Mixed identities, hereditarily separated actions and oscillation

Let V ⊆ X . We say that the word w ⁢ ( y ̄ ) ∈ F t ∗ G is explicitly oscillating in 𝑉 if w ⁢ ( y ̄ ) is non-trivial and V ∩ O w ≠ ∅ . When V = X , we just say that w ⁢ ( y ̄ ) is explicitly oscillating.

Consider the group of finitary permutations G = S fin ⁢ ( N ) . Suppose that w ⁢ ( y ̄ ) ∉ F t and w ⁢ ( y ̄ ) is in the form (2.1) with non-trivial v i . Then O w is finite.

Nevertheless, we will show in this paper how our results can be adapted to the discrete case; see for example Theorem 2.15.

Consider Thompson’s group 𝐹 with its standard action on [ 0 , 1 ] . Let w 1 ⁢ ( y ) = y ⁢ x 1 ⁢ y − 1 ⁢ x 2 ⁢ y 2 ⁢ x 1 − 1 , where 𝑦 is a variable. In the notation from the definitions, we have v 1 = x 1 − 1 , v 2 = x 2 , v 3 = x 1 and hence

Thus we see that w 1 ⁢ ( y ) is explicitly oscillating (in ( 5 8 , 1 ) ).

We introduce the following notation. For an element v ∈ G , let v 1 = v and v 0 = 1 . For every w ⁢ ( y ̄ ) given in the form w ⁢ ( y ̄ ) = u n ⁢ v n ⁢ u n − 1 ⁢ v n − 1 ⁢ … ⁢ u 1 ⁢ v 1 as in (2.1) and for every set A ⊆ X , let

and let

When w ⁢ ( y ̄ ) ∈ F t , let V w > 0 ⁢ ( A ) = { A } .

Let 𝐺 be a permutation group acting on an infinite set 𝑋.

• We say that 𝐺 separates 𝑋 if, for every finite subset Y ⊂ X , the pointwise stabilizer stab G ⁢ ( Y ) does not stabilize any point outside 𝑌.

• Assume that 𝒳 is a Hausdorff topological space, 𝐺 consists of homeomorphisms of 𝒳 and the set of fixed points Fix ⁢ ( G ) is finite. We say that 𝐺 hereditarily separates 𝒳 if, for every open and infinite subset Z ⊆ X and for every finite subset Y ⊂ Z , the subgroup stab G ⁢ ( ( X ∖ Z ) ∪ Y ) does not stabilize any point from Z ∖ ( Y ∪ Fix ⁢ ( G ) ) .

Separating actions were introduced in [1]. We mention the following observation from that paper. For every separating action of a group 𝐺 on 𝑋 and every finite Y ⊂ X , the orbits of the action of the pointwise stabilizer of 𝑌 in X ∖ Y are infinite. As a corollary, we see that, for a hereditarily separating action of 𝐺 on 𝒳 and an open and infinite subset Z ⊆ X , the action of the stabilizer stab G ⁢ ( X ∖ Z ) on Z ∖ Fix ⁢ ( G ) has only infinite orbits.

Consider ℕ as a discrete space. The symmetric group Sym ⁢ ( N ) , the group of finitary permutations S fin ⁢ ( N ) and the alternating group Alt ⁢ ( N ) ≤ S fin ⁢ ( N ) then hereditarily separate ℕ.

Observe that Thompson’s group 𝐹 is hereditarily separating with respect to its standard action on [ 0 , 1 ] . Indeed, suppose that 𝑍 is an open subset of [ 0 , 1 ] , Y : = { t 1 , t 2 , … , t s } ⊂ Z and t ∈ Z ∖ ( Y ∪ { 0 , 1 } ) . There is some non-trivial dyadic segment [ p , q ] ⊆ Z containing 𝑡 such that Y ∩ [ p , q ] = ∅ . Obviously, x [ p , q ] , 0 ∈ stab F ⁢ ( ( [ 0 , 1 ] ∖ Z ) ∪ Y ) ∖ stab F ⁢ ( { t } ) .

The action of any weakly branch group on the boundary space of the corresponding infinite rooted tree is also hereditarily separating. To define this class, we will follow [4, 9]. We consider a group 𝐺 that isometrically acts on some rooted tree 𝑇. The vertices in 𝑇 that are at the same distance from the root are said to be at the same level. We say that the action of the group 𝐺 on 𝑇 is spherically transitive if 𝐺 acts transitively on each level of 𝑇. For any vertex t ∈ T , we define its rigid stabilizer to be the set of all elements from 𝐺 that stabilize T ∖ T t pointwise, where T t is the natural subtree hanging from 𝑡. A group 𝐺 is called weakly branch if it acts spherically transitively on some rooted tree 𝑇 so that the rigid stabilizer of every vertex is non-trivial. The boundary of a tree 𝑇, denoted by ∂ T , consists of the infinite branches starting at the root. There is a topology on ∂ T where the base of open sets is determined by subtrees T t . A weakly branch group obviously acts on the boundary ∂ T by homeomorphisms (in fact, by isometries with respect to the natural metric).

To see the statement of this example, we use an argument from [1]. Fix a weakly branch group 𝐺 and the corresponding tree 𝑇. Let X : = ∂ T . To see that 𝐺 hereditarily separates 𝒳, suppose that 𝑍 is an open subset of 𝒳,

and x ∈ Z ∖ Y . Without loss of generality, assume that 𝑡 is the vertex of the infinite ray 𝑥 such that if any y ∈ X contains 𝑡, then y ∈ Z . Let 𝑘 be the level of 𝑡. Now choose a level k ′ ≥ k such that the vertices in the infinite rays x 1 , … , x ℓ , x at the k ′ -th level are all distinct. Let t 0 be the vertex of 𝑥 at level k ′ . Let 𝑆 be the stabilizer of t 0 in 𝐺 and let 𝑅 be the rigid vertex stabilizer of t 0 in 𝐺. Then 𝑆 acts spherically transitively on the infinite subtree T t 0 rooted at t 0 . Since 𝑅 is a non-trivial normal subgroup of 𝑆, we see that it cannot stabilize any infinite ray going through t 0 . In particular, there exists some g ∈ R such that g ⁢ ( x ) ≠ x . On the other hand, 𝑔 stabilizes every ray not going through t 0 . It follows that g ∈ stab G ⁢ ( ( X ∖ Z ) ∪ Y ) . This proves the statement of Example 2.11.

We say that g ̄ is distinctive for w ⁢ ( y ̄ ) and 𝑝 if all the points

are pairwise distinct.

Let 𝐺 be a group acting on a perfect Hausdorff space 𝒳 by homeomorphisms. Let w ⁢ ( y ̄ ) be a non-trivial word over 𝐺 on 𝑡 variables, y 1 , … , y t , which is reduced in F t ∗ G and non-constant (i.e. w ⁢ ( y ̄ ) ∉ G ). If 𝐺 hereditarily separates 𝒳 and w ⁢ ( y ̄ ) has a conjugate in F t ∗ G which is explicitly oscillating, then the inequality w ⁢ ( y ̄ ) ≠ 1 has a solution in 𝐺.

Moreover, assuming additionally that w ⁢ ( y ̄ ) ∈ F t or w ⁢ ( y ̄ ) is given in the form (2.1) and is explicitly oscillating, for every open set O ′ ⊆ O w , there is a distinctive tuple g ̄ = ( g 1 , … , g t ) ∈ G t with respect to w ⁢ ( y ̄ ) and some p ∈ O ′ such that, for all 𝑖 with 1 ≤ i ≤ t ,

• supp ⁢ ( g i ) ⊆ ( ⋃ V w > 0 ⁢ ( O ′ ) ) ∖ ( ⋃ V w ⁢ ( O ′ ̄ ∖ O ′ ) ) ,

• each member of V w ⁢ ( O ′ ) is g i -invariant.

Proof

Using conjugation (if necessary), we assume that if w ⁢ ( y ̄ ) ∉ F t , then w ⁢ ( y ̄ ) is an explicitly oscillating word written in the form (2.2). We preserve the notation given before the formulation. Note that, since 𝒳 is perfect, each open subset is infinite.

It is clear that we only need to prove the second part of the statement of the theorem. Fix O ′ as in the formulation and any

The existence of such a 𝑝 follows from the fact that the set O ′ ̄ ∖ O ′ is nowhere dense in 𝒳 and the action of 𝐺 is continuous. The proof of the theorem is by induction. At the 𝑘-th step (where k ≤ L n ), we will show that

• there is a tuple g ̄ = ( g 1 , … , g t ) ∈ G such that g ̄ is distinctive for ( w ) k and 𝑝.

• In the condition above, we can choose g ̄ so that, for all 𝑖 with 1 ≤ i ≤ t ,

  supp ⁢ ( g i ) ⊆ ( ⋃ V w > 0 ⁢ ( O ′ ) ) ∖ ( ⋃ V [ w ] k − 1 ⁢ ( ⋃ V w ⁢ ( O ′ ̄ ∖ O ′ ) ) )

  and each member of V w ⁢ ( O ′ ) is g i -invariant.

Indeed, let 𝑞 and O ′ be as in the formulation of the claim. Let us enumerate the elements of V w ⁢ ( O ′ ) by V m with m > 0 and inductively construct open sets O m corresponding to them. We start with O 0 = v r ε r ⋅ … ⋅ v 1 ε 1 ⁢ ( O ′ ) . If, at the 𝑚-th step, q ∈ O m − 1 ∩ V m , then we define O m = O m − 1 ∩ V m . Now suppose that q ∉ O m − 1 ∩ V m . As q ∉ ⋃ V w ⁢ ( O ′ ̄ ∖ O ′ ) , we find some open O m ⊆ O m − 1 ∖ V m containing 𝑞. After meeting all members of V w ⁢ ( O ′ ) , we obtain the final 𝑂 which is non-empty and open. If w ⁢ ( y ̄ ) ∈ F t , then we define O = O ′ , which finishes the proof of the claim.

If k = 1 , we may assume that ( w ) 1 is of the form y j ± 1 ⁢ v 1 for some 1 ≤ j ≤ t . When w ⁢ ( y ̄ ) ∈ F t , we replace v 1 by id and follow the argument below. Note that

Applying the claim above, we find in v 1 ⁢ ( O ′ ) a neighborhood 𝑂 of v 1 ⁢ ( p ) which satisfies ( † ). According to the assumptions above, when w ⁢ ( y ̄ ) ∉ F t , then we have p ′ ≠ v 1 ⁢ ( p ′ ) for all p ′ ∈ O ′ . Thus, for a non-trivial v 1 , the inequality p ≠ v 1 ⁢ ( p ) is satisfied.

Without loss of generality, suppose ( w ) 1 = y j ⁢ v 1 . Since 𝐺 hereditarily separates 𝒳, the orbit of v 1 ⁢ ( p ) with respect to

is infinite. Thus we can choose

such that f ⁢ ( v 1 ⁢ ( p ) ) ∉ { p , v 1 ⁢ ( p ) } . Defining g j : = f and extending it by any ( t − 1 ) -tuple from

we obtain a 𝑡-tuple g ̄ distinctive for 𝑝 and ( w ) 1 .

It is easy to see that, for all i ≤ t ,

Now fix i ≤ t and V ∈ V w ⁢ ( O ′ ) . If O ∩ V = ∅ , then obviously g i ⁢ ( V ) = V . On the other hand, if O ∩ V ≠ ∅ , then it follows from the construction of 𝑂 that O ⊆ V . Thus supp ⁢ ( g i ) is also a subset of 𝑉 and the condition g i ⁢ ( V ) = V is satisfied.

Let k > 1 . Assume that, for

where k − 1 = L d − 1 + s , we can find a tuple g ̄ ∈ G such that the induction hypothesis is satisfied. According to the form of ( w ) k , we consider two cases.

Case 1:

where k − 1 = L d − 1 + s , s ≥ 1 . If

then we have found an acceptable tuple g ̄ . Let us assume that p k , g ̄ = p m , g ̄ for some 0 ≤ m < k or p k , g ̄ = v m + 1 ⁢ ( p L m , g ̄ ) for some 0 ≤ m < d − 1 .

Let y j ± 1 be the first letter of ( w ) k . Replacing y j by y j − 1 and g j by g j − 1 if necessary, we may assume that u d , s + 1 = y j . Put

for g ̄ chosen at the ( k − 1 ) -th step of induction. Since, by the induction hypothesis, p ∈ O ′ and g s ± 1 ⁢ ( v r ⁢ … ⁢ v 1 ⁢ ( O ′ ) ) = v r ⁢ … ⁢ v 1 ⁢ ( O ′ ) for all r ≤ k and all s ≤ t , we see that p k − 1 , g ̄ ∈ v d ⁢ … ⁢ v 1 ⁢ ( O ′ ) . We also know that

As above, we choose a neighborhood O ⊆ v d ⁢ … ⁢ v 1 ⁢ ( O ′ ) of the point

satisfying condition ( † ). Since the action of 𝐺 is hereditarily separating, the orbit with respect to

of p k − 1 , g ̄ is infinite. Let

Since 𝑍 is finite, there exists some

taking p k − 1 , g ̄ outside 𝑍. Replacing g j by g j ⁢ f , we obtain a corrected tuple g ̄ . Since the element 𝑓 has been chosen from the stabilizer stab G ⁢ ( Y ) , the points

are the same as before. On the other hand, p k , g ̄ is distinct from all the elements

Since

we see that

This together with the induction hypothesis implies that

for the corrected g j .

Let V ∈ V w ⁢ ( O ′ ) . If O ∩ V = ∅ , then by the choice of 𝑓 and induction, we have g j ⁢ ( V ) = V . If O ∩ V ≠ ∅ , then O ⊆ V . Thus, by the choice of 𝑓 and induction, 𝑓 and the original g j (defined at Step k − 1 ) stabilize 𝑉 setwise. Thus g j ⁢ ( V ) = V . On the other hand, all elements g s for s ≠ j have not been changed and thus automatically satisfy the required conditions. This finishes the proof of Case 1.

Case 2:

where k = L d + 1 . If

then we have found an acceptable tuple g ̄ .

Assume the contrary. Suppose that y j ± 1 is the first letter of ( w ) k , and as in Case 1, we only consider the possibility u d + 1 , 1 = y j . Let u d , s = y j ′ ± 1 . Then let

Assume that v d + 1 ⁢ ( p k − 1 , g ̄ ) ∈ Y ′ . By the induction hypothesis,

By Claim M, we find some neighborhood O ⊆ v d ⁢ … ⁢ v 1 ⁢ ( O ′ ) of p k − 1 , g ̄ satisfying ( † ). Since the action of 𝐺 is hereditarily separating, the orbit of p k − 1 , g ̄ with respect to

is infinite. We replace g j ′ by some f ′ ⁢ g j ′ (or g j ′ ⁢ f ′ in the case u d , s = y j ′ − 1 ), where

and f ′ takes p k − 1 , g ̄ outside the finite set Y ′ ∪ v d + 1 − 1 ⁢ ( Y ′ ) . By the choice of O ′ and 𝑂,