Take a random finite set of integers. Here is one draw: seven points on the number line.

Now delete its second smallest element. Call that operation alpha two. In general alpha i deletes the i th smallest element.

Theorem A says this. For any k, and any tolerance epsilon, there is a finitely supported random set, always with at least k elements, whose probability law moves by less than epsilon, in total variation, under every one of the deletions alpha one through alpha k. You can delete from it, and statistically almost nothing has changed.

Why is that surprising? Start with size. Here is the distribution of how many elements the set has.

A deletion removes exactly one element, so the whole histogram slides left by one. For the two histograms to overlap almost perfectly, the size has to be spread thinly over many values: more than one over epsilon of them. That is a short argument from our own survey.

Positions are harder. Delete the smallest element, and the second smallest becomes the smallest. Every rank shifts, and the law has to absorb that as well.

The natural construction fails outright. Build the set as a random walk with independent, identically distributed gaps. Take the walk with k plus one points, delete its first point, and compare with the walk with k points. The paper proves that for any honest step distribution, these two laws are more than one quarter apart in total variation.

How do the deletions interact? Take the set two, five, seven, eleven.

Apply alpha one and then alpha two. On the slide, the map applied first is written on the right. Deleting the smallest element removes two. Then the second smallest of what remains is seven. We are left with five and eleven.

In the other order we get a different set: first five goes, then two, leaving seven and eleven. So these maps do not commute.

They do satisfy a repaired identity. Deleting the first element and then the second gives the same set as deleting the third element and then the first. Once the first element is gone, the old third element sits in position two.

In general, take i at most j. Deleting element i, and then element j of what is left, equals deleting element j plus one first, and then element i. These are the face relations. The faces of a simplex satisfy the same identities.

The paper's second theorem sounds like a different subject. Take a compact convex set, and continuous affine maps from the set to itself.

A classical theorem of Markov and Kakutani says that if the maps commute, they have a common fixed point. Without commuting, that fails in general.

Theorem B says the face relations are enough. If an infinite family of such maps satisfies them, the maps share a fixed point.

Infinite is essential. With only n maps, n at least two, there is a counterexample. Let the first n minus one maps all be the constant map to one point, and let the last be the constant map to a different point. Every relation that mentions only these n maps holds, and there is no common fixed point.

How are the two theorems connected? Both come from one object. It is called an idempotent mean: a finitely additive version of a random, very large number.

From it the paper builds a kind of random walk, and proves that deleting any entry of the walk with k plus one points gives exactly the walk with k points.

Averaging over k, and then approximating, gives Theorem A.

The same walk can be moved onto an algebraic object: the monoid S generated by symbols that obey the face relations. Its elements turn out to be finite sets of integers, and multiplying by a generator fills a hole in the set. Taking complements turns deleting into filling, and after one more averaging step the walk becomes an invariant mean on S.

A theorem of Day says that having an invariant mean is the same as the fixed point property. That property is called amenability, and it is exactly Theorem B. This lesson follows the top row.

What is a mean? Start with the uniform distribution on the first n integers.

Let n grow. The mass thins out and moves to the right. No probability distribution is the limit, because every fixed integer ends up with mass zero.

But in the space of finitely additive probability measures, a limit point does exist, by compactness. It has total mass one, and mass zero on every finite set. The authors' gloss is a random very large number.

There is a price. There is no formula for it. It is not countably additive. And when you integrate over two such variables, the order of integration matters.

The mean we want has one special property. Convolve it with itself and you get it back. A very large number plus a very large number is a very large number.

Here is why a mean can do this. The limit of uniform distributions does not change when you shift by any fixed amount. So for each fixed shift, the inner average returns the same number, and the outer average is then the average of a constant.

No real probability distribution on the positive integers can do it. The sum of two positive integers is strictly larger than either. Look at the smallest value the variable can take: the sum can never equal it.

So this property exists only at infinity, and that is where the whole construction lives.

Now build a walk. Start at zero and take steps, each with the law of our mean. The positions visited form an increasing sequence, and that sequence is our random set.

Delete one of the positions, say the third.

The step into that point and the step out of it merge into a single step. Its law is the mean convolved with itself, which is the mean again. So what remains is a walk with one step fewer, and with exactly the same kind of steps. If you delete the last point there is nothing to merge. The last step is simply dropped.

That is Theorem two point three. Deleting any entry of the walk with k plus one points gives precisely the walk with k points. No error at all. One caution: this picture is a heuristic. The real proof has to respect the order in which the means are nested, and the first deep dive goes through it.

We are not done, because a deletion changes the length of the walk. The fix is to average. Take the walks of n consecutive lengths and mix them with equal weights.

A deletion lowers every level by one.

The shifted stack agrees with the original everywhere except at the top and the bottom. So the mixture moves by at most two over n in norm, which is one over n in total variation.

Let n grow and take a limit point. The result is a mean on finite sets that is exactly invariant under the first k deletions. It is the same kind of averaging that proves the Markov Kakutani theorem, a comparison of ours.

We have an exactly invariant object, and it lives at infinity. Theorem A asks for a real, finitely supported distribution.

First, a theorem of Goldstine says that finitely supported distributions are dense among means. So there are real distributions whose deletions differ from them by something that tends to zero, in a weak sense.

Second, a trick of Mazur, which is an application of Hahn Banach, upgrades weak convergence to norm convergence after passing to convex combinations. A convex combination of finitely supported distributions is again one.

And the norm here is twice the total variation distance. That proves Theorem A.

What is lost is any construction and any rate. Compactness and Hahn Banach are used several times over. The authors conjecture an explicit model, a tower of exponentials with wildly separated scales, where a huge number plus a merely large one is still, in law, a huge number. It is verified only in small cases.

Now the algebraic reading. Let S be the monoid of all words in the symbols s one, s two, and so on, modulo the face relations. The submonoid on the first n symbols is non-amenable, for every n at least two. That is the constant map counterexample again, together with a lemma that the first n generators satisfy no hidden relations.

Yet the union of all of them, S itself, is amenable. That is Theorem B.

For groups this cannot happen, because every subgroup of an amenable group is amenable. In the authors' words, the amenability of S hides at infinity.

There is a famous neighbor. Thompson's group F has the same relations, only for i strictly less than j. Whether F is amenable has been open for about fifty years. The paper notes that S is a quotient of the positive part of F. It follows, by an inference of ours, that if F were amenable then Theorem B would hold. The converse does not follow, and the second deep dive explains why.

Before the wrap up, two questions. Pause after each one and answer it in your head. First. Why can no ordinary probability distribution on the positive integers equal its own convolution?

The answer. Let m be the smallest value it can take. A sum of two such values is at least two m, which is larger than m. So the sum never equals m, and the two laws differ at m.

Second question. Deleting a middle point of the walk uses idempotence. Deleting the last point does not. Why?

The answer. A middle point has a step into it and a step out of it, and removing the point merges them into a single step whose law is the mean convolved with itself. The last point has no step out. Its step is simply dropped, and nothing needs to merge.

One more result. Move from finite sets to infinite sequences, and let the i th map delete the i th coordinate and shift the rest left.

The paper proves that a probability law invariant under all of these maps, and ergodic, must be independent and identically distributed. Our survey found that this follows from two classical theorems on exchangeable sequences, one by Ryll Nardzewski and one by de Finetti. The paper calls its result a de Finetti type theorem and cites neither. Its proof is short and uses only the mean ergodic theorem.

Two deep dives follow. The first is the proof of Theorem A: means, why the order of integration matters, the three steps of the walk theorem, and the two approximation steps.

The second is the algebra: normal forms, why the finite pieces are non-amenable, Day's theorem, and the story of Thompson's group.