Kochen-Specker theorem

I learned from E. Kowalski’s blog a nice theorem of Kocher and Specker:

There does not exist a function f:\mathbb S^2\to\{0,1\}, such that for any x,y,z\in\mathbb S^2\subset\mathbb R^3 which are orthogonal to each other, f(x)+f(y)+f(z)=2.

The statement here was at first counter-intuitive, so I tried to prove it. I found some kind of geometrical proof, I think it’s nice (and I hope it’s right!):

Proof: Label all p\in\mathbb S^2 by f(p) and identify points of \mathbb S^2 with vectors of \mathbb R^3. For every othogonal triple one and only one of the labels is 0. Therefore, for every “zero” label (call it “north pole”), the “equator” will be made of “one”s, and the “south pole” will be a “zero”.

Consider 2 labels “zero” which have an angle between them <\pi/2, there are two intersecting (non-orthogonal!) equators of “one”s. Now while a point a moves on one of the equators, there is exactly one point b on the other equator which is orthogonal to it: as a describes the first equator, b will therefore describe the second equator. The third point c completing an orthogonal basis (a,b,c), will also describe a curve of “zero”s: it is a circumference having the two “zero”s as diameter!

Iterating the above construction it is easy to see that the set of zeroes is open. But since the set of “one”s depends continuously on the set of “zero”s (through the “equator-pole” construction), it is also open! Contraddiction (since the sphere is connected). \square

  1. Leave a comment

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: