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The project arguments are credited to Plasma AI. The exact-carrier equivalence and all-finite template interfaces consumed by the main argument are covered by the graded whole-conclusion review. That scope excludes standalone acceptance of the optional two-element template proof and unused alternatives. Propositions 1, 3, 5, 7, 8 and 9 and the carrier corollary have complete written arguments; results outside the grade's scope retain author standing. Proposition 4 uses Theorem 4 of the circle account; Proposition 2 is conditional on (R0), and Proposition 6 states a sufficient interface. Theorem 1 of the finite-cover account supplies its template, and the rigidity extension discharges the remaining field hypotheses in the examined proof. Notation follows [[research/leads/equilateral_two_scale/spectral_exclusion|the spectral account]]: $E$ is the field of algebraic numbers, $K$ its algebraic unit vectors, $\omega=e^{i\pi/3}$, and (R0) is the rigidity premise stated there.

Why the algebraic carrier stops at algebraic ratios

The side of a triangle with vertices in $E$ has algebraic square, so it is algebraic. For transcendental $x$ the carrier $E$ of the OpenAI transfer contains no equilateral triangle of side $x$. Theorem 3 of the spectral account and the algebraic-family version of the binary transfer cannot treat that scale on the carrier $E$. Two replacements are examined below; Proposition 3 shows that the second, the algebraic closure of $\mathbb Q(x)$, is exactly the carrier the problem needs.

The algebraic span of 1 and x

Let $x>0$ be transcendental and $D_x=E+Ex$. It is the smallest subgroup of $\mathbb C$ containing $E$ and $x$ that is closed under multiplication by $K$, since the ring generated by $K$ is $E$. The sum is direct.

Proposition 1. The equilateral triangles of side $1$ or $x$ with all vertices in $D_x$ are exactly the triples

$$(a+bx,\;a+u+bx,\;a+u\omega+bx)\quad\text{and}\quad (a+bx,\;a+(b+u)x,\;a+(b+u\omega)x),$$

with $a,b\in E$ and $u\in K$. So the hypergraph of these triangles on $D_x$ is the Cartesian square of the hypergraph of unit equilateral triangles on $E$, under $a+bx\mapsto(a,b)$.

Proof. A triangle of side $r$ with vertices in $D_x$ is $(p,p+v,p+v\omega)$ with $|v|=r$ and $v\in D_x$, because $\omega\in E$ and $D_x$ is an $E$-module. Write $v=a+bx$ with $a,b\in E$. Then

$$|v|^2=|a|^2+2\operatorname{Re}(a\bar b)\,x+|b|^2x^2,$$

a polynomial in $x$ with coefficients in the real algebraic numbers. If $|v|=1$, transcendence of $x$ forces $|a|=1$ and $b=0$; if $|v|=x$, it forces $a=0$ and $|b|=1$. $\square$

Corollary. For every transcendental $x$ the triangles of sides $1$ and $x$ in $D_x$ can be avoided: if $s$ is any unit-triangle-avoiding coloring of $E$, such as the half-open strip coloring of L32, then $c(a+bx)=s(a)s(b)$ avoids both sides, since on each triangle one factor is constant and the other takes both signs. Restricting a coloring of the plane to $D_x$ therefore cannot yield a contradiction from these two-scale constraints alone. This is a coloring of the countable carrier, not a counter-coloring of the whole plane.

The analog of (R0) for the carrier $D_x$ with the rotations $K$ is false. Take independent random strip colorings $s_1,s_2$ of $E$, with uniform direction and uniform offset; the law of $c(a+bx)=s_1(a)s_2(b)$ is invariant under translations by $D_x$ and rotations by $K$ and lives on avoiding labelings. Its correlation is $\kappa(a+bx)=\kappa_s(a)\kappa_s(b)$, so its spectral measure is $\sigma_s\otimes\sigma_s$ on the dual of $E\oplus E$, where $\sigma_s$ is the strip coloring's spectral measure, a sum of uniform measures on nonzero-radius circles. Indeed the strip function has zero mean over its offset period; its Fourier series has only nonzero frequencies, and averaging the direction makes each frequency uniform on a circle. In particular $\sigma_s$ has no atoms. The continuous characters of $D_x$ are the pairs with frequencies $(\xi,x\xi)$, a set of $\sigma_s\otimes\sigma_s$-measure zero because $\sigma_s$ has no atoms. Hence the label $f$ at the origin is orthogonal to the continuous vectors and fixed by rotations, while $\langle f,T_1f\rangle=\kappa_s(1)=-1/3$. In the binary transfer over $D_x$ the projection $g$ is the constant $0$ and $Q(g)=1$, so the averaging fails for this carrier: the missing input is false here, not merely unproved.

The spectral criterion just used needs no rigidity premise: a vector is continuous precisely when its spectral measure is supported on the ordinary continuous characters. In one direction this is dominated convergence. In the other, extend its continuous positive-definite coefficient from the dense carrier to $\mathbb C$, apply Bochner's theorem, and use Fourier uniqueness on the discrete carrier's compact dual.

The algebraic closure of Q(x)

Let $F_x$ be the real closure of $\mathbb Q(x)$ in $\mathbb R$, let $E_x=F_x(i)$, the algebraic closure of $\mathbb Q(x)$ in $\mathbb C$, and let $K_x$ be its unit vectors. The rotations $K_x$ include transcendental ones such as $(x+i)/(x-i)$, which mix the two summands of $D_x$, and the Cartesian-square description above is no longer an account of this carrier. In fact $K_x$ generates $E_x$ additively: $u+\bar u$ runs over $F_x\cap[-2,2]$, integer multiples give every real part, and multiplication by $i\in K_x$ gives every imaginary part. The XOR construction supplies no coloring of $E_x$; whether an avoiding coloring extends to that larger carrier is not decided by this observation.

The transfer and Theorem 3 of the spectral account run verbatim over $(E_x,K_x)$ once the analog of (R0) holds there. In the manuscript's proof of Theorem 2.3, the following comparison isolates a number-field obstruction. The graded whole-conclusion review examines the needed field hypotheses through the separate rigidity extension; the comparison is:

• Lemma 2.4 uses that $K_x$ is divisible and that $E_x=aF_x+bF_x$ for nonparallel $a,b$; Lemma 2.5 uses an invariant mean on $F_x/\mathbb Z$; Lemma 2.6 is general; Lemma 2.7 uses small $t\in F_x$ and the decomposition $t=k+k'$ with $k,k'\in K_x$, available because $F_x$ is real closed; Lemmas 3.1 and 3.2 hold for any countable divisible abelian acting group; Lemma 3.3 uses $\sqrt{AB}\in F_x$ and that equal moduli differ by an element of $K_x$; Lemma 3.5 is a statement about any function on $F_x\cap[0,\infty)$ satisfying (3.12).
• Lemma 3.4 and the completion choose a number field containing the real and imaginary parts of the rotations $q_1,\dots,q_J$ of the syndetic cover (3.16), then rotations of that field in rational general position, and clear denominators in (3.17). Over $E_x$ the $q_j$ may be transcendental and no number field contains them. If the cover can be chosen with all $q_j\in K$, the completion goes through verbatim, with $v$ ranging over $K_x$ and Lemma 3.5 applied on $F_x$. Such a cover would follow from a form of (3.13) for averages over translates $\Phi_Nv_N$ of Følner sets of the subgroup $K$, uniform in $v_N\in K_x$. The proof of (3.13) rests on Lemma 3.2, whose decoupling uses ergodicity of the relative product under the acting group, and the restriction of the action to $K$ need not keep it.

The number-field proof alone does not establish rigidity over $(E_x,K_x)$. The finite-cover theorem supplies the replacement selection, and Theorem 2 of the rigidity extension records the resulting reviewed proof.

The number-field step also suggests an alternative sufficient density condition, unused in the finite-cover proof. By (3.13), if $m(r)>0$ then the rotations $u\in K_x$ with $m(2r|\operatorname{Re}u|)\ge m(r)^2/2$ form a set of positive lower Banach density in $K_x$. If every set of rotations of positive Banach density in $K_x$ had real parts of positive upper Banach density modulo $\mathbb Z$ in the additive group $F_x$, some translation invariant mean would give the set where $m\ge m(r)^2/2$ positive weight on an interval, against Lemma 2.5, which holds for every such mean; so $m(r)=0$, and the completion of the proof gives Haar measure. Over the algebraic numbers the number-field step supplies what is needed in place of such a transfer. This real-part density assertion for $K_x$ and $F_x$ is not proved here and is not needed by the candidate argument.

A reduction to algebraic lines

The algebraic theorem (R0) does reduce the transcendental rigidity to one statement. For a character $d$ of the discrete group $E_x$ and $y\in E_x$, let $\pi_y(d)$ be the character $z\mapsto d(yz)$ of $E$, and let $B_y=\{d:\pi_y(d)\text{ is continuous on }E\}$. The restriction map is continuous between the compact dual groups, and the ordinary characters of $E$ form a Borel set by compact exhaustion of their frequency parametrization, so $B_y$ is Borel.

Proposition 2. Let $\nu$ be a finite $K_x$-invariant measure on the dual of $E_x$ giving no mass to its continuous characters, and let $y\ne0$. Then assuming (R0), $\int d(yz)\,d\nu(d)=0$ for every nonzero $z\in E$ if $\nu(B_y)=0$. Consequently, under (R0), $\nu$ is a multiple of Haar measure if and only if $\nu(B_y)=0$ for every $y\ne0$.

Proof. The image of $\nu$ under $\pi_y$ is $K$-invariant. Its part off the continuous characters of $E$ is $K$-invariant and gives them no mass, so by (R0) its Fourier coefficients vanish at nonzero $z$. More explicitly, on $L^2$ of that measure take multiplication by $d(z)$ as translation and the character rotations as $V_k$. The constant vector is rotation fixed and orthogonal to the continuous vectors by the spectral criterion above, so (R0) applies. The remaining part is the image of $\nu$ restricted to $B_y$. If $\nu(B_y)=0$ for all $y$, every coefficient of $\nu$ at a nonzero point $w=w\cdot1$ vanishes, and Fourier uniqueness gives Haar measure. Conversely Haar measure pushes forward to Haar measure, which gives the continuous characters of $E$ no mass. One way to see this last fact is to evaluate at three $\mathbb Q$-linearly independent elements of $E$: Haar measure maps to Haar measure on the three-torus, whereas bounded sets of ordinary frequencies have Lipschitz two-dimensional images there. A countable union over frequency balls still has measure zero. $\square$

For Theorem 3, or for the two-equilateral-scale transfer, at a transcendental ratio only the coefficients at modulus $1$ and $x$ matter. The sufficient zero-mass interface is this precise statement: every finite $K_x$-invariant measure $\nu$ on $\widehat{(E_x)_{\rm disc}}$ with zero mass on its ordinary continuous characters satisfies $\nu(B_1\cup B_x)=0$. Granting (R0), this supplies both required correlations. Theorem 2 of the rigidity extension gives Haar measure and hence this zero-mass conclusion, at author standing. The carrier counterexample does not refute it, because that counterexample has only the rotations $K$. The obstruction on $D_x$ is a law of exactly this excluded kind: its characters are continuous on $E$ and on $Ex$ separately but not jointly.

Two facts constrain a counterexample over $E_x$. The set $V(d)=\{y:d\in B_y\}$ is an $E$-linear subspace with $V(kd)=k^{-1}V(d)$, so its law is invariant under $K_x$. A subspace with a finite $K_x$-orbit is $K_x$-invariant, since the divisible group $K_x$ has no proper subgroup of finite index, and is then a module over the ring generated by $K_x$, which is $E_x$; so the law of $V(d)$ has atoms only at $\{0\}$ and $E_x$. Proposition 9 below shows that an invariant law on the event $V(d)=E_x$ is supported on globally continuous characters. Thus a wild invariant law has no atom there. It also gives the bound $\nu(B_y)\le\nu(D)/2$ for every nonzero $y$, where $D=\widehat{(E_x)_{\rm disc}}$. The required bound is zero; the bound alone leaves a diffuse law on proper, nonzero $E$-subspaces possible. The coefficient $\int d(w)\,d\nu$ is radial on $E_x$ and, granting (R0), on each algebraic line $yE$, coincides away from $0$ with a continuous radial positive-definite function. These two weaker deductions do not themselves prove $\nu(B_y)=0$; the rigidity extension uses the finite-cover theorem instead.

The exact carrier

For every real $x>0$ let $F_x$, $E_x$ and $K_x$ be as above; for algebraic $x$ they are $F$, $E$ and $K$. A coloring of $E_x$ avoids a side when no equilateral triangle of that side with all vertices in $E_x$ is monochromatic.

Proposition 3 (exact carrier). For every real $x>0$, some two-coloring of the plane avoids the equilateral triangles of sides $1$ and $x$ if and only if some two-coloring of $E_x$ does.

Proof. Restriction gives one direction. For the other, product compactness, as in part (b) of the binary transfer, reduces the plane to its finite subsets. Let $p_1,\dots,p_n\in\mathbb C$ and list the equilateral triangles of sides $1$ and $x$ among them, recording for each its side and the sign in $p_c-p_a=\omega^{\pm1}(p_b-p_a)$. These equations, with $|p_b-p_a|^2$ equal to $1$ or $x^2$ as recorded, form in real coordinates a polynomial system with coefficients in $\mathbb Q(\sqrt3,x)$ that has a real solution. The field $F_x$ is real closed and contains $\sqrt3$ and $x$, and the theory of real closed fields is model complete, so the inclusion $F_x\subset\mathbb R$ is elementary and the system has a solution $z_1,\dots,z_n$ with coordinates in $F_x$, that is with $z_k\in E_x$. Each recorded triple $(z_a,z_b,z_c)$ is an equilateral triangle of side $1$ or $x$ in $E_x$. A coloring of $E_x$ avoiding both sides, read at the $z_k$ and assigned to the $p_k$, makes no recorded triangle monochromatic, and the recorded triangles are all the triangles of the finite set. $\square$

So at a transcendental ratio the whole-plane question is exactly the coloring question on the countable carrier $E_x$ with its rotations $K_x$. Proposition 1 shows that the proper subcarrier $D_x$ is too small to decide it.

Proposition 4 (what an avoiding coloring would give). Let $x>0$ with $x\ne1$, and suppose some two-coloring of the plane avoids sides $1$ and $x$. Then there is a nonzero $K_x$-invariant finite measure on the dual of $(E_x)_{\rm disc}$, with no mass on its ordinary continuous characters, whose Fourier coefficient is nonzero at some point of modulus $1$ or $x$.

Proof. Restrict the coloring to $E_x$ and follow the proof of part (a) of the binary transfer with $(E_x,K_x)$ in place of $(E,K)$ and the two equilateral templates of sides $1$ and $x$. The invariant law exists by amenability of $E_x\rtimes K_x$, and the continuous-factor and sampling arguments of that page are stated for any countable dense subgroup. The proof uses (R0) only to equate $\langle f,T_zf\rangle$ with $\langle g,T_zg\rangle$ at the displacements $ur$, $ur\omega$ and $ur(\omega-1)$ of the templates, all of modulus $r\in\{1,x\}$. The residual vector $f-g$ is rotation fixed and orthogonal to the continuous vectors, so its scalar spectral measure is $K_x$-invariant and, by the spectral criterion under Proposition 1, gives no mass to the continuous characters; the cross terms vanish as in Theorem 3 of the spectral account. If its coefficients vanished at every point of modulus $1$ and $x$, the proof would produce a Borel coloring weakly avoiding both sides, contrary to Theorem 4 of the [[research/leads/equilateral_two_scale/circle_exclusion|circle account]]. $\square$

Consequently the analog of (R0) for $(E_x,K_x)$, even restricted to shifts of modulus $1$ and $x$, would settle the ratio $x$; by Proposition 2, granting (R0), it suffices that such measures give no mass to $B_1\cup B_x$. For algebraic $x$ the proposition contradicts (R0) itself, which is the algebraic case recorded in the circle account.

Patterns realizable at transcendental ratios

A pattern is a finite vertex set with a list of ordered triples $(a,b,c)$, each labeled with a side $s\in\{1,t\}$ and a sign. A realization at $t>0$ assigns points $z_a\in\mathbb C$ with $z_c-z_a=\omega^{\pm1}(z_b-z_a)$ and $|z_b-z_a|=s$ for every listed triple. A pattern is non-colorable when every two-coloring of its vertices makes a listed triple monochromatic. Let $S_H\subset(0,\infty)$ be the set of $t$ at which $H$ is realizable.

Proposition 5. (a) Each $S_H$ is a finite union of points and intervals with real algebraic endpoints; so if $S_H$ contains a transcendental $x$, it contains an open interval about $x$. (b) For $x\ne1$, no two-coloring of the plane avoids sides $1$ and $x$ exactly when some non-colorable pattern is realizable at $x$; when $x$ is transcendental such a pattern is then realizable on an open interval about $x$, so the set of ratios at which no avoiding coloring exists contains a neighborhood of each of its transcendental members. (c) If $t_n\to x\ne1$ and at each $t_n$ some non-colorable pattern with at most $N$ vertices is realizable, then one is realizable at $x$.

Proof. (a) $S_H$ is defined by an existential formula of ordered fields with real algebraic coefficients, so by Tarski and Seidenberg it is semialgebraic over the real algebraic numbers; its isolated points and interval endpoints are real algebraic. (b) By product compactness an avoiding coloring exists exactly when every finite set has one. A finite set with no avoiding coloring, with its list of triangles, is a non-colorable pattern realized at $x$. Conversely the points of a realization of a non-colorable pattern form a finite set none of whose colorings avoids both sides, since listed triples are genuine triangles of side $1$ or $x$. The rest follows from (a). (c) There are finitely many patterns with at most $N$ vertices, so one, $H$, is realizable for infinitely many $n$; a non-colorable pattern has a non-colorable connected component, so $H$ may be taken connected. Translate realizations so that one vertex is at $0$. Connectedness bounds every point by $N\max(1,t_n)$, so a subsequence converges, and the limit realizes $H$ at $x$. $\square$

Part (c) gives a second sufficient condition for a transcendental ratio: uniformly bounded witness sizes at nearby algebraic ratios. Granting (R0), witnesses exist at every algebraic ratio by the circle account, but the route through (R0) is not effective and gives no size bound, and L32 shows that sizes must grow as the ratio tends to $1$. The candidate full proof via the rigidity extension supplies a finite witness at every $x\ne1$; part (a) then gives a common witness on a neighborhood of each fixed transcendental $x$. This is an existence deduction, with no effective size bound supplied. The bounded-witness route is not used to prove the rigidity extension.

A combinatorial form of the number-field step

The proof of Theorem 2.3 in the release uses number fields only in Lemma 3.4 and the completion. The completion needs, for the syndetic cover $q_1,\dots,q_J$ of (3.16), finitely many finite sets of rotations, each with $1$ and $i$ in its rational span, one of which lies in $A_0v^{-1}$ for every $v$. The following property supplies such sets directly.

Template property. A finite set $Q=\{q_1,\dots,q_J\}\subset K_x$ has the template property when there is a finite $\Sigma\subset K_x$ such that, for every map $j:\Sigma\to\{1,\dots,J\}$, both $1$ and $i$ lie in the $\mathbb Q$-linear span of $\{q_{j(s)}s:s\in\Sigma\}$.

Proposition 6. Suppose every finite subset of $K_x$ has the template property, and that Lemmas 2.4 to 2.7, 3.1 to 3.3 and 3.5 of the release hold over $(E_x,K_x)$, as the inspection under the algebraic closure of $\mathbb Q(x)$ above indicates. Then every $K_x$-invariant probability measure on the dual of $(E_x)_{\rm disc}$ giving no mass to its continuous characters is Haar measure.

Proof. Follow the completion of the release with the terminal factor and radial function $m$ on $F_x$. If $m(r)>0$, the set $A_0$ of the release is syndetic in $K_x$ with some cover $q_1,\dots,q_J$, by the release's argument from (3.13) along every Følner sequence. Take $\Sigma$ for this cover. For $v\in K_x$ and $s\in\Sigma$ choose $j(s)$ with $q_{j(s)}sv\in A_0$. The set $T_v=\{q_{j(s)}s\}$ is one of at most $J^{|\Sigma|}$ sets, each with $1$ and $i$ in its rational span. Clearing denominators gives one integer $D_0\ge1$ and one bound $M_0$ with $D_0\cdot1=\sum_{k\in T}a_kk$, $a_k\in\mathbb Z$, $\sum|a_k|\le M_0$, for every possible $T$; since the real span of each $T$ contains $1$ and $i$, (3.18) holds with a common $c_0>0$. The rest of the completion runs verbatim: Lemma 3.5 on $F_x$ gives $w(2rD_0|\operatorname{Re}v|)\le4M_0H$ when $|\operatorname{Re}v|<c_0/D_0$; every $t\in F_x\cap[-1,1]$ is a real part of an element of $K_x$, because $F_x$ is real closed; so $m$ is bounded below on $F_x\cap(0,2rc_0)$, contradicting Lemma 2.5 over $F_x$. Hence $m$ vanishes off $0$, the measure is Haar in the ergodic case, and the ergodic decomposition gives the general case. $\square$

Theorem 1 of the finite-cover account supplies this property for every finite subset of $K_x$. The rigidity extension checks the source-lemma hypotheses and composes this interface with Proposition 4 and the circle account. These consumed interfaces are covered by the whole-conclusion grade. The property is an algebraic statement about unit vectors of $E_x$, with no measure theory. It holds trivially when all the $q_j$ lie in one coset $q_1K$: take $\Sigma=q_1^{-1}S$ with $S$ from Lemma 3.4 in a number field containing the algebraic ratios $q_j/q_1$. In general it can be reduced to algebraic coefficients. If a finite $\Pi\subset K_x$ has, for every map $l:\Pi\to\{1,\dots,J\}$, both $1$ and $i$ in the $E$-linear span of $\{q_{l(p)}p\}$, choose a conjugation-stable number field $N=N_0(i)$ containing all the finitely many coefficients, take $S\subset K\cap N$ from Lemma 3.4 with $|S|>J([N:\mathbb Q]-1)$, and put $\Sigma=\{ps:p\in\Pi,s\in S\}$. For any $j:\Sigma\to\{1,\dots,J\}$, pigeonhole within each block $pS$ gives a value $l(p)$ taken on a rational basis $q_{l(p)}pS'$ of $q_{l(p)}pN$, so the rational span contains $\sum_pq_{l(p)}pN$, which contains $1$ and $i$.

The property is invariant under $Q\mapsto gQ$, with $\Sigma\mapsto g^{-1}\Sigma$, so a two-element set may be taken as $\{1,q\}$.

Proposition 7. Every two-element subset of $K_x$ has the template property.

Proof. Take $Q=\{1,q\}$. If $q\in K$ the one-coset case applies. Otherwise $q$ is transcendental, since an algebraic unit vector lies in $K$, and $E(q)$ is a rational function field; write $T$ for $q$ in it. Choose $b_1,b_2\in E$ with $0<|b_k|<1$ and $b_1/\bar b_1\ne b_2/\bar b_2$, put $a_k=1/\bar b_k$ and $B_k=(T-b_k)/(1-\bar b_kT)=-a_k(T-b_k)/(T-a_k)$, a unit vector because $|q|=1$. Let

$$\Pi=\{1,\;T^{-1},\;B_1,\;B_2,\;T^{-1}B_1,\;T^{-1}B_2,\;B_1B_2\}.$$

We show that for every choice of $p$ or $Tp$ for each $p\in\Pi$, the $E$-span of the choices contains $1$. Otherwise some $E$-linear $\lambda$ on the span $W$ of $1,T,T^{-1},(T-a_1)^{-1},(T-a_2)^{-1}$, which contains $\Pi$ and $T\Pi$, has $\lambda(1)=1$ and kills every choice. The choices at $1$ and $T^{-1}$ force $\lambda(T)=\lambda(T^{-1})=0$. If $\lambda(B_k)\ne0$, the choices at $B_k$ and $T^{-1}B_k$ force $\lambda(TB_k)=\lambda(T^{-1}B_k)=0$; but

$$TB_k-a_k^2T^{-1}B_k=-a_k\bigl(T+(a_k-b_k)-a_kb_kT^{-1}\bigr),$$

so $\lambda(1)=0$, as $a_k\ne b_k$. Hence $\lambda(B_1)=\lambda(B_2)=0$. Write $B_1B_2=a_1a_2+\sum_kR_k/(T-a_k)$ and $B_k=-a_k+r_k/(T-a_k)$, with $r_k=a_k(b_k-a_k)$ and $R_k=r_kB_{k'}(a_k)$, where $k'$ is the other index. Cancelling the poles,

$$B_1B_2-\sum_k\frac{R_k}{r_k}B_k =\frac{a_1a_2(b_2-b_1)}{a_1-a_2},\qquad TB_1B_2-a_1a_2T-\sum_k\frac{a_kR_k}{r_k}B_k =\frac{a_1a_2(a_1b_1-a_2b_2)}{a_2-a_1},$$

using $r_k+a_k^2=a_kb_k$ and $B_{k'}(a_k)=a_{k'}(a_k-b_{k'})/(a_{k'}-a_k)$. Both constants are nonzero, since $b_1\ne b_2$ and $a_kb_k=b_k/\bar b_k$. So either choice at $B_1B_2$ forces $\lambda(1)=0$, a contradiction. The $E$-span of every choice therefore contains $1$ and $i$, and the reduction to algebraic coefficients gives the template property. $\square$

The proof uses elements with a pole at the zero of $q$, which a template needs: if every element of $\Pi$ were regular there, the evaluation at that point would kill every product $q\,p$ while taking the value $1$ at $1$. Proposition 7 settles the completion only when the syndetic cover has two elements, which the release's argument does not provide. The general property has an equivalent form without covers.

Proposition 8. The following are equivalent: (a) every finite subset of $K_x$ has the template property; (b) every syndetic subset of $K_x$ spans $E_x$ over $E$; (c) for no $E$-linear functional $\lambda$ on $E_x$ with $\lambda(1)=1$ is the set $\{u\in K_x:\lambda(u)=0\}$ syndetic in $K_x$.

Proof. (a) implies (b): let $K_x=\bigcup_jq_j^{-1}C$ and take $\Sigma$ for these $q_j$. For $v\in K_x$ choose $j(s)$ with $q_{j(s)}sv\in C$; from a rational relation $1=\sum_sa_sq_{j(s)}s$ we get $v\in\operatorname{span}C$, and $K_x$ spans $E_x$. (b) implies (c), since that set spans a subspace of $\ker\lambda$, which misses $1$. (c) implies (a): if a finite $Q=\{q_1,\dots,q_J\}$ fails the template property, it fails the algebraic-coefficient form, so for every finite $\Pi$ some choice $p\mapsto q_{j(p)}p$ has $1$ outside its $E$-span. Enumerate $K_x$; the choice sequences on initial segments whose linear systems $\lambda(1)=1$, $\lambda(q_{j(p)}p)=0$ are consistent form an infinite, finitely branching tree, and König's lemma gives an infinite branch. A linear system all of whose finite subsystems are consistent is consistent, so some $\lambda$ on $E_x$ has $\lambda(1)=1$ and, for every $p\in K_x$, some $j$ with $\lambda(q_jp)=0$; then $\{u\in K_x:\lambda(u)=0\}$ is syndetic with cover $Q$. $\square$

Theorem 1 of the finite-cover account proves condition (c), and hence all three conditions. Proposition 7 is an explicit two-element instance. The direct attempt to iterate its fresh-pole eliminations requires additional spanning identities when the cover has three or more elements; no such identities are assumed here. The general proof instead uses Cauchy-transform branches in a finite function field and a finite-dimensional quotient.

Two constraints on any proof are known. The device of Lemma 3.4 itself cannot be transplanted: a finite-dimensional $E$-subspace $W$ of $E_x$ contains unit vectors on only finitely many $E$-lines, because on a smooth projective model of a finitely generated conjugation-stable field containing $W$ a unit vector $z$ has $z\bar z=1$, so its divisor is anti-invariant and lies between the pole bound of $W$ and its conjugate, which allows finitely many divisors, each determining $z$ up to a constant. And the real closure is essential: on the rational carrier $E(x)$ with its unit vectors, the field embedding fixing $E$ and sending $x$ to another real transcendental $x_0$ keeps every unit vector unimodular, so the characters $z\mapsto\exp(i\operatorname{Re} (\bar\zeta\sigma(z)))$, with $\zeta$ radial, form an invariant law that gives no mass to the continuous characters and has nonzero coefficient at $1$. On $E_x$ no such embedding exists, since an embedding keeping $K_x$ unimodular commutes with conjugation, preserves the order of $F_x$, and is therefore the identity on $F_x$.

Remark (geometric means). Over $(E_x,K_x)$ the radial function of the completion also satisfies $m(A)m(B)\le m(A^\theta B^{1-\theta})$ for $A,B\in F_x^{\ge0}$ and rational $\theta\in[0,1]$, which a replacement for Lemma 3.4 may use. For $\theta=k/n$, factor $a+bu^n$ with $|a|=A$, $|b|=B$ over its roots and replace $n-k$ of the factors $u-\zeta$ by $1-\bar\zeta u$, which has the same modulus on the unit circle; the resulting polynomial has the same modulus as $a+bu^n$ at every $u\in K_x$ and outer coefficient of modulus $A^\theta B^{1-\theta}$, up to the choice of which factors are flipped. The average over $u$ of $m(|a+bu^n|)$ tends to $m(A)m(B)$, by the pair criterion (3.3) with exponent $n$, while by (3.2) the average for the flipped polynomial is at most $m$ at the modulus of any one coefficient. For $n=2$ this is the inequality $m(A)m(B)\le m(\sqrt{AB})$ implicit in the proof of Lemma 3.3.

Invariant laws continuous on every algebraic line

Write $D=\widehat{(E_x)_{\rm disc}}$, let $C\subset D$ be its ordinary continuous characters, and retain $B_y$ and $V(d)$ from Proposition 2. The set $C$ is Borel by the compact-exhaustion argument used there. Since $E_x$ is countable, the event $C_{\rm all}=\bigcap_{y\in E_x}B_y=\{d:V(d)=E_x\}$ is Borel too. It is invariant under $K_x$, and $C\subset C_{\rm all}$.

Proposition 9. Let $\nu$ be a finite $K_x$-invariant Borel measure on $D$. Then

$$\nu(C_{\rm all}\setminus C)=0, \qquad 2\nu(B_y)\le\nu(D)+\nu(C)\quad(y\ne0).$$

In particular, if $\nu(C)=0$, then $\nu(B_y)\le\nu(D)/2$ for every nonzero $y$. These conclusions do not use (R0).

Proof of the first assertion. Normalize the restriction of $\nu$ to $C_{\rm all}$, if it is nonzero, to a probability $\rho$. For every $d\in C_{\rm all}$ there is a unique $E$-linear map $\Lambda_d:E_x\to\mathbb C$ such that

$$d(y)=\exp(i\operatorname{Re}\Lambda_d(y)).$$

Indeed, continuity on $yE$ determines the unique real frequency $\ell_d(y)$ with $d(ty)=\exp(it\ell_d(y))$ for $t\in\mathbb Q$. Uniqueness on $\mathbb Q$ proves additivity of $\ell_d$, and continuity on $yE$ proves its linearity over the real algebraic numbers. Consequently $\Lambda_d(y)=\ell_d(y)-i\ell_d(iy)$ is $E$-linear. The converse formula for $d$ follows by taking $t=1$. Each coordinate $\Lambda_d(y)$ is Borel: restriction to $yE$ is Borel and the inverse of its ordinary-frequency parametrization is Borel, by compact exhaustion. Invariance of $\rho$ therefore gives

$$\Lambda_d(u)\ \stackrel{\rm law}=\ \Lambda_d(1) \qquad(u\in K_x).$$

For a positive rational $q$ and $z\in E_x$ with $|z|\le2q$, real closedness gives $z=q(u+v)$ for some $u,v\in K_x$. For $z\ne0$ one may take $e=z/|z|$, $a=|z|/(2q)$ and $u=e(a+i\sqrt{1-a^2})$, $v=e(a-i\sqrt{1-a^2})$; for $z=0$ take $u=1$, $v=-1$. Thus, for every $\varepsilon>0$,

$$\rho\{|\Lambda_d(z)|>\varepsilon\} \le2\rho\{|\Lambda_d(1)|>\varepsilon/(2q)\}.$$

The right side tends to zero as $q\downarrow0$, uniformly for $|z|\le2q$. Hence the Fourier coefficient $\phi(z)=\int d(z)\,d\rho(d)$ is continuous at zero in the usual topology on $E_x$: use $|e^{it}-1|\le\min(2,|t|)$ and the displayed bound. Positive definiteness makes $\phi$ uniformly continuous, so it extends to a continuous positive-definite function on $\mathbb C$. Bochner's theorem and Fourier uniqueness on $D$ identify $\rho$ with the pushforward of its ordinary frequency measure. It is therefore supported on $C$, proving the first assertion.

A subspace observation. For any probability law on $E$-linear subspaces $V$ of the countable space $E_x$, if $\mathbb P(z\in V)\ge\delta$ for every $z\in E_x$, then $\mathbb P(V=E_x)\ge\delta$. To prove this, fix $z_1,\ldots,z_n$ and choose $N$ distinct $t\in E$. Unless all $z_j$ belong to $V$, the vector polynomial $z_1+t z_2+\cdots+t^{n-1}z_n$ in $E_x/V$ has at most $n-1$ zeros. For example, apply an $E$-linear functional nonzero on one of its coefficients. Averaging the membership indicators gives

$$\delta\le \mathbb P(z_1,\ldots,z_n\in V)+\frac{n-1}{N}.$$

Let $N\to\infty$ and then exhaust $E_x$ by the finite lists.

Proof of the bound. Normalize $\nu$ to have mass one, and write $\alpha=\nu(B_1)=\mathbb P(1\in V(d))$. Every unit belongs to $V(d)$ with probability $\alpha$. The same two-unit decomposition, with $q$ chosen large enough for any prescribed $z$, and linearity of $V(d)$ give $\mathbb P(z\in V(d))\ge2\alpha-1$. The subspace observation and the first assertion now give $\nu(C)\ge\nu(C_{\rm all})\ge2\alpha-1$; when $\alpha\le1/2$ the desired bound is immediate. For general $y\ne0$, push $\nu$ forward by $d\mapsto(z\mapsto d(yz))$. This commutes with rotations, preserves the mass on $C$, and sends $B_y$ to $B_1$. Finally restore the total mass. $\square$

There is a quantitative consequence for the proposed binary transfer. In an invariant law on two-colorings avoiding unit equilateral triangles, let $f\in\{\pm1\}$ be the color at the origin and $\nu_f$ its translation spectral measure on $D$. It has mass one, is $K_x$-invariant, and has coefficient $-1/3$ at distance one. Assuming (R0), its part off $B_1$ has zero coefficient there. The part on $B_1$, restricted to $E$, is an ordinary radial frequency measure. The bound $J_0(t)>-403/1000$ from Lemma 1 of the spectral account therefore implies

$$\nu_f(B_1)\ge\frac{1000}{1209}, \qquad \nu_f(C)\ge\frac{791}{1209}.$$

Thus, under (R0), the squared norm of the projection of $f$ onto globally continuous vectors is at least $791/1209$. A positive continuous part does not preserve a zero monochromatic-triangle count: the residual coefficients at distances $1$ and $x$ still need to vanish for the quadratic transfer. Proposition 9 supplies neither that vanishing nor a bound tending to zero. The zero-mass assertion after Proposition 2 is supplied by Theorem 2 of the rigidity extension, through the stronger finite-cover argument; it does not follow just from this quantitative bound.