Comments on: Proposals (Tim Gowers): Polynomial DHJ, and Littlewood’s problem

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By: gowers

gowers — Wed, 02 Dec 2009 09:25:23 +0000

The answer to Problem 5 is yes, but that is with the dyadic distance in each coordinate (that is, , where is the first place where the two binary expansions differ). For the ordinary Euclidean distance I do not know the answer — that’s Problem 3.

By: Kristal Cantwell

Kristal Cantwell — Wed, 25 Nov 2009 05:47:28 +0000

According to a comment by Tim Gowers on his blog the answer to the question about points on the cube is yes. I think this is problem 5 on his blog. Here is a link to the comment with the solution to the problem:

http://gowers.wordpress.com/2009/11/17/problems-related-to-littlewoods-conjecture-2/#comment-4377

By: Kristal Cantwell

Kristal Cantwell — Fri, 20 Nov 2009 21:42:23 +0000

The above doesn’t work as I can add vectors to get one coordinate zero making the product zero. I don’t think there is a lattice that is going to work.

By: Kristal Cantwell

Kristal Cantwell — Fri, 20 Nov 2009 21:10:42 +0000

I think I can solve the problem about points in the cube. One starts with this basis (2,1,-2,) (2,-2,1)(1/4,1/2,1/,2) normalize it then scale it down by a factor of 1/n^1/3 then construct it by a factor of two then we should be able to fit about 8n of these points in the cube and the closest we can get two points is where one point difference from another by one coordinate and then the product of the difference of the coordinates will be much larger than 10^-14 n^3 if I am counting the zeros correctly

By: Gil Kalai

Gil Kalai — Fri, 20 Nov 2009 12:10:03 +0000

Dear Tim, yes, I overlooked the distance definition for a short time but then realized it cannot be the ususal distance and looked back in the post. I agree that problem (2) is very interesting! (In any direction it might go!)And if the answer is no it may be, as you explained, harder than proving LP but still could be useful.

By: gowers

gowers — Fri, 20 Nov 2009 10:13:06 +0000

It is important to clarify that the problem numbered (2) is not itself Littlewood’s problem but rather a related problem. Secondly, the “distance” is not what one might think. In fact, it is not even a distance. We define to be (Obviously if we were using the normal Euclidean distance then we could get points and that would be best possible.)

Also, I find the problem (2) interesting in itself, so am not put off by the possibility that there might exist such a set of points even if the Littlewood conjecture is true. It would even, in a modest way, shed some light on the Littlewood conjecture, since it would demonstrate that the problem was “genuinely number-theoretic”. I say “in a modest way” because I get the impression that the experts more or less take for granted that it is a problem in number theory, so a conclusion of that kind would not change the way people think. But it would also provide some kind of explanation for why the problem is hard.