Let R be a region on a hypersphere. Each point A of the hypersphere is associated with a vector pointing to A and with origin at the centre of the hypersphere. So let me identify each point with a vector. I have the constraint that every pair of vectors in R is not orthogonal. What is the maximal region R satisfying this constraint? I guess that this region, up to rotations, is given by two opposite hypercones with 90 degrees angular aperture. This is intuitive, but the proof is not so simple to me.
$\begingroup$
$\endgroup$
12
-
$\begingroup$ When you say that you want "the" maximal region, it seems to imply that such a region is unique. I don't think that there is a unique such R... $\endgroup$– Manny ReyesOct 18, 2011 at 21:01
-
2$\begingroup$ Could you define "region"? Presumably a region is not just an arbitrary set, but rather must have some connectivity properties...? $\endgroup$– Joseph O'RourkeOct 18, 2011 at 21:01
-
$\begingroup$ I guess that the region is unique up to rotations, but one could just ask what is the maximal area. Concerning the properties of the region, I require only that it is measurable and has an area defined. The region is not necessarily connected. $\endgroup$– AlmOct 18, 2011 at 21:20
-
$\begingroup$ Keeping the dimension at 3 for discussion, there is not only the spherical cap of angular width just shy of 90 degrees, but also the nonnegative orthant minus three points. And these are connected examples. There are nonconnected examples that take up even more of the sphere's surface area. I leave it to those more facile with higher dimensions to generalize these examples. Gerhard "Ask Me About System Design" Paseman, 2011.10.18 $\endgroup$– Gerhard PasemanOct 18, 2011 at 21:46
-
4$\begingroup$ I think this is known as the "double cap conjecture", and in particular would imply better estiamtes in the "Borsuk problem". The person to ask is probably Gil Kalai. See e.g. gilkalai.wordpress.com/2009/05/22/… $\endgroup$– Guillaume AubrunOct 19, 2011 at 14:55
|
Show 7 more comments