The semaphore codes attached to a Turing machine via resets and their various limits
aa r X i v : . [ m a t h . G R ] A p r The semaphore codes attached to a Turing machine via resets andtheir various limits
John Rhodes
Department of Mathematics, University of California, Berkeley,CA 94720, U.S.A.email: [email protected], [email protected]
Anne Schilling
Department of Mathematics, University of California, Davis,One Shields Ave., Davis, CA 95616-8633, U.S.A.email: [email protected]
Pedro V. Silva
Centro de Matem´atica, Faculdade de Ciˆencias, Universidade do Porto,R. Campo Alegre 687, 4169-007 Porto, Portugalemail: [email protected] 15, 2018 k -resets toobtain ( − ω )-resets. We mention how this opens new avenues to attack the P versusNP problem. In our previous paper [9], we developed algebraic foundations centered around the prime decompo-sition theory for finite semigroups and finite automata (see [5], [6] and [8, Chapter 4]). This analysisfocused on the right zero component action, when the corresponding pseudovariety contains all finitetransformation semigroups (
X, S ) (or automata) with the property that there exists some k ≥ s · · · s k (with s i ∈ S ) is a constant map (or reset) on X . Such finite automatawere called k -reset graphs in [9, Section 3] and their elementary properties were studied, using thelattice of right congruences on the finite free objects (De Bruijn semigroups), in [9, Sections 5 and 6].Semaphore codes, which are well known in the literature (see [2]), were proved [9, Sections 4 and7] to be in bijection with special right congruences and provide a lower approximation to any right1ongruence with the same hitting time to constant. Thus in many applications right congruencescan be replaced by semaphore codes. Except for Section 4 on semaphore codes, all the material in [9]up to and including Section 7 is restricted to finite codes and automata. Finally, in [9, Section 8], anatural random walk on any (finite or infinite) semaphore code was constructed and its stationarydistribution plus hitting time to constant were computed.In this paper we do the following: first we reveal a main application we have in mind [7] byintroducing the infinite and finite semaphore codes associated to a Turing machine via resets (seeSection 2). Then Sections 3 and 4 take the profinite limits of k -reset graphs yielding ( − ω )-resetgraphs.We consider the pseudovariety D and left infinite words, but by duality we have analogous resultsfor the pseudovariety K and right infinite words. We need both versions for studying Turing machines.Generalizing the finite case from [9], we study right congruences and special right congruences inbijection with infinite semaphore codes and the natural action in Sections 5 and 6 and obtain anapproximation theory as in the finite case of k -resets, but for ( − ω )-resets. In the final Section 7, wemake some more remarks on relating Section 2 to Sections 3-6, and the next paper on attacking Pversus NP. Acknowledgements
We are indebted to Jean-Camille Birget for good advice and patient reading. We thank also BenjaminSteinberg and Nicolas M. Thi´ery for discussions.The first author thanks the Simons Foundation–Collaboration Grants for Mathematicians fortravel grant
In this section we present a new viewpoint on Turing machines centered in the concept of resets andtheir associated semaphore codes.
For details on Turing machines, the reader is referred to [4].Let us assume that T = ( Q, A, Γ , q , F, δ ) is a (deterministic) Turing machine, where: • Q is the (finite) state set; • A is the (finite) input alphabet; • Γ is the (finite) tape alphabet (containing A and the blank symbol B ); • q ∈ Q is the initial state; • F ⊆ Q is the set of final states; • δ : Q × Γ → Q × (Γ \ { B } ) × { L, R } is the (partial) transition function.2hen we write Ω = Γ ∪ (Γ × Q ). To make notation lighter, we shall denote ( X, q ) ∈ Γ × Q by X q .To avoid confusion with powers of X , we stipulate that from now on symbols such as q, q ′ , q i will bereserved to denote states and never integers.We define two homomorphisms tape : Ω ∗ → Γ ∗ and heads : Ω ∗ → ( N , +) bytape( X ) = tape( X q ) = X, heads( X ) = 0 , heads( X q ) = 1for all X ∈ Γ and q ∈ Q . Now we define the set of all legal words byLeg( T ) = B ∗ { w ∈ Ω ∗ | tape( w ) ∈ { , B } (Γ \ { B } ) ∗ { , B } , heads( w ) ≤ } B ∗ . Note that Leg( T ) is closed under reversal and factors. The illegal words are the elements of thecomplement Ω ∗ \ Leg( T ), which is the ideal having as generating set all the words of the form • X q uX q ′ • Y B n Y ′ • Y B m B q B k Y ′ • Y q B n Y ′ • Y ′ B n Y q • B q B n Y • Y B n B q where X , X ∈ Γ; Y, Y ′ ∈ Γ \ { B } ; u ∈ Γ ∗ ; n ≥ m, k ≥ one-move mapping β : Leg( T ) → Leg( T ) as follows. Given w ∈ Leg( T ), then β ( w )is intended to be obtained from w by performing one single move of T on a tape with content w ; if T admits no such move from w (in particular, if heads( w ) = 0), we set β ( w ) = w . In all cases except(A) and (B) below, the interpretation of β ( w ) is clear and | β ( w ) | = | w | . The following two casesdeserve extra clarification:(A) if w = X q w ′ and δ ( q, X ) = ( . . . , . . . , L );(B) if w = w ′ X q and δ ( q, X ) = ( . . . , . . . , R ).In these cases, we interpret β ( w ) as β ( Bw ) (respectively β ( wB )), falling into the general case. Wesay that legal words of types (A) and (B) are β - singular . Note that | β ( w ) | = | w | + 1 in the β -singularcase. Note that, by padding the input sequences with sufficiently many B ’s before and after, cases(A) and (B) never occur.For every n ≥
0, we denote by β n the n -fold composition if β . We define also a partial mapping β ( n ) : Ω ∗ × Ω × Ω ∗ → Ωas follows. 3et u, v ∈ Ω ∗ and X ∈ Ω. If uXv ∈ Leg( T ), then β ( n ) ( u, X, v ) is the symbol replacing X atthe designated position in the tape after applying β n times. If uXv / ∈ Leg( T ), then β ( n ) ( u, X, v ) isundefined.We say that T is legal-halting if, for every u ∈ Leg( T ), the sequence ( β n ( u )) n is eventuallyconstant. This implies that the sequence ( β ( n ) ( u, X, v )) n is also eventually constant for all u, v ∈ Ω ∗ and X ∈ Ω such that uXv ∈ Leg( T ). We write β ω ( u ) = lim n →∞ β n ( u ) , β ( ω ) ( u, X, v ) = lim n →∞ β ( n ) ( u, X, v ) . Note that, from a formal viewpoint, a Turing machine which halts for every input is not necessarilylegal-halting (since not every legal word arises from an input configuration), but it can be madelegal-halting with minimum adaptations.
We say that r ∈ Ω ∗ is a right reset if β ( n ) ( t rt , X, t ) = β ( n ) ( t ′ rt , X, t )for all t , t ′ , t , t ∈ Ω ∗ , X ∈ Ω and n ≥ t rt Xt , t ′ rt Xt ∈ Leg( T ). Let RRes( T )denote the set of all right resets of T .Dually, r ∈ Ω ∗ is a left reset if β ( n ) ( t , X, t rt ) = β ( n ) ( t , X, t rt ′ )for all t , t , t , t ′ ∈ Ω ∗ , X ∈ Ω and n ≥ t Xt rt , t Xt rt ′ ∈ Leg( T ). Let LRes( T )denote the set of all left resets of T . Lemma 2.1
RRes( T ) and LRes( T ) are ideals of Ω ∗ containing all the illegal words. Proof . We prove the claim for right resets.If r is illegal then t rt Xt is always illegal and so r ∈ RRes( T ) trivially. In particular, RRes( T )is nonempty.Let r ∈ RRes( T ) and let x, y ∈ Ω ∗ . Suppose that xry / ∈ RRes( T ). Then there exist t , t ′ , t , t ∈ Ω ∗ , X ∈ Ω and n ≥ t xryt Xt and t ′ xryt Xt are legal and β ( n ) ( t ( xry ) t , X, t ) = β ( n ) ( t ′ ( xry ) t , X, t ) . Rewriting this inequality as β ( n ) (( t x ) r ( yt ) , X, t ) = β ( n ) (( t ′ x ) r ( yt ) , X, t ) , we deduce that r / ∈ RRes( T ), a contradiction. Thus xry ∈ RRes( T ) and so RRes( T ) is an ideal ofΩ ∗ . (cid:3) Lemma 2.2 If u ∈ Ω ∗ \ B ∗ , then Bu ∈ RRes( T ) and uB ∈ LRes( T ) . Proof . It is easy to see that Bu is a right reset since the only legal words of the form tBut ′ mustarise from t ∈ B ∗ . Similarly, uB is a left reset. (cid:3) emma 2.3 (i) β (RRes( T )) ⊆ RRes( T ) .(ii) β (LRes( T )) ⊆ LRes( T ) . Proof . We prove the claim for right resets.Let r ∈ RRes( T ). We may assume that r is legal, X q occurs in r and δ ( X, q ) is defined. Let t , t ′ , t , t ∈ Ω ∗ , X ∈ Ω and n ≥ t β ( r ) t Xt and t ′ β ( r ) t Xt are legal.Suppose first that r is not β -singular. Then t rt Xt and t ′ rt Xt are also legal, and β ( n ) ( t β ( r ) t , X, t ) = β ( n +1) ( t rt , X, t ) = β ( n +1) ( t ′ rt , X, t ) = β ( n ) ( t ′ β ( r ) t , X, t ) . Thus we may assume that r is β -singular. Suppose first that r = X q r ′ and δ ( q, X ) = ( p, Y, L ).Then β ( r ) = B p Y r ′ . Since t B p Y r ′ t Xt and t ′ B p Y r ′ t Xt are legal, it is easy to check that t Brt Xt and t ′ Brt Xt are legal as well. Hence β ( n ) ( t β ( r ) t , X, t ) = β ( n +1) ( t Brt , X, t ) = β ( n +1) ( t ′ Brt , X, t ) = β ( n ) ( t ′ β ( r ) t , X, t ) . Finally, suppose that r = r ′ X q and δ ( q, X ) = ( p, Y, R ). Then β ( r ) = r ′ Y B p . Since t r ′ Y B p t Xt and t ′ r ′ Y B p t Xt are legal, it is easy to check that t rBt Xt and t ′ rBt Xt are legal as well.Hence β ( n ) ( t β ( r ) t , X, t ) = β ( n +1) ( t rBt , X, t ) = β ( n +1) ( t ′ rBt , X, t ) = β ( n ) ( t ′ β ( r ) t , X, t ) . Therefore β ( r ) is a right reset in any case. (cid:3) Given an alphabet X , we define the suffix order on X ∗ by u ≤ s v if v ∈ X ∗ u. Dually, we define the prefix order ≤ p .We say that S ⊆ X ∗ is a (right) semaphore code if S is a suffix code (i.e. an antichain for thesuffix order) and SX ⊆ X ∗ S . By [9, Proposition 4.3], S is a semaphore code if and only if S is theset of ≤ s -minimal elements in some ideal I E X ∗ , denoted by Iβ ℓ .Dually, S ⊆ X ∗ is a left semaphore code if S is a prefix code (i.e. an antichain for the prefixorder) and XS ⊆ SX ∗ . Then S is a left semaphore code if and only if S is the set of ≤ p -minimalelements in some ideal of X ∗ .We describe now the semaphore code RSC( T ) defined by the right resets. It consists of allminimal right resets for the suffix order. If 1 / ∈ RRes( T ) (a trivial case), then RSC( T ) consists ofall right resets Xz with X ∈ Ω and z ∈ Ω ∗ such that z / ∈ RRes( T ). In particular, z must be a legalword.Similarly, the left semaphore code LSC( T ) consists of all minimal left resets for the prefix order.If 1 / ∈ LRes( T ), then LSC( T ) consists of all left resets zX with z ∈ Ω ∗ and X ∈ Ω such that z / ∈ LRes( T ). In particular, z must be a legal word.Note also that every right reset contains some s ∈ RSC( T ) as a suffix. Dually, every left resetcontains some s ∈ LSC( T ) as a prefix.From now on, we assume that T is legal-halting. The output function ϕ T is the restriction of thepartial function β ( ω ) : Ω ∗ × Ω × Ω ∗ to (RSC( T ) ∪ { } ) × Ω × (LSC( T ) ∪ { } ).5 roposition 2.4 Let T be a legal-halting Turing machine. Then β ω is fully determined by the outputfunction ϕ T . Proof . Clearly, β ω is fully determined by β ( ω ) . Let u, v ∈ Ω ∗ and X ∈ Ω. If uXv is illegal,then β ( ω ) ( u, X, v ) is undefined, hence we may assume that uXv ∈ Leg( T ). It follows that also BuXvB ∈ Leg( T ) and β ( ω ) ( Bu, X, vB ) = β ( ω ) ( u, X, v ).If u ∈ B ∗ , then β ( ω ) ( Bu, X, vB ) = β ( ω ) (1 , X, vB ). If u / ∈ B ∗ , then Bu ∈ RRes( T ) by Lemma 2.2and so Bu = xr for some x ∈ Ω ∗ and r ∈ RSC( T ). It follows that β ( ω ) ( Bu, X, vB ) = β ( ω ) ( r, X, vB ),so in any case we have β ( ω ) ( Bu, X, vB ) = β ( ω ) ( r, X, vB ) (2.1)for some r ∈ RSC( T ) ∪ { } .Now if v ∈ B ∗ , then β ( ω ) ( r, X, vB ) = β ( ω ) ( r, X,
1) = ϕ T ( r, X, v / ∈ B ∗ . Then vB ∈ LRes( T ) by Lemma 2.2 and so vB = r ′ x ′ for some r ′ ∈ LSC( T ) and x ∈ Ω ∗ . It follows that β ( ω ) ( r, X, vB ) = β ( ω ) ( r, X, r ′ ) = ϕ T ( r, X, r ′ ), so in view of (2.1) we have that β ( ω ) ( u, X, v ) = β ( ω ) ( Bu, X, vB ) is determined by ϕ T . Therefore β ω is determined by ϕ T . (cid:3) For every ℓ ≥
0, we define the cofinite idealsRRes ℓ ( T ) = RRes( T ) ∪ Ω ℓ Ω ∗ , LRes ℓ ( T ) = LRes( T ) ∪ Ω ℓ Ω ∗ . Note that RRes( T ) = \ ℓ ≥ RRes ℓ ( T ) , LRes( T ) = \ ℓ ≥ LRes ℓ ( T ) . (2.2)Since β (Ω ℓ ) ⊆ Ω ℓ ∪ Ω ℓ +1 , it follows from Lemma 2.3 that: Lemma 2.5 (i) β (RRes ℓ ( T )) ⊆ RRes ℓ ( T ) .(ii) β (LRes ℓ ( T )) ⊆ LRes ℓ ( T ) . The semaphore code RSC ℓ ( T ) consists of all minimal elements of RRes ℓ ( T ) for the suffix order.Equivalently, RSC ℓ ( T ) = (RSC( T ) ∩ Ω ≤ ℓ ) ∪ Ω(Ω ℓ − \ RRes( T )) . (2.3)Dually, the left semaphore code LSC ℓ ( T ) consists of all minimal elements of LRes ℓ ( T ) for the prefixorder, or equivalently LSC ℓ ( T ) = (LSC( T ) ∩ Ω ≤ ℓ ) ∪ (Ω ℓ − \ LRes( T ))Ω . (2.4)Therefore RSC ℓ ( T ) ∪ LSC ℓ ( T ) ⊆ Ω ≤ ℓ . 6 .5 A context-free example We present now a very elementary example just to illustrate the notation and ideas. Further researchwill include much more complicated examples.Let A = { a, b } and L = { a n b n | n ≥ } , one of the classical examples of a (deterministic) context-free language which is not rational. The language L is accepted by the Turing machine T depictedby q Y | Y R (cid:2) (cid:2) B | ZL (cid:15) (cid:15) q a | XR / / Y | Y R o o q Y | Y R / / a | aR (cid:2) (cid:2) b | Y L (cid:15) (cid:15) q Y | Y R (cid:2) (cid:2) b | Y L ~ ~ ⑦⑦⑦⑦⑦⑦⑦⑦⑦⑦⑦⑦⑦⑦⑦⑦ q q a | aL Z Z X | XR O O q Y | Y L Z Z a | aL o o X | XR ` ` ❅❅❅❅❅❅❅❅❅❅❅❅❅❅❅❅ where q is the initial state and q the unique final state.In state q we can only read a or Y . In the first case, a is replaced by X and we change to state q . Then we move right across other possible a ’s until we reach the first b and replace it by Y to goto state q . If we have done this routine before, we may have to move across older Y ’s – taking usinto state q . From state q , we intend to move left until we reach X , which means going through Y ’s and then a ’s (if there are some left – state q ). So we are back at state q and we repeat theprocedure. If we have replaced all the a ’s, we are supposed to read Y at state q , then move to theright end of the tape (state q ) reading only Y ’s. If we have succeeded on reaching the blank B , thenwe accept the input moving to the final state q .It is easy to check that T is legal-halting. Indeed, B can be read at most once, and any longenough sequence of transitions must necessarily involve replacing a by X or b by Y , which are bothirreversible changes.We use the notation introduced in Section 2.1 for an arbitrary Turing machine. We claim thatΩ ∗ \ RRes( T ) = a ∗ { b, Y } ∗ ∪ ( S i =1 , , a ∗ a q i a ∗ { b, Y } ∗ ) ∪ ( S i =1 , a ∗ Y ∗ b q i { b, Y } ∗ ) ∪ ( S i =1 , , a ∗ Y ∗ Y q i { b, Y } ∗ ) . (2.5)Let m, n ≥ u ∈ { b, Y } ∗ containing n b ’s. Then β ( ω ) ( a q a n · a m u, b,
1) = Y = b = β ( ω ) ( a m u, b, , hence a m u / ∈ RRes( T ). 7ssume now that i ∈ { , , } , m, n, k ≥ u ∈ { b, Y } ∗ contains k b ’s. Then β ( ω ) ( Xa k +1 · a m a q i a n u · b, b,
1) = Y = b = β ( ω ) ( a m a q i a n u · b, b, , hence a m a q i a n u / ∈ RRes( T ).Assume next that i ∈ { , } , m, n, k ≥ u ∈ { b, Y } ∗ contains k b ’s. Then β ( ω ) ( Xa k +1 · a m Y n b q i u, b,
1) = Y = b = β ( ω ) ( a m Y n b q i u, b, , hence a m Y n b q i u / ∈ RRes( T ).Finally, assume that i ∈ { , , } , m, n, k ≥ u ∈ { b, Y } ∗ contains k b ’s. Then β ( ω ) ( Xa k +2 · a m Y n Y q i u · b, b,
1) = Y = b = β ( ω ) ( a m Y n Y q i u · b, b, , hence a m Y n Y q i u / ∈ RRes( T ).To prove the converse, we must prove that any other word is necessarily a right reset. We makeextensive use from RRes( T ) being an ideal of Ω ∗ .Consider first r ∈ { B, X, Z } ∪ { B q , X q , Z q | q ∈ Q } . Suppose that t , t ′ , t , t ∈ Ω ∗ , P ∈ Ω are such that both t rt P t , t ′ rt P t ∈ Leg( T ). Let n ≥
0. If β ( n ) ( t rt , P, t ) = P , then it is easy to see that t ∈ Γ ∗ and has no influence in the computation.Thus β ( n ) ( t rt , P, t ) = β ( n ) ( t ′ rt , P, t ) and so r ∈ RRes( T ).Thus we only need to discuss words w ∈ Ω ∗ such that tape( w ) ∈ { a, b, Y } ∗ . We consider nextthe word ba . Consider t bat P t ∈ Leg( T ). If t P t / ∈ Γ ∗ , then t is irrelevant to the computationof β ( n ) ( t bat , P, t ). If t P t ∈ Γ ∗ , then β ( n ) ( t bat , P, t ) = P necessarily. It follows that ba ∈ RRes( T ).Similarly, Y a, b q a, ba q , Y q a, Y a q ∈ RRes( T ) for every q ∈ Q . Hence we have reduced the problemto words w ∈ Ω ∗ such that tape( w ) ∈ a ∗ { b, Y } ∗ .Since the transition function is undefined for these pairs, we have { a q , a q , a q , b q , b q , b q , b q , b q , Y q , Y q } ⊆ RRes( T ) . Also a q ∈ RRes( T ) because T moves to the right and will never get to the left of the new X .Similarly, Y q , Y q ∈ RRes( T ). To complete the proof of (2.5), it suffices to show that bY ∗ b q i ∪ bY ∗ Y q j ⊆ RRes( T ) (2.6)for i = 1 , j = 1 , ,
3, because any word not containing such a factor has already been establishedto be or not to be a right reset.Indeed, in neither case the head of T can pass to the left of the first b , so (2.6) and therefore(2.5) hold as claimed.Similarly, we compute the left resets, in fact we obtainLRes( T ) = RRes( T ) . (2.7)8 .5.3 Semaphore codes In view of (2.5), it is straightforward to check thatRSC( T ) = (Ω \ { a, a q , a q , a q } ) a + { b, Y } ∗ ∪ (Ω \ { a, b, Y, a q , a q , a q , b q , b q , Y q , Y q , Y q } ) { b, Y } ∗ ∪ ( S i =1 , , (Ω \ { a } ) a ∗ a q i a ∗ { b, Y } ∗ ) ∪ ( S i =1 , (Ω \ { a } ) a + Y ∗ b q i { b, Y } ∗ ) ∪ ( S i =1 , (Ω \ { a, Y } ) Y ∗ b q i { b, Y } ∗ ) ∪ ( S i =1 , , (Ω \ { a } ) a + Y ∗ Y q i { b, Y } ∗ ) ∪ ( S i =1 , , (Ω \ { a, Y } ) Y ∗ Y q i { b, Y } ∗ ) . To compute the intersection RSC( T ) ∩ Leg( T ), we replace the 5 last occurrences of Ω by Γ.Similarly,LSC( T ) = a ∗ (Ω \ { a, b, Y, a q , a q , a q , b q , b q , Y q , Y q , Y q } ) ∪ a ∗ Y + (Ω \ { b, Y, b q , b q , Y q , Y q , Y q } ) ∪ a ∗ Y ∗ b { b, Y } ∗ (Ω \ { b, Y } ) ∪ ( S i =1 , , a ∗ a q i a ∗ (Ω \ { a, b, Y } )) ∪ ( S i =1 , , a ∗ a q i a ∗ { b, Y } + (Ω \ { b, Y } )) ∪ ( S i =1 , a ∗ Y ∗ b q i { b, Y } ∗ (Ω \ { b, Y } )) ∪ ( S i =1 , , a ∗ Y ∗ Y q i { b, Y } ∗ (Ω \ { b, Y } )) . To compute the intersection LSC( T ) ∩ Leg( T ), we replace the 4 last occurrences of Ω by Γ. ℓ In view of (2.3) and (2.4), we can easily compute easily RSC ℓ ( T ) and LSC ℓ ( T ) making use of thecomputations performed in Sections 2.5.2 and 2.5.3. For general background on free pro- D semigroups, see [8, Sections 3.1 and 3.2].Let A be a finite nonempty alphabet. We denote by A − ω the set of all left infinite words on A ,that is, infinite sequences of the form · · · a a a with a i ∈ A . If u ∈ A + , we denote the left infiniteword · · · uuu by u − ω .The free semigroup A + acts on the right of A − ω by concatenation: given x = · · · a a a ∈ A − ω and u = a ′ · · · a ′ n ∈ A + , we define xu = · · · a a a a ′ · · · a ′ n ∈ A − ω . Given x ∈ A + ∪ A − ω and y ∈ A − ω , we define also xy = y . Together with concatenation on A + , thisdefines a semigroup structure for A + ∪ A − ω . 9he suffix (ultra)metric on A + ∪ A − ω is defined as follows. Given x, y ∈ A + ∪ A − ω , let lcs( x, y )be the longest common suffix of x and y , and define d ( x, y ) = ( −| lcs ( x,y ) | if x = y, x ∈ A + ∪ A − ω and δ >
0, we write B δ ( x ) = { x ∈ A + ∪ A − ω | d ( x, x ) < δ } for the open ball of radius δ around x .If S ∈ D is endowed with the discrete topology and ϕ : A → S is a mapping, then there exists aunique continuous homomorphism Φ : A + ∪ A − ω → S such that the diagram A ϕ / / (cid:127) _ (cid:15) (cid:15) SA + ∪ A − ω Φ : : ✉✉✉✉✉ commutes. This characterizes ( A + ∪ A − ω , d ) as the free pro- D semigroup on A . We shall denote itby Ω A ( D ). It is well known that Ω A ( D ) is a complete and compact topological semigroup.We remark that for a general pseudovariety V , the metric considered for free pro- V semigroupsis the profinite metric, but in the particular case of D we can use this alternative metric that equatesto the normal form. ( − ω ) -reset graphs We consider now A -graphs with possibly infinite vertex sets. For general concepts in automatatheory, the reader is referred to [1].A left infinite path in an A -graph Γ = ( Q, E ) is an infinite sequence of the form · · · a −→ q a −→ q a −→ q such that ( q i +1 , a i , q i ) ∈ E and q i ∈ Q for every i ≥
1. Its label is the left infinite word · · · a a a ∈ A − ω . We write · · · x −→ q to denote a left infinite path with label x ending at q .An A -graph Γ = ( Q, E ) is: • deterministic if ( p, a, q ) , ( p, a, q ′ ) ∈ E implies q = q ′ ; • complete if for all p ∈ Q and a ∈ A there exists some edge ( p, a, q ) ∈ E ; • strongly connected if, for all p, q ∈ Q , there exists a path p u −→ q in Γ for some u ∈ A ∗ ; • ( − ω ) -deterministic if · · · x −→ q, · · · x −→ q ′ paths in Γ ⇒ q = q ′ holds for all q, q ′ ∈ Q and x ∈ A − ω ; 10 ( − ω ) -complete if every x ∈ A − ω labels some left infinite path in Γ; • ( − ω ) -trim if every q ∈ Q occurs in some left infinite path in Γ; • a ( − ω ) -reset graph if it is ( − ω )-deterministic, ( − ω )-complete and ( − ω )-trim.We denote by RG( A ) the class of all ( − ω )-reset A -graphs.If Γ = ( Q, E ) ∈ RG( A ), then Q induces a partition A − ω = [ q ∈ Q A − ωq , where A − ωq denotes the set of all x ∈ A − ω labelling some path · · · x −→ q in Γ. Moreover, A − ωq = ∅ forevery q ∈ Q . Proposition 4.1
Let Γ ∈ RG( A ) . Then Γ is deterministic and complete. Proof . Write Γ = (
Q, E ) and suppose that ( p, a, q ) , ( p, a, q ′ ) ∈ E . Since Γ is ( − ω )-trim, there existssome left infinite path · · · x −→ p for some x ∈ A − ω . Hence there exist left infinite paths · · · xa −→ q and · · · xa −→ q ′ , and since Γ is ( − ω )-deterministic, we get q = q ′ . Therefore Γ is deterministic.Let p ∈ Q and a ∈ A . Since Γ is ( − ω )-trim, there exists some left infinite path · · · x −→ p for some x ∈ A − ω . Now xa ∈ A − ω and Γ being ( − ω )-complete implies that there exists some path · · · xa −→ q in Γ, which we may factor as · · · x −→ q ′ a −→ q. Since Γ is ( − ω )-deterministic, we get q ′ = p , hence ( p, a, q ) ∈ E and Γ is complete. (cid:3) We recall now the preorder ≤ introduced in [9, Section 3]. Given A -graphs Γ , Γ ′ , we write Γ ≤ Γ ′ if there exists a morphism Γ → Γ ′ . Lemma 4.2
Let A be a finite nonempty alphabet and let Γ , Γ ′ ∈ RG( A ) with Γ ≤ Γ ′ ≤ Γ . Then Γ ∼ = Γ ′ . Proof . Let ϕ : Γ → Γ ′ and ψ : Γ ′ → Γ be morphisms. Write Γ = (
Q, E ) and Γ ′ = ( Q ′ , E ′ ). It iseasy to see that A − ωq ⊆ A − ωqϕ , A − ωq ′ ⊆ A − ωq ′ ψ for all q ∈ Q and q ′ ∈ Q ′ . Hence A − ωq ⊆ A − ωqϕψ . Since Γ is ( − ω )-trim, we have A − ωq = ∅ . Since Γis ( − ω )-deterministic, we get q = qϕψ . Similarly, q ′ = q ′ ψϕ , hence ϕ and ψ are mutually inversebijections and therefore mutually inverse A -graph isomorphisms. (cid:3) Let [Γ] denote the isomorphism class of Γ. Similarly to [9, Section 3],[Γ] ≤ [Γ ′ ] if Γ ≤ Γ ′ defines a preorder on RG( A ) / ∼ =. Moreover, Lemma 4.2 yields: Corollary 4.3
Let A be a finite nonempty alphabet. Then ≤ is a partial order on RG( A ) / ∼ = . Right congruences on A − ω Since xy = y for all x ∈ Ω A ( D ) and y ∈ A − ω , it follows that A − ω is the minimum ideal of Ω A ( D ).Following the notation introduced in [9, Section 2.2], we denote by RC( A − ω ) the lattice of rightcongruences on A − ω (with respect to the right action of Ω A ( D )).We say that ρ ∈ RC( A − ω ) is closed if ρ is a closed subset of A − ω × A − ω for the product metric d ′ (( x, y ) , ( x ′ , y ′ )) = max { d ( x, x ′ ) , d ( y, y ′ ) } , where d denotes the suffix metric on A − ω . Given x , y ∈ A + ∪ A − ω and δ >
0, we write B δ (( x , y )) = { ( x, y ) ∈ ( A + ∪ A − ω ) | d ′ (( x, y ) , ( x , y )) < δ } . By [8, Exercise 3.1.7], this implies that xρ is a closed subset of A − ω for every x ∈ A − ω . The nextexample shows that the converse fails. Example 5.1
Let A = { a, b } and let w = . . . a ba ba bab. (5.1) For all x, y ∈ A − ω , let xρy if x = wu, y = wv with | u | = | v | or x = y. Then ρ ∈ RC( A − ω ) and xρ is closed for every x ∈ A − ω , but ρ is not closed. Indeed, it is easy to see that, given x ∈ A − ω , there is at most one word u ∈ A ∗ such that x = wu . We call this a w -factorization of x . Hence ρ is transitive and it follows immediately that ρ ∈ RC( A − ω ). The uniqueness of the w -factorization implies also that xρ is finite (hence closed) forevery x ∈ A − ω . However, lim n →∞ ( wa n , wb n ) = ( a − ω , b − ω ) / ∈ ρ. Since ( wa n , wb n ) ∈ ρ for every n ≥
1, then ρ is not closed.We denote by CRC( A − ω ) (respectively ORC( A − ω )) the set of all closed (respectively open) rightcongruences on A − ω .We consider CRC( A − ω ) (partially) ordered by inclusion. Similarly to [9, Section 5], we can relateCRC( A − ω ) with RG( A ).Given ρ ∈ RC( A − ω ), the Cayley graph Cay( ρ ) is the A -graph Cay( ρ ) = ( A − ω /ρ, E ) defined by E = { ( uρ, a, ( ua ) ρ ) | u ∈ A − ω , a ∈ A } . Lemma 5.2
Let ρ ∈ RC( A − ω ) .(i) For every x ∈ A − ω , there exists a left infinite path · · · x −→ xρ in Cay ( ρ ) .(ii) Cay( ρ ) is ( − ω ) -complete and ( − ω ) -trim. Proof . (i) Write x = · · · a a a with a i ∈ A . For every n ≥
1, write x n = · · · a n +2 a n +1 a n . Then · · · a −→ x ρ a −→ x ρ a −→ x ρ = xρ is a left infinite path in Cay( ρ ) labeled by x .(ii) By part (i). (cid:3) emma 5.3 Let ρ ∈ CRC( A − ω ) . Then:(i) If · · · x −→ q is a left infinite path in Cay( ρ ) , then q = xρ .(ii) Cay( ρ ) ∈ RG( A ) . Proof . (i) Assume that q = yρ with y ∈ A − ω . Write x = · · · a a a with a i ∈ A . For every n ≥ u n = a n · · · a . Then there exists some path y n ρ u n −→ yρ in Cay( ρ ) for some y n ∈ A − ω . Hence y n u n ∈ yρ . Since x = lim n →∞ u n = lim n →∞ y n u n and ρ closed implies yρ closed, we get x ∈ yρ , hence xρ = yρ = q .(ii) By part (i), Cay( ρ ) is ( − ω )-deterministic. By Lemma 5.2(ii), Cay( ρ ) is both ( − ω )-completeand ( − ω )-trim, therefore Cay( ρ ) ∈ RG( A ). (cid:3) We discuss next open right congruences, relating them in particular with the right congruenceson A k . Given x ∈ A − ω , let xξ k denote the suffix of length k of x . For σ ∈ RC( A k ), let b σ be therelation on A − ω defined by x b σy if ( xξ k ) σ ( yξ k ) . It is immediate that b σ ∈ RC( A − ω ).On the other hand, given ρ ∈ RC( A − ω ) and k ≥
1, we define a relation ρ ( k ) on A k by uρ ( k ) v if ( A − ω u × A − ω v ) ∩ ρ = ∅ . We denote by ρ [ k ] the transitive closure of ρ ( k ) .The next example shows that ρ ( k ) needs not to be transitive, even in the closed case. Example 5.4
Let A = { a, b } and let w be given by (5.1). For all x, y ∈ A − ω , let xρy if { x, y } = { wa u, wbau } for some u ∈ A ∗ or { x, y } = { wb av, wb v } for some v ∈ A ∗ or x = y. Then ρ ∈ CRC( A − ω ) but ρ (2) is not transitive. Indeed, by the uniqueness of the w -factorization remarked in Example 5.1, ρ turns out to betransitive and therefore a right congruence.We sketch the proof that ρ is closed. Let ( x, y ) ∈ ( A − ω × A − ω ) \ ρ . Then x = y . Write u = lcs( x, y ). We consider several cases:Case I: { x, y } = { zb au, z ′ b u } .Then either z = w or z ′ = w . We may assume that z = w . Let k ≥ w / ∈ B − k ( z ). Itis easy to see that B − k − −| u | (( x, y )) ∩ ρ = ∅ .Case II: { x, y } = { za u, z ′ bau } .Then either z = w or z ′ = w . We may assume that z = w . Let k ≥ w / ∈ B − k ( z ). Itis easy to see that B − k − −| u | (( x, y )) ∩ ρ = ∅ .Case III: all the remaining cases. 13t is easy to see that B − −| u | (( x, y )) ∩ ρ = ∅ .Therefore ρ is closed.Now ( wa , wba ) ∈ ρ yields ( a , ba ) ∈ ρ (2) , and ( wb a, wb ) ∈ ρ yields ( ba, b ) ∈ ρ (2) , However,( a , b ) / ∈ ρ (2) , hence ρ (2) is not transitive.The following lemma compiles some elementary properties of ρ ( k ) and ρ [ k ] . The proof is left tothe reader. Lemma 5.5
Let A be a finite nonempty alphabet, ρ ∈ RC( A − ω ) and k ≥ . Then:(i) ρ ( k ) ∈ RC( A k ) if and only if ρ ( k ) is transitive;(ii) ρ [ k ] ∈ RC( A k ) ;(iii) ρ ⊆ \ n ≥ d ρ [ n ] . We discuss next some alternative characterizations for open right congruences. We recall thedefinition of k -reset graph from [9, Section 3].We say that u ∈ A ∗ is a reset word for a deterministic and complete A -graph Γ = ( Q, E ) if | Qu | = 1. This is equivalent to say that all paths labeled by u end at the same vertex. Let Res(Γ)denote the set of all reset words for Γ.We say that Γ is a k -reset graph if A k ⊆ Res(Γ). We denote by RG k ( A ) the class of all stronglyconnected deterministic complete k -reset A -graphs. Proposition 5.6
Let A be a finite nonempty alphabet and ρ ∈ RC( A − ω ) . Then the following con-ditions are equivalent:(i) ρ is open;(ii) xρ is an open subset of A − ω for every x ∈ A − ω ;(iii) ρ = b σ for some σ ∈ RC( A k ) and k ≥ ;(iv) there exists some k ≥ such that ρ ( k ) is transitive and ρ = d ρ ( k ) ;(v) Cay( ρ ) ∈ RG k ( A ) for some k ≥ ;(vi) ρ is closed and has finite index. Proof . (i) ⇒ (ii). Let x ∈ A − ω . Since ( x, x ) ∈ ρ , there exists some δ > B δ (( x, x )) ⊆ ρ .Since B δ (( x, x )) = B δ ( x ) × B δ ( x ), we get B δ ( x ) ⊆ xρ and so xρ is open.(ii) ⇒ (vi). Let x, y ∈ A − ω be such that ( x, y ) / ∈ ρ . Since xρ and yρ are open, there exists some δ > B δ ( x ) ⊆ xρ and B δ ( y ) ⊆ yρ . If x ′ ∈ xρ and y ′ ∈ yρ , then ( x, y ) / ∈ ρ yields ( x ′ , y ′ ) / ∈ ρ .Hence ( B δ ( x ) × B δ ( y )) ∩ ρ = ∅ , and so B δ (( x, y )) ∩ ρ = ∅ . Thus the complement of ρ is open and so ρ is closed.On the other hand, { xρ | x ∈ A − ω } is an open cover of A − ω and so admits a finite subcover since A − ω is compact. Therefore ρ has finite index.(vi) ⇒ (v). By Lemma 5.3(ii), we have Cay( ρ ) ∈ RG( A ). Hence Cay( ρ ) is deterministic andcomplete by Proposition 4.1. 14et x, y ∈ A − ω . Since Cay( ρ ) is ( − ω )-trim, there exists a left infinite path · · · z −→ yρ in Cay( ρ ).Since ρ has finite index, we may factor this path as · · · z ′ −→ wρ u −→ wρ v −→ yρ with u = ε . On the other hand, since Cay( ρ ) is complete and ρ has finite index, there exist m ≥ p ≥ xρ u m −→ x ′ ρ u p −→ x ′ ρ in Cay( ρ ). It follows that there exist two paths · · · u − ω −→ wρ, · · · u − ω −→ x ′ ρ and so wρ = x ′ ρ since Cay( ρ ) is ( − ω )-deterministic. Thus there exists a path xρ u m −→ wρ v −→ yρ and so Cay( ρ ) is strongly connected.Suppose now that Cay( ρ ) / ∈ RG k ( A ) for every k ≥
1. Let P denote the set of pairs of distinctvertices in Cay( ρ ). Then ∀ k ≥ ∃ u k ∈ A k ∃ ( p, q ) ∈ P ∃ paths · · · u k −→ p, · · · u k −→ q in Cay( ρ ) . Since P is finite, one of the pairs ( p, q ) must repeat infinitely often. Hence there exists some ( p, q ) ∈ P such that ∀ k ≥ ∃ u k ∈ A ≥ k ∃ paths · · · u k −→ p, · · · u k −→ q in Cay( ρ ) . Since Ω A ( D ) is compact, we may replace ( u k ) k by some convergent subsequence. Let x = lim k →∞ u k .Since ( | u k | ) k is unbounded, we have x ∈ A − ω .Write p = x p ρ with x p ∈ A − ω . Since Cay( ρ ) is ( − ω )-trim, there exists some left infinite path · · · y k u k −−−→ x p ρ for some y k ∈ A − ω . By Lemma 5.2(i), there exists a path · · · y k u k −−−→ ( y k u k ) ρ in Cay( ρ ).Since Cay( ρ ) is ( − ω )-deterministic, we get ( y k u k ) ρ = x p ρ , hence y k u k ∈ x p ρ . Since ρ closed implies x p ρ closed and x = lim k →∞ u k = lim k →∞ y k u k , we get x ∈ x p ρ . By Lemma 5.2(i), there exists a path · · · x −→ xρ = x p ρ = p in Cay( ρ ). Similarly, thereexists some path · · · x −→ q . Since p = q , this contradicts Cay( ρ ) being ( − ω )-deterministic. ThereforeCay( ρ ) ∈ RG k ( A ) for some k ≥ ⇒ (iv). Assume that Cay( ρ ) ∈ RG k ( A ) for some k ≥
1. We show that xξ k = yξ k ⇒ xρy (5.2)holds for all x, y ∈ A − ω . Indeed, by Lemma 5.2(i), there exists left infinite paths · · · x −→ xρ, · · · y −→ yρ in Cay( ρ ). Since xξ k = yξ k ∈ A k ⊆ Res(Cay( ρ )), we get xρ = yρ and so (5.2) holds.15uppose now that u, v, w ∈ A k are such that uρ ( k ) vρ ( k ) w . Then there exist some x, y, y ′ , z ∈ A − ω such that ( xu ) ρ ( yv ) and ( y ′ v ) ρ ( zw ). Then ( yv ) ξ k = v = ( y ′ v ) ξ k and (5.2) yields ( yv ) ρ ( y ′ v ). Thus( xu ) ρ ( zw ) by transitivity and so uρ ( k ) w . Therefore ρ ( k ) is transitive.Now it follows from Lemma 5.5 that d ρ ( k ) is well defined and ρ ⊆ d ρ ( k ) .Conversely, let ( x, y ) ∈ d ρ ( k ) . Then ( xξ k , yξ k ) ∈ ρ ( k ) and so there exist x ′ , y ′ ∈ A − ω such that( x ′ ( xξ k ) , y ′ ( yξ k )) ∈ ρ . Since ( x ′ ( xξ k )) ξ k = xξ k , it follows from (5.2) that ( x ′ ( xξ k )) ρx . Similarly,( y ′ ( yξ k )) ρy and we get xρy by transitivity. Therefore d ρ ( k ) ⊆ ρ as required.(iv) ⇒ (iii). In view of Lemma 5.5(i).(iii) ⇒ (i). Let ( x, y ) ∈ ρ = b σ and let ( x ′ , y ′ ) ∈ B − k (( x, y )). Then x ′ ξ k = xξ k and y ′ ξ k = yξ k .Hence xρy ⇒ ( xξ k ) σ ( yξ k ) ⇒ ( x ′ ξ k ) σ ( y ′ ξ k ) ⇒ x ′ ρy ′ and so B − k (( x, y )) ⊆ ρ . Therefore ρ is open. (cid:3) The following example shows that closed is required in condition (vi).
Example 5.7
Let A = { a, b } and let ρ be the relation on A − ω defined by xρy if b occurs in both x, y or in none of them. Then ρ is a right congruence of index 2 on A − ω but it is not closed. Indeed, it is immediate that ρ is a right congruence of index 2. Since a − ω = lim n →∞ b − ω a n , ρ isnot closed.We say that ρ ∈ RC( A − ω ) is profinite if ρ is an intersection of open right congruences. Sinceopen right congruences are closed by Proposition 5.6, it follows that every profinite right congruence,being the intersection of closed sets, is itself closed. We denote by PRC( A − ω ) the set of all profiniteright congruences on A − ω .Given a graph Γ = ( Q, E ) and k ≥
1, we define a relation µ ( k )Γ on Q by pµ ( k )Γ q if there exist paths · · · u −→ p, · · · u −→ q in Γ for some u ∈ A k . Let µ [ k ]Γ denote the reflexive and transitive closure of µ ( k )Γ . Then µ [ k ]Γ is an equivalence relation on Q . Proposition 5.8
Let A be a finite nonempty alphabet and ρ ∈ RC( A − ω ) . Then the following con-ditions are equivalent:(i) ρ is profinite;(ii) ρ is an intersection of countably many open congruences;(iii) ρ = \ k ≥ c ρ [ k ] ;(iv) \ k ≥ µ [ k ]Cay( ρ ) = id . Proof . (i) ⇒ (iii). Assume that ρ = ∩ i ∈ I τ i with τ i ∈ ORC( A − ω ) for every i ∈ I .We have ρ ⊆ ∩ k ≥ c ρ [ k ] by Lemma 5.5(iii). To prove the opposite inclusion, we show that ∀ i ∈ I ∃ k ≥ c ρ [ k ] ⊆ τ i . (5.3)16ndeed, it follows from Proposition 5.6 that there exist some k ≥ σ i ∈ RC( A k ) such that τ i = b σ i . We claim that τ ( k ) i ⊆ σ i . (5.4)Assume that ( u, v ) ∈ τ ( k ) i . Then there exist x, y ∈ A − ω such that ( xu, yv ) ∈ τ i = b σ i . Hence( u, v ) = (( xu ) ξ k , ( yv ) ξ k ) ∈ σ i and (5.4) holds.Since ρ ⊆ τ i implies ρ ( k ) ⊆ τ ( k ) i , it follows that ρ ( k ) ⊆ σ i and so ρ [ k ] ⊆ σ i since σ i is transitive.Thus c ρ [ k ] ⊆ b σ i = τ i and (5.3) holds.Therefore \ k ≥ c ρ [ k ] ⊆ ∩ i ∈ I τ i = ρ as required.(iii) ⇒ (ii). By Lemma 5.5(ii), ρ [ k ] ∈ RC( A k ) for every k ≥
1, hence c ρ [ k ] is open by Proposition5.6 and we are done.(ii) ⇒ (i). Trivial.(iii) ⇒ (iv). Write µ ( k ) = µ ( k )Cay( ρ ) and µ [ k ] = µ [ k ]Cay( ρ ) . By Lemma 5.5(ii), ρ [ k ] ∈ RC( A k ) for every k ≥
1, hence c ρ [ k ] is open (and therefore closed) by Proposition 5.6. Therefore ρ is closed and soCay( ρ ) ∈ RG( A ) by Lemma 5.3(ii).Let x, y ∈ A − ω be such that xρ = yρ . Suppose that ( xρ, yρ ) ∈ µ [ k ] . Then there exist z , . . . , z n ∈ A − ω such that z = x , z n = y and ( z i − ρ, z i ρ ) ∈ µ ( k ) for i = 1 , . . . , n . For i = 1 , . . . , n , there existpaths z ′ i − ρ u i −→ z i − ρ, z ′′ i ρ u i −→ z i ρ in Cay( ρ ) for some u i ∈ A k and z ′ i − , z ′′ i ∈ A − ω .Hence z i − ρ = ( z ′ i − u i ) ρ and z i ρ = ( z ′′ i u i ) ρ , yielding( z i − ξ k ) ρ ( k ) u i ρ ( k ) ( z i ξ k )and so ( z i − ξ k ) ρ [ k ] ( z i ξ k ). Now ( xξ k ) ρ [ k ] ( yξ k ) follows by transitivity, hence ( x, y ) ∈ c ρ [ k ] . Since xρ = yρ implies ( x, y ) / ∈ d ρ [ m ] for some m ≥ xρ, yρ ) / ∈ µ [ m ] and so (iv)holds.(iv) ⇒ (iii). By Lemma 5.5(iii), we have ρ ⊆ ∩ k ≥ c ρ [ k ] . Conversely, let ( x, y ) ∈ ∩ k ≥ c ρ [ k ] . Foreach k , we have ( xξ k , yξ k ) ∈ ρ [ k ] , hence there exist u , . . . , u n ∈ A k such that u = xξ k , u n = yξ k and ( u i − , u i ) ∈ ρ ( k ) for i = 1 , . . . , n . For i = 1 , . . . , n , there exist z i − , z ′ i ∈ A − ω such that( z i − u i − , z ′ i u i ) ∈ ρ . Write also x = z ′ u and y = z n u n .By Lemma 5.2(i), there exist paths · · · z ′ i u i −−→ ( z ′ i u i ) ρ, · · · z i u i −−→ ( z i u i ) ρ in Cay( ρ ) for i = 0 , . . . , n , hence(( z i − u i − ) ρ, ( z i u i ) ρ ) = (( z ′ i u i ) ρ, ( z i u i ) ρ ) ∈ µ ( k ) . z u ) ρ, ( z n u n ) ρ ) ∈ µ [ k ] . Since (( z ′ u ) ρ, ( z u ) ρ ) ∈ µ ( k ) , we get( xρ, yρ ) = (( z ′ u ) ρ, ( z n u n ) ρ ) ∈ µ [ k ] . Since k is arbitrary, it follows from condition (iv) that xρ = yρ , hence ∩ k ≥ c ρ [ k ] ⊆ ρ as required. (cid:3) Every open right congruence on A − ω is trivially profinite and we remarked before that everyprofinite right congruence is necessarily closed. HenceORC( A − ω ) ⊆ PRC( A − ω ) ⊆ CRC( A − ω ) . We show next that these inclusions are strict if | A | > k ≥
1, let ρ k be the relation on A − ω defined by xρ k y if xξ k = yξ k . It is easy to check that ρ k ∈ ORC( A − ω ) for every k ≥
1. Since ∩ k ≥ ρ k = id , it follows that theidentity congruence is profinite, while it is clearly not open.To construct a closed non profinite right congruence is much harder. We do it through thefollowing example. Example 5.9
Let A = { a, b } . Given u, v ∈ A k , write u < v if u = u ′ aw and v = v ′ bw for some w ∈ A ∗ . Let u ( k )1 < · · · < u ( k )2 k be the elements of A k , totally ordered by < . Let p < p < · · · be theprime natural numbers. For every n ∈ N , let w n = . . . a p n ba p n ba p n b. Let ρ ∈ RC( A − ω ) be generated by the relation R = { ( w p ik u ( k ) i , w p ik u ( k ) i +1 ) | k ≥ , ≤ i < k } ∪ { ( b − ω a, a − ω b ) } . Then ρ is closed but not profinite. We start by showing that w p ik A ∗ ∩ w p i ′ k ′ A ∗ = ∅ implies ( k = k ′ and i = i ′ ) (5.5)for all k, k ′ , i, i ′ ≥
1. Indeed, suppose that w p ik u = w p i ′ k ′ v for some u, v ∈ A ∗ . By definition of w n , w p ik u has only finitely many factors of the form ba m b , and the leftmost must be ba pik b . Since w p ik u = w p i ′ k ′ v , we get ba pik b = ba pi ′ k ′ b and so p ik = p i ′ k ′ . Therefore k = k ′ and i = i ′ , and (5.5) holds.Write R ′ = { ( xu, yu ) | ( x, y ) ∈ R ∪ R − , u ∈ A ∗ } . Let x ∈ A − ω . We show thatthere exists at most one y ∈ A − ω such that ( x, y ) ∈ R ′ . (5.6)18his is obvious if x ∈ b − ω aA ∗ ∪ a − ω bA ∗ , hence we may assume that x ∈ w p ik A ∗ for some k ≥ ≤ i < k . In view of (5.5), we must have { x, y } = { w p ik u ( k ) i v, w p ik u ( k ) i +1 v } for some v ∈ A ∗ , and k, i are uniquely determined. Since w p ik / ∈ w p ik A + , also u ( k ) i , u ( k ) i +1 and v areuniquely determined. Thus (5.6) holds.Suppose that x R ′ y R ′ z with x = y = z . Since R ′ is symmetric, (5.6) yields x ′ = z ′ . It followsthat R ′ ∪ id is an equivalence relation, indeed the smallest right congruence containing R . It followsthat R ′ ∪ id = ρ. Moreover, each ρ -class contains at most two elements.We prove now that ρ is closed. Let ( x, y ) ∈ ( A − ω × A − ω ) \ ρ . Then x = y , hence we may assumewithout loss of generality that x = x ′ av and y = y ′ bv with v ∈ A ∗ . Let m = max { i ≥ | b i < s x ′ , a i < s y ′ } . Note that the above set is bounded, otherwise x ′ = b − ω and y ′ = a − ω , yielding( x, y ) = ( b − ω av, a − ω bv ) ∈ ρ, a contradiction. Write x ′ = x ′′ b m and y ′ = y ′′ a m .For j = 0 , . . . , | v | , write v = v j v ′ j with | v j | = j . Then a m bv j is the successor of b m av j in theordering of A m +1+ j , hence we may write b m av j = u ( m +1+ j ) i j , a m bv j = u ( m +1+ j ) i j +1 for some 1 ≤ i j < m +1+ j . It follows that x = x ′′ u ( m +1+ j ) i j v ′ j , y = y ′′ u ( m +1+ j ) i j +1 v ′ j (5.7)Let m j = min {| lcs( x ′′ , w p ijm +1+ j ) | , | lcs( y ′′ , w p ijm +1+ j ) |} . Note that m j is a well-defined natural number, otherwise x ′′ = w p ijm +1+ j = y ′′ and( x, y ) = ( w p ijm +1+ j u ( m +1+ j ) i j v ′ j , w p ijm +1+ j u ( m +1+ j ) i j +1 v ′ j ) ∈ ρ, a contradiction.Let p = max { m , . . . , m | v | } + m + 1 + | v | . We show that B − p (( x, y )) ∩ ρ = ∅ . (5.8)19uppose that ( z , z ) ∈ B − p (( x, y )) ∩ ρ . Since p > | v | , we have av < s z and bv < s z . Bymaximality of m , and since p > m + 1 + | v | , we have either ab m av < s z or ba m bv < s z . Hence wemust have z = w p ik u ( k ) i v ′ , z = w p ik u ( k ) i +1 v ′ (5.9)for some v ′ , k ≥ ≤ i < k . Clearly, | v ′ | ≤ | v | , hence we must have v ′ = v ′ j for j = | v | − | v ′ | .We have x = x ′′ b m av j v ′ j , hence b m av j v ′ j < s z . Similarly, y = y ′′ a m bv j v ′ j yields a m bv j v ′ j < s z .Suppose that k < m + 1 + j . Since | lcs( x, z ) | > m + 1 + | v | , it follows from (5.9) that w p ik endswith a b . Similarly, | lcs( y, z ) | > m + 1 + | v | implies that w p ik ends with an a , a contradiction. Hence k ≥ m + 1 + j .Suppose now that k > m + 1 + j . By maximality of m , we must have one of the following cases: • ab m av j ≤ s u ( k ) i and a m bv j ≤ s u ( k ) i +1 ; • b m av j ≤ s u ( k ) i and ba m bv j ≤ s u ( k ) i +1 .Any of these cases contradicts u ( k ) i +1 being the successor of u ( k ) i for the ordering of A k , hence k = m + 1 + j and we may write z = w p im +1+ j u ( m +1+ j ) i v ′ j , z = w p im +1+ j u ( m +1+ j ) i +1 v ′ j . Since d ( z , x ) < − m j − m − −| v ′ | = 2 − m j − m − − j −| v ′ j | , it follows from (5.7) that i = i j and | lcs( x ′′ , w p ijm +1+ j ) | > m j . Similarly, | lcs( y ′′ , w p ijm +1+ j ) | > m j , contradicting the definition of m j .Thus (5.8) holds and so ( A − ω × A − ω ) \ ρ is open. Therefore ρ is closed.Let k ≥
1. Since ( w p ik u ( k ) i , w p ik u ( k ) i +1 ) ∈ ρ , we have ( u ( k ) i , u ( k ) i +1 ) ∈ ρ ( k ) for every 1 ≤ i < k . Since u ( k )1 = a k and u ( k )2 k = b k , it follows that ( a k , b k ) ∈ ρ [ k ] and so ( a − ω , b − ω ) ∈ c ρ [ k ] . Since k is arbitrary,we get ( a − ω , b − ω ) ∈ \ k ≥ c ρ [ k ] . However, ( a − ω , b − ω ) / ∈ R ′ ∪ id = ρ, hence ρ = ∩ k ≥ c ρ [ k ] and so ρ is not profinite by Proposition 5.8. A − ω To avoid trivial cases, we assume throughout this section that A is a finite alphabet containing atleast two elements.Given P ⊆ A ∗ , we define a relation τ P on A − ω by: xτ P y if x = y or x, y ∈ A − ω u for some u ∈ P. Lemma 6.1
Let P ⊆ A ∗ . Then τ P is an equivalence relation on A − ω . roof . It is immediate that τ P is reflexive and symmetric. For transitivity, we may assume that x, y, z ∈ A − ω are distinct and x τ P y τ P z . Then there exist u, v ∈ P such that u < s x, y and v < s y, z .Since u and v are both suffixes of y , one of them is a suffix of the other. Hence either u < s x, z or v < s x, z . Therefore τ P is transitive. (cid:3) If we consider left ideals, being a right congruence turns out to be a special case:
Proposition 6.2
Let L E ℓ A ∗ . Then the following conditions are equivalent:(i) τ L ∈ RC( A − ω ) ;(ii) τ L ∈ PRC( A − ω ) ;(iii) L E A ∗ ;(iv) ( Lβ ℓ ) A ⊆ A ∗ ( Lβ ℓ ) .(v) Lβ ℓ is a semaphore code. Proof . (i) ⇒ (iv). We may assume that | A | >
1. Let u ∈ L and a ∈ A . Take b ∈ A \ { a } . Then( a − ω u, b − ω u ) ∈ τ L and by (i) we get ( a − ω ua, b − ω ua ) ∈ τ L . It follows that ua has some suffix in L ,hence LA ⊆ A ∗ L = L and so ( Lβ ℓ ) A ⊆ LA ⊆ L = A ∗ ( Lβ ℓ ) . (iv) ⇒ (iii). We have LA = A ∗ ( Lβ ℓ ) A ⊆ A ∗ ( Lβ ℓ ) = L. It follows that LA ∗ ⊆ L . Since L E ℓ A ∗ , we get L E A ∗ .(iii) ⇒ (ii). By Lemma 6.1, τ L is an equivalence relation. Let x, y ∈ A − ω be such that xτ L y . Wemay assume that there exists some u ∈ L such that u < s x, y . Since L E A ∗ , we have ua ∈ L and ua < s xa, ya yields ( xa, ya ) ∈ τ L . Therefore τ L ∈ RC( A − ω ).Let ( x, y ) ∈ ( A − ω × A − ω ) \ τ L . Then x = y . Let u = lcs( x, y ) and let m = | u | + 1. We claim that( xξ m ) τ [ m ] L = { xξ m } . (6.1)Indeed, suppose that ( xξ m , v ) ∈ τ ( m ) L and v = xξ m . Then there exist z, z ′ ∈ A − ω such that( z ( xξ m ) , z ′ v ) ∈ τ L . Since v = xξ m , then z ( xξ m ) and z ′ v must have a common suffix w ∈ L ,and | w | < m . But then w ≤ s u , yielding u ∈ L and xτ L y , a contradiction. Thus ( xξ m , v ) ∈ τ ( m ) L implies v = xξ m , and so (6.1) holds.Suppose that ( x, y ) ∈ d τ [ m ] L . Then ( xξ m , yξ m ) ∈ τ [ m ] L , hence xξ m = yξ m by (6.1), contradicting m > | u | . Thus ( x, y ) / ∈ d τ [ m ] L and so ∩ k ≥ d τ [ m ] L ⊆ τ L . Hence τ L = ∩ k ≥ d τ [ m ] L by Lemma 5.5(iii), and so τ L is profinite by Proposition 5.8.(ii) ⇒ (i). Trivial.(iv) ⇔ (v). By Lemma [9, Lemma 4.1], since Lβ ℓ is always a suffix code. (cid:3)
21e say that ρ ∈ RC( A − ω ) is a special right congruence on A − ω if ρ = τ I for some I E A ∗ . Inview of Proposition 6.2, this is equivalent to say that ρ = τ S for some semaphore code S on A . Wedenote by SRC( A − ω ) the set of all special right congruences on A − ω .The next result characterizes the open special right congruences. Recall that a suffix code S ⊂ A ∗ is said to be maximal if S ∪ { u } fails to be a suffix code for every u ∈ A ∗ \ S . Proposition 6.3
Let I E A ∗ . Then the following conditions are equivalent:(i) τ I ∈ ORC( A − ω ) ;(ii) Iβ ℓ is a finite maximal suffix code;(iii) A ∗ \ I is finite. Proof . (i) ⇒ (ii). Let u, v ∈ Iβ ℓ be distinct. Then(( A − ω u ) × ( A − ω v )) ∩ τ I = ∅ . Since τ I has finite index by Proposition 5.6, it follows that the suffix code Iβ ℓ is finite.Suppose now that Iβ ℓ ∪ { u } is a suffix code for some u ∈ A ∗ \ ( Iβ ℓ ). It is easy to see that notwo elements of A − ω u are τ I equivalent, a contradiction since τ I has finite index. Therefore Iβ ℓ is amaximal suffix code.(ii) ⇒ (iii). Let m denote the maximum length of the words in Iβ ℓ . Suppose that v ∈ A ∗ \ I has length > m . It is straightforward to check that Iβ ℓ ∪ { v } is a suffix code, contradicting themaximality of Iβ ℓ . Thus A ∗ \ I ⊆ A ≤ m and is therefore finite.(iii) ⇒ (i). We have τ I ∈ RC( A − ω ) by Proposition 6.2.Let m ≥ A ∗ \ I ⊆ A ≤ m . Then xξ m +1 = yξ m +1 ⇒ xτ I y holds for all x, y ∈ A − ω and so τ I has finite index. Since τ I is profinite (and therefore closed) byProposition 6.2, it follows from Proposition 5.6 that τ I is open. (cid:3) The proof of [9, Lemma 7.4] can be adapted to show that inclusion among left ideals determinesinclusion for the equivalence relations τ L : Lemma 6.4
Let | A | > and L, L ′ E ℓ A ∗ . Then τ L ⊆ τ L ′ ⇔ L ⊆ L ′ . Note that Lemma 6.4 does not hold for | A | = 1, since | A − ω | = 1.Similarly, we adapt [9, Proposition 7.6]: Proposition 6.5
Let | A | > . Then:(i) τ I ∩ J = τ I ∩ τ J and τ I ∪ J = τ I ∪ τ J for all I, J E A ∗ ;(ii) SRC( A − ω ) is a full sublattice of RC( A − ω ) ;(iii) the mapping I ( A ) → SRC( A − ω ) I τ I is a lattice isomorphism. ρ ∈ RC( A − ω ) and C ∈ A − ω /ρ , we say that C is nonsingular if | C | >
1. If C is nonsingular,we denote by lcs( C ) the longest common suffix of all words in C . We define • Λ ρ = { lcs( C ) | C ∈ A − ω /ρ is nonsingular } , • Λ ′ ρ = { lcs( x, y ) | ( x, y ) ∈ ρ, x = y } . Lemma 6.6
Let ρ ∈ RC( A − ω ) . Then:(i) A ∗ Λ ρ = A ∗ Λ ′ ρ ;(ii) Λ ′ ρ E r A ∗ ;(iii) A ∗ Λ ′ ρ E A ∗ . Proof . (i) Let C ∈ A − ω /ρ be nonsingular and let w = lcs( C ). By maximality of w there exist a, b ∈ A distinct and x, y ∈ A − ω such that xaw, ybw ∈ C . Thus w = lcs( xaw, ybw ) and soΛ ρ ⊆ Λ ′ ρ . (6.2)Therefore A ∗ Λ ρ ⊆ A ∗ Λ ′ ρ .Conversely, let ( x, y ) ∈ ρ with x = y . Then xρ is nonsingular and lcs( xρ ) is a suffix of lcs( x, y ),hence Λ ′ ρ ⊆ A ∗ Λ ρ and so A ∗ Λ ρ = A ∗ Λ ′ ρ .(ii) Let u ∈ Λ ′ ρ and a ∈ A . Then u = lcs( x, y ) for some ( x, y ) ∈ ρ with x = y . Then ( xa, ya ) ∈ ρ .Since lcs( xa, ya ) = ua , we get ua ∈ Λ ′ ρ . Therefore Λ ′ ρ E r A ∗ .(iii) Clearly, A ∗ Λ ′ ρ E ℓ A ∗ . Now we use part (ii). (cid:3) Given ρ ∈ RC( A − ω ), we write Res( ρ ) = Res(Cay( ρ )) . We refer to the elements of Res( ρ ) as the resets of ρ . Lemma 6.7
Let ρ ∈ RC( A − ω ) . Then:(i) Res( ρ ) E A ∗ ;(ii) if ρ is closed, then Res( ρ ) = { w ∈ A ∗ | ( xw, yw ) ∈ ρ for all x, y ∈ A − ω } . Proof . (i) Immediate.(ii) Let w ∈ Res( ρ ) and x, y ∈ A − ω . By Lemma 5.2(i), there exist paths · · · xw −→ ( xw ) ρ, · · · yw −→ ( yw ) ρ in Cay( ρ ). Now w ∈ Res( ρ ) yields ( xw ) ρ = ( yw ) ρ .Now let w ∈ A ∗ \ Res( ρ ). Then there exist paths p w −→ q and p ′ w −→ q ′ in Cay( ρ ) with q = q ′ . SinceCay( ρ ) is ( − ω )-trim by Lemma 5.2(ii), there exist left infinite paths · · · x −→ p, · · · y −→ p ′ in Cay( ρ ), hence paths · · · xw −→ q, · · · yw −→ q ′ . Since ρ is closed, it follows from Lemma 5.3(i) that ( xw ) ρ = q = q ′ = ( yw ) ρ and we are done. (cid:3) Proposition 6.8
Let | A | > , ρ ∈ RC( A − ω ) and I E A ∗ . Then:(i) ρ ⊆ τ I ⇔ Λ ρ ⊆ I ⇔ Λ ′ ρ ⊆ I ;(ii) if ρ is closed, then τ I ⊆ ρ ⇔ I ⊆ Res( ρ ) . Given R ⊆ A − ω × A − ω , we denote by R ♯ the right congruence on A − ω generated by R , i.e. theintersection of all right congruences on A − ω containing R .Given ρ ∈ RC( A − ω ), we denote by NS( ρ ) the set of all nonsingular ρ -classes.We can now prove several equivalent characterizations of special right congruences. Proposition 6.9
Let | A | > and ρ ∈ RC( A − ω ) . Then the following conditions are equivalent:(i) ρ ∈ SRC( A − ω ) ;(ii) lcs : NS( ρ ) → A ∗ is injective, Λ ρ is a suffix code and ∀ x ∈ A − ω ∀ w ∈ Λ ρ ( xw ) ρ ∈ NS( ρ ); (6.3) (iii) ρ = τ A ∗ Λ ρ ;(iv) ρ = τ A ∗ Λ ′ ρ ;(v) ρ = τ ♯L for some L E ℓ A ∗ . Proof . (i) ⇒ (ii). By a straightforward adaptation of the proof of (i) ⇒ (ii) in [9, Proposition 7.10],we check that lcs : NS( ρ ) → A ∗ is injective and Λ ρ is a suffix code.Now let x ∈ A − ω and w ∈ Λ ρ . Then w = lcs( yρ ) for some yρ ∈ NS( ρ ). By (6.2), we maywrite w = lcs( y ′ , y ′′ ) for some y ′ , y ′′ ∈ yρ distinct. Since ρ = τ I , it follows that w ∈ I , hence( xw, y ) , ( xw, y ′ ) ∈ τ I = ρ . Since y ′ = y ′′ , it follows that either xw = y ′ or xw = y ′′ , so in any case( xw ) ρ ∈ NS( ρ ) as required.(ii) ⇒ (iii). Write I = A ∗ Λ ρ . If ( x, y ) ∈ ρ and x = y , then lcs( xρ ) ∈ Λ ρ ⊆ I is a suffix of both x and y , hence ( x, y ) ∈ τ I . Thus ρ ⊆ τ I .Conversely, let ( x, y ) ∈ τ I . We may assume that x = y , hence there exists some w ∈ Λ ρ such that w < s x, y . Hence (6.3) yields xρ, yρ ∈ NS( ρ ).Suppose that lcs( xρ ) = w . Then lcs( xρ ) < s w or w < s lcs( xρ ), contradicting Λ ρ being a suffixcode. Hence lcs( xρ ) = w . Similarly, lcs( yρ ) = w . Since lcs : NS( ρ ) → A ∗ is injective, we get xρ = yρ .Thus ρ = τ I .(iii) ⇔ (iv). By Lemma 6.6(i).(iii) ⇒ (v). Write L = A ∗ Λ ρ . By (iii), we have τ ♯L = ρ ♯ = ρ . Since L E A ∗ by Lemma 6.6, (iv)holds.(v) ⇒ (i). Let I = LA ∗ E A ∗ . Since L ⊆ I , it follows from Lemma 6.4 that τ L ⊆ τ I , hence ρ = τ ♯L ⊆ τ ♯I = τ I by Proposition 6.2.Conversely, let ( x, y ) ∈ τ I . We may assume that x = y . Then there exist factorizations x = x ′ w and y = y ′ w with w ∈ I . Write w = zw ′ with z ∈ L . Then ( x ′ z, y ′ z ) ∈ τ L and so( x, y ) = ( x ′ w, z ′ w ) = ( x ′ zw ′ , y ′ zw ′ ) ∈ τ ♯L = ρ. Thus τ I ⊆ ρ as required. (cid:3) roposition 6.10 Let | A | > and ρ ∈ CRC( A − ω ) . Then the following conditions are equivalent:(i) ρ ∈ SRC( A − ω ) ;(ii) lcs : NS( ρ ) → A ∗ is injective, Λ ρ is a suffix code and ∀ x ∈ A − ω ∀ w ∈ Λ ρ ( xw ) ρ ∈ NS( ρ ); (iii) ρ = τ A ∗ Λ ρ ;(iv) ρ = τ A ∗ Λ ′ ρ ;(v) ρ = τ ♯L for some L E ℓ A ∗ ;(vi) ρ = τ Res( ρ ) ;(vii) Λ ρ ⊆ Res( ρ ) ;(viii) Λ ′ ρ ⊆ Res( ρ ) ;(ix) whenever p aw −→ q, p ′ bw −→ q, p ′′ w −→ r (6.4) are paths in Cay( ρ ) with a, b ∈ A distinct, then q = r . Proof . (i) ⇔ (ii) ⇔ (iii) ⇔ (iv) ⇔ (v). By Proposition 6.9.(i) ⇒ (vi). If ρ = τ I for some I E A ∗ , then I ⊆ Res( ρ ) by Proposition 6.8(ii). Since Res( ρ ) E A ∗ by Lemma 6.7(i), then Proposition 6.8(ii) also yields τ Res( ρ ) ⊆ ρ = τ I , hence Res( ρ ) ⊆ I by Lemma 6.4. Therefore I = Res( ρ ).(vi) ⇒ (vii) ⇔ (viii). By Lemma 6.7(i), Res( ρ ) E A ∗ . Now we apply Proposition 6.8(i).(viii) ⇒ (i). We have A ∗ Λ ′ ρ , Res( ρ ) E A ∗ by Lemmas 6.6(iii) and 6.7(i). It follows from Proposition6.8 that τ Res( ρ ) ⊆ ρ ⊆ τ A ∗ Λ ′ ρ . Since Λ ′ ρ ⊆ Res( ρ ) yields A ∗ Λ ′ ρ ⊆ Res( ρ ) and therefore τ A ∗ Λ ′ ρ ⊆ τ Res( ρ ) by Lemma 6.4, we get ρ = τ Res( ρ ) ∈ SRC( A − ω ).(viii) ⇒ (ix). Consider the paths in (6.4). By Lemma 5.2(ii), there exist left infinite paths · · · x −→ p, · · · x ′ −→ p ′ in Cay( ρ ), hence ( xaw, x ′ bw ) ∈ ρ by Lemma 5.3(i) and so w = lcs( xaw, x ′ bw ) ∈ Λ ′ ρ ⊆ Res( ρ ) . Thus q = r as required.(ix) ⇒ (viii). Let w ∈ Λ ′ ρ . Then w = lcs( x, y ) for some ( x, y ) ∈ ρ such that x = y . We may write x = x ′ aw and y = y ′ bw with a, b ∈ A distinct. By Lemma 5.2(i), there exist in Cay( ρ ) paths of theform · · · x ′ −→ p aw −→ xρ, · · · y ′ −→ p ′ bw −→ yρ = xρ. Now (ix) implies that w ∈ Res( ρ ). (cid:3)
25e can now prove that not all open right congruences are special, even for | A | = 2: Example 6.11
Let A = { a, b } and let σ be the equivalence relation on A defined by the followingpartition: { a , aba, ba } ∪ { bab, a b } ∪ { ab } ∪ { b a } ∪ { b } . Then b σ ∈ ORC( A ) \ SRC( A ) . Indeed, it is routine to check that σ ∈ RC( A ), hence ρ = b σ ∈ ORC( A − ω ) by Proposition 5.6.Since lcs( a − ω ρ ) = a and lcs(( b − ω a ) ρ ) = b a , then Λ ρ is not a suffix code and so ρ / ∈ SRC( A − ω ) byProposition 6.9.Let ρ ∈ RC( A − ω ) and let ρ = ∨{ τ ∈ SRC( A − ω ) | τ ⊆ ρ } ,ρ = ∧{ τ ∈ SRC( A − ω ) | τ ⊇ ρ } . By Proposition 6.5(ii), we have ρ, ρ ∈ SRC( A − ω ). Proposition 6.12
Let | A | > and ρ ∈ CRC( A − ω ) . Then:(i) ρ = τ Res( ρ ) ;(ii) ρ = τ A ∗ Λ ρ = τ A ∗ Λ ′ ρ . Proof . (i) By Lemma 6.7(i), we have Res( ρ ) E A ∗ . Now the claim follows from Proposition 6.8(ii).(ii) Similarly, we have A ∗ Λ ρ = A ∗ Λ ′ ρ E A ∗ by Lemma 6.6(iii), and the claim follows from Propo-sition 6.8(i). (cid:3) The straightforward adaptation of [9, Example 7.14] shows that the pair ( ρ, ρ ) does not univocallydetermine ρ ∈ RC( A − ω ), even in the open case: Example 6.13
Let A = { a, b } and let σ, σ ′ be the equivalence relations on A defined by the followingpartitions: { a , aba, ba } ∪ { bab, a b } ∪ { ab } ∪ { b a } ∪ { b } , { a , b a, ba } ∪ { bab, a b } ∪ { ab } ∪ { aba } ∪ { b } . Let ρ = b σ and ρ ′ = b σ ′ . Then ρ, ρ ′ ∈ ORC( A − ω ) , ρ = ρ ′ and ρ = ρ ′ . This same example shows also that ρ does not necessarily equal or cover ρ in SRC( A − ω ). Indeed,in this case we have Res( ρ ) = A ∗ A ∪ { a , ab } ⊂ I ⊂ A + \ { b, b } = A ∗ Λ ρ for I = A ∗ A ∪ { a , ab, ba } E A ∗ . By Lemma 6.4, we get ρ ⊂ τ I ⊂ ρ. Conclusion and future work
We enter now into random walks on infinite semigroups. The most sophisticated approach is describedin [3]. We use profinite limits (see [8]) as an alternative approach, as developed in Sections 3-6.Indeed, if I , I , . . . is a sequence of ideals in A ∗ with I = ∩ k ≥ I k , let J k β ℓ the semaphorecode determined by the ideal J k = I ∩ . . . ∩ I k . Whenever k ≥ m , we may define a mapping ϕ km : J k β ℓ → J m β ℓ by setting uϕ km to be the unique suffix of u in J m β ℓ . It is routine to check that: • ϕ km is onto; • ϕ km preserves the action of A ∗ on the right; • ( ϕ km ) constitutes a projective system of surjective morphisms with respect to this action; • Iβ ℓ is the projective limit of this system.In view of (2.2), each Turing machine T provides an instance of this setting when I k = RRes k ( T )and I = RRes( T ). Moreover, each ideal RRes k ( T ) is cofinite and τ RRes ( T ) is a profinite congruenceon A − ω , indeed the intersection of the open congruences τ RRes k ( T ) .Using the left-right duals of Sections 3-6, we have similar results for LRes( T ) and the sequence(LRes k ( T )).In a subsequent paper, we intend to characterize polynomial time Turing machines in this frame-work, including the natural semaphore codes action and the action of β ( n ) and β ( ∞ ) . The approachwill constitute a variation of [10]: we will need to consider certain metrics that will give the sametopology as in Sections 3-6, but conditions involving the metrics will take us from the realm oftopology into that of geometry. References [1] J. Berstel,
Transductions and context-free languages , Teubner, Stuttgart, 1979. 10[2] J. Berstel, D. Perrin and C. Reutenauer,
Codes and automata , Encyclopedia of Mathematicsand its Applications 129, Cambridge University Press, Cambridge, 2010. 1[3] G. Hognas and A. Mukherjea,
Probability measures on semigroups , Springer Series in Probabilityand its Applications, Springer, 2011. 27[4] J. E. Hopcroft and J. D. Ullman,
Introduction to automata theory, languages, and computation ,Addison-Wesley, 1979. 2[5] K. Krohn and J. Rhodes,
Algebraic theory of machines. I. Prime decomposition theorem forfinite semigroups and machines , Trans. Amer. Math. Soc. (1965) 450–464. 1[6] J. Rhodes,
Applications of automata theory and algebra. Via the mathematical theory of com-plexity to biology, physics, psychology, philosophy, and games , With an editorial preface byChrystopher L. Nehaniv and a foreword by Morris W. Hirsch. World Scientific Publishing Co.Pte. Ltd., Hackensack, NJ, 2010. xviii+274 pp. 1277] J. Rhodes and P. V. Silva,
Turing machines and bimachines , Theoret. Comput. Sci. (2008),no. 1-3, 182–224. 2[8] J. Rhodes and B. Steinberg,
The q -theory of finite semigroups , Springer Monographs in Math-ematics, Springer, 2009. 1, 9, 12, 27[9] J.Rhodes, A. Schilling and P. V. Silva, Random walks on semaphore codes and delay de Bruijnsemigroups , preprint arXiv:150903383. 1, 2, 5, 11, 12, 14, 21, 22, 24, 26[10] J. Rhodes and P. Weil,
Algebraic and topological theory of languages , RAIRO Theoret. Infor-matics Appl.29