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Given $n$ unit vectors in $\mathbb{R}^n$ s.t. $0 \leq u\cdot v<1$ for all pair of distinct vectors $u,v$. These vectors span a $d$-dimensional subspace s.t. $d< n$. We conjecture that it is possible to partition the $n$ vectors into $d$ groups such that all the vectors within the same group are pairwise non-orthogonal. It trivially holds when $d\in\{1,2,n-1,n-2\}$. However, we have not been able to show for general $d$. Does the conjecture hold for any $d< n$? If yes, how to prove it? Any thoughts/hints would be appreciated. Thanks in advance.

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By "mutually non-orthogonal" do you mean "pairwise non-orthogonal" for the vectors in the $d$ sets, or do you mean that no vector in the set is orthogonal to ALL of the others? – Geoff Robinson Jul 12 '12 at 10:04
I mean pairwise non-orthogonal. Sorry for the confusion. – Pawan Aurora Jul 12 '12 at 10:08
Your question is very related to Borsuk's conjecture, (I am sure you know it, but want to say just in case not.) – Anton Petrunin Jul 12 '12 at 12:44
Thanks for pointing that out. I must admit that I was not aware of it. – Pawan Aurora Jul 12 '12 at 14:19
The statement is of course true for $d=n$. It seems more elegant to include that case. – Will Sawin Jul 12 '12 at 15:03

1 Answer 1

up vote 3 down vote accepted

Kahn--Kalai' counterexample to Borsuk's conjecture is a collection of vertices of unit cube.

Think about this cube as it siitting in an affine hyperplane of $\mathbb R^{n+1}$, so that the origin projects to the center of the cube. Project this cube centrally to the unit sphere. For right choice of parameter you get a counterexample to your statement.

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