The core lesson left us here. Take a finite set of integers. Alpha i deletes its i-th smallest element. Alpha two removes five from this set.

Theorem A gives a random finite set whose law barely changes under the first k deletions. Theorem B gives a common fixed point for continuous affine maps of a compact convex set that satisfy one family of relations.

This lesson is the algebra between them. The relations define a monoid, S. Its elements turn out to be finite sets. An invariant mean on those sets is what Theorem B needs. Then we set S beside a famous open problem, Thompson's group F.

Take the set two, five, seven, eleven, thirteen. Products read right to left, like functions. Delete the second smallest, then the third smallest of what remains. Five goes, then eleven.

Now delete the fourth smallest first, then the second. Eleven goes, then five. Same result.

The reason is rank bookkeeping. Once rank i is gone, everything above drops one rank. So rank j afterwards was rank j plus one before. That is the face relation, for every i at most j, equality included.

A simplex is a set of vertices. Deleting the i-th vertex gives the opposite face, and two deletions reach this green edge in either order.

A monoid is a set with an associative product and an identity, like words under concatenation. A presentation adds relations. With b a equals a b, every word sorts into a's then b's.

The facial monoid S has one letter per positive integer, with the face relations.

Orient each relation as a rule. If an adjacent pair is out of order, or equal, swap it and add one to the index that moved right. Here three passes one and becomes four, then five. Then the equal pair becomes one, two.

Now the indices strictly increase and no rule applies. This flat normal form is a finite set: one, two, five.

A word can be rewritten in several places, and every order must end at the same word. Newman's lemma asks for two checks.

First, rewriting must stop. Indices grow, so the paper counts, at each position, the later indices that are not larger. On our example that vector falls from two one to zero zero, in dictionary order.

Second, every fork must rejoin. Rewrites interfere only when they overlap, on three letters. The right pair first gives the top path. The left pair first gives the bottom path. They meet.

So every element of S has one flat normal form: a finite set of positive integers.

Now multiply. Take the set one, three, four, six. Its holes are two, five, seven and onward.

Put s two on the left and rewrite. Its index rises each time it passes an element that is not larger, and it stops at five, the second hole. Left multiplication by s i fills the i-th hole. The paper says direct verification. The pushing argument is ours.

Complement inside the interval up to one past the maximum: holes and elements swap. We get two, five, seven, a pin headed set: its maximum minus one is missing.

Deleting its second element is filling the second hole, as long as the maximum is untouched.

A monoid is amenable if it has an invariant mean: an average of every bounded function, unchanged by left shifts.

Equivalently, each finite list of elements has a finitely supported distribution that it barely moves. On the integers, shifting a uniform window of length n changes two cells out of n.

Equivalently, continuous affine maps of a compact convex set share a fixed point. Day proved these equivalent. Theorem A has the second shape, Theorem B the third.

The free group on two generators fails. Shift the a branch by a inverse, and it covers everything outside the a inverse branch. Those two carry mass one. So do the b branches. Total, two.

The other lesson built walk means on k point sets, where deletion lowers the level by one. They live on pin headed sets, since a very large last step is never one. The complement turns them into means mu k with that property for hole filling.

Sigma i shifts this row one box left.

Average the first n levels. Under sigma i the interior cancels and, by our check, about two i boundary terms of weight one over n survive. Any limit point mu is invariant under every sigma i.

S is the finite sets, and left multiplication fills holes, so mu is a left invariant mean on S. Day's theorem gives the fixed point.

Keep the first n generators and their relations. Pick two different points x and y in a compact convex set. Let generator n be the constant y, and the others the constant x.

Every relation holds, since on both sides the outer map has index below n. No common fixed point exists, so this truncated monoid is not amenable.

One gap remains: S might force extra relations. These two words in the first three generators are equal, and the flat rewriting passes through generator four.

The paper reverses the rules. Indices only fall, rewriting stays in the truncated monoid, and the one fork rejoins.

So no hidden relations, and the truncated monoid is non amenable.

S is an increasing union of non amenable pieces, and S is amenable. The paper says its amenability hides at infinity.

One relation shows how. Generator n times generator one equals generator one times generator n plus one. Under the constant maps the left side is y and the right side is x. That action cannot extend to the next generator. This check is ours.

The paper notes this cannot happen for groups. A subgroup of an amenable group is amenable: cosets tile the group, and one point per coset restricts a mean. A monoid has no cosets. That gloss is ours.

For monoids, sitting inside something amenable proves nothing.

In S, s one to the power m equals s one, s two, up to s m: fill the first hole m times.

Let the opposite monoid act on the right, and let s one fix p. Then p times s i equals p times s one to the i, which is p. Everything fixes p.

Furlner sets are asymmetric. Frey's theorem gives sets A where s A barely pokes out of A.

The reverse fails: a fifth of A is missing from s one A or s two A. The paper cites Klah vay. In our reading, s one merges the distinct points s one x and s two x.

Thompson's group F consists of piecewise linear maps of the interval, presented by the face relations with i strictly below j.

S adds the case i equals j, so it is a quotient of F plus, the monoid with Thompson's presentation.

The paper calls amenability of F a notorious open problem. From the literature, not the paper: F has no free subgroup on two generators, so the usual obstruction is absent, and no announced proof has stood.

Inference runs one way. F is amenable exactly when F plus is, and amenability passes to quotients. So an amenable F implies Theorem B, and a non amenable S would have settled it. The spelling out is ours.

The paper says its proof can evoke echoes of Moore's work. The survey's reading: in twenty fifteen Moore showed that an idempotent mean on all bracketings would be invariant under F. The generator re-brackets, and idempotence absorbs it.

In twenty nineteen he proved no such mean exists. Under addition of integers they exist. There, in the survey's reading, deleting a point merges two steps, and the result is S.

Also, F plus embeds in a semidirect product of S with a free abelian monoid that records the exponents S forgets. That product is amenable.

For groups that would settle F. In our reading, for monoids it proves nothing, as the truncated monoid inside S showed.