Journal of Algebra 332 (2011) Contents lists available at ScienceDirect. Journal of Algebra.

Transcription

1 Journal of Algebra 332 (2011) 1 13 Contents lists available at ScienceDirect Journal of Algebra Enumerating alindromes and rimitives in rank two free grous Jane Gilman a,,1, Linda Keen b,2 a Deartment of Mathematics, Rutgers University, Newark, NJ 07079, United States b Deartment of Mathematics, CUNY, Lehman College and Graduate Center, Bronx, 10468, United States article info abstract Article history: Received 27 February 2008 Availableonline18February2011 Communicated by Martin Liebeck Keywords: Palindromes Primitive element Free grou Rank two Kleinian grous Fuchsian grous Markoff Numbers Farey tesselation Conjugacy classes Let F be a free grou of rank two. An element of F is rimitive if it, along with another grou element, generates the grou. If F = A, B, then a word W (A, B), in A and B, is a alindrome if it reads the same forwards and backwards. It is known that in a rank two free grou, for any fixed set of two generators a rimitive element will be conjugate either to a alindrome or to the roduct of two alindromes, but known iteration schemes for all rimitive words give only a reresentative for the conjugacy class. Here we derive a new iteration scheme that gives either the unique alindrome in the conjugacy class or exresses the word as a unique roduct of two unique alindromes that have already aeared in the scheme. We denote these words by E /q where /q is rational number exressed in lowest terms. We rove that E /q is a alindrome if q is even and the unique roduct of two unique alindromes if q is odd. We rove that the air (X, Y ) (or (X 1, Y 1 )) generates the grou if and only if X is conjugate to E /q and Y is conjugate to E r/s where s rq =1. This imroves a reviously known result that held only for q and rs both even. The derivation of the enumeration scheme also gives a new roof of the known results about rimitive words Elsevier Inc. All rights reserved. * Corresonding author. addresses: gilman@rutgers.edu, jgilman@nsf.gov (J. Gilman), linda.keen@lehman.cuny.edu (L. Keen). 1 Research suorted in art by Rutgers Research Council, Yale University, & NSF. 2 Research suorted in art by PSC-CUNY research grant /$ see front matter 2011 Elsevier Inc. All rights reserved. doi: /j.jalgebra

2 2 J. Gilman, L. Keen / Journal of Algebra 332 (2011) Introduction Let F = A, B be a rank two free grou. It is well known that u to conjugacy and inverses, rimitive elements in a rank two free grou can be indexed by the rational numbers and that airs of rimitive words that generate the grou can be obtained by the number theory of the Farey tessellation of the hyerbolic lane. It is also well known that u to conjugacy, a rimitive word can always be written as either a alindrome or a roduct of two alindromes and that certain airs of alindromes will generate the grou [1,20]. In this aer we give new roofs of the above results. The roofs yield a new enumerative scheme for conjugacy classes of rimitive words, still indexed by the rationals (Theorem 2.1). We denote the words reresenting each conjugacy class by E /q. In addition to roving that the enumeration scheme gives a unique reresentative for each conjugacy class containing a rimitive, we rove that the words in this scheme are either alindromes or the canonically defined roduct of a air of alindromes that have already aeared in the scheme and thus give a new roof of the alindrome/roduct result. Further we show that if X and Y are rimitive, then u to relacing one and/or both by their inverse, there are words E /q and E r/s resectively conjugate to X and Y, and these generate the grou if and only if q rs =1 (Theorem 2.2). This imroves the revious known result that held only for airs of alindromes. In this aer, we use continued fractions and the Farey tessellation of the hyerbolic lane to find the enumeration scheme and to rove that it actually enumerates all rimitives and all rimitive airs. Pairs of rimitives that generate the grou and their relation to continued fractions and the Farey tessellation of the hyerbolic lane arise in the discreteness algorithm for PSL(2, R) reresentations of two generator grous [4,5,11,13]. This new enumerative scheme is useful in extending discreteness criteria to PSL(2, C) reresentations where the hyerbolic geometry of alindromes lays an imortant role [8,10]. 2. The main result We are able to state and use our main result with very little notation, only the definition of continued fractions. Namely, we let and q be relative rime integers ositive integers. Write q = a 0 + a a 2 + a a 1 k =[a 0, a 1,...,a k ] where the a i are integers with a j > 0, j = 1...k, a 0 0. Enumeration scheme for ositive rationals. Set E 0/1 = A 1, E 1/0 = B, and E 1/1 = BA 1. Suose /q has continued fraction exansion [a 0, a 1,...,a k 1, a k ]. Consider the two rationals defined by the continued fractions [a 0, a 1,...,a k 1 ] and [a 0, a 1,...,a k 1, a k 1]. One is smaller than /q and the other is larger; call the smaller one m/n and the larger one r/s so that m/n < /q < r/s. The induction ste in the scheme is given by Case 1. q odd: Case 2. q even: E /q = E r/s E m/n. E /q = E m/n E r/s.

3 J. Gilman, L. Keen / Journal of Algebra 332 (2011) We have a similar scheme for negative rationals described in Section 4. With both schemes we can state our main result as Theorem 2.1 (Enumeration of rimitives by rationals). U to conjugacy and taking formal inverses, the rimitive elements of a two generator free grou can be enumerated by the rationals using continued fraction exansions. The resulting rimitive words are cyclically reduced and denoted by E /q so that: For q even, E /q is a alindrome. It is the unique alindrome in its conjugacy class. For q odd, E /q is a roduct of alindromes that have already aeared in the scheme; that is, E /q = E m/n E r/s and both E m/n and E r/s are alindromes. Remark 2.1. Note that although there are several ways a word in the q odd conjugacy class can be factored as roducts of alindromes, in this theorem we secifically choose the unique factorization for E /q that makes the enumeration scheme work. In addition we have Theorem 2.2. Let {E /q } denote the words in the enumeration scheme for rationals. If (/q, /q ) satisfies q q =1, theair(e /q, E /q ) generates the grou. Conversely, the air of elements (X, Y ) generates the grou if and only if there is a rational /q such that X is conjugate to E /q and a rational /q such that Y is conjugate to E /q,where q q =1. These theorems will be roved in Section 4. In order to rove the theorems and the related results we need to review some terminology and background. 3. Preliminaries The main object here is a two generator free grou which we denote by F = A, B. A word W = W (A, B) F is, of course, an exression of the form A m 1 B n 1 A m2 B n r (1) for some set of 2r integers m 1,...,m r,n 1,...,n r with m 2,...,m r,n 1,...,n r 1 non-zero. Definition 1. An element of F is rimitive if there is another element V such that W and V generate F. V is called a rimitive associate of W and the unordered air W and V is called a air of rimitive associates or a rimitive air for short. If F = A, B, a rimitive element W = W (A, B) is also referred to as a rimitive word when we are thinking of it as a word in A and B. There are various versions of the following theorem in the literature, for examle in [2,3,7,14,15, 17 19]. We use a version convenient for what we need below. Theorem 3.1. AwordW(A, B) F is rimitive if and only if after freely reducing the word and interchanging A and B and and q, it is conjugate to a word of the form B ɛ A n 1 B ɛ A n (2) where ɛ =±1, allofthen j have the same sign, and n j 1, 1 j <, n j n j+1 1, and jistaken modulo. The exonents satisfy i=1 n i = q with and q relatively rime and 0 /q < 1. Moreover, to every rational /q there is a unique rimitive word of this form.

4 4 J. Gilman, L. Keen / Journal of Algebra 332 (2011) 1 13 The elements of the ordered set {n 1,...,n } are called the rimitive exonents of the rimitive word. Note that the rimitive exonents are defined only u to cyclic ermutation. In the next subsections we summarize terminology and facts about the Farey tessellation and continued fraction exansions for rational numbers. Details and roofs can be found in [21 23]. See also [24] Preliminaries: the Farey tessellation In what follows when we use r/s to denote a rational number, we assume that r and s are integers with s > 0, and that r and s are relatively rime, that is, that (r, s) = 1. We let Q denote the rational numbers, and we identify the rationals with oints on the extended real axis on the Riemann shere. We use the notation 1/0 to denote the oint at infinity. We need the concet of Farey addition for fractions. Definition 2. If q, r Q with s qr =1, the Farey sum is s q r s = + r q + s. Two fractions are called Farey neighbors if s qr = 1. When we write q r s = +r q+s we tacitly assume the fractions are Farey neighbors. Remark 3.1. It is a simle comutation to see that both airs of fractions ( q, q r s ) and ( are Farey neighbors if (/q, r/s) are, and that if q < r s,then q < q r s < r s. q r s, r s It is easy to calculate that the Euclidean distance between finite Farey neighbors is strictly less than one unless they are adjacent integers. This imlies that unless one of the fractions is 0/1, both neighbors have the same sign. One creates the Farey diagram in the uer half-lane by marking each fraction by a oint on the real line and joining each air of Farey neighbors by a semi-circle orthogonal to the real line. The oint here is that because of the above roerties none of the semi-circles intersect in the uer half lane. This gives a tessellation of the hyerbolic lane where the semi-circles joining a air of neighbors, together with the semi-circles joining each member of that air to the Farey sum of the air, form an ideal hyerbolic triangle. The tessellation is called the Farey tessellation and the vertices are recisely the oints that corresond to rational numbers. A vertex corresonding to /q is labeled v /q.itistermedeven or odd according to the arity of q. See Fig. 1. The Farey tessellation is invariant under the semi-grou generated by z z + 1 and z 1/z. Fix any oint ζ on the ositive imaginary axis. Given a fraction,, there is an oriented hyerbolic q geodesic γ connecting ζ to. We assume γ is oriented so that moving from ζ to a ositive rational q is the ositive direction. This geodesic γ will intersect some number of triangles. ) Definition 3. The Farey level or the level of /q, denoted by Lev(/q), is the number of triangles traversed by γ.

5 J. Gilman, L. Keen / Journal of Algebra 332 (2011) Fig. 1. The Farey tesselation with the curve γ. Note that the definition is indeendent of the choice of ζ and that our definition imlies Lev(/q) = Lev( /q). The geodesic γ will enter any given triangle along an edge. This edge will connect two vertices and γ will exit the triangle along an edge connecting one of these two vertices and the third vertex of the triangle, the new vertex. Since γ is oriented, the edge through which γ exits a triangle is either a left edge or a right edge deending uon whether the edge γ cuts is to the right or left of the new vertex. Definition 4. We determine a Farey sequence for inductively by choosing the new vertex of the next q triangle in the sequence of triangles traversed by γ. The sequence ends at /q. Given /q, we can find the smallest rational m/n and the largest rational r/s that are neighbors of /q. These neighbors have the roerty that they are the only neighbors with lower Farey level. That is, m/n < /q < r/s and Lev(/q)>Lev(m/n), Lev(/q)>Lev(r/s), and if u/v is any other neighbor Lev(/q)<Lev(u/v). Definition 5. We call the smallest and the largest neighbors of the rational /q the distinguished neighbors or the arents of /q.

6 6 J. Gilman, L. Keen / Journal of Algebra 332 (2011) 1 13 Note that we can tell whether a distinguished neighbor r/s is smaller or larger than /q by the sign of rq s. We emhasize that we have two different and indeendent orderings of the rational numbers: the ordering as rational numbers and the artial ordering by level. Our roofs will often use induction on the level of the rational numbers involved as well as the order relations as rational numbers among arents and grandarents. Remark 3.2. It follows from Remark 3.1 that if m/n and r/s are the arents of /q, then any other neighbor of /q is of the form m+t n+qt or r+t s+qt for some ositive integer t. The neighbors of are recisely the set of integers. Finally we note that we can describe the Farey sequence of /q by listing the number of successive left or right edges of the triangles that γ crosses where a left edge means that there is one vertex of the triangle on the left of γ and two on the right, and a right edge means there is only one vertex of γ on the right. This will be a sequence of integers, the left-right sequence ±(n 0,n 1,...,n t ) where the integers n i, i > 0 are all ositive and n 0 is ositive or zero. The sign in front of the sequence is ositive or negative deending on the orientation of γ Preliminaries: continued fractions Farey sequences are related to continued fraction exansions of ositive fractions; they can also be related to exansions of negative fractions. We review the connection in art to fix our notation recisely. We do not use the classical notation of [12] for negative fractions, but instead use the notation of [9,18,21,22,25] which is standardly used by mathematicians working in Kleinian grous and three manifolds. This notation reflects the symmetry about the imaginary axis in the Farey tessellation which lays a role in our alications. This symmetry is built in to the semi-grou action on the tessellation. For /q 0write q = a 0 + a a 2 + a a 1 k =[a 0, a 1,...,a k ] where a j > 0, j = 1...k, a 0 0. For 0 n k set n q n =[a 0, a 1,...,a n ]. Remark 3.3. The continued fraction of a rational is ambiguous; the continued fractions [a 0, a 1,...,a n ] and [a 0, a 1,...,a n 1, 1] both reresent the same rational. Therefore, if we have [a 0, a 1,...,a n 1, 1] we may relace it with [a 0, a 1,...,a n 1 + 1]. Remark 3.4. Note that if q 1 has continued fraction exansion [a 0, a 1,...,a n ],then q has exansion [0, a 0,...,a n ] while if q < 1 has continued fraction exansion [0, a 1,...,a n ],then q has exansion [a 1, a 2,...,a n ]. Remark 3.5. The distinguished neighbors or arents of /q have continued fractions [a 0, a 1...,a k 1 ] and [a 0, a 1,...,a k 1, a k 1]. The Farey sequence contains the aroximants n as a subsequence. They can be comuted recursively from the continued fraction for /q as q n follows:

7 J. Gilman, L. Keen / Journal of Algebra 332 (2011) = a 0, q 0 = 1 and 1 = a 0 a 1 + 1, q 1 = a 1, j = a j j 1 + j 2, q j = a j q j 1 + q j 2, j = 2,...,k. One can calculate from these recursion formulas that the aroximants are alternately to the right and left of /q. The oints of the Farey sequence between j and j+1 have continued fraction exansions q j q j+1 [a 0, a 1,...a j + 1], [a 0, a 1,...,a j + 2],..., [a 0, a 1,...a j + a j+1 1]. Thus a j is the number of oints on the same side of /q before juming to the other side. We extend the continued fraction notion to negative fractions by defining the continued fraction of q < 0 to be the negative of that for. That is, by setting q q = [a 0, a 1...,a k ]=[ a 0, a 1..., a k ] where q =[a 0, a 1,...a k ]. We also set Lev(/q) = Lev( /q ). In [12] the continued fraction [a 0, a 1,...,a k ] of /q < 0 is defined so that a 0 < 0isthelargest integer in /q and [a 1,...a k ]=/q a 0 is the continued fraction of a ositive rational. With this notation the symmetry about the origin which lays a role in our alications is lost. We note that for any air of neighbors, unless one of them is 0/1 or 1/0, they both have the same sign and thus have equal a 0 entries. Since we almost always work with neighbors the difference between our notation and the classical one does not lay a role. It is not difficult to check that the continued fraction exansion and the left-right sequence of a rational agree. 4. Enumerating rimitives: alindromes and roducts In the enumeration scheme we define here, the grou element B 1 never aears. It is therefore clear that either a word or its inverse is in the scheme, but both cannot be. We first work with ositive rationals. We do this merely for ease of exosition and to simlify the notation. We then indicate the minor changes needed for negative rationals. Enumeration scheme for ositive rationals. Set E 0/1 = A 1, E 1/0 = B, and E 1/1 = BA 1. If /q has continued fraction exansion [a 0, a 1,...,a k 1, a k ], consider the arent fractions [a 0, 1,...,a k 1 ] and [a 0,...,a k 1, a k 1] (see Remark 3.5). Choose labels m/n and r/s for the arents so that m/n < /q < r/s. Set Case 1. q odd: E /q = E r/s E m/n. Case 2. q even: E /q = E m/n E r/s. Note that in Case 1 (q odd) the word indexed by the larger fraction is on the left and in Case 2 (q even) it is on the right.

8 8 J. Gilman, L. Keen / Journal of Algebra 332 (2011) 1 13 Enumeration scheme for negative rationals. Now assume /q < 0. We use the reflection in the imaginary axis to obtain the enumeration scheme. The reflection sends A 1 to A. This reverses the order of the distinguished neighbors on the real line. Set E 0/1 = A and E 1/0 = B. These are trivially alindromes. At the next level we have E 1/1 = BA. To give the induction scheme: we assume m/n, r/s are the distinguished neighbors of /q and they satisfy m/n > /q > r/s, /q = (m + r)/(n + s) and set Case 1. q odd: E /q = E r/s E m/n. Case 2. q even: E /q = E m/n E r/s. Note that now in Case 1 (q odd) the word indexed by the larger fraction is on the right and in Case 2 (q even) it is on the left. Proofs of Theorems 2.1 and 2.2 Before we give the roofs we note that there are other enumeration schemes for relating rational numbers to rimitive words given, for examle, in [3,6,15]. Some of these also use Farey neighbors to find rimitive airs. In general, these schemes roduce a different word in the conjugacy class of, or the inverse conjugacy class of, our E /q. Theorem 2.2 tells us that words in the enumeration scheme labeled with neighboring Farey fractions give rise to rimitive airs. Note that although cyclic ermutations are obtained by conjugation, we cannot necessarily simultaneously conjugate a rimitive air coming from the other enumeration schemes to get to the corresonding rimitive air coming from Theorem 2.2. Here by simultaneously we mean conjugate the two elements of the rimitive air by the same single element of the grou. Proof of Theorem 2.1. The roof uses the connection between continued fractions and the Farey tessellation. We call a rational /q or the word E /q odd if q is odd and even otherwise. We observe that in every Farey triangle with vertices m/n, /q, r/s one of the vertices is odd and the other two are even. To see this simly use the fact that mq n, s rq,ms nr are all congruent to 1 modulo 2. (This also gives the equivalence of arity cases for q and the + q used by other authors.) In a triangle where q is even, it may be that the smaller distinguished neighbor is even and the larger odd or vice versa and we take this into account in discussing the enumeration scheme. We note that in general if X and Y are alindromes, then so is (XY) t X for any ositive integer t. We give the roof assuming /q > 0. The roof roceeds by induction on the Farey level. The idea behind the roof is that each rational has a air of arents (distinguished neighbors) and each arent in turn has two arents so there are at most four grandarents to consider. The arents and grandarents may not all be distinct. The cases considered below corresond to the ossible ordering of the grandarents as rational numbers and also the ossible orders of their levels. To deal with negative rationals we use the reflection in the imaginary axis. The reflection sends A 1 to A. We again have distinguished neighbors m/n and r/s, and using the reflection our assumtion is m/n > /q > r/s. In the statement of the theorem, m/n is now the larger neighbor. Using our definition of the Farey level of /q as the Farey level of /q, the roof is exactly the same as for ositive rationals. If /q is odd, by induction we get the roduct of distinguished neighbor alindromes.

9 J. Gilman, L. Keen / Journal of Algebra 332 (2011) If /q is even we need to show that we get alindromes. The set u shows that we have alindromes for level 0, ({0/1, 1/0}) and the correct roduct for level 1, {1/1}. Assume the scheme works for all rationals with level less than N and assume Lev(/q) = N. Since m/n, r/s are distinguished neighbors of /q both their levels are less than N. Suose /q =[a 0, a 1,...,a k ].ThenLev(/q) = k 0 a j = N, and by Remark 3.5 the continued fractionsofthearentsof/q are Assume we are in the case where q is even. Suose first that so that we have [a 0, a 1,...,a k 1 ] and [a 0, a 1,...,a k 1, a k 1]. Lev(r/s)>Lev(m/n) m/n =[a 0, a 1,...,a k 1 ] and r/s =[a 0, a 1,...,a k 1, a k 1]. (3) Then the smaller distinguished neighbor of r/s is m/n and the larger distinguished neighbor is The smaller distinguished neighbor of m/n is w/z =[a 0, a 1,...,a k 2]. (4) and the larger distinguished neighbor is u/v =[a 0,...,a k 2, a k 1 1] (5) If rs is odd we have, by the induction hyothesis x/y =[a 0,...,a k 2 ]. (6) and E r/s = E w/z E m/n E /q = E m/n E r/s = E m/n (E w/z E m/n ) which is a alindrome. If mn is odd we have, by the induction hyothesis, and by Eqs. (3), (4), (5) and (6) so that E m/n = E x/y E u/v E r/s = E (a k 1) m/n E x/y = (E x/y E u/v ) (a k 1) Ex/y is a alindrome. E /q = E m/n E r/s = (E x/y E u/v ) a k E x/y

10 10 J. Gilman, L. Keen / Journal of Algebra 332 (2011) 1 13 If we have Lev(r/s)<Lev(m/n), m/n =[a 0, a 1,...,a k 1, a k 1] and r/s =[a 0, a 1,...,a k 1 ]. (7) Then the larger distinguished neighbor of m/n is r/s and the smaller distinguished neighbor is The larger distinguished neighbor of r/s is w/z =[a 0, a 1,...,a k 2]. (8) and the smaller distinguished neighbor is x/y =[a 0,...,a k 2, a k 1 1] (9) If mn is odd we have, by the induction hyothesis u/v =[a 0,...,a k 2 ]. (10) and E m/n = E r/s E w/z E /q = E m/n E r/s = E r/s (E w/z E r/s ) which is a alindrome. If rs is odd we have, by the induction hyothesis and by Eqs. (7), (8), (9) and (10) so that E r/s = E x/y E u/v E m/n = E u/v E (a k 1) r/s = E u/v (E x/y E u/v ) (a k 1) E /q = E m/n E r/s = E u/v (E x/y E u/v ) a k is a alindrome. It is clear that the E /q words are rimitive because they are roducts of associate rimitive airs and such a roduct, together with either element of the original rimitive air form a new rimitive air. We need, however, to show that we obtain an E /q word corresonding to every conjugacy class of rimitive words in F (or to the class of its inverse) and that this reresentative of the conjugacy class is the unique alindrome in the class when the word is even. We work with ositive rationals between 0 and 1 and note that our comments can easily be adjusted to aly to all rationals. We observe that in generating the words E /q for the ositive rationals between 0 and 1, there is no cancelation because the A s all aear with negative exonents and the B s all aear with

11 J. Gilman, L. Keen / Journal of Algebra 332 (2011) exonent one. Thus there is no difference between concatenation and multilication followed by free reduction. It follows that the E /q have minimal length in their conjugacy class. In fact, every E /q with 0 /q < 1 begins and ends with a ower of A 1 and the owers of A 1 are all searated by a factor of B with exonent +1. It is easy to check this for all words of level u through three. The rest follows by induction on the level of /q: ife /q = E r/s E m/n with the levels of m/n and r/s greater than three, the statement holds by the induction hyothesis. Assume that one of the factors is at a level smaller than three. If E /q is a alindrome, whatever its level, it must start and end with an A 1. The only time the factor E m/n or E r/s will not begin and end with an A 1 is if it is E 1/1. In that case we must have q even, E m/n = E 1/1 = BA 1 and E r/s beginning and ending with A 1 terms. Since the E /q are rimitive words, each is conjugate to one of the form of Eq. (2) in Theorem 3.1. Again because there is no cancelation, the ordered set of exonents for the A 1 terms that aear in E /q are the rimitive exonents, {v 1,...,v } for some integer > 0 and there is a q with i=1 v i = q. As there is no cancelation when we concatenate E r/s E m/n, the exonents sums of the A 1 terms add as do those of the B terms. Thus we must have = and q = q. We have thus established a bijection between the conjugacy classes of rimitives (u to inverse) and the rimitive words E /q. (See also [9].) We claim that the alindromes we obtain are unique in their conjugacy class. We assume E /q is a alindrome so that q is even. Again, we write things as if /q is in [0, 1] and leave it to the reader to make the adjustments for other rationals. As above, we have only factors of A 1 and B. Let V be an arbitrary word in F written in the form of Eq. (2). Suose W = VE /q V 1 is any word in the conjugacy class of E /q. We consider all of the ossible cases: if V begins or ends with a ower of B, W will have owers of B other than +1. If V is a single ower of A, then W will not be a alindrome. Assume V has an initial (or terminal) segment that coincides with the inverse of an initial (or terminal) segment of E /q. Then either W is of a form that does not have minimal length, V is not a alindrome or VE /q V 1 = E /q. Proof of Theorem 2.2. The roof of the first statement is by induction on the maximum of the levels of /q and /q. Again we roceed assuming /q > 0; reflecting in the imaginary axis we obtain the roof for negative rationals. At level 1, the theorem is clearly true: (A 1, B), (A 1, BA 1 ) and (BA 1, B) are all rimitive airs. Assume now that the theorem holds for any air both of whose levels are less than N and let (/q, /q ) be a air of neighbors with Lev(/q) = N. Letm/n, r/s be the distinguished neighbors of /q and assume m/n < /q < r/s. Thenm/n, r/s are neighbors and both have level less than N so that by the induction hyothesis (E m/n, E r/s ) is a air of rimitive associates. It follows that all of the airs (E m/n, E m/n E r/s ), (E m/n, E r/s E m/n ), (E m/n E r/s, E r/s ) and (E r/s E m/n, E r/s ) are airs of rimitive associates since we can retrieve the original air (E m/n, E r/s ) from any of them. Since E /q = E r/s E m/n or E /q = E m/n E r/s we have roved the theorem if /q is one of the distinguished neighbors. If /q is not one of the distinguished neighbors, then either /q = (t + m)/(tq + n) for some t > 0or /q = (t + r)/(tq + s) for some t > 0. Assume for definiteness /q = (t + m)/(tq + n); the argument is the same in the other case. Note that the airs /q,(j + m)/( jq + n) are neighbors for all j = 1,...,t. We have already shown E m/n, E /q is a air of rimitive associates. The argument above alied to this air shows that E m/n, E (+m)/(q+m) is also a air of rimitive associates. Alying the argument t times roves the theorem for the air E /q, E /q. We have established that all airs E /q, E /q with q q =1 are rimitive airs.

12 12 J. Gilman, L. Keen / Journal of Algebra 332 (2011) 1 13 Conversely, suose the air (X, Y ) generates the grou with X conjugate to E /q and Y conjugate to E /q. By Theorem 7.1 of [6], arts 8 and 9, there is a rational number m/n associated to X and a rational number r/s associated to Y such that these rationals are Farey neighbors (that is, ms rn = 1). The rimitive exonents of X and Y are determined by the continued fraction exansions of m/n and r/s: m is the exonent sum of the A terms in X, r is the sum of the exonents of the A terms in Y and m + n and r + s are resectively the minimal word lengths of the conjugacy classes. Since X is conjugate to E /q and Y is conjugate to E /q, as we saw above at the end of the roof of Theorem 2.1, this imlies m =,n = q, r = and s = q so that q q =1. An immediate corollary is Corollary 4.1. The scheme of Theorem 2.1 also gives a scheme for enumerating only rimitive alindromes and a scheme for enumerating only rimitives that are canonical alindromic roducts. 5. Examles Here we comute some examles of words in the enumeration scheme: Fraction Parents Parity E /q Parental roduct Simlified 1/2 0/1 1/1 even E 1/2 A 1 BA 1 A 1 BA 1 2/1 1/1 1/0 even E 2/1 BA 1 B BA 1 B 1/3 1/2 0/1 odd E 1/3 A 1 BA 1 A 1 A 1 BA 2 2/5 1/3 1/2 even E 2/5 A 1 BA 2 A 1 BA 1 A 1 BA 3 BA 1 1/4 1/3 0/1 even E 1/4 A 1 A 1 BA 2 A 2 BA 2 2/7 1/4 1/3 even E 2/7 A 2 BA 2 A 1 BA 2 A 2 BA 3 BA 2 We illustrate how to find the word E 31/9. The continued fraction is 31/9 =[3, 2, 4]; note that the distinguished Farey neighbors are [3, 2]=7/2 and [3, 2, 3]=24/7. We form the following words indicating E 31/9 and its neighbors in boldface E 1/1 = BA 1, E 2/1 = BA 1 B, E 3/1 = E 1/0 E 2/1 = B BA 1 B, E 4/1 = E 3/1 E 0/1 = BBA 1 B B, E 7/2 = E 3/1 E 4/1 = BBA 1 B BBA 1 BB, E 10/3 = E 3/1 E 7/2 = B 2 A 1 B 3 A 1 B 3 A 1 B 2, E 17/5 = E 7/2 E 10/3 = B 2 A 1 B 3 A 1 B 2 B 2 A 1 B 3 A 1 B 3 A 1 B 2, E 24/7 = E 17/5 E 7/2 = B 2 A 1 B 3 A 1 B 2 B 2 A 1 B 3 A 1 B 3 A 1 B 2 B 2 A 1 B 3 A 1 B 2, E 31/9 = E 7/2 E 24/7 = B 2 A 1 B 3 A 1 B 2 B 2 A 1 B 3 A 1 B 2 B 2 A 1 B 3 A 1 B 3 A 1 B 2 B 2 A 1 B 3 A 1 B Farey diagram visualization We can visualize the relation between rimitive airs and neighboring rationals using the Farey diagram. Suose the rimitive air (A, B) corresonds to (r/s, /q). Note that we have done this so that the Farey level of /q is greater than that of r/s and r/s is the arent of /q with lowest Farey level. (The other arent is (r )/(q s) and corresonds to A 1 B.) Draw the curve γ from a oint on the imaginary axis to /q. If /q is ositive, orient it toward /q; if /q is negative, orient it towards the imaginary axis. The left-right sequence and the continued fraction for /q are the same. Traversing the curve in the other direction reverses left and right. The symmetry about the imaginary axis is reflected in our definition of negative continued fractions.

13 J. Gilman, L. Keen / Journal of Algebra 332 (2011) Given two rimitive airs (A, B) and (A B ) such that B corresonds to /q and B corresonds to /q draw the curves γ and γ. We can find the sequence to go from (A, B) to (A, B ) by traversing γ from /q to the imaginary axis and then traversing γ to /q. We can also draw an oriented curve from /q to /q and read off the left-right sequence along this curve to get the continued fraction that gives (A, B ) as words in (A, B) directly. We have Corollary 6.1. Given any two sets of rimitive airs, (A, B) and (A, B ) there is a continued fraction containing either only ositive or only negative integers that connects one air to the other. Acknowledgments We thank Vidur Malik whose thesis [16,17] suggested that we look at alindromes. We also thank the referee who careful reading and helful suggestions has done much to imrove the aer. References [1] Bardakov, Srilrain, Tolytykh, On the alindromic and rimitive widths in a free grou, J. Algebra 285 (2005) [2] P. Buser, K.-D. Semmler, The geometry and sectrum of the one-holed torus, Comment. Math. Helv. 63 (2) (1988) [3] P. Cohen, W. Metzler, B. Zimmermann, What does a basis of F (a, b) look like?, Math. Ann. 257 (4) (1981) [4] Jane Gilman, Informative Words and Discreteness, Contem. Math., vol. 421, 2007, [5] Jane Gilman, Algorithms, comlexity and discreteness criteria in PSL(2, C), J. Anal. Math. 73 (1997) [6] Jane Gilman, Linda Keen, Word sequences and intersection numbers, in: Comlex Manifolds and Hyerbolic Geometry, Guanajuato, 2001, in: Contem. Math., vol. 311, Amer. Math. Soc., Providence, RI, 2002, [7] Jane Gilman, Linda Keen, Cutting sequences and alindromes, in: Geometry of Riemann Surfaces, in: London Math. Soc. Lecture Note Ser., vol. 368, 2010, [8] Jane Gilman, Linda Keen, Discreteness criteria and the hyerbolic geometry of alindromes, Conform. Geom. Dyn. 13 (2009) 76 90, arxiv: [9] Lifting from PSL(2, R) to the free grou F = a, b, in rearation. [10] Hyerbolic Geometry and Core Geodesics, in rearation. [11] J. Gilman, B. Maskit, An algorithm for 2-generator Fuchsian grous, Michigan Math. J. 38 (1991) [12] G.H. Hardy, E.M. Wright, An Introduction to the Theory of Numbers, Clarendon Press, Oxford, [13] Yicheng Jiang, Polynomial comlexity of the Gilman Maskit discreteness algorithm, Ann. Acad. Sci. Fenn. Math. 26 (2) (2001) [14] Christian Kassel, Christohe Reutenauer, Sturmian morhisms, the braid grou B 4, Christoffel words and bases of F 2, Ann. Mat. Pura Al. (4) 186 (2) (2007) [15] Linda Keen, Caroline Series, Pleating coordinates for the Maskit embedding of Teichmüller sace for a unctured torus, Toology 32 (4) (1993) [16] Vidur Malik, Curves generated on surfaces by the Gilman Maskit algorithm, PhD thesis, Rutgers University, Newark, NJ [17] Vidur Malik, Primitive words and self-intersections of curves on surfaces generated by the Gilman Maskit discreteness algorithm, in: The Tradition of Ahlfors-Bers. V, in: Contem. Math., vol. 510, Amer. Math. Soc., Providence, RI, 2010, [18] Wilhelm Magnus, Abraham Karrass, Donald Solitar, Combinatorial Grou Theory: Presentations of Grous in Terms of Generators and Relations, Interscience Publishers, John Wiley & Sons, Inc., New York, London, Sydney, [19] R.P. Osborne, H. Zieschang, Primitives in the free grou on two generators, Invent. Math. 63 (1) (1981) [20] Adam Piggott, Palindromic rimitives and alindromic bases in the free grou of rank two, J. Algebra 304 (1) (2006) [21] Caroline Series, Non-euclidean geometry, continued fractions and ergodic theory, Math. Intelligencer 4 (1982) [22] Caroline Series, The modular surface and continued fractions, J. Lond. Math. Soc. (2) 31 (1985) [23] Caroline Series, The geometry of Markoff Numbers, Math. Intelligencer 7 (3) (1985) [24] I.M. Vinogradov, An Introduction to the Theory of Numbers, English translation by H. Poova, Pergamon Press, London, New York, 1955, MR# [25] David J. Wright, Searching for the cus, in: Saces of Kleinian Grous, in: LMS Lecture Notes, vol. 329, Cambridge Univ. Press, 2004,