For every finitely generated free group we construct an explicit left order
extending the lexicographic order on the free monoid generated by the positive
letters. The order is defined by a left, free action on the orbit of 0 of a
free group of piecewise linear homeomorphisms of the line. The membership in
the positive cone is decidable in linear time in the length of the input word.
The positive cone forms a context-free language closed under word reversal.Comment: 4 figure

Sunic, Zoran

Comptes Rendus Mathematique

English

arXiv

Šunić, Zoran

Numérisation de Documents Anciens Mathématiques

C. R. Acad. Sci. Paris, Ser. I 351 (2013) 507–511Contents lists available at SciVerse ScienceDirectC. R. Acad. Sci. Paris, Ser. Iwww.sciencedirect.comGroup Theory/Dynamical SystemsExplicit left orders on free groups extending the lexicographicorder on free monoidsOrdres à gauche explicites sur les groupes libres étendant l’ordrelexicographique sur les monoïdes libresZoran Šunić 1Dept. of Mathematics, Texas A&M University, MS-3368, College Station, TX 77843-3368, USAa r t i c l e i n f o a b s t r a c tArticle history:Received 2 April 2013Accepted after revision 1 July 2013Available online 9 August 2013Presented by the Editorial BoardTo my teacher and friend, FrosinaMitruševa, on the occasion of her birthday.For every finitely generated free group, we construct an explicit left order extendingthe lexicographic order on the free monoid generated by the positive letters. The orderis defined by a left, free action on the orbit of 0 of a free group of piecewise linearhomeomorphisms of the line. The membership in the positive cone is decidable in lineartime in the length of the input word. The positive cone forms a context-free languageclosed under word reversal.© 2013 Académie des sciences. Published by Elsevier Masson SAS. All rights reserved.r é s u m éPour tout groupe libre fini engendré, nous construisons explicitement un ordre à gauchequi étend l’ordre lexicographique sur le monoïde libre engendré par les lettres positives.Cet ordre est défini par une action à gauche, libre, sur l’orbite de 0 d’un groupe libred’homéomorphismes de la droite linéaires par morceaux. L’appartenance au cône positifest décidable en temps linéaire par rapport à la longueur du mot. Le cône positif forme unlangage non contextuel fermé par image miroir.© 2013 Académie des sciences. Published by Elsevier Masson SAS. All rights reserved.Version française abrégéeUn groupe G est ordonnable à gauche s’il existe un ordre total  sur G compatible avec la multiplication à gauche,i.e., pour tous les éléments f , g et h de G , l’inégalité f  g entraîne hf  hg . Il est bien connu depuis les années 1940que le groupe libre Fk de rang k est ordonnable à gauche (et, justement, bi-ordonnable). Pourtant, la plupart des preuvesprécédentes sont non constructives ou trop compliquées (souvent à cause d’une volonté de généraliser ultérieurement).Peut-être la construction la plus explicite connue à l’heure actuelle d’un ordre sur les groupes libres est donnée parl’approche de Magnus–Bergman [1], basée sur le plongement de Magnus [2] du groupe libre Fk = F (Σk) dans l’anneaudes séries formelles à coefficients entiers en variables non commutatives dans Σk = {s1, . . . , sk}. Les monômes sur Σk sontordonnés par longueur lexicographique et un élément u de Fk est declaré positif si et seulement si le coefficient du monômeminimal (different de 1) dans la série formelle réprésentant u dans le plongement de Magnus est positif.E-mail address: sunic@math.tamu.edu.1 This material is based upon work supported by the National Science Foundation under Grant No. DMS-1105520.1631-073X/$ – see front matter © 2013 Académie des sciences. Published by Elsevier Masson SAS. All rights reserved.http://dx.doi.org/10.1016/j.crma.2013.07.001508 Z. Šunić / C. R. Acad. Sci. Paris, Ser. I 351 (2013) 507–511Pour tout groupe libre finiment engendré, nous construisons explicitement un ordre à gauche qui étend l’ordre lexico-graphique sur le monoïde libre engendré par les lettres positives. Cet ordre est défini par une action à gauche, libre, surl’orbite de 0 d’un groupe libre d’homéomorphismes de la droite linéaires par morceaux. L’appartenance au cône positif estdécidable en temps linéaire par rapport à la longueur du mot en comptant directement les sous-mots de longueur 2 d’uncertain type (voir le critère de positivité dans le théorème 0.2). Le cône positif forme un langage non contextuel fermé parimage miroir.On divise le cercle S1 = R/Z = [0,1]/{0 = 1} en sept arcs de même longueur, notés 1′ = [0,1/7), a′ = [1/7,2/7), b′ =[2/7,3/7), c′ = [3/7,4/7), C ′ = [4/7,5/7), B ′ = [5/7,6/7), A′ = [6/7,1), et on définit trois homéomorphismes du cerclelinéaires par morceaux (avec un nombre fini de singularités) conservant l’orientation a, b, et c comme dans la Fig. 1 àgauche. Les inverses de a, b et c, dénotés par A, B , et C , respectivement, sont representés dans la moitié droite de la Fig. 1.Il est évident que les inclusions (1) et (2) sont satisfaites. Par conséquent, F = F3 = 〈a,b, c〉 est libre de rang 3. De plus,F agit librement, par une action à gauche, sur le sous-ensemble F 1′ du cercle (notons que F 1′ a une mesure de Lebesgueégale à 1). En particulier, il agit librement sur l’orbite de 0.On relève les sous-intervalles du cercle S1 à des sous-ensembles de la droite R suivant la projection R → R/Z = S1,et on relève les applications a, b et c à des homéomorphismes de la droite linéaires par morceaux conservant l’orientation(avec un nombre fini de singularités dans chaque sous-ensemble compact) comme dans la Fig. 2. On ne change pas denotation pour des sous-ensembles ou des homéomorphismes relevés.Notons que le groupe F relevé est encore libre, agissant librement sur l’orbite de 0. Par conséquent, ceci définit un ordreà gauche sur F en posant g > h si et seulement si g(0) > h(0) (de manière équivalente, si et seulement si h−1 g(0) > 0).Un critère simple pour la positivité est obtenu en « traçant » l’action sur 0, c’est-à-dire en calculant, pour tout mot réduit u,la distance signée, avec une erreur plus petite que 1/2, entre u(0) and 0.Proposition 0.1. Soit w : F3 →R la fonction poids définie par :w(u) = #{des sous-mots de u de la forme cB, c A, ou b A} − #{des sous-mots de u de la forme Cb, Ca, ou Ba}+ 12{1, si u se termine par une lettre positive, i.e., par une des lettres a,b, c,−1, si u se termine par une lettre negative, i.e., par une des lettres A, B, C,0, si u est le mot trivial,où u est un mot réduit sur {a,b, c}. Alors u(0) > 0 si et seulement si w(u) > 0.En suivant une construction analogue (en divisant S1 en 2k + 1 morceaux, pour k  2, et en définissant k homéomor-phismes de S1 linéaires par morceaux s1, . . . , sk , et ainsi de suite), on peut facilement établir le résultat suivant.Théorème 0.2. Soit Σk = {s1, . . . , sk}, avec k  2. Pour i = 1, . . . ,k, posons Si = s−1i . On peut définir un ordre à gauche sur le groupelibre Fk = F (Σk) en étendant l’ordre lexicographique sur le monoide libre Σ∗k basé sur l’ordre s1 < s2 < · · · < sk sur l’alphabet deslettres positives, comme ci-dessous. Soit w : Fk → R la fonction poids définie par :w(u) = #{des sous-mots de u de la forme s j Si, pour j > i} − #{des sous-mots de u de la forme S j si, pour j > i}+ 12{1, si u se termine par une lettre positive si, i = 1, . . . ,k,−1, si u se termine par une lettre negative Si, i = 1, . . . ,k,0, si u est le mot trivial,où u est un mot réduit sur Σk. Alors, l’ensemble Pk = {u ∈ Fk | w(u) > 0} est un cône positif de Fk (i.e., u > v dansFk ⇔ w(v−1u) > 0).1. Left orders on free groups extending the lexicographic orderA group G is left orderable if there exists a linear order  on G that is compatible with the left multiplication, i.e.,for all elements f , g and h in G , if f  g , then hf  hg . It has been known at least since the 1940s that the free groupFk of rank k is left orderable (and, in fact, bi-orderable, admitting an order that is compatible with both the left and theright multiplication simultaneously). In two of his papers [6,5] related to the subject, Neumann mentions that, in additionto himself, several other authors have stated this fact in published or unpublished works, including Tarski, G. Birkhoff,Shimbireva, and Iwasawa (despite his laudable effort to give credit to all, he was unaware of the simultaneous work ofVinogradov [8]). However, most of the early proofs are nonconstructive or too involved (often because of an attempt forgreater generality).Perhaps the most explicit, currently known, construction of an order on free groups is given by the Magnus–Bergmanapproach [1], based on the Magnus embedding [2] of the free group Fk = F (Σk) into the ring of formal power series withinteger coefficients in noncommuting variables from Σk = {s1, . . . , sk}. The monomials over Σk are ordered by short-lex andan element u from Fk is declared positive if and only if the coefficient in front of the smallest monomial (different from 1)in the power series representing u under the Magnus embedding is positive.Z. Šunić / C. R. Acad. Sci. Paris, Ser. I 351 (2013) 507–511 509Fig. 1. Generators of F in PLF+(S1).For every finitely generated free group, we construct an explicit left order extending the lexicographic order on thefree monoid generated by the positive letters. The order is defined by a left, free action on the orbit of 0 of a free groupof piecewise linear homeomorphisms of the line. The membership in the positive cone is decidable in linear time in thelength of the input word by straightforward counting of subwords of length 2 of a certain type (see the positivity criterionin Theorem 1.2). The positive cone forms a context-free language closed under word reversal.We provide details only for the case of F3, the free group of rank 3. The general case is analogous.Subdivide the circle S1 = R/Z = [0,1]/{0 = 1} into seven arcs of the same length, denoted 1′ = [0,1/7), a′ = [1/7,2/7),b′ = [2/7,3/7), c′ = [3/7,4/7), C ′ = [4/7,5/7), B ′ = [5/7,6/7), A′ = [6/7,1), and define three piecewise linear, orientationpreserving homeomorphisms a, b, and c of the circle (with finitely many breaks) as in the left half of Fig. 1. The inverses ofa, b and c are denoted by A, B , and C , respectively, and are shown in the right half of Fig. 1. It is evident that:a(S1 \ A′) ⊆ a′, b(S1 \ B ′) ⊆ b′, c(S1 \ C ′) ⊆ c′, (1)A(S1 \ a′) ⊆ A′, B(S1 \ b′) ⊆ B ′, C(S1 \ c′) ⊆ C ′. (2)Therefore, F = F3 = 〈a,b, c〉 is free of rank 3. Moreover, F acts freely, through a left action, on the subset F 1′ of the circle(note that F 1′ has Lebesgue measure 1). In particular, it acts freely on the orbit of 0.We lift the subintervals from the circle S1 to subsets of the line R along the projection R → R/Z = S1, and we lift themaps a, b and c to piecewise linear, orientation preserving the homeomorphisms of the line (with finitely many breaks onevery compact subset) as in Fig. 2. We do not change the notation for the lifted subsets or homeomorphisms.Note that the lifted group F is still free, acting freely on the orbit of 0. Therefore, a left order is defined on F by declaringg > h if and only if g(0) > h(0) (equivalently, if and only if h−1 g(0) > 0). A simple criterion for positivity is obtained by“tracing” the action on 0, i.e., by calculating, for reduced group words u, the signed distance, within error smaller than 1/2,between u(0) and 0. Of course, since the action is given explicitly and the orbits of rational points are rational, u(0) can becalculated exactly, but that takes longer, obscures the features of the order, and is not necessary.Proposition 1.1. Define a weight function w : F3 → R byw(u) = #{of subwords of u of the form cB, c A, or b A} − #{of subwords of u of the form Cb, Ca, or Ba}+ 12{1, if u ends in a positive letter, i.e., one of the letters a,b, c,−1, if u ends in a negative letter, i.e., one of the letters A, B, C,0, if u is the trivial word,where u is a reduced group word over {a,b, c}. Thenu(0) > 0 if and only if w(u) > 0.Indeed, for any nontrivial word u, the point u(0) is in exactly one of the intervals510 Z. Šunić / C. R. Acad. Sci. Paris, Ser. I 351 (2013) 507–511Fig. 2. Generators of F in PLF+(R).. . . , (−2,−1), (−1,0), (0,1), (1,2), . . . .We claim that w(u) represents the midpoint of the interval to which u(0) belongs. By definition, the action of F is a leftaction, hence the last letter of u acts first on 0 and pushes it into the interval (0,1) if the letter is positive, or to (−1,0)if the letter is negative, and the part of the formula for w related to the last letter of u records this as 1/2 or −1/2. Fromhere on, we trace the action of u (from right to left) on the obtained point, but only record jumps into the next intervalup (by adding 1 in our weight) and jumps into the next interval down (by subtracting 1 in our weight). Negative lettersnever produce jumps up (see the right half of Fig. 2) and positive letters never produce jumps down (see the left half ofFig. 2). By examining the action of the positive letters in the left half of Fig. 2, we see that jumps up occur exactly whenthe letter c is applied to a point in the regions B ′ and A′ , which is precisely when a subword of the form cB or c A occursin u, or when the letter b is applied to a point in the region A′ , which is precisely when a subword of the form b A occursin u. We exclude the possibility of applying the letter c to the region C ′ , b to the region B ′ , and a to the region A′ , becauseu is a reduced word. Similarly, by examining the action of the negative letters in the right half of Fig. 2, we see that jumpsdown occur exactly when the letter C is applied to a point in the regions b′ and a′ , which is precisely when a subwordof the form Cb or Ca occurs in u, or when the letter B is applied to a point in the region a′ , which is precisely when asubword of the form Ba occurs in u. We exclude the possibility of applying the letter C to the region c′ , B to the region b′ ,and A to the region a′ , because u is a reduced word. We also exclude the possibility of applying any of the negative lettersto the region 1′ , since the point 0 will never return to the region 1′ under the action of a nontrivial word u, after, in thevery first step, the last letter of u moves it from there.We claim that the given order extends the usual lexicographic order on the free monoid M3 = {a,b, c}∗ based ona < b < c. All we need to verify is that, for all words v1, v2, v3 in M3, e < av1 < bv2 < cv3, i.e., we need to verify thatw(av1), w(v−11 Abv2), w(v−12 Bcv3) > 0. Since the weight of each of these words is 1/2, the claim is correct.In fact, it is possible to see that the order, restricted to words in M3 = {a,b, c}∗ is lexicographic just by looking at theleft half of Fig. 2. Namely, for u in M3, u(0) is trapped in the interval [0,4/7), and, for all words v1, v2, v3 in M3, we have0 < av1(0) < bv2(0) < cv3(0), since, on the interval [0,4/7) the entire graph of the function a is above 0 and below theminimum of the function b, and the entire graph of the function b is below the minimum of the function c.Following an analogous construction (subdividing S1 into 2k + 1 pieces, for k  2, defining k piecewise linear homeo-morphisms s1, . . . , sk of S1, and so on), we may easily establish the following result.Z. Šunić / C. R. Acad. Sci. Paris, Ser. I 351 (2013) 507–511 511Theorem 1.2. Let Σk = {s1, . . . , sk}, for some k  2. For i = 1, . . . ,k, denote Si = s−1i . A left order on the free group Fk = F (Σk)extending the lexicographic order on the free monoid Σ∗k based on the order s1 < s2 < · · · < sk on the alphabet of positive letters maybe defined as follows. Define a weight function w : Fk →R by:w(u) = #{of subwords of u of the form s j Si, for j > i} − #{of subwords of u of the form S jsi, for j > i}+ 12{1, if u ends in any positive letter si, i = 1, . . . ,k,−1, if u ends in any negative letter Si, i = 1, . . . ,k,0, if u is the trivial word,where u is a reduced group word over Σk, and declare that the set:Pk ={u ∈ Fk∣∣ w(u) > 0}is the positive cone of Fk (i.e., u > v in Fk ⇔ w(v−1u) > 0).The construction of the k homeomorphisms s1, . . . , sk of S1 that generate a free group acting freely on the orbit of 0 andthat can be used to prove Theorem 1.2 relies on the general approach for producing quasi-Schottky groups in [7]. They maybe defined as follows. For i = 1, . . . ,k, let si : S1 → S1 be the homeomorphism of the circle given by:si = s0(x + i − 12k + 1)+ i2k + 1 , where s0(x) ={12k x, 0  x <2k2k+1 ,2kx − (2k − 1), 2k2k+1  x < 1.The membership problem for the positive cone Pk is rather easy and can be solved in linear time in the length of theinput word. If we count the relevant subwords as we read, we can calculate the weight and tell if a word is in the positivecone by the time we finish reading the word.It is known that the positive cone of a left order of a free group cannot be finitely generated as a monoid (this canbe deduced from the work of McCleary [3], but was the first explicit proof is given by Navas [4]), which implies that itcannot be a regular language of the form Y ∗ for some finite set Y of group words. On the other hand, it is apparent thatthe positive cone Pk is a context free language over Σ±k (indeed, a push down automaton with a single stack can easilyestablish if the number of subwords of the form s j Si , with j > i in a given word u is greater than, smaller than, or equal tothe number of subwords of the form S j si , with j > i, and can take into account the last letter of u in case of a tie).Another interesting feature of the positive cone Pk is that it is closed under word reversal. Indeed, for a given word u,with w(u) > 0, we have w((uR)−1) = −w(u) < 0, where uR denotes the word reversal of u, since the transformation (uR)−1just exchanges the positive and negative letters (while keeping them in the same order as in u), and the effect of this onthe weight is to exactly exchange all positive and all negative contributions. Since w((uR)−1) < 0, we must have w(uR) > 0.The left order provided in Theorem 1.2 is not two-sided (unlike the Magnus–Bergman order, which is). In fact, it is clearthat no order extending the lexicographic order can be two-sided (this is because s1 < s1s1, but s1s2 > s1s1s2).Variations of the construction presented above lead to other explicitly stated orders on free groups, not necessarilyextending the lexicographic order (for instance, in case of F2, add 1 to the weight for every subword of the form ab or aB ,subtract 1 for every subword of the form B A or Ba, and add plus or minus 1/2; depending on the last letter). In fact, Cantorsets of left orders on Fk may be constructed by stringing together variations of the above construction (subsequent work).AcknowledgementsI would like to thank Warren Dicks and Tullio Ceccherini-Silberstein for their kind help.References[1] George M. Bergman, Ordering coproducts of groups and semigroups, J. Algebra 133 (2) (1990) 313–339.[2] Wilhelm Magnus, Beziehungen zwischen Gruppen und Idealen in einem speziellen Ring, Math. Ann. 111 (1) (1935) 259–280.[3] Stephen H. McCleary, Free lattice-ordered groups represented as o-2 transitive l-permutation groups, Trans. Am. Math. Soc. 290 (1) (1985) 69–79.[4] Andrés Navas, On the dynamics of (left) orderable groups, Ann. Inst. Fourier (Grenoble) 60 (5) (2010) 1685–1740.[5] B.H. Neumann, On ordered division rings, Trans. Am. Math. Soc. 66 (1949) 202–252.[6] B.H. Neumann, On ordered groups, Am. J. Math. 71 (1949) 1–18.[7] Zoran Šunić, Free subgroups acting properly discontinuously, Topol. Appl. 160 (10) (2013) 1108–1114.[8] A.A. Vinogradov, On the free product of ordered groups, Mat. Sb. N.S. 25 (67) (1949) 163–168.

Explicit left orders on free groups extending the lexicographic order on free monoids

Zoran Šunić

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Explicit left orders on free groups extending the lexicographic order on free monoids

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