This is the fifth in a series of six posts I am writing about Szemerédi’s theorem. In the previous post I proved that the Sz property lifts through weak mixing extension and in this post I will prove that the Sz property lifts through compact extensions.

To briefly recall our setup, is a measure preserving system, a -subalgebra is called a *factor* if for every also ; and we say that a factor is Sz if for every nonnegative bounded function measurable in satisfying and every we have

If are factors of we call the pair an *extension*. This extension is called compact if for every we can remove a set of arbitrarily small measure from its support in a way that for every there exists and functions such that for every and there exists such that

All this notions are better explained in the previous posts on this series.

The purpose of this post is to prove the following theorem:

Theorem 1If is a compact extension of a measure preserving system and is Sz, then also is Sz.

I will present one and a half proofs of this theorem. The first (one half) uses the van der Waerden theorem and is somewhat easier. It has the advantage that it was extended to prove polynomial Szemerédi’s theorem as well as other generalizations. The second proof, obtained by Furstenberg, Katznelson and Ornstein, avoids the use of the coloristic result, while also giving uniform limits.

The first proof is just one half proof because it does not give uniform Cesàro limits, (although one can get uniform limits even when using van der Waerden’s theorem, that is exactly what Furstenberg achieved in his original proof). For this reason, the first proof is not necessary in the proof of Szemerédi’s theorem, or if one is only interested in Szemerédi’s theorem, but not in getting uniform limits, then the second proof is not necessary.

In a vague way, the first proof uses the fact that multiplying an arithmetic progression by a constant gives another arithmetic progression (i.e. arithmetic progressions are invariant under multiplication), and the second proof uses the fact that adding an arithmetic progression with a constant gives another arithmetic progression (i.e. arithmetic progressions are invariant under addition).

** — 1. First proof, using van der Waerden’s theorem — **

In this section we assume that we are under the assumptions of Theorem 1. To prove this theorem we will need the van der Waerden Theorem to which I presented a proof in a previous post. Recall that .

Theorem 2(van der Waerden) Let . There exists some such that for any coloring of in colors there exists a monochromatic arithmetic progression of length .

Let such that and ; and let . We will show that

The first step is to perform some simplifications. Let be such that if has then and let be such that and is conditionally compact. Note that if (1) holds for , since then it also holds for . Therefore we can (and will) assume, without loss of generality that is itself conditionally compact. We can also assume that after multiplying it, if necessary, by a constant.

Let and denote by . Note that also and a.e.(to see why we have such bound consider the set and use the definition of conditional expectation). Let .

Applying the compactness hypothesis we can find and be such that for each we have

Let . Observe that , hence . Now apply the van der Waerden Theorem to find so that each -coloring of contains an arithmetic progression of length . By assumption we get that satisfies (1) with instead of . Let be the in (1) with and , let and let . We have

and thus

Interpreting what we have so far, for each and , we have many (actually ) points in the orbit of that fall inside , and hence in that portion of the orbit is bounded away from .

Now fix , let and consider the -coloring of induced by the orbit . More precisely the coloring is the map from to some that satisfies . By the choice of we can find some and such that and for each . Note that for each fixed (and ), the set of those such that for all is in , so we can choose the functions to be measurable on .

Since there are only at most possible ways to choose , we can find a subset in such that and some constants and such that

Using the conditional triangular inequality we also have that

Finally note that for each we also have that .

We recall this elementary identity, already used in the previous post:

This time we will take (yes, this is independent of ) and . Now using the Conditional Cauchy-Schwartz inequality we can bound (the conditional expectation of) each of the products inside the sum. Recall that and that the norm is bounded by the norm. Let and let .

Using Jensen inequality for conditional expectation we can get a lower bound on the first product in the identity (2):

We now put together the estimates into (2). Using the triangular inequality and recalling the condition on we get

Since and we deduce from this that

Recall that was defined such that . Thus for every there exists some such that

Note that while depends on , there are only finitely many choices for .

Taking the Cesàro average we conclude that

** — 2. Second proof, with uniform limits — **

In this section I present another proof of Theorem 1. This proof was obtained by Furstenberg, Katznelson and Ornstein and avoids the use of van der Waerden’s theorem. While this doesn’t seem to be very important, since van der Waerden’s theorem is significantly easier to prove that Szemerédi’s theorem, this allows more flexibility when trying to mimic this proof to other situations where a coloring result may not be available. Besides, this second proof can be applied to uniform Cesàro limits.

Let be nonnegative and such that . We can assume without loss of generality that is itself conditionally compact, as done in the first proof. Thus we know that the orbit is totally bounded in the norm. In particular, for any fixed , almost every and , if is large enough subset, than for some distinct we have . We will carefully choose some finite subset which does *not* satisfy this property and is maximal. More precisely, given (and keeping fixed), let

For instance, if is a singleton, then . Note that . Intuitively speaking, is the set of points for which is not “good enough” to cover the orbit of . We want a set of points for which is not good enough, but if we add any other integer to , than the new set is “good”. Thus we define:

Lemma 3Almost every is in some

*Proof:* We now will make precise the claims above. Let and be such that for each and almost every there is some such that

Let and be a finite subset. For each and there exists some that satisfies (4) with (at least for almost every , we will ignore the set of measure where this may not occur). Thus we can associate each with a sequence in . Since there are only possible such sequences, if there must be some pair with the same sequence. Using the conditional triangular inequality we conclude that for any we have .

This means that, if , the set has measure . Since whenever is a singleton, we deduce that every is in some .

Let and denote by . Also, let . Let . Observe that , hence . Since there are only countably many finite subsets of , we can find a finite set such that . Since , for all and each pair there exists some such that . Thus we can define a function such that is such that . The point here is that has a finite image, and hence there exists a set o positive measure and a map such that for all we have and hence is independent of .

Lemma 4Let and (in particular assume this intersection is non-empty). Then there exists s.t. we have .

*Proof:* Since for each we have , we have that for any pair there exists independent of such that and hence . In particular we can take and we have

On the other hand, , so for any with . We will make . If , then and hence, taking , the conclusion of the lemma is trivially true. Otherwise and hence there exists some pair such that for all we have

If neither nor equal , then and with . But this would contradict (5). Thus either or and we are done.

Now we can finish the proof. By assumption we get that satisfies (1). Let be the in (1) with , let and let . We have

and thus

For each and we can apply Lemma 4 and find . We now use identity 2 with and , and arguing as above we have

While depends on , it can only take finitely many possible values. Thus this dependency is lost if we simply add over all . Since and we can then integrate and deduce that

Thus we showed that for each there is some such that “gives” multiple recurrence for . Let and . Let be such that . Then we have

Finally, taking the uniform Cesàro limit we conclude: