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 ∈ (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 ∈ 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 ∈ 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 ∈ 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 | 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 | 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 | x1 x2x1 = x3, x1 x3x1 = x4, x0 x1x0 = xn∀n > 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 | 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 ϕ : H → Z is surjective. Then ker(ϕ) = [H, H]. We say an element f ∈ 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 ϕ : F → Z by ϕ(f ) = (a, b) where f has slope 2 to the right of 0 and slope 2b to the left of 1. Then ϕ is a group homomorphism. The map ϕ is surjective since ϕ(A) = (−1, 1) and ϕ(B) = (0, 1). So by the Lemma, ker(ϕ) = [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. ∀ 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 ∈ H to be the subset of [0, 1] on which h is not x 7→ x. Similarly for the support of h ∈ H in Ik . Let h1 = [f , g] ∈ [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 ∈ 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, . . . , ω, ω + 1, ω + 2, . . . , ω + ω = ω · 2, ω · 2 + 1, . . . , ω2,..., 3 ω ω , . . . , ω , . . . , 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 6∈ EG0 since F is infinite and nonabelian. Assume that α > 0 and that F 6∈ 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 6∈ 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 ∈ EGα−1. Then N is nontrivial since F 6∈ 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.