Raigorodskii’s Theorem: Follow Up on Subsets of the Sphere without a Pair of Orthogonal Vectors

raigor

Andrei Raigorodskii

(This post follows an email by Aicke Hinrichs.)

In a previous post we discussed the following problem:

Problem: Let A be a measurable subset of the d-dimensional sphere S^d = \{x\in {\bf R}^{d+1}:\|x\|=1\}. Suppose that A does not contain two orthogonal vectors. How large can the d-dimensional volume of A be?

Setting the volume of the sphere to be 1, the Frankl-Wilson theorem gives a lower bound (for large d) of  1.203^{-d},
2) The double cap conjecture would give a lower bound (for large d) of 1.414^{-d}.

A result of A. M. Raigorodskii from 1999 gives a better bound of 1.225^{-d}. (This has led to an improvement concerning the dimensions where a counterexample for Borsuk’s conjecture exists; we will come back to that.) Raigorodskii’s method supports the hope that by considering clever configurations of points instead of just \pm 1-vectors and applying the polynomial method (the method of proof we described for the Frankl-Wilson theorem) we may get closer to and perhaps even prove the double-cap conjecture.

What Raigorodskii did was to prove a Frankl-Wilson type result for vectors with 0,\pm1 coordinates with a prescribed number of zeros. Here is the paper.

Now, how can we beat the 1.225^{-d} record???

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One Response to Raigorodskii’s Theorem: Follow Up on Subsets of the Sphere without a Pair of Orthogonal Vectors

  1. Gil Kalai says:

    In a later paper, Raigorodskii improves the Borsuk bound
    still further, but the “forbidden” scalar products are different
    from 0 so it does not seem to have baring on the problem of avoiding pairs of orthogonal vectors.

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