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A MEALY MACHINE WITH POLYNOMIAL GROWTH OF IRRATIONAL DEGREE 7 0 0 2 LAURENTBARTHOLDIANDILLYAI.REZNYKOV n a Abstract. Weconsider averysimpleMealymachine(two non-trivialstates J over a two-symbol alphabet), and derive some properties of the semigroup it 2 generates. It is an infinite, finitely generated semigroup, and we show that 1 the growth function of its balls behaves asymptotically like nα, for α = 1+ log2/log1+√5;thatthesemigroupsatisfiestheidentityg6=g4;andthatits 2 ] latticeoftwo-sidedidealsisachain. R G . h 1. Introduction t a Algebraicobjectsmaybe definedbyuniversalproperties,likethe identitiesthey m satisfy,orapresentationasaquotientoftwofreeobjects. Theymayalsobedefined [ bytheiractiononsimplerobjects,viz,foranalgebra,asendomorphismsofavector 2 space;for a group,as permutations of a set, andfor a semigroup,as mappings of a v set. 3 Ifthe semigrouptobe definedis infinite, itnaturallymustactonaninfinite set; 0 extraconditionsmustbeimposedontheactiontoensurethatitremainsdescribable 2 6 (byafinitesequenceofmathematicalsymbols,say). Aspeciallyinterestingclassof 0 semigroups appear by requesting that the infinite set be the set X of words over ∗ 5 a finite alphabet, and that the action be given by a finite-state automaton. 0 The growth function γ(ℓ) of a finitely generated semigroup Γ — the number of / h semigroupelements that can be obtained as products of at most ℓ generators — is t an important invariant of the semigroup. It depends on the chosen generating set, a m but its asymptotics do not. This function is at most exponential. If it is bounded by a polynomial, then Γ is of polynomial growth; following [6], its Gelfand-Kirillov : v dimension (abbreviated GK dimension) is then defined as the infimum of those α Xi such that γ(ℓ)/ℓα does not converge to 0. The “Bergman gap” [8, Theorem 2.5] asserts that a semigroup may have GK r a dimension 0, 1, or 2; by a result by Warfield [11], there exist semigroups of GK ≥ dimensionβ foranyβ 2. BelovandIvanov[3,4]constructfinitelypresentedsemi- ≥ groups with non-integer GK dimension. Shneerson [10] constructs relatively free (i.e.,freerelativetoanidentity)semigroupsofintermediategrowth(asymptotically mlogm). His proof involves Fibonacci numbers, as does the present construction. This paper describes the semigroup Γ(I) generated by the 3-state automaton I in Figure 1 (see 3 for the definition of a semigroup generated by an automaton), § and proves that its GK dimension is irrational. A much more precise statement appears in Theorem 2.4. The main purpose of this paper is to show that a very simple-minded finite- state automaton can produce a semigroup with a highly unusual growth pattern, Date:4January2007; compiledFebruary1,2008. 1 2 LAURENTBARTHOLDIANDILLYAI.REZNYKOV 0 1 0 0 → → 1 0 %%/(.)f-*,+ 0→0 //' &!s%"$# 99%%' &!e%"$#mmqq → 1 0 1 1 → → Figure 1. The automaton I asymptotically nα for an irrational α. Other exotic types of growth of automata havealsobeendiscovered;forexample,oftypee√n in[2],andoftypenlog(n)/2logm forinteger m,in [9]. It is intriguingthat inallthese cases the determinationofthe growth function relies on enumerations of partitions with certain constraints. 1.1. Plan. Thenextsectiongivesanautomata-freeversionofthe semigroupstud- ied in this paper, along with a presentation of its main results. Section 3 gives the necessary definitions of semigroups generated by automata. Statements whose proofaretoolongtofitsmoothlyinthetextarestatedasunderlined. Theirproofs appear in Section 5. 1.2. Notation. All actions are written on the right in this paper. The identity is written e. We use = for equality of group elements, and for graphical (letter- ≡ by-letter)equalityof wordsrepresentingthese elements. We denote by X the free ∗ monoid on the set X. The length of a word w is denoted by w ; this is also the k k length of a minimal word representing a semigroup element. The integers are written Z, and the naturals (containing 0) are written N. Con- gruence is written , and the ‘mod 2’ operation (remainder after division by 2) is ≡ written n%2 as in the C programming language. Sequences of the form f f ...f , sometimes simply written f ...f , appear a a+2 b a b throughout the paper. They are taken to be e if a > b, and f f f ...f f a a+2 a+4 b 2 b − otherwise. 1.3. Acknowledgments. Weareverygratefultotheanonymousrefereewhosug- gested substantial improvements to the paper, in particular a clarification of the role of rewriting systems. 2. Main results Consider the following transformations s,f of the integers Z: (1) x 1 if x is odd, xs = − (x+1 if x is even; x 2n 1 if there exists n 0 such that x 3 2n (mod 2n+2), − − ≥ ≡ · xf = x+3 2n 1 if there exists n 0 such that x 2n (mod 2n+2),  · − ≥ ≡  1 if x=0. − Note that f is uniquely defined, because every x Z is either 0 or of the form 2ny ∈ for unique n 0 and odd y, with either y 1 or y 3 modulo 4. ≥ ≡ ≡ A MEALY MACHINE WITH POLYNOMIAL GROWTH OF IRRATIONAL DEGREE 3 2.1. The semigroup. LetF bethesemigroupgeneratedbythetransformationss andf. Define furthermorethe elements f of F by f =s, f =f, andinductively n 1 2 f = f f for n 3. For example, f = sf, f = fsf, f = sf2sf. These n n 2 n 1 3 4 5 − − ≥ wordsaresometimescalledthe“Fibonaccisequence”—see 2.2fortheconnection. § Recall [5] that a rewriting system for a semigroup Γ generated by a set Q is a set of equations, called rules, of the form ℓ r, with ℓ,r Q . An elementary ∗ → ∈ reduction in a wordw Q is the replacementof a subwordequal to the left-hand ∗ ∈ sideofarulebytheright-handside;ifnoelementaryreductionispossible,theword is reduced. A rewriting system is terminating if there is no w Q to which an ∗ ∈ infinite sequence of elementary reductions can be applied; andit is confluent if the reduced word obtained after applying as many elementary reductions as possible does not depend on the choice of elementary reductions. It is complete if it is terminating and confluent; the set of reduced words is then in bijection with the semigroup, through the natural evaluation map Q Γ, and is called a normal ∗ → form for Γ. For n 1, define words r and r over the alphabet s,f as follows: ≥ n n′ { } r =s2, r =e, 1 1′ (2) r =f f2, n n+1 n (3) r =f (f f f f )f . n′ n%2+1 n%2+5 n%2+7··· n−1 n+1 n In particular, r = sf3 and r = sf, and r = f f2 and r = f . For n 3, we 2 2′ 3 4 3 3′ 4 ≥ have r = r 4. k n′k k k− Theorem 2.1. The semigroup F is infinite, and admits as presentation s,f r =r for all n 1 , h | n n′ ≥ i Furthermore, after the relations sw=sw (which occur for even n) are replaced by ′ the equivalent relations w =w , this presentation is a complete rewriting system. ′ 2.2. A normal form. Let Φ denote the sequence of Fibonacci numbers, defined n by Φ =Φ =1 and Φ =Φ +Φ for n 3. Then 1 2 n n 2 n 1 − − ≥ Theorem 2.2. Every element g of F admits a unique representation as a word of the form (4) w =sǫf f f f , g i1 i2··· im··· in for some n 0, some ǫ 0,1 , and some indices i ...,i satisfying 1 n ≥ ∈{ } 3 i ,i +1<i ,...,i +1<i >i > >i 1. 1 1 2 m 1 m m+1 n ≤ − ··· ≥ We call i the maximal index of w . m g When spelled out in the generating set s,f , this representation of g is essen- { } tially minimal: if i is even, then w is the unique minimal-length representation 1 g of g, while if i is odd, then an initial s2 must be cancelled to obtain the unique 1 minimal representation of g. Its length (see 3.3) is § g =( 1)i1ǫ+Φ + +Φ . k k − i1 ··· in By a slight extension of the definition of rewriting system, let us admit rules of the form ℓ r that mean that the subword ℓ may be replaced in w by the word ∧ → r if it is a prefix of w. We will actually consider the following rewriting system; it is onaninfinite generatingset,but its rules aremuchsimpler,since their left-hand sides have length at most 3. 4 LAURENTBARTHOLDIANDILLYAI.REZNYKOV Theorem 2.3. On the generating set f : i 1 , the semigroup F admits a i { ≥ } complete rewriting system with rules (N1) f2 e 1 −→ (N2) f f f (a 1) a a+1 a+2 −→ ≥ (N3) f2 f f (a 3) a −→ a−2 a+1 ≥ (N4) f f2 f (a 2) a 2 −→ a ≥ (N5) f f f 2 1 3 ∧ −→ (N6) f f f f f (a 1) a+1 a a+3 a+3 a+2 −→ ≥ (N7) f f f f f f (a 1) a+2 a a+3 a a+3 a+2 −→ ≥ (N8) f f f f f f2 f2 f f a+p a a+q −→ a+p−2··· a 2 ··· a−2 a−1 a+q p 1terms − (a 2,p 1,q 2 even) | {z } ≥ ≥ ≥ (N9) f f f f f f2 f2f (a 1,p 3,q 3 odd) a+p a a+q −→ a+p−2··· a+2 2 ··· a a+q ≥ ≥ ≥ p 3terms − (N10) f f f f f f f (p 1,q 2 even). 1+p 1 1+q |1+p 2{z4 q} −→ ··· ≥ ≥ From now on, we identify g with w , and call the word w the normal form of g g g. We will never use a different notation for a semigroup element and its normal form. 2.3. Growth. Using this minimal representation,the (ball) growth function of F, γ(ℓ)=# g F : g ℓ , { ∈ k k≤ } may be quite precisely estimated; namely Theorem 2.4. There are constants C,D > 0 such that the growth function of F satisfies Cℓα γ(ℓ) Dℓα ≤ ≤ for all ℓ N, where ϕ = (1 + √5)/2, and α = 1+ log2/logϕ 2.4401, and C = ∈2√5α , and D = 2√5α . ≈ √5ϕ2(2ϕ 1)(2ϕ)α √5ϕ2(2ϕ 1) Therefore,−the Gelfand-Kirillov dime−nsion of F is α. Experimentalcomputationsindicate thatactuallythe functionγ(ℓ)/ℓα doesnot converge, but oscillates between 0.201 and 0.205, reaching its maxima at ≈ ≈ Fibonacci numbers. 2.4. Identities. Wehavenotdeterminedthecompletesetofidentitiessatisfiedby F; nevertheless, we know Theorem 2.5. The semigroup F satisfies the identity g6 =g4. 2.5. The ideal structure of F. We describe in this subsection the quotient and idealstructureofF. First,wemayconsiderforalln NthequotientofF,denoted W , acting as transformations on Xn. Let us denote∈, for n N, n ∈ f f f if n is odd, 3 5 n+2 z = ··· n (f4f6 fn+2 if n is even. ··· Then we have A MEALY MACHINE WITH POLYNOMIAL GROWTH OF IRRATIONAL DEGREE 5 Theorem 2.6. (1) The semigroup W is presented as follows: n Wn =hs,f rk =rk′,1≤k ≤n+2, and szn =fzn =zn,fn+2fn+1 =zni. (2) The elements of W may be described as all normal forms f f f n 1··· im··· it whosemaximalindexi is<n+2,normalformsoftheformz f f m n im−1··· it for n i > > i , and those normal forms of maximal index n+2 m 1 t ≥ − ··· that include neither z nor f f . n n+2 n+1 (3) The semigroup W has order n 2k+2Φ 2n 2n+1Φ +2. n k=1 k− − n We identify for the remainder of thPis section the quotient semigroup Wn with the set of normal forms of its elements, which then form a subset of F. Corollary 2.7. The Hausdorff dimension (in the sense of 3.6) of F is 0. § Proof. Using the count of elements in W , we have n n 4 (2ϕ)n+1 1 #W 2+ 2k+2Φ − (2ϕ)n+1, n k ≤ ≤ √5 2ϕ 1 ≤ k=1 − X so log#W log(2ϕ)n+1 Hdim(F)=liminf n =liminf =0. (cid:3) n log#End(Xn) n log22(2n 1) →∞ →∞ − Corollary 2.8. Let g be an arbitrary element of F with maximal index n 3. ≥ Then its trace (in the sense of 3.5) satisfies § 22 n if g includes z or f f in its normal form, − n 2 n n 1 τ(g)= − − (23−n otherwise. The ideal In =Ffn+3F coincides with F2−n, as defined in 3.5. § Theorem 2.9. All two-sided ideals of F are of the form Ff F, and equivalently n of the form F as defined in 3.5. They therefore form a unique chain. ξ § RecallthatifI isanidealofF,thequotientF/I isthesemigroupwhoseelements are the equivalence classes of the congruence I I ∆, where ∆ is the identity × ∪ relation (= diagonal) of F. Proposition2.10. For n 2, the quotient F/I has order n 2k+2Φ 2n+1 ≥ n k=1 k− − 2n+1Φ +3. It is obtained from W by identifying together all idempotents of rank n n P 1, i.e. all maps f :Xn Xn defined by f (x ...x )=p, for all p Xn. p p 1 n → ∈ Proof. ByTheorem2.6,thequotientF/I isaquotientofW . Thetransformation n n f induces a transformation of rank 1 on Xn, and its left multiples, which are n+3 identified in F/I , are the maps f in the statement of the Proposition. (cid:3) n p 3. Definitions Afinite-stateautomaton,alsocalledMealymachine,iscomprisedofthefollowing data: a set X, the alphabet; a set Q, the states; a function τ : X Q Q, the × → transition; a function π :X Q X, the output. In this section we suppose that × → the alphabet is X = 0,...,d ; we will later specialize to d=1. { } Such an automaton is usually represented, as in the figure above, as a graph. A The statesarerepresentedbyvertices,andthereis anedgefromvertexq tovertex r, labelled i o, whenever τ(i,q)=r and π(i,q)=o. → 6 LAURENTBARTHOLDIANDILLYAI.REZNYKOV 3.1. Action. States q Q of yield transformations of X , which are defined ∗ ∈ A simultaneously as follows: (5) (i ...i )q =π(i ,q)(i ...i )τ(i1,q). 1 n 1 2 n These transformations generate a semigroup Γ( ) = Q , called the semigroup A h i generated by . A 3.2. Decomposition. Yet another way of describing the automaton is via its A semigroup decomposition. Given g Γ, it yields by (5) a transformation π of X g ∈ byrestrictiontolength-1words;and,foralli X,atransformationiX∗ iπgX∗, ∈ → again by restriction. By (5), the composition X∗ iX∗ iπgX∗ X∗ is again → → → an element of Γ, which we denote by g . We write i (6) φ(g)= g ,...,g π 0 d g ≪ ≫ for the decomposition of g Γ, with g Γ and π : X X. Multiplication of i g ∈ ∈ → such decompositions obeys the rule (7) g ,...,g π h ,...,h π = g h ,...,g h π π . ≪ 0 d≫ g≪ 0 d≫ h ≪ 0 πg(0) d πg(d)≫ g h It is therefore sufficient to know the decomposition of generators, and these are determined by the transition and output functions τ,π :X Q Q of , by × → A φ(q)= τ(0,q),...,τ(d,q) (i π(i,q) i). ≪ ≫ 7→ ∀ 3.3. Metrics. On the semigroup Γ generated by a set Q, define the “norm” g =min ℓ N: g =q q , q Q i . 1 ℓ i k k { ∈ ··· ∈ ∀ } (we have gh g + h , which justifies calling it a norm.) The ball growth function iskthekf≤uncktikon γk:Nk N defined by → γ(ℓ)=# g Γ: g ℓ ; { ∈ k k≤ } it measures the volume growth of balls in the discrete normed space Γ. 3.4. Contraction. An automaton is contracting [7] if there are constants C and η <1 such that whenever φ(g)= g ,...,g π , we have 0 d g ≪ ≫ g η g +C. i k k≤ k k Contractingautomataaremoststudied,inpartbecausecontractiongivesanatural strategy of proof by induction. 3.5. Traces. Let g be a transformation of X given by an automaton. Then for ∗ all n N we have #g(Xn+1) #X#g(Xn), so the limit ∈ ≤ #g(Xn) τ(g):= lim [0,1] n #Xn ∈ →∞ exists. We call it the trace of g. The following facts are easily checked: τ(gh) min τ(g),τ(h) ; • ≤ { } τ(g)=1 if and only if g is invertible. • Therefore,ifΓis the semigroupgeneratedbyanautomaton,wemaydefine forany ξ [0,1] a two-sided ideal Γ = g Γ: τ(g) ξ . We have Γ =Γ if and only if ξ ξ ∈ { ∈ ≤ } ξ =1. A MEALY MACHINE WITH POLYNOMIAL GROWTH OF IRRATIONAL DEGREE 7 3.6. Hausdorff dimension. Let Γ be a semigroup acting on a rooted tree X . ∗ There is another metric on Γ, defined by d(g,h)=exp( max n: vg =vh for all v Xn ), − { ∈ } withtheconventionthatexp( )=0. ThisturnsΓintoametricspaceofdiameter −∞ at most 1. The Hausdorff dimension of Γ is defined as the Hausdorff dimension of thismetricspace. LetEnd(Xn)denotethesetofprefix-preservingmapsXn Xn. → If #X =d+1, then #End(Xn)=((d+1)d+1)((d+1)n 1)/d. − Then the Hausdorff dimension of Γ is given by the formula log#Γ n Hdim(Γ)=liminf , n log#End(Xn) →∞ where Γ is the quotient of Γ acting as a transformation semigroup of Xn; see [1] n for an analogous definition in the case of profinite groups. 4. The automaton I and the semigroup F We now consider the automaton I from Figure 1. This automaton has three states f,s,e and a two-letter alphabet X = 0,1 . { } Consider the following transformations σ,ζ of X: σ(i)=1 i, ζ(i)=0. − Then the decomposition of the states are φ(f)= s,f ζ, φ(s)= e,e σ, φ(e)= e,e . ≪ ≫ ≪ ≫ ≪ ≫ Clearly e acts as the identity transformation, and s is invertible, of order 2. 4.1. Action on integers. We show now that the semigroup defined in Section 2 by its action on integers is the semigroup generated by an automaton. Theorem 4.1. The semigroups F and Γ(I) are isomorphic. Proof. The actionofF = s,f givenin(1)extends to acontinuousactionofF on the 2-adics Z . Consider thhe foillowing bijection Θ between X and Z : 2 ∞ 2 ∞ Θ(x x ...)= (1 x )2i 1. 1 2 i − − i=1 X Let us denote temporarily Γ(I)= s˜,f˜. The theorem will follow from Θ(xs˜)= h i Θ(x)s and Θ(xf˜)=Θ(x)f for all x X . ∞ ∈ Consider therefore x=x x X , and write y =x x .... If x =0, then 1 2 ∞ 2 3 1 ···∈ Θ(x)s =Θ(0y)s =(1+2Θ(y))s =2Θ(y)=Θ(1y)=Θ(xs˜), while if x =1, then 1 Θ(x)s =Θ(1y)s =(2Θ(y))s =1+2Θ(y)=Θ(0y)=Θ(xs˜). Let next n N be maximal such that x = =x =1. If n= , then 1 n ∈ ∪{∞} ··· ∞ Θ(x)f =Θ(11...)f =0f = 1=Θ(00...)=Θ(xf˜); − otherwise, x=1...10x x ...; write y =x x .... If x =0, then n+2 n+3 n+3 n+4 n+2 Θ(x)f =(2n+2n+1+2n+2Θ(y))f =2n+1 1+2n+2Θ(y)=Θ(0...001y)=Θ(xf˜), − 8 LAURENTBARTHOLDIANDILLYAI.REZNYKOV while if x =1, then 1 Θ(x)f =(2n+2n+2Θ(y))f =2n+2 1+2n+2Θ(y)=Θ(0...000y)=Θ(xf˜). − (cid:3) 4.2. Fibonacci sequence. From now on, we identify the semigroup F with Γ(I) and use the notation F for both. Recall the definition of the following elements of F: f = s, f = f, and f = f f for n 3, and the Fibonacci numbers Φ 1 2 n n 2 n 1 n − − ≥ definedbyΦ =Φ =1andΦ =Φ +Φ forn 3. Wenotethat f =Φ 1 2 n n 2 n 1 n n for all n 1. Set ϕ=(1+√5)/2; rec−all that−Φn ϕ≥n/√5. k k ≥ ≈ Notethatifnisevenand 6,thenf startswithfsfs,whileifnisoddand 5 n ≥ ≥ thenf startswith sffs. The followingstatements areeasily provenby induction: n Lemma4.2. (1)Forallk n 2withk n (mod 2)wehavef f (f f f ); n k k+1 k+3 n 1 ≤ − ≡ ≡ ··· − in particular, f is a prefix of f . k n (2) For all k >1 we have f2 f f . k+1 ≡ k−1 k+2 4.3. Some relations. By (7), we have (8) g ,g ζ h ,h ζ = g h ,g h ζ 0 1 0 1 0 0 1 0 ≪ ≫ ≪ ≫ ≪ ≫ forallg ,h F,acalculationthatwewilluserepeatedly;sobydirectcomputation i i ∈ Lemma 4.3. In F the following relations hold: s2 =e; f3 =f. Proof. If x=x ...x , then xs2 =((1 x )x ...x )s =x, so s2 =e. Then 1 n 1 2 n − φ(f2)= s,f ζ s,f ζ = s2,fs ζ, ≪ ≫ ≪ ≫ ≪ ≫ φ(f3)= s2,fs ζ s,f ζ = s3,fs2 ζ = s,f ζ =f. (cid:3) ≪ ≫ ≪ ≫ ≪ ≫ ≪ ≫ 4.4. Contraction. The proofs will ultimately all rely on some form of induction on the length of representations of semigroup elements as words over s,f . { } Lemma 4.4. The semigroup F is contracting. Proof. Considerg F,andwrite itas awordw ofminimallengthw ...w ; write 1 n ∈ also φ(g)= g ,g π . Then, by Lemma 4.3, there cannot be more than two f’s 0 1 g ≪ ≫ inarowinw,soeverygroupofthreelettersw w w containsatleastan 3k+1 3k+2 3k+3 s, which will contribute no letter to g nor g . The other two letters contribute at 0 1 mostonelettereachtoeachofg andg . Intotal, g 2( g +2),andsimilarly for g . 0 1 k 0k≤ 3 k k (cid:3) 1 k k Note, in fact, that the contraction ratio η may be chosen as 1/ϕ, with a little more care. This becomes apparent in the next result. Lemma 4.5. The decomposition of f satisfies n φ(f )= e,e σ, φ(f )= s,f ζ, 1 2 ≪ ≫ ≪ ≫ φ(f )= f,s ζ, φ(f )= f ,f2 ζ, 3 4 3 ≪ ≫ ≪ ≫ φ(f )= f ,f f f ζ if n 3, 2n 2n 1 4 6 2n 2 ≪ − ··· − ≫ ≥ φ(f )= f ,f (f f f ) ζ if n 2. 2n+1 2n 2 5 7 2n 1 ≪ ··· − ≫ ≥ A MEALY MACHINE WITH POLYNOMIAL GROWTH OF IRRATIONAL DEGREE 9 Proof. The first four claims may be checked directly; we note that if φ(g) = g ,g ζ, then φ(sg)= g ,g ζ. Next, by induction, 0 1 1 0 ≪ ≫ ≪ ≫ φ(f )=φ(f )φ(f )= f ,f f f ζ f , ζ 2n 2n 2 2n 1 2n 3 4 6 2n 4 2n 2 − − ≪ − ··· − ≫ ≪ − ∗≫ = f f ,f f f f ζ 2n 3 2n 2 4 6 2n 4 2n 2 ≪ − − ··· − − ≫ as claimed, where stands for an irrelevant element of F; and ∗ φ(f )=φ(f )φ(f )= f ,f (f f ) ζ f , ζ 2n+1 2n 1 2n 2n 2 2 5 2n 3 2n 1 − ≪ − ··· − ≫ ≪ − ∗≫ = f f ,f (f f f ) ζ. (cid:3) 2n 2 2n 1 2 5 2n 3 2n 1 ≪ − − ··· − − ≫ 5. Proofs WestartbyfindingrelationsinF;wethenshowthattheyyieldasimplenormal form for elements of F. 5.1. Relations. Recallthatforn 1wedefinedin(2,3)wordsr andr overthe ≥ n n′ alphabet s,f by r =s2, r =e, and { } 1 1′ r =f f2 f f3 ( f f f if n 3), n n+1 n ≡ n−1 n ≡ n+1 n−2 n+1 ≥ rn′ =fn%2+1(fn%2+5fn%2+7···fn−1fn+1)fn. Note that r , for odd n 3, can be obtained from r by removing the first four n′ ≥ n letters; indeed (9) fsfsr f f f (f f )f f f (f f f )f f f2(f f )f n′ ≡ 4 1· 2 6··· n+1 n ≡ 4 3 6 8··· n+1 n ≡ 4 5 8··· n+1 n f f (f f )f f f2 f f f f3 r . ≡ 6 5 8··· n+1 n ≡···≡ n−3 n−2 n+1 n ≡ n−1 n ≡ n Similarly, r for evenn 4 can be obtained fromr by removingthe first three n′ ≥ n letters and replacing them by s; indeed, writing r =sr , n′ n′′ sffr f f (f f )f f f (f f f )f f f2(f f )f (10) n′′ ≡ 3 2 5··· n+1 n ≡ 3 2 5 7··· n+1 n ≡ 3 4 7··· n+1 n f f2(f f )f f f2 f f f f3 r . ≡ 5 6 9··· n+1 n ≡···≡ n−3 n−2 n+1 n ≡ n−1 n ≡ n Let e be the word obtained from f by deleting its first two symbols. n n Lemma 5.1. If n 4, then we have φ(f )= f ,e ζ. n n 1 n 1 ≥ ≪ − − ≫ Proof. Assume first that n is even. By Lemma 4.2 we have f =f f f ...f , n 1 3 4 6 n 2 − − so e = f f f , and the Lemma holds by Lemma 4.5. Similarly, if n is n 1 4 6 n 2 − ··· − odd then e = f f f by Lemma 4.2, and again the Lemma holds by n 1 2 5 n 2 Lemma 4.5. − ··· − (cid:3) Lemma 5.2. In F the relations r =r hold for all n 1. n n′ ≥ Proof. The cases n 2 are covered in 4.3, since we have r sf3 = sf r . Let ≤ 2 ≡ ≡ 2′ us therefore assume n 3. We follow the notation of Lemma 5.1. We have the ≥ decomposition φ(r )= f f2 ,e f2 ζ = r ,fsr ζ. n ≪ n n−1 n n−1≫ ≪ n−1 n′−1≫ Suppose first that n is even. Then by (8) we have φ(rn′ )=φ(f1(f5f7···fn+1)fn)=≪e4,f4≫ζ≪f6,∗≫ζ···≪fn,∗≫ζ≪fn−1,∗≫ζ =≪f2(f6f8···fn)fn−1,f4f6···fnfn−1≫ζ =≪rn′−1,fsrn′−1≫ζ, 10 LAURENTBARTHOLDIANDILLYAI.REZNYKOV where stands for an element of F that is not relevant to the calculation. If n is ∗ odd, then similarly φ(r )=φ(f f f f )= f ,f ζ f , ζ f , ζ f , ζ n′ 2 6··· n+1 n ≪ 1 2≫ ≪ 5 ∗≫ ···≪ n ∗≫ ≪ n−1 ∗≫ = f f f f ,f f f f ζ = r ,fsr ζ. ≪ 1 5··· n n−1 2 5··· n n−1≫ ≪ n′−1 n′−1≫ We have r =r by induction, so r =r . (cid:3) n 1 n′ 1 n n′ − − 5.1.1. More relations. We will ultimately show that r =r is a complete set of { n n′} relationsforF;however,wewillfirstdescribemorerelations,thatareconsequences of these but allow much faster simplifications. Lemma 5.3. Consider a,b N with a 3 and a b+2. Then ∈ ≥ ≥ f f2 =f . a b a Proof. We proceedby induction onb. Since a b+2,the wordf ends with f ; a b+2 ≥ it therefore suffices to show that f f2 = f . We also note that the statement b+2 b b+2 holds for b 2, since then f f2 =f and f f2 =f by Lemma 4.3. Then for b 3 ≤ 3 1 3 4 2 4 ≥ we have fb+2fb2 =fbfb+1fb2 =fbrb =fbrb′ =f f f f f f f b 1+b%2 5+b%2 7+b%2 b 1 b+1 b ··· − =f f2 f f f f f f f b 1+b%2 2+b%2 4+b%2 7+b%2 9+b%2··· b−1 b+1 b =f f2 f f f f f f b 2+b%2 3+b%2 7+b%2 9+b%2··· b−1 b+1 b =f f f f f f f b 3+b%2 7+b%2 9+b%2 b 1 b+1 b ··· − = =f f f f =f f2 f2 f f =f f f =f . (cid:3) ··· b b−3 b+1 b b b−3 b−2 b−1 b b b−1 b b+2 Lemma 5.4. For all n 2 we have ≥ r =(f2f2f2 f2 f2 )f . n′ 1 2 3 ··· n−3 n−2 n+1 Proof. If n=2 then r =f and the Lemma holds. If n 3, then 2′ 3 ≥ r =f f f f f n′ 1+n%2 5+n%2··· n−1 n+1 n =f2 f2 f f f f f 1+n%2 2+n%2 3+n%2 7+n%2··· n−1 n+1 n = =f2 f2 f2 f2 f2 f f . (cid:3) ··· 1+n%2 2+n%2 3+n%2··· n−3 n−2 n−1 n Lemma 5.5. The equality f4 =f2 holds in F for n 4, and f5 =f3 holds in F n n ≤ n n for all n 1. ≥ Proof. The firststatementfollowsfromthe definitionofr . It suffices to checkthe n second one for n 5. Using Lemmata 5.4 and 5.3, we have ≥ f5 f (f f )f3 f f r =f f r =f f f2f2f2 f2 f2 f n ≡ n n−2 n−1 n ≡ n n−2 n n n−2 n′ n n−2 1 2 3 ··· n−3 n−2 n+1 = =f f f2 f2 f =f r f2 f =f f2f2 f2 f3 f ··· n n−2 n−3 n−2 n+1 n n′−3 n−2 n+1 n 1 2 ··· n−5 n−2 n+1 = =f f f =f f f f =f3. (cid:3) ··· n n−2 n+1 n n−2 n−1 n n Lemma 5.6. For n 8 we have ≥ f f f f f f =f f f f f n 2 n 3 n 5 n 7 n 2 n n 2 n 3 n 5 n 3 n − − − − · − − − − · − =f f f f =f f f =f . n 2 n 3 n 4 n n 2 n 5 n n − − · − − · −

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