Quasitriangular Structure of Myhill–Nerode Bialgebras

Abstract

In computer science the Myhill–Nerode Theorem states that a set L of words in a finite alphabet is accepted by a finite automaton if and only if the equivalence relation ${\sim }_L$, defined as $x{\sim }_{L}y$ if and only if $xz\in L$ exactly when $yz\in L,\forall z$, has finite index. The Myhill–Nerode Theorem can be generalized to an algebraic setting giving rise to a collection of bialgebras which we call Myhill–Nerode bialgebras. In this paper we investigate the quasitriangular structure of Myhill–Nerode bialgebras.

1. Introduction

Let ${\Sigma }_0$ be a finite alphabet and let ${\widehat{\Sigma }}_0$ denote the set of words formed from the letters in ${\Sigma }_0$. Let $L\subseteq {\widehat{\Sigma }}_0$ be a language, and let ${\sim }_L$ be the equivalence relation defined as $x{\sim }_{L}y$ if and only if $xz\in L$ exactly when $yz\in L,\forall z\in {\widehat{\Sigma }}_0$. The Myhill–Nerode Theorem of computer science states that L is accepted by a finite automaton if and only if ${\sim }_L$ has finite index (cf. [1, 1, Chapter III, §9, Proposition 9.2], [2, §3.4, Theorem 3.9]). In [3, Theorem 5.4] the authors generalize the Myhill–Nerode theorem to an algebraic setting in which a finiteness condition involving the action of a semigroup on a certain function plays the role of the finiteness of the index of ${\sim }_L$, while a bialgebra plays the role of the finite automaton which accepts the language. We call these bialgebras Myhill–Nerode bialgebras.

The purpose of this paper is to investigate the quasitriangular structure of Myhill–Nerode bialgebras.

By construction, a Myhill–Nerode bialgebra B is cocommutative and finite dimensional over its base field. Thus B admits (at least) the trivial quasitriangular structure $(B,1\otimes 1)$. We ask: does B (or its linear dual $B^{*}$) have any non-trivial quasitriangular structures?

Towards a solution to this problem, we construct a class of commutative Myhill–Nerode bialgebras and give a complete account of the quasitriangular structure of one of them. We begin with some background information regarding algebras, coalgebras, and bialgebras.

2. Algebras, Coalgebras and Bialgebras

Let K be an arbitrary field of characteristic 0 and let A be a vector space over K with scalar product $ra$ for all $r\in K$, $a\in A$. Scalar product defines two maps $s_1:K\otimes A\to A$ with $r\otimes a\mapsto ra$ and $s_2:A\otimes K\to A$ with $a\otimes r\mapsto ra$, for $a\in A$, $r\in K$. Let $I_A:A\to A$ denote the identity map. A K-algebra is a triple $(A,m_A,{\eta }_A)$ where $m_A:A\otimes A\to A$ is a K-linear map which satisfies

$$m_A(I_A\otimes m_A)(a\otimes b\otimes c)=m_A(m_A\otimes I_A)(a\otimes b\otimes c)$$

(1)

and ${\eta }_A:K\to A$ is a K-linear map for which

$$m_A(I_A\otimes {\eta }_A)(a\otimes r)=ra=m_A({\eta }_A\otimes I_A)(r\otimes a)$$

(2)

for all $r\in K$, $a,b,c\in A$. The map $m_A$ is the multiplication map of A and ${\eta }_A$ is the unit map of A. Condition (1) is the associative property and Condition (2) is the unit property.

We write $m_A(a\otimes b)$ as $ab$. The element $1_A={\eta }_A\left(1_K\right)$ is the unique element of A for which $a{1}_A=a=1_{A}a$ for all $a\in A$. Let $A,B$ be algebras. An algebra homomorphism from A to B is a K-linear map $\varphi :A\to B$ such that $\varphi \left(m_A(a_1\otimes a_2)\right)=m_B(\varphi \left(a_1\right)\otimes \varphi \left(a_2\right))\text{for}\text{all}{a}_1,a_2\in A,\text{and}\varphi \left(1_A\right)=1_B.$ In particular, for A to be a subalgebra of B we require $1_A=1_B$.

For any two vector spaces V, W let $\tau :V\otimes W\to W\otimes V$ denote the twist map defined as $\tau (a\otimes b)=b\otimes a$, for $a\in V$, $b\in W$. For K-algebras $A,B$, we have that $A\otimes B$ is a K-algebra with multiplication

$$m_{A\otimes B}:(A\otimes B)\otimes (A\otimes B)\to A\otimes B$$

defined by

$$\begin{array}{ccc}\hfill m_{A\otimes B}((a\otimes b)\otimes (c\otimes d))& =& (m_A\otimes m_B)(I_A\otimes \tau \otimes I_B)(a\otimes (b\otimes c)\otimes d)\hfill \\ & =& (m_A\otimes m_B)((a\otimes c)\otimes (b\otimes d))=ac\otimes bd\hfill \end{array}$$

for $a,c\in A$, $b,d\in B$. The unit map ${\eta }_{A\otimes B}:K\to A\otimes B$ given as

$${\eta }_{A\otimes B}\left(r\right)={\eta }_A\left(r\right)\otimes 1_B$$

for $r\in K$.

Let C be a K-vector space. A K-coalgebra is a triple $(C,{\Delta }_C,{ϵ}_C)$ in which ${\Delta }_C:C\to C\otimes C$ is K-linear and satisfies

$$(I_C\otimes {\Delta }_C){\Delta }_C\left(c\right)=({\Delta }_C\otimes I_C){\Delta }_C\left(c\right)$$

(3)

and ${ϵ}_C:C\to K$ is K-linear with

$$s_1({ϵ}_C\otimes I_C){\Delta }_C\left(c\right)=c=s_2(I_C\otimes {ϵ}_C){\Delta }_C\left(c\right)$$

(4)

for all $c\in C$. The maps ${\Delta }_C$ and ${ϵ}_C$ are the comultiplication and counit maps, respectively, of the coalgebra C. Condition (3) is the coassociative property and Condition (4) is the counit property.

We use the notation of M. Sweedler [4, §1.2] to write

$${\Delta }_C\left(c\right)=\sum _{\left(c\right)}{c}_{\left(1\right)}\otimes c_{\left(2\right)}$$

Note that Condition (4) implies that

$$\sum _{\left(c\right)}{ϵ}_C\left(c_{\left(1\right)}\right)c_{\left(2\right)}=c=\sum _{\left(c\right)}{ϵ}_C\left(c_{\left(2\right)}\right)c_{\left(1\right)}$$

(5)

Let C be a K-coalgebra. A nonzero element c of C for which ${\Delta }_C\left(c\right)=c\otimes c$ is a grouplike element of C. If c is grouplike, then

$$\begin{array}{ccc}\hfill c& =& s_1({ϵ}_C\otimes I_C){\Delta }_C\left(c\right)\hfill \\ & =& s_1({ϵ}_C\otimes I_C)(c\otimes c)={ϵ}_C\left(c\right)c\hfill \end{array}$$

and so, ${ϵ}_C\left(c\right)=1$. The grouplike elements of C are linearly independent [4, Proposition 3.2.1].

Let $C,D$ be coalgebras. A K-linear map $\varphi :C\to D$ is a coalgebra homomorphism if $(\varphi \otimes \varphi ){\Delta }_C\left(c\right)={\Delta }_D\left(\varphi \left(c\right)\right)$ and ${ϵ}_C\left(c\right)={ϵ}_D\left(\varphi \left(c\right)\right)$ for all $c\in C$. The tensor product $C\otimes D$ of two coalgebras is again a coalgebra with comultiplication map

$${\Delta }_{C\otimes D}:C\otimes D\to (C\otimes D)\otimes (C\otimes D)$$

defined by

$$\begin{array}{ccc}\hfill {\Delta }_{C\otimes D}(c\otimes d)& =& (I_C\otimes \tau \otimes I_D)({\Delta }_C\otimes {\Delta }_D)(c\otimes d)\hfill \\ & =& (I_C\otimes \tau \otimes I_D)({\Delta }_C\left(c\right)\otimes {\Delta }_D\left(d\right))\hfill \\ & =& (I_C\otimes \tau \otimes I_D)(\sum _{\left(c\right),\left(d\right)}{c}_{\left(1\right)}\otimes c_{\left(2\right)}\otimes d_{\left(1\right)}\otimes d_{\left(2\right)})\hfill \\ & =& \sum _{\left(c\right),\left(d\right)}{c}_{\left(1\right)}\otimes d_{\left(1\right)}\otimes c_{\left(2\right)}\otimes d_{\left(2\right)}\hfill \end{array}$$

for $c\in C$, $d\in D$. The counit map ${ϵ}_{C\otimes D}:C\otimes D\to K$ is defined as

$${ϵ}_{C\otimes D}(c\otimes d)={ϵ}_C\left(c\right){ϵ}_D\left(d\right)$$

for $c\in C$, $d\in D$.

A K-bialgebra is a K-vector space B together with maps $m_B$, ${\eta }_B$, ${\Delta }_B$, ${ϵ}_B$ for which $(B,m_B,{\eta }_B)$ is a K-algebra and $(B,{\Delta }_B,{ϵ}_B)$ is a K-coalgebra and for which ${\Delta }_B$ and ${ϵ}_B$ are algebra homomorphisms. Let $B,B^{\prime }$ be bialgebras. A K-linear map $\varphi :B\to B^{\prime }$ is a bialgebra homomorphism if ϕ is both an algebra and coalgebra homomorphism.

A K-Hopf algebra is a bialgebra H together with an additional K-linear map ${\sigma }_H:H\to H$ that satisfies

$$m_H(I_H\otimes {\sigma }_H){\Delta }_H\left(h\right)={ϵ}_H\left(h\right)1_H=m_H({\sigma }_H\otimes I_H){\Delta }_H\left(h\right)$$

(6)

for all $h\in H$. The map ${\sigma }_H$ is the coinverse (or antipode) map and property Condition (6) is the coinverse (or antipode) property. Though we will not consider Hopf algebras here, more details on the subject can be found in [5,6,7,8].

An important example of a K-bialgebra is given as follows. Let G be a semigroup with unity, 1. Let $KG$ denote the semigroup algebra. Then $KG$ is a bialgebra with comultiplication map

$${\Delta }_{KG}:KG\to KG\otimes KG$$

defined by $x\mapsto x\otimes x$, for all $x\in G$, and counit map ${ϵ}_{KG}:KG\to K$ given by $x\mapsto 1$, for all $x\in G$. The bialgebra $KG$ is the semigroup bialgebra on G.

Let B be a bialgebra, and let A be an algebra which is a left B-module with action denoted by “·”. Suppose that

$$b·\left(a{a}^{\prime }\right)=\sum _{\left(b\right)}(b_{\left(1\right)}·a)(b_{\left(2\right)}·a^{\prime })$$

and

$$b·1_A={ϵ}_B\left(b\right)1_A$$

for all $a,a^{\prime }\in A$, $b\in B$. Then A is a left B-module algebra. A K-linear map $\varphi :A\to A^{\prime }$ is a left B-module algebra homomorphism if ϕ is both an algebra and a left B-module homomorphism.

Let C be a coalgebra and a right B-module with action denoted by “·”. Suppose that for all $c\in C$, $b\in B$,

$${\Delta }_C(c·b)=\sum _{\left(c\right),\left(b\right)}{c}_{\left(1\right)}·b_{\left(1\right)}\otimes c_{\left(2\right)}·b_{\left(2\right)}$$

and

$${ϵ}_C(c·b)={ϵ}_C\left(c\right){ϵ}_B\left(b\right)$$

Then C is a right B-module coalgebra. A K-linear map $\varphi :C\to C^{\prime }$ is a right B-module coalgebra homomorphism if ϕ is both a coalgebra and a right B-module homomorphism.

Let C be a coalgebra and let $C^{*}={\text{Hom}}_K(C,K)$ denote the linear dual of C. Then the coalgebra structure of C induces an algebra structure on $C^{*}$.

Proposition 2.1 If C is a coalgebra, then $C^{*}$ is an algebra.

Proof. Recall that C is a triple $(C,{\Delta }_C,{ϵ}_C)$ where ${\Delta }_C:C\to C\otimes C$ is K-linear and satisfies the coassociativity property, and ${ϵ}_C:C\to K$ is K-linear and satisfies the counit property. The dual map of ${\Delta }_C$ is a K-linear map

$${\Delta }_C^{*}:{(C\otimes C)}^{*}\to C^{*}$$

Since $C^{*}\otimes C^{*}\subseteq {(C\otimes C)}^{*}$, we define the multiplication map of $C^{*}$, denoted as $m_{C^{*}}$, to be the restriction of ${\Delta }_C^{*}$ to $C^{*}\otimes C^{*}$. For $f,g\in C^{*}$, $c\in C$,

$$\left(fg\right)\left(c\right)=m_{C^{*}}(f\otimes g)\left(c\right)={\Delta }_C^{*}(f\otimes g)\left(c\right)=(f\otimes g)\left({\Delta }_C\left(c\right)\right)=\sum _{\left(c\right)}f\left(c_{\left(1\right)}\right)g\left(c_{\left(2\right)}\right)$$

The coassociatively property of ${\Delta }_C$ yields the associative property of $m_{C^{*}}$. Indeed, for $f,g,h\in C^{*}$, $c\in C$,

$$\begin{array}{ccc}\hfill m_{C^{*}}(I_{C^{*}}\otimes m_{C^{*}})(f\otimes g\otimes h)\left(c\right)& =& {\Delta }_C^{*}(I_{C^{*}}\otimes {\Delta }_C^{*})(f\otimes g\otimes h)\left(c\right)\hfill \\ & =& {\Delta }_C^{*}(f\otimes {\Delta }_C^{*}(g\otimes h))\left(c\right)\hfill \\ & =& (f\otimes {\Delta }_C^{*}(g\otimes h)){\Delta }_C\left(c\right)\hfill \\ & =& \sum _{\left(c\right)}f\left(c_{\left(1\right)}\right){\Delta }_C^{*}(g\otimes h)\left(c_{\left(2\right)}\right)\hfill \\ & =& \sum _{\left(c\right)}f\left(c_{\left(1\right)}\right)(g\otimes h){\Delta }_C\left(c_{\left(2\right)}\right)\hfill \\ & =& (f\otimes g\otimes h)(\sum _{\left(c\right)}{c}_{\left(1\right)}\otimes {\Delta }_C\left(c_{\left(2\right)}\right))\hfill \\ & =& (f\otimes g\otimes h)(\sum _{\left(c\right)}{\Delta }_C\left(c_{\left(1\right)}\right)\otimes c_{\left(2\right)})\text{by}\text{Condition}\text{(3)}\hfill \\ & =& \sum _{\left(c\right)}(f\otimes g){\Delta }_C\left(c_{\left(1\right)}\right)\otimes h\left(c_{\left(2\right)}\right)\hfill \\ & =& \sum _{\left(c\right)}{\Delta }_C^{*}(f\otimes g)\left(c_{\left(1\right)}\right)\otimes h\left(c_{\left(2\right)}\right)\hfill \\ & =& ({\Delta }_C^{*}(f\otimes g)\otimes h){\Delta }_C\left(c\right)\hfill \\ & =& {\Delta }_C^{*}({\Delta }_C^{*}(f\otimes g)\otimes h)\left(c\right)\hfill \\ & =& {\Delta }_C^{*}({\Delta }_C^{*}\otimes I_{C^{*}})(f\otimes g\otimes h)\left(c\right)\hfill \\ & =& m_{C^{*}}(m_{C^{*}}\otimes I_{C^{*}})(f\otimes g\otimes h)\left(c\right)\hfill \end{array}$$

In addition, the counit map of C dualizes to yield

$${ϵ}_C^{*}:K:=K^{*}\to C^{*}$$

defined as ${ϵ}_C^{*}\left(k\right)\left(c\right)=k\left(ϵ\left(c\right)\right)=kϵ\left(c\right)$. Thus we define the unit map ${\eta }_{C^{*}}$ to be ${ϵ}_C^{*}$. One can show that the counit property of ${ϵ}_C$ implies the unit property for ${\eta }_{C^{*}}$. To this end, for $f\in C^{*}$, $r\in K$, $c\in C$,

$$\begin{array}{ccc}\hfill m_{C^{*}}(I_{C^{*}}\otimes {\eta }_{C^{*}})(f\otimes r)\left(c\right)& =& {\Delta }_C^{*}(I_{C^{*}}\otimes {ϵ}_C^{*})(f\otimes r)\left(c\right)\hfill \\ & =& {\Delta }_C^{*}(f\otimes {ϵ}_C^{*}\left(r\right))\left(c\right)\hfill \\ & =& (f\otimes {ϵ}_C^{*}\left(r\right))\left({\Delta }_C\left(c\right)\right)\hfill \\ & =& \sum _{\left(c\right)}f\left(c_{\left(1\right)}\right){ϵ}_C^{*}\left(r\right)\left(c_{\left(2\right)}\right)\hfill \\ & =& \sum _{\left(c\right)}f\left(c_{\left(1\right)}\right)r\left({ϵ}_C\left(c_{\left(2\right)}\right)\right)\hfill \\ & =& r\sum _{\left(c\right)}f\left(c_{\left(1\right)}\right){ϵ}_C\left(c_{\left(2\right)}\right)\hfill \\ & =& r\sum _{\left(c\right)}{ϵ}_C\left(c_{\left(2\right)}\right)f\left(c_{\left(1\right)}\right)\hfill \\ & =& r\sum _{\left(c\right)}f\left({ϵ}_C\left(c_{\left(2\right)}\right)c_{\left(1\right)}\right)\hfill \\ & =& rf\left(\sum _{\left(c\right)}{ϵ}_C\left(c_{\left(2\right)}\right)c_{\left(1\right)}\right)\hfill \\ & =& rf\left(c\right)\text{by}\text{Condition}\text{(5)}\hfill \end{array}$$

In a similar manner, one obtains

$$m_{C^{*}}({\eta }_{C^{*}}\otimes I_{C^{*}})(r\otimes f)=rf$$

Thus $(C^{*},m_{C^{*}},{\eta }_{C^{*}})$ is an algebra. Note that ${\eta }_{C^{*}}\left(1_K\right)\left(c\right)={ϵ}_C\left(c\right),\forall c$, and so, ${ϵ}_C$ is the unique element of $C^{*}$ for which ${ϵ}_{C}f=f=f{ϵ}_C$ for all $f\in C^{*}$. ⋄

Let $(A,m_A,{\eta }_A)$ be a K-algebra. Then one may wonder if $A^{*}$ is a K-coalgebra. The multiplication map $m_A:A\otimes A\to A$ dualizes to yield $m_A^{*}:A^{*}\to {(A\otimes A)}^{*}$. Unfortunately, if A is infinite dimensional over K, then $A^{*}\otimes A^{*}$ is a proper subset of ${(A\otimes A)}^{*}$, and hence $m_A^{*}$ may not induce the required comultiplication map $A^{*}\to A^{*}\otimes A^{*}$.

There is still however a K-coalgebra arising via duality from the algebra A. An ideal I of A is cofinite if $dim(A/I)<\infty$. The finite dual $A^{\circ }$ of A is defined as

$$A^{\circ }=\{f\in A^{*}:f\left(I\right)=0\text{for}\text{some}\text{cofinite}\text{ideal}I\text{of}A\}$$

Note that $A^{\circ }$ is the largest subspace W of $A^{*}$ for which $m_A^{*}\left(W\right)\subseteq W\otimes W$.

Proposition 2.2 If A is an algebra, then $A^{\circ }$ is a coalgebra.

Proof. The proof is similar to the method used in Proposition 2.1. We restrict the map $m_A^{*}$ to $A^{\circ }$ to yield the K-linear map $m_A^{*}:A^{\circ }\to {(A\otimes A)}^{*}$. Now by [4, Proposition 6.0.3], $m_A^{*}\left(A^{\circ }\right)\subseteq A^{\circ }\otimes A^{\circ }$. Let ${\Delta }_{A^{\circ }}$ denote the restriction of $m_A^{*}$ to $A^{\circ }$. We show that ${\Delta }_{A^{\circ }}$ satisfies the coassociative condition. For $f\in A^{\circ }$, $a,b,c\in A$, we have

$$\begin{array}{ccc}\hfill (I\otimes {\Delta }_{A^{\circ }}){\Delta }_{A^{\circ }}\left(f\right)(a\otimes b\otimes c)& =& (I\otimes m_A^{*})m_A^{*}\left(f\right)(a\otimes b\otimes c)\hfill \\ & =& m_A^{*}\left(f\right)\left((I\otimes m_A)(a\otimes b\otimes c)\right)\hfill \\ & =& m_A^{*}\left(f\right)(a\otimes bc)\hfill \\ & =& f\left(m_A(a\otimes bc)\right)\hfill \\ & =& f\left(a\right(bc\left)\right)\hfill \\ & =& f\left(\right(ab\left)c\right)\hfill \\ & =& f\left(m_A(ab\otimes c)\right)\hfill \\ & =& m_A^{*}\left(f\right)(ab\otimes c)\hfill \\ & =& m_A^{*}\left(f\right)\left((m_A\otimes I)(a\otimes b\otimes c)\right)\hfill \\ & =& (m_A^{*}\otimes I)m_A^{*}\left(f\right)(a\otimes b\otimes c)\hfill \\ & =& ({\Delta }_{A^{\circ }}\otimes I){\Delta }_{A^{\circ }}\left(f\right)(a\otimes b\otimes c)\hfill \end{array}$$

For the counit map of $A^{\circ }$, we consider the dual map ${\eta }_A^{*}:A^{*}\to K^{*}:=K$. Now ${\eta }_A^{*}$ restricts to a map ${\eta }_A^{*}:A^{\circ }\to K$. We let ${ϵ}_{A^{\circ }}$ denote the restriction of ${\eta }_A^{*}$ to $A^{\circ }$. For $f\in A^{\circ }$, $r\in K$,

$${ϵ}_{A^{\circ }}\left(f\right)\left(r\right)=f\left({\eta }_A\left(r\right)\right)=f\left(r{1}_A\right)=rf\left(1_A\right)=f\left(1_A\right)\left(r\right)$$

and so, ${ϵ}_{A^{\circ }}\left(f\right)=f\left(1_A\right)$. We show that ${ϵ}_{A^{\circ }}$ satisfies the counit property. First let $s_1:K\otimes A^{\circ }\to A^{\circ }$ be defined by the scalar multiplication of $A^{\circ }$. For $f\in A^{\circ }$, $r\in K$, $a\in A$,

$$\begin{array}{ccc}\hfill s_1\left(({ϵ}_{A^{\circ }}\otimes I){\Delta }_{A^{\circ }}\left(f\right)\right)\left(a\right)& =& s_1\left(({\eta }_A^{*}\otimes I)m_A^{*}\left(f\right)\right)\left(a\right)\hfill \\ & =& ({\eta }_A^{*}\otimes I)m_A^{*}\left(f\right)\left(s_1^{*}\left(a\right)\right)\hfill \\ & =& ({\eta }_A^{*}\otimes I)m_A^{*}\left(f\right)(1\otimes a)\hfill \\ & =& m_A^{*}\left(f\right)\left(({\eta }_A\otimes I)(1\otimes a)\right)\hfill \\ & =& f\left(m_A({\eta }_A\otimes I)(1\otimes a)\right)\hfill \\ & =& f\left(a\right)\hfill \end{array}$$

In a similar manner, one obtains

$$s_2\left((I\otimes {ϵ}_{A^{\circ }}){\Delta }_{A^{\circ }}\left(f\right)\right)\left(a\right)=f\left(a\right)$$

where $s_2:A^{\circ }\otimes K\to A^{\circ }$ is given by scalar multiplication. Thus $A^{\circ }$ is a coalgebra. ⋄

Proposition 2.3 If B is a bialgebra, then $B^{\circ }$ is a bialgebra.

Proof. As a coalgebra, B is a triple $(B,{\Delta }_B,{ϵ}_B)$. By Proposition 2.1, $B^{*}$ is an algebra with maps $m_{B^{*}}={\Delta }_B^{*}$ and ${\eta }_{B^{*}}={ϵ}_B^{*}$. Let $m_{B^{\circ }}$ denote the restriction of $m_{B^{*}}$ to $B^{\circ }\otimes B^{\circ }$, and let ${\eta }_{B^{\circ }}$ denote the restriction of ${\eta }_{B^{*}}$ to $B^{\circ }$. Then the triple $(B^{\circ },m_{B^{\circ }},{\eta }_{B^{\circ }})$ is a K-algebra.

As an algebra, B is a triple $(B,m_B,{\eta }_B)$. By Proposition 2.2, $B^{\circ }$ is a coalgebra with maps ${\Delta }_{B^{\circ }}$ and ${ϵ}_{B^{\circ }}$. It remains to show that ${\Delta }_{B^{\circ }}$ and ${ϵ}_{B^{\circ }}$ are algebra homomorphisms. First observe that for $f,g\in B^{\circ }$, $a,b\in B$ one has

$$\left(fg\right)\left(a\right)=m_{B^{\circ }}(f\otimes g)\left(a\right)={\Delta }_B^{*}(f\otimes g)\left(a\right)=(f\otimes g){\Delta }_B\left(a\right)$$

and

$${\Delta }_{B^{\circ }}\left(f\right)(a\otimes b)=m_B^{*}\left(f\right)(a\otimes b)=f\left(m_B(a\otimes b)\right)=f\left(ab\right)$$

We have

$$\begin{array}{ccc}\hfill {\Delta }_{B^{\circ }}\left(fg\right)(a\otimes b)& =& \left(fg\right)\left(ab\right)\hfill \\ & =& (f\otimes g)\left({\Delta }_B\left(ab\right)\right)\hfill \\ & =& (f\otimes g)\left({\Delta }_B\left(a\right){\Delta }_B\left(b\right)\right)\hfill \\ & =& (f\otimes g)(m_{B\otimes B}({\Delta }_B\left(a\right)\otimes {\Delta }_B\left(b\right))\hfill \\ & =& m_{B\otimes B}^{*}(f\otimes g)({\Delta }_B\left(a\right)\otimes {\Delta }_B\left(b\right))\hfill \\ & =& (I\otimes \tau \otimes I)({\Delta }_{B^{\circ }}\otimes {\Delta }_{B^{\circ }})(f\otimes g)({\Delta }_B\left(a\right)\otimes {\Delta }_B\left(b\right))\hfill \\ & =& ({\Delta }_{B^{\circ }}\left(f\right)\otimes {\Delta }_{B^{\circ }}\left(g\right))(I\otimes \tau \otimes I)({\Delta }_B\otimes {\Delta }_B)(a\otimes b)\hfill \\ & =& ({\Delta }_{B^{\circ }}\left(f\right)\otimes {\Delta }_{B^{\circ }}\left(g\right))\left({\Delta }_{B\otimes B}(a\otimes b)\right)\hfill \\ & =& {\Delta }_{B\otimes B}^{*}({\Delta }_{B^{\circ }}\left(f\right)\otimes {\Delta }_{B^{\circ }}\left(g\right))(a\otimes b)\hfill \\ & =& m_{B^{\circ }\otimes B^{\circ }}({\Delta }_{B^{\circ }}\left(f\right)\otimes {\Delta }_{B^{\circ }}\left(g\right))(a\otimes b)\hfill \\ & =& \left({\Delta }_{B^{\circ }}\left(f\right){\Delta }_{B^{\circ }}\left(g\right)\right)(a\otimes b)\hfill \end{array}$$

and so ${\Delta }_{B^{\circ }}$ is an algebra map. We next show that ${ϵ}_{B^{\circ }}$ is an algebra map. For $f,g\in B^{\circ }$,

$${ϵ}_{B^{\circ }}\left(f\right)={ϵ}_{B^{\circ }}\left(f\right)\left(1\right)=f\left({\eta }_B\left(1\right)\right)=f\left(1_B\right)$$

Thus

$$\begin{array}{ccc}\hfill {ϵ}_{B^{\circ }}\left(fg\right)& =& \left(fg\right)\left(1_B\right)\hfill \\ & =& f\left(1_B\right)g\left(1_B\right)\hfill \\ & =& {ϵ}_{B^{\circ }}\left(f\right){ϵ}_{B^{\circ }}\left(g\right)\hfill \end{array}$$

and so, ${ϵ}_{B^{\circ }}$ is an algebra map. ⋄

Proposition 2.4 Suppose that B is a bialgebra that is finite dimensional over K. Then $B^{*}$ is a bialgebra.

Proof. If $dim\left(B\right)<\infty$, then $B^{\circ }=B^{*}$. The result then follows from Proposition 2.3. ⋄

Let $G=\{x_1,x_2,\cdots ,x_n\}$ be a finite semigroup with unity element $1_{KG}=x_1$, and let $KG$ denote the semigroup bialgebra. By Proposition 2.4 $K{G}^{*}$ is a bialgebra of dimension n over K. Let $\{e_1,e_2,\cdots ,e_n\}$ be the dual basis for $K{G}^{*}$ defined as $e_i\left(x_j\right)={\delta }_{i,j}$.

Proposition 2.5 The comultiplication map ${\Delta }_{K{G}^{*}}:K{G}^{*}\to K{G}^{*}\otimes K{G}^{*}$ is given as

$${\Delta }_{K{G}^{*}}\left(e_i\right)=\sum _{x_i=x_j{x}_k}{e}_j\otimes e_k$$

and the counit map ${ϵ}_{K{G}^{*}}:K{G}^{*}\to K$ is defined as ${ϵ}_{K{G}^{*}}\left(e_i\right)=e_i\left(x_1\right)={\delta }_{i,1}$.

Proof, See [7, (1.3.7)]. ⋄

Let B be a K-bialgebra. Then B is cocommutative if

$$\tau \left({\Delta }_B\left(b\right)\right)={\Delta }_B\left(b\right)$$

for all $b\in B$.

Proposition 2.6 If B is cocommutative, then $B^{\circ }$ is a commutative algebra. If B is a commutative algebra, then $B^{\circ }$ is cocommutative.

Proof. See [7, Lemma 1.2.2, Proposition 1.2.4]. ⋄

3. Quasitriangular Bialgebras

Let B be a bialgebra and let $B\otimes B$ be the tensor product algebra. Let $U(B\otimes B)$ denote the group of units in $B\otimes B$ and let $R\in U(B\otimes B)$. The pair $(B,R)$ is almost cocommutative if the element R satisfies

$$\tau \left({\Delta }_B\left(b\right)\right)=R{\Delta }_B\left(b\right)R^{-1}$$

(7)

for all $b\in B$.

If the bialgebra B is cocommutative, then the pair $(B,1\otimes 1)$ is almost cocommutative since $R=1\otimes 1$ satisfies Condition (7). However, if B is commutative and non-cocommutative, then $(B,R)$ cannot be almost cocommutative for any $R\in U(B\otimes B)$ since Condition (7) in this case reduces to the condition for cocommutativity.

Write $R={\sum }_{i=1}^n{a}_i\otimes b_i\in U(B\otimes B)$. Let

$$R^{12}=\sum _{i=1}^n{a}_i\otimes b_i\otimes 1\in B\otimes B\otimes B$$

$$R^{13}=\sum _{i=1}^n{a}_i\otimes 1\otimes b_i\in B\otimes B\otimes B$$

$$R^{23}=\sum _{i=1}^{n}1\otimes a_i\otimes b_i\in B\otimes B\otimes B$$

The pair $(B,R)$ is quasitriangular if $(B,R)$ is almost cocommutative and the following conditions hold

$$({\Delta }_B\otimes I)R=R^{13}{R}^{23}$$

(8)

$$(I\otimes {\Delta }_B)R=R^{13}{R}^{12}$$

(9)

Clearly, if B is cocommutative then $(B,1\otimes 1)$ is quasitriangular.

Let B be a bialgebra. A quasitriangular structure is an element $R\in U(B\otimes B)$ so that $(B,R)$ is quasitriangular. Let $(B,R)$ and $(B^{\prime },R^{\prime })$ be quasitriangular bialgebras. Then $(B,R)$, $(B^{\prime },R^{\prime })$ are isomorphic as quasitriangular bialgebras if there exists a bialgebra isomorphism $\varphi :B\to B^{\prime }$ for which $R^{\prime }=(\varphi \otimes \varphi )\left(R\right)$. Two quasitriangular structures $R,R^{\prime }$ on a bialgebra B are equivalent quasitriangular structures if $(B,R)\cong (B,R^{\prime })$ as quasitriangular bialgebras.

The following proposition shows that every bialgebra isomorphism $\varphi :B\to B^{\prime }$ with B quasitriangular extends to an isomorphism of quasitriangular bialgebras.

Proposition 3.1 Suppose $(B,R)$ is quasitriangular and suppose that $\varphi :B\to B^{\prime }$ is an isomorphism of K-bialgebras. Let $R^{\prime }=(\varphi \otimes \varphi )\left(R\right)$. Then $(B^{\prime },R^{\prime })$ is quasitriangular.

Proof. Note that $(\varphi \otimes \varphi )\left(R^{-1}\right)={\left((\varphi \otimes \varphi )\left(R\right)\right)}^{-1}$. Let $b^{\prime }\in B^{\prime }$. Then there exists $b\in B$ for which $\varphi \left(b\right)=b^{\prime }$. Now

$$\begin{array}{ccc}\hfill \tau {\Delta }_{B^{\prime }}\left(b^{\prime }\right)& =& \tau {\Delta }_{B^{\prime }}\left(\varphi \left(b\right)\right)\hfill \\ & =& \tau (\varphi \otimes \varphi ){\Delta }_B\left(b\right)\hfill \\ & =& (\varphi \otimes \varphi )\tau {\Delta }_B\left(b\right)\hfill \\ & =& (\varphi \otimes \varphi )\left(R{\Delta }_B\left(b\right)R^{-1}\right)\hfill \\ & =& (\varphi \otimes \varphi )\left(R\right)(\varphi \otimes \varphi ){\Delta }_B\left(b\right)(\varphi \otimes \varphi )\left(R^{-1}\right)\hfill \\ & =& (\varphi \otimes \varphi )\left(R\right){\Delta }_{B^{\prime }}\left(\varphi \left(b\right)\right){\left((\varphi \otimes \varphi )\left(R\right)\right)}^{-1}\hfill \\ & =& (\varphi \otimes \varphi )\left(R\right){\Delta }_{B^{\prime }}\left(b^{\prime }\right){\left((\varphi \otimes \varphi )\left(R\right)\right)}^{-1}\hfill \\ & =& R^{\prime }{\Delta }_{B^{\prime }}\left(b^{\prime }\right){\left(R^{\prime }\right)}^{-1}\hfill \end{array}$$

and so, $(B,R^{\prime })$ is almost cocommutative. Moreover,

$$\begin{array}{ccc}\hfill ({\Delta }_{B^{\prime }}\otimes I)\left(R^{\prime }\right)& =& ({\Delta }_{B^{\prime }}\otimes I)(\varphi \otimes \varphi )\left(R\right)\hfill \\ & =& ({\Delta }_{B^{\prime }}\otimes I)(\sum _{i=1}^n\varphi \left(a_i\right)\otimes \varphi \left(b_i\right))\hfill \\ & =& \sum _{i=1}^n{\Delta }_{B^{\prime }}\varphi \left(a_i\right)\otimes \varphi \left(b_i\right))\hfill \\ & =& \sum _{i=1}^n(\varphi \otimes \varphi ){\Delta }_B\left(a_i\right)\otimes \varphi \left(b_i\right))\hfill \\ & =& (\varphi \otimes \varphi \otimes \varphi )(\sum _{i=1}^n{\Delta }_B\left(a_i\right)\otimes b_i)\hfill \\ & =& (\varphi \otimes \varphi \otimes \varphi )({\Delta }_B\otimes I)\left(R\right)\hfill \\ & =& (\varphi \otimes \varphi \otimes \varphi )\left(R^{13}{R}^{23}\right)\hfill \\ & =& (\varphi \otimes \varphi \otimes \varphi )\left((\sum _{i=1}^n{a}_i\otimes 1\otimes b_i)(\sum _{i=1}^{n}1\otimes a_i\otimes b_i)\right)\hfill \\ & =& (\sum _{i=1}^n\varphi \left(a_i\right)\otimes 1\otimes \varphi \left(b_i\right))(\sum _{i=1}^{n}1\otimes \varphi \left(a_i\right)\otimes \varphi \left(b_i\right))\hfill \\ & =& {\left((\varphi \otimes \varphi )\left(R\right)\right)}^{13}{\left((\varphi \otimes \varphi )\left(R\right)\right)}^{23}\hfill \\ & =& {\left(R^{\prime }\right)}^{13}{\left(R^{\prime }\right)}^{23}\hfill \end{array}$$

In a similar manner one shows that

$$(I\otimes {\Delta }_{B^{\prime }})\left(R^{\prime }\right)={\left(R^{\prime }\right)}^{13}{\left(R^{\prime }\right)}^{12}$$

Thus $(B^{\prime },R^{\prime })$ is quasitriangular. ⋄

Quasitriangular bialgebras are important since they give rise to solutions of the equation

$$R^{12}{R}^{13}{R}^{23}=R^{23}{R}^{13}{R}^{12}$$

(10)

which is known as the quantum Yang–Baxter equation (QYBE). The QYBE was first introduced in statistical mechanics, see [9]. An element $R\in B\otimes B$ which satisfies (10) is a solution to the QYBE.

Certainly, the QYBE admits the trivial solution $R=1\otimes 1$, and of course, if B is commutative, then any $R\in B\otimes B$ is a solution to the QYBE. For B non-commutative, it is of great interest to find non-trivial solutions $R\in B\otimes B$ to the QYBE. We have the following result due to V. G. Drinfeld [10].

Proposition 3.2 (Drinfeld) Suppose $(B,R)$ is quasitriangular. Then R is a solution to the QYBE.

Proof. One has

$$\begin{array}{ccc}\hfill R^{12}{R}^{13}{R}^{23}& =& R^{12}(\Delta \otimes I)\left(R\right)\text{by}\text{(8)}\hfill \\ & =& (R\otimes 1)(\sum _{i=1}^n\Delta \left(a_i\right)\otimes b_i)\hfill \\ & =& \sum _{i=1}^{n}R\Delta \left(a_i\right)\otimes b_i\hfill \\ & =& \sum _{i=1}^n\tau \Delta \left(a_i\right)R\otimes b_i\text{by}\text{(7)}\hfill \\ & =& (\sum _{i=1}^n\tau \Delta \left(a_i\right)\otimes b_i)(R\otimes 1)\hfill \\ & =& (\tau \Delta \otimes I)\left(R\right)R^{12}\hfill \\ & =& (\tau \otimes I)(\Delta \otimes I)\left(R\right)R^{12}\hfill \\ & =& (\tau \otimes I)\left(R^{13}{R}^{23}\right)R^{12}\text{by}\text{(8)}\hfill \\ & =& R^{23}{R}^{13}{R}^{12}\hfill \end{array}$$

⋄

The following proposition provides necessary conditions on $R\in U(B\otimes B)$ in order for $(B,R)$ to be quasitriangular.

Proposition 3.3 Suppose $(B,R)$ is quasitriangular. Then

(i) $s_1(ϵ\otimes I)\left(R\right)=1$,

(ii) $s_2(I\otimes ϵ)\left(R\right)=1$.

Proof. For (i) one has

$$\begin{array}{ccc}\hfill (s_1\otimes I)(ϵ\otimes I\otimes I)(\Delta \otimes I)\left(R\right)& =& (s_1\otimes I)(ϵ\otimes I\otimes I)(\sum _{i=1}^n\Delta \left(a_i\right)\otimes b_i)\hfill \\ & =& (s_1\otimes I)(\sum _{i=1}^n(ϵ\otimes I)\Delta \left(a_i\right)\otimes b_i)\hfill \\ & =& \sum _{i=1}^n{s}_1(ϵ\otimes I)\Delta \left(a_i\right)\otimes b_i\hfill \\ & =& \sum _{i=1}{a}_i\otimes b_i\hfill \\ & =& R\hfill \end{array}$$

In view of Condition (8)

$$\begin{array}{ccc}\hfill R& =& (s_1\otimes I)(ϵ\otimes I\otimes I)\left(R^{13}{R}^{23}\right)\hfill \\ & =& (s_1\otimes I)(ϵ\otimes I\otimes I)\left(R^{13}\right)(s_1\otimes I)(ϵ\otimes I\otimes I)\left(R^{23}\right)\hfill \\ & =& (s_1\otimes I)(ϵ\otimes I\otimes I)(\sum _{i=1}^n{a}_i\otimes 1\otimes b_i)(s_1\otimes I)(ϵ\otimes I\otimes I)(\sum _{i=1}^{n}1\otimes a_i\otimes b_i)\hfill \\ & =& (\sum _{i=1}^nϵ\left(a_i\right)1\otimes b_i)(\sum _{i=1}^n{a}_i\otimes b_i)\hfill \\ & =& (\sum _{i=1}^{n}1\otimes ϵ\left(a_i\right)b_i)R\hfill \end{array}$$

Thus

$$1\otimes \sum _{i=1}ϵ\left(a_i\right)b_i=1\otimes 1$$

and consequently,

$$1=s_1(\sum _{i=1}^nϵ\left(a_i\right)\otimes b_i)=s_1(ϵ\otimes I)\left(R\right)$$

A similar argument is used to prove (ii). ⋄

4. Myhill–Nerode Bialgebras

In this section we review the main result of [3] in which the authors give a bialgebra version of the Myhill–Nerode Therorem. Let G be a semigroup with unity, 1 and let $H=KG$ be the semigroup bialgebra. There is a right H-module structure on $H^{*}$ defined as

$$(p\leftharpoonup x)\left(y\right)=p\left(xy\right)$$

for all $x,y\in H$, $p\in H^{*}$. For $x\in H$, $p\in H^{*}$, the element $p\leftharpoonup x$ is the right translate of p by x.

Proposition 4.1 ([3, Proposition 5.4].) Let G be a semigroup with 1, let $H=KG$ denote the semigroup bialgebra. Let $p\in H^{*}$. Then the following are equivalent.

(i) The set $\{p\leftharpoonup x:x\in G\}$ of right translates is finite.

(ii) There exists a finite dimensional bialgebra B, a bialgebra homomorphism $\Psi :H\to B$, and an element $f\in B^{*}$ so that $p\left(h\right)=f(\Psi (h\left)\right)$ for all $h\in H$.

(Note: The bialgebras of (ii) are defined to be Myhill–Nerode bialgebras.)

Proof. $\left(i\right)⟹\left(ii\right)$. Let $Q=\{p\leftharpoonup x:x\in G\}$ be the finite set of right translates. For each $u\in G$, we define a right operator $r_u:Q\to Q$ by the rule

$$(p\leftharpoonup x)r_u=(p\leftharpoonup x)\leftharpoonup u=p\leftharpoonup xu$$

Observe that the set $\{r_u:u\in G\}$ is finite with $|\{r_u:u\in G\}{|\le |Q|}^{\left|Q\right|}$. The set $\{r_u:u\in G\}$ is a semigroup with unity, $1=r_1$ under composition of operators. Indeed,

$$(p\leftharpoonup x)\left(r_u{r}_v\right)=(p\leftharpoonup xu)r_v=p\leftharpoonup xuv=(p\leftharpoonup x)r_{uv}$$

Thus $r_u{r}_v=r_{uv}$, for all $u,v\in G$. Let B denote the semigroup bialgebra on $\{r_u:u\in G\}$. Let $\Psi :H\to B$ be the K-linear map defined by $\Psi \left(u\right)=r_u$. Then

$$\Psi \left(uv\right)=r_{uv}=r_u{r}_v=\Psi \left(u\right)\Psi \left(v\right)$$

and

$$\begin{array}{ccc}\hfill {\Delta }_B(\Psi \left(u\right))& =& {\Delta }_B\left(r_u\right)\hfill \\ & =& r_u\otimes r_u\hfill \\ & =& \Psi \left(u\right)\otimes \Psi \left(v\right)\hfill \\ & =& (\Psi \otimes \Psi )(u\otimes u)\hfill \\ & =& (\Psi \otimes \Psi ){\Delta }_H\left(u\right)\hfill \end{array}$$

and so, Ψ is a homomorphism of bialgebras.

Let $f\in B^{*}$ be defined by

$$\begin{array}{ccc}\hfill f\left(r_u\right)& =& \left((p\leftharpoonup 1)r_u\right)\left(1\right)\hfill \\ & =& (p\leftharpoonup u)\left(1\right)\hfill \\ & =& p\left(u\right)\hfill \end{array}$$

Then $p\left(h\right)=f(\Psi (h\left)\right)$, for all $h\in H$, as required.

$\left(ii\right)⟹\left(i\right)$. Suppose there exists a finite dimensional bialgebra B, a bialgebra homomorphism $\Psi :H\to B$, and an element $f\in B^{*}$ so that $p\left(h\right)=f(\Psi (h\left)\right)$ for all $h\in H$. Define a right H-module action · on B as

$$b·h=b\Psi \left(h\right)$$

for all $b\in B$, $h\in H$. Then for $b\in B$, $x\in G$,

$$\begin{array}{ccc}\hfill {\Delta }_B(b·x)& =& {\Delta }_B(b\Psi \left(x\right))\hfill \\ & =& {\Delta }_B\left(b\right){\Delta }_B(\Psi \left(x\right))\hfill \\ & =& (\sum _{\left(b\right)}{b}_{\left(1\right)}\otimes b_{\left(2\right)})(\Psi \otimes \Psi ){\Delta }_H\left(x\right)\hfill \\ & =& (\sum _{\left(b\right)}{b}_{\left(1\right)}\otimes b_{\left(2\right)})(\Psi \left(x\right)\otimes \Psi \left(x\right))\hfill \\ & =& \sum _{\left(b\right)}{b}_{\left(1\right)}\Psi \left(x\right)\otimes b_{\left(2\right)}\Psi \left(x\right)\hfill \\ & =& \sum _{\left(b\right)}{b}_{\left(1\right)}·x\otimes b_{\left(2\right)}·x\hfill \end{array}$$

and

$${ϵ}_B(b·x)={ϵ}_B(b\Psi \left(x\right))={ϵ}_B\left(b\right){ϵ}_B(\Psi \left(x\right))={ϵ}_B\left(b\right){ϵ}_H\left(x\right)$$

Thus B is a right H-module coalgebra.

Now, let Q be the collection of grouplike elements of B. Since Q is a linearly independent subset of B and B is finite dimensional, Q is finite. Since B is a right H-module coalgebra with action “·”,

$${\Delta }_B(q·x)=q·x\otimes q·x$$

for $q\in Q$, $x\in G$. Thus · restricts to give an action (also denoted by “·”) of G on Q. Now for $x,y\in G$,

$$\begin{array}{ccc}\hfill (p\leftharpoonup x)\left(y\right)& =& p\left(xy\right)\hfill \\ & =& f(\Psi (xy\left)\right)\hfill \\ & =& f(\Psi (x)\Psi (y\left)\right)\hfill \\ & =& f((1_B\Psi \left(x\right))\Psi \left(y\right))\hfill \\ & =& f((1_B·x)·y)\hfill \end{array}$$

(11)

Let

$$S=\{q\in Q:q=1_B·x\text{for}\text{some}x\in G\}$$

In view of Condition (11) there exists a function

$$\varrho :S\to \{p\leftharpoonup x:x\in G\}$$

defined as

$$\varrho (1_B·x)\left(y\right)=f((1_B·x)·y)=(p\leftharpoonup x)\left(y\right)$$

Since ϱ is surjective and S is finite, $\{p\leftharpoonup x:x\in G\}$ is finite. ⋄

We illustrate the connection between Proposition 4.1 and the usual Myhill–Nerode Theorem. Let ${\widehat{\Sigma }}_0$ denote the set of words in a finite alphabet ${\Sigma }_0$. Let $L\subseteq {\widehat{\Sigma }}_0$ be a language. Suppose that the equivalence relation ${\sim }_L$ (as in the Introduction) has finite index. Then the usual Myhill–Nerode Theorem says that there exists a finite automaton which accepts L. We show how to construct this finite automaton using Proposition 4.1.

Consider $G={\widehat{\Sigma }}_0$ as a semigroup with unity where the semigroup operation is concatenation and the unity element is the empty word. Let $H=KG$ denote the semigroup bialgebra. Then the characteristic function of L extends to an element $p\in H^{*}$. Since ${\sim }_L$ has finite index, the set of right translates $\{p\leftharpoonup x:x\in G\}$ is finite [3, Proposition 2.3]. Now Proposition 4.1 (i)⟹ (ii) applies to show that there exists a finite dimensional bialgebra B, a bialgebra homomorphism $\Psi :H\to B$ and an element $f\in B^{*}$ so that $p\left(h\right)=f(\Psi (h\left)\right)$, for all $h\in H$.

This bialgebra determines a finite automaton $⟨Q,\Sigma ,\delta ,q_0,F⟩$, where Q is the finite set of states, Σ is the input alphabet, δ is the transition function, $q_0$ is the initial state, and F is the set of final states (see [2, Chapter 2] for details on finite automata.)

For the states of the automata, we let Q be the (finite) set of grouplike elements of B. For the input alphabet, we choose $\Sigma ={\Sigma }_0$. As we have seen, the right H-module structure of B restricts to an action “·” of G on Q, and so we define the transition function $\delta :Q\times {\Sigma }_0\to Q$ by the rule $\delta (q,x)=q·x$, for $q\in Q$, $x\in {\Sigma }_0$. The initial state is $q_0=1_B$, and the set of final states F is the subset of Q of the form $1_B·x$, $x\in G$ for which

$$p\left(x\right)=f(\Psi \left(x\right))=f{1}_B\Psi \left(x\right))=f(1_B·x)=1$$

By construction, the finite automaton $⟨Q,{\Sigma }_0,\delta ,1_B,F⟩$ accepts L.

5. Quasitriangular Structure of Myhill–Nerode Bialgebras

In this section we use Proposition 4.1 to construct a collection of Myhill–Nerode bialgebras. We then compute the quasitriangular structure of one of these bialgebras.

Let ${\Sigma }_0=\left\{a\right\}$ be the alphabet on a single letter a. Let ${\widehat{\Sigma }}_0=\{1,a,aa,aaa,\cdots \}$ denote the collection of all words of finite length formed from ${\Sigma }_0$. Here 1 denotes the empty word of length 0. For convenience, we shall write

$$a^i=\underset{i\text{times}}{\underbrace{aaa\cdots a}},$$

for $i\ge 0$.

Fix an integer $i\ge 0$ and let $L_i=\left\{a^i\right\}\subseteq {\widehat{\Sigma }}_0$. Then the language $L_i$ is accepted by the finite automaton given in Figure 1.

Figure 1. Finite automaton accepting $L_i=\left\{a^i\right\}$, accepting state is i.

Figure 1. Finite automaton accepting $L_i=\left\{a^i\right\}$, accepting state is i.

By the usual Myhill–Nerode Theorem, the equivalence relation ${\sim }_{L_i}$, defined as $x{\sim }_{L_i}y$ if and only if $xz\in L_i$ exactly when $yz\in L_i,\forall z$, has finite index. If $p_i:{\widehat{\Sigma }}_0\to \{0,1\}\subseteq K$ is the characteristic function of $L_i$, then ${\sim }_{L_i}$ is equivalent to the relation ${\sim }_{p_i}$ defined as: $x{\sim }_{p_i}y$ if and only if $p_i\left(xz\right)=p_i\left(yz\right),\forall z\in {\widehat{\Sigma }}_0$. Let ${\left[x\right]}_{p_i}$ denote the equivalence class of x under ${\sim }_{p_i}$. The Myhill–Nerode theorem now says that the set $\{{\left[x\right]}_{p_i}:x\in {\widehat{\Sigma }}_0\}$ is finite.

Now we consider $G={\widehat{\Sigma }}_0$ as a semigroup with unity 1 with concatenation as the binary operation. Let $H=KG$ be the semigroup bialgebra. The characteristic function $p_i$ of $L_i$ extends to an element of $H^{*}$. By [3, Proposition 2.3], the set of right translates $\{p_i\leftharpoonup x:x\in G\}$ is finite. Thus by Proposition 4.1, there exists a finite dimensional bialgebra $B_i$, a bialgebra homomorphism $\Psi :H\to B_i$, and an element $f_i\in B_i^{*}$ so that $p_i\left(h\right)=f_i(\Psi \left(h\right))$ for all $h\in H$.

In what follows, we give the bialgebra structure of the collection $\{B_i:i\ge 0\}$ and compute the quasitriangular structure of the bialgebra $B_0$.

For $i\ge 0$, the finite set of right translates of $p_i\in H^{*}$ is

$$Q_i=\{p_i\leftharpoonup 1,p_i\leftharpoonup a,p_i\leftharpoonup a^2,\cdots ,p_i\leftharpoonup a^i,p_i\leftharpoonup a^{i+1}\}$$

One finds that the set of right operators on $Q_i$ is $\{r_1,r_a,r_{a^2},\cdots ,r_{a^i},r_{a^{i+1}}\}$. Under composition, the set of right operators is a semigroup with unity $r_1$. We have, for $0\le m,n\le i+1$,

$$r_{a^m}{r}_{a^n}=\left\{\begin{array}{cc}{r}_{a^{m+n}}\hfill & if 0\le m+n\le i+1\hfill \\ r_{a^{i+1}}\hfill & if m+n>i+1\hfill \end{array}\right.$$

By construction, $B_i$ is the semigroup bialgebra on $\{r_1,r_a,r_{a^2},\cdots ,r_{a^i},r_{a^{i+1}}\}$.

5.1. Quasitriangular Structure of $B_0$

In the case $i=0$, $B_0$ is the semigroup bialgebra on $\{r_1,r_a\}$ with algebra structure defined by $r_1{r}_1=r_1$, $r_1{r}_a=r_a$, $r_a{r}_1=r_a$, $r_a{r}_a=r_a$. Let $\{e_0,e_1\}$ be the dual basis defined as $e_0\left(r_1\right)=1$, $e_0\left(r_a\right)=0$, $e_1\left(r_1\right)=0$, $e_1\left(r_a\right)=1$. Then $\{e_0,e_1\}$ is the set of minimal idempotents for $B_0^{*}$. Comultiplication on $B_0^{*}$ is given as

$${\Delta }_{B_0^{*}}\left(e_0\right)=e_0\otimes e_0$$

$${\Delta }_{B_0^{*}}\left(e_1\right)=e_0\otimes e_1+e_1\otimes e_0+e_1\otimes e_1$$

and the counit map is defined by

$${ϵ}_{B_0^{*}}\left(e_0\right)=1,{ϵ}_{B_0^{*}}\left(e_1\right)=0$$

Proposition 5.1 Let $B_0$ be the K-bialgebra as above. Then there is exactly one quasitriangular structure on $B_0$, namely, $R=1_{B_0}\otimes 1_{B_0}$.

Proof. Certainly, $1\otimes 1=1_{B_0}\otimes 1_{B_0}$ is a quasitriangular structure for $B_0$. We claim that $1\otimes 1$ is the only quasitriangular structure. Observe that there is bialgebra isomorphism $\varphi :B_0\to B_0^{*}$ defined as $\varphi \left(r_1\right)=e_0+e_1$, $\varphi \left(r_a\right)=e_0$. Thus if $(B_0,R)$ is quasitriangular, then $(B_0^{*},R^{\prime })$, $R^{\prime }=(\varphi \otimes \varphi )\left(R\right)$, is quasitriangular by Proposition 3.1. So, we first compute all of the quasitriangular structures of $B_0^{*}$. To this end, suppose that $(B_0^{*},R^{\prime })$ is quasitriangular for some element $R^{\prime }\in B_0^{*}\otimes B_0^{*}$. Since

$$B_0^{*}\otimes B_0^{*}=K(e_0\otimes e_0)\oplus K(e_0\otimes e_1)\oplus K(e_1\otimes e_0)\oplus K(e_1\otimes e_1)$$

$$R^{\prime }=w(e_0\otimes e_0)+x(e_0\otimes e_1)+y(e_1\otimes e_0)+z(e_1\otimes e_1)$$

for $w,x,y,z\in K$. By Proposition 3.3(i),

$$\begin{array}{ccc}\hfill 1_{B_0^{*}}& =& e_0+e_1\hfill \\ & =& s_1(ϵ\otimes I)(w(e_0\otimes e_0)+x(e_0\otimes e_1)+y(e_1\otimes e_0)+z(e_1\otimes e_1))\hfill \\ & =& w{e}_0+x{e}_1\hfill \end{array}$$

and so, $w=x=1$. From Proposition 3.3(ii), one also has $y=1$. Thus

$$R^{\prime }=e_0\otimes e_0+e_0\otimes e_1+e_1\otimes e_0+z(e_1\otimes e_1)$$

for $z\in K$. Now,

$$\begin{array}{ccc}\hfill (\Delta \otimes I)\left(R^{\prime }\right)& =& (\Delta \otimes I)(e_0\otimes e_0+e_0\otimes e_1+e_1\otimes e_0+z(e_1\otimes e_1))\hfill \\ & =& (e_0\otimes e_0)\otimes e_0+(e_0\otimes e_0)\otimes e_1+(e_0\otimes e_1+e_1\otimes e_0+e_1\otimes e_1)\otimes e_0\hfill \\ & +& z((e_0\otimes e_1+e_1\otimes e_0+e_1\otimes e_1)\otimes e_1)\hfill \\ & =& e_0\otimes e_0\otimes e_0+e_0\otimes e_0\otimes e_1+e_0\otimes e_1\otimes e_0+e_1\otimes e_0\otimes e_0+e_1\otimes e_1\otimes e_0\hfill \\ & +& z(e_0\otimes e_1\otimes e_1)+z(e_1\otimes e_0\otimes e_1)+z(e_1\otimes e_1\otimes e_1)\hfill \end{array}$$

(12)

Moreover,

$$\begin{array}{ccc}\hfill {\left(R^{\prime }\right)}^{13}{\left(R^{\prime }\right)}^{23}& =& (e_0\otimes (e_0+e_1)\otimes e_0+e_0\otimes (e_0+e_1)\otimes e_1+e_1\otimes (e_0+e_1)\otimes e_0\hfill \\ & +& z(e_1\otimes (e_0+e_1)\otimes e_1))·((e_0+e_1)\otimes e_0\otimes e_0+(e_0+e_1)\otimes e_0\otimes e_1\hfill \\ & +& (e_0+e_1)\otimes e_1\otimes e_0+z((e_0+e_1)\otimes e_1\otimes e_1))\hfill \\ & =& (e_0\otimes e_0\otimes e_0+e_0\otimes e_1\otimes e_0+e_0\otimes e_0\otimes e_1+e_0\otimes e_1\otimes e_1\hfill \\ & +& e_1\otimes e_0\otimes e_0+e_1\otimes e_1\otimes e_0+z(e_1\otimes e_0\otimes e_1)\hfill \\ & +& z(e_1\otimes \otimes e_1\otimes e_1))·(e_0\otimes e_0\otimes e_0+e_1\otimes e_0\otimes e_0+e_0\otimes e_0\otimes e_1+e_1\otimes e_1\otimes e_0\hfill \\ & +& e_0\otimes e_1\otimes e_0+e_1\otimes e_1\otimes e_0+z(e_0\otimes e_1\otimes e_1)+z(e_1\otimes \otimes e_1\otimes e_1))\hfill \\ & =& e_0\otimes e_0\otimes e_0+e_0\otimes e_1\otimes e_0+e_0\otimes e_0\otimes e_1+z(e_0\otimes e_1\otimes e_1)\hfill \\ & +& e_1\otimes e_0\otimes e_0+e_1\otimes e_1\otimes e_0+z(e_1\otimes e_0\otimes e_1)+z^2(e_1\otimes e_1\otimes e_1)\hfill \end{array}$$

(13)

Equations 12 and 13 yield the relation $z^2=z$. Thus either $z=0$ or $z=1$. If $z=0$, then $R^{\prime }$ is not a unit in $B_0^{*}\otimes B_0^{*}$. Thus

$$R^{\prime }=e_0\otimes e_0+e_0\otimes e_1+e_1\otimes e_0+e_1\otimes e_1=1\otimes 1$$

is the only quasitriangular structure for $B_0^{*}$.

Consequently, if $(B_0,R)$ is quasitriangular, then $(\varphi \otimes \varphi )\left(R\right)=1_{B_0^{*}}\otimes 1_{B_0^{*}}$. It follows that $R=1_{B_0}\otimes 1_{B_0}$. ⋄

5.2. Questions for Future Research

Though the Myhill–Nerode bialgebra $B_0$ has only the trivial quasitriangular structure, it remains to compute the quasitriangular structure of $B_i$ for $i\ge 1$. Moreover, the linear dual $B_i^{*}$ is a commutative, cocommutative K-bialgebra and it would be of interest to find its quasitriangular structure. Unlike the $i=0$ case, we may have $B_i\cong B_i^{*}$ (for instance, $B_1\cong B_1^{*}$) and so this is indeed a separate problem.

Suppose that L is a language of words built from the alphabet ${\Sigma }_0=\{a,b\}$. If L is accepted by a finite automaton, then by Proposition 4.1, L gives rise to a Myhill–Nerode bialgebra B (see for example, [3, §6].) By construction, B is a cocommutative K-bialgebra and hence B has at least the trivial quasitriangular structure. Are there any other structures? Note that $B^{*}$ is a commutative K-algebra. For which R (if any) is $(B^{*},R)$ quasitriangular?