An explicit counterexample to the Lagarias-Wang finiteness conjecture
aa r X i v : . [ m a t h . O C ] N ov AN EXPLICIT COUNTEREXAMPLE TO THE LAGARIAS-WANGFINITENESS CONJECTURE
KEVIN G. HARE, IAN D. MORRIS, NIKITA SIDOROV, AND JACQUES THEYS
To the memory of I. M. Gelfand
Abstract.
The joint spectral radius of a finite set of real d × d matrices is defined to be themaximum possible exponential rate of growth of long products of matrices drawn from that set.A set of matrices is said to have the finiteness property if there exists a periodic product whichachieves this maximal rate of growth. J. C. Lagarias and Y. Wang conjectured in 1995 thatevery finite set of real d × d matrices satisfies the finiteness property. However, T. Bousch andJ. Mairesse proved in 2002 that counterexamples to the finiteness conjecture exist, showing inparticular that there exists a family of pairs of 2 × A α ∗ := (cid:26)(cid:18) (cid:19) , α ∗ (cid:18) (cid:19)(cid:27) we give an explicit value of α ∗ ≃ . . . . such that A α ∗ does not satisfy the finiteness property. Introduction If A is a d × d real or complex matrix and k · k is a matrix norm, the spectral radius ρ ( A ) ofthe matrix A admits the well-known characterisation ρ ( A ) = lim n →∞ k A n k /n , a result known as Gelfand’s formula. The joint spectral radius generalises this concept to sets ofmatrices. Given a finite set of d × d real matrices A = { A , . . . , A r } , we by analogy define thejoint spectral radius ̺ ( A ) to be the quantity ̺ ( A ) := lim sup n →∞ max n k A i · · · A i n k /n : i j ∈ { , . . . , r } o , a definition introduced by G.-C. Rota and G. Strang in 1960 [45] (reprinted in [44]). Note thatthe pairwise equivalence of norms on finite dimensional spaces implies that the quantity ̺ ( A ) isindependent of the choice of norm used in the definition.The joint spectral radius has been found to arise naturally in a range of mathematical contextsincluding control and stability [1, 11, 20, 28], coding theory [37], the regularity of wavelets andother fractal structures [12, 13, 36, 42], numerical solutions to ordinary differential equations [19],and combinatorics [4, 14]. As such the problem of accurately estimating the joint spectral radiusof a given finite set of matrices is a topic of ongoing research interest [5, 18, 32, 31, 38, 40, 48, 49]. Date : October 30, 2018.2010
Mathematics Subject Classification.
Primary 15A18, 15A60; secondary 37B10, 65K10, 68R15.
Key words and phrases.
Joint spectral radius, finiteness conjecture, Sturmian sequence, balanced word, Fi-bonacci word, infinite product.Research of K. G. Hare supported, in part, by NSERC of Canada. Research of I. D. Morris supported by theEPSRC grant EP/E020801/1.
In this paper we study a property related to the computation of the joint spectral radius of a setof matrices, termed the finiteness property . A set of d × d real matrices A := { A , . . . , A r } is said tosatisfy the finiteness property if there exist integers i , . . . , i n such that ̺ ( A ) = ρ ( A i · · · A i n ) /n .The finiteness conjecture of J. Lagarias and Y. Wang [34] asserted that every finite set of d × d realmatrices has the finiteness property; a conjecture equivalent to this statement was independentlyposed by L. Gurvits in [20], where it was attributed to E. S. Pyatnitski˘ı. In special cases, thisfiniteness property is known to be true, see for example [10, 25]. The existence of counterexamplesto the finiteness conjecture was established in 2002 by T. Bousch and J. Mairesse [8], with alter-native constructions subsequently being given by V. Blondel, J. Theys and A. Vladimirov [6] andV. S. Kozyakin [29]. However, in all three of these proofs it is shown only that a certain family ofpairs of 2 × × A := (cid:18) (cid:19) , A := (cid:18) (cid:19) , and for each α ∈ [0 ,
1] let us define A α := { A , αA } . The construction of Blondel-Theys-Vladimirov [6] shows that there exists α ∈ [0 ,
1] for which A α does not satisfy the finitenessproperty. The proof operates indirectly by demonstrating that the set of all parameter values α for which the finiteness property does hold is insufficient to cover the interval [0 , ̺ ( A α ) as the parameter α is varied ina rather deep manner. This allows us to prove the following theorem: Theorem 1.1.
Let ( τ n ) ∞ n =0 denote the sequence of integers defined by τ := 1 , τ , τ := 2 , and τ n +1 := τ n τ n − − τ n − for all n ≥ , and let ( F n ) ∞ n =0 denote the sequence of Fibonacci numbers,defined by F := 0 , F := 1 and F n +1 := F n + F n − for all n ≥ . Define a real number α ∗ ∈ (0 , by (1.1) α ∗ := lim n →∞ τ F n +1 n τ F n n +1 ! ( − n = ∞ Y n =1 (cid:18) − τ n − τ n τ n +1 (cid:19) ( − n F n +1 . Then this infinite product converges unconditionally, and A α ∗ does not have the finiteness property. The convergence in both of the limits given in Theorem 1.1 is extremely rapid, being of or-der O (exp( − δφ n )) where δ > φ is the golden ratio. An explicit errorbound is given subsequently to the proof of Theorem 1.1. Using this bound we may compute theapproximation α ∗ ≃ . . . . which is rigorously accurate to all decimal places shown.We shall now briefly describe the technical results which underlie the proof of Theorem 1.1. Foreach α ∈ [0 ,
1] let us write A ( α )0 := A and A ( α )1 := αA so that A α = n A ( α )0 , A ( α )1 o . The principaltechnical question which is addressed in this paper is the following: if we are given that for somefinite sequence of values u , . . . , u n ∈ { , } , the matrix(1.2) A ( α ) u n A ( α ) u n − · · · A ( α ) u A ( α ) u is “large” in some suitable sense – for example, if its spectral radius is close to the value ̺ ( A α ) n - then what may we deduce about the combinatorial structure of the sequence of values u i , andhow does this answer change as the parameter α is varied? A key technical step in the proof ofTheorem 1.1, therefore, is to show that the magnitude of the product (1.2) is maximised when thesequence u , u , . . . , u n is a balanced word . This result depends in a rather essential manner on This is the sequence A022405 from Sloane’s On-Line Encyclopedia of Integer Sequences.
OUNTEREXAMPLE TO THE FINITENESS CONJECTURE 3 several otherwise unpublished results from the fourth named author’s 2005 PhD thesis [47], whichare substantially strengthened in the present paper.In the following section we shall introduce the combinatorial ideas needed to describe balancedwords. We are then able to state our main technical theorem, describe its relationship to previousresearch in ergodic theory and the theory of the joint spectral radius, and give a brief overviewof how Theorem 1.1 is subsequently deduced. The detailed structure of this paper is described atthe end of the following section.2.
Notation and statement of technical results
Throughout this paper we denote the set of all d × d real matrices by M d ( R ). The symbol ||| · ||| will be used to denote the norm on M d ( R ) which is induced by the Euclidean norm on R d , whichsatisfies ||| B ||| = ρ ( B ∗ B ) / for every B ∈ M d ( R ). Other norms shall be denoted using the symbol k · k . We shall say that a norm k · k on M ( R ) is submultiplicative if k AB k ≤ k A k · k B k for all A, B ∈ M ( R ). For the remainder of this paper we shall also denote ̺ ( A α ) simply by ̺ ( α ).For the purposes of this paper we define a finite word , or simply word to be sequence u = ( u i )belonging to { , } n for some integer n ≥
0. We will typically use u , v or w to represent finitewords. If u ∈ { , } n then we say that u is length n , which we denote by | u | = n . If | u | is zerothen the word u is called empty . The number of terms of u which are equal to 1 is denoted by | u | . If u is nonempty, the quantity | u | / | u | is called the 1 -ratio of u and is written ς ( u ). The twopossible words of length one shall often be denoted simply by 0 and 1. We denote the set of allfinite words by Ω.We will define an infinite word to be a sequence x = ( x i ) belonging to { , } N . We will typicallyuse x , y or z to represent infinite words. If the word can be either finite or infinite, we will typicallyuse ω . We denote the set of all infinite words by Σ, and define a metric d on Σ as follows. Given x, y ∈ Σ with x = ( x i ) ∞ i =1 and y = ( y i ) ∞ i =1 , define n ( x, y ) := inf { i ≥ x i = y i } . We now define d ( x, y ) := 1 / n ( x,y ) for all x, y ∈ Σ, where we interpret the symbol 1 / ∞ as being equal to zero.The topology on Σ which is generated by the metric d coincides with the infinite product topologyon Σ = { , } N . In particular Σ is compact and totally disconnected. For any nonempty finiteword u = ( u i ) ni =1 the set { x ∈ Σ : x i = u i for all 1 ≤ i ≤ n } is both closed and open. Since everyopen ball in Σ has this form for some u , the collection of all such sets generates the topology of Σ.We define the shift transformation T : Σ → Σ by T [( x i ) ∞ i =1 ] := ( x i +1 ) ∞ i =1 . The shift transforma-tion is continuous and surjective. We define the projection π n : Σ → Ω by π n [( x i ) ∞ i =1 ] = ( x i ) ni =1 .If u = u u . . . u n and v = v v . . . v m are finite words, then we define the concatenation of u with v as uv = u u . . . u n v v . . . v m , the finite word of length n + m . Note that if u is the emptyword then uv = vu = v for every word v . The set Ω endowed with the operation of concatenationis a semigroup.Given a word u and positive integer n we let u n denote the linear concatenation of n copies of u , so that for example u := uuuu . If u is a nonempty word of length n , we let u ∞ denote theunique infinite word x ∈ Σ such that x kn + i = u i for all integers i, k with k ≥ ≤ i ≤ n .Clearly any infinite word x ∈ Σ satisfies T n x = x for an integer n ≥ u such that x = u ∞ and | u | divides n .If u is a nonempty word, and ω is either a finite or infinite word, we say that u is a subword of ω if there exists an integer k ≥ u i = ω k + i for all integers i in the range 1 ≤ i ≤ | u | . Wedenote this relationship by u ≺ ω . Clearly u ≺ ω if and only if there exist a possibly empty word v ∈ Ω and a finite or infinite word ω ′ such that ω = auω ′ . An infinite word x is said to be recurrent if every finite subword u ≺ x occurs as a subword of x an infinite number of times. A finite orinfinite word ω is called balanced if for every pair of finite subwords u, v such that u, v ≺ ω and | u | = | v | , we necessarily have || u | − | v | | ≤
1. Clearly ω is balanced if and only if every subwordof ω is balanced. An infinite balanced word which is not eventually periodic is called Sturmian .The following standard result describes the principal properties of balanced infinite words whichwill be applied in this paper:
KEVIN G. HARE, IAN D. MORRIS, NIKITA SIDOROV, AND JACQUES THEYS
Theorem 2.1. If x ∈ Σ is balanced then the limit ς ( x ) := lim n →∞ ς ( π n ( x )) exists. For each γ ∈ [0 , , let X γ denote the set of all recurrent balanced infinite words x ∈ Σ for which ς ( x ) = γ .These sets have the following properties:(i). Each X γ is compact and nonempty.(ii). For each γ ∈ [0 , , the restriction of T to X γ is a continuous, minimal, uniquely ergodictransformation of X γ . If µ is the unique ergodic probability measure supported in X γ , then µ ( { x : x = 1 } ) = γ .(iii). If γ = p/q ∈ [0 , ∩ Q in lowest terms then the cardinality of X γ is q , and for each x ∈ X γ we have X γ = { x, T x, . . . , T q − x } . If γ ∈ [0 , \ Q then X γ is uncountably infinite. Example 2.2.
We have X / = { (00101) ∞ , (01010) ∞ , (10100) ∞ , (01001) ∞ , (10010) ∞ } .Theorem 2.1 does not appear to exist in the literature in the precise form given above, but itmay be established without difficulty by combining various results from the second chapter of [35].The key step in obtaining Theorem 2.1 is to show that x ∈ X γ if and only if there exists δ ∈ [0 , x n ≡ ⌊ ( n +1) γ + δ ⌋−⌊ nγ + δ ⌋ , or x n ≡ ⌈ ( n +1) γ + δ ⌉−⌈ nγ + δ ⌉ , see Lemmas 2.1.14and 2.1.15 of [35]. Once this identification has been made, the dynamical properties of X γ underthe shift transformation largely follow from the properties of the rotation map z z + γ definedon R / Z .Given a nonempty finite word u = ( u i ) ni =1 and real number α ∈ [0 , A ( α ) ( u ) := A ( α ) u n A ( α ) u n − · · · A ( α ) u A ( α ) u and A ( u ) := A u n A u n − · · · A u A u = A (1) ( u ) . For every x ∈ Σ, α ∈ [0 ,
1] and n ≥ A ( α ) ( x, n ) := A ( α ) ( π n ( x )) , A ( x, n ) := A ( π n ( x )) = A (1) ( x, n ) . Note that the function A ( x, n ) satisfies the cocycle relationship A ( x, n + m ) = A ( T n x, m ) A ( x, n )for every x ∈ Σ, n, m ≥ x ∈ Σ for which A ( x, n ) grows rapidly in terms of the sets X γ . To do this we must be able to specify what ismeant by rapid growth. Let us therefore say that an infinite word x ∈ Σ is a strongly extremal word for A α if there is a constant δ > |||A ( α ) ( x, n ) ||| ≥ δ̺ ( α ) n for all n ≥
1, and weaklyextremal for A α if lim n →∞ |||A ( α ) ( x, n ) ||| /n = ̺ ( α ). It is obvious that every strongly extremalword is also weakly extremal. Note also that since all norms on M ( R ) are equivalent, thesedefinitions are unaffected if another norm k · k is substituted for ||| · ||| . We shall say that r ∈ [0 , unique optimal -ratio of A α if for every x ∈ Σ which is weakly extremal for A α we have ς ( π n ( x )) → r . Note that the existence of a unique optimal 1-ratio is a nontrivial property, and isshown in Theorem 2.3. For example, if A ⊂ M ( R ) is a pair of isometries then no unique optimal1-ratio for A exists. It is not difficult to see that if A α has a unique optimal 1-ratio which isirrational, then A α cannot satisfy the finiteness property, and it is this principle which underliesthe present work as well as the work of Bousch-Mairesse [8] and Kozyakin [29].The principal technical result of this paper is the following theorem which allows us to relateall of the concepts defined so far in this section: Theorem 2.3.
There exists a continuous, non-decreasing surjection r : [0 , → [0 , ] such thatfor each α , r ( α ) is the unique optimal -ratio of A α . For each α ∈ [0 , , every element of X r ( α ) is strongly extremal for A α . Moreover, for every compact set K ⊂ (0 , there exists a constant C K > such that (2.1) C − K ≤ ρ (cid:0) A ( α ) ( x, n ) (cid:1) ̺ ( α ) n ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) A ( α ) ( x, n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ̺ ( α ) n ≤ C K OUNTEREXAMPLE TO THE FINITENESS CONJECTURE 5 whenever α ∈ K , x ∈ X r ( α ) and n ≥ . Conversely, if x ∈ Σ is a recurrent infinite word which isstrongly extremal for A α then x ∈ X r ( α ) , and if x ∈ Σ is any infinite word which is weakly extremalfor A α then (1 /n ) P n − k =0 dist( T k x, X r ( α ) ) → .Remark . The definition of a strongly extremal infinite word is similar to the one previouslyproposed by V. S. Kozyakin [30], whereas the definition of a weakly extremal infinite word issimilar to a definition used previously by the fourth named author [47]. In both instances theinfinite word is simply referred to as ‘extremal’.
Remark . Note that balanced/Sturmian words (and measures) arise as optimal trajectories invarious optimisation problems – see, e.g., [7, 9, 23, 24].
Remark . A less general version of parts of Theorem 2.3 was proved in [47].The structure of the paper is as follows: Sections 3 and 4 deal with important preliminaries,such as general properties of joint spectral radius and of balanced words. In Section 5 we showthat every strongly extremal infinite word is balanced. In Section 6 we introduce an importantauxiliary function S defined as the logarithm of the exponent growth of the norm of an arbitrarymatrix product taken along balanced words with a fixed 1-ratio. In Section 7 we apply resultsfrom preceding sections to prove Theorem 2.3. Finally, in Section 8 we deduce Theorem 1.1 fromTheorem 2.3. Section 9 contains some open questions and conjectures.We believe it is worth describing here briefly how Theorem 2.3 leads to Theorem 1.1. Once wehave established the existence of such a function r , we may take any irrational γ and concludethat any element α of the preimage r − ( γ ) is a counterexample to the finiteness conjecture (sinceany weakly extremal word must be aperiodic).To construct a specific counterexample, we take γ = −√ and choose the Fibonacci word u ∞ as a strongly extremal word for this 1-ratio. Recall that u ∞ = lim n u ( n ) , where u (1) = 1 , u (2) = 0and u ( n +1) = u ( n ) u ( n − for n ≥
2. Now consider the morphism h : Ω → M ( R ) such that h (0) = A , h (1) = A . Denote B n := h ( u ( n ) ); we thus have B n +1 = B n B n − . One can easily showthat tr ( B n ) = τ n , the sequence described in Theorem 1.1. To obtain explicit formulae for α ∗ , weshow that the auxillary function S introducted in Section 6 is differentiable at γ = −√ and that − log α ∗ = S ′ (cid:16) −√ (cid:17) . We then compute this derivative, which will yield (1.1).3. General properties of the joint spectral radius and extremal infinite words
We shall begin with some general results concerning the joint spectral radius. The followingcharacterisation of the joint spectral radius will prove useful on a number of occasions:
Lemma 3.1.
Let α ∈ [0 , and let k · k be any submultiplicative matrix norm. Then: ̺ ( α ) = inf n ≥ max n kA ( x, n ) k /n : x ∈ Σ o = sup n ≥ max n ρ ( A ( x, n )) /n : x ∈ Σ o . Proof.
We review some arguments from [2, 12]. Fix α ∈ [0 ,
1] and a matrix norm k · k , and define ̺ + n ( α, k · k ) := max n(cid:13)(cid:13)(cid:13) A ( α ) i · · · A ( α ) i n (cid:13)(cid:13)(cid:13) : ( i , . . . , i n ) ∈ { , } n o = max n(cid:13)(cid:13)(cid:13) A ( α ) ( x, n ) (cid:13)(cid:13)(cid:13) : x ∈ Σ o and ̺ − n ( α ) := max n ρ (cid:16) A ( α ) i · · · A ( α ) i n (cid:17) : i , . . . , i n ∈ { , } o = max n ρ (cid:16) A ( α ) ( x, n ) (cid:17) : x ∈ Σ o . Clearly each ̺ + n ( α, k · k ) is nonzero, and ̺ + n + m ( α, k · k ) ≤ ̺ + n ( α, k · k ) ̺ + m ( α, k · k ) for every n, m ≥ ̺ + n ( α, k · k ) we obtainlim n →∞ ̺ + n ( α, k · k ) /n = inf n ≥ ̺ + n ( α, k · k ) /n . In particular the limit superior in the definition of ̺ ( α ) is in fact a limit. A well-known result ofBerger and Wang [2] implies thatlim n →∞ ̺ + n ( α, k · k ) /n = lim sup n →∞ ̺ − n ( α ) /n , KEVIN G. HARE, IAN D. MORRIS, NIKITA SIDOROV, AND JACQUES THEYS which in particular implies that the value ̺ ( α ) is independent of the choice of norm k · k . Finally,note that if ρ (cid:16) A ( α ) i · · · A ( α ) i n (cid:17) = ̺ − n ( α ) for some n , then ̺ − nm ( α ) ≥ ρ (( A i · · · A i n ) m ) = ̺ − n ( α ) m foreach m ≥
1, and hence the limit superior above is also a supremum. (cid:3)
We may immediately deduce the following corollary, which was originally noted by C. Heil andG. Strang [22]:
Lemma 3.2.
The function ̺ : [0 , → R is continuous.Proof. The first of the two identities given in Lemma 3.1 shows that ̺ is equal to the pointwiseinfimum of a family of continuous functions, and hence is upper semi-continuous. The secondidentity shows that ̺ also equals the pointwise supremum of a family of continuous functions, andhence is lower semi-continuous. (cid:3) Lemma 3.3.
For each α ∈ (0 , there exists a matrix norm k · k α such that (cid:13)(cid:13)(cid:13) A ( α ) i (cid:13)(cid:13)(cid:13) α ≤ ̺ ( α ) for i = 0 , . The matrix norms k · k α may be chosen so that the following additional property issatisfied: for every compact set K ⊂ (0 , there exists a constant M K > such that M − K k B k α ≤||| B ||| ≤ M K k B k α for all B ∈ M ( R ) and all α ∈ K .Proof. Let B = { B , . . . , B r } be any finite set of d × d real matrices and let ̺ ( B ) be its joint spectralradius. We say that B is irreducible if the only linear subspaces V ⊆ R d such that B i V ⊆ V forevery i are { } and R d . A classic theorem of N. E. Barabanov [1] shows that if B is irreduciblethen there exists a constant M B > n ≥ {k B i . . . B i n k : i j ∈ { , . . . , r }} ≤ M B ̺ ( B ) n . Note in particular that necessarily ̺ ( B ) >
0. It is then straightforward to see that if we define foreach v ∈ R d k v k B := sup n ≥ (cid:8) ̺ ( B ) − n max ||| B i · · · B i n v ||| : i j ∈ { , . . . , r } (cid:9) , where ||| · ||| denotes the Euclidean norm, then k · k B is a norm on R d which satisfies k B i v k B ≤ ̺ ( B ) k v k B for every i ∈ { , . . . , r } and v ∈ R d . It follows that the operator norm on M ( R )induced by k · k B has the property k B i k B ≤ ̺ ( B ) for each B i . More recent results due to F. Wirth[49, Thm. 4.1] and V. S. Kozyakin [33] show that the constants M B may be chosen so as to dependcontinuously on the set of matrices B , subject to the condition that the perturbed matrix familiesalso do not have invariant subspaces. It is easily shown that A α is irreducible for every α ∈ (0 , (cid:3) We immediately obtain the following:
Lemma 3.4.
For each α ∈ (0 , we have ̺ ( α ) > .Proof. Assume ̺ ( α ) ≤ α ∈ (0 , {k A ( α ) (0 n ) k α : n ≥ } ≤ {||| A n ||| : n ≥ } < ∞ . Since A n = ( n ), we have lim n →∞ ||| A n ||| =+ ∞ and therefore we must have ̺ ( α ) > (cid:3) Fix some norm k · k α which satisfies the conditions of Lemma 3.3. The following key result isa variation on part of [38, Thm 2.2]. We include a proof here for the sake of completeness. Lemma 3.5.
For each α ∈ (0 , define Z α := ∞ \ n =1 n x ∈ Σ : (cid:13)(cid:13)(cid:13) A ( α ) ( x, n ) (cid:13)(cid:13)(cid:13) α = ̺ ( α ) n o . Then each Z α is compact and nonempty, and satisfies T Z α ⊆ Z α .Proof. Fix α ∈ (0 ,
1] and define for each n ≥ Z α,n := n x ∈ Σ : (cid:13)(cid:13)(cid:13) A ( α ) ( x, n ) (cid:13)(cid:13)(cid:13) α = ̺ ( α ) n o . OUNTEREXAMPLE TO THE FINITENESS CONJECTURE 7
Clearly each Z α,n is closed. If some Z α,n were to be empty, then by Lemma 3.3 we wouldhave sup (cid:8)(cid:13)(cid:13) A ( α ) ( x, n ) (cid:13)(cid:13) α : x ∈ Σ (cid:9) < ̺ ( α ) n , contradicting Lemma 3.1. For each n ≥ Z α,n +1 ⊆ Z α,n , since if x ∈ Z α,n +1 then ̺ ( α ) n +1 = (cid:13)(cid:13)(cid:13) A ( α ) ( x, n + 1) (cid:13)(cid:13)(cid:13) α ≤ (cid:13)(cid:13)(cid:13) A ( α ) ( T n x, (cid:13)(cid:13)(cid:13) α (cid:13)(cid:13)(cid:13) A ( α ) ( x, n ) (cid:13)(cid:13)(cid:13) α ≤ ̺ ( α ) (cid:13)(cid:13)(cid:13) A ( α ) ( x, n ) (cid:13)(cid:13)(cid:13) α ≤ ̺ ( α ) n +1 using Lemma 3.3 and it follows that x ∈ Z α,n also. We deduce that the set Z α = T ∞ n =1 Z α,n isnonempty. Since each Z α,n is closed, Z α is closed and hence is compact. Finally, if x ∈ Z α,n +1 then we also have ̺ ( α ) n +1 = (cid:13)(cid:13)(cid:13) A ( α ) ( x, n + 1) (cid:13)(cid:13)(cid:13) α ≤ (cid:13)(cid:13)(cid:13) A ( α ) ( T x, n ) (cid:13)(cid:13)(cid:13) α (cid:13)(cid:13)(cid:13) A ( α ) ( x, (cid:13)(cid:13)(cid:13) α ≤ ̺ ( α ) (cid:13)(cid:13)(cid:13) A ( α ) ( T x, n ) (cid:13)(cid:13)(cid:13) α ≤ ̺ ( α ) n +1 so that T x ∈ Z α,n , and we deduce from this that T Z α ⊆ Z α . (cid:3) The remaining lemmas in this section will be applied in the proof of Theorem 2.3 to characterisethe extremal orbits of A α . Lemma 3.6.
Let α ∈ (0 , and x ∈ Σ . If x is recurrent and strongly extremal for A α , then x ∈ Z α .Proof. Let α ∈ (0 ,
1] and x ∈ Σ \ Z α , and suppose that x is recurrent. We shall show thatlim inf n →∞ ̺ ( α ) − n (cid:13)(cid:13) A ( α ) ( x, n ) (cid:13)(cid:13) α = 0 and therefore x is not strongly extremal, which proves thelemma. Since x / ∈ Z α , there exist ε > n ≥ (cid:13)(cid:13) A ( α ) ( x, n ) (cid:13)(cid:13) α < (1 − ε ) ̺ ( α ) n .Since x is recurrent, it follows that for each k ≥ r k > r k − > . . . > r >r = 0 such that (cid:13)(cid:13) A ( α ) ( T r i x, n ) (cid:13)(cid:13) α < (1 − ε ) ̺ ( α ) n for each i . By increasing k and passing toa subsequence if necessary, it is clear that we may assume additionally that r i +1 > r i + n for1 ≤ i < k . Define also r k +1 := r k + n + 1. We have (cid:13)(cid:13)(cid:13) A ( α ) ( x, r k +1 ) (cid:13)(cid:13)(cid:13) α ≤ k Y i =1 (cid:13)(cid:13)(cid:13) A ( α ) ( T r i x, n ) (cid:13)(cid:13)(cid:13) α (cid:13)(cid:13)(cid:13) A ( α ) ( T r i + n x, r i +1 − r i − n ) (cid:13)(cid:13)(cid:13) α ≤ (1 − ε ) k ̺ ( α ) r k +1 , and since k may be taken arbitrarily large we conclude thatlim inf n →∞ ̺ ( α ) − n (cid:13)(cid:13)(cid:13) A ( α ) ( x, n ) (cid:13)(cid:13)(cid:13) α = 0 , as desired. (cid:3) The following lemma is a straightforward corollary of a more general result due to S. J. Schreiber[46, Lemma 1]:
Lemma 3.7.
Let ( f n ) be a sequence of continuous functions from Σ to R such that f n + m ( x ) ≤ f n ( T m x ) + f m ( x ) for all x ∈ Σ and n, m ≥ . Then for each x ∈ Σ and m ≥ , lim inf n →∞ nm n − X k =0 f m ( T k x ) ≥ lim inf n →∞ n f n ( x ) . Lemma 3.8.
Let α ∈ (0 , and suppose that the restriction of T to Z α is uniquely ergodic, with µ being its unique T -invariant Borel probability measure. Then r := µ ( { x ∈ Σ : x = 1 } ) is theunique optimal -ratio of A α , and if x ∈ Σ is weakly extremal, then lim n →∞ n n − X k =0 dist (cid:0) T k x, supp µ (cid:1) = 0 . KEVIN G. HARE, IAN D. MORRIS, NIKITA SIDOROV, AND JACQUES THEYS
Proof.
Let M denote the set of all Borel probability measures on Σ equipped with the weak-*topology, which is defined to be the smallest topology such that µ R f dµ is continuous for everycontinuous function f : Σ → R . This topology makes M a compact metrisable space [41, Thm.II.6.4]. Let us fix α ∈ (0 ,
1] and suppose that x ∈ Σ is weakly extremal. For each n ≥ µ n := (1 /n ) P n − k =0 δ T k x ∈ M , where δ z ∈ M denotes the Dirac probability measure concentratedat z ∈ Σ. We claim that lim n →∞ µ n = µ in the weak-* topology.Applying Lemma 3.7 with f n ( x ) := log (cid:13)(cid:13) A ( α ) ( x, n ) (cid:13)(cid:13) α and noting that f n ( x ) ≤ n log ̺ ( α ) forall x and n , we obtain(3.1) lim n →∞ Z N log (cid:13)(cid:13)(cid:13) A ( α ) ( z, N ) (cid:13)(cid:13)(cid:13) α dµ n ( z ) = lim n →∞ nN n − X i =0 log (cid:13)(cid:13)(cid:13) A ( α ) (cid:0) T i x, N (cid:1)(cid:13)(cid:13)(cid:13) α = log ̺ ( α )for every N ≥
1. As in the proof of Lemma 3.5 we let Z α,N = { z ∈ Σ : kA ( α ) ( z, N ) k α = ̺ ( α ) N } foreach N ≥
1, and we recall that Z α,N +1 ⊆ Z α,N for every N . Let ν ∈ M be any limit point of thesequence ( µ n ). If f : Σ → R is any continuous function then it follows easily from the definitionof ( µ n ) that | R f dν − R f ◦ T dν | ≤ lim sup n →∞ | R f ◦ T dµ n − R f dµ n | = 0 and it follows that ν is T -invariant. For each N ≥ Z N log (cid:13)(cid:13)(cid:13) A ( α ) ( z, N ) (cid:13)(cid:13)(cid:13) α dν ( z ) = log ̺ ( α ) , and since (cid:13)(cid:13) A ( α ) ( z, N ) (cid:13)(cid:13) α ≤ ̺ ( α ) N for all z ∈ Σ it follows from this that ν ( Z α,N ) = 1. Since thisapplies for every N , and Z α,N +1 ⊆ Z α,N for every N , we deduce that ν ( Z α ) = 1. By hypothesis µ is the unique T -invariant element of M giving full measure to Z α , and it follows that ν = µ . Wehave shown that µ is the only weak-* accumulation point of the sequence ( µ n ), and since M iscompact and metrisable we deduce that lim n →∞ µ n = µ , which completes the proof of the claim.The proof of the lemma now follows easily. Let f : Σ → R be the characteristic function of theset { x ∈ Σ : x = 1 } , and note that f is continuous since this set is both open and closed. Definea further continuous function by g ( x ) := dist( x, supp µ ). Since µ n → µ we may easily derivelim n →∞ ς ( π n ( x )) = lim n →∞ n n − X i =0 f (cid:0) T i x (cid:1) = lim n →∞ Z f dµ n = Z f dµ = µ ( { x ∈ Σ : x = 1 } ) = r and lim n →∞ n n − X i =0 dist( T i x, supp µ ) = lim n →∞ Z g dµ n = Z g dµ = 0as required. The proof is complete. (cid:3) General properties of balanced words
In this short and mostly expository section we present some combinatorial properties of balancedwords which will be applied in subsequent sections. We first require some additional definitions.Given two nonempty finite words u, v of equal length, we write u < v if u strictly precedes v in the lexicographical order: that is, u < v if and only if there is k ≥ u k = 0, v k = 1,and u i = v i when 1 ≤ i < k . We define the reverse of a finite word u , which we denote by ˜ u , tobe the word obtained by listing the terms of u in reverse order. That is, if u = u u · · · u n then˜ u = u n u n − · · · u . We say that a finite word p is a palindrome if ˜ p = p . Since the reverse of theempty word is also the empty word, the empty word is a palindrome. We say that two finite words u and v of equal length are cyclic permutations of each other, and write u ≃ v , if there exist finitewords a and b such that u = ab and v = ba . For each n ≥ n .We begin by collecting together some standard results from [35]: Lemma 4.1.
Let γ ∈ (0 , and x ∈ X γ , and choose any N > max {⌈ γ − ⌉ , ⌈ (1 − γ ) − ⌉} . Thenneither N nor N is a subword of x . OUNTEREXAMPLE TO THE FINITENESS CONJECTURE 9
Proof.
Let u ≺ x with | u | = N . By [35, Prop. 2.1.10] we have γ | u | + 1 ≥ | u | ≥ γ | u | − | u | > γ ⌈ γ − ⌉ − ≥ | u | − | u | > (1 − γ ) ⌈ (1 − γ ) − ⌉ − ≥
0, so0 < | u | < | u | and u cannot be equal to 0 N or 1 N . (cid:3) Definition 4.2.
Let
W ⊂ Ω × Ω be the smallest set with the following two properties: (0 , ∈ W ;if ( u, v ) ∈ W , then ( uv, v ) ∈ W and ( u, vu ) ∈ W . We say that u ∈ Ω is a standard word if either( v, u ) ∈ W or ( u, v ) ∈ W for some v ∈ Ω. Lemma 4.3.
The set of standard words has the following properties:(i) If u is standard, with | u | = q and | u | = p , then u ∞ ∈ X p/q .(ii) For every γ ∈ [0 , there exists x ∈ X γ such that for infinitely many q ∈ N the word π q ( x ) is standard.Proof. (i). If q = 1 then the result is trivial. For q >
1, [35, Prop. 2.2.15] shows that everystandard word is balanced. If u is standard, then it is clear from the definition that u n is asubword of a standard word for every n ≥
1. In particular every u n is balanced and therefore u ∞ is balanced.(ii). Let x be the infinite word defined by x n := ⌊ γ ( n + 2) ⌋ − ⌊ γ ( n + 1) ⌋ ∈ { , } for all n ≥ characteristic word for γ . It is shown in [35, Prop 2.2.15] that x has therequired properties. (cid:3) The following result is given in the proof of [35, Prop. 2.1.3]. Note that p may be the emptyword; for example, this is true in the case w = 0011. Lemma 4.4.
Let w be a finite word which is not balanced, let u and v be subwords of w of equallength such that | u | ≥ | v | , and suppose that u, v have the minimum possible length for whichthis property may be satisfied. Then there is a palindrome p such that u = 1 p and v = 0 p . The following two results arise in the fourth named author’s PhD thesis [47]:
Lemma 4.5.
Let w be a finite word and p a palindrome, and suppose that p and p aresubwords of w . Then there is a finite word b , which may be empty, such that either p b p or p b p is a subword of w .Proof. Recall that u ≺ v means that u is a subword of v . Since 0 p p w , the only alternative is that they occur in an overlapping manner: that is, there are finitewords d, e, f such that 0 d e f ≺ w , where d e = e f = p , or similarly with 0 and 1 interchanged.Since ˜ p = p , the relation d e = e f = p implies ˜ e d = e f , and since | ˜ e | = | e | we obtain 1 = 0, acontradiction. We conclude that the words 0 p p (cid:3) Lemma 4.6.
Let u be a finite word which is not balanced. Then there exist words a, w, b suchthat awb ≺ u and one of the following two possibilities holds: either ˜ b > a and ˜ w > w , or ˜ a > b and w > ˜ w .Proof. Combining Lemmas 4.4 and 4.5 we find that there exist words p, v such that ˜ p = p andeither 0 p v p ≺ u , or 1 p v p ≺ u . In the former case we may take a := 0 p , b := p w := 0 v
1, and in the latter case we may take a := 1 p , b := p w := 1 v (cid:3) Finally, we require the following lemma which characterises those finite words for which allcyclic permutations are balanced. This result appears to be something of a “folklore theorem” inthe theory of balanced words; to the best of our knowledge, the proof which we present here isoriginal. A version of this result appears as [3, Thm 6.9]. Note that the word u := 1001 is anexample of a balanced word with the property that u ∞ is not balanced. Lemma 4.7.
Let u be a nonempty finite word. Then the following are equivalent:(i) Every cyclic permutation of u is balanced.(ii) The finite word u is balanced.(iii) The infinite word u ∞ is balanced. Proof.
It is clear that (iii) = ⇒ (ii) = ⇒ (i). To prove the implication (i) = ⇒ (ii) by we shall showthat if u is a nonempty finite word such that u is not balanced, then there is a cyclic permutationof u which is not balanced.Let us then suppose that u is a finite nonempty word such that u is not balanced. Let a, b be subwords of u of equal length such that || a | − | b | | ≥
2, and suppose that no pair of shortersubwords may be found which also has this property. Clearly we have || a | − | b | | = 2, and withoutloss of generality we shall assume that | a | = 2 + | b | . By Lemma 4.4 there exists a palindrome p such that a = 1 p b = 0 p
0, and it follows from Lemma 4.5 that | a | , | b | ≤ | u | . We maytherefore choose words c and d such that | c | = | d | = | u | − | a | = | u | − | b | and ac ≃ bd ≃ u . Since | ac | = | bd | = | u | we have | d | = 2 + | c | , and since a and b are the shortest words with thisproperty we must have | b | = | a | ≤ | c | . Now, since ac ≃ u , it is not difficult to see that every wordwhich is a subword of some cyclic permutation of u and has length at most | c | must occur as asubword of the word cac . In particular b ≺ cac , and since | b | = | a | we have either b ≺ ca or b ≺ ac .In either case we have shown that there exists a cyclic permutation of u which has both a and b as subwords, and no word with that property may be balanced. We conclude that (i) cannot nothold when (ii) does not hold, and so (i) = ⇒ (ii) as required.It is now straightforward to show that (ii) = ⇒ (iii). Let u be a finite nonempty word such that u is balanced; then every cyclic permutation of u is balanced, since the cyclic permutations of u are precisely the subwords of u with length | u | . Now, the cyclic permutations of u are preciselythe words of the form v where v ≃ u ; but since (i) = ⇒ (ii), all of these cyclic permutationsmust be balanced also. Applying the implication (i) = ⇒ (ii) again we deduce that u is balanced.Repeating this procedure inductively shows that u k is balanced for every k ≥
1, and this yields(iii). (cid:3) Relationships between balanced words and extremal orbits
The principal goal of this section is to show that for each α ∈ (0 , x ∈ Σwhich is strongly extremal for A α is balanced. We also prove some related ancillary results whichwill be applied in the following section.The following valuable lemma shows that under quite mild conditions the trace, spectral radius,Euclidean norm and smallest diagonal element of a matrix of the form A ( u ) approximate eachother quite closely. For every B ∈ M ( R ) we define d ( B ) to be the minimum modulus of thediagonal entries of B . Lemma 5.1.
Let α ∈ [0 , and N ≥ , and let u be a nonempty finite word such that N , N ⊀ u .Then, N (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) A ( α ) ( u ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ d (cid:16) A ( α ) ( u ) (cid:17) ≤
12 tr A ( α ) ( u ) ≤ ρ (cid:16) A ( α ) ( u ) (cid:17) ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) A ( α ) ( u ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . Proof.
Let m ( B ) denote the maximum of the entries of a non-negative matrix B ∈ M ( R ). Theinequalities ||| B ||| = p ρ ( B ∗ B ) ≤ p tr ( B ∗ B ) ≤ m ( B )and d ( B ) ≤
12 tr B ≤ ρ ( B ) ≤ ||| B ||| are elementary. To prove the lemma, it therefore suffices to show that m (cid:0) A ( α ) ( u ) (cid:1) ≤ N d ( A ( α ) ( u ))whenever 0 N , N ⊀ u . Since A ( α ) ( u ) ≡ α | u | A ( u ) it is clearly sufficient to consider only the case α = 1.Let us prove this inequality. We shall suppose that the final symbol occurring in u is 0, sincethe opposite case is easily dealt with by symmetry. Let n ≥ a , . . . , a n ≥ u = 0 a n a n − a n − · · · a a with n odd, or u = 1 a n a n − a n − · · · a a with n even.By hypothesis we have a k ≤ N − k . OUNTEREXAMPLE TO THE FINITENESS CONJECTURE 11
For 1 ≤ k ≤ n let us define p k q k := 1 a + 1 a + · · · + 1 a k − + 1 a k in least terms, and define also p , q − := 0 and p − , q := 1. The integers p k , q k then satisfy therecurrence relations p k = a k p k − + p k − and q k = a k q k − + q k − for all k in the range 1 ≤ k ≤ n .A well-known formula for p k /q k implies A ( u ) = A a n A a n − · · · A a = (cid:18) a n (cid:19) (cid:18) a n − (cid:19) · · · (cid:18) a (cid:19) = (cid:18) p n q n p n − q n − (cid:19) if n is odd, and A ( u ) = A a n A a n − · · · A a = (cid:18) a n (cid:19) (cid:18) a n − (cid:19) · · · (cid:18) a (cid:19) = (cid:18) p n − q n − p n q n (cid:19) if n is even (see, e.g., [16]). If n is odd then clearly d ( A ( u )) = min { p n , q n − } , and since q n = a n q n − + q n − ≤ ( a n + 1) q n − ≤ N q n − and p n /q n ≥ / ( a + 1) ≥ /N we obtain m ( A ( u )) = q n ≤ min { N p n , N q n − } < N d ( A ( u )) as required. If n is even then similarly m ( A ( u )) = q n ≤ N q n − ≤ N p n − = N d ( A ( u )). The proof is complete. (cid:3) Let a, w, b be nonempty finite words with | a | = | b | . We shall say that ( a, w, b ) is a suboptimaltriple if either ˜ a > b and w > ˜ w , or ˜ b > a and ˜ w > w . We require the following lemma due to V.Blondel, J. Theys and A. Vladimirov [6, Lemma 4.2]: Lemma 5.2.
Let w be a nonempty finite word. Then A ( ˜ w ) − A ( w ) = k ( w ) J , where k ( w ) ∈ Z and J := A A − A A = (cid:18) − (cid:19) . Moreover, k ( w ) is positive if and only if w > ˜ w , and negative if and only if w < ˜ w . The following is a slightly strengthened version of [6, Lemma 4.3]:
Lemma 5.3.
Let ( a, w, b ) be a suboptimal triple, let B , B be non-negative matrices, and let α ∈ [0 , . Then tr (cid:16) B A ( α ) ( a ˜ wb ) B (cid:17) ≥ tr (cid:16) B A ( α ) ( awb ) B (cid:17) + α | awb | d ( B ) d ( B ) . Proof.
Since tr A ( α ) ( u ) = α | u | tr A ( u ) for every finite word u it is clearly sufficient to treat onlythe case α = 1. We shall deal first with the case where ˜ a > b and w > ˜ w , the alternative casebeing similar. Since ˜ a > b we may write a = u c , b = ˜ c v for some finite words c , u and v (whichmay be empty). Note that J satisfies the relations A JA = A JA = J, A JA = (cid:18) − − − (cid:19) , A JA = (cid:18) (cid:19) , and hence by Lemma 5.2,tr ( A ( a )( A ( ˜ w ) − A ( w )) A ( b )) = k ( w )tr (cid:18) A ( u ) (cid:18) (cid:19) A (˜ v ) (cid:19) ≥ . Now, a direct calculation shows that for any non-negative matrix C ∈ M ( R ) we have tr ( B CB ) ≥ d ( B ) d ( B )tr ( C ). Since the matrix A ( a )( A ( ˜ w ) − A ( w )) A ( b ) is non-negative, we deduce thattr ( B A ( a ˜ wb ) B ) − tr ( B A ( awb ) B ) = tr ( B A ( a )( A ( ˜ w ) − A ( w )) A ( b ) B ) ≥ d ( B ) d ( B )tr ( A ( a )( A ( ˜ w ) − A ( w )) A ( b )) ≥ d ( B ) d ( B ) as required. In the case where ˜ b > a and ˜ w > w , the integer k ( w ) and the matrix A JA eachcontribute a negative sign to the product A ( a )( A ( ˜ w ) − A ( w )) A ( b )) and the same conclusion maybe reached. (cid:3) We may now prove the following two results which will allow us to characterise extremal orbitsin terms of balanced words:
Lemma 5.4.
Let ≤ pq ≤ , with the integers p and q not necessarily coprime. Suppose that | u | = q , | u | = p and (5.1) ρ ( A ( u )) = max { ρ ( A ( v )) : | v | = q and | v | = p } . Then the infinite word u ∞ is balanced.Proof. We shall begin by showing that if u has the properties described then it is balanced. Letus assume for a contradiction that u has these properties but is not balanced. By Lemma 4.6,there exists a suboptimal triple ( a, w, b ) such that awb ≺ u . Let us write u = s awbs and defineˆ u := s a ˜ wbs . By Lemma 5.3 we have tr ( A (ˆ u )) > tr ( A ( u )). Since A (ˆ u ) and A ( u ) are bothnon-negative matrices with unit determinant, it follows that ρ ( A (ˆ u )) = 12 (cid:16) tr ( A (ˆ u )) + p tr ( A (ˆ u )) − (cid:17) > (cid:16) tr ( A ( u )) + p tr ( A ( u )) − (cid:17) = ρ ( A ( u )) . Since clearly | ˆ u | = | u | and | ˆ u | = | u | this is a contradiction, so u must be balanced as required.Now, suppose that u satisfies (5.1) with | u | = p and | u | = q , and that v is a cyclic permutationof u . It is a well-known property of the spectral radius that ρ ( B B ) = ρ ( B B ) for any B , B ∈ M ( R ), and it follows from this that ρ ( A ( v )) = ρ ( A ( u )). By applying the preceding argumentto v it follows that v is also balanced. We conclude that all of the cyclic permutations of u arebalanced, and by Lemma 4.7 this implies that u ∞ is balanced as required. (cid:3) Proposition 5.5.
Let α ∈ (0 , and suppose that x ∈ Z α . Then x is balanced.Proof. To prove the proposition, let us suppose that there exists a recurrent infinite word x ∈ Z α which is not balanced. We shall then be able to deduce a contradiction, and the result follows.The general principle of the proof is that if x is recurrent and not balanced, then we can constructa word based on x along which the trace of the product A ( α ) ( x, n ) grows “too rapidly”.Fix a real number C α > C − α k B k α ≤ k B k ≤ C α k B k α for all B ∈ M ( R ). ByLemma 3.4 we have ̺ ( α ) >
1, and by Gelfand’s formula we have (cid:13)(cid:13) A ( α ) (0 n ) (cid:13)(cid:13) /nα → n → ∞ .It follows in particular that there is an integer N ≥ (cid:13)(cid:13) A ( α ) (cid:0) N (cid:1)(cid:13)(cid:13) α < ̺ ( α ) N andtherefore 0 N ⊀ z for every z ∈ Z α . Similarly we may choose N ≥ N ⊀ z for every z ∈ Z α . Let N := max { N , N } , and choose a further integer M ≥ n(cid:13)(cid:13)(cid:13) A ( α ) (cid:0) M (cid:1)(cid:13)(cid:13)(cid:13) α , (cid:13)(cid:13)(cid:13) A ( α ) (cid:0) M (cid:1)(cid:13)(cid:13)(cid:13) α o < ̺ ( α ) M C α N . If v is any subword of x , then there exists n ≥ A ( α ) ( v ) = A ( α ) ( T n x, | v | ), and since T n x ∈ Z α this implies(5.2) d (cid:16) A ( α ) ( v ) (cid:17) ≥ N (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) A ( α ) ( v ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ C α N (cid:13)(cid:13)(cid:13) A ( α ) ( T n x, | v | ) (cid:13)(cid:13)(cid:13) α = ̺ ( α ) | v | C α N , where we have used Lemma 5.1. On the other hand, for any nonempty finite word u ,(5.3) tr A ( α ) ( u ) ≤ ρ (cid:16) A ( α ) ( u ) (cid:17) ≤ (cid:13)(cid:13)(cid:13) A ( α ) ( u ) (cid:13)(cid:13)(cid:13) α ≤ ̺ ( α ) | u | . Now, since x is not balanced, it by definition has a subword which is not balanced. ApplyingLemma 5.3 to this subword we deduce that there exists a suboptimal triple ( a, w, b ) such that awb ≺ x . Define ℓ := | awb | , and fix an integer K ≥ (cid:18) α ℓ C α N M ̺ ( α ) ℓ (cid:19) K > C α N . OUNTEREXAMPLE TO THE FINITENESS CONJECTURE 13
Since x is recurrent there are infinitely many occurrences of the word awb as a subword of x , andso we may choose words s , . . . , s K +1 such that the word u (0) := s ( awb ) s ( awb ) s . . . . . . s K ( awb ) s K +1 is a subword of x . Let L := | u (0) | , and for i = 1 , . . . , K define a new word u ( i ) by reversing thefirst i explicit instances of the word w in u (0) ; that is, u (1) := s ( a ˜ wb ) s ( awb ) s . . . . . . s K ( awb ) s K +1 ,u (2) := s ( a ˜ wb ) s ( a ˜ wb ) s . . . . . . s K ( awb ) s K +1 , and so forth, up to u ( K ) := s ( a ˜ wb ) s ( a ˜ wb ) s . . . . . . s K ( a ˜ wb ) s K +1 . Note that for each i we have, by applying Lemma 5.3 i times and using (5.2),(5.4) tr A ( α ) ( u ( i ) ) ≥ tr A ( α ) ( u (0) ) ≥ d (cid:16) A ( α ) ( u (0) ) (cid:17) ≥ ̺ ( α ) L C α N , since u (0) is a subword of x . As a consequence we observe that 0 M ⊀ u ( i ) for every i , since if wewere to have 0 M ≺ u ( i ) for some i then we could obtain ̺ ( α ) L C α N ≤
12 tr A ( α ) ( u ( i ) ) ≤ ρ (cid:16) A ( α ) ( u ( i ) ) (cid:17) ≤ (cid:13)(cid:13)(cid:13) A ( α ) ( u ( i ) ) (cid:13)(cid:13)(cid:13) α ≤ (cid:13)(cid:13)(cid:13) A ( α ) (cid:0) M (cid:1)(cid:13)(cid:13)(cid:13) α .̺ ( α ) L − M < ̺ ( α ) L C α N , a contradiction. Clearly an analogous contradiction would arise if we were to have 1 M ≺ u ( i ) andwe conclude that 1 M ⊀ u ( i ) also.Now, for i = 1 , . . . , K let c ( i ) , d ( i ) be those words such that u ( i − = c ( i ) awbd ( i ) and u i = c ( i ) a ˜ wbd ( i ) . Note that | c ( i ) | + | d ( i ) | + ℓ = L for each i . Making i applications of Lemma 5.3 andusing (5.2) yields tr A ( α ) ( c ( i ) ) = tr A ( α ) ( s ( a ˜ wb ) s . . . s i − ( a ˜ wb ) s i ) ≥ tr A ( α ) ( s ( awb ) s . . . s i − ( awb ) s i ) ≥ ̺ ( α ) | c ( i ) | C α N , since the last of these words is a subword of u (0) , and u (0) is a subword of x . Since c ( i ) ≺ u ( i ) and 0 M , M ⊀ u ( i ) we have 0 M , M ⊀ c ( i ) , and by Lemma 5.1 in combination with the precedinginequality this implies(5.5) d (cid:16) A ( α ) ( c ( i ) ) (cid:17) ≥ M tr A ( α ) ( c ( i ) ) ≥ ̺ ( α ) | c ( i ) | C α N M . Equally, since d ( i ) ≺ u (0) and u (0) is a subword of x , we may apply (5.2) to obtain(5.6) d (cid:16) A ( α ) ( d ( i ) ) (cid:17) ≥ ̺ ( α ) | d ( i ) | C α N . We may now complete the proof. Combining (5.5), (5.6), and (5.3) we obtain for each iα | awb | d (cid:16) A ( α ) ( c ( i ) ) (cid:17) d (cid:16) A ( α ) ( d ( i ) ) (cid:17) ≥ α ℓ ̺ ( α ) L − ℓ C α N M ≥ α ℓ C α N M ̺ ( α ) ℓ tr A ( α ) ( u ( i − ) , and hence by Lemma 5.3,tr A ( α ) ( u ( i ) ) ≥ (cid:18) α ℓ C α N M ̺ ( α ) ℓ (cid:19) tr A ( α ) ( u ( i − ) . In combination with (5.3) and (5.4) this yields2 ̺ ( α ) L ≥ tr A ( α ) ( u ( K ) ) ≥ (cid:18) α ℓ C α N M ̺ ( α ) ℓ (cid:19) K tr A ( α ) ( u (0) ) ≥ (cid:18) α ℓ C α N M ̺ ( α ) ℓ (cid:19) K · ̺ ( α ) L C α N , contradicting our choice of K . The proof is complete. (cid:3) Study of the growth of matrix products along balanced words
In this section we analyse in detail the exponential growth rate of A ( x, n ) in the limit as n → ∞ for x ∈ X γ , investigating in particular the manner in which this value depends on γ . A constructionwith similar properties is discussed briefly in [8, § Proposition 6.1. • There exists a continuous concave function S : [0 , → R such thatfor each γ ∈ [0 , , lim n →∞ n log |||A ( x, n ) ||| = lim n →∞ n log ρ ( A ( x, n )) = S ( γ ) uniformly for x ∈ X γ . • If γ = p/q ∈ [0 , ∩ Q then S ( γ ) = q − log ρ ( A ( x, q )) for every x ∈ X γ . • The function S also satisfies inf γ ∈ [0 , S = S (0) = S (1) = 0 , sup S = S (1 /
2) = log ̺ (1) ,and S ( γ ) = S (1 − γ ) for all γ ∈ [0 , . • The function S is non-decreasing on (cid:2) , (cid:3) . The proof of Proposition 6.1 is given in the form of a sequence of lemmas. Specifically, theresult follows by combining Lemmas 6.2–6.4 and Lemma 6.6 below.
Lemma 6.2.
Let γ ∈ [0 , . Then there exists a real number S ( γ ) such that lim n →∞ n log |||A ( x, n ) ||| = lim n →∞ n log ρ ( A ( x, n )) = S ( γ ) uniformly over x ∈ X γ .Proof. In the cases γ = 0, γ = 1 the lemma is trivial, since by Theorem 2.1 the set X γ consists ofa single point which is fixed under T , and the result follows by Gelfand’s formula. To prove thelemma in the nontrivial cases we use a result due to A. Furman [17] on uniform convergence forlinear cocycles over homeomorphisms. Since in general the transformations T : X γ → X γ are nothomeomorphisms, this is achieved via an auxiliary construction.Let us fix γ ∈ (0 , X γ ⊂ { , } Z as follows: thesequence x = ( x n ) n ∈ Z ∈ { , } Z belongs to ˆ X γ if and only if there exists δ ∈ [0 ,
1] such that either x n ≡ ⌈ ( n + 1) γ + δ ⌉ − ⌈ nγ + δ ⌉ for all n ∈ Z , or x n ≡ ⌊ ( n + 1) γ + δ ⌋ − ⌊ nγ + δ ⌋ for all n ∈ Z .It follows from the discussion subsequent to the statement of Theorem 2.1 that the two-sidedsequence ( x i ) i ∈ Z belongs to ˆ X γ if and only if the one-sided sequence ( x i + k ) ∞ i =1 belongs to X γ forevery k ∈ Z . We equip ˆ X γ with the topology it inherits from the infinite product topology on { , } Z , and define ˆ T : ˆ X γ → ˆ X γ by ˆ T [( x i ) i ∈ Z ] := ( x i +1 ) i ∈ Z analogously to the definition of T . Inthe same manner as for the transformation T : X γ → X γ , one may show that ˆ T : ˆ X γ → ˆ X γ is acontinuous, uniquely ergodic transformation of a compact metrisable space.Finally, we define ˆ A : ˆ X γ × Z → M ( R ) in the following manner: given x = ( x i ) i ∈ Z ∈ ˆ X γ and n ≥
1, we define ˆ A ( x, n ) := A x n · · · A x , ˆ A ( x, − n ) := A − x − ( n − A − x − ( n − · · · A − x = ˆ A ( ˆ T − n x, n ) − ,and ˆ A ( x,
0) = I . It may be directly verified that ˆ A is continuous and satisfies the following cocyclerelation: for all x ∈ ˆ X γ and n, m ∈ Z , we have ˆ A ( x, n + m ) = ˆ A ( ˆ T n x, m ) ˆ A ( x, n ).Now let N ≥ x ∈ X γ we have 0 N , N ⊀ x . Since foreach x = ( x i ) i ∈ Z ∈ ˆ X γ we have ( x i ) ∞ i =1 ∈ X γ , it follows from this that the matrix product whichdefines ˆ A ( x, N ) is a product of mixed powers of A and A , and does not simply equal A N or OUNTEREXAMPLE TO THE FINITENESS CONJECTURE 15 A N . A simple calculation shows that this implies that for each x ∈ ˆ X γ , all of the entries of thematrix ˆ A ( x, N ) are strictly positive. We may therefore apply [17, Theorem 3] to deduce that thereexists a real number S ( γ ) such that n log k ˆ A ( x, n ) k converges uniformly to S ( γ ) for x ∈ ˆ X γ . Sinceclearly for each n ≥ n ˆ A ( x, n ) : x ∈ ˆ X γ o = {A ( x, n ) : x ∈ X γ } , this implies that n log kA ( x, n ) k converges uniformly to S ( γ ) for x ∈ X γ . Since as previouslynoted we have 0 N , N ⊀ x for all x ∈ X γ , it follows immediately from Lemma 5.1 that also n log ρ ( A ( x, n )) → S ( γ ) uniformly over x ∈ X γ . The proof is complete. (cid:3) Lemma 6.3.
The function S has the following properties:(i) Let γ = p/q ∈ [0 , , not necessarily in least terms: then S ( γ ) = q − log ρ ( A ( x, q )) for every x ∈ X p/q .(ii) Let u be a finite word such that | u | = q , | u | = p . Then S ( p/q ) ≥ q − log ρ ( A ( u )) .(iii) Let γ ∈ [0 , be irrational. Then there exist x ∈ X γ and a sequence of rational numbers ( p n /q n ) ∞ n =1 converging to γ such that S ( p n /q n ) = q − n log ρ ( A ( x, q n )) for every n ≥ .(iv) For every γ ∈ [0 , we have S ( γ ) = S (1 − γ ) .Proof. (i). By Theorem 2.1 we have T q x = x for every x ∈ X p/q , and so for every x ∈ X p/q , S ( p/q ) = lim n →∞ kq log |||A ( x, kq ) ||| = lim k →∞ kq log (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) A ( x, q ) k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 1 q log ρ ( A ( x, q )) . (ii). Clearly the set of all words v such that | v | = q and | v | = p is finite, so there existsa word v which attains the maximum value of ρ ( A ( v )) within this set. In particular we have ρ ( A ( v )) ≥ ρ ( A ( u )). By Lemma 5.4 the infinite word v ∞ ∈ Σ is balanced, and since it is clearlyrecurrent we have v ∞ ∈ X p/q by Theorem 2.1. By part (i) this implies q − log ρ ( A ( v )) = S ( p/q )as required.(iii) Let x ∈ X γ be as given by Lemma 4.3(ii), and let ( q n ) ∞ n =1 be a strictly increasing sequenceof natural numbers such that π q n ( x ) is a standard word for every n . Define p n := | π q n ( x ) | for each n ≥
1. By the definition of X γ we have p n /q n → γ . Since each π q n ( x ) is standard,[ π q n ( x )] ∞ ∈ X p n /q n for each n by Lemma 4.3(i), and by part (i) of the present lemma this implies S ( p n /q n ) = q − n log ρ ( A ( x, q n )).(iv) For each finite or infinite word ω , define ω to be the mirror image of ω , i.e., the uniqueword such that ω i = 1 if and only if ω i = 0. It is clear that x ∈ X γ if and only if x ∈ X − γ . Define R = ( ) and note that R − A R = A and R − A R = A . If x ∈ X γ and n ≥
1, then R − A ( x, n ) R = ( R − A x n R ) · · · ( R − A x R )( R − A x R ) = A ( x, n )and in particular ρ ( A ( x, n )) = ρ ( A ( x, n )). It follows easily that S ( γ ) = S (1 − γ ). (cid:3) Lemma 6.4.
The function S satisfies S (0) = inf S = 0 and S ( ) = sup S = log ̺ (1) .Proof. The reader may easily verify that(6.1) ||| A ||| = ||| A ||| = ||| A A ||| = ||| A A ||| = ρ ( A A ) / = 1 + √ . By Theorem 2.1, we have X / = { (01) ∞ , (10) ∞ } , so by Gelfand’s formula we havelim n →∞ |||A ( x, n ) ||| /n = ρ ( A A ) = ρ ( A A ) = 1 + √ x ∈ X / . Let us show that ̺ (1) = √ . In other words, we will prove thatsup n |||A ( x, n ) ||| /n : x ∈ Σ o = lim n →∞ |||A ((01) ∞ , n ) ||| /n = 1 + √ . Suppose x has a tail different from (01) ∞ . Then it must contain one of the following subwords: w = 11(01) n , w = 11(01) n , w = 00(10) n , w = 00(10) n
11 with n ≥
0. In view of mirrorsymmetry, it suffices to deal with w and w . We will show that it is possible to replace them with subwords of (01) ∞ , w ′ and w ′ respectively, in such a way that the corresponding growthexponent does not decrease.Namely, put w ′ = (10) n +1 w ′ = (10) n +2 . It is easy to see that for n ≥ A A ) n = (cid:18) F n F n − F n − F n − (cid:19) ( A A ) n = (cid:18) F n − F n − F n − F n (cid:19) , where, as above, ( F n ) ∞ n =0 is the Fibonacci sequence (with F = F = 1). Hence A ( A A ) n A = (cid:18) F n +1 F n − F n +3 F n +1 (cid:19) , whereas ( A A ) n +1 A = (cid:18) F n +2 F n +1 F n +3 F n +2 (cid:19) , i.e., A ( w ′ ) dominates A ( w ) entry-by-entry. Similarly, A ( A A ) n A = (cid:18) F n F n +2 F n +2 F n +4 (cid:19) and ( A A ) n +2 = (cid:18) F n +2 F n +3 F n +3 F n +4 (cid:19) . Thus, ̺ (1) = √ = e S ( ) , and since clearly S ( γ ) ≤ log ̺ (1) for every γ ∈ [0 ,
1] this implies thatsup S = S (1 / X contains a single point x corresponding toan infinite sequence of zeroes, and for this x we have S (0) = log ρ ( A ) = 0. Finally, since everymatrix A ( x, n ) is an integer matrix which has determinant one and is hence nonzero, every x ∈ Σhas n log |||A ( x, n ) ||| ≥ n and therefore S ( γ ) ≥ γ . (cid:3) Lemma 6.5.
The restriction of S to (0 , ∩ Q is concave in the following sense: if γ , γ , λ ∈ (0 , ∩ Q then S ( λγ + (1 − λ ) γ ) ≥ λS ( γ ) + (1 − λ ) S ( γ ) .Proof. For i = 1 , γ i = p i /q i in least terms, and let λ = k/m . Let M = max { q , q } . As aconsequence of Lemma 6.3(i) there exist finite words u (1) , u (2) ∈ Ω such that | u ( i ) | = p i , | u ( i ) | = q i and S ( γ i ) = q − i log A ( u ( i ) ) for each i .Since 0 < γ , γ < < | p i | < | q i | and therefore 0 M , M ⊀ ( u ( i ) ) ℓ for i = 1 , ℓ ≥
1. In particular, for each ℓ , ℓ ≥ u (1) ) ℓ ( u (2) ) ℓ does not have 0 M or 1 M as a subword, and hence by Lemma 5.1, ρ (cid:0) A ( u (1) ) ℓ ( u (2) ) ℓ (cid:1) ≥ d (cid:16) A ( u (1) ) ℓ ( u (2) ) ℓ (cid:17) ≥ d ( A ( u (1) ) ℓ ) d ( A (( u (2) ) ℓ ) ≥ M ρ (cid:16) A ( u (1) ) ℓ (cid:17) ρ (cid:16) A ( u (2) ) ℓ (cid:17) . Applying this inequality together with Lemma 6.3(ii), for each n ≥ S ( λγ + (1 − λ ) γ ) = S (cid:18) kp q + ( m − k ) q p mq q (cid:19) ≥ nmq q log ρ ( A (( u (1) ) nkq ( u (2) ) n ( m − k ) q )) ≥ nmq q (cid:16) log ρ ( A (( u (1) ) nkq )) + log ρ ( A (( u (2) ) n ( m − k ) q )) − log 64 M (cid:17) = kmq log ρ ( A ( u (1) )) + m − kmq log ρ ( A ( u (2) )) − log 64 M nmq q = λS ( γ ) + (1 − λ ) S ( γ ) − log 64 M nmq q . Taking the limit as n → ∞ we obtain the desired result. (cid:3) OUNTEREXAMPLE TO THE FINITENESS CONJECTURE 17
Lemma 6.6.
The function S : [0 , → R is continuous and concave.Proof. By Lemma 6.5, the restriction of S to (0 , ∩ Q is concave. Define a function e S : [0 , → R by e S ( γ ) := lim ε → sup { S ( γ ∗ ) : γ ∗ ∈ (0 , ∩ Q and | γ ∗ − γ | < ε } . Note that e S is well-defined since S is bounded by Lemma 6.4. We shall show in several stagesthat e S is continuous, concave, and equal to S throughout [0 , e S is concave. Let γ , γ , λ ∈ [0 , (cid:16) γ ( n )1 (cid:17) , (cid:16) γ ( n )2 (cid:17) and ( λ n ) belonging to (0 , γ , γ and λ , such thatlim n →∞ S (cid:16) γ ( n ) i (cid:17) = e S ( γ i ) for i = 1 ,
2. We then have e S ( λγ + (1 − λ ) γ ) ≥ lim sup n →∞ S (cid:16) λ n γ ( n )1 + (1 − λ n ) γ ( n )2 (cid:17) ≥ lim sup n →∞ λ n S (cid:16) γ ( n )1 (cid:17) + (1 − λ n ) S (cid:16) γ ( n )2 (cid:17) = lim n →∞ λ n S (cid:16) γ ( n )1 (cid:17) + (1 − λ n ) S (cid:16) γ ( n )2 (cid:17) = λ e S ( γ ) + (1 − λ ) e S ( γ )using Lemma 6.5, and e S is concave as claimed. In particular the restriction of e S to the interval(0 ,
1) is continuous (see for example [43, Thm 10.3]).We next claim that e S ( γ ) = S ( γ ) for rational values 0 < γ <
1. Given γ ∈ (0 , ∩ Q , choose asequence of rationals ( γ n ) such that γ n → γ and S ( γ n ) → e S ( γ ). If 0 < γ ≤ γ n for some n then S ( γ ) ≥ (cid:18) − γγ n (cid:19) S (0) + γγ n S ( γ n ) = γγ n S ( γ n ) , and similarly if γ n < γ < S ( γ ) ≥ (cid:18) − γ − γ n (cid:19) S ( γ n ) + (cid:18) γ − γ n − γ n (cid:19) S (1) ≥ (cid:18) − γ − γ n (cid:19) S ( γ n ) . It follows that by taking the limit as n → ∞ we may obtain S ( γ ) ≥ e S ( γ ), and the converseinequality e S ( γ ) ≥ S ( γ ) is obvious from the definition of e S . This proves the claim.We now claim that lim γ → e S ( γ ) = e S (0) = 0 = S (0) and lim γ → e S ( γ ) = e S (1) = 0 = S (1).Since S ( γ ) = S (1 − γ ) for every γ ∈ [0 ,
1] by Lemma 6.3(iv) it is sufficient to prove only the firstassertion. By Lemma 6.4 we have S (0) = inf S = 0 and therefore inf e S ≥
0. Since e S is concavethere must exist δ > e S to [0 , δ ) is monotone, and so if we can showthat lim n →∞ e S (1 /n ) = 0 then the desired result will follow. By the preceding claim it is sufficientto show that lim n →∞ S (1 /n ) = 0. For each n ≥ n ∞ ∈ X /n , so using Lemma 6.3(i) we may estimate0 ≤ S (cid:18) n (cid:19) = 1 n + 1 log ρ ( A n A ) ≤ n + 1 log tr ( A n A ) = log( n + 2) n + 1and therefore S (1 /n ) →
0. This completes the proof of the claim.To complete the proof of the lemma it suffices to show that in fact e S ( γ ) = S ( γ ) when γ isirrational. Given γ ∈ [0 , \ Q , let x ∈ X γ and ( p n /q n ) ∞ n =1 be as given by Lemma 6.3(iii). Since e S is continuous and agrees with S on the rationals, we may apply parts (iii) and (i) of Lemma 6.3to obtain S ( γ ) = lim n →∞ q n log ρ ( A ( x, q n )) = lim n →∞ S (cid:18) p n q n (cid:19) = lim n →∞ e S (cid:18) p n q n (cid:19) = e S ( γ ) , and we conclude that e S ≡ S as desired. (cid:3) To conclude the proof of Proposition 6.1, we note that the function S being non-decreasing on (cid:2) , (cid:3) follows from its concavity and the fact that max γ ∈ [0 , / S ( γ ) = S (1 / Proof of Theorem 2.3
Before commencing the proof of Theorem 2.3, we require the following simple lemma:
Lemma 7.1.
For each α ∈ [0 , we have ̺ ( α ) ≥ e S ( γ ) α γ for all γ ∈ [0 , . If α ∈ (0 , and X γ ∩ Z α = ∅ , then X γ ⊆ Z α and ̺ ( α ) = e S ( γ ) α γ .Proof. In the case α = 0, an easy calculation using Proposition 6.1 and the definition of ̺ showsthat ̺ ( α ) = ρ ( A ) = 1 = e S (0) . It is therefore clear in this case that ̺ ( α ) = e S ( γ ) α γ if and only if γ = 0. For the rest of the proof let us fix α ∈ (0 ,
1] and γ ∈ [0 , x ∈ X γ , we havelog ̺ ( α ) = lim sup n →∞ sup (cid:26) n log (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) A ( α ) ( z, n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) : z ∈ Σ (cid:27) ≥ lim n →∞ n log (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) A ( α ) ( x, n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = lim n →∞ (cid:18) n log |||A ( x, n ) ||| + ς ( π n ( x )) log α (cid:19) = S ( γ ) + γ log α so that ̺ ( α ) ≥ e S ( γ ) α γ . If x ∈ X γ ∩ Z α then by the definition of Z α we have S ( γ ) + γ log α = lim n →∞ n log (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) A ( α ) ( x, n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = lim n →∞ n log (cid:13)(cid:13)(cid:13) A ( α ) ( x, n ) (cid:13)(cid:13)(cid:13) α = log ̺ ( α )so that ̺ ( α ) = e S ( γ ) α γ , and since by Theorem 2.1 the restriction of T to X γ is minimal it is clearthat X γ ⊆ Z α . (cid:3) We also require the following lemma, which is an easy consequence of a result in [6]:
Lemma 7.2.
Let α ∈ [0 , and let u, v be nonempty finite words such that ρ ( A ( α ) ( u )) / | u | = ρ ( A ( α ) ( v )) / | v | = ̺ ( α ) . Then ς ( u ) = ς ( v ) .Proof. In [6], Blondel, Theys and Vladimirov define two nonempty finite words u, v to be essentiallyequal if there exist finite words a, b such that au ∞ = bv ∞ . In particular it is clear that if u and v are essentially equal then necessarily ς ( u ) = ς ( v ). Blondel et al. then associate to each nonemptyfinite word ω the set J ω = { α ∈ [0 ,
1] : A ( α ) ( ω ) = ̺ ( α ) | ω | } . In [6, Lemma 4.4] it is shown that if J u ∩ J v = ∅ then u and v are essentially equal. We deduce from this that if u and v are nonemptyfinite words which satisfy ρ ( A ( α ) ( u )) / | u | = ρ ( A ( α ) ( v )) / | v | = ̺ ( α ) for some fixed α ∈ [0 , α ∈ J u ∩ J v by definition; this implies that u and v are essentially equal, and therefore ς ( u ) = ς ( v ). (cid:3) Now we are ready to prove Theorem 2.3.
1. Existence of r . We shall begin by showing that for each α ∈ (0 ,
1] there exists a unique γ ∈ [0 ,
1] such that X γ ∩ Z α = ∅ . Let α ∈ (0 , Z α is compact andinvariant under T , and this implies that it contains a recurrent point (see e.g. [27, p.130]). Itfollows by Proposition 5.5 that Z α contains a recurrent balanced infinite word, and hence thereexists γ α ∈ [0 ,
1] such that X γ α ∩ Z α = ∅ . By Lemma 7.1 it follows that e S ( γ α ) α γ α = ̺ ( α ). Weclaim that γ α is the unique element of [0 ,
1] with this property. By Lemma 7.1 this further impliesthat X γ ∩ Z α = ∅ when γ = γ α .To prove this claim, let us suppose that 0 ≤ γ < γ ≤ e S ( γ ) α γ = e S ( γ ) α γ = ̺ ( α ),and derive a contradiction. Choose λ , λ ∈ [0 ,
1] such that ˜ γ := λ γ + (1 − λ ) γ and ˜ γ := λ γ + (1 − λ ) γ are both rational with γ ≤ ˜ γ < ˜ γ ≤ γ . Applying Proposition 6.1 we deduce S (˜ γ i ) + ˜ γ i log α = S ( λ i γ + (1 − λ i ) γ ) + ( λ i γ + (1 − λ i ) γ ) log α ≥ λ i ( S ( γ ) + γ log α ) + (1 − λ i )( S ( γ ) + γ log α ) = log ̺ ( α ) , and hence e S (˜ γ i ) α ˜ γ i ≥ ̺ ( α ), for i = 1 ,
2. Applying Lemma 7.1 it follows that e S (˜ γ ) α ˜ γ = e S (˜ γ ) α ˜ γ = ̺ ( α ). Let x ∈ X ˜ γ and y ∈ X ˜ γ , and let u := π q ( x ) and v := π q ( y ). By Propo-sition 6.1 we have ̺ ( α ) = ρ ( A α ( u )) / | u | = ρ ( A α ( v )) / | v | , and since ς ( u ) = ˜ γ < ˜ γ = ς ( v ) thiscontradicts Lemma 7.2. The claim is proved. OUNTEREXAMPLE TO THE FINITENESS CONJECTURE 19
Let us define r ( α ) := γ α for all α ∈ (0 , r (0) := 0. Note that ̺ (0) = ρ ( A ) = 1 = e S (0) asa consequence of Lemma 3.1 and Proposition 6.1. It follows from this and the previous argumentsthat for all α, γ ∈ [0 ,
1] we have ̺ ( α ) ≥ e S ( γ ) α γ with equality if and only if γ = r ( α ), and for all α ∈ (0 ,
1] we have X γ ∩ Z α = ∅ precisely when γ = r ( α ), in which case X r ( α ) ⊆ Z α .
2. Monotonicity of r . We now show that the function r thus defined is non-decreasing. Letus suppose that α , α ∈ [0 ,
1] with r ( α ) < r ( α ); this implies in particular that α is nonzero.By the preceding result we have ̺ ( α ) = e S ( r ( α )) α r ( α )1 > e S ( r ( α )) α r ( α )1 and similarly ̺ ( α ) = e S ( r ( α )) α r ( α )2 > e S ( r ( α )) α r ( α )2 . Consequently α r ( α ) − r ( α )1 < e S ( r ( α )) − S ( r ( α )) < α r ( α ) − r ( α )2 ,and since r ( α ) − r ( α ) > α < α . We conclude that if 0 ≤ α < α ≤ r ( α ) ≤ r ( α ) and therefore r is non-decreasing as required.
3. Continuity of r . We may now show that r is continuous. Given α ∈ (0 ,
1] let r − be the limitof r ( α ) as α → α from the left, which exists since r is monotone. For every α ∈ (0 ,
1] we have ̺ ( α ) = e S ( r ( α )) α r ( α ) . By Lemma 3.2 and Proposition 6.1, ̺ and S are continuous, so taking theleft limit at α yields e S ( r ( α )) α r ( α )0 = ̺ ( α ) = e S ( r − ) α r − . Since r ( α ) is the unique value for whichthis equality may hold we deduce that r ( α ) = r − as required. Similarly for every α ∈ [0 ,
1) thelimit of r ( α ) as α → α from the right is equal to r ( α ), and we conclude that r is continuous.Since r (0) = 0 and r (1) = 1 / r iscontinuous and monotone, we deduce that r maps [0 ,
1] surjectively onto [0 , ] as claimed.
4. 1-ratio and characterisation of extremal orbits.
It remains to show that for each α the extremal orbits of A α may be characterised in terms of X r ( α ) in the manner described by theTheorem, and that r ( α ) is the unique optimal 1-ratio of A α . In the case α = 0 it is obvious that x ∈ Σ is weakly extremal if and only if it is strongly extremal, if and only if x = 0 ∞ ∈ X , and inthis case the proof is then complete. For each α ∈ (0 , Z α , and therefore belongs to X r ( α ) by Proposition 5.5and the uniqueness property of r ( α ).To show that weakly extremal infinite words accumulate on X r ( α ) in the desired manner werequire an additional claim. Given α ∈ (0 , T -invariant Borelprobability measure whose support is contained in Z α , and that this support is equal to X r ( α ) .Indeed, let µ r ( α ) be the unique T -invariant measure with support equal to X r ( α ) , the existence ofwhich is given by Theorem 2.1. If ν is a T -invariant Borel probability measure with supp ν ⊆ Z α ,define e X := { x ∈ supp ν : x is recurrent } . It follows from the Poincar´e recurrence theorem that e X is dense in supp ν (see e.g. [27, Prop. 4.1.18]). By Proposition 5.5 every element of e X is balanced, and since r ( α ) is the unique γ ∈ [0 ,
1] for which X γ ∩ Z α = ∅ , it follows that e X ⊆ X r ( α ) . We conclude that supp ν ⊆ X r ( α ) and therefore ν = µ r ( α ) since the restriction of T to X r ( α ) is known to be uniquely ergodic, which proves the claim. By Theorem 2.1 we have µ r ( α ) ( { x ∈ Σ : x = 1 } ) = r ( α ), and we may now apply Lemma 3.8 to see that if x ∈ Σ is weaklyextremal for A α , then (1 /n ) P n − k =0 dist( x, X r ( α ) ) → ς ( π n ( x )) → r ( α ) as required.It remains only to show that for each α ∈ (0 , x ∈ X r ( α ) is strongly extremal in the strictfashion described by (2.1). Given any compact set K ⊆ (0 , N K large enoughthat N K > max {⌈ r ( α ) − ⌉ , ⌈ (1 − r ( α )) − ⌉} for every α ∈ K , and let M K > α ∈ K and x ∈ X r ( α ) . By Lemma 4.1 we have 0 N K , N K ⊀ x , and since X ⊆ Z α we have kA ( α ) ( x, n ) k α = ̺ ( α ) n for all n ≥
1. Applying Lemma 5.1 and Lemma 3.3, ̺ ( α ) n M K N K = 12 M K N K (cid:13)(cid:13)(cid:13) A ( α ) ( x, n ) (cid:13)(cid:13)(cid:13) α ≤ N K (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) A ( α ) ( x, n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ρ (cid:16) A ( α ) ( x, n ) (cid:17) ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) A ( α ) ( x, n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ M K (cid:13)(cid:13)(cid:13) A ( α ) ( x, n ) (cid:13)(cid:13)(cid:13) α ≤ M K ̺ ( α ) n < M K N K ̺ ( α ) n so that (2.1) holds with C K := 2 M K N K . In particular this shows that for each α ∈ (0 , x ∈ X r ( α ) is strongly extremal. The proof of the Theorem is complete. Proof of Theorem 1.1
Recall from Proposition 6.1 that there exists a continuous concave function S : [0 , → R suchthat for each γ ∈ [0 , S ( γ ) = lim n →∞ n log |||A ( x, n ) ||| = lim n →∞ n log ρ ( A ( x, n ))uniformly for x ∈ X γ . We saw in the course of the proof of Theorem 2.3 that the function r : [0 , → [0 , ] is characterised by the fact that ̺ ( α ) ≥ e S ( γ ) α γ for all α, γ ∈ [0 ,
1] with equalityif and only if γ = r ( α ). Readers who have skipped the proof of Theorem 2.3 may note that thischaracterisation can be deduced easily from the definition of S and the statement of Theorem 2.3.The proof of Theorem 1.1 operates by exploiting the concavity of S and the above relationshipbetween S and r to compute a value α ∗ ∈ [0 ,
1] such that r ( α ∗ ) / ∈ Q as the limit of a series ofapproximations. We begin with a result from convex analysis. Lemma 8.1.
For each γ ∈ (cid:0) , (cid:1) , we have r − ( γ ) = { α } with α ∈ (0 , if and only if S isdifferentiable at γ and S ′ ( γ ) = − log α . See Figure 1 for a graph of S ( γ ) along with the tangent line of slope α ∗ .Figure 1: Graph of S ( γ ), and tangent line at γ ≈ . . . . of slope − log(0 . . . . ) Proof.
Recall that if f : [ a, b ] → R is a concave function then η ∈ R is called a subgradient of f at z ∈ [ a, b ] if f ( y ) ≤ f ( z ) + η ( y − z ) for all y ∈ [ a, b ]. Furthermore, f is differentiable at z ∈ ( a, b )with f ′ ( z ) = η if and only if η is the unique subgradient of f at z (see for example [43, Thm 25.1]).To prove the lemma it therefore suffices to show that for each γ ∈ (0 , ), η ∈ R is a subgradientof S at γ if and only if e − η ∈ (0 ,
1] and r ( e − η ) = γ .Let us prove that this is the case. For every α, γ ∈ [0 ,
1] we have e S ( r ( α )) α r ( α ) ≥ e S ( γ ) α γ withequality if and only if γ = r ( α ). For each fixed α ∈ (0 ,
1) it now follows by a simple rearrangementthat − log α is a subgradient of S at r ( α ). Conversely, suppose that η ∈ R is a subgradient of S atsome γ ∈ (0 , ). By Proposition 6.1, S is monotone increasing on the interval [0 , ] and thereforewe must have η ≥
0. Since η is a subgradient we have e S ( γ ) − ηγ ≥ e S ( γ ) − ηγ for all γ ∈ [0 , e − η ∈ (0 ,
1] it follows that γ = r ( e − η ) as required. (cid:3) OUNTEREXAMPLE TO THE FINITENESS CONJECTURE 21
The following corollary is not needed in this paper but since it is straightforward, we believeit’s worth mentioning.
Corollary 8.2.
The function S is strictly concave on [0 , and strictly increasing on [0 , / .Proof. If S were not strictly concave, there would be an interval ( γ , γ ) such that S would belinear on this interval. Hence S ′ would be constant on ( γ , γ ) which would mean, in view of theprevious lemma, that r − ( γ ) would be constant for all γ ∈ ( γ , γ ). This contradicts r being welldefined (Theorem 2.3), whence S is strictly concave.Since S is non-decreasing, continuous and strictly concave on [0 , / (cid:3) Throughout this section we let φ := √ denote the golden ratio. Recall that a real number γ is said to be Liouville if for every k > p, q such that 0 < | γ − p/q | < /q k .A classical theorem of Liouville asserts that no algebraic number can be Liouville (see, e.g., [21,Theorem 191]). In particular φ − is not Liouville. Lemma 8.3.
Let γ ∈ [0 , ] and suppose that γ is an irrational number which is not Liouville.Then there exists a unique α ∈ [0 , such that r ( α ) = γ .Proof. By Theorem 2.3 the function r is surjective and monotone, so the set r − ( γ ) is either apoint or an interval. To show that this set cannot be an interval, we shall suppose that there exist α ∈ (0 ,
1) and ε > r ( α ) = γ for all α ∈ [ e − ε α , e ε α ], and derive a contradiction.Since γ is irrational but not Liouville, we may choose an integer k > p, q with q nonzero we have | γ − p/q | > /q k . A theorem due to the second named author [38,Thm 1.2] implies that for every r > (cid:26) ρ (cid:16) A ( α ) ( x, m ) (cid:17) m : x ∈ Σ and 1 ≤ m ≤ n (cid:27) = ̺ ( α ) + O (cid:18) n r (cid:19) in the limit as n → ∞ . In particular it follows that if n is some sufficiently large integer, thenthere exist an integer m and an infinite word x ∈ Σ such that 1 ≤ m ≤ n and(8.1) ρ (cid:16) A ( α ) ( x, m ) (cid:17) /m > (cid:18) − n k +1 (cid:19) ̺ ( α ) > e − εn − k ̺ ( α ) . Let ς ( π m ( x )) = p/q in least terms; we shall suppose firstly that pq − γ >
0, the opposite case beingsimilar. By hypothesis we have ̺ ( λα ) = e S ( r ( λα )) ( λα ) r ( α ) = e S ( γ ) ( λα ) γ = λ γ ̺ ( α ) for every λ ∈ [ e − ε , e ε ], and also pq − γ = | pq − γ | > q − k ≥ n − k . Combining this with (8.1) and Lemma 3.1we obtain ̺ ( e ε α ) ≥ ρ (cid:16) A ( e ε α ) ( x, m ) (cid:17) /m = e εp/q ρ (cid:16) A ( α ) ( x, m ) (cid:17) /m > e εp/q − εn − k ̺ ( α ) = e ε ( p/q − γ − n − k ) ̺ ( e ε α ) > ̺ ( e ε α ) , a contradiction. In the case pq − γ < ̺ (cid:0) e − ε α (cid:1) > e − εp/q − εn − k ̺ ( α ) = e ε ( γ − p/q − n − k ) ̺ (cid:0) e − ε α (cid:1) > ̺ (cid:0) e − ε α (cid:1) which is also a contradiction. The proof is complete. (cid:3) Let ( F n ) ∞ n =0 denote the Fibonacci sequence, which is defined by F := 0, F := 1 together withthe recurrence relation F n +2 := F n +1 + F n , and recall that F n = ( φ n − ( − /φ ) n ) / √ n ≥
0. Define a sequence of integers ( τ n ) ∞ n =0 by τ := 1, τ = τ := 2, and τ n +1 := τ n τ n − − τ n − for every n ≥
2. Finally, define a sequence of matrices ( B n ) ∞ n =1 by B := A , B := A and B n +1 := B n B n − for every n ≥
2. The key properties of F n , B n and τ n are summarised in thefollowing three lemmas. Lemma 8.4.
For each n ≥ the identities S ( F n − /F n ) = F − n log ρ ( B n ) and F n F n − − F n +1 F n − =( − n hold, and the value φ − lies strictly between F n − /F n and F n − /F n +1 . Proof.
Define a sequence of finite words by u (1) := 1, u (2) := 0, and u ( n +1) := u ( n ) u ( n − for every n ≥
2. Clearly we have A ( u ( n ) ) = B n for all n ≥
1. A simple induction argument shows that each u ( n ) is a standard word in the sense defined in Lemma 4.3, and that | u ( n ) | = F n , | u ( n ) | = F n − for every n ≥
2. By Lemma 4.3 and Lemma 6.3(i) we therefore have [ u ( n ) ] ∞ ∈ X F n − /F n andconsequently S ( F n − /F n ) = F − n log ρ ( A ( u ( n ) )) = F − n log ρ ( B n ) for every n ≥ F n − /F n are precisely thecontinued fraction convergents of φ − . Alternatively these results can be derived from the explicitformula for ( F n ). (cid:3) Lemma 8.5.
For each n ≥ we have tr B n = τ n .Proof. By direct evaluation the reader may obtain tr B = tr B = 2 = τ = τ and tr B = 3 = τ ,so it suffices to show that the sequence (tr B n ) satisfies the same recurrence relation as ( τ n ) for all n ≥
3. Let us write B n = (cid:18) a n b n c n d n (cid:19) for each n ≥
1. Notice that we have a n d n − b n c n = det B n = 1 for every n , and for each n ≥ B n +1 := B n B n − implies the identity (cid:18) a n +1 b n +1 c n +1 d n +1 (cid:19) = (cid:18) a n a n − + b n c n − a n b n − + b n d n − c n a n − + d n c n − c n b n − + d n d n − (cid:19) . Fix any n ≥
3. By definition we havetr B n +1 = a n +1 + d n +1 = a n a n − + b n c n − + c n b n − + d n d n − and (tr B n )(tr B n − ) = a n a n − + a n d n − + d n a n − + d n d n − , so we may compute(tr B n )(tr B n − ) − tr B n +1 = a n d n − + d n a n − − b n c n − − c n b n − = d n − ( a n − a n − + b n − c n − ) + a n − ( c n − b n − + d n − d n − ) − c n − ( a n − b n − + b n − d n − ) − b n − ( c n − a n − + d n − c n − )= a n − ( a n − d n − − b n − c n − ) + d n − ( a n − d n − − b n − c n − )= a n − + d n − = tr B n − , which establishes the required recurrence relation. (cid:3) Lemma 8.6.
There exist constants δ , δ > such that | log τ n − log ρ ( B n ) | = O (cid:0) e − δ F n (cid:1) and (cid:12)(cid:12)(cid:12)(cid:12) log (cid:18) − τ n − τ n +1 τ n (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) = O (cid:0) e − δ F n (cid:1) in the limit as n → ∞ .Proof. It is clear that F n − /F n → φ − using the formula for F n , and since S is continuous itfollows via Lemma 8.4 that F − n log ρ ( B n ) → S ( φ − ) >
0. Since det B n = 1 and B n is non-negative, the eigenvalues of B n are ρ ( B n ) and ρ ( B n ) − respectively, so for each n ≥ τ n = tr B n = ρ ( B n ) + ρ ( B n ) − , where we have used Lemma 8.5. Hence,0 ≤ log τ n − log ρ ( B n ) = log (cid:18) ρ ( B n ) + ρ ( B n ) − ρ ( B n ) (cid:19) ≤ ρ ( B n ) = O (cid:16) e − F n S ( φ − ) (cid:17) , where we have used the elementary inequality log(1 + x ) ≤ x which holds for all real x , and thisproves the first part of the lemma.It follows from this result that lim n →∞ F − n log τ n = S ( φ − ). We may therefore apply this toobtain OUNTEREXAMPLE TO THE FINITENESS CONJECTURE 23 lim n →∞ F n log (cid:18) τ n − τ n +1 τ n (cid:19) = lim n →∞ (cid:18) F n log τ n − − F n log τ n +1 − F n log τ n (cid:19) = S ( φ − )( φ − − φ −
1) = − S ( φ − ) < , from which the second part of the lemma follows easily. (cid:3) Proof of Theorem 1.1 . We will show that S ′ ( φ − ) = − log α ∗ , where α ∗ satisfies the productand limit formulas given in the statement of the Theorem. By Lemma 8.1 this implies that r ( α ∗ ) = φ − / ∈ Q , and by Theorem 2.3 this implies that A α ∗ does not satisfy the finitenessproperty.By Lemma 8.1 together with Lemma 8.3, the derivative S ′ ( φ − ) exists and is finite. UsingLemma 8.4 and Lemma 8.6, we may now compute S ′ ( φ − ) = lim n →∞ S (cid:16) F n − F n +1 (cid:17) − S (cid:16) F n − F n (cid:17) F n − F n +1 − F n − F n = lim n →∞ F n +1 log ρ ( B n +1 ) − F n log ρ ( B n ) F n − F n +1 − F n − F n = lim n →∞ F n log ρ ( B n +1 ) − F n +1 log ρ ( B n ) F n F n − − F n +1 F n − = lim n →∞ ( − n ( F n log ρ ( B n +1 ) − F n +1 log ρ ( B n ))= lim n →∞ ( − n ( F n log τ n +1 − F n +1 log τ n ) . Let us define α ∗ := e − S ′ ( φ − ) = lim n →∞ τ F n +1 n τ F n n +1 ! ( − n which yields the first of the two expressions for α ∗ . We shall derive the second expression. Let uswrite α n := ( τ F n +1 n /τ F n n +1 ) ( − n for each n ≥ α ∗ = lim n →∞ α n . Applying the recurrencerelations for ( F n ) and ( τ n ) once more, we obtain for each n ≥ α n +1 α n = (cid:16) τ F n +2 n +1 /τ F n +1 n +2 (cid:17) ( − n +1 (cid:16) τ F n +1 n /τ F n n +1 (cid:17) ( − n = τ F n +1 n +2 τ F n n +1 τ F n +2 n +1 τ F n +1 n ! ( − n = (cid:18) τ n +2 τ n +1 τ n (cid:19) ( − n F n +1 = (cid:18) τ n +1 τ n − τ n − τ n +1 τ n (cid:19) ( − n F n +1 = (cid:18) − τ n − τ n +1 τ n (cid:19) ( − n F n +1 . Since τ = τ = 2 and F = F = 1 we have α = 1. Using the formula above we may now obtainfor each N ≥ α N = α N − Y n =1 α n +1 α n = N − Y n =1 (cid:18) − τ n − τ n +1 τ n (cid:19) ( − n F n +1 . It follows from Lemma 8.6 that these partial products converge unconditionally in the limit N →∞ , and taking this limit we obtain the desired infinite product expression for α ∗ . (cid:3) Remark . The proof of Theorem 1.1 may be extended to give an explicit estimate for thedifference | α ∗ − α N | as follows. Note that for each n ≥ / ≤ F n − /F n ≤ / F − n log τ n ≥ F − n log ρ ( B n ) = S (cid:18) F n − F n (cid:19) ≥ S (cid:18) (cid:19) = log ρ ( A A )3 = log(2 + √ . On the other hand, if we define a sequence (˜ τ n ) ∞ n =1 by ˜ τ = ˜ τ = τ = τ = 2 and ˜ τ n +1 := ˜ τ n ˜ τ n − for n ≥
3, then it is clear that τ n ≤ ˜ τ n = 2 F n for every n ≥
1. Combining these estimates yields | log α N − log α ∗ | ≤ ∞ X n = N F n +1 (cid:12)(cid:12)(cid:12)(cid:12) log (cid:18) − τ n − τ n +1 τ n (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ∞ X n = N F n +1 τ n − τ n +1 τ n ≤ ∞ X n = N F n +1 F n − (2 + √ F n +2 / < C ∞ X n = N ( φ n +1 + 1) θ φ n ≤ ∞ X n = N (cid:18) (cid:19) φ n < (cid:18) (cid:19) φ N for all N ≥
3, where C := 4(2 + √ / √ . . . . , θ := (cid:18) √ φ (cid:19) φ √ = 0 . . . . It follows in particular that the value α := τ F /τ F satisfies | α ∗ − α | < − , which yieldsthe approximation given in the introduction.9. Further questions Is it true that α ∗ is irrational or transcendental? The fast rate of convergence of the sequence (cid:18) τ Fn +1 n τ Fnn +1 (cid:19) ( − n suggests that α ∗ is probably irrational; however, perhaps unexpectedly, this rateitself is not fast enough to claim this. Roughly, to apply known results (see, e.g., [39]), we need τ n to grow like A B n with A >
B >
2. Then Theorem 1 from the aforementioned paper wouldapply. In our setting however we “only” have B = φ < ∞ Y n =0 (cid:18) n (cid:19) equal to 2, despite its “superfast” convergence rate. However, a similar product ∞ Y n =0 (cid:18) n (cid:19) is indeed irrational. We conjecture that α ∗∗ = r − (1 − / √
2) (which corresponds to the substitution0 → , → α ∗ corresponding to the Fibonacci substitution 0 → , →
0) isirrational. Is r − ( γ ) always a point when γ is irrational? We know this to be true if γ is not Liouville (i.e.,for all irrational γ except a set of zero Hausdorff dimension) but the method used in Lemma 8.3is somewhat limited. We hope to close this gap in a follow-up paper. If the answer to the previous question is yes, then is it true that r − ( γ ) / ∈ Q whenever γ / ∈ Q ? This question is pertinent to a conjecture of Blondel and Jungers, which says that the finitenessproperty holds for all matrices with rational entries [26]. Our model should not, therefore, yield acounterexample to this conjecture. Is r − ( γ ) always an interval with nonempty interior when γ is rational? It was shown bythe fourth named author in his thesis [47] that r − (cid:0) (cid:1) = (cid:2) , (cid:3) , and all other known examplesindicate that the answer is positive. However proving this for a general γ ∈ Q seems like a difficultquestion. Does the set of all α such that r ( α ) / ∈ Q have zero measure? Does it have zero Hausdorffdimension? Analogues of these properties are claimed for Bousch-Mairesse’s example but proofsare not given [8].
OUNTEREXAMPLE TO THE FINITENESS CONJECTURE 25
We conjecture that the graph of r is a devil’s staircase with the plateau regions correspondingto { γ : r ( γ ) ∈ Q } – see Figure 2. Figure 2: Graph of r ( γ ) Remark . Between the time of submisssion and present, some progress has been made on someof the questions above. Interested readers are welcome to contact the authors above to find outthe current progress on these problems.
Acknowledgement
The authors are indebted to V. S. Kozyakin for his helpful remarks and suggestions.
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Department of Pure Mathematics, University of Waterloo, Waterloo, Ontario, Canada N2L 3G1.
E-mail address : [email protected] Dipartimento di Matematica, Universit`a di Roma Tor Vergata, Via della Ricerca Scientifica, 00133Roma, Italy.
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