Commutators in groups of piecewise projective homeomorphisms
aa r X i v : . [ m a t h . G R ] M a r COMMUTATORS IN GROUPS OF PIECEWISE PROJECTIVEHOMEOMORPHISMS.
JOS´E BURILLO, YASH LODHA, AND LAWRENCE REEVES
Abstract.
In [7] Monod introduced examples of groups of piecewise projective homeo-morphisms which are not amenable and which do not contain free subgroups, and in [6]Lodha and Moore introduced examples of finitely presented groups with the same prop-erty. In this article we examine the normal subgroup structure of these groups. Twoimportant cases of our results are the groups H and G . We show that the group H ofpiecewise projective homeomorphisms of R has the property that H ′′ is simple and thatevery proper quotient of H is metabelian. We establish simplicity of the commutatorsubgroup of the group G , which admits a presentation with 3 generators and 9 relations.Further, we show that every proper quotient of G is abelian. It follows that the normalsubgroups of these groups are in bijective correspondence with those of the abelian (ormetabelian) quotient. Introduction
In [7] Monod proved that the group H of piecewise projective homeomorphisms of thereal line is non-amenable and does not contain non-abelian free subgroups. This providesa new counterexample to the so called von Neumann–Day problem [8, 2]. In fact, Monodintroduced a family of groups H ( A ) for a subring A of R . In the case where A is strictlylarger than Z , they were all demonstrated to be counterexamples. The group H is thecase in which A = R .The subgroups G and G of H were introduced by Lodha and Moore in [6] as finitelypresented counterexamples. The groups G and G share many features with Thompson’sgroup F . They can be viewed as groups of homeomorphisms of the Cantor set of infinitebinary sequences, and as groups of homeomorphisms of the real line. They admit smallfinite presentations, and symmetric infinite presentations with a natural normal form[6, 5]. Further, they are of type F ∞ [5]. Viewed as homeomorphisms of the Cantor set,the elements can be represented by tree diagrams.Thompson’s group F satisfies the property that F ′ is simple, and every proper quotientof F is abelian [1]. In this article we examine the normal subgroup structure, and inparticular the commutator subgroup structure of G , G , and H ( A ) for a subring A of R ,and obtain properties similar to F . We prove the following. Theorem 1.
Let A be a subring of R . If A has units other than ± , then: (1) H ( A ) ′ = H ( A ) ′′ . The first author’s research is supported by MICINN grant MTM2014-54896-P. The second author’sresearch is supported by an EPFL-Marie Curie grant. The second author would like to thank Swiss Airfor its hospitality during a flight on which a portion of the paper was written. (2) H ( A ) ′′ is simple. (3) Every proper quotient of H ( A ) is metabelian.If the only units in A are ± , then: (1) H ( A ) ′ is simple. (2) Every proper quotient of H ( A ) is abelian. (3) All finite index subgroups of H ( A ) are normal in H ( A ) . We show the following for the finitely presented groups G and G (defined in Section 1). Theorem 2.
The group G satisfies the following: (1) G ′ is simple. (2) Every proper quotient of G is abelian. (3) All finite index subgroups of G are normal in G .The group G satisfies the following: (1) G ′′ = G ′ . (2) G ′′ is simple and G ′′ = G ′ . (3) Every proper quotient of G is metabelian. One interesting feature of this article is that although the proofs of Theorems 1 and 2 bothuse a theorem of Higman, the proofs are different in the following sense. The argumentsof the proof of Theorem 2 are intrinsic to the combinatorial model developed for G , G in [6] using continued fractions. The arguments of the proof of Theorem 1 arise in thesetting of the action of the groups H ( A ) on the real line.1. Background
Actions considered will be from the right except when we explicitly use function notationor represent elements by matrices. The conjugate of g by h is h − gh . All groups understudy here are subgroups of P P SL ( R ), the group of piecewise projective homeomor-phisms of R ∪ {∞} preserving orientation. Namely, for each element in P P SL ( R ) thereare finitely many points t , . . . , t n such that in each of the intervals ( −∞ , t ], [ t i , t i +1 ] and[ t n , ∞ ) the map is of the form t ( at + b ) / ( ct + d ) for some matrix (cid:18) a bc d (cid:19) with determinant one. The group H is the subgroup of P P SL ( R ) formed of those mapsthat stabilize infinity, and that hence give homeomorphisms of R . Observe that elementsof H have affine germs at ±∞ , that is, in the interval [ t n , ∞ ) the map is ( at + b ) /d with ad = 1, and similarly on the interval ( −∞ , t ]. Given a subring A of R , we denote by P A the set of fixed points of hyperbolic elements of P SL ( A ). Then H ( A ) is defined to bethe subgroup of H consisting of elements that are piecewise P SL ( A ) with breakpointsin P A . See [7] for details on these groups. OMMUTATORS IN GROUPS OF PIECEWISE PROJECTIVE HOMEOMORPHISMS. 3
The two Lodha–Moore groups [6] are finitely presented subgroups of H and will be denotedby G and G . The group G is the group of homeomorphisms of R generated by thefollowing three maps: a ( t ) = t + 1 b ( t ) = t if t ≤ t − t if 0 ≤ t ≤ t − t if 12 ≤ t ≤ t + 1 if 1 ≤ t c ( t ) = tt + 1 if 0 ≤ t ≤ t otherwiseAs is done with Thompson’s group F , elements will be given an interpretation in termsof maps of binary sequences and tree diagrams. A binary sequence is a (finite or infinite)sequence of and . The set of infinite binary sequences will be denoted by 2 N , and theset of finite ones by 2 < N . We will use s and t to denote finite binary sequences (with adistinctive font), and Greek letters ξ , ζ or η for infinite ones. Two binary sequences canbe concatenated as long as the first one is finite, such as ξ , s01 or s ζ .We define the binary sequence map: x : 2 N −→ N ( ξ ) · x = ξ ( ξ ) · x = ξ ( ξ ) · x = ξ and also, recursively, the pair of mutually inverse maps y : 2 N −→ N ( ξ ) · y = ( ξ · y )( ξ ) · y = ( ξ · y − )( ξ ) · y = ( ξ · y ) y − : 2 N −→ N ( ξ ) · y − = ( ξ · y − )( ξ ) · y − = ( ξ · y )( ξ ) · y − = ( ξ · y − ) . Each of these maps will give rise to a family of maps of binary sequences defined in thefollowing way. Given a finite binary sequence s , the map x s is the identity except on theinfinite sequences starting with s , where it acts as x on the tail. That is,( s ξ ) · x s = s ( ξ · x )and as the identity if the sequence does not start with s . The maps y s or y − s are definedin an analogous way. The map x s (and also similarly y s ) also admits a partial action onthe set 2 < N of finite binary sequences in the natural way: if a sequence extends s00 , s01 or s1 , we have ( su ) · x s = s ( u · x ), and x s is the identity on those sequences which areincompatible with s . The map x s is not defined on the other sequences, namely, x s is notdefined on s0 or on the prefixes of s , in particular, observe that x is not defined on thesequence . JOS´E BURILLO, YASH LODHA, AND LAWRENCE REEVES
The reason for these definitions is that they represent elements of G under an identifica-tion given by the following maps: ϕ : 2 N −→ [0 , ∞ ] ϕ ( ξ ) = 11 + ϕ ( ξ ) ϕ ( ξ ) = 1 + ϕ ( ξ ) Φ : 2 N −→ R Φ( ξ ) = − ϕ ( ˜ ξ )Φ( ξ ) = ϕ ( ξ )where ˜ ξ is the sequence obtained from ξ by replacing all symbols by and viceversa.Under these definitions, the maps a , b and c are represented by the binary sequence maps x , x and y , respectively. We have the following result (Proposition 3.1 in [6]): Proposition 1.1.
For all ξ in N we have a (Φ( ξ )) = Φ( x ( ξ )) b (Φ( ξ )) = Φ( x ( ξ )) c (Φ( ξ )) = Φ( y ( ξ ))For all details and proofs, see [6]. Hence, we can consider that the group G is the groupof maps of 2 N generated by x, x , y .The group G is defined to be the group generated by all x s and y s . The group G isgenerated by all x s and by all those y s where s is not constant, that is, s is not n nor n . We have a series of relations which are satisfied by these generators:(1) x s = x s0 x s x s1 ,(2) if t · x s is defined, then x t x s = x s x t · x s ,(3) if t · x s is defined, then y t x s = x s y t · x s ,(4) if s and t are incompatible, then y s y t = y t y s ,(5) y s = x s y s0 y − s10 y s11 .The relations (1) and (2) are part of a known presentation for Thompson’s group F givenby Dehornoy in [3]. The key relation for the Lodha–Moore groups is the relation (5),which represents algebraically the recursive definition of y given above.Finally, the relations satisfied by these generators, and in particular the Thompson’s grouprelations, allow us to obtain finite presentations. The group G is generated only by x , x and y with a set of 9 relations, whereas G is generated by x , x , y , y and y .In [6] it is shown that every element of G can be written in a standard form. Recall that G is generated by the set X ∪ Y where X = { x s : s ∈ < N } and Y = { y s : s ∈ < N } . Aword over X ∪ Y is said to be in standard form if it is of the form hy a s . . . y a n s n where h is a word over X , s j is a prefix of s i only if j ≥ i , and the a i are arbitrary,non-zero integers. It is shown in [6, Lemma 5.4] that every element of G can be writtenin standard form. The standard form is, however, not unique. OMMUTATORS IN GROUPS OF PIECEWISE PROJECTIVE HOMEOMORPHISMS. 5 Commutators for G We first study the abelianization of the group G . The partial action of F on the set of allnon-constant binary sequences is transitive. Therefore, from relation (3) it follows thatfor all non-constant s , y s is a conjugate of y . Lemma 2.1.
The map { x, x , y } → Z given by x (1 , , x (0 , , y (0 , , extends to a surjective homomorphism π : G → Z , with kernel being the commutatorsubgroup G ′ = [ G , G ] .Proof. We use the infinite presentation of G having generating set S = X ∪ Y where X = { x s : s ∈ < N } and Y = { y s : s ∈ < N , s is not a constant word } . The set of relationsis given by those relations of the form (1) to (5) above that involve only elements of S .That this gives a presentation of G is Theorem 3.3 in [6]. One could, of course, also usethe finite presentation for G on the generating set { x, x , y } (see [6]), but we shall notdo so.Noting that each element y s ∈ Y is conjugate in G to y , that x m ∈ X is conjugateto x , that x m ∈ X is conjugate to x and that x s ∈ X is conjugate to x when s isnon-constant, we consider the map from S → Z given by y s (0 , , x (1 , , x s (1 , − ,
0) if s = m for some m ≥ , ,
0) if s = m for some m ≥ , ,
0) if s is non-constantThat this map extends to a homomorphism π : G → Z can be readily seen by consideringthe effect on each of the relations (1) to (5).Since { x, x , y } projects to a generating set for the abelianization G /G ′ and the image { π ( x ) , π ( x ) , π ( y ) } is a basis for Z we conclude that π induces an isomorphism from G /G ′ to Z . (cid:3) The restriction of π to the subgroup F generated by { x, x } gives (after a change of basis)the abelianization map for Thompson’s group F , where elements are evaluated by the twogerms at ±∞ . The third component of the map π represents the total exponent for the y -generators in a word representing an element. Hence, we have the following result. Proposition 2.2.
The commutator G ′ contains exactly those elements in G which havecompact support, and which have a total exponent in the y -generators equal to zero. (cid:3) The goal of this section is to prove the first part of Theorem 2, which concerns G . Toproceed with the proof, we will use the following theorem, due to Higman. Let Γ be agroup of bijections of some set E . For g ∈ Γ define its moved set D ( g ) as the set of points x ∈ E such that g ( x ) = x . This is analogous to the support, but since a priori there isno topology on E , we do not take the closure. Theorem 2.3.
Suppose that for all α, β, γ ∈ Γ \ { Γ } , there is a ρ ∈ Γ such that thefollowing holds: γ ( ρ ( S )) ∩ ρ ( S ) = ∅ where S = D ( α ) ∪ D ( β ) . Then Γ ′ is simple. JOS´E BURILLO, YASH LODHA, AND LAWRENCE REEVES
The proof can be seen in [4].
Corollary 2.4.
Let Γ be a group of compactly supported homeomorphisms of R thatcontains F ′ . Then Γ ′ is simple. This is the consequence of the high transitivity of F ′ which ensures that the conditionsof the theorem are satisfied. We shall apply this corollary to several groups in the paper.This corollary cannot be applied directly to G , since this group contains elements (e.g., x ) whose support is all of R . So we will apply this corollary to the group G ′ , whoseelements have compact support. We conclude that G ′′ is simple. The proof of Theorem 2for G will now be complete with the following result. Proposition 2.5. G ′ = G ′′ .Proof. Consider an element g ∈ G ′ and write the element in standard form as g = hz ,where h ∈ F and z = y a s . . . y a n s n for some binary sequences s i , and with a + · · · + a n = 0. Since h has compact support,we have h ∈ F ′ = F ′′ ⊂ G ′′ . So we only need to show that z ∈ G ′′ . Note that for anygenerator x s such that s is a non constant sequence, x s ∈ G ′′ . We will make use of thisfact repeatedly in what follows.The proof proceeds by induction on k = | a | + · · · + | a n | which is clearly an even number.For the element z there will be some a i positive and some negative. Since G ′′ is normal,we can cyclically conjugate and assume that the word starts with a subword of the type y s y − t . As the starting point of the induction, just take 1 ∈ G ′′ . We just need to provethat y s y − t ∈ G ′′ and using the induction hypothesis for the rest of the word, the proofwill be complete. We have three cases.Case (1): s and t are consecutive. This just means that the corresponding intervals in R are adjacent, or that if s and t are leaves in a tree, they are consecutive in the naturalorder of the leaves.Take the word w = y y − ∈ G ′ . Construct an element f ∈ F ′ such that f wf − = y y − , which is possible because these two sequences are also consecutive and any of these canbe F ′ -conjugated to any other. Now we have that [ w, f ] = wf w − f − ∈ G ′′ , since it isthe commutator of two elements in G ′ . But clearly[ w, f ] = y y − , Now apply relation (5) to y to get[ w, f ] = x y y − y y − = x y y − . As mentioned before, we know that x ∈ F ′ ⊂ F ′′ ⊂ G ′′ , so this implies that y y − ∈ G ′′ , and these are two consecutive binary sequences. Hence, by conjugation, any y s y − t with consecutive binary sequences is in G ′′ . OMMUTATORS IN GROUPS OF PIECEWISE PROJECTIVE HOMEOMORPHISMS. 7
Case (2): s and t are not consecutive and also not comparable. Assume s < t , as theother case reduces to this by inverting the element. In that case, just write y s y − t = y s y − s y s y − s y s . . . y − s m y s m y − t such that the pairs y s y − s y s i y − s i +1 y s m y − t lie in the previous case.Case (3): s and t are comparable, so assume t = su . The case s = tu reduces to this bytaking an inverse. We apply relation (5) to y s again and find pairs which now correspondto cases (1) or (2). Cases y − s y s are cyclically permuted to y s y − s . We distinguish allfour easy cases for clarity: • u = or u = . Since y s y − su = x s y s0 y − s10 y s11 y − su , y − su cancels with one of the resultsof relation (5) applied to y s . We are left with the product of an x -generator (in G ′′ ) with a word that falls in case (1). • u = . Applying relation (5) we obtain x s y s0 y − s10 y s11 y − s1 and we apply case (1) to the pairs y s0 y − s1 y − s10 y s11 . • u = . Using relations (4) , (5) we obtain: x s y s0 y − s10 y s11 and we just need to apply case (1) to y s0 y − s10 and y − s10 y s11 . • u = , , , . Using relation (5) we obtain x s y s0 y − s10 y s11 y − su . It suffices to show that y s0 y − s10 y s11 y − su ∈ G ′′ . If u begins with a , by cyclicconjugation we obtain ( y − su y s0 )( y − s10 y s11 ). The word y − su y s0 falls in cases (1) or (2)and the word y − s10 y s11 falls in (1), so we are done. For the case where u beginswith a we express the word as a product of y s0 y − s10 and y s11 y − su , both words fallin previous case. (cid:3) Our main theorem for G has some important corollaries. Corollary 2.6.
The finite-index subgroups of G are in bijection with the finite-indexsubgroups of Z . Every subgroup H of finite index is normal, and H ′ = G ′ .Proof. If H is not normal, then take the intersection K of all its conjugates, which is nownormal. Consider K ∩ G ′ . This is a finite-index normal subgroup of G ′ , and since thisgroup is simple and infinite, it has to be that K ∩ G ′ = G ′ and hence G ′ ⊂ K ⊂ H .Since every finite-index subgroup contains G ′ , now all of them correspond to those of theabelianization map, and hence they are all normal. The last assertion is true because H ′ ⊆ G ′ , and since H ′ is characteristic in H and hence normal in G , we have that G ′ ⊆ H ′ . (cid:3) JOS´E BURILLO, YASH LODHA, AND LAWRENCE REEVES
For our next corollary we will need the following fact.
Proposition 2.7.
The center of G is trivial.Proof. Let g ∈ G be in the center of G . In particular, g commutes with integertranslations.Let I be an interval on which the action of g is not affine. Now the action of any piecewiseprojective homeomorphism near infinity is affine. We conjugate g by x + n for n ∈ N toobtain a map g ′ which is not affine on the interval n + I . By our hypothesis g = g ′ , so itfollows that g is in fact piecewise affine.Moreover, by our hypothesis it follows that the set of breakpoints of our piecewise affine g is invariant under translation, hence empty. So g is in fact an affine map. The onlyaffine maps that commute with integer translations are themselves translations, so g is ofthe form x + t for t ∈ R . Now our lemma follows from the fact that b, c do not commutewith x + t for t = 0. (cid:3) We remark that the above argument is quite general. Given any group of piecewiseprojective homeomorphisms that contains both a translation and a non translation, thenthe center of the group is trivial.So we have now the following.
Corollary 2.8.
Every proper quotient of G is abelian.Proof. Let p : G −→ Q be a proper quotient map, and let K = ker p . Since the quotientis not G , there exists x ∈ K with x = 1. Since the center is trivial, then there exists y ∈ G such that [ x, y ] = 1. But then [ x, y ] ∈ K ∩ G ′ , which is a normal subgroup of G ′ ,so it follows that G ′ ⊂ K and Q is abelian. (cid:3) Commutators for G In this section we consider the commutator subgroups G ′ and G ′′ . Recall that G isgenerated by the set { x, x , y , y , y } , that for all n ∈ N , y n is a conjugate of y and y n is a conjugate of y and that for all non-constant s , y s is a conjugate of y .From the relations of the form y s = x s y s0 y − s10 y s11 we see that, unlike the situation for G ,the elements x and x lie in the kernel of the abelianization map. In more detail, we havethe following relation in G : y = x y y − y which, combined with y = x − y x y = x − y x shows that x is in the kernel of the abelianization. Similarly x is in the kernel of theabelianization because of the relation: y = x y y − y , and observing that, in an analogous way as above, y is a conjugate of y by an elementof F , and similarly, y and y are also conjugate by an element of F . Hence when we OMMUTATORS IN GROUPS OF PIECEWISE PROJECTIVE HOMEOMORPHISMS. 9 abelianize this relation and simplify, we obtain just x = 1. That x is also in the kernelthen follows from the relation x = x x − x − which is a consequence of relation (1). Since x and x together generate F , the whole of F lies in G ′ .We obtain the following. Lemma 3.1.
The map given by: x (0 , , x (0 , , y (1 , , y (0 , , y (0 , , . extends to a surjective homomorphism G → Z , and its kernel is exactly the commutatorsubgroup G ′ .Proof. As in the proof of Lemma 2.1 we consider the infinite presentation and showthat the given map extends to a homomorphism by first noting the extension to theinfinite generating set. The infinite generating set for G we consider is S = X ∪ Y where X = { x s : s ∈ < N } and Y = { y s : s ∈ < N } . The set of relations is given by (1) to (5)above. That this gives a presentation of G is Thereom 3.3 in [6].Consider the map from S → Z given by x s (0 , , y ( − , , y s (0 , ,
0) if s = m for some m ≥ , ,
1) if s = m for some m ≥ , ,
0) if s is non-constantThat this map extends to a homomorphism π : G → Z can be readily seen by consideringthe effect on each of the relations (1) to (5).Since { y , y , y } projects to a generating set for the abelianization G/G ′ and the image { π ( y ) , π ( y ) , π ( y ) } is a basis for Z we conclude that π induces an isomorphism from G/G ′ to Z . (cid:3) Given a word in standard form hy a s . . . y a n s n , define the left y -exponent sum to be theinteger given by summing the elements of { a i : s i = n , n ≥ } . Similarly, define the right y -exponent sum and central y -exponent sum as the sums of the sets { a i : s i = n , n ≥ } and { a i : s i is non-constant } respectively. The above discussion established the following. Proposition 3.2.
The commutator subgroup G ′ consists precisely of those elements of G that have a standard form expression with left y -exponent sum, right y -exponent sum andcentral y -exponent sum all equal to zero. (cid:3) Note that elements of G ′ need not be compactly supported (e.g., x ). An element of G ′ iscompactly supported precisely when h is compactly supported and all s i are non-constant.Elements of G ′′ have compact support. In fact, we have the following. Proposition 3.3. G ′′ = G ′ Proof.
One inclusion is clear since G ′ = G ′′ ⊆ G ′′ . For the reverse inclusion recall that theaction of any piecewise projective homeomorphism is affine near infinity. The elements of G ′′ have compact support and therefore we claim they have a standard form that doesnot contain any elements of the form y s with s constant.It is shown in [5] (Section 5) that any element g ∈ G can be represented as a normal form g = f y t s ...y t n s n with the property that it does not admit potential cancellations or potentialcontractions. (These notions are defined in [5], we do not recall them here since we onlyneed a corollary of these. Also, this is proved for the group G but the same proof, lineby line, applies to G .)We use from [5] the notion of calculation of a standard form on binary sequences. This isdefined for infinite sequences in Definition 3 .
13 in [5] and the definition for finite sequencesoccurs in the paragraph after Lemma 3 .
14 in [5].Let u be a finite binary sequence that contains as a prefix s i for some 1 ≤ i ≤ n .Moreover, assume that for any sequence s j in the set { s , ..., s n } either s j ⊂ u or u s j .This holds for instance whenever u is longer than all sequences in { s , ..., s n } .Let Λ be the associated calculation of f y t s ...y t n s n on u . Since f y t s ...y t n s n does not containpotential cancellations, Λ does not contain potential cancellations. So by Lemma 3 .
15 in[5], there is a finite binary sequence u such that one can perform a sequence of moves onthe calculation Λ u to obtain a string of the form u y n for some u ∈ < N and n ∈ N \ { } .Here n is the number of occurrences of y ± in the calculation Λ u , which is positive since u contains s i as a prefix.The calculation of f y t s ...y t n s n on u u n n ... equalsΛ u n n ... By our hypothesis, upon performing moves this simplifies to u y n n n ... and finally to u n n ... Hence the action of f y t s ...y t n s n on the sequence u u n n ... does not preserve tail equiv-alence.In particular, the element g does not preserve tail equivalence on a dense subset of Supp ( y t i s i ) for any 1 ≤ i ≤ n . If g ∈ G ′′ , it is compactly supported, and hence mustpreserve tail equivalence outside a compact interval. This means that the support of each y t i s i is compact. In particular, f y t s ...y t n s n does not contain elements of the form y ± s with s constant.The total y -exponent sum of such a standard form is zero since this is true for G ′ . It thenfollows from Proposition 2.2 that G ′′ ⊆ G ′ . (cid:3) In particular G ′′ is simple. Note that G ′′ is strictly smaller than G ′ since elements of G ′′ have compact support.Recall that x and x (hence any element of F ) are in G ′ . OMMUTATORS IN GROUPS OF PIECEWISE PROJECTIVE HOMEOMORPHISMS. 11
Proposition 3.4.
The group G ′ /G ′′ is generated by the cosets of x and x . There is anisomorphism G ′ /G ′′ → Z given by the images of the generators: xG ′′ (1 , x G ′′ (0 , . Proof.
Any element of G ′ has y exponent sum equal to zero and y exponent sum equalto zero. The relations y n +1 = x n y x − n and y n +1 = x − n y x n then show that any elementof G ′ /G ′′ can be written as hy a s . . . y a n s n G ′′ with h ∈ F , each s i non-constant and the sum of the a i equal to zero. Hence, y a s . . . y a n s n ∈ G ′ by Proposition 2.2. Therefore, as G ′ = G ′′ , any element of G ′ /G ′′ can be written inthe form x m x n G ′′ with m, n ∈ Z .That there is a homomorphism sending xG ′′ to (1 ,
0) and x G ′′ to (0 ,
1) is clear fromthe above discussion. To see that this is surjective, note that there is a homomorphism G ′ → Z , given by the germs at infinity. This map must therefore be precisely G ′ /G ′′ . (cid:3) We have seen that the derived series for G is G ′′ ✁ G ′ ✁ G with G/G ′ ∼ = Z , G ′ /G ′′ ∼ = Z and G ′′ perfect. 4. Commutators for H ( A )In this section we consider the group H ( A ), where A is a subring of R . We observe abasic fact about P A . Lemma 4.1.
Let A be a subring of R . (1) If A has a unit c = ± , then ∞ ∈ P A . (2) If the only units in A are ± , then ∞ / ∈ P A .Proof. First we consider the case where A has a unit c = ±
1. Consider an affine map ofthe form t → c t . Then, this map fixes 0 and ∞ . The corresponding matrix in P SL ( A ) (cid:18) c c − (cid:19) is a hyperbolic matrix which fixes ∞ . So it follows that ∞ ∈ P A .Now we consider the case when the only units in A are ±
1. Any hyperbolic matrix thatfixes ∞ must be the the form (cid:18) u v u − (cid:19) so that u is a unit that does not equal ±
1. Since there are no such units in A , there areno such matrices in P SL ( A ), and so ∞ / ∈ P A . (cid:3) The fact that ∞ is such an important point and that it belongs to P A only in the casewhere there are nontrivial units is the reason why the proof of the theorem is split inthese two cases. So we consider first the case in which A has units other than ±
1. We define H c ( A ) as thegroup of compactly supported elements of H ( A ). Since the elements of H ( A ) are affinenear infinity, it follows that H ( A ) ′′ ⊆ H c ( A ), because two elements of the type y = ax + b have a commutator of the type y = x + k , and two of these commute. Due to the hightransitivity of the action of P SL ( Z ) on the real line, a variation of Corollary 2.4 appliesto the groups H c ( A ) and H ( A ) ′′ and therefore the groups H c ( A ) ′ and H ( A ) ′′′ are simple.From our hypothesis on A it is clear that H ( A ) ′ = H ( A ) ′′ , since all elements of H ( A ) ′′ arecompactly supported, whereas there are maps in H ′ ( A ) that are not compactly supported.For instance, consider a commutator of an integer translation with a map of the form t p t , where p = ± A . Moreover, since H ( A ) ′′ ⊆ H c ( A ) it followsthat H ( A ) ′′′ is a normal subgroup of H c ( A ) ′ and by the simplicity of these groups wededuce that H c ( A ) ′ = H ( A ) ′′′ . To establish simplicity of H ( A ) ′′ it suffices to show that H ( A ) ′′ = H c ( A ) ′ . Indeed it suffices to show that if g, h ∈ H ( A ) ′ , then [ g, h ] ∈ H c ( A ) ′ .Before we show this, we first build some generic elements of H ( A ) which will be used inthe proof. Definition 4.2.
Given any positive real number r ∈ A , there is an x ∈ (0 , such that x − x = x + r , because the graphs of t → t − t and t → t + r must intersect in (0 , . Wedefine a map: t · γ r = t if t ≤ t − t if ≤ t ≤ xt + r if x ≤ t Now the matrices associated to the maps t → t + r, t − t are (cid:18) r (cid:19) , (cid:18) − (cid:19) respectively. It follows that x is fixed by (cid:18) r (cid:19) · (cid:18) − (cid:19) − which is a hyperbolic matrix. Therefore x ∈ P A , and hence γ r ∈ H ( A ) .Now given any n ∈ Z , r ∈ A, r > , we define the map: t · γ n,r = t if t ≤ n t − n − ( t − n ) + n if n ≤ t ≤ n + xt + r if n + x ≤ t This map is obtained by conjugating γ r by the map t → t + n . Clearly, γ n,r ∈ H ( A ) . Definition 4.3.
Let r ∈ A be a negative real number. The graph of the map t → t + r meets the map t t at some number x contained in the interval [ − r, − r + 1] . We definethe map OMMUTATORS IN GROUPS OF PIECEWISE PROJECTIVE HOMEOMORPHISMS. 13 r x Figure 1.
The maps γ r and λ r . t · λ r = t if t ≤ t t if ≤ t ≤ xt + r if x ≤ t Just as in the previous definition, one checks that x ∈ P A and so λ r ∈ H ( A ) . For n ∈ Z ,the map λ n,r is obtained by conjugating λ r by t → t + n . t · λ n,r = t if t ≤ n t − n t − n ) + n if n ≤ t ≤ n + xt + r if n + x ≤ t It follows that λ n,r ∈ H ( A ) . The idea of the construction of these elements is to provide “bump” or “step” functionsbetween the identity and t + r , always within H ( A ). Our maps in H ( A ) are translations t + r near infinity, so these functions will be used to provide steps to the identity whichwill transform them into compactly supported maps. See Figure 1.Stated in general, the generic elements constructed above allow us to observe the following. Lemma 4.4.
For each r ∈ A and p ∈ R there is a f ∈ H ( A ) such that: (1) f is supported on [ y, ∞ ) for some y < p . (2) The restriction of f to ( p, ∞ ) equals addition by r . In an analogous fashion, one establishes the following.
Lemma 4.5.
For each r ∈ A and p ∈ R there is a f ∈ H ( A ) such that: (1) f is supported on ( −∞ , z ] for some z > p . (2) The restriction of f to ( −∞ , p ) equals addition by r . ff,g g,f gluedglued y yx xx y x yg Figure 2.
An example of gluing.We now provide an elementary gluing construction that allows one to build piecewiseprojective maps, by gluing pieces of piecewise projective maps provided they agree on asuitable interval. If two maps agree in an interval, we can take a hybrid of the two whichconsists of one on one side, and another in the other side. See Figure 2.The proof of this lemma is straightforward.
Lemma 4.6. (Gluing) Let f, g be an ordered pair of elements in H ( A ) and let I = [ x, y ] be such that the restrictions of f, g on [ x, y ] agree. Then there is an element h ∈ H ( A ) such that: (1) The restriction of f, h on ( −∞ , x ] agree. (2) The restrictions of g, h on [ y, ∞ ) agree. (3) The restriction of h on [ x, y ] agrees with the restrictions of both f, g on [ x, y ] . Note that in the gluing construction the order of the pair f, g is essential in determininghow the elements are glued, the resulting maps are different of we glue f, g or if we glue
OMMUTATORS IN GROUPS OF PIECEWISE PROJECTIVE HOMEOMORPHISMS. 15 g, f . The interval [ x, y ] in Lemma 4.6 will be called the gluing interval . Now we are readyto prove the main lemma for the case where A has units other than ± Lemma 4.7.
Let g, h ∈ H ( A ) ′ . Then [ g, h ] ∈ H c ( A ) ′ .Proof. The main idea of the proof is to find elements h , h , k , k ∈ H ( A ) such that:(1) [ h , h ] , [ k , k ] ∈ H c ( A ) ′ .(2) [ h , h ][ g, h ][ k , k ] ∈ H c ( A ) ′ .This will finish the proof. The construction of these elements will be done in four steps.Step 1: The maps f, g ∈ H ( A ) ′ , they are translations near infinity. This allows usto choose a sufficiently large interval [ r, s ] for which the restriction of each element of { f, g, f − , g − } to ( −∞ , r ) and ( s, ∞ ) are translations. Outside of this interval, we willglue the step functions constructed above so our maps become compactly supported.Step 2: Applying Lemma 4.4, we find elements h , h such that:(1) h , h are supported on an interval [ x, ∞ ) for x < r .(2) There is a real x < x < r such that the restrictions of h , f on [ x , r ] agree.(3) There is a real x < x < r such that the restrictions of h , g on [ x , r ] agree.(4) Let j , j be the elements obtained by gluing h , f and h , g along [ x , r ] , [ x , r ]respectively. Then [ j , j ] = [ h , h ][ f, g ].The final condition above is satisfied if the gluing intervals are sufficiently large. Sincetranslations commute, as we apply the sequence of elements of the commutator j , j , j − , j − in that order, one by one, we notice that if the gluing interval is large enough, it containsa subinterval on which each element acts like a translation, and hence the net result isthe identity map. On the right side of this piece, the commutator acts like [ f, g ], and onthe left side it acts like [ h , h ]. See Figure 3.Step 3: In this step we will do the same procedure as in step 2, but now on the right handside of the maps. Applying Lemma 4.5, we find elements k , k such that:(1) k , k are supported on an interval ( −∞ , y ).(2) There is a real s < x < y such that the restrictions of k , f on [ s, x ] agree.(3) There is a real s < x < y such that the restrictions of k , g on [ s, x ] agree.(4) Let l , l be the elements obtained by gluing f, k and g, k along [ s, x ] , [ s, x ]respectively. Then [ l , l ] = [ f, g ][ k , k ].Step 4: We glue j , l along [ r, s ] to obtain f ′ , and we glue j , l along [ r, s ] to obtain g ′ .It follows that [ f ′ , g ′ ] = [ h , h ][ f, g ][ k , k ], and since f ′ and g ′ are constructed to be in H c ( A ), we conclude that [ f ′ , g ′ ] ∈ H c ( A ) ′ .This finishes the proof of the fact that [ h , h ][ f, g ][ k , k ] ∈ H c ( A ) ′ . To prove our lemmait suffices to show that [ h , h ] , [ k , k ] ∈ H c ( A ) ′ . We will show this for [ h , h ]. Theother case is completely analogous. We assume that h , h are supported on [0 , ∞ ) for f f,gh h h ,hj j ,jg j [[[ ]]] Figure 3.
Step 2 in the proof that [ f, g ] ∈ H c ( A ) ′ . The maps f and g donot have compact support, but they are glued to the maps h and h insuch a way that the resulting maps j and j are the identity near −∞ andtheir commutator [ j , j ] agrees with [ f, g ] except that [ h , h ] has appeared.The next step is to perform this procedure also near + ∞ , to produce maps f ′ , g ′ which have compact support, and whose commutator agrees with [ f, g ]except that [ h , h ] and [ k , k ] appear, one above, and one below [ f, g ]. Theproof ends when these two latter commutators are also shown to be in H c ( A ) ′ . OMMUTATORS IN GROUPS OF PIECEWISE PROJECTIVE HOMEOMORPHISMS. 17 simplicity. (If this is not the case, we can conjugate the elements h , h by a sufficientlylarge integer translation p , and then establish that [ h p , h p ] = [ h , h ] p ∈ H ′ c .)The map t · M = t t fixes 0 and maps (0 , ∞ ) to (0 , h = [ M − h M, M − h M ] = M − [ h , h ] M is clearly in H c ( A ) ′ . In fact, the closure of the support of h is contained in an interval[0 , t ] ⊂ [0 , m ∈ H ( A ) such that m agrees with M − on the supportof h . For a sufficiently large k ∈ N , ∃ x ∈ ( t,
1) such that x + k = x · M − = x − x . Itfollows that x is fixed by (cid:18) k (cid:19) · (cid:18) − (cid:19) − which is a hyperbolic matrix. So we define m as: t · m = t if t ≤ t − t if 0 ≤ t ≤ xt + k if x ≤ t The breakpoints of m are 0 , x, ∞ . This means that m ∈ H ( A ). We remark that this isthe part of the argument where the existence of units in A other than ± m − h m = [ h , h ]. Since H c ( A ) ′ is characteristic in H c ( A ), itis invariant under conjugation by elements of H ( A ). Since h ∈ H c ( A ) ′ it follows that[ h , h ] ∈ H c ( A ) ′ as desired. (cid:3) We conclude the following.
Corollary 4.8. H ( A ) ′′ is simple. The proof of Proposition 2.7 applies to both H ( A ) and H ( A ) ′ , so the center of thesegroups is trivial. So we obtain the following. Proposition 4.9.
Every proper quotient of H ( A ) is metabelian.Proof. Let N be a normal subgroup of H ( A ). Let N = H ( A ) ′ ∩ N and N = H ( A ) ′′ ∩ N .Since H ( A ) ′′ is simple, either N is trivial or N = H ( A ) ′′ . In the former case it followsthat N is in the center of H ( A ) ′ , which means that N is trivial. This means that N isin the center of H ( A ) which means that N is trivial. So indeed it follows that either N is trivial or N contains H ( A ) ′′ . (cid:3) This finishes the proof of Theorem 1 for the case where A contains units other than ± A does not contain units other than ± H ( A ) to be translations near infinity,since the only affine maps in P SL ( A ) are translations. It follows that elements of H ( A ) ′ are compactly supported. By applying Higman’s theorem to H ( A ) ′ and H c ( A ) we obtain that the groups H ( A ) ′′ and H c ( A ) ′ are simple, and as a consequence H c ( A ) ′ = H ( A ) ′′ . Toprove our claim it suffices to show that H c ( A ) ′ = H ( A ) ′ .Recall that because of Lemma 4.1, we now have that ∞ is not in P A . It follows that forany f ∈ H ( A ), the translations on both germs at ∞ are the same. Moreover, Q ∩ P A = ∅ ,since the orbit of ∞ under the action of P SL ( Z ) is Q ∪ {∞} .We shall need an analogue of Lemmas 4.4 and 4.5 in which the resulting functions satisfythat their breakpoints that lie in R are elements of P A .Consider the hyperbolic matrix (cid:18) − − (cid:19) and the corresponding fractional linear transformation t → − t +1 t − . The fixed points p < q of this map are elements of P Z ⊂ P A and the map tends to infinity as t →
1. Moreover, p < < q < r ∈ A, r >
0, the curves of t → t + r and t → − t +1 t − meet in a real number s ∈ ( q, s ∈ P A since s is a fixed point of the hyperbolic matrix (cid:18) − r (cid:19) (cid:18) − − (cid:19) = (cid:18) − − r r − (cid:19) We define the map: t · ζ r = t if t ≤ q − t +1 t − if q ≤ t ≤ st + r if s ≤ t Upon considering inverses and conjugation of this function with integer translations, aswell as the analogous functions with non-trivial germs at −∞ , we obtain the followinganalogues of Lemmas 4.4 and 4.5. Lemma 4.10.
For each r ∈ A and p ∈ R there is a map f such that: (1) f is a piecewise P SL ( A ) homeomorphism of R . (2) f is supported on [ y, ∞ ) for some y < p . (3) The restriction of f to ( p, ∞ ) equals addition by r . (4) The breakpoints of f , besides ∞ , all lie in P A . Lemma 4.11.
For each r ∈ A and p ∈ R there is a map f such that: (1) f is a piecewise P SL ( A ) homeomorphism of R . (2) f is supported on ( −∞ , z ] for some z > p . (3) The restriction of f to ( −∞ , p ) equals addition by r . (4) The breakpoints of f , besides ∞ , all lie in P A . Now we are ready to prove the main Lemma.
Lemma 4.12.
Let A be a subring of R whose only units are ± . Let f, g ∈ H ( A ) . Then [ f, g ] ∈ H c ( A ) ′ . OMMUTATORS IN GROUPS OF PIECEWISE PROJECTIVE HOMEOMORPHISMS. 19
Proof.
The proof will follow analogous lines to the proof of Lemma 4.7. We assume forthe rest of the proof that the translation near infinity for f is t → t + c and for g is t → t + c . The main idea of the proof is to find piecewise P SL ( A ) elements h , h , k , k which are step functions which will transform f and g into compactly supported versionsof themselves by stepping down the translations t + c i to the identity. Precisely, theelements h , h , k , k will satisfy(1) [ h , h ][ k , k ] ∈ H c ( A ) ′ (2) [ f, g ][ h , h ][ k , k ] ∈ H c ( A ) ′ and this will finish the proof. We shall follow a five step procedure to construct therequired elements.Step 1: Choose a sufficiently large interval [ r, s ] so that the restriction of each element of { f, g, f − , g − } to R \ [ r, s ] is a translation. That is, all the nontranslation action for f and g happens well inside [ r, s ].Step 2: The idea now is to construct the elements h , h to provide the stepping down tothe identity right outside [ r, s ], or more precisely, to the left of r . Applying Lemma 4.10,we find elements h , h such that:(1) h , h are supported on an interval [ x, ∞ ).(2) There is a real number x , with x < x < r , such that the restrictions of h , f on[ x , r ] agree.(3) There is a real number x , with x < x < r , such that the restrictions of h , g on[ x , r ] agree.(4) Let j , j be the elements obtained by gluing h , f and h , g along [ x , r ] , [ x , r ]respectively. Then [ j , j ] = [ h , h ][ f, g ].The last condition above is satisfied if the gluing intervals are sufficiently large, just as inthe analogous case in the proof of Lemma 4.7. Observe that the elements h , h are notin H ( A ), because they have a breakpoint at ∞ . The germ at −∞ is the identity whereasthe germ at + ∞ is a translation by c or c .Step 3: Analogously to the previous step, applying Lemma 4.11, we find elements k , k which bring the germ at + ∞ down to the identity. That is:(1) k , k are supported on an interval ( −∞ , y ).(2) There is a real number y , with s < y < y , such that the restrictions of k , f on[ s, y ] agree.(3) There is a real number y , with s < y < y , such that the restrictions of k , g on[ s, y ] agree.(4) Let l , l be the elements obtained by gluing f, k and g, k along [ s, y ] , [ s, y ]respectively, then [ l , l ] = [ f, g ][ k , k ].At this point we remark that by construction, the restriction of the maps h , k on [ r, s ]equals translation by c , and the restriction of the maps h , k on [ r, s ] equals translationby c . All four maps will work as step functions, having a step outside [ r, s ] to have the appropriate identity germ at ∞ . As pointed out in step 2, these maps do not belong to H ( A ).Step 4: We glue j , l along [ r, s ] to obtain s , and glue j , l along [ r, s ] to obtain s .Step 5: We glue h , k along [ r, s ] to obtain t , and glue h , k along [ r, s ] to obtain t .We would like to emphasise that all four maps s , s , t , t do belong to H ( A ), and in fact,they belong to H c ( A ). This is because after gluing, for all four maps, both germs at ∞ are equal to the identity.Finally, by construction, from the fact that the steps have been constructed far away fromthe nontranslation part of f and g , it follows that [ s , s ] = [ h , h ][ f, g ][ k , k ] and thatthis element belongs to H c ( A ) ′ because s , s ∈ H c ( A ). Also by construction, same asbefore, we see that the supports of [ h , h ] , [ f, g ] are disjoint, so we conclude that[ s , s ] = [ h , h ][ f, g ][ k , k ] = [ f, g ][ h , h ][ k , k ] ∈ H c ( A ) . Since [ t , t ] = [ h , h ][ k , k ] ∈ H c ( A ) ′ because the maps t , t are in H c ( A ), this finishes the proof. (cid:3) It follows from similar arguments, as in the case of A with units other than ±
1, that thecenter of H ( A ) is trivial, and every proper quotient of H ( A ) is abelian. This concludesthe proof of Theorem 1.The groups H ( A ) are not finitely presented, and a presentation for these groups wouldinvolve infinitely many maps similar to the map y and their interactions. Writing downthese generators and relations would be quite complicated. Hence this makes it difficult tocompute the quotients of H ( A ) by the commutators H ( A ) ′ or H ( A ) ′′ to get its abelianiza-tion and metabelianization. We believe that, unlike the cases for G and G , it is difficultto find easy expressions for these quotients. References [1] J. W. Cannon, W. J. Floyd, and W. R. Parry. Introductory notes on Richard Thompson’s groups.
Enseign. Math. (2) , 42(3-4):215–256, 1996.[2] Mahlon M. Day. Means for the bounded functions and ergodicity of the bounded representations ofsemi-groups.
Trans. Amer. Math. Soc. , 69:276–291, 1950.[3] Patrick Dehornoy. Geometric presentations for Thompson’s groups.
J. Pure Appl. Algebra , 203(1-3):1–44, 2005.[4] Graham Higman. On infinite simple permutation groups.
Publ. Math. Debrecen , 3:221–226 (1955),1954.[5] Yash Lodha. A type F ∞ group of piecewise projective homeomorphisms. Preprint, arXiv:1408.3127.[6] Yash Lodha and Justin Tatch Moore. A finitely presented group of piecewise projective homeomor-phisms. Preprint, arXiv:1308.4250.[7] Nicolas Monod. Groups of piecewise projective homeomorphisms. Proc. Natl. Acad. Sci. USA ,110(12):4524–4527, 2013.[8] J. von Neumann. Zur allgemeinen Theorie des Masses.
Fund. Math. , 13(1):73–116, 1929.
OMMUTATORS IN GROUPS OF PIECEWISE PROJECTIVE HOMEOMORPHISMS. 21
Departament de Matem`atiques, UPC, C/Jordi Girona 1-3, 08034 Barcelona, Spain
E-mail address : [email protected] Department of Mathematics, EPFL, Lausanne, Switzerland.
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