Common Right Multiples in KP of Linear Combinations of Generators of the Positive Monoid P of Thompson s Group F

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1 International Mathematical Forum, 5, 2010, no. 32, Common Right Multiples in KP of Linear Combinations of Generators of the Positive Monoid P of Thompson s Group F John Donnelly Department of Mathematics University of Southern Indiana 8600 University Boulevard Evansville, Indiana 47712, USA jrdonnelly@usi.edu Abstract We show that any two elements of KP which can be written as a linear combination of generators of the positive monoid P of the group F have nonzero common right multiples in KP. 1 Introduction Let S be a semigroup, and let BS denote the set of all bounded, real valued functions on S. For each f BS and for each a S, define f a BS byf a x fax for each x S. We say that S is left amenable if there exists a function μ : BS R such that for all f,g BS, for each a S, and for all r R: 1. sup fx μf inf fx x S x S 2. μf a μf 3. μf + g μf+μg 4. μrf rμf. We say that a group G is amenable if it is left amenable. We say that a semigroup S has common right multiples if for each pair of elements b 1,b 2 S, there exist elements d 1,d 2 S such that b 1 d 1 b 2 d 2. One can analogously define the notion of common right multiples in a ring. A theorem of Ore states that if M is a cancellative monoid which has common right multiples, then M embeds as a submonoid into a group G such that the following two conditions are satisfied:

2 1582 J. Donnelly i For each g G, there exists x, y M such that g xy 1. ii If M has monoid presentation A R, then G is isomorphic to the group defined by the presentation A R. The group G is called the group of right fractions of M, and M is called the positive monoid of G. We define Richard Thompson s group F to be the group of right fractions of the monoid P which is given by the presentation x 0,x 1,x 2,... x n x m x m x n+1 for n>m. We define a k, m-binary forest F to be a sequence starting the count at zero of rooted binary trees such that F has a total of exactly m carets, and such that for each i k + 1, the i th tree of F is the trivial tree, which consists of just a single point. The following is an example of a 3,21-binary forest The set of all k, m-binary forests is denoted by S k,m.iff is a binary forest, then we denote the i th tree of F by τ F i. We enumerate the leaves of a binary forest F from left to right. The following example shows the enumeration on the leaves of a 5,23-binary forest Using this enumeration on the leaves of a binary forest, we define a multiplication on the set of all binary forests. Given two binary forests F and G, construct the binary forest FG by attaching the i th tree of G to the i th leaf of F. For example, we multiply a 3,4-binary forest F with a 7,15-binary forest G: F G

3 Common right multiples 1583 to get the following binary forest FG: FG With this multiplication, the set of all binary forests forms a monoid which is isomorphic to the monoid P. More specifically, for n 0, the generator x n of P is the binary forest all of whose trees are trivial except for the n th tree, which consists of a single caret. In [7], Tamari proves the following theorem. A proof is also given by Engel in [4]. Theorem 1. Let M be a left amenable, cancellative monoid, and let K be a field. Then for each pair of nonzero elements α, β KM there exist nonzero elements γ,ρ KM such that αγ βρ. Geoghegan has conjectured that the group F is an example of a finitely presented, nonamenable group which has no free subgroup on two generators [5]. In [1], Brin and Squier show that F has no free subgroup on two generators. However, the question of whether or not the group F is amenable is still open, and has resisted the efforts of mathematicians for over twenty years [2]. Thus, one can ask if for any field K, does the group ring KF have nonzero common right multiples, thereby satisfying the conclusion of Theorem 1. In Section 2, we show that the group ring KF has nonzero common right multiples if and only if the monoid ring KP has nonzero common right multiples. In Section 4 we show that any two nonzero elements α, β KP which can be written as a linear combination of generators of the positive monoid P have nonzero common right multiples in KP. Moreover, we show that that if γ,ρ KP are such that αγ βρ, then γ and ρ are linear combinations of elements from S k+1,m. 2 The Relationship Between KG and KM Throughout this section, we assume that M is a cancellative monoid which has common right multiples, that G is the group of right fractions of the monoid M, and that K is a field such that KM has no zero divisors. Using induction on H, together with the fact that G is the group of right fractions of the monoid M, one can show that if H is a finite subset of G, then there exists q M such that Hq M. The proof of the following lemma is left to the reader.

4 1584 J. Donnelly Lemma 1. Let G be the group of right fractions of a monoid M. Let z 1,...,z n KG. Then there exists q M such that for each i {1,...,n}, we have that z i q KM. Lemma 2. Let G be the group of right fractions of a monoid M. Then KG has nonzero common right multiples if and only if KM has nonzero common right multiples. Proof. Assume that KG has nonzero common right multiples. Let z 1 and z 2 be nonzero elements of KM. Since KM KG, and since KG has nonzero common right multiples, then there exist u 1,u 2 KG such that z 1 u 1 z 2 u 2 0. By Lemma 1, there exists q M such that u 1 q, u 2 q KM. Since z 1 u 1 0,z 2 u 2 0,andq 0, and since KG has no zero divisors, then z 1 u 1 q 0 and z 2 u 2 q 0. Thus, u 1 q and u 2 q are elements of KM such that z 1 u 1 qz 2 u 2 q 0. Hence, KM has nonzero common right multiples. Conversely, assume that KM has nonzero common right multiples. Let y 1 and y 2 be nonzero elements of KG. By Lemma 1, there exists b M such that y 1 b, y 2 b KM. Since y 1 b, y 2 b KM, and since KM has nonzero common right multiples, then there exist v 1,v 2 KM ZG such that y 1 bv 1 y 2 bv 2 0. Hence, KG has nonzero common right multiples. 3 Properties of S k,m The set S 0,m is essentially the set of all rooted binary trees, each of which has exactly m carets. It is well known that the number of elements in this set is the m th catalan 2m! number [6]. The following lemma is proven in [3]. m!m + 1! Lemma 3. Let m 0. i S 1,m S 0,m+1. ii For each k 1, S k+1,m S k,m+1 S k 1,m+1. k + 12m + k! Theorem 2. For each k 0, and for each m 0, S k,m m + k + 1!m!. Proof. We prove this by induction on k. Ifk 0, then we see that S 0,m 2m! m!m + 1! m + 0! m!m +0+1! Assume that k 1. It follows by Lemma 3 that S 1,m S 0,m+1. Thus, we see that S 1,m S 0,m+1 2m + 1! m + 1!m +1+1! 2m + 2! m + 1!m + 2!.

5 Common right multiples 1585 Since 22 m +1 m +1 2m +2 m +1 2m + 22m + 1!m! m + 12m + 1!m! 2m + 2!m! 2m + 1!m + 1! then which implies that 2m + 2! m + 1! 22m + 1! m! S 1,m 2m + 2! 22m + 1! m + 1! m + 1!m + 2! m!m + 2! m +1+1!m!. Now assume that there exists k 1, such that for each j {0,...,k} and for each m 0, S j,m j + 12m + j! m + j + 1!m! Since k 1, then it follows by Lemma 3 that S k+1,m S k,m+1 - S k 1,m+1. Thus, we see that S k+1,m S k,m+1 S k 1,m+1 k + 12m +1+k! k m +1+k 1! m +1+k + 1!m + 1! m +1+k 1 + 1!m + 1! k + 12m + k + 2! k2m + k + 1!m + k +2 m + k + 2!m + 1! 2m + k + 1!km + k +2m +2 m + k + 2!m!m +1 2m + k + 1!k + 2m +1 m + k + 2!m!m +1 k m +k + 1!. m +k !m!

6 1586 J. Donnelly S k+1,m 2k +2 Theorem 3. For each k 0, m S k,m k +1. Proof. We see that S k+1,m m S k,m m k+22m+k+1! m+k+2!m! k+12m+k! m+k+1!m! k + 22m + k +1 2k +2 m k + 1m + k +2 k +1. S k,m Theorem 4. For each k 0, m S 0,m k + 12k. Proof. We prove this by induction on k. When k 0, we see that S 0,m m S 0,m S k,m Assume that there exists k 1 such that m S 0,m k + 12k. Thus, by Lemma 3, together with our induction hypothesis, we see that S k+1,m m S 0,m m S k+1,m S k,m m S k,m S 0,m 2k +2 k + 12 k k + 22 k+1. k +1 S 0,m+k Theorem 5. For each k 0, 4 k. m S 0,m Proof. We prove this by induction on k. When k 0, we see that S 0,m+0 m S 0,m Assume that k 1. Then we see that S 0,m+1 m S 0,m m 2m+1! m+1!m+2! 2m! m!m+1! S 0,m m S 0,m 140. m Now assume that there exists k 1 such that S 0,m+k m S 0,m 4 k. 4m 2 +6m +2 m 2 +3m As m, then m + k. Therefore, by the base case for k 1, we see that

7 Common right multiples 1587 Hence, we see that S 0,m+k+1 m S 0,m m S 0,m+k+1 S 0,m+1 4. m S 0,m+k m S 0,m S 0,m+k+1 S 0,m+k m S 0,m+k 44 k 4 k+1. S 0,m 4 The Main Theorem The following theorem is a version of Theorem 1. However, we have reworded the statement of the theorem to fit the context here more appropriately. In particular, we do not assume that the monoid M is amenable, but instead focus on one specific nonempty, finite subset H of M. Note that if M is amenable, then the theorem can be applied to any nonempty, finite subset of M. The proof is identical to that of Theorem 1. However, we include the proof here for the sake of completeness. Theorem 6. Let M be a monoid, and let H be a nonempty, finite subset of M. Assume that H consists precisely of the z elements h 1,...,h z. Let α and β be nonzero elements of KM such that α z k1 a kh k and β z k1 b kh k, where for each k {1,...,z}, we have that a k,b k K. If there exists a nonempty, finite subset E M such that HE < 2 E, then there exist nonzero elements γ,ρ KM such that αγ βρ. Proof. Assume that there exists a nonempty, finite subset E M such that HE < 2 E. Let r E, and let t HE. Let HE consist of the elements g 1,g 2,...,g t. We assume that the order of these elements is fixed. Since HE < 2 E, then t<2r. We need to show that there exist nonzero elements γ,ρ KM such that αγ βρ. Let E consist precisely of the r elements e 1,e 2,...,e r. Again, we assume that the order of these elements is fixed. Let γ r l1 u le l and ρ r l1 w le l, where for each l {1,...,r}, the coefficients u l and w l are unknown. Let V γ denote the r 1 vector whose entries are the unknown coefficients u 1,u 2,...,u r of γ. That is, u 1 u 2 V γ. u r

8 1588 J. Donnelly Given g i HE, then there exist h ki H and e li E such that g i h ki e li. Let M γ [m i,j ] be the t r matrix whose ij th entry m i,j is given by { a ki if g i h ki e j m i,j 0 otherwise Similarly, let V ρ denote the r 1 vector whose entries are the unknown coefficients w 1,w 2,...,w r of ρ. That is, w 1 w 2 V ρ. Also, let M ρ [q i,j ] be the t r matrix whose ij th entry q i,j is given by { b ki if g i h ki e j q i,j 0 otherwise Since M γ V γ M ρ V ρ is a linear system with t equations and 2r unknowns, and since t < 2r, then it has a nontrivial solution. Thus, we have coefficients c 1,c 2,...,c r,d 1,d 2,...,d r, not all of which are zero, such that for each i {1,...,t}, a ki c j b ki d j h ki e j g i h ki e j g i w r Thus, we have elements γ,ρ KP such that γ r l1 c le l and ρ r l1 d le l, and z r z r t moreover, that αγ a k h k c l e l a k c l h k e l a ki c j g i k1 l1 i1 l1 i1 h ki e j g i t z r z r b ki d j g i b k d l h k e l b k h k d l e l βρ. i1 h ki e j g i i1 l1 k1 l1 Let k Z +. Let A k {x 0,x 1,...,x k }. Theorem 7. Let k Z +. A k S k+1,m S k,m+1. Then for each m Z, with m 0, we have that Proof. Let m Z, with m 0. Let y A k S k+1,m. Thus, there exist x i A k and z S k+1,m such that y x i z. We form the product x i z by attaching a caret λ to the forest z from above by attaching the left leaf of λ to the root of the i th tree τ z i of z, and by attaching the right leaf of λ to the root of the i +1 st tree τ z i +1of

9 Common right multiples 1589 z. Since z has a total of m carets overall, and since we get x i z by attaching a single caret to z, then x i z has a total of m + 1 carets overall. The i th and i +1 st trees τ z i and τ z i + 1, respectively, of z become subtrees of the i th tree τ xi zi ofx i z. The roots of these trees bcome the the left and right children of the root caret λ of τ xi zi. Moreover, for each j Z, with j i + 2, the j th tree τ z j ofz becomes the j 1 st tree τ xi zj 1 of x i z. Since for each b Z, with b k + 2, the b th tree τ z b ofz is trivial, then for each b Z, with b k + 1, the b th tree τ xi zb ofx i z is trivial. Therefore, x i z S k,m+1. Thus, A k S k+1,m S k,m+1. Conversely, assume that w S k,m+1. Since w has at least one caret, then at least one of the trees τ w 0,τ w 1,...,τ w k is nontrivial. Let j {0,...,k} be such that the j th tree τ w j ofw is nontrivial. Remove the top caret λ of τ w j to get a forest which is equal to x 1 j w. Since w has a total of m + 1 carets overall, and since we get w by removing a single caret from w, then x 1 j w has a total of m carets overall. The subtrees of τ w j whose roots are the left and right children of the root of τ w j become the j th and j +1 st trees τ x 1 j w j and τ x 1 j w j + 1, respectively, of x 1 j w. Also, for each i Z, with i j + 1, we have that τ x 1 j w i + 1 and τ wi. Note that if j 1, then for each i {0,...,j 1}, we have that τ x 1 j w i and τ wi. Since for each i Z, with i k +1 j +1, τ w i is trivial then for each d Z, with d k +2, τ x 1 j w d is trivial. Therefore x 1 j w is a forest which has a total of m carets overall such that for each i Z, with i k +2, τ x 1 j w i is trivial. Thus, x 1 j w S k+1,m. Since j {0,...,k}, then x j A k. Since we get w by multiplying x 1 j w on the left by x j, then w A k S k+1,m. Hence, S k,m+1 A k S k+1,m. x 1 j Theorem 8. Let k Z +. There exists m Z, with m 0, such that A k S k+1,m < 2. S k+1,m Proof. By Theorems 4 and 5, we see that S k,m+1 m S k+1,m S k,m+1 S 0,m+1 S 0,m m S 0,m+1 m S 0,m m S k+1,m k + 12 k 1 4 < 2. k k+1 Therefore, there exists b Z, with b 0, such that S k,b+1 S k+1,b < 2. By Theorem 7, we see that A k S k+1,b S k,b+1. Thus, there exists b Z, with b 0, such that A k S k+1,b < 2. S k+1,b

10 1590 J. Donnelly Let α k j1 a jx j and β k j1 b jx j, where for each j {1,...,k}, we have that a j,b j K, and x j A k. Let q S k+1,m. By Theorem 8, there exists m Z, with m 0, such that A k S k+1,m < 2 S k+1,m. Thus, by Theorem 6, there exist γ,ρ KM such that αγ βρ. Moreover, by Theorems 6 and 8, we see that γ q l1 c lz l, and ρ q l1 d lz l, where for each l {1,...,q}, we have that c j,d j K, and z l S k+1,m. References [1] M. Brin, and C. Squier, Groups of piecewise linear homeomorphisms of the real line, Invent. Math , no. 3, [2] J. W. Cannon, W. J. Floyd, and W. R. Parry, Introductory notes on Richard Thompson s groups, Enseign. Math ,no. 3-4, [3] J. Donnelly, The Ruinous-Thin Relationship for Three Element Sets, International Journal of Algebra and Computation , no. 3, [4] A. B. Engel, The homomorphic image of a T n semigroup, Semigroup Forum , no. 1, [5] S. M. Gersten, Selected problems, Combinatorial Group Theory and Topology S. M. Gersten and John Stallings, eds., Annals of Mathematics Studies, vol. 111, Princeton University Press, 1987, pp [6] R. Stanley, Enumerative Combinatorics, Volume 2, Cambridge University Press, Cambridge, [7] D. Tamari, A refined classification of semigroups leading to generalized polynomial rings with a generalized degree concept, Proceedings of the International Congress of Mathematicians, Amsterdam, , Received: November, 2009