Another false proof of nonamenability of the R. Thompson group F


False proof of amenability and non-amenability of the R. Thompson group F appear about once a year. The interesting thing is that about half of the wrong papers claim amenability and about half claim non-amenability. The latest attempt to prove non-amenability was made by Bronislaw Wajnryb and Pawel Witowicz. The idea of their argument is very nice. Take any amenable group G generated by 2 elements a,b, and a Foelner set X of the Cayley graph \Gamma of G for, say, constant 1/100. Then most vertices in the induced subgraph \Gamma_X with vertex set X will have degree 4. Embed the subgraph X into an oriented surface S, so that the complement of the set of edges of \Gamma_X in S is a union of discs (it is well known that this can always be done and the authors repeat the construction). Every vertex of \Gamma_X with degree 4 has two outgoing edges and two incoming edges labeled by a,b and we can assume that the outgoing edges counted counterclockwise (on the surface), are always labeled by a, b, a^{-1}, b^{-1} (in that order).

Then some boundary label of S \setminus \Gamma_X, read counterclockwise, is a relator of G (all relators here and below are cyclic words) and has most 2-letter subwords  from the following set  \{ab, b^{-1}a, ba^{-1}, a^{-1}b^{-1}\} called regular words. It follows (see Theorem 3 in the paper), that every amenable 2-generated group as above has a relator w=1 such that, say, more than 99 percent of all 2-letter subwords of w are regular. The authors claim (Theorem 8) that F does not have such relators. For this, the authors introduce weights \kappa of irregular 2-letter words. Say, ab^{-1} weights 24 while b^{-1}b weights 23, and bb weights sometimes 17 sometimes 15 depending on the neighbor letters. The weight \kappa(w) is the sum of weights of irregular 2-letter subwords and the number of maximal subwords containing only irregular 2-letter subwords. Let n(w) be the length of w. If 99 percent of 2-letter subwords of w are regular, then n>4\kappa, and the authors claim that it is impossible that a relator w of F satisfies this property.

The authors use the semigroup diagram representation of elements of F as in our papers with Victor Guba and the action of F on binary trees with a marked leaf. It is very easy to describe the action of each of the two standard generators (the authors denote them by a,b) on trees: basically a moves the marker one leaf to the left, and b does some changes to the right of the marker. Analyzing the proof, one can quickly come to the conclusion that the only properties of F the authors are using are the following:

(1) The group is torsion-free and non-commutative.

(2) Every relator w contains a (cyclic) subword of the form a^{-1}b^saUa^{-1}b^ta where s>0, t<0, and aUa^{-1} commutes with b in F (this immediately follows from Lemma 13 and is explained in the proof of Lemma 14).

Victor Guba noticed that the wreath product \mathbb{Z}\wr\mathbb{Z} acting in the standard way on the bi-infinite sequences of integers with a marked place (a moves the marker by 1 to the left, b increases the value of the marked integer by 1) has the same properties. To prove that, one just has to repeat the proof of Lemma 13 of the paper. It works for the action of \mathbb{Z}\wr \mathbb{Z} on marked bi-infinite sequences of numbers just as well as it works for the action of F on trees.  What is used there is that a moves the marker to the left while b does not change the configuration to the left of the marker.  Since \mathbb{Z}\wr\mathbb{Z} is amenable, the proof in the paper cannot possibly work. The concrete “fatal” mistake in the proof is in Case 3 on page 20 (the proof of Lemma 14). The authors did not notice that after they replace the subword  a^{-1}b^saUa^{-1}b^ta by the word a^{-1}b^{s-1}aUa^{-1}b^{t+1}a (which can be done since aUa^{-1} commutes with b), some subwords of the form bb have new neighbors, and so their weight may increase (by 2, see above). That leads to increasing the total weight \kappa, and so the inequality n>4\kappa is no longer true after the rewriting.

So the paper contains some nice ideas (Theorem 3, in particular), but the proof is wrong. It is interesting whether the next “proof” will claim amenability or nonamenability.  We’ll probably know it next year.

Update 1: The authors have put a new version of their paper on the arXiv (see the link above).
The general idea remains the same but the weight system has changed, it has become more complicated. The authors said that the new proof does not work for \mathbb{Z}\wr \mathbb{Z}. Of course there are many other 2-generated amenable groups. But in any case the new proof needs to be checked.

Update 2: Gili Golan (a student in Bar Ilan University) found a mistake in the new version of the paper. Here is her message:

In Lemma 18, second to last line on page 18 of the paper, it says that if p=a^{-1} then x\neq b.
The proof of this claim (under some assumptions) is given in the third paragraph on page 20, but I don’t understand why the argument works.

If we have a red maximal edge from v_4 to some vertex v_5 to the right of v_1 it is possible that removing it by applying b results in a maximal red edge from v_4 to v_1 (and from v_1 to v_5). Then multiplying by a^{-1} would get us to v_1 as required.

Update 3: Note that so far nobody could find a counterexample to the statement that F does not have a relator r=1 with the number of regular 2-letter subwords \ge .99 |r|. Note also  that there are no counterexamples to Lemma 18 either. One can construct a 2-generated non-amenable group where for every \epsilon>0, there exists a relator r with the number of regular  2-letter subwords \ge (1-\epsilon)|r|. Just take a group \langle a,b\mid [a,b]^n=w_n, n=1,2,3,...\rangle where the words w_n satisfy a small cancelation condition and n\gg |w_n|.

Update 4: A counterexample to the proof of Lemma 18 has been found by Victor Guba. The word BABabABaaBAbABaabAAAbABabAAbaaaa where A=a^{-1}, B=b^{-1} is a relator of F but the subword constructed as in the proof of Lemma 18 does not satisfy all the conditions of that lemma. Matt  Zaremsky found a subword of that word, abABaaBAbABaabAAA, which does satisfy all conditions of Lemma 18, so this is not a counterexample to Lemma 18.

Update 5: Lemma 18 is wrong. Victor Guba found a counterexample. Here is his message:

Let W=BAbABaabABAbABaabABAbaaa. The only thing one needs to check here is the fact that W is a relation in F. This is quite easy.

Now, W has only two subwords of the form Ba. For each of them, we see bA before such a subword.

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