A core foundation of the subject now known as arithmetic combinatorics (and particularly the subfield of *additive combinatorics*) are the elementary sum set estimates (sometimes known as “Ruzsa calculus”) that relate the cardinality of various sum sets

and difference sets

as well as iterated sumsets such as , , and so forth. Here, are finite non-empty subsets of some additive group (classically one took or , but nowadays one usually considers more general additive groups). Some basic estimates in this vein are the following:

Lemma 1 (Ruzsa covering lemma)Let be finite non-empty subsets of . Then may be covered by at most translates of .

*Proof:* Consider a maximal set of disjoint translates of by elements . These translates have cardinality , are disjoint, and lie in , so there are at most of them. By maximality, for any , must intersect at least one of the selected , thus , and the claim follows.

Lemma 2 (Ruzsa triangle inequality)Let be finite non-empty subsets of . Then .

*Proof:* Consider the addition map from to . Every element of has a preimage of this map of cardinality at least , thanks to the obvious identity for each . Since has cardinality , the claim follows.

Such estimates (which are covered, incidentally, in Section 2 of my book with Van Vu) are particularly useful for controlling finite sets of small doubling, in the sense that for some bounded . (There are deeper theorems, most notably Freiman’s theorem, which give more control than what elementary Ruzsa calculus does, however the known bounds in the latter theorem are worse than polynomial in (although it is conjectured otherwise), whereas the elementary estimates are almost all polynomial in .)

However, there are some settings in which the standard sum set estimates are not quite applicable. One such setting is the continuous setting, where one is dealing with bounded open sets in an additive Lie group (e.g. or a torus ) rather than a finite setting. Here, one can largely replicate the discrete sum set estimates by working with a Haar measure in place of cardinality; this is the approach taken for instance in this paper of mine. However, there is another setting, which one might dub the “discretised” setting (as opposed to the “discrete” setting or “continuous” setting), in which the sets remain finite (or at least discretisable to be finite), but for which there is a certain amount of “roundoff error” coming from the discretisation. As a typical example (working now in a non-commutative multiplicative setting rather than an additive one), consider the orthogonal group of orthogonal matrices, and let be the matrices obtained by starting with all of the orthogonal matrice in and rounding each coefficient of each matrix in this set to the nearest multiple of , for some small . This forms a finite set (whose cardinality grows as like a certain negative power of ). In the limit , the set is not a set of small doubling in the discrete sense. However, is still close to in a metric sense, being contained in the -neighbourhood of . Another key example comes from graphs of maps from a subset of one additive group to another . If is “approximately additive” in the sense that for all , is close to in some metric, then might not have small doubling in the discrete sense (because could take a large number of values), but could be considered a set of small doubling in a discretised sense.

One would like to have a sum set (or product set) theory that can handle these cases, particularly in “high-dimensional” settings in which the standard methods of passing back and forth between continuous, discrete, or discretised settings behave poorly from a quantitative point of view due to the exponentially large doubling constant of balls. One way to do this is to impose a translation invariant metric on the underlying group (reverting back to additive notation), and replace the notion of cardinality by that of metric entropy. There are a number of almost equivalent ways to define this concept:

Definition 3Let be a metric space, let be a subset of , and let be a radius.

- The
packing numberis the largest number of points one can pack inside such that the balls are disjoint.- The
internal covering numberis the fewest number of points such that the balls cover .- The
external covering numberis the fewest number of points such that the balls cover .- The
metric entropyis the largest number of points one can find in that are -separated, thus for all .

It is an easy exercise to verify the inequalities

for any , and that is non-increasing in and non-decreasing in for the three choices (but monotonicity in can fail for !). It turns out that the external covering number is slightly more convenient than the other notions of metric entropy, so we will abbreviate . The cardinality can be viewed as the limit of the entropies as .

If we have the bounded doubling property that is covered by translates of for each , and one has a Haar measure on which assigns a positive finite mass to each ball, then any of the above entropies is comparable to , as can be seen by simple volume packing arguments. Thus in the bounded doubling setting one can usually use the measure-theoretic sum set theory to derive entropy-theoretic sumset bounds (see e.g. this paper of mine for an example of this). However, it turns out that even in the absence of bounded doubling, one still has an entropy analogue of most of the elementary sum set theory, except that one has to accept some degradation in the radius parameter by some absolute constant. Such losses can be acceptable in applications in which the underlying sets are largely “transverse” to the balls , so that the -entropy of is largely independent of ; this is a situation which arises in particular in the case of graphs discussed above, if one works with “vertical” metrics whose balls extend primarily in the vertical direction. (I hope to present a specific application of this type here in the near future.)

Henceforth we work in an additive group equipped with a translation-invariant metric . (One can also generalise things slightly by allowing the metric to attain the values or , without changing much of the analysis below.) By the Heine-Borel theorem, any precompact set will have finite entropy for any . We now have analogues of the two basic Ruzsa lemmas above:

Lemma 4 (Ruzsa covering lemma)Let be precompact non-empty subsets of , and let . Then may be covered by at most translates of .

*Proof:* Let be a maximal set of points such that the sets are all disjoint. Then the sets are disjoint in and have entropy , and furthermore any ball of radius can intersect at most one of the . We conclude that , so . If , then must intersect one of the , so , and the claim follows.

Lemma 5 (Ruzsa triangle inequality)Let be precompact non-empty subsets of , and let . Then .

*Proof:* Consider the addition map from to . The domain may be covered by product balls . Every element of has a preimage of this map which projects to a translate of , and thus must meet at least of these product balls. However, if two elements of are separated by a distance of at least , then no product ball can intersect both preimages. We thus see that , and the claim follows.

Below the fold we will record some further metric entropy analogues of sum set estimates (basically redoing much of Chapter 2 of my book with Van Vu). Unfortunately there does not seem to be a direct way to abstractly deduce metric entropy results from their sum set analogues (basically due to the failure of a certain strong version of Freiman’s theorem, as discussed in this previous post); nevertheless, the proofs of the discrete arguments are elementary enough that they can be modified with a small amount of effort to handle the entropy case. (In fact, there should be a very general model-theoretic framework in which both the discrete and entropy arguments can be processed in a unified manner; see this paper of Hrushovski for one such framework.)

It is also likely that many of the arguments here extend to the non-commutative setting, but for simplicity we will not pursue such generalisations here.

** — 1. Approximate groups — **

In discrete sum set theory, a key concept is that of a *-approximate group* – a finite symmetric subset of containing the origin such that is covered by at most translates of . The analogous concept here will be that of a *-approximate group*: a precompact symmetric subset of containing the origin such that is covered by at most copies of . Such sets obey good iterated doubling properties; for instance, is covered by at most copies of for any . They can be generated from sets of small tripling:

Lemma 6Let be a precompact non-empty subset of , and let . If or , then is a -approximate group.

*Proof:* From Lemma 5 we have

(for an appropriate choice of sign) and

and thus by Lemma 4, may be covered by at most copies of , giving the claim.

** — 2. From small doubling to small tripling or quadrupling — **

As we saw above, Lemma 5 and Lemma 4 are already very powerful once one has some sort of control on triple or higher sums such as , , or . But if only controls a double sum such as or , it is a bit trickier to proceed. Here is one estimate (somewhat analogous to Proposition 2.18 from my book with Van Vu, but with slightly worse numerology):

Lemma 7Let be a precompact non-empty subset of , and let . If , then .

One can combine this lemma with Lemma 5 to obtain similar conclusions starting with a hypothesis on rather than ; we leave this to the interested reader. Of course, the conclusion can also be combined with Lemma 6; we again leave this as an exercise.

*Proof:* Write . Then , so we may find a -separated subset of with . By hypothesis, we may cover by balls with . Call a centre *popular* if contains at least differences with (counting multiplicity), and let denote the set of popular centres. Then at most of the pairs have lying in an unpopular ball , thus we have for at least pairs in . Thus, by the pigeonhole principle, there exists such that at least elements of lie in . Thus

and thus

Next, for any , we consider the set of pairs such that . We may write

for some and . By definition of , we can find distinct pairs for with such that for all . As is -separated and has diameter at most , the must be distinct in , and similarly for the . We then have

for . Each lies in and is thus lies in for some , and similarly for some . Then

and so . Also, since and the are -separated, we see that the are distinct as varies. We conclude that . On the other hand, the total number of pairs is , and any two -separated points in generate disjoint sets . We conclude that there can be at most -separated points in , thus

and thus

By Lemma 5 and (1), we conclude that

and the claim follows.

** — 3. The Balog-Szemeredi-Gowers lemma — **

One of the most difficult, but powerful, components of the elementary sum set theory is the tool now known as the *Balog-Szemerédi-Gowers lemma*, which converts control on partial sumsets (or equivalently, lower bounds on “additive energy”) to control on total sumsets, after suitable refinements of the sets. Here is one metric entropy version of this lemma.

Lemma 8 (Balog-Szemerédi-Gowers)Let and , and let be precompact subsets of . Suppose thatand

where we endow with the sup norm metric. Then there exist subsets , of respectively with

and

Again, this lemma may be usefully combined with the previous sum set estimates, much as was done in my book with Van Vu; I leave the details to the interested reader.

*Proof:* Let , be maximal -separated subsets of respectively, thus , and similarly .

By hypothesis, we can find at least quadruples which are -separated in , such that

for all such quadruples. By construction of , each such quadruple can be associated to a nearby quadruple with

and thus by the triangle inequality

Also by the triangle inequality we see that each can be associated to at most one of the quadruples , and as the are -separated, the are -separated. We conclude that there is a set of at least quadruples in obeying (2) that are -separated.

Call a pair *popular* if there are at least of the above quadruples in obeying (2) with the indicated first two coefficients. The unpopular pairs absorb at most of the quadruples, so at least of the quadruples are associated to popular pairs . On the other hand, as the quadruples are -separated, we see from (2) and the triangle inequality that for each there is at most one giving rise to a quadruple . Thus each can be associated to at most quadruples, and we conclude that the set of popular pairs has size at least . In particular this shows that .

We now apply the graph-theoretic Balog-Szemerédi-Gowers lemma (see Corollary 6.19 of my book with Van Vu) to conclude that there exists a subset of and of with

such that for every and there exist pairs such that lie in . Since was already -separated, we conclude that

Now fix and , and let be one of the above pairs. As is popular, we can thus find pairs such that

furthermore, the lie in an -separated set. Similarly, we can find pairs and pairs such that

with the and also lying in an -separated set. In particular, we see that given , uniquely determine , and uniquely determine , so a single sextuple can arise from at most one pair ; in particular, we see that sextuples are associated to each pair . Taking alternating combinations of (3), (4), (5) we see that

In particular, if and are two pairs in with at least apart, then a single sextuple can be associated to at most one of these pairs. Since the number of sextuples is at most , we conclude that there are at most pairs with -separated differences , thus as required.

## 17 comments

Comments feed for this article

19 March, 2014 at 6:44 pm

AnonymousIs there a typo in the definition of difference sets?

[Corrected, thanks - T.]19 March, 2014 at 10:18 pm

domotorpIn proof of Lemma 1, {|A|} should be {|B|}.

[Corrected, thanks - T.]19 March, 2014 at 11:50 pm

domotorpAnother typo above Def 3: odubling

[Corrected, thanks - T.]20 March, 2014 at 12:26 am

domotorpAnd at the end of Def 3 there is an extra {itemize}

[Corrected, thanks - T.]20 March, 2014 at 12:30 am

domotorpClosing parenthesis missing from para above lemma 4.

[Corrected, thanks - T.]19 March, 2014 at 11:56 pm

domotorpThe internal covering number, {N^{int}_r(E)}, might decrease in E – consider r=3 and E={0,2,4} and E’={0,4}.

[Corrected, thanks - T.]20 March, 2014 at 4:55 am

kyleI think there are some typos in the proof of lemma 1. For example, I think the maximal set that is chosen should be a subset of A, not a subset of B as stated. I also agree with the earlier comment by domotorp that the cardinality of each of these translates is |B|, not |A|.

20 March, 2014 at 4:56 am

kyleSorry, didn’t mean to post a reply under this comment.

20 March, 2014 at 4:54 am

chorasimilarityKind of being in this frame of “discretized”, but not discrete, nor continuous, in arXiv:1304.3358 there is a very simple note about a “geometric Ruzsa triangle intequality”, which does not even use group operations, maybe is of some interest here.

20 March, 2014 at 1:06 pm

PascalIs there a version of the basic results of additive combinatorics for multisets rather than sets, or even better for multisets where negative cardinalities are allowed?

If this looks like a baroque idea, note that there is a quite natural interpretation: if c_a denotes the mulitplicity of element a, we can represent a multiset by the polynomial sum_a c_a X^a. Taking a sum set A+B then amounts to multiplying the 2 corresponding polynomials.

20 March, 2014 at 3:46 pm

Terence TaoThere is a closely analogous theory to discrete sumset theory for random variables, with Shannon entropy taking the place of cardinality: see http://www.ams.org/mathscinet-getitem?mr=2647496 . For signed functions, the appropriate tool is Fourier analysis and convolution, for which there are certainly a lot of estimates available (though once one gives up positivity, many of the more combinatorial estimates in the sumset theory no longer have good analogues).

27 March, 2014 at 2:35 am

Geoff SangHi Terence,

I am a year 9 student from Western Australia, and I find your blog very fascinating. The school maths I do are easy however I know that there are many more students better than I am.

I have never done complicated maths before like the Australian Maths Olympiad (unlike you when you were my age), but now I am starting to learn the fundamentals. Proof has always been an obstacle when I do competitions (e.g. The Western Australian Junior Mathematical Olympiad). How could I practice proof and also understand a question better when doing competitions?

Thank you,

Geoff Sang

21 March, 2014 at 5:05 pm

arch1In the proof of Lemma 1 should “lie in |A + B|” be “lie in A + B”?

[Corrected, thanks - T.]25 March, 2014 at 5:59 am

arch1In the orthogonal nxn matrix example: “…let A be the matrices obtained by starting with an orthogonal matrix in On(R) and rounding each coefficient to the nearest multiple of epsilon…. This is a finite set (whose cardinality grows as [epsilon goes to 0] like a certain negative power of epsilon).”

The above recipe seems to me to produce exactly one matrix, regardless of epsilon’s value, so I must be missing something real basic. Can someone help?

[Reworded for clarity - T.]28 March, 2014 at 8:09 am

MrCactu5 (@MonsieurCactus)It looks like you are describing some sort of generalized Hamming Code? For noisy channel coding the “buzzwords” should be a certain Hammin distance apart.

At this point I am taking a step back and reading through your book with Van Vu.

29 May, 2014 at 9:15 am

AnonymousProf. Tao,

In general, we can not replace by in Ruzsa’s covering lemma without some sort of loss. That is showed in an exercise in the book of you and Prof. Vu. If are finite subsets of integers, is it true that we can cover by at most () translates of ? I did some simulations, and no counterexample was found.

Best,

Jiange

29 May, 2014 at 11:57 am

Terence TaoUnfortunately not; one can take a counterexample in a finite group such as (e.g. the one in Exercise 2.4.1 of my book with Van), and “transfer” it to the integers by considering a set such as and , where is much larger than . This modifies the covering number by a factor of 2 or so, but the failure in of the naive Ruzsa covering lemma is by an arbitrarily large constant, so it can also create an arbitrarily large failure of that covering lemma in the integers.