ${\mathbb {G}}$. Recall also that if $\phi \in \text {Idem} ({\mathbb {G}})$, we have $\phi =\phi S$ on $\mathrm {Pol}({\mathbb {G}})$, where S is the antipode on $\mathrm {Pol}({\mathbb {G}})$ [9]. We use $A'$ to denote the set of all bounded linear functionals on A.

The first lemma is a special case of [8, Lemma 3.1], and also a variation of [15, Lemma 4.3].

Lemma 3.1

Let $\phi$ be an idempotent state on a compact quantum group ${\mathbb {G}}$. For $b\in A$ define $\phi _b\left( a\right) :=\phi \left( ab\right)$ for all $a\in A$. Then we have

$$\begin{aligned} \phi \star \phi _b=\phi \left( b\right) \phi . \end{aligned}$$

For $\phi \in \text {Idem}\left( {\mathbb {G}}\right)$ set ${\mathbb {E}}_\phi ^\ell :=\left( \phi \otimes \iota \right) \varDelta$ and ${\mathbb {E}}_\phi ^r:=\left( \iota \otimes \phi \right) \varDelta$. The next lemma lists some useful properties of ${\mathbb {E}}_\phi ^\ell$ and ${\mathbb {E}}_\phi ^r$.

Lemma 3.2

Let ${\mathbb {G}}=(A,\varDelta )$ be a compact quantum group.

• (1) ${\mathbb {E}}_\phi ^\ell \left( a^*\right) ={\mathbb {E}}_\phi ^\ell \left( a\right) ^*,{\mathbb {E}}_\phi ^r\left( a^*\right) ={\mathbb {E}}_\phi ^r\left( a\right) ^*,a\in A$.

• (2) $\varDelta {\mathbb {E}}_\phi ^\ell =({\mathbb {E}}_\phi ^\ell \otimes \iota )\varDelta ,\varDelta {\mathbb {E}}_\phi ^r=(\iota \otimes {\mathbb {E}}_\phi ^r)\varDelta$.

• (3) $(\iota \otimes {\mathbb {E}}_\phi ^\ell )\varDelta =({\mathbb {E}}_\phi ^r\otimes \iota )\varDelta$.

• (4) ${\mathbb {E}}_\phi ^\ell {\mathbb {E}}_\phi ^r={\mathbb {E}}_\phi ^r{\mathbb {E}}_\phi ^\ell$.

• (5) For any $a,b\in A$,

  $$\begin{aligned} {\mathbb {E}}_\phi ^\ell (a){\mathbb {E}}_\phi ^\ell (b)= & {} {\mathbb {E}}_\phi ^\ell (a{\mathbb {E}}_\phi ^\ell (b)) ={\mathbb {E}}_\phi ^\ell ({\mathbb {E}}_\phi ^\ell (a)b), \end{aligned}$$

  (3.2)
  $$\begin{aligned} {\mathbb {E}}_\phi ^r(a){\mathbb {E}}_\phi ^r(b)= & {} {\mathbb {E}}_\phi ^r(a{\mathbb {E}}_\phi ^r(b)) ={\mathbb {E}}_\phi ^r({\mathbb {E}}_\phi ^r(a)b). \end{aligned}$$

  (3.3)

  Consequently, we have ${\mathbb {E}}_\phi ^\ell {\mathbb {E}}_\phi ^ \ell ={\mathbb {E}}_\phi ^\ell$ and ${\mathbb {E}}_\phi ^r{\mathbb {E}}_\phi ^r={\mathbb {E}}_\phi ^r$.

Proof

(1)–(4) are just straightforward computations. For (5) we prove (3.2) first. For this note that it suffices to show the first equation for any a, b the coefficients of unitary representation of ${\mathbb {G}}$. The case for general a, b follows from the linearity and density. Let $\left\{ u^\alpha _{ij},1\le i,j\le n_\alpha \right\}$ be the coefficients of the irreducible unitary representations $u^\alpha ,\alpha \in \mathrm {Irr}\left( {\mathbb {G}}\right)$. On one hand, for any $u^\alpha _{ij}$ and $u^\beta _{kl}$ we have

$$\begin{aligned} {\mathbb {E}}_\phi ^\ell (u^\alpha _{ij}){\mathbb {E}}_\phi ^\ell (u^\beta _{kl}) =\sum _{s=1}^{n_\alpha }\sum _{t=1}^{n_\beta }\phi (u^\alpha _{is})\phi (u^\beta _{kt})u^\alpha _{sj}u^\beta _{tl}. \end{aligned}$$

On the other hand,

$$\begin{aligned} {\mathbb {E}}_\phi ^\ell ({\mathbb {E}}_\phi ^\ell (u^\alpha _{ij})u^\beta _{kl}) =\sum _{r,s=1}^{n_\alpha }\sum _{t=1}^{n_\beta }\phi (u^{\alpha }_{ir})\phi (u^{\alpha }_{rs}u^{\beta }_{kt})u^\alpha _{sj}u^\beta _{tl}. \end{aligned}$$

Now by using Lemma 3.1 we have for any $b\in A$

$$\begin{aligned} \phi \left( b\right) \phi \left( u^\alpha _{is}\right) =\sum _{r=1}^{n_\alpha }\phi \left( u^\alpha _{ir}\right) \phi (u^\alpha _{rs}b),~~1\le i,s\le n_\alpha ,\alpha \in \mathrm {Irr}\left( {\mathbb {G}}\right) . \end{aligned}$$

Choosing b as $u^{\beta }_{kt}$, we have

$$\begin{aligned} {\mathbb {E}}_\phi ^\ell ({\mathbb {E}}_\phi ^\ell (u^\alpha _{ij})u^\beta _{kl})&=\sum _{r=1}^{n_\alpha }\sum _{t=1}^{n_\beta }\left( \sum _{s=1}^{n_\alpha }\phi (u^{\alpha }_{ir})\phi (u^{\alpha }_{rs}u^{\beta }_{kt})\right) u^\alpha _{sj}u^\beta _{tl}\\&=\sum _{s=1}^{n_\alpha }\sum _{t=1}^{n_\beta }\phi (u^\alpha _{is})\phi (u^\beta _{kt})u^\alpha _{sj}u^\beta _{tl} ={\mathbb {E}}_\phi ^\ell (u^\alpha _{ij}){\mathbb {E}}_\phi ^\ell (u^\beta _{kl}). \end{aligned}$$

So ${\mathbb {E}}_\phi ^\ell (a){\mathbb {E}}_\phi ^\ell (b)={\mathbb {E}}_\phi ^\ell ({\mathbb {E}}_\phi ^\ell (a)b)$. In the same way one can show that ${\mathbb {E}}_\phi ^\ell (a){\mathbb {E}}_\phi ^\ell (b) ={\mathbb {E}}_\phi ^\ell (a{\mathbb {E}}_\phi ^\ell (b))$, then (3.2) is proved. Note that by choosing $b=1_A$ in (3.2), we obtain ${\mathbb {E}}_\phi ^\ell {\mathbb {E}}_\phi ^ \ell ={\mathbb {E}}_\phi ^\ell$. The proof of (3.3) and ${\mathbb {E}}_\phi ^r{\mathbb {E}}_\phi ^r={\mathbb {E}}_\phi ^r$ is similar. $\square$

Now we define ${\mathbb {E}}_\phi :={\mathbb {E}}_\phi ^\ell {\mathbb {E}}_\phi ^r={\mathbb {E}}_\phi ^r{\mathbb {E}}_\phi ^\ell$. The map ${\mathbb {E}}_\phi$ shares the similar properties of ${\mathbb {E}}_\phi ^\ell$ and ${\mathbb {E}}_\phi ^r$, see the following lemma (1)–(4). Moreover, ${\mathbb {E}}_\phi$ commutes with the antipode S, as the following lemma (5) shows. This enables the algebra ${\mathbb {E}}_\phi (A)$ to possess nicer properties than ${\mathbb {E}}_\phi ^\ell (A)$ and ${\mathbb {E}}_\phi ^r(A)$. We shall choose ${\mathbb {E}}_\phi (A)$ to be our $A_{\phi }$.

Lemma 3.3

Let ${\mathbb {G}}=(A,\varDelta )$ be a compact quantum group.

• (1) ${\mathbb {E}}_\phi \left( a^*\right) ={\mathbb {E}}_\phi \left( a\right) ^*,a\in A$.

• (2) $\varDelta {\mathbb {E}}_\phi =({\mathbb {E}}_\phi ^\ell \otimes {\mathbb {E}}_\phi ^r)\varDelta$.

• (3) $({\mathbb {E}}_\phi \otimes {\mathbb {E}}_\phi )\varDelta =(\iota \otimes {\mathbb {E}}_\phi ^\ell )\varDelta {\mathbb {E}}_\phi =({\mathbb {E}}_\phi ^r\otimes \iota )\varDelta {\mathbb {E}}_\phi$.

• (4) ${\mathbb {E}}_\phi (a){\mathbb {E}}_\phi (b)={\mathbb {E}}_\phi (a{\mathbb {E}}_\phi (b))={\mathbb {E}}_\phi ({\mathbb {E}}_\phi (a)b),a,b\in A$. Consequently, ${\mathbb {E}}_\phi {\mathbb {E}}_\phi ={\mathbb {E}}_\phi$.

Hence $A_{\phi }:={\mathbb {E}}_\phi (A)$ is a unital $C^*$-subalgebra of A. Moreover,

• (5) $S{\mathbb {E}}_\phi ={\mathbb {E}}_\phi S$ on $\mathrm {Pol}({\mathbb {G}})$.

Proof

Again we omit the proof of (1)-(4) here. The fact that $A_{\phi }$ is a unital $C^*$-subalgebra follows directly from these properties. To prove (5), it suffices to check the equality for $u^\alpha _{ij},~ 1\le i,j\le n_\alpha ,\alpha \in \text{ Irr }({\mathbb {G}})$, the coefficients of unitary representations of ${\mathbb {G}}$. And that is a consequence of

$$\begin{aligned} S{\mathbb {E}}_\phi (u^\alpha _{ij})&=S(\phi \otimes \iota \otimes \phi )\varDelta ^{(2)}(u^\alpha _{ij})\\&=\sum _{k,l=1}^{n_\alpha }\phi (u^\alpha _{ik})\phi (u^\alpha _{lj})S(u^\alpha _{kl})\\&=\sum _{k,l=1}^{n_\alpha }\phi (u^\alpha _{ik})\phi (u^\alpha _{lj})(u^\alpha _{lk})^*, \end{aligned}$$

and

$$\begin{aligned} {\mathbb {E}}_\phi S(u^\alpha _{ij})&=(\phi \otimes \iota \otimes \phi )\varDelta ^{(2)}((u^\alpha _{ji})^*)\\&=\sum _{k,l=1}^{n_\alpha }\phi ((u^\alpha _{ki})^*)\phi ((u^\alpha _{jl})^*)(u^\alpha _{lk})^*\\&=\sum _{k,l=1}^{n_\alpha }\phi (u^\alpha _{ik})\phi (u^\alpha _{lj})(u^\alpha _{lk})^*, \end{aligned}$$

where the last equality follows from the facts that $\phi =\phi S$ on $\mathrm {Pol}({\mathbb {G}})$ and $S(u^\alpha _{lj})=(u^\alpha _{jl})^*$. $\square$

3.2 $\phi$-bi-invariant functionals

Let $\phi$ be an idempotent state on a compact quantum group ${\mathbb {G}}=(A,\varDelta )$. A functional $u\in A'$ is called $\phi$-bi-invariant if $u\phi =\phi u=u$. In this subsection we characterize the $\phi$-bi-invariant functionals. It turns out that one can transfer each $\phi$-bi-invariant functional on A to its restriction to $A_{\phi }$, preserving the norm and the *-algebra structure. See [4] for related work.

For linear functionals $\varphi _1,\varphi _2$ on $A_{\phi }$, one can also define their convolution product within $(A_\phi )'$: $\varphi _1\star \varphi _2$ is defined as $(\varphi _1\otimes \varphi _2)\varDelta _{\phi }$. And for any linear functional $\varphi$ on ${\mathbb {E}}_{\phi }(\mathrm {Pol}({\mathbb {G}}))$ one can also define its involution within $(A_\phi )'$: $\varphi ^*:=\overline{\varphi (S(\cdot )^*)}$. Still, we write $\varphi _1\varphi _2$ for short to denote $\varphi _1\star \varphi _2$. Note here that $\varphi ^*$ is well-defined because $S{\mathbb {E}}_\phi ={\mathbb {E}}_\phi S$.

We formulate the results of $\phi$-bi-invariant functionals here without the proof. Note that in the sequel we shall use $\Vert u\Vert$ to denote the norm of $u\in A'$ as a functional on A.

Lemma 3.4

Let $\phi \in Idem ({\mathbb {G}})$ and $u\in A'$. Then u is $\phi$-bi-invariant if and only if $u=u|_{A_{\phi }}{\mathbb {E}}_\phi$. In this case, the following hold:

• (1) $\Vert u\Vert =\Vert u|_{A_{\phi }}\Vert$;

• (2) u is a positive linear functional (resp. a state) on A if and only if $u|_{A_{\phi }}$ is a positive linear functional (resp. a state) on $A_\phi$;

• (3) $u^*|_{A_{\phi }}=(u|_{A_{\phi }})^*$ on ${\mathbb {E}}_{\phi }(\mathrm {Pol}({\mathbb {G}}))$;

• (4) if u and v in $A'$ are both $\phi$-bi-invariant, then $(uv)|_{A_{\phi }}=u|_{A_{\phi }}v|_{A_{\phi }}$;

• (5) $\phi =\epsilon {\mathbb {E}}_\phi =\epsilon |_{A_{\phi }}{\mathbb {E}}_\phi$.

3.3 The Plancherel triple $(A_{\phi },{\hat{A}}_p,\langle ,\rangle )$

Let ${\mathbb {G}}=(A,\varDelta )$ be a finite quantum group. Let $\phi$ be an idempotent state on ${\mathbb {G}}$, then $p=\phi$ is a group-like projection in ${\hat{A}}$. By Lemma 3.3, $A_\phi ={\mathbb {E}}_{\phi }(A)$ is a finite dimensional C*-algebra. Clearly ${\hat{A}}_p=p{\hat{A}}p$ is also a finite dimensional C*-algebra. Moreover, Lemma 3.4 implies that $(A_\phi )'={\hat{A}}_p$. The main result of this section is the following proposition.

Proposition 3.5

Let $A_\phi$ and ${\hat{A}}_p$ be as above. Then $(A_{\phi },{\hat{A}}_p,\langle ,\rangle )$ forms a Plancherel triple, where the bilinear form is inherited from the one on the pair $(A,{\hat{A}})$.

Proof

Since $(A_{\phi })'={\hat{A}}_p$, the bilinear form $\langle , \rangle$ on the pair $(A_\phi ,{\hat{A}}_p)$ is non-degenerate. Now we check that the triple $(A_{\phi },{\hat{A}}_p,\langle ,\rangle )$ verifies the conditions 1–4 of Definition 2.2. For simplicity, in the sequel we shall use $\langle , \rangle$ to denote the bilinear forms on the different pairs $(A,{\hat{A}})$, $(A_\phi ,{\hat{A}}_p)$ and their corresponding tensor products $(A\otimes A,{\hat{A}}\otimes {\hat{A}})$ and $(A_\phi \otimes A_\phi ,{\hat{A}}_p\otimes {\hat{A}}_p)$. The readers can distinguish them easily. We shall use $h:=h_A$ and ${\hat{h}}:=h_B$ to denote the Haar states on ${\mathbb {G}}$ and $\hat{{\mathbb {G}}}$, respectively.

• 1. In this case, we claim that the comultiplications on $A_{\phi }$ and ${\hat{A}}_p$ are respectively $\varDelta _{\phi }:=({\mathbb {E}}_{\phi }\otimes {\mathbb {E}}_{\phi })\varDelta |_{A_{\phi }}$ and ${\hat{\varDelta }}_p:=(p\otimes p){\hat{\varDelta }}|_{{\hat{A}}_p}(\cdot )(p\otimes p)$. Then automatically they are positive. Indeed, by definition,

  $$\begin{aligned} \langle \varDelta _{A_\phi }(a),x\otimes y\rangle =\langle a,xy\rangle =(x\otimes y)\varDelta (a),~a\in A_{\phi },x,y\in {\hat{A}}_p. \end{aligned}$$

  By Lemma 3.4, we have $x=x{\mathbb {E}}_{\phi }\text { and }y=y{\mathbb {E}}_{\phi }$. Thus $(x\otimes y)\varDelta (a)=(x\otimes y)\varDelta _{\phi }(a)$ and

  $$\begin{aligned} \langle \varDelta _{A_\phi }(a),x\otimes y\rangle =\langle \varDelta _{\phi }(a),x\otimes y\rangle ,~a\in A_{\phi },x,y\in {\hat{A}}_p. \end{aligned}$$

  Since $\langle ,\rangle$ is non-degenerate, we have $\varDelta _{A_\phi }=\varDelta _{\phi }$. This proves the claim. Similarly, one can show $\varDelta _{{\hat{A}}_p}={\hat{\varDelta }}_{p}$.

• 2. The counits on $A_{\phi }$ and ${\hat{A}}_p$ are respectively $\epsilon _{\phi }:=\epsilon |_{A_\phi }$ and ${\hat{\epsilon }}_p:={\hat{\epsilon }}|_{{\hat{A}}_p}$. Here $\epsilon$ and ${\hat{\epsilon }}$ are respectively the counits on A and ${\hat{A}}$. In fact, by definition,

  $$\begin{aligned} \epsilon _{A_\phi }=\langle \cdot ,1_{{\hat{A}}_p}\rangle =\langle \cdot ,1_{{\hat{A}}}\rangle |_{A_\phi }=\epsilon _{\phi },~ \epsilon _{{\hat{A}}_p}=\langle 1_{A_\phi },\cdot \rangle =\langle 1_{A},\cdot \rangle |_{{\hat{A}}_p}={\hat{\epsilon }}_{p}. \end{aligned}$$

  Then it is easy to check that $\epsilon _{\phi }:A_\phi \rightarrow {\mathbb {C}}$ and ${\hat{\epsilon }}_{p}:{\hat{A}}_p\rightarrow {\mathbb {C}}$ are both *-isomorphisms.

• 3. On one hand, the support projection $p_{{\hat{A}}}$ of ${\hat{\epsilon }}=\epsilon _{{\hat{A}}}$ verifies

  $$\begin{aligned} p_{{\hat{A}}} p=p p_{{\hat{A}}}={\hat{\epsilon }}(p) p_{{\hat{A}}}=\phi (1_A)p_{{\hat{A}}}=p_{{\hat{A}}}. \end{aligned}$$

  So $p_{{\hat{A}}}\in {\hat{A}}_p$. Moreover,

  $$\begin{aligned} b p_{{\hat{A}}}=p_{{\hat{A}}} b={\hat{\epsilon }}(b)p_{{\hat{A}}}={\hat{\epsilon }}_p(b)p_{{\hat{A}}},~~b\in {\hat{A}}_p. \end{aligned}$$

  Therefore the support projection of ${\hat{\epsilon }}_p$ is $p_{{\hat{A}}_p}=p_{{\hat{A}}}$. Thus the Haar functional on $A_{\phi }$ is

  $$\begin{aligned} h_{\phi }:=h_{A_\phi }=\langle \cdot ,p_{{\hat{A}}_p}\rangle =\langle \cdot ,p_{{\hat{A}}}\rangle |_{A_\phi }=h|_{A_\phi }. \end{aligned}$$

  On the other hand, by Lemma 3.3,

  $$\begin{aligned} {\mathbb {E}}_{\phi }(p_A){\mathbb {E}}_{\phi }(a) ={\mathbb {E}}_{\phi }(p_A{\mathbb {E}}_{\phi }(a)) =\epsilon ({\mathbb {E}}_{\phi }(a)){\mathbb {E}}_{\phi }(p_A) =\epsilon _{\phi }({\mathbb {E}}_{\phi }(a)){\mathbb {E}}_{\phi }(p_A),~~a\in A. \end{aligned}$$

  Similarly, ${\mathbb {E}}_{\phi }(p_A){\mathbb {E}}_{\phi }(a)=\epsilon _{\phi }({\mathbb {E}}_{\phi }(a)){\mathbb {E}}_{\phi }(p_A),a\in A.$ By choosing $a=p_A$, we obtain that ${\mathbb {E}}_{\phi }(p_A)^2=\phi (p_A){\mathbb {E}}_{\phi }(p_A)$. So $q:=\frac{1}{\phi (p_A)}{\mathbb {E}}_{\phi }(p_A)$ is a self-adjoint projection in $A_\phi$ such that

  $$\begin{aligned} qa=aq=\epsilon _{\phi }(a)q,~~a\in A_\phi . \end{aligned}$$

  That is, $p_{A_\phi }=q$ is the support projection of $\epsilon _\phi$. So the Haar functional on ${\hat{A}}_p$ is

  $$\begin{aligned} {\hat{h}}_p:=h_{{\hat{A}}_p} =\langle p_{A_\phi },\cdot \rangle =\frac{1}{\phi (p_A)}\langle {\mathbb {E}}_{\phi }(p_{A}),\cdot \rangle =\frac{1}{\phi (p_A)}\langle p_{A},\cdot \rangle |_{{\hat{A}}_p} =\frac{1}{\phi (p_A)}{\hat{h}}|_{{\hat{A}}_p}. \end{aligned}$$

  Hence the Haar functionals $h_\phi$ and ${\hat{h}}_p$ are faithful, positive and tracial.

• 4. For any $a\in A_\phi$, we claim that ${\mathscr {F}}_{A}(a)=h(\cdot a)\in {\hat{A}}_p$, i.e. ${\mathscr {F}}_{A}(a)$ is $\phi$-bi-invariant. Indeed, by (2.1),

  $$\begin{aligned} (\phi \otimes h(\cdot a))\varDelta (x)= & {} (\phi \otimes h)(\varDelta (x)(1\otimes a))\\= & {} (\phi \circ S\otimes h)((1\otimes x)\varDelta (a)),~a\in A_{\phi },x\in A. \end{aligned}$$

  Since $\phi =\phi S$ and ${\mathbb {E}}_{\phi }(a)=a$, we have

  $$\begin{aligned} (\phi \otimes h(\cdot a))\varDelta (x) =(\phi \otimes h)((1\otimes x)\varDelta (a)) =h(x{\mathbb {E}}^{\ell }_{\phi }(a)) =h(xa),~a\in A_{\phi },x\in A. \end{aligned}$$

  Hence $\phi \star h(\cdot a)=h(\cdot a)$, for all $a\in A_{\phi }$. From (2.2) and ${\mathbb {E}}^{r}_{\phi }(a)=a$ one can deduce in a similar way that $h(\cdot a)\star \phi =h(\cdot a)$. Thus ${\mathscr {F}}_{A}(a)=h(\cdot a)$ is $\phi$-bi-invariant for all $a\in A_\phi$. This proves our claim. Now for any $a\in A_\phi$ we have

  $$\begin{aligned} \langle x,{\mathscr {F}}_{A_\phi }(a)\rangle =h_{\phi }(xa)=h(xa)=\langle x,{\mathscr {F}}_{A}(a)\rangle ,~x\in A_\phi . \end{aligned}$$

  Since ${\mathscr {F}}_{A}(a)\in {\hat{A}}_p$, ${\mathscr {F}}_{A}(a)$ is the image of ${\mathscr {F}}_{A_\phi }(a)$ under the natural inclusion ${\hat{A}}_p\rightarrow {\hat{A}}$. Hence the Parsevel’s identity (2.4) on the pair $(A,{\hat{A}})$ yields

  $$\begin{aligned} {\hat{h}}_{p}({\mathscr {F}}_{A_\phi }(a_1){\mathscr {F}}_{A_\phi }(a_2)^*) ={\hat{h}}({\mathscr {F}}_{A}(a_1){\mathscr {F}}_{A}(a_2)^*) =c h(a_1 a_2^*) =c h_{\phi }(a_1 a_2^*),~a_1,a_2\in A_{\phi }. \end{aligned}$$

  Similarly, one can show that

  $$\begin{aligned} h_{\phi }({\mathscr {F}}_{{\hat{A}}_p}(b_1){\mathscr {F}}_{{\hat{A}}_p}(b_2)^*) =c' {\hat{h}}_{p}(b_1 b_2^*),~b_1,b_2\in {\hat{A}}_p, \end{aligned}$$

  for some constant $c'>0$. $\square$

4 Poisson states on compact quantum groups

Let ${\mathbb {G}}=(A,\varDelta )$ be a compact quantum group. Denote by ${\mathscr {S}}(A)$ the set of all states on A. For each $\phi \in \text {Idem} ({\mathbb {G}})$, we say that $\{\omega _t\}_{t\ge 0}$ is a convolution semigroup of functionals on A starting from $\phi$ if

• (1) $\omega _t\in A'$ for each $t\ge 0$.

• (2) $\omega _{s+t}=\omega _{s}\omega _{t}$ for all $s,t\ge 0$.

• (3) $\omega _0=\phi$.

If moreover, each $\omega _t\in {\mathscr {S}}(A)$, we call $\{\omega _t\}_{t\ge 0}$ a convolution semigroup of states starting from $\phi$. We say that the convolution semigroup of states $\{\omega _t\}_{t\ge 0}$ is norm continuous if

$$\begin{aligned} \lim \limits _{t\rightarrow 0^+}\Vert \omega _t-\phi \Vert =0. \end{aligned}$$

Recall here that $\Vert u\Vert$ denotes the norm of $u\in A'$ as a functional on A. Moreover, it is a Banach norm, since

$$\begin{aligned} \Vert uv\Vert =\Vert (u\otimes v)\varDelta \Vert \le \Vert u\Vert \Vert v\Vert ,~~u,v\in A'. \end{aligned}$$

For a $\phi$-bi-invariant functional $u\in A'$ define

$$\begin{aligned} \exp _\phi (u):=\phi +\sum _{n\ge 1}\frac{u^n}{n!}. \end{aligned}$$

Then it is easy to check that $\{\exp _\phi (tu)\}_{t\ge 0}$ form a norm continuous convolution semigroup of functionals. We aim to find sufficient and necessary conditions on u such that $\{\exp _\phi (tu)\}_{t\ge 0}$ is a convolution semigroup of states. For this we make some notations. A functional $u\in A'$ is called $\textit{Hermitian}$ if $u\left( x^*\right) =\overline{u\left( x\right) }$ for all x; it is further called conditionally positive definite with respect to$\phi$ if $u\left( x^*x\right) \ge 0$ for all x such that $\phi \left( x^*x\right) =0$. The main theorem in this section is as follows.

Theorem 4.1

Suppose that ${\mathbb {G}}=\left( A,\varDelta \right)$ is a compact quantum group. Let $\phi \in \text {Idem}\left( {\mathbb {G}}\right)$. Then for $u\in A'$, the following statements are equivalent.

• (1) $u(1_A)=0$, u is $\phi$-bi-invariant and conditionally positive definite w.r.t. $\phi$.

• (2) $u=r(v-\phi )$, where $r\ge 0$ and v is a $\phi$-bi-invariant state.

The following proposition proves Theorem 4.1 in the level of unital $C^*$-algebras. It can be considered as a special case of Theorem 4.1 when $\phi =\epsilon$ is the counit. In this case each $u\in A'$ is $\epsilon$-bi-invariant and $\phi =\epsilon$ is a character.

Proposition 4.2

Let A be a unital $C^*$-algebra with $\epsilon$ a character. Then for any non-zero bounded linear functional u on A such that $u(1_A)=0$ and $u(x^*x)\ge 0$ for all $\epsilon (x^*x)=0$, we have $u=r(v-\epsilon )$, where $r>0$ and v is a state.

Proof

Note first that $\epsilon (x^*x)=|\epsilon (x)|^2$. So $\epsilon (x^*x)=0$ if and only if $x\in \ker \epsilon =\{x:\epsilon (x)=0\}$. Let $u_0:=u|_{\ker \epsilon }$ be the restriction of u to $\ker \epsilon$. By assumption, $u_0$ is a bounded linear positive functional on the ideal $\ker \epsilon$. So it admits a unique positive linear extension $\widetilde{u_0}$ to A such that $\widetilde{u_0}|_{\ker \epsilon }=u_0$ and $\Vert \widetilde{u_0}\Vert =\Vert u_0\Vert$. Hence for any $x\in A$, we have $x-\epsilon (x)1_A\in \ker \epsilon$ and thus

$$\begin{aligned} u(x)=u\left( x-\epsilon (x)1_A\right) =u_0\left( x-\epsilon (x)1_A\right) =\widetilde{u_0}\left( x-\epsilon (x)1_A\right) =r(v-\epsilon )(x), \end{aligned}$$

where $r:=\Vert \widetilde{u_0}\Vert =\Vert u_0\Vert >0$ and $v:=\frac{1}{r}\widetilde{u_0}$ is a state. $\square$

Now we are ready to prove Theorem 4.1. The idea is to restrict the problem to $A_{\phi }$, and then transfer the decomposition from $A_\phi$ to A.

Proof of Theorem 4.1

The direction $(2)\Rightarrow (1)$ is clear. To prove $(1)\Rightarrow (2)$, suppose $u\ne 0$ and write $u=u|_{A_{\phi }}{\mathbb {E}}_\phi$ by Lemma 3.4. Note that $\epsilon |_{A_{\phi }}$ is a character on the unital $C^*$-algebra $A_{\phi }$. From the definition of u, we have $u|_{A_{\phi }}(1_{A_\phi })=0$. Moreover, for any $x\in A$ such that $\epsilon |_{A_{\phi }}\left( {\mathbb {E}}_\phi \left( x\right) ^*{\mathbb {E}}_\phi \left( x\right) \right) =0$, we have by Lemma 3.2 and Lemma 3.4 that

$$\begin{aligned} 0 =\epsilon |_{A_{\phi }}\left( {\mathbb {E}}_\phi \left( x\right) ^*{\mathbb {E}}_\phi \left( x\right) \right) =\epsilon |_{A_{\phi }}{\mathbb {E}}_\phi \left( {\mathbb {E}}_\phi \left( x\right) ^*{\mathbb {E}}_\phi \left( x\right) \right) =\phi \left( {\mathbb {E}}_\phi \left( x\right) ^*{\mathbb {E}}_\phi \left( x\right) \right) . \end{aligned}$$

Again, by Lemma 3.2 and Lemma 3.4, the conditionally positive definiteness of u with respect to $\phi$ implies

$$\begin{aligned} u|_{A_{\phi }}\left( {\mathbb {E}}_\phi \left( x\right) ^*{\mathbb {E}}_\phi \left( x\right) \right) =u|_{A_{\phi }}{\mathbb {E}}_\phi \left( {\mathbb {E}}_\phi \left( x\right) ^*{\mathbb {E}}_\phi \left( x\right) \right) =u\left( {\mathbb {E}}_\phi \left( x\right) ^*{\mathbb {E}}_\phi \left( x\right) \right) \ge 0. \end{aligned}$$

So we have by Proposition 4.2 that $u|_{A_{\phi }}=r(w-\epsilon |_{A_{\phi }})$ with $r>0$ and w a state on $A_{\phi }$. Set $v:=w{\mathbb {E}}_\phi$, then v is, by Lemma 3.4, a $\phi$-bi-invariant state on A such that

$$\begin{aligned} u =u|_{A_{\phi }}{\mathbb {E}}_\phi =r\left( w{\mathbb {E}}_\phi -\epsilon |_{A_{\phi }}{\mathbb {E}}_\phi \right) =r(v-\phi ), \end{aligned}$$

as desired. $\square$

Definition 4.3

Let ${\mathbb {G}}$ be a compact quantum group. Let $\phi \in \text {Idem}({\mathbb {G}})$. We denote by ${\mathscr {N}}_\phi ({\mathbb {G}})$ the class of functionals $u\in A'$ that satisfy the conditions (2) in Theorem 4.1. Then for each $u\in {\mathscr {N}}_\phi ({\mathbb {G}})$, $\omega :=\exp _\phi (u)$ is a state. Denote by ${\mathscr {P}}_\phi ({\mathbb {G}})$ the familly of all such states. Set ${\mathscr {P}}({\mathbb {G}}):=\bigcup _{\phi \in \text {Idem} ({\mathbb {G}})}{\mathscr {P}}_\phi ({\mathbb {G}})$. Then any $\omega \in {\mathscr {P}}({\mathbb {G}})$ is said to be of Poisson type, or a Poisson state on ${\mathbb {G}}$.

Recall that any norm continuous convolution semigroup of states $\{\omega _t\}_{t\ge 0}$ on a compact quantum group ${\mathbb {G}}=(A,\varDelta )$ can be recovered by exponentiation with a bounded generator $u:=\lim _{t\rightarrow 0^+}\frac{1}{t}(\omega _t-\omega _0)$. It is not difficult to see that $u(1_A)=0$ and u is conditionally positive definite with respect to $\omega _0$, since these hold for each $\frac{1}{t}(\omega _t-\omega _0),t>0$. Then together with Theorem 4.1 we have the following result.

Theorem 4.4

Let $\phi$ be an idempotent state on a compact quantum group ${\mathbb {G}}=(A,\varDelta )$. For any non-zero bounded linear functional $\omega$ on A such that $\omega \phi =\phi \omega =\omega$, the following are equivalent

• 1. $\omega =\omega _1$ with $\{\omega _t\}_{t\ge 0}$ a norm continuous convolution semigroup of states such that $\omega _0=\phi$;

• 2. $\omega =\exp _\phi (u)$, where $u \in A'$ is $\phi$-bi-invariant, $u(1_A)=0$, and $u(x^*x)\ge 0$ for all $x\in A$ such that $\phi (x^*x)=0$;

• 3. $\omega =\exp _\phi (u)$, where $u=r(v-\phi )$, with $r>0$ and v a $\phi$-bi-invariant state on A.

Remark 4.5

The convolution semigroup of states $\{\omega _t\}_{t\ge 0}$ on a compact quantum group ${\mathbb {G}}$ is said to be weakly continuous if $\omega _t(a)\rightarrow \omega _0(a),t\rightarrow 0^+$ for any $a\in C({\mathbb {G}})$. Clearly norm continuous convolution semigroup of states is weakly continuous. When ${\mathbb {G}}$ is a finite quantum group, the converse also holds. But for general compact quantum group ${\mathbb {G}}$, there exists weakly continuous convolution semigroup of states which is not norm continuous. In this case, the generator is unbounded.

5 Infinitely divisible states on finite quantum groups

In this section we prove the main result of the paper.

Definition 5.1

Let ${\mathbb {G}}=(A,\varDelta )$ be a compact quantum group. A state $\omega \in {\mathscr {S}}(A)$ is said to be infinitely divisible if $\omega =\omega _n^n$ for some $\omega _n\in {\mathscr {S}}(A)$ and for all $n\ge 1$. We use ${\mathscr {I}}({\mathbb {G}})$ to denote the set of all infinitely divisible states on ${\mathbb {G}}$.

Clearly Poisson states are infinitely divisible. Our main result in this section is that any infinitely divisible state on a finite quantum group is a Poisson state. From now on, unless stated otherwise, ${\mathbb {G}}=(A,\varDelta )$ always denotes a finite quantum group.

The following lemma is well-known, and the proof follows from standard arguments.

Lemma 5.2

Let $A=\oplus _{k=1}^{m}{\mathbb {M}}_{n_k}({\mathbb {C}})$ with matrix units $\{e^k_{ij}:1\le i,j\le n_k,1\le k\le m\}$. Denote the dual basis by $\{\omega ^k_{ij}\}$. Then for any $\omega =\sum _{k=1}^{m}\sum _{i,j=1}^{n_k}a^{(k)}_{ij}\omega ^{k}_{ij}$, $\omega$ is a positive linear functional if and only if $[a^{(k)}_{ij}]_{i,j=1}^{n_k}$ is positive semi-definite for each k. In this case, $\Vert \omega \Vert =\sum _{k=1}^{m}\sum _{i=1}^{n_k}a^{(k)}_{ii}$.

As a direct consequence, we have the following decomposition, which is quite easy but very helpful.

Corollary 5.3

Let $A=\oplus _{k=1}^{m}{\mathbb {M}}_{n_k}({\mathbb {C}})$ with matrix units $\{e^k_{ij}:1\le i,j\le n_k,1\le k\le m\}$. Denote the dual basis by $\{\omega ^k_{ij}\}$. Let $\omega =\sum _{k=1}^{m}\sum _{i,j=1}^{n_k}a^{(k)}_{ij}\omega ^{k}_{ij}$ such that for each k either $[a^{(k)}_{ij}]_{i,j=1}^{n_k}\ge 0$ or $[a^{(k)}_{ij}]_{i,j=1}^{n_k}\le 0$. Then $\omega _+:=\sum _{k\in \varLambda }\sum _{i,j=1}^{n_k}a^{(k)}_{ij}\omega ^{k}_{ij}$ and $\omega _-:=\sum _{k\notin \varLambda }\sum _{i,j=1}^{n_k}a^{(k)}_{ij}\omega ^{k}_{ij}$ are positive functionals on A such that

$$\begin{aligned} \omega =\omega _+ -\omega _- \text { and }\Vert \omega \Vert =\Vert \omega _+\Vert +\Vert \omega _-\Vert , \end{aligned}$$

where $\varLambda$ is the set of all the k’s such that $[a^{(k)}_{ij}]_{i,j=1}^{n_k}\ge 0$.

Another important corollary is as follows.

Corollary 5.4

Let $(A,B,\langle ,\rangle )$ be a Plancherel triple. Suppose that $A=\oplus _{\alpha =1}^{m}{\mathbb {M}}_{n_\alpha }({\mathbb {C}})$ with the matrix units $\{e^{\alpha }_{ij}:1\le i,j\le n_{\alpha },1\le \alpha \le m\}$. Then for any positive linear functional $u\in A'$, we have

$$\begin{aligned} h_B(u)\le \epsilon _B(u). \end{aligned}$$

Proof

By Lemma 5.2 we may write u as $u=\sum _{\alpha =1}^{m}\sum _{i,j=1}^{n_{\alpha }}a^{\alpha }_{ij}\widetilde{e^{\alpha }_{ij}}$, where each $[a^{\alpha }_{ij}]_{i,j=1}^{n_\alpha }$ is positive semi-definite and $\{\widetilde{e^{\alpha }_{ij}}\}$ is the dual basis of $\{e^{\alpha }_{ij}\}$. Suppost that $\widetilde{e^{\alpha _0}}=1_B$. Then by (2.7) and (2.8), we have

$$\begin{aligned} h_B(u)=a^{\alpha _0}\le \sum _{\alpha =1}^{m}\sum _{i=1}^{n_\alpha }a^{\alpha }_{ii}=\epsilon _B(u). \end{aligned}$$

$\square$

Let ${\mathbb {G}}=(A,\varDelta )$ be a finite quantum group. Then for any $u\in A'={\hat{A}}$ we denote by $\Vert u\Vert _{{\hat{A}}}$ the C*-norm of u as an element in C*-algebra ${\hat{A}}$. Recall that $\Vert u\Vert$ is the norm of u as a functional. Moreover, we have $\Vert u\Vert _{{\hat{A}}}\le \Vert u\Vert$.

Let $\phi$ be an idempotent state on finite quantum group ${\mathbb {G}}=(A,\varDelta )$. For any $u\in A'$ such that $u=u\phi =\phi u$ and $\Vert u-\phi \Vert <1$, define the logarithm ofuwith respect to$\phi$ as

$$\begin{aligned} \log _\phi (u):=-\sum _{k\ge 1}\frac{(\phi -u)^k}{k}. \end{aligned}$$

Then we have the following properties of logarithm and exponential.

Lemma 5.5

Suppose that ${\mathbb {G}}=(A,\varDelta )$ is a finite quantum group. Let $\phi$ be an idempotent state on A, then for any bounded linear functionals u, v on A such that $u=u\phi =\phi u$ and $v=v\phi =\phi v$, we have

• (1) $\exp _\phi (\log _\phi (u))=u$, if $\Vert u-\phi \Vert <1$.

• (2) $\log _\phi (\exp _\phi (u))=u$, if $\Vert u\Vert <\log 2$.

• (3) $\exp _\phi (u+v)=\exp _\phi (u)\exp _\phi (v)$ if $uv=vu$.

• (4) $\log _\phi (uv)=\log _\phi (u)+\log _\phi (v)$, if $uv=vu$ and the following holds:

  $$\begin{aligned} \Vert u-\phi \Vert<1,~~\Vert v-\phi \Vert<1,\text { and }\Vert uv-\phi \Vert <1. \end{aligned}$$

• (5) If moreover, u is a state such that

  $$\begin{aligned} \Vert u-\phi \Vert<\frac{1}{2} \text { and } \Vert u^n-\phi \Vert <\frac{1}{2} \end{aligned}$$

  for some $n\ge 1$, then

  $$\begin{aligned} \Vert u^k-\phi \Vert <\frac{1}{2}\text { for all }1\le k\le n. \end{aligned}$$

  Consequently, in such a case we have

  $$\begin{aligned} \log _\phi (u^k)=k\log _\phi (u)\text { for all }1\le k\le n. \end{aligned}$$

Proof

(1)-(4) are direct and hold for all Banach algebras. To show (5), let $u_{\phi }$ be the restriction of u to $A_{\phi }$. Then by Lemma 3.4, $u_{\phi }$ is a state on a finite-dimensional $C^*$-algebra $A_{\phi }$. Moreover,

$$\begin{aligned} \Vert u-\phi \Vert =\Vert (u_{\phi }-\epsilon _{\phi }){\mathbb {E}}_\phi \Vert =\Vert u_{\phi }-\epsilon _{\phi }\Vert , \end{aligned}$$

where $\epsilon _{\phi }$ denotes the restriction of counit $\epsilon$ of A to $A_{\phi }$. Write $A_{\phi }=\oplus _{k=1}^{m}{\mathbb {M}}_{n_k}({\mathbb {C}})$ with matrix units $\{e^k_{ij}:1\le i,j\le n_k,1\le k\le m\}$. Let $\{\omega ^k_{ij}\}$ be the dual basis. By Lemma 5.2, $u_{\phi }=\sum _{k=1}^{m}\sum _{i,j=1}^{n_k}b^{(k)}_{ij}\omega ^k_{ij}$ with $[b^{(k)}_{ij}]_{i,j=1}^{n_k}\ge 0$ and $\Vert u_{\phi }\Vert =\sum _{k=1}^{m}\sum _{i=1}^{n_k}b^{(k)}_{ii}=1$. Since $\epsilon _{\phi }$ is a character on $A_{\phi }$, there exists $k_0$ such that $n_{k_0}=1$ and $\omega ^{k_0}=\epsilon _{\phi }$. Thus $u_{\phi }-\epsilon _{\phi }=(b^{(k_0)}-1)\epsilon _{\phi } +\sum _{k\ne k_0}\sum _{i,j=1}^{n_k}b^{(k)}_{ij}\omega ^k_{ij}$ verifies the condition of Corollary 5.3 and it follows that

$$\begin{aligned} \Vert u_{\phi }-\epsilon _{\phi }\Vert =1-b^{(k_0)}+\sum _{k\ne k_0}\sum _{i=1}^{n_k}b^{(k)}_{ii} =2-2b^{(k_0)} =-2(u_{\phi }-\epsilon _{\phi })(e^{k_0}). \end{aligned}$$

So for $v_1:=u-\phi$ we have $\Vert v_1\Vert =\Vert w_1\Vert =-2w_1(e^{k_0})$, where $w_1=v_1|_{A_{\phi }}$. Similarly for $v_j:=u^j-\phi$ and $w_j:=v_j|_{A_{\phi }}$ we have

$$\begin{aligned} \Vert v_j\Vert =\Vert w_j\Vert =-2w_j(e^{k_0}),~~j\ge 1. \end{aligned}$$

(5.1)

Now we show (5) by the induction argument. Clearly it holds for $n=1$. Suppose for now that it holds for n. Set $r:=\Vert v_1\Vert$, $s:=\Vert v_n\Vert$ and $t:=\Vert v_{n+1}\Vert$. From $(v_1+\phi )(v_n+\phi )=v_{n+1}+\phi$ it follows that $v_{n+1}=v_1+v_n+v_1 v_n$ and thus by Lemma 3.4 (4) $w_{n+1}=w_1+w_n+w_1 w_n$. This, together with (5.1) and Lemma 3.4 (1), yields

$$\begin{aligned} t=r+s-2(w_1 w_n)(e^{k_0}) \ge r+s-2\Vert w_1 w_n\Vert = r+s-2\Vert v_1 v_n\Vert \ge r+s-2rs. \end{aligned}$$

Then

$$\begin{aligned} (1-2r)(1-2s)=4rs-2r-2s+1\ge 1-2t. \end{aligned}$$

By assumption, $1-2r,1-2t>0$, so $1-2s>0$. Hence u and $u^n$ verify the conditions in (4), and we obtain

$$\begin{aligned} \log _\phi (u^{n+1})=\log _\phi (u)+\log _\phi (u^n)=\log _\phi (u)+n\log _\phi (u)=(n+1)\log _\phi (u), \end{aligned}$$

where in the second equality we have used the induction for n. This finishes the proof for $n+1$ and then shows (5). $\square$

Remark 5.6

In fact, to prove (5) we have used the fact that $\Vert u+v\Vert =\Vert u\Vert +\Vert v\Vert$ for all $u,v\in {\mathscr {N}}_\phi ({\mathbb {G}})$.

Proposition 5.7

Let $\omega$ be an infinitely divisible state on a finite quantum group ${\mathbb {G}}=(A,\varDelta )$. Let $\phi$ be an idempotent state on A. Assume that there exists a sequence $\{\omega _{m_j}\}_{j\ge 0}$ of roots of states of $\omega$, with $\omega =\omega _{m_j}^{m_j}$, for all j, such that

• (1) $\{m_j\}_{j\ge 0}$ is strictly increasing;

• (2) $\omega _{m_j}=\omega _{m_j}\phi =\phi \omega _{m_j}$ for all j;

• (3) $\omega _{m_j}=\omega _{m_{j+1}}^{n_j}$ for some positive integer $n_j$, $j\ge 0$;

• (4) $\omega _{m_j}\rightarrow \phi$, as $j\rightarrow \infty$.

Then $\omega \in {\mathscr {P}}_\phi ({\mathbb {G}})$.

Proof

Assume that $\{\omega _{m_j}\}$ contains infinitely many different elements, otherwise $\omega =\phi \in {\mathscr {P}}_\phi ({\mathbb {G}})$. By (4), we can choose $j_0>0$ such that $\Vert \omega _{m_{j}}-\phi \Vert <1/2$ for all $j\ge j_0$. This inequality, together with (2), allows us to define

$$\begin{aligned} v_0:=\log _\phi (\omega _{m_{j_0}}), \text { and }v:=m_{j_0}v_0. \end{aligned}$$

Then by the definition of $\omega _{j_0}$ and Lemma 5.5 (1)(2),

$$\begin{aligned} \omega =\omega _{m_{j_0}}^{m_{j_0}} =\left( \exp _\phi (\log _\phi (\omega _{m_{j_0}}))\right) ^{m_{j_0}} =\exp _\phi (m_{j_0}v_0) =\exp _\phi (v). \end{aligned}$$

To prove $\omega \in {\mathscr {P}}_\phi ({\mathbb {G}})$, it suffices to show that $v\in {\mathscr {N}}_\phi ({\mathbb {G}})$. For this we check that v verifies Theorem 4.1 (1). Clearly, $v(1_A)=0$, since $\omega _{j_0}$ is a state. By the definition of logarithm, $v=v\phi =\phi v$. It remains to show that for any $x\in A$ such that $\phi (x^*x)=0$, we have $v_0(x^*x)\ge 0$. By (3) we have

$$\begin{aligned} \omega _{m_{j_0}}=\omega ^{N_j}_{m_j},~~j\ge j_0, \end{aligned}$$

where $N_j:=n_{j_0}\cdots n_{j-1}$. Recall that for all $j\ge j_0$, $\Vert \omega _{m_{j}}-\phi \Vert <1/2$. Thus by Lemma 5.5 (5) we have

$$\begin{aligned} \log _\phi (\omega _{m_j})=\frac{1}{N_j}\log _\phi (\omega _{m_{j_0}})=\frac{v_0}{N_j}, \end{aligned}$$

and by Lemma 5.5 (1)

$$\begin{aligned} \omega _{m_j}=\exp _\phi \left( \frac{v_0}{N_j}\right) ,~~j\ge j_0. \end{aligned}$$

The condition (1) implies that $N_j\rightarrow \infty$ as $j\rightarrow \infty$. Now for any $x\in A$ such that $\phi (x^*x)=0$, we have

$$\begin{aligned} 0\le \omega _{m_j}(x^*x) =\phi (x^*x)+\frac{v_0(x^*x)}{N_j}+\sum _{m\ge 2}\frac{v_0^m(x^*x)}{N^m_j \cdot m!} =\frac{v_0(x^*x)}{N_j}+O \left( \frac{1}{N^2_j}\right) , \end{aligned}$$

for all $j\ge j_0$. Hence

$$\begin{aligned} v_0(x^*x)+O \left( \frac{1}{N_j}\right) \ge 0,~~j\ge j_0. \end{aligned}$$

Letting $j\rightarrow \infty$, we have $v_0(x^*x)\ge 0$, which ends the proof. $\square$

As this proposition suggests, to show that an infinitely divisible state is of Poisson type, it is important to capture the idempotent state where the infinitely divisible state is “supported on”. For this we need two lemmas. The first one is an easy fact in matrix theory.

Lemma 5.8

Let $P\in {\mathbb {M}}_n({\mathbb {C}})$ be a self-adjoint projection. Let $A,B\in {\mathbb {M}}_n({\mathbb {C}})$ be such that $A=AP=PA,AB=P$ and $\Vert A\Vert \le 1$, $\Vert B\Vert \le 1$. Here $\Vert \cdot \Vert$ denote the C*-norm in matrix algebra. Then $A^*A=AA^*=P$. Consequently, if u and v are states on a finite quantum group ${\mathbb {G}}=(A,\varDelta )$ such that $u=u\phi =\phi u$ and $uv=\phi$, where $\phi$ is an idempotent state on ${\mathbb {G}}$, then $u^*u=uu^*=\phi$.

Proof

Since P is a self-adjoint projection, we may assume without loss of generality that

$$\begin{aligned} P=\begin{pmatrix} I_r&{}\quad 0\\ 0&{}\quad 0 \end{pmatrix}, \end{aligned}$$

where $I_r$ is the identity in ${\mathbb {M}}_r({\mathbb {C}})$ with $r=\text {rank}(P)$. From $A=AP=PA$ and $AB=P$ it follows

$$\begin{aligned} A=\begin{pmatrix} A_r&{}\quad 0\\ 0&{}\quad 0 \end{pmatrix},~~ B=\begin{pmatrix} B_r&{}\quad *\\ *&{}\quad * \end{pmatrix}, \end{aligned}$$

with $A_r B_r=I_r$. Note that

$$\begin{aligned} 1 =\Vert I_r\Vert =\Vert A_r B_r\Vert \le \Vert A_r\Vert \Vert B_r\Vert \le \Vert A\Vert \Vert B\Vert \le 1. \end{aligned}$$

So $\Vert A_r\Vert =\Vert B_r\Vert =1$. This is to say,

$$\begin{aligned} \Vert A_r^*A_r\Vert =1=\Vert B_r B_r^*\Vert =\Vert (A_r^*A_r)^{-1}\Vert . \end{aligned}$$

Then all the eigenvalues of $A_r^*A_r$ must be 1 and thus $A_r^*A_r=I_r$. Hence $B_r=A^*_r$ and thus $A^*A=AA^*=P$.

The remaining part follows from the facts that $p=\phi$ is a self-adjoint projection in ${\hat{A}}$ and

$$\begin{aligned} \Vert u\Vert _{{\hat{A}}}\le \Vert u\Vert =1,~~\Vert v\Vert _{{\hat{A}}}\le \Vert v\Vert =1. \end{aligned}$$

$\square$

Lemma 5.9

Let $(A,B,\langle ,\rangle )$ be a Plancherel triple. Suppose that $u\in B$ is a state on A such that $uu^*=u^*u=\epsilon _A$, where $\epsilon _A$ is the counit of A. Then u is an n-th root of $\epsilon _A$ for some $n\le \dim B$.

Proof

Suppose that $A=\oplus _{\alpha =1}^{m}{\mathbb {M}}_{n_\alpha }({\mathbb {C}})$ with the matrix units $\{e^{\alpha }_{ij}:1\le i,j\le n_{\alpha },1\le \alpha \le m\}$. Let $\{\widetilde{e^{\alpha }_{ij}}\}$ be the dual basis. Then by Lemma 5.2 we can write u as

$$\begin{aligned} u=\sum _{\alpha =1}^{m}p_\alpha \sum _{i,j=1}^{n_\alpha }a^\alpha _{ij}\widetilde{e^{\alpha }_{ij}}, \end{aligned}$$

where $p_\alpha \ge 0$, $\sum _{\alpha =1}^{m}p_\alpha =1$ and $[a^\alpha _{ij}]_{i,j=1}^{n_\alpha }$ positive semi-definite with trace one. Let $\alpha _0$ be such that $\widetilde{e^{\alpha _0}}=1_B$. Then we can write $u=p_{\alpha _0}\widetilde{e^{\alpha _0}}+v$ with $v=\sum _{\alpha \ne \alpha _0}p_\alpha \sum _{i,j=1}^{n_\alpha }a^\alpha _{ij}\widetilde{e^{\alpha }_{ij}}$ such that $\Vert v\Vert =\sum _{\alpha \ne \alpha _0}p_\alpha =1-p_{\alpha _0}$. Note that $uu^*=p^2_{\alpha _0}\widetilde{e^{\alpha _0}}+p_{\alpha _0}(v+v^*)+vv^*$. Then by (2.7), $h_B(v)=h_B(v^*)=0$. Since v and $v^*$ are both positive functionals, $vv^*(\cdot )=\langle \varDelta _A(\cdot ),v\otimes v^*\rangle$ is also a positve functional on A. Hence we have by Corollary 5.4 that

$$\begin{aligned} h_{B}(vv^*)\le \epsilon _{B}(vv^*)=|\epsilon _{B}(v)|^2=(1-p_{\alpha _0})^2. \end{aligned}$$

So we have

$$\begin{aligned} 1 =h_{B}(1_B) =h_{B}(uu^*) =p_{\alpha _0}^2+h_{B}(vv^*)\le p_{\alpha _0}^2+(1-p_{\alpha _0})^2 =1-2p_{\alpha _0}+2p_{\alpha _0}^2, \end{aligned}$$

which yields $p_{\alpha _0}(p_{\alpha _0}-1)\ge 0$. Recall that $0\le p_{\alpha _0}\le 1$, hence either $p_{\alpha _0}=0$ or $p_{\alpha _0}=1$. That is to say, either $h_{B}(u)=0$ or $u=\epsilon _A$. Since for any $n\ge 1$, $u^n$ is again a state such that $u^n u^{*n}=\epsilon _A$, we obtain, by a similar argument, that either $h_{B}(u^n)=0$ or $u^n=\epsilon _A$.

If u is not a n-th root of $\epsilon _A$ for all $1\le n\le \dim B$, then we have

$$\begin{aligned} h_{B}(u^n)=0,~~n=1,2,\dots ,\dim B. \end{aligned}$$

(5.2)

Note that we may choose $m\le \dim B$ such that u is unitary in ${\mathbb {M}}_{m}({\mathbb {C}})$. Set $P(\lambda ):=\det (\lambda I_m-u)=\sum _{i=0}^{m}a_i \lambda ^i$, then Cayley-Hamilton Theorem implies that $P(u)=0$, where $I_m$ denotes the identity matrix in ${\mathbb {M}}_{m}({\mathbb {C}})$. Since u is unitary in ${\mathbb {M}}_{m}({\mathbb {C}})$, we have $a_0=(-1)^m\det (u)\ne 0$. But by (5.2)

$$\begin{aligned} a_0h_{B}(I_m)=a_0h_{B}(I_m)+\sum _{i=1}^{m}a_ih_{B}( u^i)=h_{B}(P(u))=0. \end{aligned}$$

Thus $a_0=0$, which leads to a contradiction. So we must have $u^m=\epsilon _A$ for some $1\le m\le \dim B$. $\square$

The following proposition, gathering the main ingredients of preceding lemmas, will be used to prove Theorem 5.11.

Proposition 5.10

Let ${\mathbb {G}}=(A,\varDelta )$ be a finite quantum group with dual $\hat{{\mathbb {G}}}=({\hat{A}},{\hat{\varDelta }})$. Suppose that $u,v\in {\mathscr {S}}(A)$ and $\phi \in \text {Idem}({\mathbb {G}})$ are such that $u=u\phi =\phi u$ and $uv=\phi$. Then there exists a positive integer $m\le \dim {\hat{A}}$ such that $u^m=\phi$.

Proof

From Lemma 5.8 it follows $u^*u=uu^*=\phi$. Let $u_{\phi }$ and $\epsilon _{\phi }$ be the restrictions of u and the counit $\epsilon$ of A to $A_{\phi }$, respectively. From Proposition 3.5$(A_{\phi },{\hat{A}}_p,\langle ,\rangle )$ forms a Plancherel triple. By Lemma 3.4, $u_{\phi }$ is a state on $A_{\phi }$ such that $u_{\phi }u_{\phi }^*=\epsilon _{\phi }$. So Lemma 5.9 implies $u^{m}_{\phi }=\epsilon _{\phi }$ for some $m\le \dim \widehat{A_\phi }\le \dim {\hat{A}}$. Hence Lemma 3.4 gives $u^m=u^m_{\phi }{\mathbb {E}}_\phi =\epsilon _{\phi }{\mathbb {E}}_\phi =\phi$. $\square$

Now we are ready to prove the main result of this paper.

Theorem 5.11

Let ${\mathbb {G}}=(A,\varDelta )$ be a finite quantum group. Then ${\mathscr {P}}({\mathbb {G}})={\mathscr {I}}({\mathbb {G}})$.

The first proof

${\mathscr {P}}({\mathbb {G}})\subset {\mathscr {I}}({\mathbb {G}})$ is clear. Let $\omega \in {\mathscr {I}}({\mathbb {G}})$. We claim that for any positive integer $N\ge 2$, there exists a sequence $\{b_n\}_{n\ge 0}$ of roots of $\omega$ such that $b_0=\omega ,b_{n-1}=b^N_n,n\ge 1$. Indeed, since A is finite dimensional, the set of states $Z={\mathscr {S}}(A)$ is compact with respect to the norm topology. Thus $\prod _{j\ge 0}Z_j$, where $Z_j=Z$ for all j, is compact with respect to the product topology. Let $a_n\in Z$ be any n-th root of $\omega$ for all $n\ge 0$. Then the sequence of non-empty closed sets

$$\begin{aligned} W_k:={\overline{\bigcup }}_{j\ge k}\{a^{N^j}_{N^j}\}\times \{a^{N^{j-1}}_{N^j}\}\times \cdots \times \left\{ a_{N^j}\right\} \times \prod _{i\ge j}Z_i,~~k\ge 1, \end{aligned}$$

is decreasing: $W_1\supset W_2\supset \cdots$, and thus any finite intersection of $\{W_k\}_{k\ge 1}$ is non-empty. By compactness of $\prod _{j\ge 0}Z_j$, $\bigcap _{k\ge 1}W_k\ne \emptyset$. Hence one can choose $(b_0,b_1,\dots )\in \bigcap _{k\ge 1}W_k$ such that

$$\begin{aligned} b_0=\omega ,~~b_{n-1}=b^{N}_n,~~n\ge 1. \end{aligned}$$

This proves the claim.

Choose $N=(\dim {\hat{A}})!\ge 2$ and let $\{b_n\}_{n\ge 0}$ be as above. Since Z is compact, there exists a subsequence $\{c_j\}_{j\ge 0}$ of $\{b_i\}_{i\ge 0}$ such that $c_j$ converges to some $c\in Z$. If we fix a non-negative integer i, we have $b_i=c^{r_j}_j$ for sufficient large j and some integer $r_j\ge N\ge 2$. That is,

$$\begin{aligned} b_i=c_j c^{r_j-1}_j=c^{r_j-1}_j c_j. \end{aligned}$$

(5.3)

We can assume that $c^{r_j-1}_j$ converges to some $d_i\in Z$, otherwise consider some subsequence, since $\{c^{r_j-1}_j\}_{j\ge 0}\subset Z$. Thus letting $j\rightarrow \infty$ in (5.3), we have

$$\begin{aligned} b_i=c d_i=d_i c,~~i\ge 0. \end{aligned}$$

(5.4)

This implies $b_i\in cZ\cap Zc$ for all $i\ge 0$. From the choice of $c_j$ we have $c_j\in cZ\cap Zc$ for all $j\ge 0$. Then for any i the corresponding $c^{r_j-1}_j\in cZ\cap Zc$ for all j, which implies that $d_i\in cZ\cap Zc$ by the compactness of $cZ\cap Zc$. Now consider (5.3) for $\{c_j\}_{j\ge 0}$, instead of $\{b_i\}_{i\ge 0}$, we obtain an updated version of (5.4):

$$\begin{aligned} c_j=c d'_j=d'_j c,~~j\ge 0, \end{aligned}$$

(5.5)

where $d'_j\in cZ\cap Zc$. Letting $j\rightarrow \infty$, consider the subsequence of $\{d'_j\}_{j\ge 0}$ if necessary, one obtains

$$\begin{aligned} c=cd=dc, \end{aligned}$$

(5.6)

where $d\in cZ\cap Zc$ by the compactness of $cZ\cap Zc$. Suppose $d=ce$ for some $e\in Z$, then $d^2=dce=ce=d$, i.e. d is an idempotent state. By Proposition 5.10, we obtain $c^m=d$ for some $m\le \dim {\hat{A}}$. Since our choice of N satisfies m|N, we have

$$\begin{aligned} c^N_j\rightarrow c^N=(c^m)^{\frac{N}{m}}=d, \text { as }j\rightarrow \infty . \end{aligned}$$

Denote by $\phi$ the idempotent state d. Set $\omega _0:=\omega$ and $\omega _n:=c_n^N$ for all $n\ge 1$. Then $\omega _n\rightarrow \phi$ as n tends to $\infty$. By definition, $\{\omega _n\}_{n\ge 0}$ is a subsequence of $\{b_j\}_{j\ge 0}$, thus $\omega _{n-1}=\omega _n^{s_n}$ with $N|s_n$ for all $n\ge 1$. Moreover, from (5.5) we have

$$\begin{aligned} \omega _n=c^N_n=(c d'_n)^N=c^N d_n^{'N}=\phi d_n^{'N}=\phi (\phi d_n^{'N})=\phi \omega _n,~~n\ge 0. \end{aligned}$$

Similarly, $\omega _n=\omega _n\phi ,n\ge 0.$ Hence $\{\omega _n\}_{n\ge 0}$ verifies the conditions of Proposition 5.7, and consequently $\omega \in {\mathscr {P}}_\phi ({\mathbb {G}})$. $\square$

Before giving the second proof, we introduce the following proposition, which could be formulated and proved for a general Banach algebra.

Proposition 5.12

Let ${\mathbb {G}}=(A,\varDelta )$ be a compact quantum group, with A separable. Let $\omega$ be an infinitely divisible state on ${\mathbb {G}}$. Suppose that there exist an idempotent state $\phi$ and a sequence of $\phi$-bi-invariant roots $\{\omega _{n_k}\}_{k\ge 1}$ of $\omega$, where $\{n_k\}_{k\ge 0}$ is an increasing sequence of positive integers, such that $\omega _{n_k}^{n_k}=\omega$ for all $k\ge 1$, and

$$\begin{aligned} \sup _{k\ge 1}n_k\Vert \omega _{n_k}-\phi \Vert =M<\infty , \end{aligned}$$

(5.7)

then $\omega \in {\mathscr {P}}_\phi ({\mathbb {G}})$.

Proof

Recall that $\{\varphi \in A':\Vert \varphi \Vert \le M\}$ is compact with respect to weak* topology for each $M>0$. Then from (5.7) we have for some subsequence of $\{n_k\}_{k\ge 1}$, still denoted by $\{n_k\}_{k\ge 1}$, that $n_k(\omega _{n_k}-\phi )$ converges to an element $u\in A'$ with respect to the $\hbox {weak}^*$ topology. Then $u=\lim \limits _{k\rightarrow \infty }n_k(\omega _{n_k}-\phi )\in {\mathscr {N}}_\phi ({\mathbb {G}})$ and $\exp _\phi (u)$ is a Poisson state. It suffices to show that $\omega =\exp _\phi (u)$. Set

$$\begin{aligned} u_{n_k}:=\sum _{m\ge 2}\frac{(\omega _{n_k}-\phi )^m}{m!},~~k\ge 1. \end{aligned}$$

It is well-defined, since $\sup _{k\ge 1}\Vert \omega _{n_k}-\phi \Vert <\infty$. Moreover, from (5.7) it follows

$$\begin{aligned} \Vert u_{n_k}\Vert \le \sum _{m\ge 2}\frac{1}{m!}\left( \frac{M}{n_k}\right) ^m\le \frac{1}{n_k^2}\sum _{m\ge 2}\frac{M^m}{m!}, \end{aligned}$$

whence

$$\begin{aligned} \lim \limits _{k\rightarrow \infty }n_k\Vert u_{n_k}\Vert =0, \end{aligned}$$

and

$$\begin{aligned} \lim \limits _{k\rightarrow \infty }\left( 1+\Vert u_{n_k}\Vert \right) ^{n_k}=\lim \limits _{k\rightarrow \infty }\left( 1+\Vert u_{n_k}\Vert \right) ^{\frac{1}{\Vert u_{n_k}\Vert }\cdot n_k\Vert u_{n_k}\Vert }=1. \end{aligned}$$

Hence

$$\begin{aligned} \Vert \omega -\exp _\phi (n_k(\omega _{n_k}-\phi ))\Vert&=\Vert \omega _{n_k}^{n_k}-(\omega _{n_k}+u_{n_k})^{n_k}\Vert \\&\le \sum _{m=1}^{n_k}\left( {\begin{array}{c}n_k\\ m\end{array}}\right) \Vert \omega _{n_k}^{n_k-m}u_{n_k}^m\Vert \\&\le \sum _{m=1}^{n_k}\left( {\begin{array}{c}n_k\\ m\end{array}}\right) \Vert u_{n_k}\Vert ^m =\left( 1+\Vert u_{n_k}\Vert \right) ^{n_k}-1, \end{aligned}$$

which tends to 0 as $k\rightarrow \infty$. This shows $\omega =\exp _\phi (u)$ and finishes the proof. $\square$

The second proof of Theorem 5.11

Again, ${\mathscr {P}}({\mathbb {G}})\subset {\mathscr {I}}({\mathbb {G}})$ is clear. Let $\omega \in {\mathscr {I}}({\mathbb {G}})$. From the first proof we know that there exist an idempotent state $\phi \in Idem ({\mathbb {G}})$ and a sequence of roots $\{\omega _{n_k}\}_{k\ge 0}\subset {\mathscr {S}}(A)$ with $\{n_k\}_{k\ge 0}$ an increasing sequence of positive integers such that

$$\begin{aligned} \omega _{n_k}^{n_k}=\omega ,~~ \omega _{n_k}=\omega _{n_k}\phi =\phi \omega _{n_k},~~k\ge 0, \end{aligned}$$

and $\omega _{n_k}\rightarrow \phi$ as $k\rightarrow \infty$. Let u and $u_{n_k}$ be the restrictions of $\omega$ and $\omega _{n_k}$ to $A_{\phi }$ for all $k\ge 0$, respectively. Then from Lemma 3.4u is a state on $A_{\phi }$ such that $\{u_{n_k}\}_{k\ge 0}$ is a sequence of roots of u in ${\mathscr {S}}(A_{\phi })$ verifying

$$\begin{aligned} u_{n_k}^{n_k}=u \text { and }u_{n_k}\rightarrow \epsilon _{\phi },~~ k\rightarrow \infty , \end{aligned}$$

where $\epsilon _{\phi }$ is the counit of $A_{\phi }$. Now we repeat a calculation in Lemma 5.9. Suppose that $A=\oplus _{\alpha =1}^{m}{\mathbb {M}}_{n_\alpha }({\mathbb {C}})$ with the matrix units $\{e^{\alpha }_{ij}:1\le i,j\le n_{\alpha },1\le \alpha \le m\}$. Let $\{\widetilde{e^{\alpha }_{ij}}\}$ be the dual basis. Then by Lemma 5.2 we can write u as

$$\begin{aligned} u_{n_k}=\sum _{\alpha =1}^{m}p_{\alpha ,k}\sum _{i,j=1}^{n_\alpha }a^{\alpha ,k}_{ij}\widetilde{e^{\alpha }_{ij}} \end{aligned}$$

with $p_{\alpha ,k}\ge 0$, $\sum _{\alpha =1}^{m}p_{\alpha ,k}=1$ and $[a^{\alpha ,k}_{ij}]_{i,j=1}^{n_\alpha }$ is positive semi-definite with trace one for each k. Let $\alpha _0$ be such that $\widetilde{e^{\alpha _0}}=\epsilon _{\phi }=1_{{\hat{A}}_p}$. By Corollary 5.3 and the assumption,

$$\begin{aligned} \Vert u_{n_k}-\epsilon _{\phi }\Vert = 2(1-p_{\alpha _0,k})\rightarrow 0,~~k\rightarrow \infty . \end{aligned}$$

(5.8)

So $p_{\alpha _0,k}\rightarrow 1$ as $k\rightarrow \infty$.

Now for each $q>0$ we introduce the $L_q$-norm induced by ${\hat{h}}_{p}$. Recall that ${\hat{h}}_p=\frac{1}{\phi (p_A)}{\hat{h}}|_{{\hat{A}}_p}$ is a tracial state on ${\hat{A}}_p$. Namely, we define

$$\begin{aligned} \Vert x\Vert _q:=[{\hat{h}}_p(|x|^{q})]^{\frac{1}{q}},~~x\in {\hat{A}}_p, \end{aligned}$$

where $|x|=(x^*x)^{\frac{1}{2}}$. Then Hölder’s inequality still holds and we have

$$\begin{aligned} \Vert u\Vert _{\frac{2}{n_k}}=\Vert u_{n_k}^{n_k}\Vert _{\frac{2}{n_k}}\le \Vert u_{n_k}\Vert _{2}^{n_k},~~k\ge 0. \end{aligned}$$

(5.9)

Note that u is invertible in ${\hat{A}}_p$. Indeed, for large k there holds

$$\begin{aligned} \Vert u_{n_k}-\epsilon _{\phi }\Vert _{{\hat{A}}_p}\le \Vert u_{n_k}-\epsilon _{\phi }\Vert <1,~~k\ge 0, \end{aligned}$$

then we have that $u_{n_k}$ is invertible for large k. So is $u=u^{n_k}_{n_k}$ is also invertible in ${\hat{A}}_p$.

Following a similar calculation to that in Lemma 5.9, we obtain

$$\begin{aligned} {\hat{h}}_p(u_{n_k}u_{n_k}^*) \le p_{\alpha _0,k}^2+(1-p_{\alpha _0,k})^2. \end{aligned}$$

This, together with (5.9), yields

$$\begin{aligned} {\hat{h}}_p(|u|^{\frac{1}{n_k}}) =\Vert u\Vert _{\frac{2}{n_k}}^{\frac{2}{n_k}} \le \Vert u_{n_k}\Vert _{2}^2 ={\hat{h}}_p(u_{n_k}u_{n_k}^*) \le p_{\alpha _0,k}^2+(1-p_{\alpha _0,k})^2, \end{aligned}$$

for all k. Since $p_{\alpha _0,k}\rightarrow 1$ as $k\rightarrow \infty$, there exists $K>0$ such that for all $k\ge K$, $\frac{1}{2}\le p_{\alpha _0,k}\le 1$. Thus for all $k\ge K$, $1-p_{\alpha _0,k}\le p_{\alpha _0,k}$ and then

$$\begin{aligned} {\hat{h}}_p(|u|^{\frac{1}{n_k}}) \le p_{\alpha _0,k}^2+(1-p_{\alpha _0,k})^2 \le p_{\alpha _0,k}^2+p_{\alpha _0,k}(1-p_{\alpha _0,k}) =p_{\alpha _0,k}. \end{aligned}$$

(5.10)

Combining this with (5.8), we have

$$\begin{aligned} n_k\Vert u_{n_k}-\epsilon _{\phi }\Vert =2n_k(1-p_{\alpha _0,k}) \le 2n_k(1-{\hat{h}}_p(|u|^{\frac{1}{n_k}})) =2n_k{\hat{h}}_p(1_{{\hat{A}}_p}-|u|^{\frac{1}{n_k}}) \end{aligned}$$

for all $k\ge K$. Since u is invertible in ${\hat{A}}_p$, $n(1_{{\hat{A}}_p}-|u|^{\frac{1}{n}})$ converges to $-\log |u|$ in norm as $n\rightarrow \infty$. Then there exists a constant $M<\infty$ such that

$$\begin{aligned} \sup _{n\ge 1}n{\hat{h}}_p(1_{{\hat{A}}_p}-|u|^{\frac{1}{n}})\le M. \end{aligned}$$

Thus from Lemma 3.4 it follows that

$$\begin{aligned} \sup _{k\ge 0}n_k\Vert \omega _{n_k}-\phi \Vert =\sup _{k\ge 0}n_k\Vert u_{n_k}-\epsilon _{\phi }\Vert \le \sup _{k\ge 0}2n_k{\hat{h}}_p(1_{{\hat{A}}_p}-|u|^{\frac{1}{n_k}}) <\infty . \end{aligned}$$

Hence $\omega \in {\mathscr {P}}_\phi ({\mathbb {G}})$ by Proposition 5.12. $\square$

Remark 5.13

Both proofs rely on the capture of idempotent state where the infinitely divisible state is “supported on” and the sequence of roots converging to this idempotent state. After this the first proof aims to show that this sequence of roots can chosen to form a submonogeneous convolution semigroup (Proposition 5.7 (3)), while the idea of the second proof is derived from a general result Proposition 5.12, concerning the decay property (5.7) of this sequence of roots. The inequality (5.10) also allows us to simplify the proof of the main theorem in [17] for the finite group case.