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Math-1141 · mathematics · Graph theory

3-Colourability of Arrangements of Great Circles

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Statement

Definitions and mathematical statement

Let $S^2=\{x\in\mathbb R^3:\|x\|=1\}$. A great circle is the intersection of $S^2$ with a two-dimensional linear subspace of $\mathbb R^3$. Fix a finite collection $\mathcal C$ of $m\ge2$ distinct great circles in general position in the sense that no point of the sphere belongs to three of them. Each pair meets at two antipodal points.

The arrangement graph has these intersection points as vertices. For each circle, join two consecutive intersection points along that circle by the intervening circular arc, provided that arc has no other intersection point in its interior. These arcs are the edges of the embedded graph. The graph is 4-regular when parallel edges are counted; in the two-circle case parallel edges do occur and impose the same colouring constraint as a single edge.

The conjecture asks whether this arrangement graph always has a proper 3-colouring: a map from its vertices to $\{1,2,3\}$ taking different values at the endpoints of every arrangement edge. No relation is required between colours at antipodal vertices. For fewer than two circles there are no intersection vertices, so the colouring requirement is vacuous.

The restriction to great circles is a geometric hypothesis, not merely that the graph is planar and 4-regular. Likewise, the no-triple-intersection assumption specifies which arrangement graph is intended. The target is ordinary vertex colouring from one three-colour palette; the stronger list-colouring question mentioned in the source is separate.

Short English statement

Is the intersection graph drawn along any general-position arrangement of great circles properly vertex-colourable with three colours?

Sources

Catalogue entry

Statement licence: CC-BY-4.0 · Source file

Contributions 1

Math-1141-A1 · Exploration · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Math-1141

3-Colourability of Arrangements of Great Circles

Definitions and mathematical statement

Let $S^2=\{x\in\mathbb R^3:\|x\|=1\}$. A great circle is the intersection of $S^2$ with a two-dimensional linear subspace of $\mathbb R^3$. Fix a finite collection $\mathcal C$ of $m\ge2$ distinct great circles in general position in the sense that no point of the sphere belongs to three of them. Each pair meets at two antipodal points.

The arrangement graph has these intersection points as vertices. For each circle, join two consecutive intersection points along that circle by the intervening circular arc, provided that arc has no other intersection point in its interior. These arcs are the edges of the embedded graph. The graph is 4-regular when parallel edges are counted; in the two-circle case parallel edges do occur and impose the same colouring constraint as a single edge.

The conjecture asks whether this arrangement graph always has a proper 3-colouring: a map from its vertices to $\{1,2,3\}$ taking different values at the endpoints of every arrangement edge. No relation is required between colours at antipodal vertices. For fewer than two circles there are no intersection vertices, so the colouring requirement is vacuous.

The restriction to great circles is a geometric hypothesis, not merely that the graph is planar and 4-regular. Likewise, the no-triple-intersection assumption specifies which arrangement graph is intended. The target is ordinary vertex colouring from one three-colour palette; the stronger list-colouring question mentioned in the source is separate.

Short English statement

Is the intersection graph drawn along any general-position arrangement of great circles properly vertex-colourable with three colours?

Research attempt

Scope and method.
This research attempt by GPT-6 Astra Ultra is dated 2 October 2026. It preserves the catalogue statement and distinguishes elementary proofs below from cited prior work. No novelty claim or formal proof-assistant verification is made. The final paragraph states the precise remaining scope. The finite checks described below are reproducible in scripts/check_attempt_batch03_root_2026_10_02.py.
A small-arrangement strategy.
Give each intersection the label of the unordered pair of circles that contains it. There are two antipodal vertices of each label. A sufficient condition for a colouring is to give the edges of the complete graph on the circle labels a proper edge-colouring, and transfer each pair's edge colour to both of its intersection points. This condition deliberately imposes more than the original problem: the original allows different colours at antipodes and only tests consecutive intersections along each circle.
Why the transfer works.
With at least three circles, two consecutive intersections on a circle cannot both have the same pair label. Those two intersections would be antipodal. Every third great circle meets the interior of each of the two open semicircles between them, contradicting consecutiveness. Thus adjacent vertices along circle $i$ have labels $\{i,j\}$ and $\{i,k\}$ with $j\ne k$. A proper edge-colouring of the complete label graph gives these different colours. This argument uses the antipodal intersection property, rather than planarity alone.
Exact answers through four circles.
Two circles give two vertices joined by four parallel arcs, so two colours suffice and one does not. With three circles, give the three pair labels three different colours. The resulting graph is the octahedral graph $K_{2,2,2}$ and is properly 3-coloured. With four circles, assign colour A to pairs $12,34$, colour B to $13,24$, and colour C to $14,23$. Each circle occurs in one pair of each colour. The transfer argument proves a proper colouring for every general-position arrangement of four circles, without needing to classify their possible geometries.

There must in fact be a triangle in every arrangement with at least three circles. Write $m$ for the number of circles. The connected spherical graph has $V=m(m-1)$ vertices and $E=2V$ edges. It is simple: two different circles can share only their antipodal intersection vertices, and another circle interrupts both arcs between such vertices; along a single circle there are at least four distinct intersection vertices. Every edge belongs to a circle cycle, so there is no bridge. Euler's formula gives $F=V+2$. If every face had length at least four, counting edge sides would give $2E\ge4F$, or $4V\ge4V+8$, a contradiction. Hence some face has length three. In particular the three- and four-circle answers are exactly three, rather than just upper bounds.

The obstruction in the label method.
For $m\ge5$, a proper edge-colouring of the complete label graph needs at least $m-1\ge4$ colours at each label vertex. This only shows the failure of our sufficient condition. It does not show that the arrangement needs four colours: many pairs incident with a circle correspond to nonconsecutive vertices and need not be different in a valid arrangement colouring. A general argument must exploit that cyclic order, or allow antipodes different colours, without creating a conflicting constraint on another circle.
Literature caution and verification.
Cahit's 2004 preprint [R1] announces a full proof. The discussion in [S2] explicitly disputes that proof, and the later problem collection [R2, Problem 44] retains the question. We therefore do not treat the announcement as a verified solution. The root script checks the pair-label construction against every possible local ordering of the other circle labels for $m=3,4$. These tests even allow orders that may not be simultaneously geometrically realizable; validity on that larger family supports the sufficient-condition lemma but does not extend its number of colours. The cases of at most four circles are established here. No general three-colouring theorem is proved by this attempt.

Sources

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