I have just uploaded to the arXiv the paper “An inverse theorem for an inequality of Kneser“, submitted to a special issue of the Proceedings of the Steklov Institute of Mathematics in honour of Sergei Konyagin. It concerns an inequality of Kneser discussed previously in this blog, namely that

whenever are compact non-empty subsets of a compact connected additive group with probability Haar measure . (A later result of Kemperman extended this inequality to the nonabelian case.) This inequality is non-trivial in the regime

The connectedness of is essential, otherwise one could form counterexamples involving proper subgroups of of positive measure. In the blog post, I indicated how this inequality (together with a more “robust” strengthening of it) could be deduced from submodularity inequalities such as

which in turn easily follows from the identity and the inclusion , combined with the inclusion-exclusion formula.

In the non-trivial regime (2), equality can be attained in (1), for instance by taking to be the unit circle and to be arcs in that circle (obeying (2)). A bit more generally, if is an arbitrary connected compact abelian group and is a non-trivial character (i.e., a continuous homomorphism), then must be surjective (as has no non-trivial connected subgroups), and one can take and for some arcs in that circle (again choosing the measures of these arcs to obey (2)). The main result of this paper is an inverse theorem that asserts that this is the only way in which equality can occur in (1) (assuming (2)); furthermore, if (1) is close to being satisfied with equality and (2) holds, then must be close (in measure) to an example of the above form . Actually, for technical reasons (and for the applications we have in mind), it is important to establish an inverse theorem not just for (1), but for the more robust version mentioned earlier (in which the sumset is replaced by the partial sumset consisting of “popular” sums).

Roughly speaking, the idea is as follows. Let us informally call a *critical pair* if (2) holds and the inequality (1) (or more precisely, a robust version of this inequality) is almost obeyed with equality. The notion of a critical pair obeys some useful closure properties. Firstly, it is symmetric in , and invariant with respect to translation of either or . Furthermore, from the submodularity inequality (3), one can show that if and are critical pairs (with and positive), then and are also critical pairs. (Note that this is consistent with the claim that critical pairs only occur when come from arcs of a circle.) Similarly, from associativity , one can show that if and are critical pairs, then so are and .

One can combine these closure properties to obtain further ones. For instance, suppose is such that . Then (cheating a little bit), one can show that is also a critical pair, basically because is the union of the , , the are all critical pairs, and the all intersect each other. This argument doesn’t quite work as stated because one has to apply the closure property under union an uncountable number of times, but it turns out that if one works with the robust version of sumsets and uses a random sampling argument to approximate by the union of finitely many of the , then the argument can be made to work.

Using all of these closure properties, it turns out that one can start with an arbitrary critical pair and end up with a small set such that and are also critical pairs for all (say), where is the -fold sumset of . (Intuitively, if are thought of as secretly coming from the pullback of arcs by some character , then should be the pullback of a much shorter arc by the same character.) In particular, exhibits linear growth, in that for all . One can now use standard technology from inverse sumset theory to show first that has a very large Fourier coefficient (and thus is biased with respect to some character ), and secondly that is in fact almost of the form for some arc , from which it is not difficult to conclude similar statements for and and thus finish the proof of the inverse theorem.

In order to make the above argument rigorous, one has to be more precise about what the modifier “almost” means in the definition of a critical pair. I chose to do this in the language of “cheap” nonstandard analysis (aka asymptotic analysis), as discussed in this previous blog post; one could also have used the full-strength version of nonstandard analysis, but this does not seem to convey any substantial advantages. (One can also work in a more traditional “non-asymptotic” framework, but this requires one to keep much more careful account of various small error terms and leads to a messier argument.)

*[Update, Nov 15: Corrected the attribution of the inequality (1) to Kneser instead of Kemperman. Thanks to John Griesmer for pointing out the error.]*

## 13 comments

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15 November, 2017 at 2:14 pm

nicodeanReference 12 says “preprint”, but doesnt give a link to the preprint.

[This will be corrected in the next revision of the ms – T.]16 November, 2017 at 12:03 am

hxypqrCould this result be looked as a continuous version generate of “Cauchy-Davenport theorem”, i.e. for any nonempty set ?

16 November, 2017 at 9:03 am

Terence TaoYes, in particular one can deduce the case of Kneser’s inequality from the Cauchy-Davenport inequality by a discretisation argument.

16 November, 2017 at 3:32 am

AnonymousThe inverse theorem for the case G=R/Z was proved earlier by Candela and Roton? See Theorem 1.5 in their paper:

https://arxiv.org/pdf/1709.04501.pdf

16 November, 2017 at 9:03 am

Terence TaoThanks for this reference. It seems that I did not do an adequate literature search before finalising this paper; I will revise it to add in a number of additional references including this one.

16 November, 2017 at 7:12 am

AnonymousDear Terence Tao,

This year 2017 is coming to end.We have one more month.I have been waiting for you for over 10 years with your strongest and greatest result.I am very very far away you,but I feel to be near you every day.Math community has recently had many shocks of the Field medalist’s death.Life is short.Mathematician or vendor,farmer,finally they die for karma.when they pass away,we sometime remind them on book,or magazine or anniversary within 15minutes.that’s all.We turn back the land ,air,water.The Tao is greater than science.But ,you are lucky to have these both.So,I think you are able to do more than other people.Many times I think that you can solve to 100 best problems ,not simply 6 best problems.You certainly succeed and that time your name is the second sun.Once you goodbye forever to this world, your name is sacared to students in all exams,your name is forever

16 November, 2017 at 9:34 am

Anonymousmissing “paper” tag

[Added, thanks – T.]16 November, 2017 at 11:50 am

wanzgProf Tao,

You might also change one of tags: Kemperman’s inequality

[Changed, thanks – T.]16 November, 2017 at 2:29 pm

AnonymousP. 13, l. 4: woudl –> would

[Thanks, this will be corrected in the next revision of the ms. -T]20 November, 2017 at 2:44 pm

AnonymousEquation tag in the blog post: “3)” –> “(3)”

[Corrected, thanks – T.]23 November, 2017 at 8:11 am

AnonymousCorrections to the bibliography:

[2]: “343353” –> “343–353”

[9]: remove “p.”

[15]: “49, (1953) 40”–> “49, (1953), 40”

[16]: “1973) 148” –> “1973), 148”

[18]: “(1939). 425” –> “(1939), 425”

[22]: “(1955) 57” –> “(1955), 57”

[Thanks, this will be corrected in the next version of the ms. -T]10 December, 2017 at 9:52 am

Not CharlieWhat’s that mu operator do

[ is a measure. -T]10 December, 2017 at 3:32 pm

H.H.Holmesimagine that the golbach’s conjectur say that for any positive integer n, and π(2n) random positive integer numbers less than 2n, there are at least two numbers which the sum of them is 2n. even the numbers are not certainly prime, but the probability of existing the numbers is very close to 1. for example we have π(500) different number less than 500. then the probability of existing two of the numbers with sum of 2n is :

99.99999999999999999999479116…

and for bigger numbers it is even closer to 1.

So □