A classification of finite Euclidean Ramsey configurations

Proof. Write \mathscr B=\R\otimes_{\Q}\R, let \tau(x\otimes y)=y\otimes x, and let m_{\R}:\mathscr B\to\R be multiplication. The coordinate decomposition of W identifies W\otimes_{\Q}W with arrays Q=(Q_{\alpha\beta})_{0\leq\alpha,\beta\leq d} over \mathscr B. Under this identification, evaluation is

and the gradient matrix is the spatial block (m_{\R}(Q_{\alpha\beta}))_{1\leq\alpha,\beta\leq d}. Tensor flip becomes

Directly from the evaluation formula, E_i(Q^*)=\tau(E_i(Q)), and multiplication of Q^* transposes the multiplied array. Consequently \tfrac12(Q+Q^*) preserves every zero evaluation and any multiplied spatial block equal to I_d. This proves the implication from a matrix over B to a symmetric tensor, using the natural inclusion B\subset\mathscr B.

For the converse, we show that extending the coefficient ring from B to \mathscr B introduces no new solutions. Choose an F-basis (r_\lambda)_{\lambda\in\Lambda} of \R. Distributing tensor products over direct sums gives the free B-module decomposition

Here a coefficient x\otimes y\in B in the indicated summand represents (xr_\lambda)\otimes(yr_\mu). If an array Q over \mathscr B solves the evaluation equations and has multiplied spatial block I_d, its entries have a common finite expansion

All coefficients in the maps E_i belong to B. Thus (4) implies E_i(P^{\lambda\mu})=0 for each pair and each label. Let C^{\lambda\mu}\in\operatorname{Mat}_d(F) be the multiplied spatial block of P^{\lambda\mu}. Applying multiplication to the expansion gives

This finite linear system has coefficients and right-hand side in F. Since it is consistent over \R, row reduction over F shows that it is consistent over F. Choose c_{\lambda\mu}\in F with I_d=\sum c_{\lambda\mu}C^{\lambda\mu}. Then

is a solution over B. The individual matrices C^{\lambda\mu} need not be symmetric; the descent took place in the full space \operatorname{Mat}_d(F). If a symmetric tensor is wanted after descent, the symmetrization already proved applies to P.□

Proof. Regard

as a \Q-vector space and define the \Q-linear map

By assumption y_0=(0,\ldots,0,I_d) is outside its image. In the quotient vector space \mathcal Y/\operatorname{im}\Phi, extend the nonzero class of y_0 to a basis over \Q. Sending that basis element to 1 and the others to 0 gives a \Q-linear functional \ell:\mathcal Y\to\R such that

Write its components as \ell_i:\R\otimes_{\Q}\R\to\R and L:\Sym_d(\R)\to\R, and put h_i(x)=\ell_i(x\otimes x). Applying \ell to \Phi(w\otimes w) gives

Taking u=0 shows that

These are purely algebraic functionals; no continuity or measurability is asserted or needed.

Fix D and define H_i(z)=\sum_{j=1}^D h_i(z_j) for z\in\R^D. We claim that every ordered congruent copy (b_1,\ldots,b_s) satisfies

Indeed, equality of the pairwise distances gives equality of the Gram matrices of (a_i-a_1)_i and (b_i-b_1)_i by polarization. Thus sending each a_i-a_1 to b_i-b_1 defines an inner-product-preserving linear map on their span: any relation among the first differences has squared norm zero for the corresponding combination of the second. Since A affinely spans \R^d, there are a linear isometry Q:\R^d\to\R^D and t\in\R^D such that b_i=t+Qa_i. Writing u_j^{\mathsf T} for the jth row of Q, we have

Sum (5) over j. The finite additivity of L proves (7). Also (6) gives \sum_iH_i(z)=0 at every single point z.

Choose K=2s+1 and partition [0,2) into K half-open intervals of equal length 2/K<1/s. Color a point z by the s-tuple of interval labels of the representatives of H_i(z) modulo 2. This uses at most r_0=K^s colors, independently of D. If b_1,\ldots,b_s were monochromatic, there would be integers n_i and real numbers \varepsilon_i, with |\varepsilon_i|<2/K, such that

Summation yields

The first identity makes the error sum an odd integer, contradicting the second. This coloring avoids A in every dimension and completes the proof.□

Proof. The Laurent polynomial ring

is the group algebra of \Gamma, with X^\gamma=\prod_jX_j^{\gamma_j}. The quotient map \Gamma\to\Gamma/\Lambda_i induces a ring homomorphism whose kernel is

Indeed, its kernel consists of Laurent polynomials with zero coefficient sum on each coset. Within a coset, choose a representative \gamma_0 from the finite support; a zero-sum polynomial on that coset is a sum of terms c_\gamma(X^\gamma-X^{\gamma_0}), each in J_i.

We will find a Laurent polynomial in every J_i with prescribed terms through degree two near X=(1,\ldots,1). To use the tensor hypothesis, we pass to formal logarithmic coordinates, in which the generators X^h-1 become linear forms multiplied by units. Put \mathfrak m=(X_1-1,\ldots,X_k-1) and complete R with respect to \mathfrak m. With Z_j=X_j-1, its completion is \widehat R=\Q[[Z_1,\ldots,Z_k]]: the factors 1+Z_j are already invertible modulo each power of (Z_1,\ldots,Z_k). The formal changes of variables

identify this ring with \Q[[Y_1,\ldots,Y_k]] and identify \mathfrak m\widehat R with (Y_1,\ldots,Y_k). For each h\in\Gamma,

The final factor is a unit. Hence

Consider the quadratic polynomial

Identify \Q[Y_1,\ldots,Y_k] with the symmetric algebra of \Q^k by sending \varepsilon_j to Y_j. The map to the symmetric algebra of \Q^k/V_i has kernel generated by V_i in degree one. This follows, for example, by extending a basis of V_i to a basis of \Q^k. The tensor hypothesis therefore puts Q_2 in the ideal generated by the linear forms h\cdot Y, h\in\Lambda_i. By (11), Q_2\in J_i\widehat R for every i.

The ring R is Noetherian, being a localization of a polynomial ring over \Q. Its \mathfrak m-adic completion is therefore flat over R [23, Lemma 10.97.2, Tag 00MB]. If I=\bigcap_{i=1}^sJ_i, the injection

remains injective after tensoring with \widehat R. Consequently

and Q_2\in I\widehat R.

We now recover a finite Laurent polynomial while retaining exact membership in I. Write

as a finite sum. For each a, truncate the formal Z-expansion of \alpha_a through degree two to obtain b_a\in R with b_a-\alpha_a\in\mathfrak m^3\widehat R. Then f=\sum_af_ab_a\in I and

Write the finite Laurent expansion as f=\sum_\gamma\nu(\gamma)X^\gamma. Membership in each J_i gives (9). Finally,

Comparing the constant term, gradient, and Hessian at Y=0 gives (10).□

Proof. We will average Gaussian densities with covariance matrices running from S to S+C. We first construct a positive definite S for which S+C is positive definite and \tr(USU^{\mathsf T}) is small.

The identity UCU^{\mathsf T}=I_d implies that U has rank d. Set H=\operatorname{im}(U^{\mathsf T}) and V=\ker U, so that \R^k=H\oplus V orthogonally. The restriction of the quadratic form C to H is positive definite: if 0\ne h=U^{\mathsf T}y, then

Choose \epsilon>0 with \epsilon\tr(UU^{\mathsf T})<\eta/8. We claim that, for a sufficiently large L>0, the matrix

is positive definite and so is S+C. Here P_H,P_V are the orthogonal projections. To check the latter assertion, write

Choose L such that C_{VV}+L I_V-B^{\mathsf T}A^{-1}B>0 and use the Schur complement. If V=\{0\}, the second block is absent. Since UP_V=0, this construction also gives

Every matrix S+tC, 0\leq t\leq1, is positive definite, being a convex combination of S and S+C.

For a positive definite k\times k matrix Q, let

be the centered Gaussian density with covariance Q. Differentiating the displayed formula, with Q_t=S+tC, gives

This identity does not require C to be positive semidefinite. The covariances Q_t have uniformly bounded eigenvalues, bounded away from zero, so the densities and each of their spatial derivatives have uniform Gaussian decay. Consequently

is a nonnegative Schwartz function of integral 1, and differentiation under the integral gives

The pointwise inequality (\phi_{S+C}-\phi_S)_-\leq\phi_S and (14) imply

To make the support compact, choose a smooth function \chi with 0\leq\chi\leq1, equal to 1 on the unit ball and zero outside the ball of radius 2. Put \chi_R(x)=\chi(x/R) and b_R=\chi_Rb_*. Then \int b_R\to1. The product rule reads

The derivatives of \chi_R have bounds of orders R^{-1} and R^{-2} and are supported where R\leq\|x\|\leq2R. The Schwartz decay therefore implies

Since r\mapsto r_- is 1-Lipschitz, the corresponding weighted negative-part integrals converge as well. For sufficiently large R, we have \int b_R>1/2 and the last displayed integral is less than \eta/8. Equation (15) now gives

Taking b=b_R proves the lemma.□

Proof. Take b from Lemma 4.2. For each positive integer n, set

These functions have finite support. Quotient pushforward commutes with convolution, so every required pushforward of \mu_n is zero. More generally, for any finitely supported real function d on \Gamma, the moment identities for \nu give

Indeed, substitute \lambda=\beta+\gamma in the double sum. The terms in \beta\beta^{\mathsf T}, \beta\gamma^{\mathsf T} and \gamma\beta^{\mathsf T} vanish by the zeroth and first moments of \nu. Thus, with

the identity in (16) holds exactly for (\mu_n,\zeta_n), and Riemann sums give \zeta_n\to\int b>0.

For x=\lambda/n, Taylor expansion gives

The error is uniform in x: the third derivatives of b are bounded, and \sum_\gamma|\nu(\gamma)|\|\gamma\|^3 is finite. Moreover, the supports of both \mu_n(\lambda) and D_Cb(\lambda/n), after division of the lattice by n, lie in a single fixed ball for all n\geq1. There are O(n^k) relevant lattice points and the weight \|Ux\|^2 is bounded on that ball. The 1-Lipschitz property of the negative part therefore shows that

The second sum is a Riemann sum for a continuous compactly supported function. By (13), it follows that

Fix a sufficiently large n such that \zeta_n>0 and (17) holds strictly for (\mu_n,\zeta_n).

It remains to obtain rational coefficients. Let E\subset\Gamma be a finite set supporting d_n. For functions d supported in E, the maps

are linear, and the function

is continuous on \R^E. The two strict inequalities \zeta>0 and (17) consequently persist in an open neighborhood of d_n. Choose a rational vector d in that neighborhood and extend it by zero outside E. Put \mu=\nu*d. Both \mu and \zeta are rational. Convolution still gives zero quotient pushforwards, and (18) still gives the exact second moment (16). This proves the lemma.□

Proof of Lemma 4.1. Use the notation preceding Lemma 3.1 and the function \nu supplied by that lemma: w_j=(v_j,u_j), w(\lambda)=\sum_j\lambda_jw_j, U=(u_1\ \cdots\ u_k), \Gamma=\Z^k, and

The rational symmetric matrix C of the tensor satisfies UCU^{\mathsf T}=I_d. Apply Lemma 4.3 to obtain n,\mu,\zeta. Choose a positive integer M such that every M\mu(\lambda) is an integer. Form two finite lists of affine rows: the positive list has M\mu(\lambda)_+ repetitions of w(\lambda/n), and the negative list has M\mu(\lambda)_- repetitions.

The zero total mass of \mu shows that the two lists have the same length. For each label i,

Thus the zero pushforward on \Gamma/\Lambda_i says exactly that, after evaluating at a_i, every real coordinate value appears equally often in the two lists. The resulting evaluation vectors are coordinate permutations of one another.

Divide every affine row in both lists by \sqrt{\zeta M}. Their gradient Gram matrices become

Equations (16) and (17) give

Since B_- is positive semidefinite, each of its eigenvalues is at most its trace. Hence P=\eta I_d-B_- is positive semidefinite. Choose vectors r_1,\ldots,r_d\in\R^d with \sum_jr_jr_j^{\mathsf T}=P, for example the columns of P^{1/2}, and append the same affine rows (0,r_j) to both lists. The two Gram matrices are now (1+\eta)I_d and \eta I_d. Identical appended rows preserve equality of all coordinate multisets.

Finally divide every row in both lists by \sqrt{1+\eta}, and let f_i,g_i be their respective evaluation vectors at a_i. Their coordinate multisets remain equal. For any affine row (v,u), the difference of its evaluations at a_i and a_j is (a_i-a_j)\cdot u. The two gradient Gram identities therefore give (12) exactly, with q=\eta/(1+\eta).□

Proof. Suppose instead that every \R^D admits an r-coloring avoiding A. Consider the compact product space \{1,\ldots,r\}^{\mathcal E}. For each ordered congruent copy (b_1,\ldots,b_s) of A in \mathcal E, impose the condition that the colors of the b_i are not all equal. This is a clopen condition depending on finitely many coordinates of the product space.

Every finite collection of these conditions involves finitely many finite-support vectors. They all lie in the first D coordinate positions for some D. An avoiding coloring of this \R^D, extended arbitrarily to \mathcal E, satisfies the finite collection. Compactness therefore gives a coloring satisfying all the conditions, contrary to the hypothesis.□

Proof. Set

These numbers satisfy 0<\alpha_n<\cdots<\alpha_1=1. Partition [0,1] at these points. For each resulting interval of length l>0, apply to both paths the endomorphism of K defined by

It preserves diagonal factors and multiplies every copy-factor weight by l/\lambda. The two new paths thus have the same endpoint and weights l and \rho l.

Concatenate the interval paths in increasing interval order. The resulting endpoint is independent of which of the two paths is chosen on each interval. For a given t, choose the larger weight on the intervals before \alpha_t and the smaller weight on the remaining intervals. The total weight is

The fixed order of concatenation gives one endpoint w for all t; no commutation of factors is used.□

Proof. For fixed v, the map B\mapsto\{x:x^{-1}B\in v\} preserves finite intersections and complements. Thus (20) defines an ultrafilter uv. For u,v,w\in\beta\mathcal S, both parenthesizations of their product contain B precisely when

This proves associativity. For fixed v, the inverse image of \overline B under u\mapsto uv is \overline{\{x:x^{-1}B\in v\}}, so right translation is continuous. The assertion about \delta_\epsilon follows directly from the definition.

Choose a minimal nonempty compact left ideal L_0 in \beta\mathcal S. Such an ideal exists because every chain of nonempty compact left ideals has a nonempty intersection. For each v\in L_0, the set (\beta\mathcal S)v is a nonempty compact left ideal contained in L_0, hence

The ideal L_0 is also a compact subsemigroup.

Choose a minimal nonempty compact subsemigroup H_0\subseteq L_0. For x\in H_0, the set H_0x is compact and is a subsemigroup: (ax)(bx)=a(xb)x\in H_0x for a,b\in H_0. Minimality gives H_0x=H_0. Consequently

is nonempty, compact, and a subsemigroup. It equals H_0 by minimality, so x^2=x. In particular L_0 contains an idempotent p.

The corner G=p(\beta\mathcal S)p is a monoid with identity p. Every y\in G lies in L_0, so (21) gives p=vy for some v\in\beta\mathcal S. Since py=y,

Thus every element of G has a left inverse in G. A monoid with this property is a group: if ay=p and ba=p, then b=b(ay)=(ba)y=y, whence ya=p as well.□

Proof. Choose the strings from left to right. Before column j, let P_i be the already chosen prefix and maintain the invariant

Initially P_i=\epsilon. Put v_i=q_{i,j+1}\cdots q_{i,M}, interpreting an empty product here as \delta_\epsilon. By (20), the set

belongs to q_{i,j}. In a common column, the finite intersection \bigcap_i E_i belongs to the common ultrafilter and is nonempty; choose the same string from it for all i. In a prescribed principal column, z_{i,j}\in E_i, so choose that string. Replacing P_i by P_ix_{i,j} preserves the invariant. After the last column it says P_i^{-1}B\in\delta_\epsilon, which is precisely P_i\in B.□

Proof of Proposition 5.1. Fix r\ge2 and an arbitrary coloring c:\mathcal E\to\{1,\ldots,r\}. Choose n such that every r-coloring of \{1,\ldots,s\}^n has a monochromatic combinatorial line, by the Hales–Jewett theorem [8]; see also Shelah’s proof and primitive-recursive bound [22, p. 686 and Section 1]. We use only existence of the length. Such a line has a nonempty set of variable positions, occupied by one common letter, and fixed letters in all other positions.

In the group formulation of Leader, Russell and Walters [16, Conjecture C and the following discussion], the number of varying positions is prescribed before coloring; without that restriction, ordinary Hales–Jewett applies. Here we allow the line to choose that number and prepare all the required path weights in advance. The hypothesis and Lemma 5.3 give an endpoint w=(w_1,\ldots,w_s)\in K^s and a path of weight 1/t ending at w for every 1\le t\le n.

Color \mathcal S by c\circ\Phi, and write its color classes as B_1,\ldots,B_r.

Choose p as in Lemma 5.4. The freeness of K gives a homomorphism

where (z) is the one-letter string with entry z. Color a word i_1\cdots i_n by the unique index k for which

Every ultrafilter contains exactly one class of a finite partition, so this defines an r-coloring.

Take a monochromatic combinatorial line and let t be its number of variable positions. Figure 1 records the order of these choices.

For each of the line’s s words, expand the factor at every variable position using the chosen path ending at w with weight 1/t. A diagonal path factor contributes one common ultrafilter \psi(g). A copy factor contributes p\delta_{(b_i)}p in role i: the two p factors are common and the middle factor is principal at a prescribed one-letter string. Unexpanded fixed positions are common factors as well. The identity of the corner group is p; if an empty group product occurs, it is retained as this common factor. The expanded products therefore have the column forms required by Lemma 5.5, and all contain the same color class B_k.

Apply that lemma. It produces strings z^{(1)},\ldots,z^{(s)} in B_k. Every common factor was realized by the same string in all roles, and every varying factor has length one. Hence the resulting strings have equal lengths and their entries align. Common strings contribute zero to their pairwise squared distances. Each of the t expanded paths contributes \|a_i-a_j\|^2/t by its weight. Equation (19) consequently gives

Their images form a monochromatic congruent copy of A in \mathcal E. Lemma 5.2 now supplies a finite forcing dimension for this r. Since r was arbitrary, A is Euclidean Ramsey.□

Proof. Put r=\|a_i-a_j\|>0. Every ordered pair z,z'\in\mathcal E at distance r extends to an ordered congruent copy of A. Indeed, send the direction of a_j-a_i to (z'-z)/r, extend it to an orthonormal frame for the affine span of A, and translate to send a_i to z. The frame can be chosen inside a finite-dimensional coordinate subspace of \mathcal E.

Set x=\xi_z and y=\xi_{z'}. Both (x,y) and (y,x) therefore belong to M_{ij}, and

The diagonal generators are also invertible within M_{ij}, so this monoid is a group. It contains

and their conjugates by diagonal elements.

The graph on \mathcal E joining points at distance r is connected. To see this, subdivide any displacement into pieces of length at most 2r. The endpoints of each piece have a common point at distance r: choose an appropriate perpendicular displacement from their midpoint. An additional coordinate supplies the perpendicular direction if needed. Differences of letters along graph edges consequently generate, after taking their normal closure in K, all differences \xi_z\xi_{z'}^{-1}. That normal closure is N, since identifying all free generators gives the infinite cyclic group. The two displayed types of elements and diagonal conjugation thus give N\times N\subset M_{ij}.□

Proof. Inversion of a right fraction interchanges its numerator and denominator. If d=\Delta(k), then

These operations preserve the represented cost. Applying each operation in both directions proves inversion and conjugation invariance.

The following multiplication rule supplies the cost estimate. Given any represented right fraction uv^{-1} and g\in L_i, put h=v_i^{-1}gv_i and choose h^{[i]}=ab^{-1} of cost c_i(h)=c_i(g). Then

The resulting cost is at most the previous cost plus c_i(g). This proves product closure and subadditivity for L_i; together with inversion and conjugation it proves normality. Successively applying (23) to the single-coordinate factors of (g_i)_i proves the asserted tuple cost bound.

For the derived subgroup inclusion, we also need a multiplication rule for fractions supported on a pair of coordinates. Suppose that z=cd^{-1} is such a fraction, supported on a pair J, and let uv^{-1} be any right fraction. Since the projection of M onto J is a group, choose b\in M such that t=vb is the identity on J. The tuples t and z commute, because their supports are disjoint. Consequently

This is a right fraction of finite cost. We require no bound on the cost of the auxiliary b.

We now prove (22) by induction on the number of coordinates. The proof applies to any submonoid of K^s containing the diagonal whose pair projections are groups containing N\times N. For s=2, the whole monoid is its pair projection and contains N\times N, which proves the claim. For s\ge3, fix distinct indices i,j,k. The monoid obtained by projecting away j satisfies the same abstract hypotheses. By induction, every a\in N^{(s-3)} has a right fraction in that projection supported at i. Lift its numerator and denominator to M. Their quotient is a right fraction z_a supported on \{i,j\}, with value a at i. Projecting away k similarly gives a right fraction z_b supported on \{i,k\} with any prescribed b\in N^{(s-3)} at i.

The four factors of z_a z_b z_a^{-1}z_b^{-1} are pair-supported right fractions; inverses remain fractions by interchanging numerator and denominator. Repeated use of (24) makes their product a right fraction. Its only possibly nonidentity coordinate is i, where its value is [a,b]. Thus [a,b]\in L_i. Since L_i is a subgroup, it contains the subgroup generated by all these commutators, namely N^{(s-2)}.□

Proof. We first arrange membership in L, then make the costs small. Start with w_x=\xi_x, represented with degree one; its differences w_xw_y^{-1} belong to N.

Suppose the differences on an orbit O belong to N^{(r)}. Choose x_0\in O and right-multiply by the diagonal element w_{x_0}^{-1}. After this operation, every value on O belongs to N^{(r)}, and every difference on every orbit is unchanged. Replace the resulting tuple by

using a fixed order on H. For x\in O, the list (h x)_{h\in H} visits every point of O with the same multiplicity |H|/|O|. The products in (27) therefore have the same image in the abelian group N^{(r)}/N^{(r+1)} for every x\in O. Their differences belong to N^{(r+1)}.

This procedure preserves any normal-subgroup condition already obtained on another orbit. Indeed, if w_xw_y^{-1}\in G there for a normal subgroup G\lhd K, corresponding factors of the two products have the same image in K/G. Their products have the same image as well. Process the finitely many orbits, increasing the derived depth on each to s-2. Every degree stays positive, and we obtain membership in L for all within-orbit differences. When s=2, the initial tuple already has this property.

Fix one orbit O. Right-normalize at x_0\in O once more, so all values on O lie in L, and write B=\max_{x\in O}c(w_x). Put h=|H|. If h=1, all orbits are singletons and (26) already holds. Hence suppose h\ge2. Repeatedly perform the operation

For each x\in O, one factor in the product is a=w_{x_0}. Write the product as P a Q. Then

All new values on O still lie in L, and their represented degree is h times the preceding degree. Thus, after k repetitions starting from degree m_0, the maximum difference cost divided by degree is at most

Here the factor two uses c(w_xw_y^{-1})\le c(w_x)+c(w_y). The bound tends to zero.

It remains to check that treating this orbit does not spoil a bound on another orbit. If a_jb_j^{-1}\in L for 1\le j\le h, conjugation invariance and subadditivity give

For completeness, write A_j=a_1\cdots a_j and B_j=b_1\cdots b_j, with A_0=B_0=1. The identity

proves (30) inductively. Consequently the product in (28) multiplies the maximum difference cost on any other orbit by at most h, while multiplying the degree by exactly h. The appended common right factor does not change differences. Their cost divided by degree cannot increase. We may therefore achieve the bound \delta on one orbit after another; there are only finitely many orbits.□

Proof. Embed \R^\ell into \mathcal E, let H be its full coordinate permutation group, acting trivially on the remaining coordinates, and let Z be the union of the H-orbits of the f_i and g_i. It is finite, and f_i,g_i lie in the same orbit for each i. Apply Lemma 6.3 with the given \delta and let m>0 be the degree of its represented tuple w.

Restricting the same expression to the labels f_i gives the endpoint D_+=(w_{f_i})_{i=1}^s of a path of weight m. Restricting it to the labels g_i gives the endpoint D_-=(w_{g_i})_{i=1}^s of a path of weight qm. Indeed, each counted factor acts by a coordinate permutation, so it preserves the corresponding distances; diagonal factors stay diagonal.

For each i, put r_i=w_{f_i}w_{g_i}^{-1}. By (26), r_i\in L and c(r_i)\le\delta m. The ith component of D_+^{-1}D_- is

so it too belongs to L and has cost at most \delta m. Lemma 6.2 supplies represented u,v\in M such that

Thus D_+u=D_-v. Writing a=\deg u and b=\deg v, the two resulting paths have weights

□

Completion of the sufficiency implication in Theorem 1.1. Assume the matrix condition of the theorem. By Proposition 2.1, the tensor condition (3) holds. For every integer n\ge1, Lemma 4.1 supplies two images as in Proposition 6.4 with 0<q<1/(2n). Apply that proposition with \delta=1/(2sn). It gives equal-endpoint paths whose weight ratio is strictly less than 1/n. Proposition 5.1 now implies that A is Ramsey. All choices are made after fixing n; no bound uniform in n on the degrees or the number of coordinates is required.□

Proof. A singleton is spherical. Otherwise use the notation and a matrix P from Theorem 1.1, and put H=m_F(P), entrywise. Multiplying the evaluation equations gives

Thus all points lie on the sphere with center -\ell/2 and squared radius \|\ell\|^2/4-c. The radius is positive because the set has at least two distinct points. No symmetry of the constant row and column of H is needed.□

Proof. Singletons are immediate. Let A=\{a_1,\ldots,a_s\}\subset\R^d, with s\ge2, affinely span \R^d, and suppose an isometric embedding maps a_i to y_i in a finite transitive set Y\subset\R^n. Write H for the finite group of distance-preserving permutations of Y. For each h\in H, the map a_i\mapsto h(y_i) extends to an affine isometry. Let w_{hj}\in W=\R^{1+d} be its jth affine coordinate row, so that e_i(w_{hj})=(h(y_i))_j. The gradient Gram matrix of these n rows is I_d.

Let c_{hj}=((h(y_1))_j,0)\in W, a constant affine row. The symmetric rational tensor

has G(T)=I_d. For each i, the multiset (h(y_i))_{h\in H} contains every point of Y exactly |H|/|Y| times. Consequently

and (e_i\otimes e_i)(T)=0. Proposition 2.1 and Theorem 1.1 now give the Ramsey property.□

Proof. The singleton case is immediate, so assume s\ge2. Write N=(d+1)^2. Sphericity gives a real matrix with spatial block I_d and quadratic values zero at every p_i. These requirements are a linear system over F, so a solution exists over F as well, by Gaussian elimination. Denote such a matrix, written as a vector in F^N, by h.

Let D\in\operatorname{Mat}_{s\times N}(B) be the tensor evaluation matrix, whose entry in row i and column (\alpha,\beta) is p_{i\alpha}\otimes p_{i\beta}. Lift h to b\in B^N by applying x\mapsto x\otimes1 entrywise. The assumed row independence allows us to select s columns of D whose square submatrix M has

Let \iota:B^s\to B^N insert a vector into these selected coordinates, so that D\iota=M. Multiplying by \det M allows us to cancel the tensor evaluation error Db through the selected columns: the adjugate identity gives

Since m_F(D)h=0, multiplication gives m_F(z)=\delta h. Multiplying z by \delta^{-1}\otimes1 therefore produces a solution of the criterion in Theorem 1.1. This argument does not require \det M to be a unit of B.□

Proof. Work in the plane of the circle and extend the set, if necessary, to five distinct points on that circle. Let F be the coordinate field of the resulting five-point set. Fix one of these points a_i, partition the other four into two pairs, and take the product of affine equations for the two lines through those pairs. No three circle points are collinear, so this quadratic polynomial vanishes at the other four points and is nonzero at a_i. Its coefficients lie in the coordinate field F. Rescaling the five resulting polynomials shows that quadratic evaluation onto F^5 is surjective. Its five rows are therefore independent, and Proposition 7.3 makes the five-point set Ramsey. A monochromatic copy of that set contains a copy of the original set.□

Proof. The four points are distinct and lie on the unit circle, so Corollary 7.4 makes K_a Ramsey. Leader, Russell and Walters proved that K_a does not embed in any finite transitive Euclidean set [15, Corollary 2].□

Proof. The parametrization lies on the unit circle and is injective, since t_i=(a_i)_2/(1+(a_i)_1). Thus the points are distinct, affinely span \R^2, and have coordinate field F=\Q(t_1,\ldots,t_9). Algebraic independence identifies this field with a rational function field. Consequently B=F\otimes_\Q F is the localization of \Q[T_1,\ldots,T_9,U_1,\ldots,U_9] obtained by inverting all nonzero polynomials in the T variables alone and all nonzero polynomials in the U variables alone. In particular, B is a domain and embeds in \Q(T_1,\ldots,T_9,U_1,\ldots,U_9).

Put

Over this fraction field, the nine tensor evaluation equations have coefficient matrix D with rows (p_\alpha(T_i)p_\beta(U_i))_{0\le\alpha,\beta\le2}. We claim that \det D\ne0. To check the claim, specialize the nine pairs (T_i,U_i) to the grid \{0,1,-1\}^2, with both coordinates ordered as 0,1,-1. All denominators in the displayed matrix remain nonzero. The resulting matrix is the ordinary Kronecker product C\otimes C, where

Its determinant is 2^6=64, proving the claim. This evaluates the specific determinant expression; no specialization of the entire localized ring B is asserted.

Thus the only solution of D\,\operatorname{vec}(P)=0 over the fraction field is P=0. Injectivity of the embedding of B shows that the same holds over B. Such a matrix cannot have multiplied spatial block I_2, so Theorem 1.1 excludes the Ramsey property.□

Proof. First, 0.11<t<0.12 and t is transcendental. Indeed, the rational partial sums t_m=p_m/q_m, with q_m=10^{m!}, satisfy

If t had an irreducible polynomial P\in\Z[T] of degree d, then P(t_m)\ne0 and |P(t_m)|\ge q_m^{-d}. A bound C\ge1 for |P'| on [0,1] would give q_m^{-d}<2Cq_m^{-(m+1)}, a contradiction for large m. The rotations R_+ and R_- have angles \theta=2\arctan t and -\theta, where 0<\theta<\pi/4. Their differences from each other and from zero are nonzero and smaller than \pi/2 in absolute value. Hence the three squares are disjoint. They lie on the unit circle, and Q_0 already affinely spans the plane.

The coordinate field of X is F=\Q(t): its coordinates include u,v, and t=v/(1+u). On this field define the derivation

It satisfies \partial u=-v and \partial v=u. Write R_0=I_2, label the points by x_{\epsilon,q}=R_\epsilon q for \epsilon\in\{+,-,0\} and q\in Q_0, and assign weights \beta_{+,q}=\beta_{-,q}=1, \beta_{0,q}=-2. For these twelve labels put p_i=(1,x_i)^{\mathsf T}. We have the three augmented moment identities

Here the block sizes are 1 and 2. To verify them, use \sum_{q\in Q_0}q=0 and \sum_{q\in Q_0}qq^{\mathsf T}=2I_2. With J=\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right), the derivatives are \partial R_+=R_+J, \partial R_-=-R_-J, and \partial R_0=0. The first spatial moment is 2(1+1-2)I_2=0; the mixed moment is 2J-2J=0; and the derivative moment is 2I_2+2I_2=4I_2. The constant and vector entries vanish by the same square symmetries.

Define the \Q-linear map \Phi:F\otimes_\Q F\to F by \Phi(a\otimes b)=(\partial a)(\partial b). For every P\in\operatorname{Mat}_3(B), the product rule and (31) give

For completeness, expand each entry of P into pure tensors a\otimes b. In \partial(p_{i\alpha}a)\partial(p_{i\beta}b), the terms involving \partial a or \partial b vanish after summing with \beta_i by the first two moment identities and the transpose of the second. The remaining term is ab(\partial p_{i\alpha}) (\partial p_{i\beta}), and the third identity gives exactly the right side of (32).

A matrix satisfying the classification criterion would make the left side zero and the right side 8. This contradiction proves the claim. The argument uses a derivation only on F, and imposes no regularity condition on the colorings in the Ramsey definition.□