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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 (0,a,b)(0,a,b) of distinct points of C\mathbb C, possibly collinear; its copies are (p,  p+ua,  p+ub)(p,\;p+ua,\;p+ub) with p∈Cp\in\mathbb C and ∣u∣=1|u|=1. Let T={(0,aj,bj):j∈J}\mathcal T=\{(0,a_j,b_j):j\in J\} be a countable family. A two-coloring c:C→{−1,+1}c:\mathbb C\to\{-1,+1\} avoids T\mathcal T if no copy of any member is monochromatic; a measurable cc weakly avoids T\mathcal T if for each jj 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 B(0,n)×S1B(0,n)\times S^1 is zero for every positive integer nn. Changing colors on a plane-null set preserves this condition: for each fixed uu, hitting that set at any of the three vertices is a null event in pp. 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 T\mathcal T has algebraic side lengths.

  • (a) If some two-coloring of the plane avoids T\mathcal T, then some Borel two-coloring weakly avoids T\mathcal T, granting (R0).
  • (b) If some measurable two-coloring weakly avoids T\mathcal T, then some two-coloring avoids T\mathcal T. This direction needs neither (R0) nor algebraic sides.

Two-scale extension. For Problem 173, the same proof applies to T\mathcal T consisting of equilateral sides 1,x1,x for every real x>0x>0. Use L=Fx(i)L=F_x(i), where FxF_x is the real closure of Q(x)\mathbb Q(x) in R\mathbb R, 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 LL in place of EE. This yields the transfer for every real ratio. Theorem 4 of the circle account then excludes avoidance when x≠1x\ne1. 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 a,b,c∈[−1,1]a,b,c\in[-1,1] put Q(a,b,c)=1+ab+ac+bcQ(a,b,c)=1+ab+ac+bc. Then

Q(a,b,c)=12((1+a)(1+b)(1+c)+(1−a)(1−b)(1−c))≥0,Q(a,b,c)=\tfrac12\bigl((1+a)(1+b)(1+c)+(1-a)(1-b)(1-c)\bigr)\ge0,

with equality exactly when one of a,b,ca,b,c equals −1-1 and one equals +1+1. On signs, Q/4Q/4 is the indicator that a,b,ca,b,c are equal. So a [−1,1][-1,1]-valued field whose three pair correlations on a triangle sum to −1-1 has Q=0Q=0 almost surely there, and any rounding that keeps +1+1 and −1-1 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 (0,0,0)(0,0,0) gives a monochromatic triple. That projection changes each pair correlation from −1/3-1/3 to 00, 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 E⊂CE\subset\mathbb C, with Koopman operators TzT_z. Define HcH_c as in (R0). It is closed, since ∥Tzv−v∥2≤2∥v−w∥2+∥Tzw−w∥2\|T_zv-v\|_2\le2\|v-w\|_2+\|T_zw-w\|_2, and it contains constants. Conjugation and Lipschitz composition preserve HcH_c. Products of two bounded members belong to HcH_c, by the product inequality in L2L^2.

Let Fc\mathcal F_c consist of the measurable sets AA with 1A∈Hc\mathbf1_A\in H_c, modulo completion. Complements and finite unions stay in Fc\mathcal F_c by the preceding properties; increasing unions do by L2L^2 convergence. Thus it is a sigma-algebra. For a real h∈Hch\in H_c, the Lipschitz functions min⁡(1,max⁡(0,n(h−t)))\min(1,\max(0,n(h-t))) converge in L2L^2 to 1{h>t}\mathbf1_{\{h>t\}}. Every real member of HcH_c is therefore Fc\mathcal F_c-measurable, and the same follows for complex members by real and imaginary parts. Conversely the simple Fc\mathcal F_c-measurable functions belong to HcH_c and are dense in L2(Fc)L^2(\mathcal F_c). Hence Hc=L2(Fc)H_c=L^2(\mathcal F_c) and its orthogonal projection is conditional expectation. In particular it sends real functions valued in [−1,1][-1,1] to functions with the same bounds.

For y∈Cy\in\mathbb C and zn∈Ez_n\in E tending to yy, the sequence TznhT_{z_n}h is Cauchy in L2L^2 for h∈Hch\in H_c, because the norm of its difference at n,mn,m is ∥Tzn−zmh−h∥2\|T_{z_n-z_m}h-h\|_2. Its limit defines TyhT_yh, independently of the approximating sequence. The group law, unitarity and strong continuity extend by limits. Realness and interval bounds are closed L2L^2 conditions, so TyT_y preserves them. No pointwise action of C\mathbb C on the probability space is asserted.

For a real g∈Hcg\in H_c with −1≤g≤1-1\le g\le1 there is a jointly Borel field G(y,η)∈[−1,1]G(y,\eta)\in[-1,1] representing TygT_yg for every fixed yy, after choosing a bounded Borel representative of gg. To see this, choose Borel maps qn:C→Eq_n:\mathbb C\to E with countable range and uniformly small ∣qn(y)−y∣|q_n(y)-y|, so that ∥Tqn(y)g−Tyg∥2≤2−n\|T_{q_n(y)}g-T_yg\|_2\le2^{-n} for every yy. Such maps exist by density of EE and uniform continuity of the orbit of gg. Realize Gn(y,η)=Tqn(y)g(η)G_n(y,\eta)=T_{q_n(y)}g(\eta) by the original countable action. The maps GnG_n are jointly Borel and bounded. For each fixed yy the sum of the L1L^1 norms of Gn+1(y,⋅)−Gn(y,⋅)G_{n+1}(y,\cdot)-G_n(y,\cdot) is finite, so the sequence converges almost surely. Define GG to be this limit where it exists and 00 otherwise. It is jointly Borel, and its section at every fixed yy represents the L2L^2 limit TygT_yg. The exceptional subset of the probability space may depend on yy; Tonelli, not a common pointwise action, handles it.

Proof of (a)

Rotate each member so that aj>0a_j>0; with algebraic sides, aj,bj∈Ea_j,b_j\in E, the field of algebraic numbers, and KK denotes its algebraic unit vectors. Let Ω⊂{±1}E\Omega\subset\{\pm1\}^E consist of the labelings with no monochromatic triple (p,p+uaj,p+ubj)(p,p+ua_j,p+ub_j), p∈Ep\in E, u∈Ku\in K, j∈Jj\in J. It is closed, contains the restriction of the given coloring, and is invariant under η↦η(k ⋅+z)\eta\mapsto\eta(k\,\cdot+z). Amenability of E⋊KE\rtimes K gives an invariant probability μ\mu on Ω\Omega, as in the manuscript's Proposition 4.2. Define TzT_z, VkV_k, f(η)=η(0)f(\eta)=\eta(0) and the projection PP onto the continuous vectors HcH_c as in the proof of Theorem 3 of the spectral account. Put g=Pfg=Pf.

Bounds. The continuous-factor argument above shows that gg is real with −1≤g≤1-1\le g\le1, and extends the translations on HcH_c to a strongly continuous unitary representation of C\mathbb C with −1≤Tyg≤1-1\le T_yg\le1 for every yy.

Correlations. As in Theorem 3, (R0) gives $\langle f,T_zf\rangle=\langle g,T_zg\rangle$ for all nonzero z∈Ez\in E. Let κ(y)=⟨g,Tyg⟩\kappa(y)=\langle g,T_yg\rangle, continuous on C\mathbb C. For u∈Ku\in K every η∈Ω\eta\in\Omega satisfies η(0)η(uaj)+η(0)η(ubj)+η(uaj)η(ubj)=−1\eta(0)\eta(ua_j)+\eta(0)\eta(ub_j)+\eta(ua_j)\eta(ub_j)=-1, and the three displacements uajua_j, ubjub_j, u(bj−aj)u(b_j-a_j) are nonzero. Integrating,

1+κ(uaj)+κ(ubj)+κ(u(bj−aj))=0(u∈K),1+\kappa(ua_j)+\kappa(ub_j)+\kappa(u(b_j-a_j))=0\qquad(u\in K),

and by continuity and density of KK in S1S^1 this holds for every unit uu.

Pointwise vanishing. For p∈Cp\in\mathbb C and unit uu the three functions G1=TpgG_1=T_pg, G2=Tp+uajgG_2=T_{p+ua_j}g, G3=Tp+ubjgG_3=T_{p+ub_j}g take values in [−1,1][-1,1], and ∫Q(G1,G2,G3) dμ\int Q(G_1,G_2,G_3)\,d\mu is the left side above, because ⟨Tyg,Ty′g⟩=κ(y′−y)\langle T_yg,T_{y'}g\rangle=\kappa(y'-y). Since Q≥0Q\ge0, Q(G1,G2,G3)=0Q(G_1,G_2,G_3)=0 μ\mu-almost surely.

A measurable sample. Use the jointly Borel field constructed above. For each fixed (p,u)(p,u) its three sections represent the three vectors in the preceding paragraph. Tonelli then gives, for every j∈Jj\in J and every integer n≥1n\ge1,

∫Ω∫∣p∣<n∫S1Q(G(p,η),G(p+uaj,η),G(p+ubj,η)) dσ(u) dp dμ(η)=0.\int_\Omega\int_{|p|<n}\int_{S^1} Q\bigl(G(p,\eta),G(p+ua_j,\eta),G(p+ub_j,\eta)\bigr) \,d\sigma(u)\,dp\,d\mu(\eta)=0 .

These are countably many conditions, so one η0\eta_0 makes every inner double integral vanish. Put c′(y)=+1c'(y)=+1 if G(y,η0)≥0G(y,\eta_0)\ge0 and c′(y)=−1c'(y)=-1 otherwise; c′c' is Borel. For almost every (p,u)(p,u) and every jj the three values of G(⋅,η0)G(\cdot,\eta_0) include +1+1 and −1-1, which c′c' keeps, so the copy is not monochromatic. Thus c′c' weakly avoids T\mathcal T. □\square

Proof of (b)

Let cc weakly avoid T\mathcal T, and let A∗A^* and B∗B^* be the points of density one of {c=1}\{c=1\} and {c=−1}\{c=-1\}. They are disjoint and their union is conull. Suppose pp, p+uajp+ua_j, p+ubjp+ub_j all lie in A∗A^*. For small h>0h>0 choose p′p' uniformly in the disk of radius hh about pp and, independently, u′u' uniformly on an arc about uu of angular width h/(1+∣aj∣+∣bj∣)h/(1+|a_j|+|b_j|). Each of p′+u′ajp'+u'a_j and p′+u′bjp'+u'b_j has a density at most 1/(πh2)1/(\pi h^2), supported in the disk of radius 2h2h about p+uajp+ua_j, respectively p+ubjp+ub_j. Density one at the three points makes the probability that all three random points lie in {c=1}\{c=1\} tend to one. The law of (p′,u′)(p',u') is absolutely continuous, which contradicts weak avoidance. The same holds for B∗B^*. 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 A∗∪B∗A^*\cup B^*, where the density classes color it with no monochromatic copy. Each avoidance condition is closed in {±1}C\{\pm1\}^{\mathbb C} and involves three coordinates, so product compactness gives a coloring of the whole plane avoiding T\mathcal T. □\square

This yields an arbitrary avoider, not a measurable completion of the density classes on the omitted null set.