Thompson's group F is nonamenable

The explanation

Here is one move in Thompson's group F. Take the interval from 0 to 1 and cut it into three pieces. The first half shrinks to half its length. The next quarter slides left by 1/4. The last quarter doubles in length. So 1/2 lands on 1/4, 3/4 lands on 1/2, and 1 stays at 1. Call this move A.

Every cut point of A is a dyadic fraction, meaning a fraction whose bottom number is a power of 2. Every stretch factor is a power of 2, such as 1/2, 1 or 2. Thompson's group F is the collection of all such moves with finitely many pieces. It is a group: doing one move after another gives another move, and every move can be undone. Order matters. Let B leave the left half alone and do a half-size copy of A on the right half. Then A after B sends 3/4 to 3/8. B after A sends 3/4 to 1/2.

Now a question about balance, first in a simpler group. Take the whole numbers and one move: add 1. The set 1 to 100, shifted by 1, becomes 2 to 101. Only two numbers differ between the old set and the new one: 1 and 101. That is 2 out of 100, or 2 percent. For the set 1 to 1000 it drops to 0.2 percent. So the whole numbers have finite sets that barely notice the move.

A group is called amenable if it has a fair way to average every bounded function on it. Fair means the average does not change when you shift the function by any move. This idea goes back to von Neumann and Day. A criterion of Følner from 1955 links it to the example above. Suppose a group is amenable. Then any finite list of moves has finite sets that each move barely shifts. The share that changes can be made as small as you like.

This paper proves that Thompson's group F is not amenableSource 1. Its argument fixes one finite list of moves in F. It then claims a uniform bound. Take any nonempty finite set of moves. At least one move on the list shifts a fixed positive share of itSource 2. No set can push that share toward zero, the way 2 percent fell to 0.2 percent. By Følner's criterion, F cannot be amenable.

The question is old. Thompson introduced F in 1965. Geoghegan conjectured in 1979 that F is not amenable. The usual test does not help. A group that contains a free subgroup on two moves cannot be amenable. Such a subgroup is a pair of moves that obey no shortcut rules when combined. Brin and Squier showed in 1985 that F contains none. Yet F also lies outside the simplest family of amenable groups, called elementary amenable. Neither fact decides the question.

Finite sets were the main way in, and they proved slippery. Moore showed that any barely shifted sets in F would have to be enormous, growing like a tower of powers. Complete answers were claimed in both directions. Shavgulidze claimed amenability, and Moore identified errors in that approach. Moore withdrew his own amenability claim after Akhmedov identified an error in it. Akhmedov's 2021 version claims nonamenability through what he calls a height-function criterion.

The paper plays two facts against each other. The first is a 1983 result of Benyamini and Sternfeld. It concerns a space with infinitely many independent directions, and the ball of points within distance 1 of its centre. That ball has a map into itself that moves every point by at least a fixed distanceSource 3. The map also never stretches distances by more than a fixed factor. The paper fixes a large number of small parent pieces of the interval. It gives each dyadic cutting of the interval a color, which is a point in that ball. A cutting's color comes from applying the map to the average color of its zoomed-in views of the parent piecesSource 4. If F had barely shifted sets, averaging colors over one would give a nearly fixed point of the mapSource 5. That contradicts the fixed minimum distance. The list of moves is chosen before the set, so no set can dodge itSource 6.

How sure is this? A computer proof checker called Lean has checked a formal statement. It rules out a fair average on F, the standard group of dyadic piecewise-linear moves of the interval. That Lean statement gives no explicit size for the shifted share. It names no particular list of moves. The paper itself states a bound in terms of its own constantsSource 7. The paper also draws two consequences, about special representations of F and about random networks built on it. Those rely on separate companion papers, not on this proof, and lie outside the Lean checkSource 8. No specialists have published a review yet.

What it doesn't show. This does not give an explicit size for the shifted share in its Lean statement. Its two consequences rest on companion papers.