MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Last year a paper on the arXiv (Akhmedov) claimed that Thompson's group $F$ is not amenable, while another paper, published in the journal "Infinite dimensional analysis, quantum probability, and related topics" (vol. 12, p173-191) by Shavgulidze claimed the exact opposite, that $F$ is amenable. Although the question of which, if either, was a valid proof seemed to be being asked by people, I cannot seem to find a conclusion anywhere and the discussion of late seems to have died down considerably. From what I can gather, Shavgulidze's paper seems unfixable, while the validity of Akhmedov's paper is undecided (although it may of course be decided now).

So, does anyone know if either of these papers is valid?

share|cite|improve this question
Or both, perhaps??? – Gil Kalai Jun 22 '10 at 9:42
@Gil - only if Edward Nelson succeeds in his program of proving PA inconsistent... :P – David Roberts Dec 9 '11 at 1:58
I worked on this question as a grad student in the 1990's at Cornell. A lifetime ago! I went into business but occasionally check for updates out of curiosity. It is a complex problem! J birge – user48541 Mar 21 '14 at 5:42
There will be a workshop on Thompson's group in May. – Narutaka OZAWA Mar 21 '14 at 8:26
An update, which I don't think has been remarked upon here: Akhmedov has withdrawn his claim of having a proof that $F$ is non-amenable. See his comments on arXiv:1310.4395, which conclude with: "It is hard to say if there is any chance for the approach to succeed." – user64255 Dec 23 '14 at 16:53
up vote 46 down vote accepted

While I did not participate in most of the checking of Shavgulidze's argument, I can offer the following partial account of the situation. I am told the paper was correct except for a lemma (or sequence of them) claiming that a sequence of auxiliary measures had certain properties. These were Borel measures on the $n$-simplex (one for each $n$). I believe it was shown that the original proposed auxiliary sequence of measures did not have one of the two properties. Shavgulidze proposed other sequences of measures. The most recent attempt that I am aware of (which was presented during his 2010 trip to the US mentioned by Mark Sapir in the above comment) involved the direct construction of Folner sets for the action of $F$ on the finite subsets of dyadic rationals (see the next paragraphs). The details were somewhat sparse and the definitions involved many unspecified numerical parameters, but it appeared to be the case that these sets could not be Folner in the necessary sense (see below for a clarification of "necessary sense"). This is because they would likely both contradict the iterated exponential lower bound on the Folner function which I have demonstrated and because they appear to violate the qualitative properties which I have demonstrated that Folner sets of trees must have (see the pre-print on my webpage; the qualitative condition appears in lemma 5.7, noting that marginal implies measure 0 with respect to any invariant measure).

Meanwhile I was able to provide a direct elementary proof that the existence of such a sequence having these properties implied the amenability of $F$. In fact the proof gives an explicit procedure for constructing (weighted) Folner sets from the sequence of measures satisfying the hypotheses mentioned above. A note containing the details was circulated to a few people around the time of Shavgulidze's visit to Vanderbilt. While I am reluctant to speak for anyone else (including the author), it appears to me that after the dust had settled (which took a considerable amount of time), the problem with the proof seems to have at least some of its roots in the following observation (which I now include for the sake of prosperity). $F$ acts on the finite subsets of the dyadic rations (let's call this set $\mathcal{D}$) by taking the set-wise image (here I am utilizing the piecewise linear function model of $F$). Now let $\mathcal{T}$ denote the finite subset of $[0,1]$ which contain $0$ and $1$ and are such that any consecutive pair is of the form $p/2^q,(p+1)/2^q$ (for natural numbers $p,q$). $F$ only acts partially on $\mathcal{T}$: the action $T \cdot f$ is defined if $f'$ is defined on the complement of $T$ in $[0,1]$ (there may be other cases when $T \cdot f$ is in $\mathcal{T}$, but let's restrict the domain of the action as above). The full action of $F$ on $\mathcal{D}$ is amenable. The point here is that the action of the standard generators on the sets $\{0,1-2^{-n},1\}$ is the same for large enough $n$ and thus we can build Folner sets as in a $\mathbb{Z}$ action. The amenability of partial action of $F$ on $\mathcal{T}$ is, on the other hand, equivalent to the amenability of $F$ (this is well known, but see the preprint above to see this spelled out in the present jargon).

Now here is the catch, if we also require that the invariant measure/Folner sets for the action of $F$ on $\mathcal{D}$ to concentrate on sets of mesh less than $1/16$, then one again arrives at an equivalent formulation of the amenability of $F$. The author was aware of the need for the mesh condition, but (in the most recent example) arranged it only in a modification after the fact (which interferes with invariance).

Incidentally the hypotheses on the sequence of measures mentioned above are a condition requiring that the measures concentrate on sets of arbitrarily small mesh as $n$ tends to infinity and a condition which is an analog of translation invariance.

I apologize if this borders on ``too much information.''

[Added 1/28/2011] Shavgulidze's 1/14/2011 posting to the ArXiv is essentially a more detailed version of what he was saying in notes, seminars, and private communication in January 2010 during his visit to the US mentioned in Mark Sapir's post above. The present note is still sufficiently vague and full of sufficiently many errors (many typographical in nature) that it is hard (or easy, if you like) to say explicitly which line of the proof is incorrect. It is possible, however to point to places where crucial details are missing and where there are certainly going to be errors (specifically the problems will be on page 11, if not elsewhere as well). The comments from my answer above still apply equally well to the present version. It appears that the present version (or any perturbation of it) still would violate the lower bound on the growth of the Folner function which I have established. The present version still totally ignores that the combinatorial statements on page 11 themselves readily imply the amenability of F, without the involvement of any analytical concepts.

[Added 2/3/2011] Details on what is incorrect with Shavgulidze's proof of the amenablity of $F$ can be found here.

[Added 10/3/2012] Well, well, well: now I'm in the position of having announced a proof that $F$ is amenable only to have an error be found. The error was finally found by Azer Akhemedov after being overlooked for roughly 4 weeks by myself and 9 or more people who had checked the proof and found no problems. The basic strategy of the proof still may be valid: it began by considering an extension of the free binary system $(\mathbb{T},*)$ on one generator to the finitely additive probability measures on this system: $$\mu * \nu (E) = \int \int \chi_E(s * t) d \nu (t) d \mu (s).$$ It was shown (correctly) that any idempotent measure is $F$-invariant (there is a natural way of identifying $\mathbb{T}$ with the positive elements of $F$). The difficulty came in constructing the idempotent measure. A version of the Kakutani Fixed Point Theorem was used to construct approximations $K_{\mathcal{B},k,n}$ to the set of idempotent measures. The error occurs in attempting to intersect these compact families of measures. In the proof, it was claimed that the parameter $k$ could be stablized along the an ultrafilter (Lemma 4.13 in the most recent version), allowing one to take a directed intersection of nonempty compact sets. This lemma is likely false and at least is not proved as claimed. One may still be able to argue that a relevant intersection of these approximations is nonempty and hence that there is an idempotent. This seems to require new ideas though.

share|cite|improve this answer
Thanks for the update! – Andrés E. Caicedo Jan 28 '11 at 18:12
@Justin: good luck with your work on this difficult question! – tweetie-bird Oct 4 '12 at 13:59
It looks like $\mu^\mu=\mu$ is impossible for the free 1-generated magma $\mathbb T$. Indeed, suppose that such a measure $\mu$ exists. Then for any homomorphic image of $\mathbb T$ should exists an idempotent measure. Using SAGE I was found that magma $\{0,1,2,3\}$ with multiplication matrix $$ \left(\begin{array}{rrrr} 1 & 3 & 2 & 2 \\ 1 & 2 & 1 & 3 \\ 0 & 2 & 3 & 3 \\ 0 & 2 & 1 & 2 \end{array}\right) $$ has no idempotent measure. It is also generated by $0$: $0^0=1$, $1^0=3$, $3^0=2$. So, $\mathbb T$ does not have an idempotent measure. Am I right? – Lev Glebsky Nov 23 '12 at 23:53
I am wrong as it was explained to mi by Justin. – Lev Glebsky Dec 1 '12 at 0:42

Igor Pak told me about this site. The sitiuation with Thompson group is this. There is a counterexample to the main statement of Akhmedov's paper currently in arXiv. The counterexample (by Victor Guba) exists for almost a year. The paper is unreadable, so it is hard to find a concrete place where the mistake is. Shavgulidze's paper(s) was much more readable. It has "Lemma 5" whose proof was found wrong by Matt Brin. Whether the statement of Lemma 5 is correct, is not clear. Most probably it is wrong. Shavgulidze gave 8 talks in the US this January (2 in Texas A&M, 4 in Vanderbilt, one is Cornell and one in Bingamton). He presented an alternative proof (trying to avoid "Lemma 5"). During his talks a big difficulty was found, mostly by Justin Moore from Cornell, but also by Brin and others. In fact several objections about the proof were expressed. One of them is that the new proof of Shavgulidze in fact produces Foelner sets in F, whose sizes seem to be much smaller than the sizes predicted by a recent result of Justin Moore. Shavgulidze said that he would address these concerns after he returns to Moscow and we have not heard from him since. So currently we are back on square one. It is not clear whether F is amenable or not, and how to approach the proof.

share|cite|improve this answer
Just to clarify: "unreadable" refers to the original paper, not to counterexample paper, right? Also, what is the status of Justin Moore's approach? – Victor Protsak Jun 22 '10 at 11:42
"Unreadable" refers to Akhmedov's paper (it is my opinion, you may want to try reading it by yourself). The counterexample was communicated to Akhmedov by Guba a year ago and did not appear as a preprint. It is essentially elementary. Guba also gave a counterexample to the previous version of Akhmedov's paper (it also did not appear as a preprint but at least Alhmedov substituted the previous version by the new one as a result of that counterexample). – Mark Sapir Jun 22 '10 at 20:10
P.S. There are very nice notes of Matt Brin's seminar on Shavgulidze's proof:… The mistake in Lemma 5 is discovered on page 39 of the notes. – Mark Sapir Jun 23 '10 at 6:57
About the status of Justin Moore's paper it is better to ask him. As far as I know both Matt Brin and Victor Guba now say that the paper is OK. But I do not know if the paper is finally accepted in a journal. Also Guba found an easier way to prove the results of the paper which is a good sign (it is better to have two proofs than none). – Mark Sapir Jun 23 '10 at 21:40
To clarify Mark's comment in the context of my new annoucement, in this refers to my paper on the lower bound for the Folner function for F. That paper has now been accepted in Groups, Geometry, and Dynamics. – Justin Moore Sep 9 '12 at 11:59

The following blog post by Danny Calegari (and its links) seems to constitute the best online source of information on this matter:

share|cite|improve this answer
See also Tom Leinster's answer,… – Victor Protsak Jun 2 '10 at 14:24
I have come across both of these links, and in some ways the first was my reason for asking this question. The conversation dies down in the middle of January (5 months ago, and this all kicked off about a year ago) having established that Akhmedov's paper needed closer scrutiny. – ADL Jun 3 '10 at 13:35

Shavgulidze's back.

A new "proof" is now on the arxiv.

The first part contains what was correct in the previous preprint. In the last 5 pages he tries to give a suitable proof for the group $F$. The point is always the same: he looks for a good density on the space of partitions of the interval. The style is always the same: a few pages full of maths, with no explication.

share|cite|improve this answer
and horrendous typesetting ... – Yemon Choi Jan 17 '11 at 20:02
See the addendum to my answer above. – Justin Moore Jan 28 '11 at 17:34

My understanding is that the situation has not changed much. A group of mathematicians at Binghamton University had been investigating Shavgulidze's argument, and they found a flaw which Shavgulidze has not addressed. As of now they are still waiting for Shavgulidze to respond.

I haven't heard anything about the Akhmedov paper recently.

share|cite|improve this answer
Adding to this here is a webpage for the Thompson groups seminar at Binghamton led by Matt Brin: – hypercube Jun 2 '10 at 19:12

[Update 25.11.2012]

Azer Akhemedov found an example which contradicts Theorem B*. Withdrawal of the preprint is already scheduled.

share|cite|improve this answer
I think we should wait for closer examination of the paper before saying that the problem is definitely solved. – Yemon Choi Dec 9 '11 at 1:35
More pointedly, skimming through the paper, it claims to show that the quotient of $F$ by some canonical normal subgroup $H_F$ is non-amenable. But then since the commutator subgroup of $F$ is simple, this would force $H_F$ to be trivial, and it seems unclear why that should be the case... – Yemon Choi Dec 9 '11 at 2:07
The main result of the preprint in question claims that if G is an amenable group acting by orientation preserving homeomorphisms on R, then there is a "projectively G-invariant" measure on R (action of any element on the measure only dilates the measure). There is a counterexample to this with a group in which every element has a fixed point. The author says that adding a hypothesis that G has at least one fixed point free element will fix the theorem (and still apply to F). Details were promised. We will have to wait and see. – Matt Brin Jan 7 '12 at 15:09
Unfortunately, Moore has just withdrawn this preprint. – Lior Silberman Oct 2 '12 at 8:21
@Yemon Choi, you are absolutely right about overconfidence of my statement. It was an emotional impulse because Prof. Levon Beklaryan is my father and the problem of the classification theorem for groups of homeomorphisms of the line waited for the solutions for 20 years. I didn't edit my post because I thought that it would be unethical towards the participants of the community. My name is Armen and I am PhD in MSU and specialize in differential equations. – Mariarty Oct 2 '12 at 21:09

protected by Todd Trimble Dec 23 '14 at 20:22

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.