Laws with constants in Thompson's group 𝐹

Definition 1.

Let G be a group, H⊆G, and let 𝔽 be a free group freely generated by y1,…,yz. Let w⁢(y1,…,yz,g1,…,gv) be a reduced word in the group G∗𝔽 with variables y1,…,yz and constants g1,…,gv∈G∖{1}, where at least one variable and one constant appears. We call w⁢(y1,…,yz,g1,…,gv)≡1 a law with constants (LC) satisfied in H if w⁢(h1,…,hz,g1,…,gv)=1 for any h1,…,hz∈H.

Definition 2.

The LCw≡u is a semigroup law with constants (SLC) if both w and u do not contain any variable with negative power. Furthermore, if both w and u do not have any constant, then w≡u is called a semigroup law.

Lemma 1.

If a group G satisfies a law or LC, then it satisfies the LC w≡u with only one variable, where u starts with a variable and w starts with a constant or is trivial.

Proof.

Let g∈G∖{1}. If G satisfies a law or LC with variables y1,…,yz, then, after substituting any yi by yi⁢g⁢yi and possibly cancelling by some powers of y, we get a (nontrivial) LC with one variable y.

Let h1,h2∈G, h1≠h2. Suppose now that G satisfies the LCh1⁢w≡h2⁢u with one variable y, where w,u do not start with constants. If h2-1⁢h1⁢w≡u is reduced, then it is an LC satisfying the desired conditions. Otherwise, we can cancel h2-1⁢h1⁢w≡u only if w=1 and u ends with y⁢h2-1⁢h1. Then, after cancelling, we get 1≡u′, where u′ starts and ends with y. If u′ includes any nontrivial constant, then 1≡u′ is an LC. Otherwise, we get a group of finite exponent which satisfies the SLCyn⁢g≡g⁢yn. ∎

Lemma 2.

If a group G satisfies a semigroup law or SLC, then it satisfies the SLC w≡u with only one variable, where u starts with a variable and w starts with a constant or is trivial.

Proposition 1.

If a group G satisfies a semigroup law, then any LC satisfied in G is equivalent to an SLC or vanishes.

Proof.

If G is of finite exponent n, then, for any variable yi, putting yi-1:-yin-1 in any LC, we get an equivalent SLC. Now let G be not of finite exponent. If G satisfies any semigroup law, then we can assume that it is of the form y1⁢w≡y2⁢u for some words w,u. Its equivalent form y2-1⁢y1≡u⁢w-1 allows us to “move” all variables with positive powers to the left in any LC. Hence we see that the LCw¯⁢u¯-1≡1 equivalent to the SLCw¯≡u¯ (if the words w¯,u¯ are different; otherwise, we get 1≡1). ∎

Proposition 2.

The following statements hold.

1. A group with nontrivial center satisfies an SLC.

2. A direct product of groups satisfying an LC satisfies an LC.

3. The free product G∗H of two nontrivial groups, where H is torsion-free, does not satisfy any LC.

Proof.

(1) If c∈Z⁢(G)∖{1}, then G satisfies c⁢y≡y⁢c.

(2) Let G satisfy the LCw and H satisfy the LCu. We can assume that w and u do not have any common variable. We identify G×{1}, {1}×H with their isomorphic copies G,H. Then [w,u] is a reduced word, and it is easy to see that it is a law satisfied in G×H.

(3) Let α1,…,αk be nonzero integers and a1,…,ak∈G∗H. Assume that

is an LC and h∈H is non-torsion, and let g∈G∖{1}. Then we can choose α∈ℤ such that, in a1⁢(hα)α1⁢a2⁢(hα)α2⁢⋯⁢ak⁢(hα)αk, each section from H does not disappear. ∎

Corollary 1.

A nonabelian free group does not satisfy any LC, but their direct product with a nontrivial abelian group does.

Proposition 3.

If g∈F has the defragmentation g=g1⁢…⁢gn, then some roots of the elements g1,…,gn are the generators of cyclic components of the decomposition of the centralizer above. The components of this decomposition, which are isomorphic to F, are just the groups of the form F[a,b], where [a,b] is one of the subintervals [pi-1,pi]⊆[0,1], which are stabilized pointwise by g.

The center of F is trivial, and hence the center of any F[a,b] is trivial.

Proposition 4.

For any f,f′∈F, we have supp⁡(ff′)=(f′)-1⁢(supp⁡(f)).

Theorem 1.

Consider the standard action of Thompson’s group F on [0,1]. Suppose we are given two pairwise disjoint open dyadic subintervals

and assume that p1<p2. Fix any nontrivial h1∈FI¯1 and h2∈FI¯2, and denote

Then the word w:-[w-,w+] is a law with constants in F.

Proof.

It is easy to see that w cannot be reduced to a constant. We claim that, for any g∈F satisfying w-⁢(g)≠1, the word w+⁢(g) is equal to the identity.

It follows from Proposition 3 that if the supports of any two given elements from F do not intersect, then the commutator of these two elements is trivial. Hence we see that w-⁢(g)=[h1g,h2]≠1 implies supp⁡(h1g)∩supp⁡(h2)≠∅. By Proposition 4, this implies

It follows that g⁢(p2)<q1. Since any g∈F is an increasing function, we obtain the sequence of implications

This shows that supp⁡(h1)∩g-1⁢(supp⁡(h2))=∅. Hence, using Proposition 4 and Proposition 3 similarly as above, we see that w+⁢(g)=1. This proves that, for any g∈F, either w-⁢(g)=1 or w+⁢(g)=1. ∎

Theorem 2.

Consider the standard action of Thompson’s group F on [0,1]. Suppose we are given four pairwise disjoint closed dyadic subintervals

and assume p1<p2<p3<p4. Then, for any nontrivial h1∈FI1, h2∈FI2, h3∈FI3 and h4∈FI4, let

Then the word w obtained from [w14,w23] by reduction in Z∗F (we treat the variable y as a generator of Z) is a law with constants in F.

Proof.

It is easy to see that w cannot be reduced to a constant. We claim that, for any g∈F satisfying w14⁢(g)≠𝑖𝑑, the word w23⁢(g) is equal to the identity.

We repeat the argument of the proof of Theorem 1. It follows from Proposition 3 that if the supports of the given two elements from F do not intersect, then the commutator of these two elements is trivial. Hence w14⁢(g)=[h1g,h4]≠𝑖𝑑 implies supp⁡(h1g)∩supp⁡(h4)≠∅. By Proposition 4, this implies

It follows that g⁢(p4)<q1. Since any g∈F is an increasing function, we obtain the sequence of implications

This shows that supp⁡(h3)∩g-1⁢(supp⁡(h2))=∅. Hence, using Proposition 4 and Proposition 3, we see similarly as above that w23⁢(g)=𝑖𝑑. This proves that, for any g∈F, either w14⁢(g)=𝑖𝑑 or w23⁢(g)=𝑖𝑑. ∎

Lemma 3.

If F satisfies (2.1), then

Proof.

We substitute y in (2.1) by xr with sufficiently large r.

The number of xn’s in w⁢(x0,…,xl,xr) with indexes greater than l is equal to ∑i=1γpi and, in u⁢(x0,…,xl,xr), the corresponding number is equal to ∑j=0δqj. If r is large enough, then these numbers are preserved in their normal forms obtained by relations from (1.1). The rest follows from the uniqueness of normal forms. ∎

Corollary 2.

If LC (2.1) is satisfied in F, then the sums of powers of any variable in w and u are equal.

Proof.

We just substitute the chosen variable by xr and other variables by the identity element. The rest of the proof runs as in Lemma 3. ∎

Theorem 3.

Thompson’s group F does not satisfy any SLC.

Proof.

Let us assume that F satisfies an SLC. From Lemmas 2 and 3, it follows that F satisfies an SLC of form (2.1), where ∑i=1γpi=∑j=0δqj and all pi,qj are nonnegative.

Suppose that w1⁢(x0,…,xl)=x0α. Then we substitute y by x1⁢y in (2.1), multiply both sides of (2.1) by x1-1 and rewrite new constants between variables y to normal forms. Then the first obtained constant is x1-1⁢w1⁢(x0,…,xl)⁢x1; its normal form is

and hence we can assume that w1⁢(x0,…,xl) in (2.1) contains xl for some l>0.

Let us choose a sufficiently large integer r, substitute y by x0-r in (2.1) and “move” all the inserted x0’s to the right in both sides of (2.1) using relations from (1.1). We obtain the relation

If we leave only w1⁢(x0,x1,…,xl) on the left side of the relation, we obtain

During the normalization of any element of F, the number of x0’s never increases, and xn’s with nonzero indexes may change maximally by the length of this element. So if r is large enough, then, in the normal form of u¯⁢(x0,…,xl+r⁢∑j=1γqj), the smallest positive index of xn will be greater than l. It means that relation (2.2) cannot occur because w1⁢(x0,x1,…,xl) contains xl – a contradiction. ∎