Circular colouring the orthogonality graph
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Definitions and mathematical statement
Let $\mathcal O$ be the graph whose vertices are the one-dimensional linear subspaces of $\mathbb R^3$. Two distinct lines $L,M$ are adjacent when they are perpendicular, meaning $x\cdot y=0$ for nonzero representatives $x\in L$ and $y\in M$. Scaling representatives, including by negative numbers, does not change this condition. Consequently opposite directions on the unit sphere correspond to one vertex, rather than two.For integers $p\ge2q\ge2$, a circular $(p,q)$-coloring of $\mathcal O$ is a map $c:V(\mathcal O)\to\{0,\ldots,p-1\}$ such that $q\le|c(L)-c(M)|\le p-q$ whenever $L\perp M$. Define $\chi_c(\mathcal O)$ as the infimum of all ratios $p/q$ for which such a coloring exists. A coloring is an arbitrary function; no continuity, measurability, or definability is imposed.
The question is whether $\chi_c(\mathcal O)=4$. In particular, since an ordinary proper four-coloring gives the upper bound four, the disputed direction asks whether every rational ratio $p/q<4$ is impossible. It would suffice to produce, for each such ratio, a finite configuration of lines admitting no $(p,q)$-coloring, by the compactness principle for finite-color constraints.
The target is circular chromatic number, which may be nonintegral, and not merely the ordinary chromatic number. Ruling out ordinary three-colorings alone does not establish the desired lower bound of four for circular colorings.
Short English statement
Is the circular chromatic number of the graph of perpendicular lines through the origin in three-dimensional space equal to four?Sources
- S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 3235 (OPG-2040), OpenGarden. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
- S2 Open Problem Garden, Circular colouring the orthogonality graph. Original problem and discussion. http://www.openproblemgarden.org/op/circular_colouring_the_orthogonality_graph.
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