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I've been writing some linear algebra problems with colleagues, and the following question occurred to us:

Let $S^2$ denote the unit sphere in $\mathbb{R}^3$. Does there exist a partition $S^2=\bigsqcup_\alpha B_\alpha$ such that each $B_\alpha$ is an orthonormal basis for $\mathbb{R}^3$?

The analogous question for the unit circle in the plane is easily affirmative, but that approach doesn't seem to work here.

Dustin G. Mixon
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  • The three-sphere is a Lie group, so it can be can be partitioned into orthonormal bases for $\mathbb{R}^4$. What about $S^5$? (I have no ideas about $S^2$.) – Sam Nead Sep 30 '21 at 16:18
  • How does the Lie group structure help? $S^1$ is a Lie group too but the only decomposition into bases that I can think of is still patched together in the sense that the bases do not form a differentiable (or even continuous) curve. – M. Winter Sep 30 '21 at 22:14
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    @M.Winter It seems like the quaternion structure of $S^{3}$ gives a partition into an orthonormal basis. Define an equivalence relation on $S^{3}$ where $x\simeq y$ iff $x=ay$ where $a\in{1,i,j,k,-1,-i,-j,-k,}$. Then for each equivalence class $P$ containing $x$, $P$ can be partitioned down further into two orthonormal bases, namely ${x,ix,jx,kx},{-x,-ix,-jx,-kx}.$ – Joseph Van Name Oct 01 '21 at 00:38
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    I am curious why this question is interesting? what is its application? – C.F.G Oct 01 '21 at 07:53
  • The partitions of $S^{2}$ into orthonormal bases are in a canonical one-to one correspondence with the functions $f:S^{2}\rightarrow S^{2}$ that satisfy the following functional equations: $f(z)\times(z\times f(z))=z,(z\times f(z))\times z=f(z),f(f(z))=z\times f(z),f(z\times f(z))=z.$ – Joseph Van Name Oct 04 '21 at 04:52

1 Answers1

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Using the Axiom of Choice, yes you can.

To get such a partition, start by enumerating all the points of the sphere with order type $\mathfrak{c}$ (the least ordinal number with the same cardinality as the sphere): say $\langle p_\alpha :\, \alpha < \mathfrak{c} \rangle$ is such an enumeration. Now we define our partition elements one at a time, via a transfinite recursion of length $\mathfrak{c}$. At stage $\alpha$ of the recursion, suppose we've already selected, at previous stages of the recursion, some bases $\{ B_\xi :\, \xi < \alpha \}$ that will be in our partition. Now consider the point $p_\alpha$. There are two possibilities: either we already put $p_\alpha$ into one of the $B_\xi$'s for some $\xi < \alpha$, or we didn't. In the first case, we do nothing at stage $\alpha$ of the recursion: formally, we could define $B_\alpha = \emptyset$ in this case. In the second case, we choose an orthonormal basis $B_\alpha$ that contains $p_\alpha$, and that is disjoint from everything we put into our partition at an earlier stage. This is possible there is a $\mathfrak{c}$-sized collection $\mathcal C$ of orthonormal bases, any two of which intersect only in $p_\alpha$ (just rotate a basis a bit around $p_\alpha$); because $\bigcup_{\xi < \alpha}B_\xi$ contains $\leq 3 \cdot |\alpha| < \mathfrak{c}$ points, one of the bases in $\mathcal C$ contains no points from $\bigcup_{\xi < \alpha}B_\xi$. We choose some such basis to be $B_\alpha$. In the end, $\{ B_\alpha :\, \alpha < \mathfrak{c} \} \setminus \{\emptyset\}$ is the desired partition.

Will Brian
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    For the record, I would still be interested to know if there is a more hands-on way to do this. For example, is there a Borel partition of the sphere into orthonormal bases? (I can't seem to think of one.) – Will Brian Sep 30 '21 at 20:47
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    The argument of Will Brian is analogous to the proof in Komjáth and Totik, , Problems and Theorems in Classical Set Theory, Chapter 13, Problem 13, that $\mathbb{R}^3$ can be partitioned into unit circles. I believe it is still open whether this result can be proved without the Axiom of Choice. – Richard Stanley Oct 03 '21 at 14:18
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    Re Richard Stanley's comment, see https://mathoverflow.net/questions/28647/is-it-possible-to-partition-mathbb-r3-into-unit-circles – aorq Oct 04 '21 at 00:21