# Seemingly elementary geometric problem in R^3 which requires the axiom of choice

While playing with what I called "quantum matching", the following problem arose: construct a map $F$ from the unit sphere $S_2$ in $R^3$ to itself such that $F(X)$ is orthogonal to $X$ plus has one of the properties:

1. Strong: if 3 vectors $X_1,X_2, X_3$ are independent then 3 vectors $F(X_1), F(X_2), F(X_3)$ are also independent.

2. Weak: if 3 vectors $X_1,X_2, X_3$ are orthogonal then 3 vectors $F(X_1), F(X_2), F(X_3)$ are independent.

It is not hard to prove that such maps with the Strong property don't exist. I proved the existence of such maps satisfying the Weak property, but my proof uses the Axiom of Choice and some version of the Continuum Hypothesis. Granted the maps I am after supposed to be bad, at least discontinuous.

My question:

Are things like Axiom of Choice really needed here?

The discussion and proofs are in http://arxiv.org/abs/quant-ph/0201022

• Are you using $S_4$ for the two dimensional unit sphere in $\mathbb{R}^3$? That is strange. Oct 23, 2013 at 18:01
• Sorry, just a typo. Oct 23, 2013 at 18:10
• I couldn't find the proof of the impossibility with Strong property in your arxiv manuscript. Could you please elaborate it or giving a reference?
– Name
Oct 23, 2013 at 18:50
• Must such an $F$ be discontinuous? Oct 23, 2013 at 20:43
• @AndrejBauer: If $F(X)$ is orthogonal to $X$ then $F$ is a unit tangent vector field on $S_2$. The hairy ball theorem asserts that it cannot be continuous. Oct 23, 2013 at 21:52