# Equal endpoints

*Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).*

The two configurations of [Two-scale configurations](two-scale-configurations.md) give paths in the free group \(K\) of [The path criterion](the-path-criterion.md) with weights in a small ratio, but their endpoints differ. This lesson corrects the endpoints at a cost that is small compared with the larger weight and deduces the classification theorem. The corrections are paths of unit copies; a description of the group elements that such corrections can produce, through the derived series of a subgroup of \(K\), together with an averaging over coordinate permutations, makes their cost small.

Notation: \(A=\{a_1,\dots,a_s\}\) with \(s\ge2\); \(\mathcal E=\mathbb R^{(\mathbb N)}\); \(K\) the free group on the \(\xi_z\), \(z\in\mathcal E\); \(\Delta g=(g,\dots,g)\in K^s\); copy factors \(X_b=(\xi_{b_1},\dots,\xi_{b_s})\). For \(g\in K\), \(g^{[i]}\in K^s\) has \(g\) in coordinate \(i\) and \(1\) elsewhere. The derived series of a group \(N\) is \(N^{(0)}=N\) and \(N^{(r+1)}=[N^{(r)},N^{(r)}]\), the subgroup generated by commutators \([a,b]=aba^{-1}b^{-1}\).

## 1. Corrections

Let \(M\subseteq K^s\) be the submonoid generated by the diagonal \(\Delta K\) and the copy factors \(X_b\) of copies \(b\) of \(A\) at scale one, that is, with \(\|b_i-b_j\|=\|a_i-a_j\|\). An element of \(M\) is given by an *expression*, a path of factors of weight zero or one; its *degree* is the number of copy factors. If two paths have endpoints \(D_+,D_-\) and
\[
D_+^{-1}D_-=uv^{-1},\qquad u,v\in M,
\]
then \(D_+u=D_-v\): appending the expressions of \(u\) and \(v\) gives two paths with the same endpoint, with added weights \(\deg u\) and \(\deg v\).

Let \(N\) be the kernel of the homomorphism \(K\to\mathbb Z\) with \(\xi_z\mapsto1\). For \(i\ne j\) let \(M_{ij}\subseteq K\times K\) be the image of \(M\) under projection to the coordinates \(i,j\).

**Lemma 1.1.** For every \(i\ne j\), \(M_{ij}\) is a group containing \(N\times N\).

*Proof.* Put \(r=\|a_i-a_j\|>0\). *Every pair at distance \(r\) extends to a copy.* Take \(A\subset\mathbb R^d\) with affine span \(\mathbb R^d\), and let \(z,z'\in\mathcal E\) with \(\|z-z'\|=r\). Extend \(e=(a_j-a_i)/r\) to an orthonormal basis \(e,e_2,\dots,e_d\) of \(\mathbb R^d\), and choose orthonormal vectors \(f=(z'-z)/r,f_2,\dots,f_d\) in \(\mathcal E\); there is room, since \(\mathcal E\) has infinite dimension. The linear map \(R\) with \(e\mapsto f\) and \(e_m\mapsto f_m\) preserves inner products, and \(b_l=z+R(a_l-a_i)\) is a copy with \(b_i=z\) and \(b_j=z'\).

*A group.* So, with \(x=\xi_z\) and \(y=\xi_{z'}\), both \((x,y)\) and \((y,x)\) lie in \(M_{ij}\), and
\[
\begin{aligned}
(x,y)^{-1}&=(x^{-1},x^{-1})(y,x)(y^{-1},y^{-1})\\
&\in M_{ij} .
\end{aligned}
\]
Diagonal elements have their inverses in \(M_{ij}\) as well. A monoid generated by elements whose inverses it contains is a group.

*The subgroup \(N\times N\).* \(M_{ij}\) contains \((x,y)(y^{-1},y^{-1})=(xy^{-1},1)\) and \((x,y)(x^{-1},x^{-1})=(1,yx^{-1})\), and their conjugates by diagonal elements. The graph on \(\mathcal E\) joining points at distance \(r\) is connected: divide a segment into pieces of length \(\delta\le2r\); the endpoints \(p,q\) of a piece have a common neighbour \(m+\sqrt{r^2-\delta^2/4}\,e'\), where \(m\) is the midpoint and \(e'\) a unit vector orthogonal to \(q-p\). Along a path \(z=p_0,p_1,\dots,p_k=z''\) in this graph, \(\xi_z\xi_{z''}^{-1}\) is the product of the \(\xi_{p_l}\xi_{p_{l+1}}^{-1}\). So the normal subgroup generated by the \(\xi_z\xi_{z'}^{-1}\) with \(\|z-z'\|=r\) contains every \(\xi_z\xi_{z''}^{-1}\). The quotient of \(K\) by the latter is generated by the image of any one generator, and maps onto \(\mathbb Z\) under \(\xi_z\mapsto1\); a cyclic group mapping onto \(\mathbb Z\) maps isomorphically, so that normal subgroup is exactly \(N\). Hence \(N\times\{1\}\) and \(\{1\}\times N\) lie in the group \(M_{ij}\). \(\square\)

For each \(i\), let \(L_i\) be the set of \(g\in K\) such that \(g^{[i]}=uv^{-1}\) for some \(u,v\in M\), a *right fraction*. Its *cost* is \(\deg u+\deg v\) for chosen expressions, and \(c_i(g)\) is the least cost of a right fraction equal to \(g^{[i]}\).

**Lemma 1.2.** Each \(L_i\) is a normal subgroup of \(K\), and
\[
\begin{gathered}
c_i(g^{-1})=c_i(g),\qquad c_i(kgk^{-1})=c_i(g),\\
c_i(gh)\le c_i(g)+c_i(h)
\end{gathered}
\]
for \(g,h\in L_i\) and \(k\in K\). A tuple \((g_1,\dots,g_s)\) with \(g_i\in L_i\) equals a right fraction of cost at most \(\sum_ic_i(g_i)\). Moreover \(N^{(s-2)}\subseteq L_i\) for every \(i\).

*Proof.* *Inverses and conjugates.* The inverse of \(uv^{-1}\) is \(vu^{-1}\), of the same cost. If \(g^{[i]}=uv^{-1}\) and \(d=\Delta k\), then \((kgk^{-1})^{[i]}=(du)(dv)^{-1}\), of the same cost, since \(d\) has degree zero. Applying these in both directions gives the two equalities.

*A multiplication rule.* Let \(uv^{-1}\) be any right fraction and \(g\in L_i\). Put \(h=v_i^{-1}gv_i\), where \(v_i\) is the \(i\)th coordinate of \(v\); then \(v^{-1}g^{[i]}v=h^{[i]}\), and \(h\in L_i\) with \(c_i(h)=c_i(g)\). Choose \(h^{[i]}=ab^{-1}\) of cost \(c_i(g)\). Then
\[
\begin{aligned}
uv^{-1}g^{[i]}&=uh^{[i]}v^{-1}=ua\,b^{-1}v^{-1}\\
&=(ua)(vb)^{-1},
\end{aligned}
\]
of cost at most the cost of \(uv^{-1}\) plus \(c_i(g)\). Starting from \(g^{[i]}\) and multiplying by \(h^{[i]}\) gives closure under products and subadditivity, so \(L_i\) is a normal subgroup. Starting from the empty fraction and multiplying successively by \(g_1^{[1]},\dots,g_s^{[s]}\) gives the bound for tuples.

*A rule for fractions on two coordinates.* Let \(z=cd^{-1}\) be a right fraction whose coordinates are \(1\) outside a pair \(J=\{i,j\}\), and let \(uv^{-1}\) be any right fraction. Since the projection of \(M\) to \(J\) is a group (Lemma 1.1), there is \(b\in M\) whose projection to \(J\) is the inverse of that of \(v\). Then \(t=vb\) is \(1\) on \(J\), so \(t\) and \(z\) commute, and \(v^{-1}=bt^{-1}\) gives
\[
\begin{aligned}
uv^{-1}z&=ubt^{-1}z=ubzt^{-1}\\
&=(ubc)(td)^{-1},
\end{aligned}
\]
a right fraction. Its cost is finite; no bound on the degree of \(b\) is needed.

*The derived series.* We prove, by induction on \(s\ge2\), the following for every submonoid \(M'\subseteq K^s\) that contains the diagonal and whose projections to pairs of coordinates are groups containing \(N\times N\): every \(a\in N^{(s-2)}\) has \(a^{[i]}\) equal to a right fraction in \(M'\). The two rules above hold for \(M'\) with the same proofs. For \(s=2\), \(M'\) is its own pair projection and contains \(N\times N\). For \(s\ge3\), fix distinct \(i,j,k\). Projecting away the coordinate \(j\) gives a monoid in \(K^{s-1}\) with the same properties; by induction, every \(a\in N^{(s-3)}\) has a right fraction there equal to \(a^{[i]}\). Lifting its two factors to \(M'\) gives a right fraction \(z_a\) whose coordinates are \(1\) outside \(\{i,j\}\) and \(a\) at \(i\). In the same way, projecting away \(k\) gives \(z_b\), with coordinates \(1\) outside \(\{i,k\}\) and \(b\in N^{(s-3)}\) at \(i\). The four fractions \(z_a,z_b,z_a^{-1},z_b^{-1}\) are supported on pairs, so by the pair rule their product \(z_az_bz_a^{-1}z_b^{-1}\) is a right fraction. Its coordinate \(i\) is \([a,b]\); at \(j\) and at \(k\) it is \(1\), because \(z_b\) and \(z_a\) respectively are \(1\) there; elsewhere it is \(1\). So \([a,b]\in L_i\), and since \(L_i\) is a subgroup it contains \(N^{(s-2)}=[N^{(s-3)},N^{(s-3)}]\). For \(M'=M\), Lemma 1.1 gives the hypotheses. \(\square\)

Put \(L=N^{(s-2)}\), a normal subgroup of \(K\), and \(c(g)=\max_ic_i(g)\) for \(g\in L\). By Lemma 1.2, \(c\) is finite, invariant under inversion and conjugation by \(K\), and subadditive.

## 2. Averaging over coordinate permutations

Let \(H\) be a finite group of permutations of coordinates, acting on \(\mathcal E\) by isometries, and let \(Z\subset\mathcal E\) be a finite \(H\)-invariant set. Consider tuples \(w=(w_x)_{x\in Z}\in K^Z\) given as products of *diagonal factors* \((g)_{x\in Z}\) and *orbit factors* \((\xi_{hx})_{x\in Z}\) with \(h\in H\); the *degree* is the number of orbit factors. Right multiplication by a diagonal tuple does not change the degree, and neither does replacing \(w\) by \(x\mapsto w_{hx}\), which turns the orbit factor of \(g\) into that of \(gh\).

**Lemma 2.1.** For every \(\delta>0\) there is such a tuple \(w\), of degree \(m>0\), with
\[
w_xw_y^{-1}\in L,\qquad c(w_xw_y^{-1})\le\delta m
\]
whenever \(x\) and \(y\) lie in the same \(H\)-orbit.

*Proof.* *Membership in \(L\).* Start with \(w_x=\xi_x\), of degree one; all differences \(w_xw_y^{-1}\) lie in \(N\). Suppose the differences on an orbit \(O\) lie in \(N^{(r)}\). Choose \(x_0\in O\) and multiply on the right by the diagonal tuple \(w_{x_0}^{-1}\); this does not change differences, and makes all values on \(O\) lie in \(N^{(r)}\). Then replace the tuple by
\[
w'_x=\prod_{h\in H}w_{hx},
\]
with a fixed order on \(H\). For \(x\in O\), the list \((hx)_{h\in H}\) passes through every point of \(O\) exactly \(|H|/|O|\) times, so all the products \(w'_x\), \(x\in O\), have the same image in the abelian group \(N^{(r)}/N^{(r+1)}\): their differences lie in \(N^{(r+1)}\). If the differences on another orbit already lie in a normal subgroup \(G\), corresponding factors \(w_{hx},w_{hy}\) of \(w'_x,w'_y\) have the same image in \(K/G\), and so do the products. Treating the orbits one at a time, and raising the depth on each to \(s-2\), gives all differences within orbits in \(L\) (for \(s=2\) the starting tuple already has this property). The degree stays positive.

*Small costs.* Fix an orbit \(O\) and a point \(x_0\in O\). Right multiplication by the diagonal tuple \(w_{x_0}^{-1}\) makes all values on \(O\) lie in \(L\); let \(B\) be the largest cost \(c(w_x)\), \(x\in O\). Put \(h=|H|\). If \(h=1\), every orbit is a single point and there is nothing to prove; let \(h\ge2\). Replace the tuple by
\[
w'_x=\Bigl(\prod_{g\in H}w_{gx}\Bigr)w_{x_0}^{-1} .
\]
For \(x\in O\), one factor of the product is \(a=w_{x_0}\); writing the product as \(PaQ\), we get \(w'_x=P\,(aQa^{-1})\) and
\[
c(w'_x)\le c(P)+c(Q)\le(h-1)B ,
\]
since \(P\) and \(Q\) together are products of \(h-1\) values of cost at most \(B\). The new values on \(O\) lie in \(L\), and the degree is multiplied by \(h\). After \(k\) repetitions starting at degree \(m_0\), the largest difference cost on \(O\) divided by the degree is at most
\[
\frac{2B}{m_0}\Bigl(\frac{h-1}h\Bigr)^k ,
\]
using \(c(w_xw_y^{-1})\le c(w_x)+c(w_y)\); this tends to zero.

*Other orbits are not spoiled.* If \(a_jb_j^{-1}\in L\) for \(j=1,\dots,h\), then, with \(A_j=a_1\cdots a_j\), \(B_j=b_1\cdots b_j\) and \(A_0=B_0=1\),
\[
\begin{gathered}
A_jB_j^{-1}=g_j\,A_{j-1}B_{j-1}^{-1},\\
g_j=A_{j-1}(a_jb_j^{-1})A_{j-1}^{-1},
\end{gathered}
\]
so by conjugation invariance and subadditivity
\[
\begin{aligned}
&c\bigl((a_1\cdots a_h)(b_1\cdots b_h)^{-1}\bigr)\\
&\quad\le\sum_{j=1}^hc(a_jb_j^{-1}).
\end{aligned}
\]
On any other orbit, the operation therefore multiplies the largest difference cost by at most \(h\), while it multiplies the degree by exactly \(h\); the common right factor does not change differences. So the ratio does not increase there. Making the ratio at most \(\delta\) on one orbit after another proves the lemma. \(\square\)

## 3. Equal endpoints and the proof of the classification

**Proposition 3.1.** Let \(q>0\) and \(f_i,g_i\in\mathbb R^\ell\) satisfy \(\|f_i-f_j\|^2=\|a_i-a_j\|^2\) and \(\|g_i-g_j\|^2=q\|a_i-a_j\|^2\) for all \(i,j\), with each \(g_i\) a coordinate permutation of \(f_i\). For every \(\delta>0\) there are two paths with the same endpoint and weights \(\lambda>0\) and \(\mu\ge0\) such that \(\mu/\lambda\le q+s\delta\).

*Proof.* View \(\mathbb R^\ell\) as the first \(\ell\) coordinates of \(\mathcal E\), let \(H\) be the group of all permutations of these coordinates, and let \(Z\) be the union of the \(H\)-orbits of all \(f_i\) and \(g_i\). It is finite, and \(f_i,g_i\) lie in the same orbit. Take \(w\) of degree \(m\) from Lemma 2.1.

Restrict the expression of \(w\) to the labels \(f_1,\dots,f_s\). An orbit factor becomes \((\xi_{hf_1},\dots,\xi_{hf_s})\), a copy factor of weight one, because \(h\) is an isometry; diagonal factors stay diagonal. So \(D_+=(w_{f_i})_i\) is the endpoint of a path of weight \(m\). Restricted to the labels \(g_i\), the orbit factors are copy factors of weight \(q\), and \(D_-=(w_{g_i})_i\) is the endpoint of a path of weight \(qm\).

Put \(r_i=w_{f_i}w_{g_i}^{-1}\); by Lemma 2.1, \(r_i\in L\) and \(c(r_i)\le\delta m\). The \(i\)th coordinate of \(D_+^{-1}D_-\) is \(w_{f_i}^{-1}w_{g_i}=w_{f_i}^{-1}r_i^{-1}w_{f_i}\), which lies in \(L\) and has cost at most \(\delta m\). By Lemma 1.2, \(D_+^{-1}D_-=uv^{-1}\) with \(\deg u+\deg v\le s\delta m\), and \(D_+u=D_-v\). The two paths have weights \(\lambda=m+\deg u>0\) and \(\mu=qm+\deg v\), and
\[
\frac\mu\lambda\le\frac{qm+s\delta m}m=q+s\delta .\qquad\square
\]

**Proof of the classification theorem.** Let \(A\) have at least two points and affine span \(\mathbb R^d\). If \(A\) is Ramsey, the tensor condition holds by Proposition 2.1 of [A colouring obstruction](a-colouring-obstruction.md), and so does the matrix condition, by Proposition 2.1 of the first lesson. Conversely, assume the matrix condition; by Proposition 2.1 of the first lesson the tensor condition holds. Let \(n\ge1\). Theorem 3.1 of [Two-scale configurations](two-scale-configurations.md) with \(\eta=1/(2n)\) gives configurations as in Proposition 3.1 with \(q=\eta/(1+\eta)<1/(2n)\). Proposition 3.1 with \(\delta=1/(2sn)\) gives two paths with a common endpoint and weight ratio at most \(q+1/(2n)<1/n\). By the path criterion (Proposition 1.1 of [The path criterion](the-path-criterion.md)), \(A\) is Ramsey. \(\square\)

All choices depend on \(n\); no bound uniform in \(n\) on degrees or dimensions is needed.

## 4. Exercises

**4.1** (easy). Check the identity \((x,y)^{-1}=(x^{-1},x^{-1})(y,x)(y^{-1},y^{-1})\) in \(K\times K\).

**4.2** (medium). For \(s=3\), describe \(L=N^{(1)}\) and explain in words why a single coordinate can only be corrected by commutators.

**4.3** (easy). Show that in a monoid every element with a left and a right inverse has them equal.

**4.4** (medium). In Lemma 2.1 with \(|H|=2\) and one orbit \(\{x,y\}\), start from values \(w_x=\alpha\), \(w_y=1\) with \(\alpha\in L\) and \(c(\alpha)=B\). Compute the tuple after one contraction step and its cost-to-degree ratio.

## 5. Solutions

**4.1.** The coordinates of the right side are \(x^{-1}yy^{-1}=x^{-1}\) and \(x^{-1}xy^{-1}=y^{-1}\).

**4.2.** \(L\) is the commutator subgroup of \(N\). Projecting a correction to two coordinates is unrestricted (Lemma 1.1), but to change one coordinate alone, with the other two fixed, the proof combines fractions supported on \(\{i,j\}\) and on \(\{i,k\}\), and their commutator is what survives at \(i\) while cancelling at \(j\) and \(k\).

**4.3.** If \(ba=1\) and \(ac=1\), then \(b=b(ac)=(ba)c=c\).

**4.4.** With \(x_0=y\), the product \(\prod_{g\in H}w_{gx}\) is \(w_xw_y=\alpha\) for \(x\) (with the order identity first) and \(w_yw_x=\alpha\) for \(y\); multiplying by \(w_y^{-1}=1\) gives \(w'_x=w'_y=\alpha\). The difference is \(1\), of cost zero, while the degree doubles: one step already reaches ratio zero on this orbit.

## References

- [OpenAI-ER] OpenAI, *A classification of finite Euclidean Ramsey configurations*, OpenAI Math Release preprint, 23 September 2026. https://github.com/openai/math/tree/main/preprints/A-classification-of-finite-Euclidean-Ramsey-configurations-September-23-2026