False proof of amenability and non-amenability of the R. Thompson group 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 generated by 2 elements , and a Foelner set of the Cayley graph of for, say, constant . Then most vertices in the induced subgraph with vertex set will have degree 4. Embed the subgraph into an oriented surface , so that the complement of the set of edges of in is a union of discs (it is well known that this can always be done and the authors repeat the construction). Every vertex of with degree 4 has two outgoing edges and two incoming edges labeled by and we can assume that the outgoing edges counted counterclockwise (on the surface), are always labeled by (in that order).

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

The authors use the semigroup diagram representation of elements of as in our papers with Victor Guba and the action of 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 ) on trees: basically moves the marker one leaf to the left, and does some changes to the right of the marker. Analyzing the proof, one can quickly come to the conclusion that the only properties of the authors are using are the following:

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

(2) Every relator contains a (cyclic) subword of the form where , and commutes with in (this immediately follows from Lemma 13 and is explained in the proof of Lemma 14).

Victor Guba noticed that the wreath product acting in the standard way on the bi-infinite sequences of integers with a marked place ( moves the marker by 1 to the left, 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 on marked bi-infinite sequences of numbers just as well as it works for the action of on trees. What is used there is that moves the marker to the left while does not change the configuration to the left of the marker. Since 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 by the word (which can be done since commutes with ), some subwords of the form have new neighbors, and so their weight may increase (by 2, see above). That leads to increasing the total weight , and so the inequality 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 . 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 then .

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 to some vertex to the right of it is possible that removing it by applying results in a maximal red edge from to (and from to ). Then multiplying by would get us to as required.

**Update 3: **Note that so far nobody could find a counterexample to the statement that does not have a relator with the number of regular 2-letter subwords . Note also that there are no counterexamples to Lemma 18 either. One can construct a 2-generated non-amenable group where for every , there exists a relator with the number of regular 2-letter subwords . Just take a group where the words satisfy a small cancelation condition and .

**Update 4: **A counterexample to the proof of Lemma 18 has been found by Victor Guba. The word where is a relator of 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, , 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 . The only thing one needs to check here is the fact that is a relation in . This is quite easy.

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