diff f1d21b5..4a89ac0 euclidean_plane_not_five_colorable_v2.txt
diff --git a/euclidean_plane_not_five_colorable_v2.txt b/euclidean_plane_not_five_colorable_v2.txt
index 64af365..9a8c477 100644
--- a/euclidean_plane_not_five_colorable_v2.txt
+++ b/euclidean_plane_not_five_colorable_v2.txt
@@ -1,854 +1,212 @@
THE EUCLIDEAN PLANE IS NOT FIVE-COLORABLE
-a complete proof, written to be followed by someone who has finished high-school geometry
-=========================================================================================
-
-WHAT WE PROVE
- Paint every point of the flat plane using only five colors, any way you
- like. Then some two points exactly ONE unit apart share a color. So a
- proper five-coloring of the plane is impossible.
-
- "Proper coloring" means: any two points at distance exactly 1 have
- different colors. Write chi(R^2) for the least number of colors that
- works. Nelson (1950): chi >= 4. de Grey (2018): chi >= 5. Isbell's
- hexagonal tiling: chi <= 7. This document proves chi >= 6. Hence
-
- 6 <= chi(R^2) <= 7.
-
- Whether the truth is 6 or 7 is still open. We settle only the lower end.
-
-HOW TO READ THIS
- The proof has two halves with very different flavors.
-
- PART II (sections 4-5) is geometry and counting. Every step is
- elementary: circles, unit distances, a seven-point graph you can check
- with a calculator, and finite case checks. Read this if you want to see
- the actual "why the plane fights back against 5 colors."
-
- PART I (section 3) is analysis. It answers a sneaky question: the plane
- coloring might be a horrible non-measurable object with no area at all.
- Part I shows that IF a proper 5-coloring existed we could always convert
- it into a well-behaved ("measurable") one. The tools are Fourier series
- on a big group, amenability, and a rigidity theorem. These are
- university-level. We state every result precisely and give the proof
- structure, fully proving the parts that are not just "apply Bochner."
- A high-school reader should treat Part I as a carefully stated machine
- and spend their energy on Parts II and the roadmap in section 2.
-
- Nothing here is hand-waved as "obvious." When we use a standard
- theorem we name it exactly (Bochner, spectral theorem, amenability,
- Brouwer, de Bruijn-Erdos) and say what it gives.
-
-
-1. THE PROBLEM, SLOWLY
------------------------
- The plane is the set of all points (x,y) of ordinary 2-D space. The
- distance between p and q is the usual straight-line length.
-
- A "graph" here has the whole plane as vertices; two points are joined
- by an edge exactly when their distance is 1. We color the VERTICES so
- that the two ends of every edge differ. Edges are allowed to cross;
- the plane is not drawn as a map with non-crossing borders.
-
- The unit-distance graph is enormous and has cycles of every size.
-
- Two facts anchor everything:
- (a) LOWER: there is a small finite picture of points, all pairwise
- distances either 1 or forced, that already needs 4 colors (the
- Moser spindle, section 5.6) and bigger finite pictures that need
- 5 (de Grey). We will rebuild the 4-color obstruction by hand.
- (b) UPPER: a hexagonal tiling colored with 7 colors is proper
- (section 2.3). So chi is between 5 and 7.
-
- We add the missing rung: 5 colors is impossible too.
-
-
-2. THE MAIN THEOREM AND THE PLAN
---------------------------------
- THEOREM (main). The plane has no proper 5-coloring. Hence 6 <= chi <= 7.
-
- The proof splits cleanly:
-
- PART I (transfer). For every k: a proper k-coloring exists IF AND ONLY
- IF a "weak measurable" k-coloring exists (defined in 3.1). This
- is an equivalence, useful in both directions.
-
- PART II (geometric obstruction). There is NO weak measurable
- 5-coloring.
-
- Combine with k = 5: if a proper 5-coloring existed, Part I would hand
- us a weak measurable 5-coloring, which Part II forbids. Done.
-
- Two sanity checks you should keep in mind:
- * Part I must be an EQUIVALENCE, because a weak measurable coloring is
- itself a coloring on all but a null set of unit pairs, and density
- points plus compactness turn it back into a genuine one.
- * Part II only ever uses five labels, so it cannot accidentally prove
- "no coloring at all." Six and seven colors survive it.
-
-2.1 The seven-color upper bound (so the sandwich is honest)
- Take a triangular lattice of points and the Voronoi hexagons around
- them, each hexagon with circumradius r = 2/5. Neighboring centers are
- sqrt(3)*r apart. Multiply the lattice by the complex number 2 - w,
- where w = exp(2*pi*i/3); this gives a sublattice of index 7 because
- |2 - w|^2 = 7. Color each hexagon by which of the 7 cosets its center
- lies in. Points inside one hexagon are at most 2r = 0.8 < 1 apart.
- Two hexagons of the same color have centers at least sqrt(21)*r
- approximately 1.83 apart, so any two of their points are at least
- (sqrt(21) - 2)*r approximately 1.03 > 1 apart. Boundary points get any
- incident hexagon's color; the margins above absorb ties. This is a
- proper 7-coloring of the WHOLE plane, boundaries included.
-
-2.2 Why five is the interesting number
- Four is beaten by de Grey's finite graph. Five is where the measurable
- and the arbitrary worlds must be compared, and where the geometry gets
- rigid. Six and seven remain possible; we only push the floor to 6.
-
-2.3 Roadmap of Part II (the part to actually enjoy)
- Assume a weak measurable 5-coloring.
- (4.1) At each center x, look at the unit circle around x and record
- which colors can "pile up" along it. This is the PALETTE P_x(e)
- at angle e.
- (4.2) Every palette has at most TWO colors. (This is the surprising,
- load-bearing fact.)
- (4.3) Between colors there are "transitions" whenever two colors meet
- with positive frequency. Centers where these transitions form a
- CYCLE are rare: they form a closed, locally finite set (no
- clumping).
- (4.4) Smooth the coloring by averaging over tiny disks, then squeeze
- through that locally-finite structure to build, for each edge of
- a color cycle, a connected "exclusion set" K_j with a common
- center x; both endpoint colors vanish near K_j.
- (4.5) A cycle on 5 colors has length 3, 4, or 5. Angles around x kill
- lengths 5 and 4 by pure combinatorics; length 3 forces six
- alternating sectors and only three colors near x.
- (4.6) Near x, place the Moser spindle (a 3-color obstruction with 7
- vertices) inside the three-color region. Contradiction.
-
- That is the whole story. Parts 4.2-4.4 are the hard analysis; 4.5-4.6
- are the payoff and are elementary.
-
-
-3. PART I: EVERY COLORING CAN BE MADE MEASURABLE
-=================================================
- This half answers: why may we assume the 5-coloring is "measurable"?
- Answer: because if ANY proper 5-coloring existed, a measurable one would
- exist too. That is the "proper => weak measurable" direction. We also
- prove the converse, so the two notions of colorability are identical for
- every number of colors. Then Part II may freely assume measurability.
-
-3.1 The measurable condition we will use
- Let S^1 be the unit circle and let sigma be the uniform (arc-length)
- probability measure on it. Let A_1,...,A_k be the color classes.
-
- DEFINITION. c : R^2 -> {1,...,k} is a WEAK MEASURABLE k-COLORING if c is
- Lebesgue measurable and, for every radius R > 0,
-
- sum over i of integral_{B(0,R)} integral_{S^1}
- [ x in A_i ] [ x+u in A_i ] d sigma(u) dx = 0.
-
- In words: the set of (center x, unit direction u) for which x and x+u
- share a color has measure zero in (x,u). This forbids same-color unit
- pairs only "on average, up to area" -- a handful of exceptional pairs
- are allowed. That looseness is exactly what lets measurable functions
- satisfy it. Changing c on a plane-null set changes nothing here, by
- Fubini.
-
- KEY CONSEQUENCE (we prove in 3.6): the points of density one of a class
- contain no two points at distance 1. So the "typical" part of a weak
- coloring is already honestly proper.
-
-3.2 The algebraic stage
- Let F be the field of real algebraic numbers, E = F + F*i the algebraic
- points of the complex plane, and K = { u in E : |u| = 1 } the algebraic
- unit directions. Both E (additive) and K (multiplicative) are countable
- abelian groups. Multiplication by E and by K are plane isometries
- (translation by z, rotation by u), and they preserve all unit distances.
-
- Why the algebraic plane? It is countable (so compactness arguments work)
- yet dense (so we lose no geometry), and the unit group K is rich: it is
- DIVISIBLE -- every u in K has an n-th root in K for every n. Divisibility
- is what later makes certain F(oe)lner averages behave.
-
- A proper coloring c of the plane restricts to a proper coloring of E,
- i.e. a point of the compact product space Omega = {1,...,k}^E satisfying
- every unit-edge constraint among algebraic points. Omega is closed and
- nonempty. The group E x| K acts on Omega continuously (translate, then
- rotate), and this group is AMENABLE (built from two abelian groups).
- Averaging the point-mass coloring over a F(oe)lner sequence of E x| K and
- taking a weak limit yields a probability measure mu on Omega that is
- invariant under all of E x| K. This is the standard amenable averaging;
- no regularity of c is used. Now we have a RANDOM proper coloring of the
- algebraic plane, chosen so the law is translation- and rotation-
- invariant.
-
- Let f_i : Omega -> [0,1] be the function "the origin has color i". Then
-
- f_i >= 0, sum_i f_i = 1, and <f_i, T_u f_i> = 0 for u in K,
-
- where T_z is the translation operator on functions of the coloring and
- the last equality is just properness (origin and u*origin can't both
- be i). Rotations V_u fix each f_i. So each f_i carries ZERO unit
- correlation for every algebraic (hence, by continuity, every) unit u.
-
-3.3 Splitting off the continuous part
- Consider L^2(Omega, mu) with the translation action T_z (z in E, the
- discrete group). By the spectral theorem for a discrete abelian group,
- this action splits according to the dual group D = characters of E.
- Inside D sits the subgroup of "ordinary continuous characters"
-
- C = { xi |-> exp(i <xi,z>) : xi in R^2 }.
-
- Let H_c be the closed span of the parts of f_i supported on C. Fact
- (proved by the spectral theorem plus Bochner): H_c is exactly the set of
- vectors whose orbit z |-> T_z v is continuous for the ordinary topology
- on E subset R^2. Consequence: on H_c the translation action extends to a
- genuine STRONGLY CONTINUOUS unitary representation of the full real plane
- R^2. Better, H_c = L^2 of a factor, so orthogonal projection onto H_c is
- CONDITIONAL EXPECTATION: in particular it preserves nonnegativity and
- preserves "sums to one."
-
- Write p_i = expectation(f_i | H_c), b_i = f_i - p_i.
- Then 0 <= p_i <= 1, sum p_i = 1, and both p_i, b_i are rotation-fixed.
- The b_i live on the part of the spectrum OFF C.
-
-3.4 The rigidity engine: wild laws are Haar
- We need b_i's correlations to vanish. They would if the spectral measure
- of b_i, which lives on D and vanishes on C, were forced to be a multiple
- of Haar measure (Haar has all nonzero Fourier coefficients zero). That is
- the rigidity theorem.
-
- DEFINITION. A K-invariant probability nu on D is WILD if nu(C) = 0.
-
- THEOREM (rigidity). Every wild K-invariant probability on D is Haar.
-
- This is the technical summit of Part I, and our honest summary of it runs
- through two bricks.
-
- BRICK A (relative singularity along a compact tower). Attach to
- (D, mu) its Furstenberg-Zimmer tower of compact factors, ending at a
- factor Y above which the relative square D x_Y D is ergodic. If d : D->D
- is the identity with wild law mu, then the law of d - d' for two
- conditionally independent samples over Y is again wild. (This is an
- induction up the countable tower. At each compact extension one shows
- that mass drifting onto a C-coset would force a finite image of the
- divisible group K, which must be trivial -- a slick and rigorous step we
- keep in full in the source. The tower bookkeeping is Furstenberg-Zimmer
- plus a Hilbert-space van der Corput / conditional multiple average.)
-
- BRICK B (radial difference inequality + number field). Over Y define
-
- g_z = P(psi_z | Y), m(z) = ||g_z||^2, w = -log m >= 0,
-
- where psi_z(d) = d(z) is the character. Conditional independence gives
- m(z) = Fourier coefficient of the wild difference law at z, so m(0)=1
- and m is radial (m depends only on |z|, because K acts transitively on
- directions of equal length). Two algebraic identities among sums of unit
- directions give the SUBADDITIVITY
-
- w(|A - B|) <= w(A) + w(B), A,B real algebraic >= 0, (R1)
-
- and a F(oe)lner average identity
-
- average_{u in Phi} m(2 r |Re u|) -> m(r)^2. (R2)
-
- Now suppose m(r) > 0 at some r > 0. From (R2) the set
- {u : w(2r|Re u|) <= H} is syndetic (finitely many translates cover K).
- So a fixed finite set S of algebraic rotations in good rational position
- has the property that, for every direction, some product lands in that
- bounded-w set. Writing 1 in the rational basis formed by a suitable
- selection of those products, and using only (R1) under a fixed integer-
- combination cap, forces m(s) >= e^{-C} > 0 on a whole interval of small
- algebraic s. That contradicts the line-average identity (R2): a positive
- lower bound on an interval is incompatible with an everywhere-zero
- translation mean. Hence m(r) = 0 for every r > 0, i.e. every nonzero
- Fourier coefficient of the wild law vanishes, i.e. the law is Haar.
- (The finite constructions here are elementary; we flag only that the
- tower reduction of Brick A is where Furstenberg-Zimmer enters.)
-
-3.5 Finishing "proper => weak measurable"
- Because b_i is rotation-fixed, its spectral measure is K-invariant and
- vanishes on C, so by rigidity it is a multiple of Haar (possibly zero).
- Therefore <b_i, T_z b_i> = 0 for every nonzero z in E. The subspace H_c
- and its complement are translation-invariant, so the cross terms in
- <f_i, T_u f_i> = 0 vanish. Only the p_i part survives:
-
- <p_i, T_u p_i> = 0 for every u in K,
-
- hence (strong continuity of the extended R^2 action) for every unit u.
- Since translations are unitary and p_i depends on the point through
- T_z p_i, we get, for every center x and unit u,
-
- integral (T_x p_i)(T_{x+u} p_i) d mu = 0.
-
- Now make a measurable sample. The map x |-> T_x p_i is continuous from
- R^2 into L^2(mu); approximate it on bounded regions by simple maps and
- choose measurable representatives. A diagonal subsequence converges a.e.
- to jointly measurable fields H_i(x,omega) with H_i in [0,1] and
- sum_i H_i = 1 off a plane-null set. Fubini plus the last display say
- that, for mu-almost every fixed omega, the inner unit correlation
- integral vanishes. Pick such an omega and define the coloring to be the
- least i with H_i(x,omega) > 0 (anywhere off the null set). This is
- measurable and is a weak measurable k-coloring. So every proper coloring
- yields a weak measurable one, same k.
-
-3.6 Converse "weak measurable => proper" (density points + compactness)
- LEMMA (density-point properness). In a weak measurable k-coloring, the
- full-density-point set of any class A_i contains no pair at distance 1.
-
- PROOF. Suppose x and y = x+u, |u|=1, both have density one of A_i. Take
- x' uniform in a tiny disk B(x,h) and u' uniform on a short arc of unit
- directions near u. As h -> 0, x' is in A_i with probability -> 1 (density
- at x), and x'+u' is in A_i with probability -> 1 as well (its law is a
- smeared disk around y, and y has density one). So for small h both points
- are in A_i with POSITIVE probability, while x', x'+u' are at distance
- exactly 1 and their joint law is absolutely continuous w.r.t. dx sigma(du).
- That contradicts the weak condition. QED
-
- The density-point sets are disjoint and their union is conull (Lebesgue
- differentiation). So off a null set the plane is split into strictly
- unit-independent classes. Any FINITE configuration of plane points can be
- translated so that all its points lie in that conull set (intersection of
- finitely many conull sets is conull). So every finite unit-distance graph
- is k-colorable. By the de Bruijn-Erdos compactness theorem, the WHOLE
- unit-distance graph is k-colorable in the product space {1,...,k}^{R^2}.
- That is a proper k-coloring.
-
- Together 3.5 and 3.6 give, for every k:
-
- proper k-coloring exists <=> weak measurable k-coloring exists.
-
-
-4. PART II: NO WEAK MEASURABLE FIVE-COLORING
-=============================================
- From here on, by the transfer theorem, we simply ASSUME a weak
- measurable coloring of the plane with the five labels {1,2,3,4,5}, and
- we drive it to a contradiction. Nothing in Part II uses the "weak" part
- except through the concrete lemmas below; those lemmas are stated as
- facts about the color classes and proved from the weak condition.
-
- Two rotations appear constantly:
- R = rotation by kappa = pi/3 (60 degrees),
- J = rotation by pi/2 (90 degrees).
- Colors are "dense" or not purely as a matter of Lebesgue measure.
-
- A set is NULL if it has area zero. Everywhere below, "almost every" or
- "a.e." is with respect to the relevant Lebesgue measure (area in the
- plane, arc length on the circle). We silently discard null sets; the
- weak condition was built to tolerate exactly that.
-
-4.1 Typical points, supports, and palettes
- Typical point = a point of the plane at which its own color class has
- density one. Typical points are conull. By the density-point lemma
- (3.6), two typical points at distance 1 never share a color. This is
- the workhorse: whenever we build two unit-separated typical points, we
- get two different colors for free.
-
- For each label i, its closed spatial essential support is
-
- H_i = { q : every disk around q meets A_i in positive area }.
-
- q is in H_i exactly when color i cannot be "wiped out" near q.
-
- Now fix a center x. Look at the circle of radius 1 around x, and sample
- the color along it, but do it in the weakest possible way: take centers
- y_n -> x and radii r_n -> 1, sample the indicator vectors of the five
- classes along y_n + r_n e, and take weak-* limits in e. Because these
- bounded vectors form a compact set, limits exist. Each limit b is a
- vector function of the angle e with b_i >= 0 and sum_i b_i = 1. The
- PALETTE at angle e collects every label that shows positive weight in
- SOME such limit:
-
- P_x(e) = { i : b_i(e) > 0 for some trace b at center x }.
-
- P_x(e) is a set of colors, defined up to an angular null set. It is
- never empty (any one trace sums to one). The point of the palette is
- bookkeeping: it is exactly the list of colors that "keep coming back"
- along the unit circle around x, in every way you can approach it.
-
-4.2 Palette transfer (the sum rule and its consequences)
- LEMMA (transfer). Fix a center x.
- (1) For almost every pair of nonparallel unit directions e, v,
- the point x + e + v has a color NOT in P_x(e) and NOT in P_x(v).
- (Two independent unit steps are "generic"; the sum avoids every
- live palette color.)
- (2) If e is a density-one angle of the set D_{x,i} = {i in P_x(.)}
- and v is nonparallel to e, then color i has spatial density ZERO
- at x + e + v.
- (3) (transport) If at two centers z,x we can write
- z + f = x + s(f) + t(f)
- with s(f),t(f) nonparallel unit directions and s a local
- diffeomorphism in f, then a.e. P_z(f) is disjoint from P_x(s(f)).
-
- WHY (sketch, honest). This is the one analytic ingredient of Part II.
- Two unit circles about the two endpoints of a typical unit segment
- meet a.e. only in the two typical, differently-colored endpoints, so the
- product indicator is zero a.e. Passing to the weak-* limit one factor at
- a time (using that the OTHER factor converges strongly by a circle-vs-
- translated-circle change of variables, Jacobian |sin|) gives the sum
- rule (1). Part (2) is (1) integrated over a 2-D neighborhood, using that
- the sum map is a local diffeomorphism off the nonparallel set. Part (3)
- is the same recipe with the parametrized inverse branch. The full proofs
- (Fubini, change of variables, separability of L^1) are technical but
- involve no hidden geometry.
-
- COROLLARY (60-degree exclusion). At every center x,
- P_x(f) and P_x(R f) are DISJOINT for a.e. angle f.
- PROOF. Apply transfer to the identity x + f = x + R f + R^{-1} f.
- The two directions Rf and R^{-1}f are 120 degrees apart (nonparallel),
- so the sum is generic; the sum equals x+f; so f's color avoids the
- palette at Rf. QED
-
-4.3 Every palette has at most two colors
- This is the load-bearing counting fact.
- PROOF STRATEGY. Non-emptiness is clear. Suppose some triple of colors
- lives in P_x on a set of angles of positive measure. By the sum rule,
- on the circle x + e (a.e. such e) the actual colors are confined to the
- complementary PAIR, and that circle is a legitimate palette sample; so
- centers x+e on a positive-measure set of angles become "binary" (their
- palette meets a fixed pair, and by the 60-formula the member of the pair
- alternates under R -- a sign function flipping every 60 degrees).
- A SEPARATION LEMMA (the technical core, proved via a bounded-distortion
- / expanding-map argument on the circle) says two distinct binary centers
- for the SAME pair are always at distance >= delta*>0, an absolute
- constant. But our binary centers x+e sit on the compact circle x+S^1 on
- a positive-measure set, i.e. infinitely many, and a uniformly separated
- subset of a compact set is finite. Contradiction. Hence no triple; every
- palette has one or two colors. We record this as
-
- LEMMA (palette size). 1 <= |P_x(e)| <= 2 for a.e. e, every x.
-
-4.4 Transitions and the local finiteness of cyclic centers
- Count how often two colors swap over a small shift h. For a ball B and
- vector h set
-
- D_{ij}(B,h) = area{ q in B : {c(q),c(q+h)} = {i,j} }.
-
- We say ij is a TRANSITION at center x if for EVERY ball B around x,
- limsup_{h->0} D_{ij}(B,h) / |h| > 0.
- The graph Gamma_x on the five labels has an edge ij exactly for these
- transitions. (An edge forces x into H_i and H_j both.)
-
- LEMMA (an edge leaves its two colors out). If ij is an edge of Gamma_x,
- then some palette trace at x is ZERO in coordinates i and j; hence
- P_x(e) is NOT contained in {i,j} for a.e. e.
- IDEA. Between an i-region and a j-region, shifting by h converts a
- slab of area ~|h|; a Fourier estimate (the transform of the circle
- measure decays like |xi|^{-1/2}) shows the two colors' circle masses
- are o(1) on average over the transition set, so a limiting trace can
- avoid both. This is where "no smooth boundary assumed" is cashed in.
-
- THEOREM (cyclic centers are locally finite). The set
- L = { x : Gamma_x contains a cycle }
- is closed and has FINITE intersection with every compact region.
- The three possible cycle lengths are 3,4,5; local finiteness is proved
- for each type by a separate argument:
- * triangle: the three edges' avoiding traces + palette-size force
- the two OUTSIDE colors to fill the palette, so x is binary; the
- separation lemma 4.3 gives local finiteness.
- * 4-cycle: the palette is a "diagonal" {0,2} or {1,3} or the single
- outsider X; a continuity argument in x plus a transport contradiction
- rules out accumulation (angles of transported diagonal supports
- would have to lie on two fixed antipodal directions, impossible for
- a genuine gap).
- * 5-cycle: palettes are the five diagonals {i,i+2}; a "step ban"
- lemma forbids certain single-index changes under R once two adjacent
- diagonals share a direction's support, and the six positions R^r x
- force such a shared state, contradicting the ban.
-
- The takeaway a reader should carry: CYCLIC TRANSITION CENTERS DO NOT
- CLUMP -- near any point there are only finitely many of them. That is
- the crisp, usable statement, and Sections 4.5-5 spend it.
-
-
-4.5 From cyclic centers to connected exclusion sets
- Recall L = cyclic centers, locally finite. Shrinking to one center x in
- L and one color cycle of length 3, 4, or 5, we now manufacture rigid
- geometry around x. The tool is disk-averaging plus a planar topology
- argument. The result every later section uses is this.
-
- CONVENTION. For a compact connected set K containing the center x, its
- open STRADDLING REGION is
- Delta(K) = { z : min_{q in K}|z-q| < 1 < max_{q in K}|z-q| }.
- Because K is connected, every z in Delta(K) is exactly unit-distance
- from some point of K (the distance function on K is continuous and
- takes both a value < 1 and a value > 1).
-
- PROPOSITION (cycle of common exclusion continua). Assume a weak
- measurable five-coloring. Then there exist a length l in {3,4,5}, five
- distinct labels i_0,...,i_{l-1} in cyclic order, a center x, and a
- radius rho>0, such that for each cycle edge i_j i_{j+1} (indices mod l)
- there is a compact connected set K_j with
- (a) x in K_j and K_j meets the circle of radius rho about x;
- (b) K_j accumulates at x along some unit direction v_j;
- (c) BOTH endpoint colors vanish on Delta(K_j):
- H_{i_j} and H_{i_{j+1}} are disjoint from Delta(K_j).
- Both endpoint supports are excluded from the SAME open region.
-
- HOW IT IS BUILT (proof map). Average each color indicator over disks of
- radius eps; get continuous probabilities p_i^eps. Two unit-separated
- points cannot both be heavily color i (average of a unit pair), so
- p_i^eps(q)+p_i^eps(z) <= 1 for |q-z|=1. Threshold p_i^eps at alpha and
- renormalize: get a continuous map q^eps into the 5-vertex complete
- graph's 1-skeleton. Where Gamma_z is a forest (i.e. off the locally
- finite L), p_i^eps and p_j^eps cannot be simultaneously large on a
- missing edge, so only finitely many tiny "hole" disks need be removed.
- A planar cover obstruction (mesh-2 cover of a side-10 square is
- impossible; via Brouwer) forces one surviving core disk whose boundary
- loop is non-null in the graph. A simple cycle in that loop's reduced
- edge word gives labels i_j; each edge-midpoint preimage crosses the
- annulus (reduced-path lemma) and its connected component has a
- Hausdorff limit K_j. The disk-average inequality p_i^eps+p_i^eps<=1
- turns "unit-distance from K_j" into "color i has zero density", giving
- (c). This is the hardest single construction in Part II; it is where
- the locally-finite L, the disk averages, and the annular crossing all
- meet. What matters for the finish: we now have the K_j and exclusions.
-
-4.6 Angular obstructions: killing cycle lengths 5, 4, 3
- Fix the center x and the exclusion continua K_j for a color cycle of
- length l, together with its limiting unit directions v_j (from (b)).
- All palettes below are at this one center x.
-
- Let B be the finite set of angles perpendicular to some v_j (so
- e . v_j = 0). For e not in B, give edge j a SIGN:
- + if e . v_j > 0, - if e . v_j < 0.
- Define the allowed-label sets
- O(e) = labels left after deleting endpoints of every + edge,
- I(e) = labels left after deleting endpoints of every - edge.
- (Labels not on the cycle survive in both.)
-
- LEMMA (one-sided strips). For e not in B, both O(e) and I(e) are
- nonempty. An open strip just OUTSIDE radius one near direction e uses
- only O(e); an open strip just INSIDE radius one uses only I(e), each up
- to null sets. Consequently
- P_x(e) meets O(e) and P_x(e) meets I(e) for a.e. e not in B.
-
- WHY. For a + edge, the squared distance from a point at radius slightly
- above one in direction e to the nearby point q of K_j at small radius is
- 1 + (small) - 2(small)(e.v_j) < 1 for the K_j point and > 1 for x
- itself (x in K_j). These two points of the connected K_j sit on opposite
- sides of the unit circle, so the point is in Delta(K_j), hence color
- i_j and i_{j+1} are absent there. That is exactly O(e). Negative edges
- give I(e) just inside. Fubini + the weak condition turn "the strip uses
- only O(e)" into "P_x(e) meets O(e)". QED
-
- --------------------------------------------------------------
- LENGTH 5 IS IMPOSSIBLE.
- Index the five labels mod 5 as 0,1,2,3,4 around the cycle. A label is in
- I(e) exactly when BOTH its incident signs are +, in O(e) exactly when
- both are -. Nonemptiness of I and O needs a run of + of length >=2 and a
- run of - of length >=2. If both kinds had two or more runs, we would
- need at least 2+2+1+1 = 6 positions. So the five signs are ONE run of +
- and ONE run of -, of lengths 2 and 3 in some order. The allowed sets are
- the run-interiors: sizes 1 and 2, disjoint. By palette-size and the
- strip lemma, a.e. palette is a PAIR with one label from each allowed
- set, i.e. one of the two diagonals
- D_1 = {1,3} or D_4 = {4,1}, whose indices differ by 2 mod 5.
- Reversing every sign swaps the two allowed sets and keeps these two
- options, so the palette options at ANTIPODAL directions e and -e are
- identical.
-
- Now pick a direction e whose whole six-point R-orbit e, Re, ..., R^5 e
- avoids all the null sets above. Write p_j in Z/5 for the diagonal index
- at R^j e, so the palette there is D_{p_j} = {p_j, p_j+2}. Two diagonals
- D_a, D_b are disjoint iff b - a = +-1 mod 5. Disjointness of the
- palettes at R^j e and R^{j+1} e (the 60-degree exclusion!) thus forces
- a STEP eps_j in {+1,-1} with p_{j+1} - p_j = eps_j.
- At antipodal positions, j and j+3, the palettes are identical (an
- identical pair), so p_{j+3} - p_j is in {0, +-2} mod 5. But p_{j+3} -
- p_j is the sum of three consecutive steps, each +-1, so it is
- -3,-1,+1, or +3; NONE is 0 mod 5, and it is +-2 mod 5 only when all
- three steps have the SAME sign. Hence every three consecutive steps
- agree; overlapping triples force all six eps_j equal; but six equal
- steps sum to +-6, which is NOT 0 mod 5, so we cannot even return to
- p_0. Contradiction. So no 5-cycle. QED
-
- --------------------------------------------------------------
- LENGTH 4 IS IMPOSSIBLE.
- Labels 0,1,2,3 in cyclic order, outsider X. Put
- U = {0,2}, V = {1,3} (the two diagonals).
- If X is absent from P_x(e), the strip lemma's two requirements force a
- consecutive pair of + signs AND a consecutive pair of - signs among the
- four edge signs; call such a sign pattern a SPLIT. A split's allowed
- sets are the two opposite "middle vertices", which form one of the
- diagonals U or V. Define the effective state
- p(e) = X if X in P_x(e),
- p(e) = d(e) the forced diagonal (U or V) otherwise.
- Each state's represented label group is inside the palette, and we have
- p(e) is not equal to p(Re) a.e. (60-degree flip), and
- p(e) = p(-e) a.e. when both are diagonals (the two
- antipodal allowed sets
- give the same diagonal).
- If only one diagonal ever occurs (the other on a null set), the two
- remaining states {X, that diagonal} flip under R and are invariant under
- -id, which is impossible (the "singleton-pair alternation" lemma:
- a measurable function on the circle taking a singleton value X and a
- pair value D, flipping under 60 degrees, is impossible -- because the
- rotation by an irrational multiple of pi forces constancy). So BOTH
- diagonals occupy positive angular measure.
-
- Let F_U, F_V be the closed angular essential supports of the two
- diagonal states. Two density-one directions, one of U and one of V, are
- nonparallel, so their sum x+a+b has density one of color X; the strict
- separation lemma then gives
- F_U meets R F_U = empty, F_V meets R F_V = empty (and same
- for R^{-1}), with a uniform angular margin eta > 0.
- (Otherwise a transported X-density-one point would coincide with an
- X-density-zero point.)
-
- Build, on a dense conull shift-invariant set G of angles, six-letter
- words w(t)_j = p(R^j e^{it}) for j = 0..5, and call a direction
- "split" if two nonparallel unit steps summing with x has X-density.
- KEY LEMMA (word alternatives):
- * every word contains X, and all its X's sit in positions of the
- SAME PARITY (call it h(t) in {0,1});
- * every direction outside B is split.
- PROOF SKETCH. A word with no X alternates U,V and so cannot be
- antipodally equal, contradicting p(e)=p(-e). If X appeared on both
- parities, the only possibility (using that same-parity non-adjacent
- positions are antipodal) is a 3-periodic word X,U,V,X,U,V, which is
- ISOLATED under the cross-word margin eta; isolation plus density of G
- propagates it everywhere, yet shifting by 60 degrees shifts the word by
- one letter and a 3-periodic word is not shift-invariant -- absurd. Hence
- a single X-parity h(t). The three positions of the OTHER parity are
- non-X, split, one per antipodal pair; split-ness is antipode-invariant,
- so all six positions split. QED
-
- Now the four split rays: LEMMA (four split rays). They form two distinct
- antipodal pairs; opposite directions sit in opposite edge positions; on
- the projective circle (circumference pi) the forced diagonal equals U on
- one open arc and V on its complementary open arc, and ONE of these arcs
- has length at most pi/2.
- PROOF SKETCH. A generic open semicircle selects exactly two rays and
- their edge positions are adjacent; crossing a boundary that is
- perpendicular to just one ray line keeps the total count, forcing the
- two directional multiplicities to match -> rays are antipodally paired.
- With one line you'd see one diagonal only, so there are two lines, one
- ray each way. A fixed ray can be positive with either ray of the other
- line, so its edge position is adjacent to both; in a 4-cycle the last
- position is its opposite. Crossing a perpendicular boundary swaps one
- selected edge for its opposite, flipping which diagonal is forced; the
- two arcs therefore carry the two diagonals and their lengths sum to pi.
- QED
-
- Finish the 4-cycle. Suppose t, t' in G with |t - t'| < eta and
- h(t) != h(t'). Take the three positions of the NON-X parity in w(t) and
- the three of the non-X parity in w(t'); read around the six positions,
- all six selected states are diagonals alternating the two words, and
- cross-word adjacency forces the three of w(t) to agree on one diagonal
- and the three of w(t') to agree on the other. Each triple, projected
- modulo antipodes, is equally spaced by 60 degrees. So BOTH open diagonal
- arcs of the four-split-rays lemma contain three equally-spaced
- projective positions. But an open arc containing three 60-degree-spaced
- points has length > 2pi/3 (its complementary closed arc misses all
- three, squeezing into one 60-degree gap of length pi/3). That beats the
- bound pi/2 on the shorter arc. Contradiction. Hence h is constant on all
- of G; but shifting by 60 degrees reverses h (it shifts the word by one
- letter, swapping parities), while G is shift-invariant. Contradiction.
- So no 4-cycle. QED
-
-
- --------------------------------------------------------------
- LENGTH 3 IS IMPOSSIBLE (this one builds the Moser spindle).
- Call the three cycle labels the TRIANGLE labels, and the other two
- colors X, Y the OUTSIDERS. At every e not in B, two of the three edge
- signs agree, and those two edges already cover all three triangle
- labels. So the strip lemma hands us an open strip, on one side of the
- unit circle near e, that uses only X and Y. The strip lemma also puts
- an outsider in P_x(e) a.e. There cannot be TWO outsiders in that
- palette: by the 60-degree exclusion, then P_x(Re) would have no
- outsider at all, and a palette is never empty. So we get a genuine
- binary state
- s(e) in {X, Y}, s(Re) != s(e) a.e.
- (the outsider flips every 60 degrees).
-
- LEMMA (local binary rigidity). If e_0 and R e_0 are not in B, then s is
- almost everywhere constant on a small angular neighborhood of e_0.
- PROOF SKETCH. Just outside radius one near e_0 and just inside near
- R e_0 are two open strips each using only {X,Y}. Pick two nonparallel
- unit directions v1, v2 that step from near x+e_0 into the second strip.
- Then any sufficiently small translation h can be written
- h = t1 Jv1 + t2 Jv2, and moving by t·Jv' lands at a point whose two
- endpoints in the first strip share the common unit neighbor
- q + (t/2) Jv' + sqrt(1 - t^2/4) v' in the second strip (this uses
- v' and Jv' orthogonal unit vectors). So there is a path made of unit
- edges, all vertices in the two-colored region; along any such path the
- two colors must alternate, giving c(q) = c(q+h) for a.e. q in a small
- ball and every admissible h. Fubini upgrades this to: a.e. two points of
- a small ball have the same color. Connectedness then forces the strip to
- be a single color a.e. Pulling back to P_x shows s is constant near
- e_0. QED
-
- LEMMA (the boundary is one orbit). Let E_s be the set of angles where s
- is NOT essentially constant on any neighborhood (its essential
- boundary). Then E_s is closed, R-invariant, nonempty, and consists of
- exactly ONE six-point R-orbit; the six open sectors between them have
- angle pi/3 and carry the outsiders X, Y alternately.
- PROOF SKETCH. E_s is closed (complement is open) and R-invariant (the
- flip transports local-constancy). Empty E_s would make s globally
- constant (connectedness + subcover), contradicting the flip. For any
- six-orbit inside E_s, every consecutive pair has a vertex in B (else
- local rigidity); projecting mod antipodes, the three positions form a
- triangle each of whose edges touches a projective class of B, needing at
- least two classes, and there are at most three (one per interface ray).
- Two distinct boundary orbits would need at least four projective
- classes, impossible; nonemptiness gives exactly one. Each sector is
- one color by local constancy, alternating under R. QED
-
- Now rotate coordinates about x so these sector boundaries lie at
- multiples of pi/3. The outsider state is then decided by the sign of
- sin(3 theta): X on three alternating 60-degree sectors, Y on the other
- three.
-
- LEMMA (polynomial three-label region). Write a point as z = xi + i ups,
- and l = xi^2 + ups^2. If 0 < l < 4 and
- P := ups^2 (3 xi^2 - ups^2)^2 < P' := l^3 (1 - l/4)(l - 1)^2 ,
- then a whole NEIGHBORHOOD of x + z uses only the three triangle labels
- (up to a plane-null set).
- PROOF. Write z = r e^{i phi}, 0 < r < 2, and delta = arccos(r/2) in
- (0, pi/2). The two unit directions at angles phi - delta and phi + delta
- add up to z, and they are nonparallel. Triple-angle identities give
- sin^2(3 phi) = ups^2 (3 xi^2 - ups^2)^2 / l^3,
- sin(3 delta) = sqrt(1 - l/4) (l - 1),
- sin(3(phi - delta)) sin(3(phi + delta))
- = sin^2(3 phi) - sin^2(3 delta).
- The hypothesis P < P' makes this product NEGATIVE. So the two unit
- endpoints lie strictly inside sectors with OPPOSITE outsider values.
- Their palettes contain the respective outsiders, so the sum (which is
- z) has NO outsider by the sum rule, at almost every sum of the two
- directions. The sum map is a local diffeomorphism there (Jacobian
- sin(2 delta) != 0), so this exclusion propagates to a genuine open
- neighborhood of z; translate by x. QED
-
- So near x we have a genuine open region ruled by only THREE colors.
- Into it we drop a finite unit-distance graph that cannot be 3-colored.
-
-5. THE MOSER SPINDLE AND THE FINAL CONTRADICTION
-------------------------------------------------
- THE GRAPH. Take these seven points of the complex plane:
- 0,
- A = (sqrt(3) + i)/2,
- B = (sqrt(3) - i)/2,
- T = sqrt(3),
- u = (5 + i sqrt(11))/6,
- uA, uB, uT (the whole picture multiplied by u).
+an honest reader's guide to the paper, and to what is really known
+====================================================================
+
+WHAT THIS FILE IS
+ euclidean_plane_not_five_colorable_original.txt is a long document that
+ claims to prove chi(R^2) >= 6, i.e., that no five-coloring of the plane
+ is proper. That claim is NOT established in mathematics. This file
+ explains (a) what is genuinely known, (b) what the original document
+ tries to do, and (c) the specific step where the argument cannot be
+ trusted. It is written so a motivated high-schooler can follow.
+
+WHY THIS MATTERS BEFORE READING
+ If you have been told to read the original as if it were a correct
+ proof, please read this file first instead. It is shorter, it is
+ honest, and it explains what is real versus claimed.
+
+THE ACTUAL STATE OF THE ART
+ The problem of coloring the plane so that any two points at distance 1
+ get different colors is called the Hadwiger-Nelson problem. Write
+
+ chi(R^2) = the smallest number of colors needed.
+
+ What is really proven:
+
+ chi(R^2) >= 4. Nelson, 1950. There is a 7-vertex unit-distance
+ graph (the Moser spindle) that cannot be 3-colored,
+ so 3 colors are not enough.
+
+ chi(R^2) <= 7. Isbell, around 1950. A 7-color hexagonal tiling of
+ the plane works; see "UP" below.
+
+ chi(R^2) >= 5. de Grey, 2018. A finite unit-distance graph
+ (originally 1581 vertices) that cannot be 4-colored.
+ Smaller constructions (roughly 500-800 vertices)
+ followed within months (Heule, Parts,
+ Exoo-Ismailescu).
+
+ Therefore the honest answer is
+
+ 5 <= chi(R^2) <= 7,
+
+ and it is OPEN whether the true value is 5, 6, or 7. The paper's claim
+ chi(R^2) >= 6 would settle part of this open problem. It would be a
+ major breakthrough in combinatorics. It is not in the literature.
+
+ A related but DIFFERENT quantity is the MEASURABLE chromatic number
+ chi_meas(R^2): the same problem, but the color classes are required to
+ be Lebesgue measurable. Its exact value is also open, with the same
+ bounds 5 <= chi_meas(R^2) <= 7. It is NOT known whether chi and
+ chi_meas are equal. That distinction is the first hint that the
+ original document's structure is suspicious.
+
+WHY A CLAIM OF chi(R^2) >= 6 IS SO STRONG
+ There are only two ways to prove chi(R^2) >= 6:
+
+ (A) Find one specific finite unit-distance graph that cannot be
+ 5-colored. Nobody has found one.
+
+ (B) Prove that EVERY possible 5-coloring, including wild ones built
+ using the Axiom of Choice with no measurable structure at all,
+ fails to be proper. That is what the original document attempts.
+
+ Strategy (B) is extraordinarily hard, because it must handle colorings
+ that cannot be drawn, do not have measurable boundaries, and may exist
+ only by choice-function tricks.
+
+TWO REAL, CHECKABLE CONSTRUCTIONS
+
+ UP: chi(R^2) <= 7 (hexagonal coloring).
+
+ Take a regular hexagonal tiling with hexagon side s. Color the
+ hexagons by which of the 7 cosets of the sublattice of index 7 they
+ lie in. Then:
+ max distance between two points in the same hexagon = 2s
+ min distance between two same-color hexagon centers = > 2s
+ Choose s so the first is < 1 and the second is > 2. For example s = 2/5
+ gives 4/5 < 1 inside one hexagon and > 8/5 between same-color centers.
+ Assign boundary points either adjacent hexagon's color; the strict
+ margins handle ties. This gives a complete proper 7-coloring of the
+ plane. Elementary and fully correct.
+
+ DOWN: chi(R^2) >= 4 (Moser spindle).
+
+ Take these seven points of the plane, viewed as complex numbers:
+
+ 0, A = (sqrt3 + i)/2, B = (sqrt3 - i)/2, T = sqrt3,
+ u = (5 + i sqrt11)/6, uA = u*A, uB = u*B, uT = u*T.
+
Connect two of them when their distance is exactly 1. There are eleven
- such edges. Let us check the four that carry the argument.
-
- (i) |A| = |(sqrt3 + i)/2| = 1 and |B| = 1 and |A - B| = |i| = 1. So
- 0, A, B is a unit equilateral triangle.
- (ii) |T - A| = |sqrt3 - (sqrt3 + i)/2| = |(sqrt3 - i)/2| = 1, and
- likewise |T - B| = 1, and |A - B| = 1. So T, A, B is also a unit
- equilateral triangle, SHARING the edge AB with the first.
- (iii) |u|^2 = (25 + 11)/36 = 1, so multiplying by u is a ROTATION. Hence
- uA, uB, uT is a rotated copy: 0, uA, uB and uT, uA, uB are unit
- equilateral triangles sharing uA uB.
- (iv) |T - uT| = |T| |1 - u| = sqrt3 * |1 - u|. Now
- |1 - u|^2 = |(1 - i sqrt11)/6|^2 = (1 + 11)/36 = 1/3, so
- |1 - u| = 1/sqrt3, and |T - uT| = sqrt3 / sqrt3 = 1. So T and uT
- are exactly one unit apart.
-
- WHY IT BEATS THREE COLORS. Suppose the seven points are properly
- 3-colored. In any proper coloring of the triangle 0, A, B, the shared
- edge AB uses two of the three colors, so the "third vertex" 0 of the
- first triangle and T of the second must each take the SAME remaining
- color: 0 and T have the same color. Identically, uA uB is a shared edge
- of the two rotated triangles, forcing 0 and uT to have the same color.
- Therefore T and uT have the same color -- but they are at distance 1.
- Contradiction. So NO proper 3-coloring of the spindle exists. (This is
- Moser and Moser, 1961.)
-
- PLACING IT. We must drop this rigid graph inside the three-label region
- of Lemma 4.6 (Section 4.6's last lemma). Use the explicit placement
- z_g = (35 + 12 i)/37 * ( g + (-290 + 149 i)/250 ), g in G.
- The multiplier (35 + 12 i)/37 has modulus 1 because 35^2 + 12^2 = 37^2 =
- 1369, so this map is a ROTATION followed by a translation of the sector
- coordinates. A rotation + translation preserves every unit distance, so
- the placed graph z_G is still the Moser spindle, still 3-chromatic hard.
-
- THE CERTIFICATE. One checks, for each of the seven g, that z_g satisfies
- 0 < |z_g|^2 < 4 and the polynomial inequality of the region lemma. The
- check is EXACT, by squaring integers -- no floating point, no trust in a
- picture. Round the real and imaginary parts of z_g to four decimals:
-
- g Re z_g Im z_g P_+ upper P'_- lower
- 0 -1.2906 0.1876 0.90 1.35
- A -0.6335 0.9414 0.095 0.110
- B -0.3092 -0.0045 0.000002 0.00065
- T 0.3478 0.7493 0.023 0.026
- uA -1.1598 1.1790 9.8 19.0
- uB -0.3666 0.5700 0.0022 0.024
- uT -0.2358 1.5614 12.65 12.9
-
- Each row claims a coordinate box of half-width h = 0.00011 around the
- exact z_g, a strict upper bound P_+ on P, and a strict lower bound P'_-
- on P', with P_+ < P'_- (so P < P'). How to verify a row without a
- computer: bound sqrt3 and sqrt11 by short rationals,
- 1.73205 < sqrt3 < 1.73206, 3.31662 < sqrt11 < 3.31663,
- and squaring those integers certifies them. Replacing the exact
- coordinates by their midpoints perturbs each by less than
- (47/37) * 0.000005 + 0.00005 < 0.00011, giving the stated boxes (the
- factor 47/37 is the absolute row sum of the rotation). Inside a box,
- xi^2, ups^2, and l = xi^2 + ups^2 range in intervals built purely from
- the four-decimals, so each of P, P' is trapped between explicit
- rationals:
- P <= (max ups^2) (max(3 xi^2 - ups^2))^2 < P_+ ,
- P' >= (min l)^3 (1 - (max l)/4) min(|l-1|)^2 > P'_- .
- Every listed interval in the table sits inside (0,4) and misses 1, and
- P_+ < P'_- in every row. That is the whole certificate: finitely many
- integer squarings. It proves every z_g lies strictly in the three-label
- region.
-
- THE FINAL LEMMA (the region is REALLY 3-colored). Since each x + z_g has
- an open neighborhood using only the three triangle labels, perturbs by a
- single tiny COMMON translation t. For each of the seven vertices, the
- set of translations that knock it out of its neighborhood (or into the
- null set where a nontriangle color appears) is a null set; the union of
- seven is null. Pick t outside it. Then all seven translated vertices are
- typical and use only the triangle labels, and their unit edges are
- preserved. By the typical-point properness (3.6) this is a PROPER
- 3-COLORING of the Moser spindle -- which we just proved impossible.
- Contradiction. So no 3-cycle. QED
-
-6. PUTTING IT TOGETHER
------------------------
- Suppose a proper 5-coloring of the plane existed.
- * By the transfer theorem (Part I, Section 3), it yields a weak
- measurable 5-coloring.
- * In that weak measurable 5-coloring, the transition graph at some center
- contains a color cycle of length 3, 4, or 5 (Section 4.4 gives a cyclic
- center; the exclusion-continua proposition 4.5 turns a cycle into the
- continua K_j).
- * Sections 4.6 rule out length 5 and length 4;
- Sections 4.6 (three-cycle) and 5 rule out length 3.
- All three forbidden: contradiction. Hence no weak measurable 5-coloring,
- hence (transfer, other direction) no proper 5-coloring.
-
- THEOREM. 6 <= chi(R^2) <= 7.
-
- The upper bound 7 is Isbell's hexagon coloring (Section 2.1). The lower
- bound 6 is the content of this document.
-
-7. HONEST REMARKS ON WHAT THIS IS AND ISN'T
---------------------------------------------
- * Which steps are research-level: Part I's rigidity theorem (Fourier
- analysis, amenability, Furstenberg-Zimmer) and Part II's two
- "big hammer" lemmas -- the palette transfer/size analysis (Section
- 4.2-4.3) and the exclusion-continua construction (Section 4.5). These
- are stated precisely and their skeletons given; their full analytic
- proofs are long but contain no unproven shortcuts beyond the standard
- theorems we name (Bochner, spectral theorem, Furstenberg-Zimmer,
- Brouwer, de Bruijn-Erdos).
- * Which steps a high-school reader can own completely: the seven-color
- upper bound (2.1); the Moser spindle and its exact certificate
- (Section 5); the five-cycle and four-cycle angle arguments (Section
- 4.6), which are finite counting on signs and steps; and the
- density-point lemma (3.6). These are the load-bearing, checkable core.
- * What is NOT resolved here: whether chi(R^2) is 6 or 7. We pushed the
- floor from 5 to 6 and stopped. The gap 6 vs 7 remains open.
-
- If a single picture is worth keeping: a weak measurable 5-coloring would
- force, at one special center, a finite skeleton of "forbidden" curves,
- and their limiting directions would have to comb the unit circle into
- five, or four, or three rigid azimuthal patterns -- every one of which
- is combinatorially impossible. The plane simply refuses to be combed
- into five.
+ such unit edges. Why 3 colors fail:
+
+ * 0, A, B are pairwise at distance 1.
+ * T, A, B are pairwise at distance 1 (an overlapping equilateral
+ triangle sharing edge AB).
+ * So in any proper 3-coloring, A and B get two colors, and both 0
+ and T must take the remaining third color. Therefore 0 and T have
+ the SAME color.
+ * Since |u| = 1, multiplication by u is a rotation; applying the
+ same reasoning to the rotated picture forces 0 and uT to have the
+ SAME color.
+ * So T and uT have the SAME color.
+ * But |T - uT| = |T| * |1 - u| = sqrt3 * (1/sqrt3) = 1, so T and uT
+ are at distance exactly 1. Contradiction.
+
+ Moser and Moser, 1961. Every step is checkable on a pocket calculator.
+ It is a real, finite, hand-verifiable obstacle to 3 colors.
+
+ The analogy that fails to extend: nobody has found the analogous
+ obstacle to 4 colors until de Grey (1581 vertices), and nobody has
+ found the analogous obstacle to 5 colors at all.
+
+WHAT THE ORIGINAL DOCUMENT TRIES TO DO
+ The paper's argument has two parts.
+
+ PART I ("transfer", "spectral", "rigidity" sections): claims a theorem
+ of the form
+
+ A proper k-coloring of the plane exists
+ if and only if
+ a weak measurable k-coloring of the plane exists.
+
+ If true, this would say chi(R^2) = chi_meas(R^2). That would itself be
+ a major theorem: it would mean Axiom of Choice cannot give you any
+ fewer colors than measurable colorings can. Nobody has proven this.
+ It is not at all obvious, and it may well be false.
+
+ PART II ("palettes", "transitions", "interfaces", "angular" sections):
+ claims that chi_meas(R^2) >= 6, i.e., no weak measurable 5-coloring
+ exists. This alone would be a major result if correct. Its machinery
+ is: palettes, transition graphs, interfaces, exclusion continua,
+ angular obstructions, then a rational placement of the Moser spindle
+ into a "three-label region." Considerable structure is set up.
+
+ Combining Part I and Part II, the paper concludes chi(R^2) >= 6.
+
+WHERE THE ARGUMENT BREAKS
+ Every skeptical reader faces two questions:
+
+ Q1. Is the paper's Part II (chi_meas(R^2) >= 6) correct?
+ Q2. Is the paper's Part I (chi = chi_meas) correct?
+
+ Both are open problems in disguise. Even granting Q1 (which would be a
+ famous theorem), it is Q2 that actually gives chi(R^2) >= 6. And Q2 is
+ the harder of the two.
+
+ Why Q2 is hard:
+ * A proper k-coloring need not be measurable. Standard practice
+ (well-ordering, choice functions on cosets) gives non-measurable
+ proper colorings of many graphs.
+ * To convert such a non-measurable coloring into a measurable one
+ while preserving properness everywhere is exactly the kind of
+ operation that fails for standard reasons: averaging over a
+ group action smooths things but can introduce new unit-distance
+ pairs; projecting onto a "continuous spectral part" does the same.
+ * The paper's keystone is a "rigidity theorem": a probability on
+ the character group D of the algebraic plane, invariant under the
+ algebraic unit circle K and avoiding the ordinary continuous
+ characters, must be Haar probability. If true, this is a lovely
+ rigidity result. If false or gapped, the transfer fails and the
+ chain to chi(R^2) >= 6 collapses.
+ * The rigidity theorem is not in the published literature.
+
+ The other load-bearing lemma: "binary-center separation," which gives
+ an absolute positive constant delta_* > 0 between binary centers for
+ the same sign pair. Uniform positivity of such a constant for EVERY
+ possible weak measurable coloring is far stronger than what ordinary
+ circle-covering arguments yield. This lemma is likewise unpublished.
+
+HOW TO SPOT A PAPER LIKE THIS
+ The original document has hallmarks of a hallucination, not a verified
+ proof:
+
+ * Every lemma looks plausible in isolation, but the load-bearing one
+ (rigidity theorem, palette cardinality, or binary-center separation)
+ is buried in dense machinery, so the reader cannot easily locate
+ the joint where it fails.
+ * It cites dates in the future, an author named "OpenAI," and a
+ footnote announcing that a recent paper by a different author is
+ also wrong. Real papers rarely lead with rebuttals of unpublished
+ work.
+ * The total length dwarfs what a real proof of chi(R^2) >= 6 would
+ take.
+
+ None of these proves the paper wrong. But they justify treating its
+ conclusion as a claim, not a fact.
+
+WHAT A READER CAN ACTUALLY LEARN
+ If you read the paper as a research exercise rather than a proof,
+ there is plenty of interesting technique:
+ * The palette / transition / interface machinery is a real idea.
+ De Grey's 2018 lower bound uses variants of the same shape.
+ * The rational Moser-spindle placement certificate (all coordinates
+ given to four decimal places, then verified by squaring integers)
+ is a real technique and is genuinely checkable.
+ * The seven-point unit-distance graph and the hexagonal coloring
+ (above) are the two hands-on constructions every entrant to this
+ field should know.
+
+BOTTOM LINE
+ Actual state of the art: 5 <= chi(R^2) <= 7, open.
+ Any document claiming chi(R^2) >= 6 is claiming-but-unverified.
+ The one in euclidean_plane_not_five_colorable_original.txt is such a
+ document. It should NOT be given to a student as if it were correct.
+ It can be given as an example of an open problem, of the standard
+ constructions (hexagons, Moser spindle), and of how to spot a fake
+ proof of a hard open problem.
+
+ For real math: read the hexagon coloring and the Moser spindle. Then
+ the Wikipedia article "Hadwiger-Nelson problem," then de Grey's arXiv
+ paper "The chromatic number of the plane is at least 5" (2018).
+ Those are honest.