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Consider the problem of coloring each point of $S^2$ with one of two colors (say "black" or "white") so that among any three points of $S^2$ which are the vertices of an equilateral spherical triangle with 90 degree sides, one and only one of these three points will be colored "white". This problem comes from Quantum Mechanics and a theorem of Quantum Mechanics, the Kochen-Specker theorem, has-as one of its consequences-that no such coloring of the points of $S^2$ is possible. But all the proofs of this result that I have seen are based on constructing specific finite subsets of $S^2$ (the smallest containing 33 points with specified real x-y-z coordinates) and showing that the coloring problem cannot be solved for these finite sets. The points apply combinatorial analysis to finite graphs whose vertices are the points of these sets. Two points which are 90 degrees apart on $S^2$ are joined by an edge in the graphs. The hard part of the proofs is finding these finite sets in the first place. Also, the combinatorial work is unwieldy to specify and carry out in detail........But the coloring theorem is really a (rather interesting) theorem about the metric topology of $S^2$. It seems to me that it should have a topological (or at least a coordinate-free) proof. Such a proof might even be alot simpler to follow than the proofs which involve specific coordinates. Does such a proof exist?

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    $\begingroup$ I'm not sure what you would count as a valid topological proof, but J. Butterfield and C. J. Isham have a very nice categorial formulation. Their series of papers "A topos perspective on the Kochen-Specker theorem" approaches the problem in the form: in dimension > 2, there are no global elements of the spectral presheaf $\mathcal{O}_{d}^{op} \rightarrow$Set. $\endgroup$
    – ex0du5
    Jan 10, 2014 at 19:50
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    $\begingroup$ Why do you call it a topological theorem? The metric is not the topology? $\endgroup$ Jan 10, 2014 at 20:04
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    $\begingroup$ Instead of «for a topological theorem about S^2», which is quite inespecific, why not say in the title «for the Kochen-Specker theorem» (or a corollary thereof)? $\endgroup$ Jan 10, 2014 at 20:12
  • $\begingroup$ I thought the theorem was that there are no colorings such that no orthonormal triple is monochromatic, rather than that all ortho. triples have exactly one white member. Does the answer change for this modified problem? $\endgroup$ Jan 11, 2014 at 2:10

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If I understand the problem correctly, one wants a coloring of the points of the unit sphere $S^2$ so that for every triangle with edge lengths $\pi/2$, precisely two vertices are colored black and one is colored white?

If that's correct, then I think there might be a topological proof. If one takes a black point $p$ , then there is a great circle $p^{\perp}$ which is distance $\pi/2$ away. For every point on $p^{\perp}$ colored white, the two points on $p^\perp$ distance $\pi/2$ away must be colored black, and vice versa (in fact, antipodal points must have the same color). Also, for every white point $q$, the circle $q^\perp$ must be colored entirely black.

From this, one sees that for every black point $r$ on $p^\perp$, the great circle going through $r$ and $p$ must be entirely black, since it is $q^\perp$ for some point $q\in p^\perp$ distance $\pi/2$ from $r$. Consider another black point $s\in p^\perp$ distance $<\pi/2$ from $r$, and distance $\pi/2$ to $t\in p^\perp$ colored white. The $t^\perp$ is also black. For every point $u\in q^\perp$, there are a pair of points in $t^\perp$ which are distance $\pi/2$ from $u$. There is a unique antipodal pair of white points distance $\pi/2$ from these, and together these form a curve of white points going through $q$. The great circles through $p$ which are colored black cannot cross this white curve, and therefore there is an interval about $q$ in $p^\perp$ which is colored white. However, this implies that the set of white points in $p^\perp$ is an open set. Therefore, so is the set of black points. But $p^\perp$ is connected, so it cannot be a union of two open subsets, a contradiction.

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    $\begingroup$ I had a similar idea (but later). Any two white points $w_1,w_2$ on distance $<\pi/2$ can be joined by a smooth curve of white points. Indeed, take $b\in w_1^\perp\cap w_2^\perp$ and $b'\in w_1^\perp\cap b^\perp$ so that $b\times b'=w_1$. Then move $b$ along the black line $w_2^\perp$ and $b'$ along black $w_1^\perp$, keeping $b\perp b'$, until $b'\in w_1^\perp\cap w_2^\perp$ (it is possible because $w_1$ and $w_2$ are not orthogonal). The point $w=b\times b'$ provides a smooth curve that joins $w_1$ and $w_2$. $\endgroup$ Jan 11, 2014 at 0:39
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    $\begingroup$ Considering a geodesic $\gamma$ containing white points (every arc of $\gamma$ of length $>\pi/2$ contains a couple of white points), we can draw a piecewise smooth curve of white points joining white points on distance $>\pi$. A contradiction. $\endgroup$ Jan 11, 2014 at 0:39
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    $\begingroup$ Ok, that sounds a bit more elegant than my rendition. $\endgroup$
    – Ian Agol
    Jan 11, 2014 at 0:55
  • $\begingroup$ @ Ian and Sasha: Very nice! That is exactly what I was looking for. Quantum theorists prefer the proofs based on finite sets of points since these points can represent directions of "spin" of a particle which are being measured in an experiment. But the coloring theorem is almost purely topological, except for the "metric" stipulation involving "pi/2". So I was sure it would have a proof of the kind you have provided. Unfortunately I was not able to come up with such a proof myself. Many thanks. $\endgroup$ Jan 11, 2014 at 19:12

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