Throughout this post, by *projective plane* we mean the set of all lines through the origin in .

**Definition**Say that a subset of the projective plane is

*octahedral*if all lines in pass through the closure of two opposite faces of a regular octahedron centered at the origin.

**Definition**Say that a subset of the projective plane is

*weakly octahedral*if every set such that is octahedral.

**Conjecture**Suppose that the projective plane can be partitioned into four sets, say and such that each set is weakly octahedral. Then each is octahedral.

Also, see the posting on mathoverflow.

There is an equivalent definition of the "weakly octahedral" condition which may be useful.

**Lemma**A subset of the projective plane is weakly octahedral if for any three lines in and any three vectors and which span these lines, we have where is the standard (dot) inner product on .

The fact that and partition the projective plane seems to be important. Here is an example of a weakly octahedral set that is not octahedral: Fix any vector and let be the set of all lines which are spanned by vectors which meet at an angle strictly less than .

This question came up while working on another problem posted to this site: Circular colouring the orthogonality graph. It is possible that a solution to the problem stated here can be applied to solve this problem. Moreover, it may be useful in proving that the real orthogonality graph (defined in the other posting) has (essentially) only one proper -colouring.

## Bibliography

* indicates original appearance(s) of problem.