Throughout this post, by projective plane I mean the set of all lines through the origin in $\mathbb{R}^3$.

Side Note: If there are more standard definitions for any of the ideas presented here, please let me know.

Definition: Say that a subset $S$ of the projective plane is octahedral if all lines in $S$ pass through the closure of two opposite faces of a regular octahedron centered at the origin.

Definition: Say that a subset $S$ of the projective plane is weakly octahedral if every set $S'\subseteq S$ such that $|S'|=3$ is octahedral.

Equivalent Definition: Say that a subset $S$ of the projective plane is weakly octahedral if for any three lines in $S$ and any three vectors $x, y$ and $z$ which span these lines, we have $$\langle x,y\rangle \cdot\langle x,z\rangle \cdot\langle y,z\rangle \geq 0$$ where $\langle\cdot,\cdot\rangle$ is the standard (dot) inner product on $\mathbb{R}^3$.

Now, here is my question.

Question: Suppose that the projective plane can be partitioned into four sets, say $S_1,S_2,S_3$ and $S_4$ such that each set $S_i$ is weakly octahedral. Then is it necessarily true that each $S_i$ is octahedral?

Note: The fact that $S_1,S_2,S_3$ and $S_4$ partition the projective plane seems to be important. I believe that this is an example of a weakly octahedral set that is not octahedral: Fix any vector $x$ and let $S$ be the set of all lines which are spanned by vectors which meet $x$ at an angle strictly less than $\frac{\pi}{4}$.

This question came up while I was working on a graph colouring problem where the vertex set is the projective plane. For the specific problem, follow this link: http://www.openproblemgarden.org/?q=op/circular_colouring_the_orthogonality_graph

I've also posted the problem to the Open Problem Garden, as the answer seems to be unknown: http://www.openproblemgarden.org/op/partitioning_the_projective_plane

  • $\begingroup$ Ifyou are right about S not being octahedral, why not make it one of the four parts of a weakly octahedral partitioning? $\endgroup$ – The Masked Avenger Aug 26 '13 at 4:06
  • $\begingroup$ And how would you define the other three parts? :) $\endgroup$ – Jon Noel Aug 26 '13 at 4:13
  • $\begingroup$ Start with an octahedral partition, and pick a central line l from one of them (like (1,1,1) is for the positive orthant). Then use l to define S, taking away lines from the other three parts and putting them in S. $\endgroup$ – The Masked Avenger Aug 26 '13 at 4:24
  • $\begingroup$ My imagery assumes S is a superset of one of the parts. If S is not, then something else is needed. $\endgroup$ – The Masked Avenger Aug 26 '13 at 4:27
  • $\begingroup$ @TMA I don't think that this works. The angle between (1,1,1) and (1,0,0) is more than pi/4. Moreover, if you modify the set S from above to use an angle greater than pi/4, then it will no longer be weakly octahedral. Simply take a plane P through the origin containing the original vector x. Then there are vectors y and z in P with an angle of pi/2 + epsilon between them for some small epsilon. The set S' = {x,y,z} is not octahedral, and therefore S is not weakly octahedral. $\endgroup$ – Jon Noel Aug 28 '13 at 18:38

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