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The polymath1 project has just uploaded to the arXiv the paper “A new proof of the density Hales-Jewett theorem“, to be submitted shortly. Special thanks here go to Ryan O’Donnell for performing the lion’s share of the writing up of the results, and to Tim Gowers for running a highly successful online mathematical experiment.

I’ll state the main result in the first non-trivial case for simplicity, though the methods extend surprisingly easily to higher (but with significantly worse bounds). Let be the size of the largest subset of the cube that does not contain any combinatorial line. The density Hales-Jewett theorem of Furstenberg and Katznelson shows that . In the course of the Polymath1 project, the explicit values

were established, as well as the asymptotic lower bound

(actually we have a slightly more precise bound than this). The main result of this paper is then

Theorem.( version) .

Here is the inverse tower exponential function; it is the number of times one has to take (natural) logarithms until one drops below 1. So it does go to infinity, but extremely slowly. Nevertheless, this is the first explicitly quantitative version of the density Hales-Jewett theorem.

The argument is based on the density increment argument as pioneered by Roth, and also used in later papers of Ajtai-Szemerédi and Shkredov on the corners problem, which was also influential in our current work (though, perhaps paradoxically, the generality of our setting makes our argument *simpler* than the above arguments, in particular allowing one to avoid use of the Fourier transform, regularity lemma, or Szemerédi’s theorem). I discuss the argument in the first part of this previous blog post.

I’ll end this post with an open problem. In our paper, we cite the work of P. L. Varnavides, who was the first to observe the elementary averaging argument that showed that Roth’s theorem (which showed that dense sets of integers contained at least one progression of length three) could be amplified (to show that there was in some sense a “dense” set of arithmetic progressions of length three). However, despite much effort, we were not able to expand “P.” into the first name. As one final task of the Polymath1 project, perhaps some readers with skills in detective work could lend a hand in finding out what Varnavides’ first name was? *Update, Oct 22:* Mystery now solved; see comments.

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