Department of Mathematics/Informatics Institute, Auburn University Montgomery, P.O. Box 244023, Montgomery, AL 36124, USA
Received: 20 June 2012; in revised form: 15 July 2012 / Accepted: 17 July 2012 / Published: 24 July 2012
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 , defined as if and only if exactly when , 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.
Let be a finite alphabet and let denote the set of words formed from the letters in . Let be a language, and let be the equivalence relation defined as if and only if exactly when . The Myhill–Nerode Theorem of computer science states that L is accepted by a finite automaton if and only if 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 , 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 . We ask: does B (or its linear dual ) 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 for all , . Scalar product defines two maps with and with , for , . Let denote the identity map. A K-algebra is a triple where is a K-linear map which satisfies
and is a K-linear map for which
for all , . The map is the multiplication map of A and is the unit map of A. Condition (1) is the associative property and Condition (2) is the unit property.
We write as . The element is the unique element of A for which for all . Let be algebras. An algebra homomorphism from A to B is a K-linear map such that In particular, for A to be a subalgebra of B we require .
For any two vector spaces V, W let denote the twist map defined as , for , . For K-algebras , we have that is a K-algebra with multiplication
for , . The unit map given as
Let C be a K-vector space. A K-coalgebra is a triple in which is K-linear and satisfies
and is K-linear with
for all . The maps and 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
Note that Condition (4) implies that
Let C be a K-coalgebra. A nonzero element c of C for which is a grouplike element of C. If c is grouplike, then
and so, . The grouplike elements of C are linearly independent [4, Proposition 3.2.1].
Let be coalgebras. A K-linear map is a coalgebra homomorphism if and for all . The tensor product of two coalgebras is again a coalgebra with comultiplication map
for , . The counit map is defined as
for , .
A K-bialgebra is a K-vector space B together with maps , , , for which is a K-algebra and is a K-coalgebra and for which and are algebra homomorphisms. Let be bialgebras. A K-linear map 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 that satisfies
for all . The map 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 denote the semigroup algebra. Then is a bialgebra with comultiplication map
defined by , for all , and counit map given by , for all . The bialgebra 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
for all , . Then A is a left B-module algebra. A K-linear map 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 , ,
Then C is a right B-module coalgebra. A K-linear map 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 denote the linear dual of C. Then the coalgebra structure of C induces an algebra structure on .
Proposition 2.1If C is a coalgebra, then is an algebra.
Proof. Recall that C is a triple where is K-linear and satisfies the coassociativity property, and is K-linear and satisfies the counit property. The dual map of is a K-linear map
Since , we define the multiplication map of , denoted as , to be the restriction of to . For , ,
The coassociatively property of yields the associative property of . Indeed, for , ,
In addition, the counit map of C dualizes to yield
defined as . Thus we define the unit map to be . One can show that the counit property of implies the unit property for . To this end, for , , ,
In a similar manner, one obtains
Thus is an algebra. Note that , and so, is the unique element of for which for all . ⋄
Let be a K-algebra. Then one may wonder if is a K-coalgebra. The multiplication map dualizes to yield . Unfortunately, if A is infinite dimensional over K, then is a proper subset of , and hence may not induce the required comultiplication map .
There is still however a K-coalgebra arising via duality from the algebra A. An ideal I of A is cofinite if . The finite dual of A is defined as
Note that is the largest subspace W of for which .
Proposition 2.2If A is an algebra, then is a coalgebra.
Proof. The proof is similar to the method used in Proposition 2.1. We restrict the map to to yield the K-linear map . Now by [4, Proposition 6.0.3], . Let denote the restriction of to . We show that satisfies the coassociative condition. For , , we have
For the counit map of , we consider the dual map . Now restricts to a map . We let denote the restriction of to . For , ,
and so, . We show that satisfies the counit property. First let be defined by the scalar multiplication of . For , , ,
In a similar manner, one obtains
where is given by scalar multiplication. Thus is a coalgebra. ⋄
Proposition 2.3If B is a bialgebra, then is a bialgebra.
Proof. As a coalgebra, B is a triple . By Proposition 2.1, is an algebra with maps and . Let denote the restriction of to , and let denote the restriction of to . Then the triple is a K-algebra.
As an algebra, B is a triple . By Proposition 2.2, is a coalgebra with maps and . It remains to show that and are algebra homomorphisms. First observe that for , one has
and so is an algebra map. We next show that is an algebra map. For ,
and so, is an algebra map. ⋄
Proposition 2.4Suppose that B is a bialgebra that is finite dimensional over K. Then is a bialgebra.
Proof. If , then . The result then follows from Proposition 2.3. ⋄
Let be a finite semigroup with unity element , and let denote the semigroup bialgebra. By Proposition 2.4 is a bialgebra of dimension n over K. Let be the dual basis for defined as .
Proposition 2.5The comultiplication map is given as
Let B be a bialgebra and let be the tensor product algebra. Let denote the group of units in and let . The pair is almost cocommutative if the element R satisfies
for all .
If the bialgebra B is cocommutative, then the pair is almost cocommutative since satisfies Condition (7). However, if B is commutative and non-cocommutative, then cannot be almost cocommutative for any since Condition (7) in this case reduces to the condition for cocommutativity.
Write . Let
The pair is quasitriangular if is almost cocommutative and the following conditions hold
Clearly, if B is cocommutative then is quasitriangular.
Let B be a bialgebra. A quasitriangular structure is an element so that is quasitriangular. Let and be quasitriangular bialgebras. Then , are isomorphic as quasitriangular bialgebras if there exists a bialgebra isomorphism for which . Two quasitriangular structures on a bialgebra B are equivalent quasitriangular structures if as quasitriangular bialgebras.
The following proposition shows that every bialgebra isomorphism with B quasitriangular extends to an isomorphism of quasitriangular bialgebras.
Proposition 3.1Suppose is quasitriangular and suppose that is an isomorphism of K-bialgebras. Let . Then is quasitriangular.
Proof. Note that . Let . Then there exists for which . Now
and so, is almost cocommutative. Moreover,
In a similar manner one shows that
Thus is quasitriangular. ⋄
Quasitriangular bialgebras are important since they give rise to solutions of the equation
which is known as the quantum Yang–Baxter equation (QYBE). The QYBE was first introduced in statistical mechanics, see . An element which satisfies (10) is a solution to the QYBE.
Certainly, the QYBE admits the trivial solution , and of course, if B is commutative, then any is a solution to the QYBE. For B non-commutative, it is of great interest to find non-trivial solutions to the QYBE. We have the following result due to V. G. Drinfeld .
Proposition 3.2(Drinfeld) Suppose is quasitriangular. Then R is a solution to the QYBE.
Proof. One has
The following proposition provides necessary conditions on in order for to be quasitriangular.
Proposition 3.3Suppose is quasitriangular. Then
Proof. For (i) one has
In view of Condition (8)
A similar argument is used to prove (ii). ⋄
4. Myhill–Nerode Bialgebras
In this section we review the main result of  in which the authors give a bialgebra version of the Myhill–Nerode Therorem. Let G be a semigroup with unity, 1 and let be the semigroup bialgebra. There is a right H-module structure on defined as
for all , . For , , the element is the right translate of p by x.
Proposition 4.1([3, Proposition 5.4].) Let G be a semigroup with 1, let denote the semigroup bialgebra. Let . Then the following are equivalent.
(i) The set of right translates is finite.
(ii) There exists a finite dimensional bialgebra B, a bialgebra homomorphism , and an element so that for all .
(Note: The bialgebras of (ii) are defined to be Myhill–Nerode bialgebras.)
Proof. . Let be the finite set of right translates. For each , we define a right operator by the rule
Observe that the set is finite with . The set is a semigroup with unity, under composition of operators. Indeed,
Thus , for all . Let B denote the semigroup bialgebra on . Let be the K-linear map defined by . Then
and so, Ψ is a homomorphism of bialgebras.
Let be defined by
Then , for all , as required.
. Suppose there exists a finite dimensional bialgebra B, a bialgebra homomorphism , and an element so that for all . Define a right H-module action · on B as
for all , . Then for , ,
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 “·”,
for , . Thus · restricts to give an action (also denoted by “·”) of G on Q. Now for ,
In view of Condition (11) there exists a function
Since ϱ is surjective and S is finite, is finite. ⋄
We illustrate the connection between Proposition 4.1 and the usual Myhill–Nerode Theorem. Let denote the set of words in a finite alphabet . Let be a language. Suppose that the equivalence relation (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 as a semigroup with unity where the semigroup operation is concatenation and the unity element is the empty word. Let denote the semigroup bialgebra. Then the characteristic function of L extends to an element . Since has finite index, the set of right translates 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 and an element so that , for all .
This bialgebra determines a finite automaton , where Q is the finite set of states, Σ is the input alphabet, δ is the transition function, 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 . 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 by the rule , for , . The initial state is , and the set of final states F is the subset of Q of the form , for which
By construction, the finite automaton 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 be the alphabet on a single letter a. Let denote the collection of all words of finite length formed from . Here 1 denotes the empty word of length 0. For convenience, we shall write
Fix an integer and let . Then the language is accepted by the finite automaton given in Figure 1.
Finite automaton accepting , accepting state is i.
Finite automaton accepting , accepting state is i.
By the usual Myhill–Nerode Theorem, the equivalence relation , defined as if and only if exactly when , has finite index. If is the characteristic function of , then is equivalent to the relation defined as: if and only if . Let denote the equivalence class of x under . The Myhill–Nerode theorem now says that the set is finite.
Now we consider as a semigroup with unity 1 with concatenation as the binary operation. Let be the semigroup bialgebra. The characteristic function of extends to an element of . By [3, Proposition 2.3], the set of right translates is finite. Thus by Proposition 4.1, there exists a finite dimensional bialgebra , a bialgebra homomorphism , and an element so that for all .
In what follows, we give the bialgebra structure of the collection and compute the quasitriangular structure of the bialgebra .
For , the finite set of right translates of is
One finds that the set of right operators on is . Under composition, the set of right operators is a semigroup with unity . We have, for ,
By construction, is the semigroup bialgebra on .
5.1. Quasitriangular Structure of
In the case , is the semigroup bialgebra on with algebra structure defined by , , , . Let be the dual basis defined as , , , . Then is the set of minimal idempotents for . Comultiplication on is given as
and the counit map is defined by
Proposition 5.1Let be the K-bialgebra as above. Then there is exactly one quasitriangular structure on , namely, .
Proof. Certainly, is a quasitriangular structure for . We claim that is the only quasitriangular structure. Observe that there is bialgebra isomorphism defined as , . Thus if is quasitriangular, then , , is quasitriangular by Proposition 3.1. So, we first compute all of the quasitriangular structures of . To this end, suppose that is quasitriangular for some element . Since
for . By Proposition 3.3(i),
and so, . From Proposition 3.3(ii), one also has . Thus
for . Now,
Equations 12 and 13 yield the relation . Thus either or . If , then is not a unit in . Thus
is the only quasitriangular structure for .
Consequently, if is quasitriangular, then . It follows that . ⋄
5.2. Questions for Future Research
Though the Myhill–Nerode bialgebra has only the trivial quasitriangular structure, it remains to compute the quasitriangular structure of for . Moreover, the linear dual is a commutative, cocommutative K-bialgebra and it would be of interest to find its quasitriangular structure. Unlike the case, we may have (for instance, ) and so this is indeed a separate problem.
Suppose that L is a language of words built from the alphabet . 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 is a commutative K-algebra. For which R (if any) is quasitriangular?
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