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

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???