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The construction and proofs are credited to Plasma AI. The graded whole-conclusion review covers the stated conditional forward direction, the reverse implication and the two-scale extension at every real ratio. The forward direction of the algebraic-family theorem is conditional on premise (R0) of [[research/leads/equilateral_two_scale/spectral_exclusion|the spectral account]]. The two-scale extension below uses the separate prose rigidity proof examined in that review. The continuous-factor and sampling arguments are supplied here. They adapt Lemma 4.1 and Proposition 4.2 of the OpenAI plane-coloring manuscript, read at proof depth and examined at their stated use in this argument.
Statement
A triangle here is an ordered triple of distinct points of , possibly collinear; its copies are with and . Let be a countable family. A two-coloring avoids if no copy of any member is monochromatic; a measurable weakly avoids if for each the set of $(p,u)\in \mathbb C\times S^1$ whose copy is monochromatic is null for area times arc length. Equivalently its measure in is zero for every positive integer . Changing colors on a plane-null set preserves this condition: for each fixed , hitting that set at any of the three vertices is a null event in . Reflection is not built into this ordered-template convention; when reflected copies are required, include the conjugate template as well. This preserves countability and algebraic side lengths.
Theorem. Suppose every member of has algebraic side lengths.
- (a) If some two-coloring of the plane avoids , then some Borel two-coloring weakly avoids , granting (R0).
- (b) If some measurable two-coloring weakly avoids , then some two-coloring avoids . This direction needs neither (R0) nor algebraic sides.
Two-scale extension. For Problem 173, the same proof applies to consisting of equilateral sides for every real . Use , where is the real closure of in , and its unit group. Corollary 3 of the rigidity extension provides exactly the nonzero pair-correlation identity used in the proof. All the factor, sampling and threshold arguments below concern a countable dense carrier and work with in place of . This yields the transfer for every real ratio. Theorem 4 of the circle account then excludes avoidance when . The graded whole-conclusion review covers this composition. The recorded standing of the separate algebraic Lean premise (R0) is unchanged.
Why triangles and two colors
For put . Then
with equality exactly when one of equals and one equals . On signs, is the indicator that are equal. So a -valued field whose three pair correlations on a triangle sum to has almost surely there, and any rounding that keeps and produces a nonmonochromatic triple. Only pair correlations enter. For four points, or for three or more colors on a triangle, the monochromatic indicator has terms of degree above two, which the projection below does not preserve.
Ordinary conditional expectation does not preserve avoidance of general hyperedges. The uniform law on the six nonmonochromatic sign triples has all three conditional means zero over the trivial factor, and rounding gives a monochromatic triple. That projection changes each pair correlation from to , so it is not of the pair-preserving kind used here and does not test this mechanism.
Continuous factors and measurable representatives
Let a standard probability space carry a measure-preserving action of the countable dense additive subgroup , with Koopman operators . Define as in (R0). It is closed, since , and it contains constants. Conjugation and Lipschitz composition preserve . Products of two bounded members belong to , by the product inequality in .
Let consist of the measurable sets with , modulo completion. Complements and finite unions stay in by the preceding properties; increasing unions do by convergence. Thus it is a sigma-algebra. For a real , the Lipschitz functions converge in to . Every real member of is therefore -measurable, and the same follows for complex members by real and imaginary parts. Conversely the simple -measurable functions belong to and are dense in . Hence and its orthogonal projection is conditional expectation. In particular it sends real functions valued in to functions with the same bounds.
For and tending to , the sequence is Cauchy in for , because the norm of its difference at is . Its limit defines , independently of the approximating sequence. The group law, unitarity and strong continuity extend by limits. Realness and interval bounds are closed conditions, so preserves them. No pointwise action of on the probability space is asserted.
For a real with there is a jointly Borel field representing for every fixed , after choosing a bounded Borel representative of . To see this, choose Borel maps with countable range and uniformly small , so that for every . Such maps exist by density of and uniform continuity of the orbit of . Realize by the original countable action. The maps are jointly Borel and bounded. For each fixed the sum of the norms of is finite, so the sequence converges almost surely. Define to be this limit where it exists and otherwise. It is jointly Borel, and its section at every fixed represents the limit . The exceptional subset of the probability space may depend on ; Tonelli, not a common pointwise action, handles it.
Proof of (a)
Rotate each member so that ; with algebraic sides, , the field of algebraic numbers, and denotes its algebraic unit vectors. Let consist of the labelings with no monochromatic triple , , , . It is closed, contains the restriction of the given coloring, and is invariant under . Amenability of gives an invariant probability on , as in the manuscript's Proposition 4.2. Define , , and the projection onto the continuous vectors as in the proof of Theorem 3 of the spectral account. Put .
Bounds. The continuous-factor argument above shows that is real with , and extends the translations on to a strongly continuous unitary representation of with for every .
Correlations. As in Theorem 3, (R0) gives $\langle f,T_zf\rangle=\langle g,T_zg\rangle$ for all nonzero . Let , continuous on . For every satisfies , and the three displacements , , are nonzero. Integrating,
and by continuity and density of in this holds for every unit .
Pointwise vanishing. For and unit the three functions , , take values in , and is the left side above, because . Since , -almost surely.
A measurable sample. Use the jointly Borel field constructed above. For each fixed its three sections represent the three vectors in the preceding paragraph. Tonelli then gives, for every and every integer ,
These are countably many conditions, so one makes every inner double integral vanish. Put if and otherwise; is Borel. For almost every and every the three values of include and , which keeps, so the copy is not monochromatic. Thus weakly avoids .
Proof of (b)
Let weakly avoid , and let and be the points of density one of and . They are disjoint and their union is conull. Suppose , , all lie in . For small choose uniformly in the disk of radius about and, independently, uniformly on an arc about of angular width . Each of and has a density at most , supported in the disk of radius about , respectively . Density one at the three points makes the probability that all three random points lie in tend to one. The law of is absolutely continuous, which contradicts weak avoidance. The same holds for . This is the argument of the manuscript's Lemma 4.3 with three points.
Every finite set of points has a translate inside the conull set , where the density classes color it with no monochromatic copy. Each avoidance condition is closed in and involves three coordinates, so product compactness gives a coloring of the whole plane avoiding .
This yields an arbitrary avoider, not a measurable completion of the density classes on the omitted null set.