The Euclidean plane is not five-colorable

Deduction of Theorem 1.1. A proper five-coloring would give a weak measurable five-coloring by Theorem 1.3, contrary to Theorem 1.4.

For the upper bound, use the hexagonal construction described by Hadwiger [13]. Take Voronoi hexagons of a triangular lattice, each with circumradius r=2/5. Their centers have nearest-neighbor distance \sqrt3r. Write the lattice as a scaled copy of \Z+\Z\omega, where \omega=e^{2\pi i/3}. Multiplication by 2-\omega gives a similar sublattice of index seven, since |2-\omega|^2=7. Color centers by its seven cosets, and give each point the color of its hexagon, assigning boundary points to any incident hexagon. Distances inside one hexagon are at most 2r<1. Centers of distinct same-color hexagons are at least \sqrt{21}r apart, so points in those hexagons are at least (\sqrt{21}-2)r>1 apart. This is a proper seven-coloring, including all boundaries.□

Proof. Theorem 1.1 and graph compactness [4, Theorem 1] give a finite unit-distance graph H that is not five-colorable. Set m=|E(H)| and \delta=1/m. Every five-coloring of H has at least one monochromatic edge. Averaging this count with any of the measures in (i) gives mp_5^\mu(c)\ge1, by isometry invariance and finite additivity; this is the finite-graph bound of Gwyn and Stavrianos [12, Lemma 2.10 and Theorem 2.11]. The periodic and asymptotic needle frequencies in (ii) are bounded below by the minimum monochromatic-edge fraction of H, hence by 1/m, by Bourgeat et al. [2, Theorems 1–2]. Borel representatives give the same needle frequencies by Fubini; for a periodic coloring choose them in one cell and extend periodically. The graph and the positive constant here are existential.□

Proof. If a coset d+C has positive mass, its K-orbit is finite, since the cosets in that orbit are disjoint and have the same mass. The action on this finite orbit gives a finite image of the divisible abelian group K. Such an image is trivial: a finite abelian group cannot be divisible unless it is trivial. Hence every u\in K fixes the coset, and (u-1)d\in C. Taking u\ne1 and applying the inverse scalar shows d\in C, contrary to \nu(C)=0. The difference assertion now follows by integrating \nu(d+C)=0 against \nu(d); the relation d'-d\in C is Borel.

Restriction to the line aF is a continuous map from D to \widehat{F_{\mathrm{disc}}}. The ordinary continuous characters on F are exactly t\mapsto e^{i\lambda t}, \lambda\in\R. Their parametrization is continuous and injective, with Borel image by compact exhaustion, as for j. This proves the Borel assertion for L_a.

If aF,bF are nonparallel, then a,b form an F-basis of E. A character continuous on both lines has the form

for real \lambda,\eta. The corresponding real coordinate change is invertible and continuous, so this character belongs to C. Thus L_a\cap L_b\subset C for nonparallel lines. Rotations give infinitely many distinct unoriented lines uaF, and their events have the same \nu-measure. Any finite collection of them is pairwise disjoint modulo \nu-null sets. Their common measure must therefore be zero.□

Proof. On L^2(\nu) consider the unitary representation of the additive group F given by

First, this representation has no nonzero vector that is strongly continuous for the usual topology on F. Suppose that v is such a vector. Put

It is positive definite and uniformly continuous on F, since

It therefore extends continuously to \R. The extension remains positive definite, by approximation of each finite set of real arguments by elements of F.

Apply the ordinary Bochner theorem to this continuous positive-definite function on \R; see Bekka et al. [1, Theorem D.2.2]. It is important here that the coefficient has first been extended to \R; no local compactness of F in its usual topology is assumed. Push its finite positive representing measure to \widehat{F_{\mathrm{disc}}} using the continuous-character parametrization. Its Fourier coefficients agree with those of the restriction-map pushforward of |v|^2\nu. Uniqueness of Fourier coefficients for finite measures on the compact character group identifies these two measures. Indeed, the characters form a unital conjugation-closed algebra separating its points, so their uniform density in the continuous functions gives this uniqueness. The pushforward of |v|^2\nu is thus concentrated on the ordinary continuous characters, so |v|^2\nu is concentrated on L_a. Lemma 2.4 gives \nu(L_a)=0, and hence v=0.

Now form the weak integral

Translation invariance of I gives

and consequently

as s\in F tends to zero. Thus V_\phi is strongly continuous and must be zero by the preceding argument. Pairing with the constant function proves (2.4).□

Proof. The difference in (2.5) is bounded in absolute value by |s|. The objective is therefore finite without a first-moment assumption. It is Lipschitz in s and strictly convex, because s\mapsto\sqrt{1+|s-t|^2} is strictly convex for every t.

Choose R>0 with p=\lambda(\overline{B(0,R)})>1/2. On this ball the integrand is at least |s|-R-\sqrt{1+R^2}, and elsewhere it is at least -|s|. Hence

Coercivity and continuity give a minimizer, and strict convexity gives uniqueness.

For each fixed s, the integrand is bounded and continuous in t, so \lambda\mapsto J_\lambda(s) is Borel. The infimum over rational points equals the minimum. Enumerate \Q^2 and, for each positive integer n, choose its first point whose objective is less than that infimum plus 1/n. These choices are measurable. For a fixed \lambda they converge to the unique minimizer: coercivity confines all sufficiently accurate choices to a compact set, and every cluster point minimizes J_\lambda. Their limit is therefore the Borel map L.

If \lambda' is the law of t-r for t of law \lambda, then an identity of bounded differences gives

Thus L(\lambda')=L(\lambda)-r. Orthogonal covariance follows from J_{O_*\lambda}(Os)=J_\lambda(s). These identities prove both asserted equivariances.□

Proof. All relative difference laws in the statement are K-invariant, by equivariance of conditional kernels. We use transfinite induction in its universal form: at each ordinal the assertion ranges over all ambient ergodic standard systems, equivariant maps with wild law, and towers satisfying (2.6). This allows an earlier induction assertion to be applied below to a new factor system and a new equivariant map.

At the trivial factor the assertion is Lemma 2.4. Suppose that it holds at every earlier ordinal, and that it fails at a noninitial stage Y of a tower in a system X. Write \kappa_y for the conditional law of d over y\in Y.

Positive-mass cosets and their centers. The function

is Borel in (y,z), by Borel parameter integration. This uses only that C is Borel, not that it is closed. The function x\mapsto r(y(x),d(x)) is invariant and has integral \nu_Y(C)>0. Ergodicity makes it a positive constant a almost everywhere.

For almost every y, therefore, \kappa_y-almost every sampled coset has mass a. Such cosets are disjoint and there are only finitely many; they exhaust the probability. Their number m satisfies ma=1, and is consequently independent of y. Thus \kappa_y is concentrated on m distinct cosets of mass 1/m each. None is C almost surely, because the original marginal law gives C measure zero.

On the Borel set r(y,z)>0, define the probability kernel \lambda_{y,z} on \R^2 by

for Borel A\subset\R^2, interpreting the integrand as zero off C. The Borel inverse of j makes this a measurable kernel. Set

If z is replaced by z+j(r_0), the kernel is translated by -r_0; Lemma 2.6 shows that b(y,z) is unchanged. It is therefore constant on each positive-mass coset, and remains in that coset. The same lemma and the orthogonal action of K on frequency parameters show

on the relevant conull sets.

Let Z be the factor of X generated by Y and b(x)=b(y(x),d(x)). It is a standard ergodic factor. The map b:Z\to D is equivariant and has wild law. Its conditional law \beta_y over Y is uniform on the m distinct centers of the positive-mass cosets. We have converted the hypothetical failure of singularity into a finite set of equivariant centers over each fiber. The earlier induction hypothesis and the line average will give selected real algebraic frequencies with small conditional collision probabilities on a common fiber. Compact approximation, however, will force many pairs of unit-direction phase lists to match one pair of lists. Writing each selected frequency as a sum of its two directions then places its phases near at most m^2 products, forcing a larger collision probability.

A finite list of phases. For each y, consider the unordered list of m phases \psi_1 at the points of \beta_y, retaining repetitions when phases coincide. Encode this list by the vector H(y)\in\C^m of its elementary symmetric functions. The vector is measurable: its power sums are m\int\psi_1^n\,d\beta_y, and Newton’s identities recover the elementary symmetric functions. It is bounded.

The encoding is a continuous injection of the compact space \T^m/S_m into \C^m. On unordered lists use maximum matching distance,

The inverse encoding is uniformly continuous for this metric. Equivariance of \beta_y and \psi_1(kd)=\psi_k(d) show that U_kH encodes the list of phases at \psi_k.

Choose \rho>0 so small that uniform-circle probability of \{|z-1|\le5\rho\} is less than 1/(16m^2). There are an integer N\ge1 and a constant c>0 with the following property: if a probability \omega on \T satisfies

then

Indeed, choose a continuous majorant of the smaller arc that takes values in [0,1] and vanishes outside the larger arc. Approximate it uniformly within 1/(64m^2) by a real trigonometric polynomial of degree at most N. The constant coefficient is within that error of its circle integral. Choose c so that c times the sum of the absolute values of its nonconstant coefficients is less than 1/(32m^2). The integral against \omega of the majorant is then less than 1/(8m^2), which proves (2.7). Negative moments are controlled by the positive ones because \omega is a positive measure.

Choose s>0 such that two encoding vectors at distance less than s have phase lists matching within \rho. Put \tau=s/100, and fix \gamma>0 such that

Return to an earlier factor. The rotations of H admit vector-valued approximations of the form

with fixed bounded \C^m-valued functions H_r on Y and coefficients over some earlier factor W. At a successor this follows from (2.6), applied to the finitely many coordinates and combining their data. At a limit, first approximate H in L^2, within \gamma/2, by bounded functions on an earlier successor factor. Such approximation follows from generation of the limit factor, for example by conditional expectations. Its error remains unchanged under rotation. Then use (2.6) at that successor, with its predecessor as W, to obtain a further error less than \gamma/2. Combining coordinates gives (2.9). The coordinate count, number of basis functions, their uniform bounds, and B are all fixed independently of k.

The earlier tower through W is also a tower of factors of Z: Z contains Y and hence every predecessor of Y. All measures, actions, limit joins, and approximation properties of those factors are unchanged. The universal earlier induction assertion therefore applies in Z to the map b. It says that the difference law of b over W is wild. Its Fourier coefficient at a\in E is

by conditional independence. Lemma 2.5 applies to this coefficient function.

Fixing the number of tests before choosing a small parameter. At any fixed fiber of W, the two coefficient arrays for two approximations in (2.9) belong to a fixed bounded subset of \C^{2q}. Partition this subset into a finite number L of bins so fine that, within one bin, the corresponding approximating functions differ pointwise by at most \tau for each of the two vectors. For example, coordinate diameters less than

suffice. The number L is independent of the fiber and of the chosen rotations. Fix an integer M\ge2 so large that

All approximation data have been fixed before this choice.

There is a nonzero t\in F with |t|<1/M for which every value

is as small as we prescribe. To see this, choose a nonnegative \phi\in C_c((-1/M,1/M)) with positive integral. Each indicated nonnegative test has zero \phi-weighted I-integral by (2.4). Their finite sum does too. If that sum were bounded below by a prescribed positive number at every point where \phi>0, positivity of I would contradict I(\phi)>0. Thus some such point makes their sum, and hence every test, as small as required. Choosing this bound below one also ensures t\ne0, since p(0)=1.

Set x_j=jt. Choose the prescribed smallness so that Markov’s inequality and a finite union bound give, on a set of W-probability greater than 0.99, all of the bounds

More explicitly, the probability of failure of any one bound is at most p(n(x_j-x_l))/c^2, by (2.10); there are finitely many bounds. Apply (2.7) to the conditional law of \psi_{x_j-x_l}(b). On the same set of fibers,

A common coefficient bin on a good fiber. Each |x_j|<1 is a sum of two elements of K. One explicit choice is

The square roots belong to F, and x_j=k_j+k'_j. Apply the fixed approximations (2.9) to U_{k_j}H and U_{k'_j}H for every j.

Let e_j^+(w),e_j^-(w) be the respective conditional squared L^2 errors on the fiber of Y over w. Their global integrals are less than \gamma^2. If at least 0.1M indices fail to have both conditional L^2 errors less than \tau, then

Markov’s inequality and (2.8) bound the probability of this event by

In particular, on a set of fibers of probability greater than 0.99, more than 0.9M indices have both conditional errors less than \tau.

Fix a fiber w satisfying this property and all of (2.11), and avoid the null sets for the finitely many identities and bounds used here. Bin the two coefficient arrays at this w as above. There is a subset J of the good indices in a single bin with |J|>10m^2. Fix j_0\in J. For either type of encoding the conditional L^2 distance from index j to index j_0 is at most 3\tau: the two approximation errors each cost less than \tau, and the approximants differ pointwise by at most \tau.

For a uniform j\in J and a conditional point y over w, the probability that either encoding distance is at least s is at most

Thus, with probability greater than 0.95, both phase lists for index j match their respective representative lists for j_0 within \rho. Let q(y) be the fraction of indices in J for which both matches hold. We have

The collision contradiction. Fix such a fiber point y and any point b_0 in the support of \beta_y. For every matching index j, the two phases \psi_{k_j}(b_0) and \psi_{k'_j}(b_0) are within \rho of entries of their respective representative lists. Consequently

is within 2\rho of one of the at most m^2 products of entries from those two lists. The matchings need not identify the same center in the two lists; allowing all m^2 products accounts for this.

Assign each of the q(y)|J| matching indices to one such product. By the sum-of-squares inequality, the number of ordered pairs assigned to the same product is at least q(y)^2|J|^2/m^2. Removing diagonal pairs discards at most |J| pairs. Hence, for independent uniform j,l\in J,

This bound holds for every possible center b_0, so it can be integrated first over \beta_y and then over the conditional law of y given w. Jensen’s inequality and (2.12) give a lower bound strictly greater than

But (2.11) bounds the conditional collision probability for every distinct pair in the original set of M indices by 1/(4m^2). Averaging those bounds over the selected J gives an upper bound smaller than 1/(4m^2). Choosing J after fixing w is harmless because all those pair bounds hold simultaneously at that fiber. This contradiction completes the induction step, and hence the universal transfinite induction.□

Proof. We first construct a strict extension Y' of any factor Y whose relative product is not ergodic. An invariant kernel on that product will provide an equivariant finite-rank space on almost every fiber. Its bounded basis generates Y' and supplies the approximation property (2.6). We then iterate this enlargement.

Suppose that Y has already been constructed and that X\times_YX is not ergodic. Choose a bounded invariant nonconstant kernel L(x,x') on this relative product. Its conditional expectations onto either coordinate are invariant functions on X, so both are the constant \int L\,d(\mu\mathbin{\times_Y}\mu). Subtract this constant. The result is still nonzero, and its two coordinate conditional expectations are zero.

For almost every y, let A_y be the integral operator with kernel L on the fiber Hilbert space \mathcal H_y=L^2(\mu_y). These are Hilbert–Schmidt operators, with

Invariance of the kernel makes the operator field equivariant under the unitary maps between fibers. For a sufficiently small fixed \delta>0, the spectral projection

is nonzero on a positive-measure set of fibers. Its rank is finite and bounded above by \norm{L}_\infty^2/\delta. The rank is invariant on Y, and Y is ergodic as a factor of X. It is therefore a constant integer d\geq1 almost everywhere. The range of Q_y is orthogonal to the fiber constants.

Here are the measurability details for this construction. Work with standard Borel models and fiber-supported conditional probabilities. Choose a countable algebra generating the Borel structure of X and its simple functions with complex-rational coefficients. Their restrictions are dense in each L^2(\mu_y). Conditional integration against the bounded kernel represents the images of these sections under A_y and A_y^* by measurable functions on X; the same holds for polynomials in A_y^*A_y. On their common bounded spectral interval, there are uniformly bounded polynomials converging pointwise to \ind_{(\delta,\infty)}: first choose bounded continuous approximations to this indicator, with value zero at \delta, and then uniformly approximate them by polynomials. Spectral calculus gives convergence in each fiber Hilbert norm. Domination gives convergence in global L^2(X) as well. A subsequence converges in fiber norm for almost every y, identifying the global limit with the projected section. Thus the projected dense sections are measurable. Their fiber scalar products make rank and linear-independence tests measurable. Selecting the first available independent section at each step and applying fiberwise Gram–Schmidt gives measurable fiber-orthonormal sections e_1,\ldots,e_d spanning the range of Q_y.

For each k\in K the sections U_ke_i differ from the e_i by a unitary coefficient matrix over Y. Consequently \sum_{i=1}^d|e_i(x)|^2 is invariant on X. Its integral is d, so ergodicity gives

In particular, the sections are bounded. Adjoin them to Y to generate an invariant factor Y'. This is a strict extension: every e_i has conditional mean zero over Y and has unit fiber norm, and hence is not Y-measurable.

Bounded polynomials in the e_i,\bar e_i, with bounded coefficients from Y, are dense in L^2(Y'). Indeed these functions form a bounded unital conjugation-closed algebra generating Y', and approximation first by functions of finitely many generators and then by polynomials gives the density. A rotation of a fixed polynomial expands into finitely many fixed bounded monomials in the e_i,\bar e_i. Its coefficients are functions on Y, uniformly bounded independently of the rotation: the original coefficient sup norms are preserved, and the entries of the unitary matrices have modulus at most one. For a bounded H on Y', choose one polynomial approximation in L^2. Rotating preserves its approximation error. The finite expansion of the rotated polynomial proves (2.6) for Y'\to Y.

Starting at the trivial factor, repeat this step whenever the relative product is not ergodic, and take generated joins at limit ordinals. Every countable stage is a standard factor modulo null sets. The process stops at a countable ordinal. Otherwise, at every successor through all countable ordinals, choose a unit vector in the new factor orthogonal to the preceding factor. Nestedness makes these vectors pairwise orthogonal. There would be \aleph_1 such vectors, contradicting separability of L^2(X). At the stopping stage the relative product is ergodic by the stopping rule.□

Proof. Put \mathcal R=X\times_YX and denote its conditional expectation onto Y by P_{\mathcal R}. First prove (3.3). With v_0=v-Pv and w_0=w-Pw, its difference equals P(v_0U_{u^d}w_0). Conditional independence gives

The integral of w_0\otimes\bar w_0 is \int_Y|Pw_0|^2\,d\nu=0. The mean ergodic theorem on the ergodic system \mathcal R, applied through the surjective power map, makes the average of the right side tend to zero. This proves the pair criterion along every Følner sequence.

The same criterion holds on \mathcal R over Y. To see this first for simple tensors v=a_1\otimes a_2 and w=b_1\otimes b_2, use

Subtract the analogous product of the factor parts. The difference is a sum of two terms, each containing one error from (3.3) on X and one uniformly bounded factor. Its averaged squared L^2(Y) norm therefore tends to zero. Linearity gives the result for finite sums of tensors. Such sums can approximate any bounded function on \mathcal R in L^2, with a common sup bound: use conditional expectations onto finite partitions generated by rectangles in the two coordinates. Approximate one of v,w at a time; conditional-expectation contraction and measure preservation bound the error uniformly in u. The pair criterion follows for all bounded functions on \mathcal R.

We next prove the following vector assertion on either X or \mathcal R. On either system use P for its expectation onto Y. If n_1,\ldots,n_l are distinct nonzero integers, the v_j are bounded, and Pv_{j_*}=0 for some j_*, then

for every Følner sequence. Both systems are ergodic and satisfy the pair criterion, so the same induction proves the assertion on each. By rescaling we may assume \norm{v_j}_\infty\leq1.

For l=1, the assertion is mean ergodicity and \int v_1=\int Pv_1=0. Suppose it is proved for l-1, and set

For each fixed h, invariance of the integral gives

The l-1 shifted exponents in the last expression are distinct and nonzero. Expand each shifted F_{j,h} into PF_{j,h} and its centered remainder. The induction hypothesis makes the average of every term with a centered shifted factor tend to zero in L^2; multiplying by the fixed bounded F_{l,h} does not affect that conclusion after integration. The remaining shifted product is Y-measurable, so the unshifted F_{l,h} may also be replaced by PF_{l,h} inside its integral. All the conditional factors are bounded by one. Hence, for every Følner sequence,

This bound also covers j_*=l: bound that unshifted factor in L^1, then in L^2, and the others in L^\infty. If j_*<l, use the same bound on its shifted factor and measure preservation. The centered factor is therefore not lost by moving one exponent to zero.

The pair criterion with v=\bar v_{j_*}, w=v_{j_*}, and d=n_{j_*}\ne0 shows that averages of a(h)^2 tend to zero along every Følner sequence; the same is true of a(h) by Cauchy–Schwarz. If (\Psi_M) is any chosen Følner sequence, it follows that

Indeed, a subsequence of bad translates would itself be a Følner sequence, because K is abelian, contradicting the every-sequence vanishing just proved.

To finish the induction, fix a finite nonempty shift set S\subset K. The difference between the averages of V_u and of |S|^{-1}\sum_{h\in S}V_{hu} tends to zero in norm by the Følner property. Jensen’s inequality and expansion in pairs of shifts give

For the second inequality substitute u'=h'u and apply (3.7) along the translated Følner sequence. Now take S=\Psi_M. The last expression is bounded by \sup_t\operatorname{Avg}_{h\in\Psi_M}a(ht) and tends to zero by (3.8). The limits are taken first in N for fixed finite S, and only then in M. This proves (3.5).

Finally, expand the product in (3.2) by writing each v_j=Pv_j+(v_j-Pv_j). Consider a term W_u=\prod_jU_{u^{n_j}}w_j with at least one Pw_j=0. Conditional independence gives

The corresponding tensor is centered over Y, since P_{\mathcal R}(w_j\otimes\bar w_j)=|Pw_j|^2. In this scalar integral we may add one common integer to all exponents, by measure preservation, and choose it so that none of them is zero. Apply (3.5) on \mathcal R. The average of (3.9) tends to zero. There are finitely many centered terms, so the squared norm of their sum has vanishing average as well. The term with no centered factor is already Y-measurable and equals the product of projections in (3.2). This proves the lemma.□

Proof. First take A,B>0 and put L=\sqrt{AB}\in F. For every u\in K,

For the second equality, factor the last expression as u((A-B)+L(u-u^{-1})) and use u-u^{-1}=2i\operatorname{Im}u.

On the first side, \psi_{A-Bu^2}=\psi_AU_{u^2}\psi_{-B}. By (3.2) the projection of this product may be replaced, in averaged squared L^2 error, by g_AU_{u^2}g_{-B}. Both have L^2 norm at most one, so their squared norms differ in averaged absolute value by a quantity tending to zero. Indeed the absolute difference is at most twice their L^2 distance, and then Cauchy–Schwarz applies to the finite average. Ergodicity of Y and surjectivity of the square map now give

Here the intervening squared norm is \int_Y|g_A|^2U_{u^2}|g_{-B}|^2\,d\nu.

On the second side of (3.14), apply (3.2) with exponents 0,1,2. The projected character is approximated by

Its squared norm is at most \norm{g_{A-B}}_2^2=m(|A-B|), since the two outside factors have modulus at most one. Equal moduli give equal m-values, so (3.15) implies m(A)m(B)\leq m(|A-B|). Taking negative logarithms proves (3.12), including zero values of m with the stated extended-real convention. If A=0 or B=0, the inequality is an identity because w(0)=0.

For (3.13), use ru+r\bar u=2r\operatorname{Re}u and the distinct exponents 1,-1. The same approximation reduces the average to squared norms of U_ug_r\,U_{u^{-1}}g_r. Those squared norms equal

Mean ergodicity on Y through the surjective power map u\mapsto u^{-2} gives the limit m(r)^2. Radiality replaces the real argument by its absolute value, proving (3.13).□

Proof. For t\in N_0 put

Choose a \Q-basis of N_0. In the rational coordinates of t, the coordinates of u(t) in a \Q-basis of N are rational functions over \Q. This follows, for example, by inverting the matrix for multiplication by 1-it. Its denominator is nonzero for all parameters in N_0, since t is real.

No nonzero \Q-linear functional \ell:N\to\Q vanishes on all u(t). Suppose one did. For every fixed a\in N_0, the rational function \ell(u(ha)) vanishes at every rational h, and hence is identically zero. Its formal expansion at h=0 comes from

The first three coefficients imply \ell(1)=\ell(ia)=\ell(a^2)=0 for every a\in N_0. Squares span N_0 over \Q, as 4a=(a+1)^2-(a-1)^2. Thus \ell vanishes on N_0+iN_0=N, a contradiction. This is a calculation in rational function coordinates and formal series; it uses no usual-topology continuity of \ell.

For any proper rational subspace of N, choose a nonzero rational linear functional annihilating it. Its value on u(t) has a nonzero numerator polynomial in the rational coordinates of t. A finite union of such proper subspaces therefore cannot contain all u(t): the product of their nonzero numerator polynomials is nonzero, and a nonzero polynomial over the infinite field \Q cannot vanish on all of rational affine space. Inductively choose each new element outside the spans of all subsets of at most d_0-1 earlier elements. Every subset of at most d_0 chosen elements remains independent. This constructs finite sets of every required size.□

Proof. Keep every partial sum as its residue in [0,p). A positive input p_j<p has cost at most H, while the input p has zero residue. The inverse of a nonzero residue b is p-b and satisfies w(p-b)\leq w(p)+w(b). Thus each signed input \pm p_j has a residue of cost at most 2H; the zero cases cause no difficulty.

For two current residues a,b\in[0,p), if a+b<p, use three successive applications of the difference inequality:

All arguments are nonnegative. If a+b>p, the new residue is a-(p-b), whose cost is at most w(a)+w(p)+w(b). If a+b=p, its cost is zero. Adding a signed input therefore increases the current cost by at most 4H. Expand the integer combination into at most M signed inputs and start from residue zero. Its final residue is z itself, proving the bound.□

Proof of Theorem 2.3. First let \mu be ergodic and wild, and use the factor and radial function constructed above. Suppose that m(r)>0 for some r\in F, r>0. Choose H>0 finite so that e^{-H}<m(r)^2. The set

is syndetic. Indeed, if no finite translates cover K, then for every finite S\subset K some translate Sv avoids A_0. Apply this to the sets of a Følner sequence and choose the avoiding translate separately for each set. Those translates still form a Følner sequence, whereas all their averaged m-values are at most e^{-H}. This contradicts (3.13). Hence there are q_1,\ldots,q_J\in K such that

Choose a number field N_0\subset F containing the real and imaginary parts of every q_j, and put N=N_0(i) and d_0=[N:\Q]. Lemma 3.4 supplies a finite S\subset K\cap N of size greater than J(d_0-1) in rational linear general position. For each v\in K, apply (3.16) to all the points sv, s\in S. Pigeonhole gives an index j and d_0 elements s for which q_jsv\in A_0. Their coefficients k=q_js form a \Q-basis of N, since multiplication by q_j is an invertible \Q-linear map. Call such a basis a good selection for v.

There are only finitely many possible selections. Clearing the rational denominators in the basis expression of 1, choose one positive integer D_0 such that, for each possible selection \mathcal T,

where the finite bound M_0 is common to all selections. There is also a common c_0>0 such that

for every selection. To prove this, for a fixed selection the maximum is a continuous function of the ordinary unit direction v. It never vanishes: the rational span of the selection contains 1 and i, so vanishing would imply both \operatorname{Re}v=0 and \operatorname{Re}(iv)=0. Take a positive minimum on the compact unit circle and then the minimum over the finitely many selections. We may and do take c_0\leq1.

Now let v\in K satisfy |\operatorname{Re}v|<c_0/D_0, and choose a good selection \mathcal T for it. Its nonnegative numbers

satisfy w(p_k)\leq H and p\geq2rc_0. Taking real parts of (3.17), and inserting the signs of the real parts and then the overall sign, expresses

as an integer combination of the p_k with coefficient length at most M_0. All these numbers lie in F. Lemma 3.5 therefore gives

Every t\in F\cap[-1,1] is the real part of an element of K, namely t+i\sqrt{1-t^2}. For every s\in F\cap(0,2rc_0) apply (3.19) with t=s/(2rD_0). It follows that

Choose a nonnegative \phi\in C_c(\R) supported in this open interval with positive integral. Positivity of the functional I and its agreement with Lebesgue integration on continuous test functions give

This contradicts (2.4) for the wild law \nu_Y, with a=1, in view of (3.11). This step uses no ordinary-measure measurability of the radial function: I is defined on all the bounded, boundedly supported functions in question.

Thus m(r)=0 for every positive r\in F, and radiality gives m(z)=0 for all nonzero z\in E. Consequently

Fourier uniqueness on the compact abelian group D identifies \mu with Haar probability.

Finally, an arbitrary wild invariant probability has an ergodic decomposition for the countable group K on the standard compact space D. Since C is invariant Borel and has measure zero, almost every ergodic component also gives C measure zero. The ergodic case just proved makes each such component Haar probability. Their integral is therefore Haar probability as well. This proves the theorem.□

Proof. The subspace is closed. Indeed, if v is close in L^2 to a continuous vector w, then

It contains the constants and is invariant under the translations.

Apply the spectral theorem for the discrete abelian group E, with dual group D. For v\in L^2, write \lambda_v for its finite scalar spectral measure, so

If \lambda_v(C^c)=0, the second integral tends to zero as z\to0 in the usual topology, by dominated convergence and the continuous-character parametrization.

Conversely, suppose v\in\mathcal H_c. Its coefficient F_v(z)=\langle v,T_zv\rangle is uniformly continuous on E:

It extends to a continuous positive-definite function on \R^2. Positive definiteness follows by approximating each finite list of real-plane points by points of E and passing to the limit in its defining quadratic form. Bochner’s theorem supplies a finite measure on \R^2 representing this extension. Its pushforward under j has the same Fourier coefficients on D as \lambda_v, so Fourier uniqueness identifies the two measures. Thus \lambda_v(C^c)=0. The representation result is the abelian spectral theorem (also called the Stone–Naimark–Ambrose–Godement theorem); it and Bochner’s theorem are given in Bekka et al. [1, Theorems D.3.1 and D.2.2]. The spectral theorem is applied to the discrete locally compact group E, while Bochner’s theorem is applied to \R^2. This identifies \mathcal H_c with the range of the spectral projection on the Borel set C.

To obtain the factor assertion, it is essential that the original T_z are Koopman operators. They commute with complex conjugation and with pointwise composition. For bounded v,w\in\mathcal H_c,

so vw\in\mathcal H_c. For a Lipschitz function \Phi:\C\to\C,

In particular, radial projection of v onto a bounded complex disk is still in \mathcal H_c. Such projections converge to v in L^2 as the disk radius tends to infinity.

Let \mathcal F_c be the completed factor generated by the bounded members of \mathcal H_c. Those members form a bounded unital star-algebra. Its L^2 closure is L^2(\mathcal F_c): for a finite list of its bounded generators, polynomials in their coordinates and conjugates approximate continuous functions on the compact range by Stone–Weierstrass; continuous functions are dense in L^2 of the corresponding finite-dimensional distribution. Finite-coordinate measurable functions are dense in the generated factor, by the monotone-class theorem. The closedness already proved therefore gives L^2(\mathcal F_c)\subset\mathcal H_c. Conversely, the bounded truncations of every v\in\mathcal H_c are \mathcal F_c-measurable and converge to v in L^2. Hence \mathcal H_c=L^2(\mathcal F_c). Its orthogonal projection is conditional expectation.

For x\in\R^2, choose z_n\in E with z_n\to x. For v\in\mathcal H_c,

Define T_xv by this limit. The definition is independent of the approximating sequence, and strong limits together with the original group law give the group law on \R^2, inverses, and strong continuity. The operators are unitary. The positive cone in L^2 is closed, so T_x preserves positivity, and T_x1=1. No pointwise action of \R^2 is needed for these conclusions.

Finally, for u\in K, covariance gives

Thus rotations preserve \mathcal H_c; their inverses do as well, so its orthogonal projection commutes with them.□

Proof. An invariant law on proper labelings. Restrict the given coloring to E. In the compact metrizable space \{1,\ldots,k\}^E, let \Omega be the closed set of proper labelings of the unit-distance graph on E. It is nonempty. The countable group E\rtimes K acts continuously on \Omega; translations and rotations preserve the unit-edge constraints. This group is amenable, being an extension of two abelian groups [1, Theorem G.2.1 and Proposition G.2.2(ii)]. The Følner criterion [1, Theorem G.5.1], applied to successive finite subsets of this countable group, gives a finite-set Følner sequence. Averages of a point mass over a finite-set Følner sequence have a weakly convergent subsequence, and every limit is invariant. This gives an invariant Borel probability \mu on \Omega.

Let f_i be the indicator that the origin has label i. Choose the translation convention so T_zf_i is the indicator of label i at z. The rotations about the origin fix each f_i, and properness gives

The continuous factor and zero correlations. Use Lemma 4.1 to write

Then 0\le p_i\le1 and \sum_i p_i=1. Both p_i and b_i are rotation-fixed. The scalar translation spectral measure of b_i is K-invariant: covariance gives equality of its Fourier coefficients at z and uz, and Fourier uniqueness gives invariance of the measure. It gives C measure zero, by the spectral-subspace assertion of the lemma. If it is nonzero, normalize it and apply Theorem 2.3. The scalar spectral measure is therefore a multiple of Haar measure, including the zero multiple. It follows that

The subspace \mathcal H_c and its orthogonal complement are translation invariant, so both cross terms in (4.1) vanish. Consequently

The algebraic unit directions are dense in the usual unit circle, as is seen already from the rational tangent-half-angle parametrization. Strong continuity of the extended operators therefore gives this identity for every usual unit direction u\in S^1. For every x\in\R^2, unitarity and the group law then give

The functions here are nonnegative equivalence classes. This calculation uses no pointwise realization or multiplicativity of the extended \R^2 action.

A measurable sample. The continuous map x\mapsto T_xp_i takes values in the separable space L^2(\mu). On a bounded spatial domain, approximate it by finite-valued simple maps in the L^2(\mu) norm, choosing the integrals of their squared approximation errors to be summable. Choose measurable representatives for their finitely many values. The resulting jointly measurable functions converge in the product L^2 space. A subsequence converges in L^2(\mu) for almost every x, so its product-space limit represents T_xp_i at almost every x. Applying this on a countable exhaustion, or on its disjoint measurable annuli, gives a jointly measurable field H_i(x,\omega) on \R^2\times\Omega, correct as an L^2(\mu) representative for almost every x. Since the probability model is standard, Borel versions may be used. Taking real parts and clamping each H_i to [0,1] preserves this representation property and makes all the fields everywhere nonnegative.

There is a plane-null set N outside which all these representations are correct. For almost every x, their sum is one for \mu-almost every \omega. Moreover, on every bounded spatial domain,

The assertion about x+u follows by translation invariance of Lebesgue measure for each fixed u, followed by Fubini. Thus (4.2) applies to the chosen representatives for almost every pair (x,u). Tonelli gives, for every positive integer m,

Choose one sample \omega for which all the inner integrals vanish, and for which \sum_iH_i(x,\omega)=1 for almost every x. There are only countably many conditions here. At every point where the sum is one, assign the least index with positive field; on the remaining plane-null set assign any index. This is measurable. A same-color pair outside that null set would have positive product of the corresponding fields, which occurs only on a dx\,d\sigma(u)-null set. Hitting the spatial exceptional set at either endpoint is also a product-null event. This proves the weak measurable condition.□

Proof. Suppose x and y=x+u, with \abs{u}=1, both have density one of A_i. For h>0 small, choose x' uniformly in B(x,h) and independently choose u' uniformly on an arc of positive angular measure satisfying \abs{u'-u}<h. Density at x gives \mathbb P(x'\notin A_i)=o(1) as h\downarrow0. The distribution of x'+u' is a mixture of translated uniform disk distributions. Its density is bounded by 1/(\pi h^2), and its support is in B(y,2h). Therefore

For small h the probability that both points lie in A_i is positive. But the pair-parameter distribution is absolutely continuous with respect to dx\,d\sigma(u), and the two points have distance one. This contradicts the weak condition.

Lebesgue differentiation makes the union of the density-point sets conull. Two different color classes, being disjoint up to null sets, cannot both have density one at the same point, so these sets are disjoint.□

Proof of Theorem 1.3. The forward implication is Proposition 4.2. For the reverse implication, Lemma 4.3 gives a conull set partitioned into strictly unit-independent density-point classes. Any fixed finite configuration of plane points has a translate entirely inside that conull set: the permissible translation vectors are a finite intersection of conull sets. Its unit-distance graph therefore has a proper k-coloring.

In the compact product space \{1,\ldots,k\}^{\R^2}, each unit-edge condition is closed and depends on two coordinates. Every finite family of these conditions involves a finite configuration and is satisfiable by the preceding paragraph. The finite intersection property, using product compactness in ZFC, produces a coloring satisfying every unit-edge condition. This is the graph compactness principle of [4, Theorem 1]; the argument just given specifies its application to the unrestricted plane. It proves (1.1).□

Proof. Fix b\in\mathcal W_x, with realizing samplings (y_n,r_n). For every n, the point y_n+r_ne is typical for almost every e. The second point y_n+r_ne+v is typical for almost every pair (e,v): in angular coordinates the map

has Jacobian of absolute value r_n\abs{\sin(\arg v-\arg e)}, which is nonsingular off a product-null set. On countably many smaller coordinate patches away from that set, change of variables pulls back any plane-null set to a product-null set. The two typical points have distance one. Lemma 4.3 therefore gives

We need the following strong convergence for the second factor:

Here is a change-of-variables proof which also covers arbitrary measurable color boundaries. For \delta>0, discard the angular pairs with \abs{\sin(\arg v-\arg e)}\le\delta. Their measure tends to zero with \delta. On the remaining set, for large n, the Jacobian is bounded below by \delta/2. The multiplicity of G_n there is at most two: an inverse image of a specified sum amounts to an intersection of the circle of radius r_n with a translated circle of radius one. A coincident-circle case can occur only at the antiparallel, hence discarded, directions. The same assertions hold for G(e,v)=x+e+v. Their images lie in one fixed compact plane region for all large n.

It follows that on the retained angular set the pushforwards of angular product measure have densities bounded by a constant depending only on \delta, uniformly in n. Approximate a_i on a compact neighborhood of those images by a continuous function taking values in [0,1], in plane L^1. The density bounds control its composition errors uniformly. For the continuous approximant the compositions converge uniformly, since y_n\to x and r_n\to1. First making the approximation error small, then letting n\to\infty, and finally letting \delta\downarrow0, proves (5.5).

Set B_n(e)=a_i(y_n+r_ne) and Q(e,v)=a_i(x+e+v). The error in replacing the second factor of (5.4) by Q, after integration, is at most the L^1 error in (5.5). Weak-* convergence of B_n then applies to the fixed L^1(S) test \int_S Q(e,v)\,d\sigma(v). We obtain

Both factors are nonnegative, so their product is zero almost everywhere. Taking the countable generating trace families in Definition 5.1, for all five labels, proves exclusion of P_x(e). Interchanging e,v proves exclusion of P_x(v), and gives (5.2). In particular this passage used a strongly convergent factor against a weak-* limit, not a product of two weak-* limits.

Next fix an angular density-one point e of D_{x,i} and any nonparallel direction v. The map

is a smooth local diffeomorphism near (e,v). Choose real angular coordinates near e,v and an inverse branch on a plane neighborhood of x+e+v. For small h, the inverse image under this branch of B(x+e+v,h) lies in a product of angular intervals of length O(h) about e,v, and its change-of-variables factors are bounded. On this branch, (5.2) implies that a point of A_i has first coordinate outside D_{x,i}, apart from a spatial null set. Therefore

The constants can depend on the fixed nonparallel pair. This establishes assertion 2 for each such v. It is a two-dimensional neighborhood argument and does not restrict (5.2) to a possibly exceptional single-angle fiber.

Finally fix the centers and representation in assertion 3, and work first on a smaller compact arc on which one inverse branch of the two-angle sum map is available. Let a trace b\in\mathcal W_z be realized by (z_n,r_n). Applying this inverse branch to z_n+r_nf-x gives perturbed representations

where s_n,t_n are nonparallel and converge to s,t in C^1 on the smaller arc. Shrinking the arc if necessary, s is a diffeomorphism with derivative bounded away from zero, and the s_n have uniformly bounded inverse derivatives for large n.

For D_i=D_{x,i}, these facts imply

To verify it, approximate \ind_{D_i} on a common image interval by a continuous function in angular L^1. Change of variables and the inverse-derivative bounds control the errors of composition uniformly for s_n and for s. Uniform convergence of s_n handles the continuous approximant.

Almost every f with s_n(f)\in D_i has s_n(f) a density-one point of D_i. Indeed the set of exceptions in D_i is angular null by differentiation, and s_n pulls it back to a null set. Assertion 2 makes the density of A_i at z_n+r_nf zero for all these f. The sampling (z_n,r_n) is typical on its circle, so for almost every such f its actual color cannot be i. Consequently

Use (5.6) to replace the second factor strongly in L^1, and then use weak-* convergence of the first factor. Nonnegativity gives

Taking a countable generating family of traces at z and the five labels proves (5.3) on this arc. A countable cover by smaller arcs proves it on the stated domain.□

Proof. Use x+f=x+Rf+R^{-1}f in Lemma 5.2. The endpoint directions are nonparallel and the first endpoint map is a rotation.□

Proof. Suppose that x,y are distinct binary centers for the same pair, and put d=|y-x|. We will derive a contradiction whenever 0<d<\delta_*, with \delta_* chosen below independently of the coloring, the pair, and the centers. Rotate angular coordinates so that y-x=d lies on the positive real axis.

The maps forced by palette transport. For d<1/2, the vector d+e^{it} has length strictly between zero and two. Define real lifts

where the logarithm is the branch near 1. These formulas give the exact representation

Its two unit directions are nonparallel. Their angular difference is twice an angle strictly between zero and \pi/2, and stays away from both parallel cases for sufficiently small d.

The functions \alpha_\pm are smooth degree-one lifts: \alpha_\pm(t+2\pi)=\alpha_\pm(t)+2\pi. Their derivatives converge uniformly to 1 as d\to0, so for small d they induce orientation-preserving circle diffeomorphisms. We may therefore apply Lemma 5.2 locally, with either endpoint as its first direction, and cover the angular circle by finitely many such domains. It follows that

Both binary selections belong to the same fixed pair of labels. Consequently,

Since \alpha_- is nonsingular for angular Lebesgue measure, s_x is invariant almost everywhere under H=\alpha_+\circ\alpha_-^{-1}.

By (5.8), s_x is also invariant under translation by 2\kappa. Define

Then s_x\circ F=s_x almost everywhere. More generally, (5.8) gives s_x(t+j\kappa)=(-1)^j s_x(t) almost everywhere, and hence

almost everywhere. The changes of variables in these identities are legitimate because all the maps involved are diffeomorphisms.

Uniform expansions of the six maps. For a periodic function, an O_{C^2}(d^2) remainder means that its value and its first two derivatives in t are bounded by a fixed constant times d^2, uniformly in t. Taylor expansion gives

All the bounds are uniform: the logarithm, square root, and arccosine in these formulas range in fixed smooth domains for d<1/2. The inverse maps also depend smoothly on d, with uniformly bounded derivatives on a period.

For the composition, put

Then

and therefore

Thus

Among six phases spaced by \pi/3, the maximum sine and the maximum cosine are each at least \sqrt3/2. Absorbing the quadratic remainders, we obtain absolute constants c_0,C_0,d_1>0, with d_1<1/2, such that whenever 0<d<d_1, for every t,

The map attaining the first maximum need not attain the second. We will use the derivative maximum to enlarge small intervals, and the displacement maximum later to extend a conull interval.

Choose once and for all

and assume 0<d<\delta_*. In particular every F_j and its inverse is Lipschitz, since

Enlarging density intervals with bounded distortion. This is an adaptation of Sullivan’s expanding-action and bounded-distortion strategy; see Deroin–Kleptsyn–Navas [6, Section 2.1]. The argument below replaces the usual minimality step by explicit small-displacement overlap propagation when extending the conull interval. Let

It is measurable and 2\pi-periodic, and has measure \pi in every period. The identities already proved say that F_j^{-1}(A) and A differ by a null set. Because F_j and its inverse preserve null sets, the same holds for F_j(A) and A, and for every finite composition of the maps. There is no difficulty from choosing the compositions adaptively: there are only countably many finite words in the six maps and their inverses. Equivalently, one may discard the union of the images of the finitely many exceptional null sets under this countable group, obtaining a common invariant conull set on which all the membership identities hold.

Take an interval I_0 of positive length \ell_0<L. Given I_r with length \ell_r<L, choose F_{j_r} whose derivative at the midpoint of I_r is at least 1+c_0d, and put I_{r+1}=F_{j_r}(I_r). For every t\in I_r,

Writing \lambda=1+c_0d/2>1, we therefore have \ell_{r+1}\ge\lambda\ell_r. The process stops after finitely many steps, at the first N for which \ell_N\ge L, and

Let G=F_{j_{N-1}}\circ\cdots\circ F_{j_0}. For each elementary map,

Applying the chain rule at two points of I_0, whose images before step r lie in I_r, gives

The geometric growth bounds the sum independently of N:

Consequently,

In particular,

uniformly in the initial interval, the number of steps, and 0<d<\delta_*.

The invariance of A, together with preservation of null sets, implies that G(I_0)\setminus A and G(I_0\setminus A) differ by a null set. Change of variables and the distortion bound therefore give

This is the only measure comparison needed; the maps need not preserve Lebesgue measure.

Since A has positive measure, choose a Lebesgue density point of A and a sequence of centered intervals shrinking to it. Their relative complements in A tend to zero. Apply the construction to each interval. The resulting intervals J_n have lengths in [L,(1+C_0d)L], and

Translate J_n by an integer multiple of 2\pi so that its left endpoint lies in [0,2\pi). This leaves the preceding measure statement unchanged. Passing to a subsequence, their two endpoints converge to those of an interval J=[a,b] of length at least L. Endpoint convergence gives |J_n\mathbin{\triangle}J|\to0, and hence

We have obtained an interval of fixed positive length on which A is conull.

Extending a conull interval. Suppose A is conull on [a,b], with b-a\ge L. By the first inequality in (5.10), choose j so that

The image F_j([a,b]) is also conull in A. Moreover,

Thus this image overlaps [a,b], and their union contains [a,F_j(b)], on which A is conull. Its right endpoint has advanced by at least c_0d. This argument concerns geometric endpoints only; no membership in A is asserted at either endpoint.

Repeat, retaining the left endpoint a. After finitely many steps the conull interval has length at least 2\pi. Periodicity then forces A to have full measure in a period, contrary to the measure \pi forced by (5.8). This contradiction excludes 0<|y-x|<\delta_* and proves the lemma.□

Proof. Nonemptiness follows from any one trace in \mathcal W_x, whose coordinates are nonnegative and sum to one. Suppose the upper bound fails on a positive-measure set of angles. There are only finitely many triples of labels, so some fixed triple \mathcal A\subset\{1,\ldots,5\} is contained in P_x(e) on a positive-measure angular set B.

By (5.2) and Fubini, for almost every e\in B, the actual coloring on the circle

uses only the complementary pair \{1,\ldots,5\}\setminus\mathcal A almost everywhere. For almost every such e, this exact radius-one circle sampling is also typical. Indeed the map (e,v)\mapsto x+e+v is nonsingular off a product-null set, so the inverse image of the plane-null set of atypical points is product-null; Fubini then gives the asserted slices.

For each of these e, the actual sampling at center x+e, radius exactly one, is itself a trace in \mathcal W_{x+e}, using a constant sequence. Its coordinates sum to one and are supported in the complementary pair. Hence the maximal palette at x+e intersects that pair almost everywhere. By (5.7), it has a unique member of the pair almost everywhere, alternating at an R-step. Thus x+e is binary for that same fixed pair, even if other traces add further labels to the maximal palette.

The good angles still have positive measure, so they give infinitely many distinct binary centers on the compact circle x+S. Lemma 5.5 makes these centers uniformly separated. A uniformly separated subset of a compact metric space is finite, by a finite cover with sufficiently small balls. This contradiction proves the upper bound.□

Proof. Write

The estimate we need, for every fixed bounded ball B and every label s, is

Here and in (6.1), the vector h may approach zero in any direction.

Choose a boundedly supported g\in L^2(\R^2) agreeing with a_s on a ball containing B+S+h for all sufficiently small h, as well as B+S. The Fourier transform of normalized circle measure satisfies

For completeness, after rotating coordinates the relevant integral is a constant multiple of \int_0^{2\pi}\exp(ir\cos t)\,dt, where r is proportional to |\xi|. For r\ge1, removing the arcs on which |\sin t|\le r^{-1/2} costs O(r^{-1/2}). On each remaining interval, integration by parts using the derivative -ir\sin t gives endpoint terms O(r^{-1/2}), while the integral of |\cos t|/(r\sin^2t) is also O(r^{-1/2}). The bound for r\le1 is immediate.

By Plancherel, the square of the global L^2 norm of \mathcal S(g-g(\,\cdot+h)), divided by |h|, is bounded by the integral of |\widehat g(\xi)|^2 times

This multiplier is uniformly bounded: split at |\xi|=|h|^{-1}. For each fixed \xi it tends to zero as h\to0. Dominated convergence therefore proves (6.3); the cutoff is fixed before h tends to zero.

For a fixed displacement h, almost every root q has both q,q+h typical, and both exact unit-circle samplings centered at these points typical on their circles. At a transition from color s to color t, the first circle avoids s and the second avoids t, by properness of the coloring on typical points. Thus the mass of color t on the first circle equals \mathcal S(a_t-a_t(\,\cdot+h))(q). For a transition between i and j in either order, the first-circle mass of these two colors together is consequently bounded by

Fix a ball B about x. Since ij\in\Gamma_x, there is a constant c_B>0 and arbitrarily small nonzero h for which the transition set in B has area at least c_B|h|. Cauchy–Schwarz and (6.3) show that the average of the preceding mass over this transition set is at most

Now take balls B_m of radii tending to zero about x. After fixing B_m and its constant c_{B_m}, choose h_m sufficiently small and then a transition root q_m\in B_m from the preceding full-measure set, whose first-circle mass of i,j is less than 1/m. This order of choice requires no lower bound on c_{B_m} uniform in m. The circle samples at q_m have radius one; a weak-* subsequential limit belongs to \mathcal W_x and has zero mass in the two specified coordinates. Since it is nonnegative, sums to one, and is supported within P_x almost everywhere, it also proves (6.2).□

Proof. The density conclusion of Lemma 5.2, following (5.2), gives density zero of each of the four represented labels at the nonparallel sum. Their complement, the singleton X, therefore has density one there.

We prove the uniform assertion by contradiction; the fixed-center case follows by using a constant sequence. If it fails, choose pairs in the supports of B_n with chordal distances tending to one. Each support point is approximable by density-one directions of B_n. We may thus choose density directions b_n,d_n in those states such that, after a subsequence,

Choose a small angular interval I with \sigma(A\cap I)>0, whose closure is separated from the directions parallel to any of w,b,d. Such an interval exists because only finitely many directions are excluded and A has positive measure.

For a\in I, the implicit function theorem gives a unit direction c_n(a), smoothly depending on a, with

Indeed, at w_n=w and c_n(a)=a the squared-length equation is satisfied and its derivative in the angle of c_n(a) is -2w\cdot Ja\ne0. Shrinking I around a density point of A if necessary makes this construction uniform on its closure.

The maps c_n have uniformly bounded inverse derivatives. It follows that \ind_{A_n}\circ c_n\to\ind_A in L^1(I): the difference between \ind_{A_n}\circ c_n and \ind_A\circ c_n is controlled by the L^1 convergence under change of variables; continuous approximation of \ind_A and c_n\to\id treat the remaining difference. Together with \ind_{A_n}\to\ind_A on I, this shows that the set of a\in I for which both a and c_n(a) lie in A_n has positive measure for large n. Excluding null sets, choose such an a for which both are density-one directions. The choice of I ensures that (a,b_n) and (c_n(a),d_n) are nonparallel. The points

both have density one of color X, yet their distance is |w_n+a-c_n(a)|=1. This contradicts properness of density-one points. At a single center, compactness of the support turns exclusion of distance-one pairs into the asserted strict margin.□

Proof. Put

The ratio \gamma/\pi is irrational. Otherwise e^{i\gamma} would be a root of unity, making 2\cos\gamma=5/3 a rational algebraic integer, hence an integer, a contradiction.

There is a positive-measure set of angles t whose states at t and t+\gamma are equal. If not, the state would flip almost everywhere under \gamma, and hence be invariant under 2\gamma. A measurable function invariant under an irrational circle rotation is constant almost everywhere: its nonconstant Fourier coefficients vanish, since the corresponding rotation eigenvalues differ from one. This would contradict the flip under \kappa.

Choose t from the positive-measure equality set, also imposing all finitely many needed R-alternation identities and density-direction conditions. For j=0,1,2, form the points

They have distance |1-e^{i\gamma}|=1/\sqrt3 from x and form an equilateral triangle of side one. In each sum the two nonparallel endpoints have opposite states: 2\kappa preserves the state and \pi=3\kappa flips it. The density consequence of Lemma 5.2 therefore gives density one at each z_j to the union of the two labels outside \{X\}\cup D. A uniform common translation in a sufficiently small disk puts all three vertices in that union and in the typical set with positive probability; the three failure probabilities tend to zero by density one. The resulting unit equilateral triangle would be properly colored with two labels, a contradiction.□

Proof. At each y\in L choose an avoiding trace b_y^{ij} for each of its four specified edges. If k belongs to the group represented by a state s, then

To see this, fix a generic direction in state s. If the palette has another member besides k, some cycle edge contains that member but does not contain k. For a diagonal state, choose an edge incident to its other member; for k=X, choose an edge incident to the possible additional label. If no other member exists, any cycle edge suffices. The corresponding trace avoids its edge endpoints and is supported within the palette, so its k-coordinate equals one. There are only finitely many chosen traces, allowing one common conull set of directions for this argument.

Let y_n\to x in L. Along any subsequence, pass further to weak-* limits of the three state indicators and all four chosen traces. Denote the former limits by f_X,f_U,f_V. The trace limits belong to \mathcal W_x by its closed-graph property. Inequality (6.5) passes to the weak-* limits, and its nonnegative right-hand side shows that wherever f_s>0, every label represented by s belongs to P_x, almost everywhere. Two different state groups would require at least three palette labels, contradicting Proposition 5.6. Since the f_s are nonnegative and sum to one, exactly one equals one at almost every direction. Its group determines precisely the canonical state w_x: a diagonal already fills both palette slots, while a positive singleton state requires X\in P_x. Thus f_s=\ind_{\{w_x=s\}} almost everywhere.

Weak-* convergence of indicators to an indicator implies L^1 convergence: for indicators f_n,f, test against 1 and f in \int|f_n-f|=\int f_n+\int f-2\int f_nf. Every subsequence has a further subsequence with this same L^1 limit, which proves the claimed continuity.□

Proof. Set z=x+a. There is a trace in \mathcal W_z avoiding D_i. In fact, in each neighborhood of a the state-i set has positive measure. By (5.2), the two-angle sum map, and Fubini, almost every direction in this set has a typical actual unit-circle sampling at the center x plus that direction, avoiding D_i almost everywhere. Choose such directions tending to a and pass to a weak-* limit. The same construction gives, separately, a trace at z avoiding D_{i+1}. Consequently, almost everywhere,

On the other hand, (5.3) applied to z+f=x+f+a gives P_z(f)\cap P_x(f)=\varnothing almost everywhere. Its first endpoint map is f\mapsto f, a local angular diffeomorphism; its second endpoint is constant, which is permitted. The excluded parallel directions f=\pm a form a null set.

Let X be the label outside D_i\cup D_{i+1}. Whenever w_x(f)=i, disjointness from P_x(f)=D_i leaves only D_{i+1}\cup\{X\} available to P_z(f), and the avoiding trace for D_{i+1} forces X\in P_z(f). The same reasoning applies when w_x(f)=i+1. A step in either order between these states would therefore put X into both P_z(f) and P_z(Rf), contrary to (5.7). The statements used hold on a common conull set, also after pulling back by R, proving the global step ban.□

Proof. Translate q and z by the same vector uniformly distributed in B(0,\epsilon). Almost every such translation puts both points in the typical set. They cannot then both have color i; averaging proves (7.2). Thus this inequality has no exceptional unit pairs. For the last assertion, every neighborhood of q eventually contains B(q_n,\epsilon_n), and the stated lower bound makes its intersection with A_i have positive area.□

Proof. The negation of (6.1), and nonnegativity of D_{ij}, give a ball B with D_{ij}(B,h)=o(|h|). More explicitly, for all sufficiently small r>0 put

For independent uniform vectors h_1,h_2\in B(0,\epsilon), expand the product of the two disk averages and integrate over q\in B'. When \epsilon<\dist(\overline{B'},\R^2\setminus B), the change of root r=q+h_1 bounds the area where c(q+h_1)=i and c(q+h_2)=j by

For h_1=h_2 the simultaneous event is empty. Fubini now gives (7.3), with a bound uniform over the two shifts.

Take a square grid with side \ell=c_\alpha\epsilon, where c_\alpha>0 is fixed small enough that the Lipschitz oscillation of either probability on a square is at most \alpha/2. Select all squares touching the displayed violation set in B''. Their number N_\epsilon is finite, and for small \epsilon each lies in B' and has both probabilities at least \alpha/2 throughout. Disjointness of their interiors and (7.3) imply

Cover every selected closed square by the open disk with the same center and radius \ell. These disks strictly contain the squares and have total radii N_\epsilon\ell=o(1).□

Proof. Fix a small \delta>0. Start with the original core disks B(x,\delta), x\in L_0, and put

This is compact, and every \Gamma_z, z\in M_\delta, is a forest, including any isolated vertices. For each nonedge of \Gamma_z, Lemma 7.3 supplies a ball on which the corresponding simultaneous threshold violations admit finite open disk covers with total radii tending to zero as \epsilon\to0. There are finitely many nonedges, so choose one radius s_z>0 working for all of them. More explicitly, choose it small enough to have the transition estimate on B(z,3s_z), the integral estimate on B(z,2s_z), and the disk covers for violations in B(z,s_z). The covers include the violation points in the interiors of their disks.

Choose finitely many balls B(z,s_z) covering M_\delta, and let \lambda_\delta>0 be a Lebesgue number of this cover on M_\delta. For each selected ball, mark all the disks covering violations for its forest’s nonedges. These additional disks form a finite family whose total radii m_{\delta,\epsilon} tend to zero for this fixed \delta. At a point of M_\delta outside all marked disks, the positive coordinates of (7.4) form a clique in the forest of every selected ball containing the point. They therefore number at most two, and a two-element set is a forest edge. This proves the required local graph-valued assertion.

We next merge disks until their closures are pairwise disjoint. Whenever two current disks have intersecting closures, replace them by an open disk strictly containing both closures, with radius at most the sum of their radii plus a prescribed positive error. An enclosing disk with radius at most the sum exists by choosing its center on the line of centers; in the containment case the larger disk suffices. A slight enlargement makes containment strict. Each merge reduces the finite number of disks, so the sum of all enlargement errors can be made less than any specified \eta>0, using the initial number as a bound on the number of merges.

A final disk is called core if its merger class contains an original core disk. Every final boundary avoids every original disk interior. The total final radii are at most

and each final non-core radius is at most m_{\delta,\epsilon}+\eta. Take \delta_n\downarrow0. After fixing \delta_n, its finite cover, and its Lebesgue number, choose \epsilon_n and the merger errors so that

omitting the preceding-\epsilon condition at n=1. The total final radii now tend to zero.

For large n, no final disk can contain two distinct points of L_0: its diameter is smaller than their minimum positive separation. Every original core disk belongs to some final disk, so this gives exactly one core disk for each point. Since a disk of radius r containing x lies in B(x,2r), the positive distances between the finitely many points of L_0 and \partial Q give the asserted positions of the core disks.

Consider a final non-core disk whose closure meets Q. Its vanishing diameter puts its entire closed disk inside Q' for large n. Moreover, its closure is disjoint from every final core closure, and these final core disks contain all the original core disks. It therefore lies outside all original core interiors, hence in M_{\delta_n}. Its diameter is less than \lambda_{\delta_n}, so its entire closed disk lies in one selected forest ball. Its boundary avoids all original marked interiors. The original map q^{\epsilon_n} thus maps that boundary into the selected forest, and its connected boundary image lies in one component tree, possibly a single vertex. A finite tree is contractible, so the boundary map extends continuously over the closed disk with values in that tree. If the disk crosses \partial Q, make the extension on the whole disk and restrict it to Q.

Outside all final open disks, every point of Q belongs to M_{\delta_n} and avoids all original marked disks, so the original map is graph-valued there. Paste it to the finitely many tree extensions. Their closed disks have disjoint closures and agree with the original map on their boundaries, so finite pasting proves (7.5). Empty disk lists require no operation.□

Proof. Compactness gives a finite subcover U_1,\ldots,U_N with the same bounds. Choose z_a\in U_a. There is a continuous partition of unity \phi_a with \phi_a(z)>0 only when z\in U_a; explicitly, use \dist(z,Q\setminus U_a) and normalize their sum. None of the U_a equals Q, since its diameter would exceed two, so these distance functions are well defined. Set

Convexity gives F(Q)\subset Q. At most two weights are positive at any point, so the image is contained in the finite union of segments [z_a,z_b] and singleton vertices. Also

For any y\in Q with \dist(y,\partial Q)>2, the continuous map z\mapsto y+z-F(z) takes Q into Q, since its image lies in the closed radius-two disk about y. The Brouwer fixed point theorem, in its planar form [14, Theorem 1.9], therefore supplies a fixed point, and at that point F(z)=y. Thus F(Q) contains the nonempty open central region \{y\in Q:\dist(y,\partial Q)>2\}, impossible for a finite union of segments.□

Proof. Otherwise, along an infinite subsequence all the core boundary loops are null homotopic. Extend each such loop over its disk in \mathscr G, and paste to obtain a continuous map \widetilde g_n:Q\to\mathscr G, still equal to q^{\epsilon_n} outside all holes. For each i\in\mathcal I, take the components of the relatively open set

The square and its relatively open subsets are locally path connected; hence these components are open and path connected. They form a cover of Q of multiplicity at most two, since every point of \mathscr G has at most two positive coordinates.

We claim that the mesh of this component cover is at most 1+o(1). If not, after restricting to a subsequence there are a fixed constant c>0, a fixed label i, and paths in U_{n,i} joining points at distance greater than 1+c. Replace either endpoint lying in a hole by a boundary point on the path. The path must exit that disk because the other endpoint is distant, and the replacement changes the endpoint by at most the disk’s diameter. If both endpoints require replacement, their disks are distinct for large n. On every hole boundary the original map is preserved. Thus the replacement endpoints a_n,b_n satisfy

and remain at distance at least 1+c-o(1). Pass to limits a_n\to a, b_n\to b in Q, with |a-b|>1. Lemma 7.2 gives a\in H_i.

Choose a sufficiently small fixed open plane ball U about a. For all large n and every z\in U, the distances to a_n and b_n are strictly below and strictly above one, respectively. Continuity along the path supplies a point at exact distance one from z. If z\notin E_n, every such crossing point is outside all closed hole disks by (7.6). At this point \widetilde g_n=q^{\epsilon_n} and its ith coordinate is positive, so p_i^{\epsilon_n}>\alpha. The pointwise inequality (7.2) therefore gives p_i^{\epsilon_n}(z)\le1-\alpha.

It follows that

The convergence uses local L^1 convergence of disk averages and (7.6). This contradicts a\in H_i, which requires positive A_i-measure in U. The mesh claim follows. For large n the mesh is therefore at most two, contradicting Lemma 7.5. This argument also rules out the possibility of having no core disks in Q.□

Proof. An edge path is reduced when consecutive edges never reverse each other, and a reduced closed edge path is cyclically reduced when its last and first edges do not reverse each other either. Paths between vertices, up to homotopy fixing their endpoints, have unique reduced edge representatives. One way to see this is to construct the covering tree based at a vertex v: its vertices are reduced paths starting at v, and its edges append one edge and then reduce. Deleting the last edge gives the unique path toward the empty word, so this covering graph is a tree. The map recording the endpoint is a bijection on each incident-edge star and hence a covering map. Lifting a path to this tree and contracting it to the unique simple path between its lifted endpoints proves existence and uniqueness of the reduced representative. These contractions apply to arbitrary continuous paths: their compact images in the locally finite covering tree lie in finite subtrees.

Move a loop’s basepoint to a vertex along a path, reduce its edge word, and successively trim inverse first and last edges. The result is cyclically reduced and is empty precisely when the loop is null homotopic. Moving the basepoint, or following the basepoint during a free homotopy, changes the based class by conjugation.

We record how conjugation affects a nonempty cyclically reduced word. First reduce the conjugating path P in PWP^{-1}. At the last edge of P, cancellation against one end of W shortens P and moves one edge of W to its other end. This only rotates W cyclically. Cancellation at both ends of W at once is impossible because W is cyclically reduced. If there is no cancellation at either end, the conjugating path is simply removed when the whole closed word is cyclically trimmed. Induction on the length of P shows that conjugation leaves the cyclically reduced circuit unchanged up to cyclic rotation.

A nonzero power of W is already cyclically reduced: concatenate copies of W for a positive power, or copies of its reversed inverse for a negative power. Its cyclic reduction therefore uses every edge of W. On the other hand, G\setminus\{m\} deformation retracts onto the graph with the edge containing m deleted: contract its two remaining half-open stubs to their endpoint vertices. A loop avoiding m consequently has a reduced circuit using no such edge, and cannot be freely homotopic to a nonzero power of W.

Finally, take a shortest nonempty subloop between repeated vertices of W. It has no repeated interior vertex. Its length is not one, since G has no loop edges, and is not two, since that would be an immediate reversal in a simple graph. Thus it is a simple cycle.□

Proof. Every ray from x meets \partial B once, at a positive radius r(\theta) depending continuously on the angle. Radial interpolation between r(\theta) and \rho identifies \mathcal A with a closed annulus. In particular, winding number about x identifies its fundamental group with \Z, with the counterclockwise inner boundary as generator.

Put Z=g^{-1}(\{m\}). This is a compact metric space. Suppose no component of Z meets both boundaries, and write I=Z\cap\partial B and O=Z\cap\partial B(x,\rho) for its compact inner and outer contact sets. We first obtain a disjoint compact partition Z=Z_1\sqcup Z_2 with I\subset Z_1 and O\subset Z_2.

In any compact metric space, a point’s component is the intersection of all of its clopen neighborhoods. Indeed, this intersection contains the component. If it were disconnected, its two nonempty compact parts could be enclosed in disjoint open sets. A finite intersection of clopen neighborhoods already lies in their union, by compactness of the complement of that union. Intersecting this clopen set with the open set containing the original point gives a smaller clopen neighborhood omitting the other part, a contradiction. The intersection is therefore connected and equals the component.

Thus, for each point of I and each point of O, some clopen neighborhood of the first omits the second. Compactness of O permits a finite intersection producing a clopen neighborhood of the first point disjoint from all of O. Compactness of I then permits a finite union containing all of I and still disjoint from O. Take this union as Z_1 and its complement in Z as Z_2. If either I or O is empty, use Z_1=\varnothing or Z_1=Z, respectively. This proves the partition assertion.

The set

is compact, lies in the open outer disk, and is disjoint from Z_2. Choose a sufficiently fine translated square grid, with x on no grid line, and take every closed grid cell meeting C_1. Their finite union V lies in the open outer disk and avoids Z_2, by the positive compact separations. Also C_1\subset\operatorname{int}(V): at a point of C_1 on a grid edge or vertex, all adjacent cells are included. Consequently \partial V lies in \mathcal A and avoids Z_1\cup Z_2=Z.

Orient each selected square boundary counterclockwise and cancel shared edges. The remaining edges lie on \partial V and form a balanced finite directed chain, so they decompose into closed polygonal loops. Their total winding about x is one. Indeed, winding is additive before and after cancellation, and exactly one selected square contains x in its interior; its winding is one and that of every other selected square is zero. At least one of the resulting loops has nonzero winding, say d. Its image under g avoids m, but the annulus calculation makes that image freely homotopic to W^d. This contradicts Lemma 7.7.□

Proof of Proposition 7.1. Use the maps of Lemma 7.4 and the non-null loops from Lemma 7.6. Since L_0\cap Q is finite, pass to a subsequence with a fixed x\in L_0\cap\operatorname{int}(Q) whose core boundary loop is non-null at every index. Choose \rho>0 so small that \overline{B(x,\rho)}\subset\operatorname{int}(Q) and contains no other point of L_0. Decreasing \rho if necessary, the other core closed disks are disjoint from \overline{B(x,\rho)} for large n, while \overline{B_{n,x}}\subset B(x,\rho). The map g_n is therefore defined on the closed annulus

The inner disk contains x strictly; its possibly eccentric position causes no difficulty for the radial annulus coordinates used in Lemma 7.8.

Let W_n be a nonempty cyclically reduced circuit representing the image of the counterclockwise inner boundary. By Lemma 7.7, it contains a simple cycle. There are only finitely many simple cycles in the complete graph on five vertices. Pass to a subsequence with one chosen cycle, of length \ell\in\{3,4,5\}, and label its vertices i_0,\ldots,i_{\ell-1} in cyclic order. For each of its edge midpoints m_j, Lemma 7.8 gives a compact connected component

meeting both annulus boundaries.

Take Hausdorff limits along a common further subsequence for the finitely many indices j:

The nonempty compact subsets of a compact metric space are compact in the Hausdorff metric. For completeness, finite nets give total boundedness of this hyperspace, and for a Hausdorff-Cauchy sequence the set of limits of convergent point selections is its nonempty compact Hausdorff limit, proving completeness. Connectedness also passes to the limit: a disconnection of K_j into two nonempty compact parts gives disjoint separated neighborhoods containing every sufficiently close K_{n,j}, with each neighborhood met, contradicting connectedness of K_{n,j}.

Each K_{n,j} meets \partial B_{n,x}. These contact points tend to x, because the disk contains x and its diameter tends to zero. Hence x\in K_j. Outer-boundary contact and compactness likewise give K_j\cap\partial B(x,\rho)\ne\varnothing. In particular each K_j is nontrivial.

We prove the support exclusion for either endpoint i\in\{i_j,i_{j+1}\}. At a point of K_{n,j} outside all holes, g_n=q^{\epsilon_n} has value m_j, whose two endpoint coordinates are 1/2. Formula (7.4) therefore gives p_i^{\epsilon_n}>\alpha there.

Fix any closed ball U of positive radius contained in \Delta(K_j). The strict inequalities defining the straddling region hold with a positive uniform margin on U. For every z, either distance extremum changes by at most d_{\rm H}(K_{n,j},K_j) when K_j is replaced by K_{n,j}. Thus, for all sufficiently large n, the distances from every z\in U to points of K_{n,j} straddle one. The continuous image of the connected set K_{n,j} under q\mapsto|z-q| is an interval, so some point of K_{n,j} is at exact distance one from z.

For z\in U\setminus E_n, that point is outside every closed hole disk. The preceding endpoint probability bound and (7.2) give p_i^{\epsilon_n}(z)\le1-\alpha. Exactly as in (7.7),

Thus A_i has zero measure in U. Every point of the open set \Delta(K_j) lies in the interior of such a ball, and consequently has a neighborhood of zero A_i-measure. By the definition of H_i, this proves H_i\cap\Delta(K_j)=\varnothing for both endpoint labels.

Finally, connectedness and the two contact conditions imply

Choose points in K_j at radii \rho/k, k\ge2, and pass to a subsequence of their unit directions converging to some v_j\in S. Relabel these points as q_{j,k}. They supply the required sequences and complete the proof.□

Proof. For a positive edge sign, the identity

is strictly less than one for some sufficiently large fixed n. By continuity, the strict inequality holds when e varies in a small angular neighborhood and its radius increases slightly above one. Such points have distance greater than one from x\in K_j and less than one from q_{j,n}\in K_j; they lie in \Delta(K_j). Proposition 7.1 excludes both endpoint essential supports there. If the sign is negative, the displayed distance is greater than one for large n, and the same argument at radii slightly below one gives the inner strip. There are finitely many edges, so the angular neighborhoods and positive radial widths can be intersected. An empty allowed set would exclude every color from a nonempty open strip, contrary to the measurable partition.

Cover S\setminus\mathcal B by countably many smaller arcs on each of which all signs are constant and both strip restrictions hold with positive widths. For each such arc, polar-coordinate Fubini gives the relevant restriction almost everywhere angularly for almost every radius in its permitted interval. Impose no condition from that arc outside its permitted interval; its failures then exclude only a null set from the whole radial half-line. Intersect these countably many full-measure radial conditions with the full-measure condition of typical sampling at center x. Choose radii from this intersection tending to one from above. For each fixed arc the outer restriction eventually holds along the entire sequence. A weak-* subsequential trace in \mathcal W_x has zero coordinates for the excluded labels on that arc. Its coordinates sum to one, and its positive coordinates belong to P_x almost everywhere, so it supplies an outer allowed palette member. Countably many arcs suffice. A separate sequence tending to one from below proves the inner assertion. The two sides need not use the same trace.□

Proof. Index its five labels modulo five. A label belongs to I(e) exactly when its two incident signs are positive, and to O(e) exactly when both are negative. Their nonemptiness requires a pair of consecutive signs of each kind. If there were two or more runs of each kind, the two required runs of length at least two and the other two nonempty runs would occupy at least six positions. Thus the five signs consist of one positive run and one negative run, of lengths two and three in some order.

The allowed sets are the internal vertices of these runs, of sizes one and two. They are disjoint. The palette size bound (5.12) and (8.1) imply that almost every palette is a pair, containing one vertex from each allowed set. Put D_i=\{i,i+2\}. After a cyclic relabeling, the length-two run occupies edges 0,1, with middle vertex 1, and the length-three run occupies edges 2,3,4, with internal vertices 3,4. The two possible palettes are therefore D_1=\{1,3\} and D_4=\{4,1\}, whose indices differ by two modulo five. Reversing every sign interchanges the two allowed sets and preserves these options. Consequently the palette options at antipodal directions are identical.

Choose a direction on a full six-position R-orbit for which all the preceding almost-everywhere assertions and (5.7) hold. Write p_j\in\Z/5\Z for the index of its palette D_{p_j} at R^je. Two diagonals D_a,D_b are disjoint exactly when b-a=\pm1 modulo five. Thus successive indices make steps \varepsilon_j\in\{1,-1\}, including the closing step from position five to position zero. For antipodal positions j,j+3, the common two-element option set allows index difference only 0 or \pm2. But the possible sums of three steps are -3,-1,1,3; none is zero modulo five, and a sum is \pm2 modulo five only if all three steps have the same sign. Every consecutive triple of steps must therefore be constant. Overlapping triples force all six signs to agree, whereas six equal steps do not close modulo five. This is the desired contradiction.□

Proof. A word without X alternates U,V by adjacent inequality, contradicting the antipodal equality in (8.2). Suppose that X occurs on both parities. Any pair of its opposite-parity positions must be antipodal, since otherwise they are adjacent on the six-cycle. There are exactly two such occurrences: fixing one on each parity prevents any additional occurrence on either parity. Cyclically shift them to positions zero and three. Positions one and two are distinct non-X states, and their antipodes four and five have the same respective values. Thus the word is a three-periodic permutation of X,U,V.

Such a word is isolated under the cross-word restriction. In any compatible nearby word, each of its positions zero and three has neighbors U,V in the original word; it must therefore be X. The remaining four positions are non-X by their own adjacent inequality. If the original positions one and two are U,V, the cross-word restrictions against position two and position one, respectively, force the new positions one and two to be U,V; the other block is forced in the same way. Hence the new word is identical.

Any two points of a dense subset of the circle can be joined by a finite chain within that subset whose steps have angular distance less than \eta: subdivide a connecting arc into steps smaller than \eta/3 and approximate its intermediate vertices by points of the dense set. Isolation would therefore propagate this word throughout G. But w(t+\kappa)_j=w(t)_{j+1}, and a three-periodic permutation word is not equal to its one-position shift. Since G is shift invariant, this is impossible.

It follows that the nonempty set of X positions has a unique parity, denoted h(t)\in\Z/2\Z. The three positions of the other parity are non-X, hence split. They contain one representative from each antipodal pair of positions. Since splitting is antipodally invariant, all six positions are split. In particular almost every direction is split. Signs are locally constant outside the finite set \mathcal B. A nonsplit direction there would have a nonsplit open neighborhood of positive measure, which is impossible.□

Proof. Every generic open semicircle \{v:e\cdot v>0\} selects exactly two rays, counted with multiplicity, and the selected edge positions are adjacent. At a boundary perpendicular to one unoriented ray line, cross the boundary without crossing any other such boundary. The selected multiplicity on that line changes from the number of its rays in one direction to the number in the opposite direction; all other selections stay fixed. Equality of the total count before and after shows that these two multiplicities coincide. Thus all rays are antipodally paired.

If only one unoriented line occurs, it has two rays in each direction. The selected edge positions are a fixed adjacent pair or its complement. These two split patterns give the same diagonal. The other diagonal could not occur on a set of positive measure, contrary to our assumption. There are therefore two distinct lines, with one ray in each of their two directions.

A specified ray on one line can be positive simultaneously with either ray on the other line: the corresponding two nonopposite unit directions fit in an open semicircle. Its edge position is consequently adjacent to both positions from the other line. In a four-cycle the remaining position is its opposite, so its antipodal ray occupies the opposite edge position. This applies to both lines.

Modulo antipodes, the two perpendicular boundaries divide the angular circle into two open arcs. The signs, up to global reversal, are constant on each arc. Crossing a boundary replaces one selected edge by its opposite while keeping the selected edge from the other line. The two middle vertices of the resulting split are the other diagonal. Thus the forced diagonal takes the two different values on the two arcs. Their lengths sum to \pi, which proves the last assertion.□

Proof. Suppose t,t'\in G are less than \eta apart and h(t)\ne h(t'). Select the three positions opposite to the X parity in w(t), and the other three positions opposite to the X parity in w(t'). All selected states are diagonals; around the six positions they alternate between the two words. Cross-word adjacent states differ. Since only U,V are available, all three selected entries of the first word have one diagonal value, and all three of the second have the other value.

For either word the three selected directions, projected modulo antipodes, are spaced equally by \kappa. Each is a non-X direction, so the forced diagonal there is its selected value. Both of the open diagonal arcs in Lemma 8.4 would therefore contain three equally spaced projective positions. An open arc containing such three positions has length greater than 2\pi/3: its closed complementary arc avoids the three points, so lies strictly inside one of the three gaps and has length less than \pi/3. This contradicts the diagonal arc of length at most \pi/2.

Thus h agrees at all sufficiently close points of G. The dense-set chain argument from Lemma 8.3 makes it constant on G. Yet shifting by \kappa shifts every word by one position and reverses this parity. The shift invariance of G gives a contradiction.□

Proof. Let \mathcal U,\mathcal V be the open strips restricted to X,Y near e_0,Re_0, respectively. Each approaches radius one from its prescribed side, and these sides may differ. Put z_0=x+e_0. There are two nonparallel unit directions v'_1,v'_2 such that z_0+v'_j\in\mathcal V. To verify this, rotate so e_0=1 and consider

At (1,\pi/3), F=0 and \partial_\theta F=\sqrt3\ne0. The implicit function theorem permits r to move onto either desired radial side while \theta stays in the angular arc of \mathcal V. This gives a point strictly inside \mathcal V at unit distance from z_0. Openness then gives an open set of such unit directions, containing two nonparallel choices.

Choose a smaller connected polar product strip \mathcal U'\subset\mathcal U, still approaching radius one on an angular neighborhood of e_0, close enough to z_0 that q+v'_1,q+v'_2\in\mathcal V for every q\in\mathcal U'. We claim that c is almost everywhere constant on \mathcal U'.

Fix a point of \mathcal U'. On a sufficiently small ball around it, all starting points q and sufficiently small shifts h admit the following construction inside \mathcal U and \mathcal V. Since Jv'_1,Jv'_2 form a basis, write h=t_1Jv'_1+t_2Jv'_2, with coefficients linear in h. Move first by t_1Jv'_1 and then by t_2Jv'_2. For either move, starting at q' and using t=t_j, v'=v'_j, its two endpoints in \mathcal U have the common unit neighbor

The two distances are one because v' and Jv' are orthogonal unit vectors. Smallness keeps \abs{t}<2 and every constructed point in the stated open set. The combined path has four unit edges, with repetitions allowed.

Fix such an h. Every vertex of the path is a translation of q, so for almost every starting point all five vertices are typical and have colors in \{X,Y\}. Typical-point properness makes the colors alternate along the path. Consequently c(q)=c(q+h) for almost every q in the small ball, for each fixed permissible h. All path maps are jointly measurable in (q,h): the coefficients are linear and the square-root terms continuous. Fubini gives this equality for almost every (q,h). In a still smaller ball, every ordered pair (q,q') has q'-q within the permissible shift range; the invertible change (q,h)\mapsto(q,q+h) shows that almost every pair in that ball has equal colors. Choosing one typical first point by Fubini proves local essential constancy of c.

The sets of points of \mathcal U' with local essential color X, or local essential color Y, are disjoint open sets covering \mathcal U'; they are disjoint because overlapping open neighborhoods have positive area. Connectedness forces just one of these sets to be nonempty. A countable subcover of monochromatic neighborhoods then shows that \mathcal U' uses one color almost everywhere.

Finally choose typical radial samplings at x tending to one from the side of \mathcal U'. Polar Fubini ensures that the samples equal this color almost everywhere on its angular arc. A weak-* trace puts that outsider in P_x almost everywhere on the arc. It is the unique outsider there, so s is essentially constant near e_0, as asserted.□

Proof. Let E_s be the set of directions at which s is not almost everywhere constant on any open neighborhood. Its complement is open, so E_s is closed. If it were empty, local essential constancy and connectedness of the circle, followed by a countable subcover, would make s globally constant almost everywhere. The flip under R excludes this. That same almost-everywhere flip transports local essential constancy in both directions, so E_s is R-invariant.

Consider any six-position orbit contained in E_s. By Lemma 8.6, every consecutive pair in it has at least one vertex in \mathcal B. Because \mathcal B is antipodally symmetric, project the orbit modulo antipodes. Its three positions form a triangle, and every edge has an endpoint among the projective classes of \mathcal B. At least two such classes are needed. There are at most three projective classes in \mathcal B, one for each interface ray. Distinct R-orbits have disjoint projective positions: a shared class means a shared direction or its antipode, and the antipode is already the third rotation in the orbit. Thus two distinct boundary orbits would require at least four projective classes in \mathcal B. There can be at most one orbit, and nonemptiness supplies exactly one.

On each of the six complementary open sectors, local essential constancy and connectedness give one essential value. Rotation maps each sector to the next and flips that value, proving the assertion.□

Proof. Write z=r e^{i\phi}, with 0<r<2, and let \delta=\arccos(r/2)\in(0,\pi/2). The unit directions with angles \phi-\delta and \phi+\delta sum to z, and they are nonparallel. Triple-angle identities give

Also,

Equation (8.4) makes this product negative. Both endpoints therefore lie strictly inside sectors, with opposite outsider values. Take open arc neighborhoods of the endpoints staying in their respective sectors. Their palettes contain the respective outsiders almost everywhere. Equation (5.2) excludes both outsiders at almost every sum of the two directions. The sum map is a local diffeomorphism at this nonparallel pair, with Jacobian \sin(2\delta)\ne0. On a smaller coordinate neighborhood its image is open and contains z, and change of variables transports the product-null exceptions to a spatial null set. Translating by x gives the claim.□

Proof. Lemmas 8.8 and 8.9 give, for each of the seven points x+z_g, an open neighborhood restricted almost everywhere to the triangle labels. Choose \rho>0 so that each point remains in its corresponding neighborhood after any common translation t\in B(0,\rho). For each vertex, the translations that put it outside the typical set, or in the null subset where a nontriangle label occurs, form a null set. The union of these seven exceptional translation sets is null. Choose t outside that union. All seven translated vertices are then typical and use the three triangle labels. Their unit edges are preserved by the placement and common translation, and typical-point properness gives a proper three-coloring of the graph in (8.5). The two shared-edge triangle pairs proved that this graph has no such coloring.□

Proof of Theorem 1.4. If a weak measurable five-coloring existed, Proposition 7.1 would produce an interface cycle of length three, four, or five with the common-center and exclusion properties used above. Propositions 8.2, 8.5, and 8.10 exclude all three possibilities. Thus no weak measurable five-coloring exists.□

Proof of Lemma 8.9. Table 1 gives coordinate centers (\xi_0,\upsilon_0), each within h=0.00011 of the corresponding exact coordinate of z_g, and strict upper and lower polynomial bounds.

Here are explicit checks of the boxes and polynomial bounds. Squaring rational endpoints proves

Put a_*=1.732055, b_*=3.316625, and \epsilon=0.000005. Replacing the two radicals by these midpoints changes their product by less than

The coordinates of the three products in (8.5) are

For each coordinate of every g, the midpoint substitution therefore has error at most \epsilon: the largest product error is less than 0.000026/6<\epsilon, and the other coefficients give this bound directly. The rotation in (8.6) has absolute row sum 47/37. For an exact check of the displayed centers, let \widetilde z_g denote the placed point after this rational midpoint substitution. Direct substitution gives the residuals

Every residual numerator has magnitude less than 35520000=710400000000/20000, proving that the midpoint-substituted coordinates have the stated four-decimal roundings, with error less than 0.00005. Thus every final coordinate error is bounded by

For completeness, the remaining interval calculations reduce to the small rational factors in Table 2. Define

All listed absolute coordinate centers exceed h, so these intervals enclose \xi^2,\upsilon^2,l. In each row the factors in the second table satisfy

These claims require only squaring integers: with M=10^5, m=M\abs{\xi_0}, n=M\abs{\upsilon_0}, the four squared-coordinate endpoints are

Thus every entry is verifiable using the displayed coordinate integers, without evaluation of an irrational number.

Every interval (a,b) lies in (0,4) and avoids one. The polynomial factors are therefore bounded by

The last inequalities are direct rational products of the second table, with the target bounds in the first. For example, its final row gives

The other six rows give the corresponding displayed strict bounds by the same two products. The squared-radius intervals also lie respectively in

Finally P_+<P'_- in every row of Table 1. This proves (8.4) and the radius conditions for all seven vertices.□