# Preprint 'On strongly summable ultrafilters'

18 Jun 2010I have just uploaded a preprint titled On strongly summable ultrafilters to the arXiv. Let me give a short account of what it’s about.

In the preprint I extend a theorem orginally due to Neil Hindman and Dona Strauss. The theorem shows that certain ultrafilters can only be written as sums in the (so-called) trivial fashion. On the one hand, this property is quite unique and I find it algebraically fascinating. On the other, the existence of the ultrafilters in question is independent of ZFC, so set theoretic interests are immediate. Let’s start with the ultrafilters.

## Strongly summable ultrafilters

Among the idempotent ultrafilters on $\mathbb{N}$ a certain type is quite special. To define it, we only need to know what an *FS-set* is.

For a sequence $\mathbf{x}$ in $\mathbb{N}$, the set of all (distinct) finite sums is

Then $p\in \beta \mathbb{N}$ is called **strongly summable** if it has a base of FS-sets.

By the Galvin-Glazer Theorem, any set in an idempotent ultrafilter contains an FS-set. The difference to arbitrary idempotent ultrafilters is that strongly summables have such a set in the ultrafilter itself! So in measure theoretic terms, not only does every measure 1 set contain an FS-set, but one of measure 1. As a comparison, the very important *minimal* idempotents can never be strongly summable; in fact strongly summable ultrafilters are at the other end of the spectrum — they are what is called strongly right maximal idempotents.

In other words, what selective ultrafilters are for Ramsey’s Theorem, strongly summable ultrafilters are for Hindman’s Theorem. Unlike idempotent ultrafilters strongly summable ultrafilters might not exist, since their existence implies the existence of P-points in $\beta \mathbb{N}$. But under rather weak assumptions (like CH or weak forms of Martin’s Axiom) they do exist.

## Union Ultrafilters

There is an equivalent notion, union ultrafilters, see here. In the preprint, as always, it’s hard to speak about the one without the other.

Denote the non-empty, finite subsets of $\omega$ by $\mathbb{F}$ (with semigroup operation $\cup$). In what follows a sequence $\mathbf{s}$ in $\mathbb{F}$ is always assumed to have pairwise disjoint elements.

The **FU-set (generated by $\mathbf{s}$)** is the set of all finite unions, i.e.,

A **union ultrafilter** on $\mathbb{F}$ is an ultrafilter with a base of FU-sets.

### Trivial sums

As mentioned, the main result in the preprint is about writing ultrafilters as sums. Any idempotent ultrafilter can be written as a sum in many ways, most trivially as $p+ p = p$. However, since $\beta \mathbb{N}$ is a left ideal in $\beta \mathbb{Z}$, there is another way by means of integers.

If $p \in \beta \mathbb{N}$ is idempotent and $z \in \mathbb{Z}$, then

This is simply due to the fact that the integers still commute with everything in $\beta \mathbb{Z}$.

Let’s say that an (idempotent) ultrafilter $p$ has **the trivial sums property** if this is the only way to write is as a sum, i.e.,

This property fascinates me for many reasons. One easy but equally fascinating consequence is that the maximal group of such an idempotent $p$, i.e., the maximal subgroup of $\beta \mathbb{N}$ with identity $p$ (which is the same as the union of all such subgroups), is just $p + \mathbb{Z}$. This is fascinating since it is the minimal case — the famous Theorem by Yevhen Zelenyuk showed that there are no finite subgroups in $\beta \mathbb{N}$.

In contrast, minimal idempotent ultrafilters have huge maximal groups. In fact, those always include a copy of the free group on $2^{2^\omega}$-many generators! That is mind-bogglingly big, so in comparison the integers are really very minimal. And on top of everything it is also open whether any ultrafilters with a maximal group isomorphic to $\mathbb{Z}$ exist in ZFC…

### Using union ultrafilters

I gave a simplified version of the main theorem here. For this consider the very natural map $f: \mathbb{F} \rightarrow \mathbb{N}, s \mapsto \sum_{i \in s} 2^i$.

**Theorem** If $u$ is a union ultrafilter, then $f(u)$ has the trivial sums property.

In fact, the theorem in the preprint does a bit more. You see, such an $f(u)$ contains $ FS ( \mathbf{x} ) $ for a sequence $\mathbf{x}$ with **disjoint binary support** — simply because that’s what happens to pairwise disjoint sets under $f$.

Now if you take any other divisible sequence $\mathbf{a}$, i.e., with $a_n \vert a_{n+1}$ for all $n$ (and assume for convenience that $a_0 = 1$), then just as in the binary situation, we have a unique representation for every natural number by means of $\mathbf{a}$.

Then we can formulate the full theorem from the preprint.

**Theorem** If $p$ is strongly summable and contains an FS-set for a sequence with disjoint support in the $\mathbf{a}$-representation, then $p$ has the trivial sums property.

## About the result

The original theorem by Neil Hindman and Dona Strauss [doi] (also to be found in their book, chapter 12) showed that the trivial sums property holds for strongly summable ultrafilters with special properties. These included the restriction that the strongly summable ultrafilters must have a base of FS-sets coming from divisible sequences. In particular, these strongly summables will contain the FS-set for a sequence with disjoint support for a divisible sequence (just for the sequence itself), i.e., the result in the preprint entails the original result.

To see that it is a little bit more general, you need to know another concept, *additive isomorphism*. To keep it short, let’s just state it for strongly summable ultrafilters.

Two strongly summable ultrafilters $p_0, p_1$ are **additively isomorphic** if there are $ {FS(} \mathbf{x} ) \in p_ 0$, $ {FS(} \mathbf{y} ) \in p_ 1$, such that the map $ \sum_ {i\in s} x_ i \mapsto \sum_ {i \in s} y_ i$ is well defined and maps $p_0$ to $p_1$ (so in particular, $p_ 0 \cong p_ 1$ in the usual sense).

In their original paper [doi], Neil Hindman and Dona Strauss give a beautiful example of a strongly summable ultrafilter that does not concentrate on FS-sets from divisible sequences (this is done using unordered union ultrafilters) — and no additively isomorphic copy does. However, every strongly summable is additively isomorphic to one as in the result from the preprint. Hence the result is a little bit stronger.

(Un)fortunately, the trivial sums property is not transferred via an additive isomorphism, so one open question is whether all strongly summable ultrafilters have it. Although I included some evidence in the preprint towards believing that all strongly summables have it, I hope that it is not the case, simply because I hope that a counterexample would be a new kind of strongly summable ultrafilter. Of course, the big question is whether any other idempotents have this property! In particular, if there might exist such examples in ZFC.

## About the proof

The proof in the preprint follows the same strategy as the proof of the original theorem by Neil Hindman and Dona Strauss. So the main task (or rather the initial observation that got me thinking about their proof again) was to overcome some of the restrictions on the strongly summable ultrafilters. Besides the ones already mentioned, there was another technical condition in the original result, a strange combinatorial property which I prefer to think about in terms of union ultrafilters.

An ultrafilter $u$ on $\mathbb{F}$ is **special** if for every infinite $L\subseteq \omega$ there exists $X \in u$ such that $L \setminus \bigcup X$ is infinite.

In other words, $u$ is able to ‘miss’ any infinite set by a large amount. The counterpart for strongly summable ultrafilters (which was already considered by Hindman and Strauss) is more complicated to write down, but suffice it to say that the two are equivalent. The key is that for union ultrafilters this property is not special at all.

**Theorem** Every union ultrafilter is special.

The proof is not very complicated, it involves a standard parity argument that often appears when arguing with union ultrafilters (and then you just have to look carefully to see that you’re done).

But now this strange and difficult extra condition from the original result by Neil Hindman and Dona Strauss is suddenly available for free.

The final result is derived in essentially two steps. There is a bit of technicality involved, but I think nothing really complicated (to read).

Fixing some divisible sequence $\mathbf{a}$, let’s call the support of a number in the representation via $\mathbf{a}$ the **$\mathbf{a}$-support**. Then it roughly proceeds as follows.

- Show trivial sums for the ultrafilters that contain the set of multiples for each member of $\mathbf{a}$.
- This essentially reflects from a simple observation: if two elements with $\mathbf{a}$-support sufficiently far apart have their sum in an FS-set with disjoint $\mathbf{a}$-support, then they must both be in the FS-set already.

- Finally show that any sum equal to the strongly summable is essentially only possible by integer translates of ultrafilters that contain these sets of multiples.

In this tricky final part the specialness condition comes into play. It is used to create an FS-set that has a lot of ‘holes’ in the $\mathbf{a}$-support. This makes it impossible for a single number to translate infinitely many numbers into that FS-set (as happens in a sum of ultrafilters all the time) simply because you would need too much ‘carrying over’ when translating/adding.

Well, of course that’s only a vague description of the proof, but the preprint should have enough details.

## Questions

The most important questions I have already asked. Does every strongly summable have the trivial sums property and are their other ultrafilters with it, maybe even in ZFC? A result by Neil Hindman and Dona Strauss shows that all strongly summable ultrafilters have a maximal group isomorphic to the integers. It is still unkown if any other ultrafilters carry this property. So I think there is still a lot to be learned about this property.