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The Banach–Tarski Paradox and Amenability Lecture 25: Thompson's Group F

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The Banach–Tarski Paradox and Amenability Lecture 25: Thompson's Group F The Banach{Tarski Paradox and Amenability Lecture 25: Thompson's Group F 30 October 2012 Thompson's groups F , T and V In 1965 Richard Thompson (a logician) defined 3 groups F ≤ T ≤ V Each is finitely presented and infinite, and Thompson used them to construct groups with unsolvable word problem. From the point of view of amenability F is the most important. The group F is not elementary amenable and does not have a subgroup which is free of rank ≥ 2. We do not know if F is amenable. The groups T and V are finitely presented infinite simple groups. The commutator subgroup [F ; F ] of F is finitely presented and ∼ 2 simple and F =[F ; F ] = Z . Definition of F We can define F to be the set of piecewise linear homeomorphisms f : [0; 1] ! [0; 1] which are differentiable except at finitely many a dyadic rationals (i.e. rationals 2n 2 (0; 1)), and such that on intervals of differentiability the derivatives are powers of 2. Examples See sketches of A and B. Lemma F is a subgroup of the group of all homeomorphisms [0; 1] ! [0; 1]. Proof. Let f 2 F have breakpoints 0 = x0 < x1 < ··· < xn = 1. Then f (x) = a1x on [0; x1] where a1 is a power of 2, so f (x1) = a1x1 is a dyadic rational. Hence f (x) = a2x + b2 on [x1; x2] where a2 is a power of 2 and b2 is dyadic rational. By induction f (x) = ai x + bi on [xi−1; xi ] with ai a power of 2 and bi a dyadic rational. It follows that f −1 2 F and that F is closed under composition. First properties of F Lemma F is nonabelian and torsion free. Proof. A does not commute with B. Consider a nontrivial f 2 F . Then there is a smallest dyadic rational x1 ≥ 0 such that f (x1) = x1 and on some interval [x1; x2], f is differentiable with slope not equal to 1. Consider powers of f on [x1; x2]. Lemma F has a subgroup isomorphic to F × F. Proof. Draw picture. Lemma F has a free abelian subgroup of rank 2, and a free abelian subgroup of infinite rank. Finite and infinite presentations for F Consider the finite presentation −1 −1 F = hx0; x1 j x1 x2x1 = x3; x1 x3x1 = x4i −1 −2 2 3 −3 where x2 = x0 x1x0, x3 = x0 x1x0 and x4 = x0 x1x0 and the infinite presentation −1 P = hxn; n ≥ 0 j xi xj xi = xj+1 if i < ji To see these presentations are equivalent, note that for all n > 1 −(n−1) (n−1) xn = x0 x1x0 so F is equivalent to F 0 defined by −1 −1 −(n−1) (n−1) hxn; n ≥ 0 j x1 x2x1 = x3; x1 x3x1 = x4; x0 x1x0 = xn8n > 1i Use induction to show that these relations imply −1 xi xj xi = xj+1 if i < j. Finite presentation for F Define a function φ from the group −1 −1 G = hx0; x1 j x1 x2x1 = x3; x1 x3x1 = x4i to F by φ(x0) = A, φ(x1) = B. Then φ is a group isomorphism. The proof uses a normal form for the elements of G and of F . Thus Thompson's group F has a finite presentation. In particular F is generated by A and B. Commutator subgroup of F Lemma 2 Let H be a 2-generator group and suppose ' : H ! Z is surjective. Then ker(') = [H; H]. We say an element f 2 F is trivial in neighbourhoods of 0 and 1 if f has slope 20 = 1 to the right of 0 and to the left of 1. Theorem [F ; F ] consists of all elements of F that are trivial in ∼ 2 neighbourhoods of 0 and 1. Also, F =[F ; F ] = Z . Proof. 2 a Define ' : F ! Z by '(f ) = (a; b) where f has slope 2 to the right of 0 and slope 2b to the left of 1. Then ' is a group homomorphism. The map ' is surjective since '(A) = (−1; 1) and '(B) = (0; 1). So by the Lemma, ker(') = [F ; F ]. Subgroups of F Theorem Every nonabelian subgroup of F contains a free abelian group of infinite rank. Corollary F does not contain a nonabelian free group of rank 2. Let H be a subgroup of F generated by elements f and g such that fg 6= gf . Let I1;:::; In be the closed intervals in [0; 1] with nonempty interiors s.t. 8 1 ≤ k ≤ n if x is an endpoint of Ik then f (x) = g(x) = x, and if x is an interior point of Ik then either f (x) 6= x or g(x) 6= x. We claim that for all k, the endpoints of Ik are cluster points of the H-orbit of every interior point of Ik . Let x be in the interior of Ik and let y be the greatest lower bound of the H-orbit of x. If y is not the left endpoint of Ik then wlog f (y) 6= y, so f (y) < y or f −1(y) < y. But then there is a neighbourhood of y such that the image of this neighbourhood under either f or f −1 is less than y, so the H-orbit of x contains points less than y, a contradiction. Thus y is the left endpoint of Ik . Similarly for least upper bounds. Subgroups of F Define the support of h 2 H to be the subset of [0; 1] on which h is not x 7! x. Similarly for the support of h 2 H in Ik . Let h1 = [f ; g] 2 [H; H]. Then h1 is trivial in neighbourhoods of the endpoints of I1. Then since the left endpoint of I1 is a cluster point of the H-orbit of any interior point of I1, there is an h 2 H −1 such that the support in I1 of h2 = hh1h is disjoint from the support of h in I1. So we can find an infinite sequence of functions h1; h2; h3;::: in H whose supports in I1 are mutually disjoint. So [hi ; hj ] is trivial on I1 for all i; j. If [hi ; hj ] 6= 1 then there is a k > 1 on which [hi ; hj ] is nontrivial. Repeat the argument for [hi ; hj ] on Ik . After finitely many repetitions we get an infinite sequence of functions in H which generate a free abelian subgroup. Elementary amenability Theorem Thompson's group F is not elementary amenable. We use results of Chou. Define EG0 to be the collection of all finite and abelian groups. For each ordinal α i.e. 0; 1; 2; : : : ; !; ! + 1;! + 2;:::;! + ! = ! · 2;! · 2 + 1;:::;!2;:::; 3 ! ! ;:::;! , . , if α is a successor define EGα to be the collection of amenable groups obtained from groups in EGα−1 by the following processes: I forming group extensions I taking directed unions and if α is a limit define [ EGα = EGβ β<α By a result of Chou, it suffices to show that F is not in EGα for any ordinal α. Thompson's group F is not elementary amenable We will use transfinite induction. Let P(α) be the statement that F is not in EGα. To prove P(α) for all ordinals α, it suffices to prove: I P(0) and I P(α) follows from [P(β) for all β < α] Now F 62 EG0 since F is infinite and nonabelian. Assume that α > 0 and that F 62 EGβ for each ordinal β < α. If α is a limit ordinal then by definition of EGα we are done. So suppose α is a successor and F 62 EGα−1. We want to show F cannot be constructed from groups in EGα−1 as a group extension or directed union. Thompson's group F is not elementary amenable The group F is not a directed union of groups in EGα−1, because F is finitely generated. To show F is not obtained as a group extension from groups in EGα−1 we will use the following theorem which is proved using normal forms: Theorem Every proper quotient of F is abelian. Suppose there is a short exact sequence 1 −! N −! F −! Q ! 1 with N; Q 2 EGα−1. Then N is nontrivial since F 62 EGα−1. So Q is a proper quotient. Hence by the theorem above, Q is abelian. Thus [F ; F ] ⊂ N. Since [F ; F ] consists of elements of F which are trivial in neighbourhoods of 0 and 1, we can find a copy of F in N. By a result of Chou, subgroups of groups in EGα−1 are in EGα−1, so F is in EGα−1, a contradiction..
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