Partitioning the Projective Plane

Partitioning the Projective Plane

Definitions and mathematical statement

Write $\mathbb{RP}^2$ for the set of one-dimensional linear subspaces of $\mathbb R^3$. Its elements are unoriented lines through the origin. For an orthonormal basis $e=(e_1,e_2,e_3)$ define \[ O_e=\{\mathbb R x:x\ne0,\quad x=t_1e_1+t_2e_2+t_3e_3,\quad t_1,t_2,t_3\ge0\}. \] A set of lines $S$ is octahedral when $S\subseteq O_e$ for some basis $e$. This agrees with the source's geometric definition: every line meets the closed face $\operatorname{conv}\{e_1,e_2,e_3\}$ and its opposite in the regular octahedron with vertices $\pm e_i$. Zero coordinates are allowed, so face edges and vertices are included.

Call $S$ weakly octahedral if each three-element subset of $S$ is contained in some $O_e$, where the basis may depend on the triple. The conjecture states that if \[ \mathbb{RP}^2=S_1\mathbin{\dot\cup}S_2\mathbin{\dot\cup}S_3 \mathbin{\dot\cup}S_4 \] and every $S_i$ is weakly octahedral, then every $S_i$ is octahedral. Thus for each part there should be one orthonormal basis working for all its lines simultaneously. Different parts may use different bases.

The partition must cover every projective line, with disjoint parts; no measurability or closedness assumption is imposed. The local triple condition alone, for an arbitrary subset not occurring as a part of such a partition, is not the assertion. Directions are identified up to all nonzero scalar multiples, so the choice of orientation of representatives cannot affect either condition.

Short English statement

If the real projective plane is partitioned into four sets whose every triple fits in an octahedral face pair, must each entire set fit in one?

Research attempt

Scope and source check.

The original posting by Jonathan Noel [S2], checked 1 October 2026, explicitly warns that the four-part partition hypothesis matters. We investigate the obstruction by proving the acute-cap example and a compactness reduction. Neither result assumes that the parts are measurable, and neither purports to construct a counterexample partition.

Three acute vectors fit in an orthant.

Let $x,y,z$ be distinct unit representatives with pairwise nonnegative inner products. Relabel them so that $c=\langle y,z\rangle$ is the largest of the three products, and put $a=\langle x,y\rangle$, $b=\langle x,z\rangle$. Since $0\le a,b,c\le1$, we have $c\ge ab$. For distinct unoriented lines, $a<1$. Choose $e_1=x$, $e_2=(y-ax)/\sqrt{1-a^2}$, and choose the sign of the remaining perpendicular unit vector $e_3$ so that $\langle z,e_3\rangle\ge0$. In this orthonormal frame the three coordinate rows are \[ (1,0,0),\quad(a,\sqrt{1-a^2},0),\quad \left(b,\frac{c-ab}{\sqrt{1-a^2}},\langle z,e_3\rangle\right). \] Every entry is nonnegative. Degenerate linear dependence simply makes the last entry zero. Hence these three lines are octahedral.

An infinite weakly octahedral set with no global frame.

Fix a unit vector $u$ and let $C$ be the open spherical cap of vectors $x\in S^2$ with $\langle x,u\rangle>1/\sqrt2$. Let $S$ be their unoriented lines. Any two vectors in $C$ have angle strictly below $\pi/2$, by adding their angles from $u$. Thus every triple fits in an orthant by the preceding calculation, and $S$ is weakly octahedral. Suppose all its lines belonged to a single $O_e$. The connected cap $C$ would lie in the union of the positive and negative closed orthants for that frame. Those orthants are disjoint on the unit sphere, so connectedness puts the whole cap in one of them; reversing all three axes makes it positive. Each coordinate functional must therefore be nonnegative throughout $C$. This forces $\langle e_i,u\rangle\ge1/\sqrt2$ for every $i$: otherwise a vector at angle less than $\pi/4$ from $u$, tilted away from $e_i$, has negative $i$th coordinate. But orthonormality gives \[ 1=\|u\|^2=\sum_{i=1}^3\langle e_i,u\rangle^2\ge3/2, \] a contradiction. This reproduces, with explicit verification, the obstruction reported by [S2]; it is not asserted to be a new counterexample.

A finite-witness reduction.

For each projective line $\ell$ let $F_\ell\subset O(3)$ consist of frames whose positive or negative orthant contains a unit representative of $\ell$. It is closed: for a fixed representative it is a union of two sets defined by three weak continuous inequalities. The space $O(3)$ is compact, since it is closed and bounded in the space of real $3$ by $3$ matrices. If every finite subset of $S$ has a common frame, the family $\{F_\ell:\ell\in S\}$ has the finite intersection property. Compactness gives a frame for all of $S$. Consequently every non-octahedral set has a finite non-octahedral subset. For a weakly octahedral set that witness must have at least four lines; sets of at most two lines fit by choosing signs with nonnegative inner product.

Verification and remaining gap.

The cap calculation checks both orientation invariance and zero-coordinate boundary conventions. The compactness argument does not upgrade triplewise intersection to finite intersection: the cap proves that such an upgrade is false. A proof of the catalogue conjecture must use the other three parts to exclude every finite obstruction inside a part. No extension of the cap to a four-part weakly octahedral partition has been found here.

Final status.

UNRESOLVED. The output is a proved local obstruction and a finite-witness reduction, not a proof or disproof of the partition assertion. Completion estimate: 5%.

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 3088 (OPG-56328), OpenGarden collection. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 Open Problem Garden, Partitioning the Projective Plane. Original problem and discussion; the mathematical formulation below is expanded from the supplied source record. http://www.openproblemgarden.org/op/partitioning_the_projective_plane.