% Hexagon source adaptation of the published Zenodo edition. % Changes are limited to packaging, bibliography rendering, navigation and archive links. % Mathematical formulas and their order are preserved; see the local integrity record. \PassOptionsToPackage{expansion=false,nopatch=footnote}{microtype} \documentclass[11pt,a4paper]{article} \usepackage[T1]{fontenc} \usepackage[utf8]{inputenc} \usepackage[english]{babel} \usepackage{lmodern,microtype} \usepackage[a4paper,margin=26mm,headheight=14pt]{geometry} \usepackage{amsmath,amssymb,amsthm,mathtools,mathrsfs} \usepackage{booktabs,tabularx,array,needspace} \usepackage{enumitem} \usepackage{fancyhdr} \usepackage[hidelinks,unicode]{hyperref} \usepackage{bookmark} \bookmarksetup{depth=2,open,openlevel=1} \hypersetup{pdftitle={Selective normality and algebraic rigidity of multiplicative traces},pdfauthor={Brice Pouly},pdfsubject={Selective completion of completely multiplicative sign functions}} \pagestyle{fancy}\fancyhf{} \fancyhead[L]{\small Selective normality and algebraic rigidity} \fancyhead[R]{\small Brice Pouly}\fancyfoot[C]{\thepage} \renewcommand{\headrulewidth}{0.3pt} \setlength{\parskip}{3pt plus 1pt minus 1pt} \setlength{\emergencystretch}{2em} \setcounter{tocdepth}{1} \hypersetup{bookmarksdepth=2} \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{definition}\newtheorem{definition}[theorem]{Definition} \theoremstyle{remark}\newtheorem{remark}[theorem]{Remark} \newcommand{\N}{\mathbb N}\newcommand{\Z}{\mathbb Z}\newcommand{\Q}{\mathbb Q}\newcommand{\F}{\mathbb F} \newcommand{\E}{\mathbb E}\newcommand{\Prob}{\mathbb P}\newcommand{\M}{\mathcal M_{\pm}} \newcommand{\D}{\mathcal D}\newcommand{\Pd}{\mathcal P_{d_0}}\newcommand{\one}{\mathbf 1} \newcommand{\TV}{d_{\mathrm{TV}}}\newcommand{\Law}{\mathcal L} \DeclareMathOperator{\Gal}{Gal}\DeclareMathOperator{\Fix}{Fix}\DeclareMathOperator{\res}{res} \DeclareMathOperator{\im}{im}\DeclareMathOperator{\bl}{bl}\DeclareMathOperator{\osc}{osc} \DeclareMathOperator{\Norm}{N}\DeclareMathOperator{\rank}{rank} \newcommand{\abs}[1]{\lvert#1\rvert}\newcommand{\bigabs}[1]{\left|#1\right|} \title{Selective normality and algebraic rigidity\\of multiplicative traces} \author{Brice Pouly} \date{\small Research and writing assisted by GPT-6 Astra} \begin{document} \maketitle \begin{abstract} A completely multiplicative sign function may be changed so that two quadratic traces remain identical at every index while a third becomes normal. We classify this freedom for the traces $n^2+1$, $n^2+2$ and $n^2-2$. After two traces and a finite prefix have been protected, a monic irreducible polynomial can be made normal from every prescribed source exactly when the field generated by one of its roots contains neither protected quadratic field. In the complementary case, every value of the polynomial is already fixed. A single decidable set of primes of relative density zero supports a continuous, relatively computable completion treating all admissible polynomials jointly, with uniform bounds for each fixed observation family. The proof combines weighted root counts with a sieve for moments whose order increases with the observation scale, followed by a selection using exact finite normalisers. We also give a sufficient condition on prime availability, explicit sparser reservoirs, and a divergent reservoir on which some sources have no completion with a convergent ordinary mean, despite convergence of all fixed word laws in probability. For larger protected families, the obstruction is governed by odd products of quadratic square classes. An explicit example distinguishes certificates for every numerical value from a rational norm identity in the variable. \end{abstract} \begingroup \small\setlength{\parskip}{0pt} \tableofcontents \endgroup % Original source: sections/01_introduction.tex \section{Which observations can change after two norms are protected?} \label{sec:introduction} Suppose that every value of a completely multiplicative sign function is prescribed on the two families \[ a^2+2b^2,\qquad |a^2-2b^2|, \] where only non-zero values are considered. These are infinitely many constraints, and they include the traces $n^2+2$ and $n^2-2$. They do not, however, determine the statistics of the remaining trace $n^2+1$. Starting from any source, one can preserve both families exactly and arrange that every sign word of length $H$ on the remaining trace has frequency $2^{-H}$, for every fixed $H$. There is a simple reason why changes are possible. A prime $q\equiv5\pmod8$ is inert in both $\Q(\sqrt2)$ and $\Q(\sqrt{-2})$. Its valuation in either protected norm is even. Changing the sign at $q$ is therefore invisible to both norms. The same prime has two simple roots on $n^2+1$, so it can act on that trace. The difficulty is not to find one invisible change. It is to distribute enough such changes across all large observation scales, without treating the values of a multiplicative function as independent. The algebraic boundary is already visible in two quartics. The identity \begin{equation}\label{eq:intro-quartic} n^4+1=(n^2+1)^2-2n^2 \end{equation} forces every value on $n^4+1$ to be preserved. By contrast, the trace of \[ \Phi_5(n)=n^4+n^3+n^2+n+1 \] can be made normal by the same type of completion. Both polynomials are irreducible over $\Q$. The distinction is not irreducibility alone, nor degree, but the quadratic subfields of the field generated by a root. \subsection{Sources, traces and empirical laws} Write $\M$ for the completely multiplicative functions from $\N_{\ge1}$ to $\{-1,1\}$. Such a function takes the value one on squares and extends to positive rationals by $f(a/b)=f(a)f(b)$. Throughout, \[ \D=\{-1,-2,2\},\qquad Q_d(T)=T^2-d. \] We choose a target $d_0\in\D$ and protect the other two traces. A \emph{compatible completion} of a source $f_0\in\M$, with prefix $B\ge1$, means a function $g\in\M$ satisfying \begin{align} g(m)&=f_0(m) &&(1\le m\le B),\label{eq:prefix}\\ g(Q_e(n))&=f_0(Q_e(n)) &&(e\in\D\setminus\{d_0\},\ n\in\Z,\ Q_e(n)>0). \label{eq:protected-traces} \end{align} The source is already defined everywhere. Completion refers to realising its protected data in a class with prescribed statistics, not to filling in undefined source values. For a finite-valued sequence $X(n)$ and a fixed starting point $n_0$, set \[ \Law_N(X)=\frac1N\sum_{n=n_0+1}^{n_0+N}\delta_{X(n)}. \] We choose $n_0$ large enough that all polynomial arguments used in that law are positive. It is fixed throughout each limit. For the three quadratic traces we use $n_0=1$. A sign sequence is \emph{normal} if its words of every fixed length $H$ have limiting law $U_H$, the uniform measure on $\{-1,1\}^H$. Total variation has the convention \[ \TV(\mu,\nu)=\sup_A|\mu(A)-\nu(A)|=\frac12\sum_x|\mu\{x\}-\nu\{x\}| \] on a finite space. All logarithms without a subscript are natural. Here is the representative selective completion statement. The more general polynomial conclusion below will use a different, common reservoir. \begin{theorem}[Two norms intact, one trace normal]\label{thm:quadratic-intro} Fix $d_0\in\D$ and $B\ge1$. There is a decidable set $E$ of primes, chosen before the source, with \[ \frac{\#(E\cap[1,x])}{\pi(x)}\longrightarrow0, \] and a deterministic map $f_0\mapsto g=f_0\omega_{f_0}$ such that $\omega_{f_0}(q)=1$ outside $E$, \eqref{eq:prefix} holds, and \begin{equation}\label{eq:all-norms} g(|a^2-eb^2|)=f_0(|a^2-eb^2|) \quad(e\ne d_0,\ a,b\in\Q,\ a^2-eb^2\ne0). \end{equation} The map is continuous in the product topology of prime signs and computable relative to $f_0$. Let \[ X_g^H(n)=(g(Q_{d_0}(n+a)))_{0\le a0$ depending on the fixed lists and on $d_0,B$, but not on $f_0$, \begin{equation}\label{eq:universal-rate} \sup_{f_0\in\M}\TV\bigl(\Law_N(X_g,Z_{f_0}), U_k\otimes\Law_N(Z_{f_0})\bigr) \le C\exp(-c\sqrt{\log\log N}) \end{equation} for all sufficiently large $N$. The same output works for every such list. It preserves the norms \eqref{eq:all-norms} and the prefix. \end{theorem} The coefficients, degrees and number of observations in \eqref{eq:universal-rate} are fixed before $N$ tends to infinity. Joint words are covered when the translated polynomials actually read are distinct. Algebraic repetitions are not independent coordinates: Section~\ref{sec:universal} describes their exact finite-dimensional constraint. Monicity is part of the classification, not a dispensable convention for arbitrary leading coefficients; the example in Section~\ref{subsec:monicity} shows the constant phase that can otherwise remain free. An empty mark list means that $Z_{f_0}$ takes its value in a singleton. \subsection{Why the reservoir is part of the theorem} The set of allowed primes is an algebraic object. A sufficiently well-placed subset can supply the statistical freedom, but infinite total harmonic mass alone does not suffice. For a set $E$ contained in the exclusive prime class, the condition \begin{equation}\label{eq:availability-intro} \sum_{\substack{q\in E\\ \log\log x{\raggedright\arraybackslash}p{.29\linewidth}>{\raggedright\arraybackslash}X>{\raggedright\arraybackslash}X@{}} \toprule Result and fixed data & Dependence and conclusion & Uniformity and effectivity\\ \midrule Proposition~\ref{prop:as}: the reservoir and a countable family of weighted tests are fixed before the fair signs. & Almost-sure convergence at all large prefixes. The full-measure event may depend on the source and tests. & The moment estimates are uniform in the fixed bounded weights. This argument alone does not select computable signs.\\[5pt] Theorems~\ref{thm:quadratic-intro}--\ref{thm:classification} and Proposition~\ref{prop:selector}: the support and task templates precede the input. & One algorithm maps each $f_0$ to its own $\omega_{f_0}$. Polynomial mark values are read from that input. & For each fixed task, the finite bound is independent of $f_0$. Selection is relatively computable; no useful numerical onset is supplied.\\[5pt] Proposition~\ref{prop:probability-only}: a divergent reservoir, a source and a fixed word test precede the signs. & Convergence in probability and every finite moment, not a conclusion about every sample path. & The supremum over sources is outside the expectation. No convergent realisation is selected by this assertion.\\[5pt] Theorem~\ref{thm:bad-reservoir}: $E$ precedes $f_\star$; the base $f$ is then constructed. & All subsets $A\subseteq E$ and finite prime changes are covered after $f$ is fixed. Their witness sizes can differ. & The calendar is source-independent, and $f$ is relatively computable from $f_\star$. No practical complexity bound is asserted.\\ \bottomrule \end{tabularx} \caption{Quantifiers for completion and obstruction. A computable finite error bound and a proof that it tends to zero are distinct from a supplied usable modulus of convergence.} \label{tab:quantifiers} \end{table} \subsection{Relation to previous work} The precursor \cite{PoulyNoise}, Theorem~1.1 and Proposition~8.1, preserves polynomial traces while regularising affine observations by selective prime noise, starting from a prescribed source. Its Corollary~5.3 already protects the whole family of sums of two squares. Its Theorems~6.1--6.2 describe a polynomial factor channel under non-selective noise. The present question is which polynomial observations can be regularised \emph{inside the exact fibre of two protected norms}. We identify that fibre, classify its frozen irreducibles, and construct a common sparse support with a relatively computable, source-dependent selection. These are static conclusions; they do not replace the precursor's simultaneous guarantees at all positive times. The factor-parity mechanism is recalled locally in Section~\ref{subsec:product-channel}. \paragraph{Normality and three random models.} Fish~\cite{Fish} proves normality for independent fair prime signs extended completely multiplicatively. Najnudel~\cite{Najnudel}, Proposition~6.3 in the cited arXiv edition, obtains, for each fixed word of length $k$ in $q$-th roots of unity, almost-sure frequency $q^{-k}+O(N^{-1/2+\varepsilon})$ for every $\varepsilon>0$. The case $q=2$ is the completely multiplicative sign model. These full-prime models do not impose our exact protection or our common sparse support, so the rates cannot be compared without their different constraints. For polynomial observations, it is important to distinguish that sign model, sometimes called \emph{extended Rademacher}, from two other models. A \emph{Steinhaus} function has independent uniform unit-circle values at the primes and is extended completely multiplicatively. A \emph{squarefree-supported Rademacher} function has independent fair prime signs, is extended multiplicatively on squarefree integers, and vanishes on the others. Our multipliers are completely multiplicative signs, random only on a specified prime set before selection; they are not squarefree-supported functions. Klurman--Shkredov--Xu~\cite{KSX}, Theorem~1.1 and Proposition~1.4, connect polynomial sums in the Steinhaus model with Gaussian limits and multiplicative product equations. Wang--Xu~\cite{WangXu}, Theorem~1.1 and Section~1.3, obtain asymptotics for all fixed even moments, for polynomials with at least two distinct complex roots, through equalities of products of polynomial values. Their error constants depend on the fixed moment order. Chinis--Shala~\cite{ChinisShala}, Theorem~1.2, prove a central limit theorem in the squarefree-supported Rademacher model for irreducible quadratics and products of at least two distinct linear factors, under the condition that no prime square divides every polynomial value. Their Section~1.4 separately discusses the completely multiplicative model and the corresponding square-product counts. The squarefree-supported theorem must not be read as a theorem for that different model. For the latter model, de la Bret\`eche--Wang--Xu~\cite{DBWX}, Theorem~1.2, prove a Gaussian limit for $N^{-1/2}\sum_{n\le N}g(P(n))$ when $P$ is separable, has positive leading coefficient and total degree at least two, and every irreducible factor has degree at most three. Our conclusion concerns $N$-normalised averages and fixed word frequencies in a protected fibre. Its allowance of admissible irreducibles of arbitrary fixed degree is not an extension of that central limit theorem. Section~\ref{sec:moments} explains the different parity count and precision required here. \paragraph{Sparse completion and the inherited tools.} Klurman--Mangerel--Ter\"av\"ainen~\cite{KMT}, Theorem~2.1 in arXiv:2304.05344v2, give vanishing affine correlations for a completely multiplicative function negative on a prime-density-zero set of infinite harmonic mass. Thus sparsity and affine normality alone are not new conclusions. Nor does multiplication by such a fixed function give a completion for every prescribed source: if the source equals the multiplier, their product is identically one. Our quantifier is a source-dependent choice, with a uniform error estimate for the selector, rather than a single multiplier valid for all sources chosen afterwards. The algebraic inputs are classical norm theory and the Frobenius interpretation of polynomial roots; see \cite{Milne,NIST,BerendBilu,EKT}. The harmonic input is the fixed-extension Chebotarev--Mertens theorem of Arango-Pi\~neros--Keliher--Keyes~\cite{AKK}, Theorem~A. We use Selberg's upper-bound sieve in the form recorded in \cite{Kedlaya}, Section~13.2, and keep the dependence on the increasing moment order explicit. The selection uses the method of conditional expectations, with exact finite normalisers and a paid tail after every decision. Constructive Borel--Cantelli methods have a general antecedent in Galatolo--Hoyrup--Rojas~\cite{GHR}; the selection needed here is proved directly. The private-prime modification used in Section~\ref{sec:obstruction} has a direct antecedent in Klurman~\cite{Klurman}, Lemma~2.1 and the proof of Proposition~1.1. Its role here is within a calendar that treats all exposed sign configurations, producing persistent witnesses for every subset of one reservoir. Products of norm groups also have a substantial literature: Pollio--Rapinchuk~\cite{PollioRapinchuk} provide local-global context, not a theorem contradicted by our comparison of numerical specialisations with identities over $\Q(T)$. The contribution is the connection between the exact protected fibre, the weighted supply of roots, and a common completion from arbitrary sources, together with its reservoir and collective-rigidity boundaries. None of the conclusions is a new cancellation statement for an unmodified Liouville function. % Original source: sections/02_fibres.tex \section{The exact fibre and the field of a root}\label{sec:fibres} We first determine every prime sign that can vary under the protected data. This prevents the probabilistic construction from silently changing the fibre it is meant to occupy. \subsection{Invisible primes determine the whole fibre} Quadratic reciprocity gives the following exclusive classes: \begin{center} \begin{tabular}{@{}cll@{}} \toprule Target $d_0$ & Free primes $\Pd$ & Protected absolute norms\\ \midrule $-1$ & $q\equiv5\pmod8$ & $a^2+2b^2$, $|a^2-2b^2|$\\ $-2$ & $q\equiv3\pmod8$ & $a^2+b^2$, $|a^2-2b^2|$\\ $2$ & $q\equiv7\pmod8$ & $a^2+b^2$, $a^2+2b^2$\\ \bottomrule \end{tabular} \end{center} Every prime in the indicated class has two simple roots on the target quadratic and none on either protected quadratic. \begin{lemma}[Exact fibre]\label{lem:fibre} For $\omega\in\M$, the following are equivalent: $\omega$ equals one on every positive value of the two protected traces; and $\omega(q)=1$ at every prime outside $\Pd$. Each such $\omega$ also equals one on every non-zero absolute rational norm $|a^2-eb^2|$ with $e\ne d_0$. Imposing the prefix \eqref{eq:prefix} removes exactly the free primes at most $B$. \end{lemma} \begin{proof} If $q\in\Pd$ and $e\ne d_0$, then $e$ is not a square modulo $q$. For an integer pair of non-zero norm, \begin{equation}\label{eq:inert-valuation} v_q(a^2-eb^2)=2\min\{v_q(a),v_q(b)\}, \end{equation} with $v_q(0)=+\infty$. After dividing the coordinates by their common power of $q$, further divisibility would give a square root of $e$ modulo $q$. The valuation is therefore even. Clearing rational denominators multiplies the norm by a square, proving the sufficiency and the stronger norm assertion. For necessity, the protected traces determine the sign at two: use $Q_{-1}(1)=2$ when $-1$ is protected, and otherwise $Q_2(2)=2$. If three is not exclusive, $Q_{-2}(1)=3$ is protected. Suppose inductively that all smaller non-exclusive prime signs have been determined, and let $p\ge5$ be non-exclusive. At least one protected quadratic has a centred root $1\le r\le(p-1)/2$. For that $e$, \[ 04rD$, let $\rho_q=\rho_P(q)$. The proportion of tuples modulo $q^2$ having total valuation exactly one is \begin{equation}\label{eq:local-density} g_r(q)=\frac{r\rho_q}{q}\left(1-\frac1q\right) \left(1-\frac{\rho_q}{q}\right)^{r-1}. \end{equation} If $\rho_q>0$, then \begin{equation}\label{eq:local-bounds} 04rD}n^{-2}\le rD$, we obtain \begin{equation}\label{eq:G-mass} \log G_r(z)\ge-\log2+r\Delta_{P,E}(T,r)-rD. \end{equation} Take the $r$-th root in \eqref{eq:finite-sieve}, use $z^8\le T^{1/2}$, and use $(a+b)^{1/r}\le a^{1/r}+b^{1/r}$. This proves \eqref{eq:growing-bound}. \end{proof} This estimate explains both cutoffs. Primes below a multiple of $r$ cannot be used with the same local bound, while the primes above $T^{c/r}$ need not fit the truncated sieve product. It is the root-weighted mass \emph{between} these cutoffs, not the total mass up to $T$, that will make the moment norm small. % Original source: sections/04_availability.tex \section{From available primes to all large prefixes}\label{sec:availability} We now pass from moment bounds at a size to a single completion valid at every large size. On the scale $T_j=2^j$, a grid with polynomially many points in $j$ will suffice. Taking the moment order comparable to $\log j$ lets us pay for this grid without prescribing a decay rate for the moment norm: once the ratio in Markov's inequality is below $1/8$, raising it to the chosen order gives a summable bound over all grids. The full prime class also admits a faster logarithmic rate using fixed high moments; we treat that regime separately. \subsection{A sufficient condition with no prescribed rate} For a fixed separable polynomial $P$, consider the root-weighted condition \begin{equation*}\label{eq:WP} \sum_{\substack{q\in E\\\log\log x\eta\right) \le (2b_{P,j}/\eta)^{r_j}. \] For fixed $\eta>0$, the ratio is at most $1/8$ eventually. Since $r_j\ge2\log_2(j+1)$, the bound is at most $(j+1)^{-6}$. Its sum over the grids and over $j$ converges. Borel--Cantelli controls every sufficiently late grid point. Between neighbouring points the average of terms bounded by one changes by $O(j^{-2}+T_j^{-1})$. This gives \eqref{eq:weighted-as} at every size. Intersect over rational $\eta>0$ and then over the countable collection of tests. \end{proof} The weights are fixed before the prime signs. Choosing $w_n=\omega(P(n))$ after those signs would make the average equal to one and is excluded. The probability-one event in Proposition~\ref{prop:as} may depend on the input and the prescribed marks. \subsection{The exact Fourier step for marked laws} For distinct polynomial observations $R_1,\ldots,R_k$, let \[ X(n)=(g(R_i(n)))_{i\le k},\qquad Z(n)\in\mathcal Z, \] where $\mathcal Z$ is finite and $Z$ is fixed before the random signs. Assume the products $P_J=\prod_{i\in J}R_i$, for non-empty $J\subseteq\{1,\ldots,k\}$, are separable and satisfy \eqref{eq:WP}. With $g=f_0\omega$, put \[ A_{J,v}(N)=\sum_{n=n_0+1}^{n_0+N} \one_{\{Z(n)=v\}}f_0(P_J(n))\omega(P_J(n)). \] The identity \[ \one_{\{X=x\}}=2^{-k}\prod_{i=1}^k(1+x_iX_i) \] gives the finite, exact expansion \begin{equation}\label{eq:marked-fourier} \Law_N(X,Z)\{(x,v)\}-2^{-k}\Law_N(Z)\{v\} =\frac{2^{-k}}N\sum_{\varnothing\ne J\subseteq[k]}x_J A_{J,v}(N), \qquad x_J=\prod_{i\in J}x_i. \end{equation} Therefore \begin{equation}\label{eq:marked-tv} \TV\bigl(\Law_N(X,Z),U_k\otimes\Law_N(Z)\bigr) \le\frac1{2N}\sum_{J\ne\varnothing,\ v\in\mathcal Z}|A_{J,v}(N)|. \end{equation} Proposition~\ref{prop:as} applies to each sum, with deterministic weights $\one_{\{Z(n)=v\}}f_0(P_J(n))$. This proves the marked normality conclusion without any assumed convergence of the law of $Z$. For the target quadratic and its distinct shifts, each product $P_J$ has $2|J|$ simple roots at every exclusive prime outside a finite exceptional set. The condition \begin{equation*}\label{eq:A} \mathfrak a_E(x):=\sum_{\substack{q\in E\\\log\log x0. \end{equation} This includes the full exclusive class for a product of distinct admissible irreducibles. With $r$ fixed, \eqref{eq:local-density} gives \[ \prod_{\substack{q\le x\\q\text{ a sieve prime}}}(1-g_r(q))^{-1} \asymp_{P,r}(\log x)^{r\kappa}. \] Select primes only up to $x=z^a$ with a sufficiently small fixed $a>0$. The random-product argument in Lemma~\ref{lem:growing-bound} retains at least half of the product mass below $z$. Thus $G_r(z)\gg_{P,r}(\log z)^{r\kappa}$. Taking $z=\lfloor N^{1/16}\rfloor$ in \eqref{eq:finite-sieve} yields \begin{equation}\label{eq:fixed-moment} \E\bigabs{N^{-1}\sum w_n\omega(P(n))}^{2k} \ll_{P,k}(\log(2N))^{-2k\kappa}+N^{-1/2}. \end{equation} \begin{corollary}[Fixed-order rates]\label{cor:full-rate} Under \eqref{eq:kappa-general}, for each $0<\beta<\kappa$ and each fixed bounded deterministic weight sequence, almost surely \[ \sum_{n=n_0+1}^{n_0+N}w_n\omega(P(n)) =O_{\omega,P,w,\beta}\bigl(N(\log(2N))^{-\beta}\bigr). \] Countably many fixed tests may be imposed simultaneously. \end{corollary} \begin{proof} Choose an integer $A>\beta$. In $[2^j,2^{j+1}]$, use a grid with gaps at most $2^j/j^A+1$ and $O(j^A)$ points. Then choose $k$ such that $2k(\kappa-\beta)>A+2$. Markov applied to \eqref{eq:fixed-moment}, with threshold $j^{-\beta}$, gives a summable union bound over all grid points and $j$. Interpolation costs $O(j^{-A})$. Rational choices of $\beta$ approaching $\kappa$ give all the asserted exponents. \end{proof} For the target quadratic word, the smallest coefficient is $\kappa=1/2$. Hence the full-class marked completion has rate $O((\log N)^{-\beta})$ for every $\beta<1/2$, almost surely for each fixed source. A fixed family of positive-slope integer affine forms, pairwise non-proportional, may be adjoined to a fixed family of distinct target quadratic shifts. A non-empty moment with $a$ affine factors and $b$ quadratic factors has $a+2b$ simple roots at every good exclusive prime, and coefficient $(a+2b)/4$. The common rate is therefore $O((\log N)^{-\beta})$ for every $\beta<1/4$ when affine coordinates are allowed. Proportional forms, for example $n$ and $2n$, must be excluded because their signs have a fixed multiplicative relation. All fixed progressions of origins can also be included. In $n\equiv a\pmod M$, substitute $n=Mt+a$; primes dividing the fixed $M$ are the only additional exceptions. Normalise by the number of origins in that progression, and intersect the countably many resulting probability-one events. This gives no uniformity in moving moduli. Ordinary convergent word frequencies imply the corresponding logarithmic frequencies by partial summation. The logarithmic conclusion here is positive; the obstruction later concerns only ordinary means. The full-class and sparse constructions serve different purposes. The former gives a faster almost-sure rate. Fair bits on all free primes change a prime set of relative density $1/8$ almost surely, whereas the deterministic sparse constructions below change a prime-density-zero set. Neither statement entails the other with the same cost and quantifiers. % Original source: sections/05_selection.tex \section{Choosing the completion from finite exact costs}\label{sec:selection} An almost-sure argument need not give a computable sign assignment. We instead use the method of conditional expectations, with the constructive statistical context provided by \cite{GHR}. For a task $a$ at scale $j$, the non-negative cost $F_{a,j}$ will penalise non-zero character sums, while an exact arithmetic count $D_{a,j}>0$ will bound its expectation independently of the input. Thus $F_{a,j}/D_{a,j}$ has expected cost at most one. These ratios, not unknown constants from Chebotarev or the sieve, are what the algorithm computes. The asymptotic estimates enter later, to show that bounded cost forces the required empirical law. The remaining difficulty is infinite: each prime sign must be decided without evaluating every future task. A summable positive combination of the normalised costs makes a finite rational comparison sufficient, because the entire omitted tail can still be bounded after conditioning on all earlier decisions. \subsection{Tasks and their normalisers} Assume that membership in the infinite set $E$ is decidable. A task consists of a non-empty finite list of positive polynomial observations $R_1,\ldots,R_k$ ($k\ge1$), a fixed starting point $n_a$, and a finite-alphabet mark $Z_{f_0}(n)$ computable from finitely many values of the input $f_0$ at each $n$. In our applications the marks are fixed finite lists of original polynomial values. The list defining a task is fixed; its actual signs depend on $f_0$. Only non-empty observation lists are enumerated. The mark list may be empty, in which case the mark alphabet is a singleton. Thus every task has at least one non-empty character $J$, including when no marks are requested. Choose an effective enumeration of the tasks, indexed by $a\ge2$. For the quadratic completion these are simply the horizons $H=a-1$, with the target word and the original three-trace word. For the common polynomial completion they enumerate all pairs consisting of a non-empty finite observation list and a possibly empty finite mark list of monic non-constant integer polynomials. Positivity beyond a computable starting point follows from monicity, using a bound in terms of the coefficients. This enumeration need not decide which lists are admissible. Only the tasks for which the moment hypotheses hold will carry a limiting conclusion. Use $T_j,r_j,\mathcal G_j$ from Section~\ref{sec:availability}. For $J\ne\varnothing$ set $P_J=\prod_{i\in J}R_i$, and define \begin{align} A_{J,v}(t)&=\sum_{n=n_a+1}^{n_a+t} \one_{\{Z_{f_0}(n)=v\}}f_0(P_J(n))\omega(P_J(n)),\label{eq:task-sum}\\ F_{a,j}(\omega;f_0)&=\sum_{t\in\mathcal G_j}\sum_{J\ne\varnothing}\sum_v A_{J,v}(t)^{r_j},\label{eq:task-energy}\\ D_{a,j}&=|\mathcal G_j|\sum_{J\ne\varnothing} C_{P_J,E,r_j}(T_j). \label{eq:task-normalizer} \end{align} Each collision count uses the task's starting point $n_a$. The finite parity count remains meaningful for every polynomial in the task enumeration, including non-separable products; separability is required only when a sieve estimate is subsequently applied. For task $a=1$, add control of the multiplier on the ordinary integers: \begin{equation}\label{eq:mask-task} F_{1,j}(\omega)=\sum_{t\in\mathcal G_j} \left(\sum_{n=2}^{t+1}\omega(n)\right)^{r_j},\qquad D_{1,j}=|\mathcal G_j|C_{T,E,r_j}(T_j), \end{equation} where the subscript $T$ in $C$ denotes the identity polynomial and the starting point is one. All energies are non-negative because $r_j$ is even. Their normalisers are positive, since tuples paired into equal indices contribute to each parity count. Expanding a moment and using prime-sign orthogonality gives \begin{equation}\label{eq:task-expectation} 0\le\E F_{a,j}(\omega;f_0)\le D_{a,j}. \end{equation} There is no extra factor for the number of mark states in this inequality. For each tuple in the expansion, the sum over $v$ of the product of the mark indicators is one if all its indices have the same mark and zero otherwise. The remaining source weights have absolute value at most one. A prefix count at $t\le T_j$ is bounded by the full count at $T_j$. Both $D_{a,j}$ and the finite polynomial $F_{a,j}$ in the free signs are exactly computable. One may factor all the finitely many polynomial values, record their free-prime parity vectors, and expand the powers. The expectation of a monomial conditional on previously fixed prime signs is zero unless every still-free sign has even exponent. The calculation uses integers and rational numbers, not an unknown constant in an asymptotic estimate. Its running time is not claimed to be practical. \paragraph{One finite decision.} Take the four values $n^2+1=5,10,17,26$ for $n=2,3,4,5$. In this finite illustration, let $x=\omega(5)$ and $y=\omega(13)$ be independent fair signs, with every other multiplier sign equal to one. From the constant source the sum is $2x+1+y$. Its squared cost $F=(2x+1+y)^2$ has expectation $6$, and \[ \E(F/6\mid x=1)=\frac53,\qquad \E(F/6\mid x=-1)=\frac13. \] The cheaper choice is $x=-1$. From the source $f_0(m)=(-1)^{v_{17}(m)}$, the sum is instead $2x-1+y$ and the cheaper choice is $x=1$, with the same normaliser. Uniform cost bounds therefore need not mean source-independent sign choices. This calculation illustrates one finite comparison, not a run of the universal selector. Its infinitely many requirements are controlled by the full potential below. \subsection{A tail estimate that survives every decision} Set \begin{equation}\label{eq:potential} U_{f_0}(\omega)=\sum_{a,j\ge1}\frac{2^{-a}}{j(j+1)} \frac{F_{a,j}(\omega;f_0)}{D_{a,j}}. \end{equation} The two sums of weights equal one, and \eqref{eq:task-expectation} gives $\E U_{f_0}\le1$, uniformly in $f_0$. For the finite rectangle \[ 1\le a\le R,\qquad 1\le j\le2^R-1, \] the omitted weight is \begin{equation}\label{eq:rectangle-tail} 1-(1-2^{-R})^2\le2^{1-R}. \end{equation} After fixing $s$ signs, the cylinder has probability $2^{-s}$. By positivity, the conditional expectation of the omitted part of \eqref{eq:potential} is therefore at most \begin{equation}\label{eq:conditional-tail} 2^s2^{1-R}. \end{equation} This estimate pays for all future tasks at once, including those not yet evaluated. \begin{proposition}[Irrevocable computable selection]\label{prop:selector} There is a deterministic procedure choosing every sign on $E$, with \begin{equation}\label{eq:potential-bound} U_{f_0}(\omega_{f_0})\le\frac54. \end{equation} Every selected sign is obtained from finitely many oracle queries to $f_0$. The map $f_0\mapsto\omega_{f_0}$ is continuous and computable relative to the input. \end{proposition} \begin{proof} Enumerate $E$ in increasing order. At decision $s\ge1$, use the finite rectangle with $R=2s+3$. Compute its two rational conditional expectations corresponding to the two possible signs at the next prime. Choose the smaller one, choosing $+1$ in a tie. After this decision, the tail bound \eqref{eq:conditional-tail} for each child is at most \[ \varepsilon_s=2^{-s-2},\qquad \sum_{s\ge1}\varepsilon_s=\frac14. \] Let $C_{s-1}$ be the true conditional expectation of the whole potential before the decision. The average of the two children's true costs is $C_{s-1}$. The smaller finite cost is at most that average, since each omitted tail is non-negative. Adding the selected child's omitted tail gives $C_s\le C_{s-1}+\varepsilon_s$. Thus $C_s\le5/4$ for every $s$. Each finite partial sum of \eqref{eq:potential} depends on finitely many signs. Once these have all been decided, its realised value is at most the current conditional expectation of $U_{f_0}$, by positivity. It is at most $5/4$. Increasing the finite partial sums proves \eqref{eq:potential-bound}; no probabilistic choice of an uncomputed infinite realisation is used. The rectangle and every arithmetic computation at a decision are finite. Each output sign consequently uses finitely many input values, and its value is never revised. This proves relative computability and continuity in the product topology. For an ordinary computable input, all required oracle queries can be computed, so the output is computable. \end{proof} The method also permits a finite pilot without discarding future requirements. Let $P_{f_0}\ge0$ be a computable finite energy with $\E P_{f_0}\le1$, and fix a positive rational $\delta\le1$. Apply the same selection to $(1-\delta)P_{f_0}+\delta U_{f_0}$. Its total cost is at most $5/4$, and therefore $U_{f_0}\le5/(4\delta)$. The pilot is always evaluated; when it alone determines an early decision, the universal part must still be omitted with the conditional tail \emph{paid}. The limiting conclusions remain unchanged, with constants allowed to depend on $\delta$. \subsection{A finite uniform bound for the selected output} The potential estimate gives, for each task, \begin{equation}\label{eq:energy-bound} F_{a,j}(\omega_{f_0};f_0)\le\frac54\,2^a j(j+1)D_{a,j}. \end{equation} Let $L_a=(2^k-1)|\mathcal Z|$ be the number of non-empty character--mark pairs. For a grid point $t$, Hölder's inequality and \eqref{eq:marked-tv} bound the marked total variation by \[ \frac{L_a}{2t}F_{a,j}(\omega_{f_0};f_0)^{1/r_j}. \] At $T_j/2\le N\le T_j$, compare with a grid point and also compare the two empirical mark laws. This yields the explicit finite bound \begin{equation}\label{eq:finite-uniform-bound} \TV\bigl(\Law_N(X_g,Z_{f_0}),U_k\otimes\Law_N(Z_{f_0})\bigr) \le\min(1,\mathcal B_{a,j}), \end{equation} where \begin{equation}\label{eq:Ba} \mathcal B_{a,j}= \frac{L_a}{T_j}\left(\frac54\,2^a j(j+1)D_{a,j}\right)^{1/r_j} +\frac8{(j+1)^2}+\frac8{T_j}. \end{equation} The two final terms bound the interpolation of the joint empirical measure and its moving product target. All quantities on the right are independent of the input $f_0$. The bound is computable from finite integers; it may be larger than one at accessible sizes. Computing this bound at one scale is distinct from computing an index after which every later bound lies below a requested tolerance. The proof below establishes convergence under the moment hypotheses, but supplies no useful numerical modulus of that convergence. If the moment norm \eqref{eq:bj} tends to zero for every $P_J$ in the fixed task, then $\mathcal B_{a,j}\to0$. Indeed $D_{a,j}^{1/r_j}/T_j$ is at most \[ |\mathcal G_j|^{1/r_j}(2^k-1)^{1/r_j} \max_{J\ne\varnothing} b_{P_J,j}, \] and the $r_j$-th roots of the polynomial factors in $j$ remain bounded because $r_j\asymp\log j$. We have therefore proved a uniform bound over all inputs of this selector, not merely an almost-sure existence statement. In particular, if $E\subseteq\Pd\cap(B,\infty)$ is decidable and satisfies \eqref{eq:A}, all target quadratic tasks satisfy the moment hypotheses. Together with Lemma~\ref{lem:fibre}, Proposition~\ref{prop:selector} proves the computable selective completion under \eqref{eq:A}. It works for all horizons, with each horizon fixed in its own limit. No procedure here decides \eqref{eq:A} for an arbitrary supplied membership predicate. \subsection{What sparse changes actually cost} Let \[ A(f_0)=\{q:g(q)\ne f_0(q)\}\subseteq E. \] Whenever the identity-polynomial task has vanishing moment norm, \eqref{eq:mask-task} and the same argument give $N^{-1}\sum_{n=2}^{N+1}\omega_{f_0}(n)\to0$, uniformly in the input. Consequently \begin{equation}\label{eq:integer-disagreement} \frac1N\#\{2\le n\le N+1:g(n)\ne f_0(n)\}\longrightarrow\frac12. \end{equation} The marked target law with $H=1$, projected onto the old and new target signs, similarly gives disagreement density $1/2$ on the target trace. \begin{proposition}[Infinite harmonic cost of the actual changes]\label{prop:actual-cost} For each output having \eqref{eq:integer-disagreement}, \[ \sum_{q\in A(f_0)}\frac1q=\infty. \] If $E$ has relative prime density zero, the actual changes still have relative prime density zero. \end{proposition} \begin{proof} Suppose the reciprocal sum over $A=A(f_0)$ were finite. Truncating $\omega$ to negative signs only at $q\in A$, $q\le Y$, changes its mean on the ordinary integers by at most $2\sum_{q\in A,q>Y}1/q$. For a finite set of primes, the Chinese remainder theorem and truncation of their valuations give mean \[ \prod_{\substack{q\in A\\q\le Y}} (1-q^{-1})\sum_{j\ge0}(-1)^j q^{-j} =\prod_{\substack{q\in A\\q\le Y}}\frac{q-1}{q+1}. \] Under the assumed summability this product tends to a strictly positive number, contradicting the zero mean of $\omega$. Prime density zero follows simply from $A\subseteq E$. \end{proof} Thus a small proportion of changed prime coordinates is not a small perturbation of the integer sequence. The compactness consequence is also precise: the continuous image of the compact space $\M$ is a compact family of outputs, uniformly normal for each tested observation family, and its projection onto the two protected traces covers all data compatible with a source. This concerns freedom after protecting two traces, not stable reconstruction from all three. % Original source: sections/06_sparse.tex \section{Explicit reservoirs of successively smaller harmonic mass}\label{sec:sparse} The availability theorem does not require a positive prime density or an asymptotic formula for the reservoir. Explicit regular examples nevertheless show how far it reaches. We first thin one exclusive class by a slowly growing gap in \emph{prime rank}; the same construction can be iterated to any fixed finite depth. Enumerate the exclusive primes as $q_1B\}. \end{equation} The target class and the retained ranks are different indices. The construction uses only integer operations. At depth one the retained ranks begin \[ 1,2,4,6,8,11,14,17,20,23,\ldots; \] at depth two they begin \[ 1,2,6,10,16,22,28,34,40,46,\ldots. \] Write $\ell_1(x)=\log x$ and $\ell_{k+1}(x)=\log(\ell_k(x))$ at sufficiently large arguments. Thus $\ell_k$ denotes an iterated natural logarithm, whereas $\bl$ is the integer bit length used in the rank construction. \begin{proposition}[Counting, harmonic mass and availability]\label{prop:iterated-reservoir} For fixed $m\ge1$, with $c_m=(\log2)^m/4$, \begin{align} \#(E_m\cap[1,x])&\sim c_m\frac{x}{\ell_1(x)\ell_2(x)\cdots\ell_{m+1}(x)},\label{eq:iter-count}\\ \sum_{\substack{q\in E_m\\q\le x}}\frac1q&\sim c_m\ell_{m+2}(x). \label{eq:iter-mass} \end{align} These reservoirs are decidable, have relative prime density zero, and satisfy \eqref{eq:A}. \end{proposition} \begin{proof} For $\sqrt R\le n\le R$, each fixed $B_k(n)$, $k\ge2$, is $B_k(R)(1+o(1))$ uniformly. The gaps $b_m(n)$ in this range are consequently $b_m(R)(1+o(1))$. At most $\sqrt R$ retained ranks precede it, which is negligible compared with $R/b_m(R)$. Thus \[ \#\{j:a_j\le R\}\sim\frac{R}{b_m(R)},\qquad b_m(R)\sim\frac{\ell_2(R)\cdots\ell_{m+1}(R)}{(\log2)^m}. \] The prime number theorem in the fixed residue class modulo eight gives $\#\{j:q_j\le x\}\sim x/(4\log x)$; this is the fixed-modulus instance of \cite{NIST}, equation (27.12.8), or \cite{Kedlaya}, Theorem~4.12. Substituting this value for $R$ gives \eqref{eq:iter-count}, because all iterated logarithms from the second onwards are asymptotic to the corresponding ones at $x$. Partial summation, using \[ \int^x\frac{dt}{t\ell_1(t)\cdots\ell_{m+1}(t)} =\ell_{m+2}(x)+O(1), \] proves \eqref{eq:iter-mass}. The removed finite prefix does not affect either equivalent. At the lower cutoff in \eqref{eq:A}, \[ \sum_{\substack{q\in E_m\\q\le\log\log x}}\frac1q \sim c_m\ell_{m+4}(x)=o(\ell_{m+2}(x)). \] Subtracting proves availability. Membership can be decided by enumerating the primes of the fixed class and the increasing sequence of retained ranks up to the queried prime. \end{proof} For target-word moments $P_J$, the root number is $\rho=2|J|$ outside finitely many primes. At the cutoffs of Lemma~\ref{lem:growing-bound}, with $r=r_j$, \begin{equation}\label{eq:iter-annulus} \sum_{\substack{q\in E_m\\ V_j0$ and $x_P$ such that \begin{equation}\label{eq:strong-WP} \sum_{\substack{q\in E\\\log\log x0$ if necessary. Indeed $\log\log x_j=\log\log T_j-O_{P_J}(\log\log\log T_j)$, and raising the lower cutoff costs only a bounded amount. Lemma~\ref{lem:growing-bound} gives \begin{equation}\label{eq:universal-moment} \frac{C_{P_J,E,r_j}(T_j)^{1/r_j}}{T_j} \ll_J\exp(-c_J\sqrt{\log\log T_j}). \end{equation} The second term of that lemma is smaller than the displayed quantity. Apply \eqref{eq:Ba} to the fixed task. The finite list of $J$'s permits a common positive $c$, and the other $r_j$-th roots remain bounded. The interpolation terms are smaller as well. This proves \eqref{eq:universal-rate}, uniformly in $f_0$. The potential covers all the tasks, so the same output works for all lists. Its support lies above $B$ in $\Pd$, giving the prefix and both norms by Lemma~\ref{lem:fibre}. The continuity and computability statements are Proposition~\ref{prop:selector}. Distinct translates of a monic non-constant polynomial are distinct polynomials: their coefficients of next-highest degree differ. Translating preserves irreducibility and the root field. Thus the result for all finite distinct lists gives normality of the trace of each admissible irreducible. This proves the implication \textup{(i)}$\Rightarrow$\textup{(iii)} of Theorem~\ref{thm:classification}. Proposition~\ref{prop:norm-factor} gives the converse and the frozen assertion, and Section~\ref{sec:fibres} proved \textup{(i)}$\Leftrightarrow$\textup{(ii)}. The theorem is complete. For this common selector, the identity task and the mark given by each original admissible trace also show disagreement density $1/2$ on the integers and on each such trace. Proposition~\ref{prop:actual-cost} applies to its actual changed primes. The deterministic selector is not expected to choose exactly half of the reserved prime signs; only the inclusion of its changed set in $E$ is needed for prime density zero. \subsection{Products retain their square relations}\label{subsec:product-channel} The factor-parity representation and the corresponding sign-subgroup channel occur in \cite{PoulyNoise}, Theorems~6.1--6.2, for non-selective noise. Here the algebraic classification removes the frozen irreducibles from the free coordinates, and the common selector supplies the remaining factor signs inside the protected fibre. The finite Fourier mechanism is inherited; the selection and the classification of active factors provide its selective form. The restriction to distinct irreducible coordinates explains the law of general monic observations. Fix monic non-constant $R_1,\ldots,R_k\in\Z[T]$ and factor them into monic irreducibles. List the distinct admissible factors as $P_1,\ldots,P_s$, and let \[ B=(b_{ia})\in\F_2^{k\times s} \] record their multiplicities modulo two. The other irreducible factors are frozen, and even exponents contribute one to a sign. Define the sign subgroup \[ V_B=\left\{\left((-1)^{(Bv)_i}\right)_{i\le k}:v\in\F_2^s\right\} \subseteq\{-1,1\}^k. \] If $s=0$, this is the singleton consisting of the all-positive vector. \begin{corollary}[The finite polynomial channel]\label{cor:product-channel} For the common selector, the empirical vector of multiplier signs \[ (\omega_{f_0}(R_i(n)))_{i\le k} \] becomes uniform on $V_B$, asymptotically independent of any fixed finite collection of original polynomial marks. The comparison is uniform over $f_0$. Equivalently, the output vector $(g(R_i(n)))_i$ is asymptotically obtained by multiplying its original source vector coordinatewise by a uniform element of $V_B$, while retaining the empirical original marks. \end{corollary} \begin{proof} If $s=0$, every factor is frozen or has even exponent, so the multiplier vector is identically the all-positive vector and the comparison is exact. Assume henceforth that $s\ge1$. Include the original signs $f_0(P_a(n))$ among the marks in Theorem~\ref{thm:classification}. Multiplication of the new independent uniform factor signs by these marked source signs leaves a uniform cube, so the multiplier factor vector is uniform and independent of the marks. Apply the group homomorphism represented by $B$. A uniform vector on a finite group maps to the uniform vector on its image, because all fibres have the same cardinality. Total variation contracts under this common measurable map. Frozen factors and squares make no contribution to the multiplier. Including the original $R_i$ signs among the marks gives the output formulation. \end{proof} For example, if $P,Q$ are distinct admissible irreducibles, the three multiplier observations at $P,Q,PQ$ have four equiprobable states, not eight: their product is always one. The original empirical vector need not itself converge, so its randomised coset need not have a fixed limiting law. Monic factorisation avoids an additional rational leading-coefficient phase, as illustrated in Section~\ref{subsec:monicity}. No classification for arbitrary non-monic observation families is being substituted for the stated one. % Original source: sections/08_obstruction.tex \section{Divergent reservoirs with no convergent completion}\label{sec:obstruction} The preceding constructions have more than infinite harmonic mass: they keep supplying usable primes at the scales where high moments are needed. We now show that this distinction changes the existence of a fixed completion, not just its available convergence rate. For a prime set $A$, let $\beta_A\in\M$ be negative exactly at the primes in $A$. Write \[ d_1(f,h)=\sum_{q:f(q)\ne h(q)}\frac1q,\qquad M_{d,f}(N)=\frac1N\sum_{n=2}^{N+1}f(Q_d(n)), \] and $\osc_N M=\limsup_N M(N)-\liminf_N M(N)$ for a bounded real sequence. \begin{theorem}[A reservoir without a convergent completion]\label{thm:bad-reservoir} Fix $d_0\in\D$, an integer $B\ge1$, and a positive rational $\varepsilon$. There exists a decidable set $E\subseteq\Pd\cap(B,\infty)$, independent of a reference source, with \begin{equation}\label{eq:bad-reservoir-size} \#(E\cap[1,x])\ll\frac{x}{\log x\log\log x},\qquad \sum_{q\in E}\frac1q=\infty, \end{equation} such that for every $f_\star\in\M$ there is $f\in\M$ satisfying \begin{equation}\label{eq:bad-source-close} d_1(f,f_\star)<\varepsilon,\qquad f(m)=f_\star(m)\ (m\le B),\qquad f(q)=f_\star(q)\ (q\in E), \end{equation} and, simultaneously for every $A\subseteq E$ and every finite set of primes $F$, \begin{equation}\label{eq:bad-oscillation} \osc_N M_{d_0,f\beta_A\beta_F}(N)\ge\frac12. \end{equation} The map $f_\star\mapsto f$ can be chosen continuous and relatively computable. For $F=\varnothing$, all the completions $f\beta_A$ share the prefix and both protected norms of $f$. \end{theorem} The protected norms in the last sentence belong to $f$, not necessarily to $f_\star$. Constructing the obstructive base is a separate modification from subsequently completing it on $E$. The theorem does not exclude normal completion using the \emph{whole} exclusive class. \subsection{Nevertheless, every fixed word law converges in probability} \begin{proposition}[What harmonic divergence alone gives]\label{prop:probability-only} Let $E\subseteq\Pd$ have infinite reciprocal sum, with no regularity or computability assumption. For fair independent prime signs on $E$, every fixed horizon $H$ satisfies \begin{equation}\label{eq:prob-tv} \lim_{N\to\infty}\sup_{f\in\M} \E\,\TV\bigl(\Law_N(X_{f\omega}^H),U_H\bigr)^2=0. \end{equation} The convergence also holds in every finite positive moment and in probability. For any deterministic mark sequence $Z(n)$ in a fixed finite alphabet, the analogous conclusion holds for $\Law_N(X_{f\omega}^H,Z)$ compared with $U_H\otimes\Law_N(Z)$. \end{proposition} \begin{proof} Let $P$ be a fixed product of distinct translates of $Q_{d_0}$. At all but finitely many primes in $E$, it has a fixed positive number $\rho$ of simple roots. The second-moment version of \eqref{eq:moment-count} bounds the squared norm of a weighted average by $C_{P,E,2}(N)/N^2$. Fix a finite set $F\subseteq E$ of good primes, and put $m_F=\prod_{q\in F}q^2$. A pair counted in $C_{P,E,2}$ must avoid total valuation one at all primes in $F$. The forbidden local density is \[ g_2(q)=\frac{2\rho}{q}(1-q^{-1})(1-\rho/q) =\frac{2\rho}{q}+O_P(q^{-2}). \] On the full period modulo $m_F$, the proportion of surviving pairs is $\prod_{q\in F}(1-g_2(q))$. The two-coordinate residue count gives \begin{equation}\label{eq:finite-F-bound} \frac{C_{P,E,2}(N)}{N^2} \le\prod_{q\in F}(1-g_2(q)) +\frac{2m_F}{N}\left(1+\frac{m_F}{N}\right). \end{equation} First let $N\to\infty$ with $F$ fixed, and only then let $F$ increase. Infinite harmonic mass makes the product tend to zero. The bound was independent of the deterministic weights. For a measure $\mu$ on the sign cube, Fourier inversion and Cauchy--Schwarz give \[ \TV(\mu,U_H)^2\le\frac14\sum_{\varnothing\ne J\subseteq[H]}|\widehat\mu(J)|^2. \] Apply the second-moment estimate to the finitely many products $P_J$ in these Fourier coefficients. This proves \eqref{eq:prob-tv}. For marks, use the weights $f(P_J(n))\one_{\{Z(n)=v\}}$ in \eqref{eq:marked-tv}, then Cauchy--Schwarz over the finitely many pairs $(J,v)$. All total variations lie between zero and one, so convergence in the second moment implies convergence in every finite positive moment and in probability. \end{proof} The supremum over $f$ in \eqref{eq:prob-tv} is outside the expectation. It does not permit an input chosen after the signs. More importantly, convergence in probability on the full sequence of sizes does not imply a fixed realisation converging on that full sequence. An elementary model isolates the issue. At level $j$, list all words $v\in\{0,1\}^j$ and use the indicator that an infinite bit sequence begins with $v$. Each indicator has probability $2^{-j}$, so the resulting ordered sequence tends to zero in probability. Every bit sequence nevertheless meets an indicator equal to one at every level, as well as zeros. The construction below realises this moving-exception mechanism inside a fibre of multiplicative completions. \subsection{Private primes can create a lasting discrepancy} We recall the arithmetic resource, including its cost. The method has an antecedent in \cite{Klurman}, Lemma~2.1; the following argument records the precise range used here. \begin{lemma}[A supply of private large prime factors]\label{lem:private} Fix $\tau\ge4$, $R\ge2$ and $d\in\D$. Then \begin{equation}\label{eq:private-count} \#\{R+2\le n\le N+1:P^+(Q_d(n))>\tau N\} \ge(1/2-o_{\tau,R}(1))N. \end{equation} For all sufficiently large $N$, each indicated value has exactly one prime factor exceeding $\tau N$, that prime has valuation one, and it divides no other value of the same trace at indices $2,\ldots,N+1$. \end{lemma} \begin{proof} Put $V_N=(N+1)^2+2$. Summing the logarithms of all values gives \[ \sum_{n=2}^{N+1}\log Q_d(n)=2N\log N+O(N). \] At an odd split prime, the two roots are simple and lift to every prime power. Thus \[ \sum_{n=2}^{N+1}v_q(Q_d(n)) \le\frac{2N}{q-1}+\frac{2\log V_N}{\log q}. \] Odd inert primes contribute nothing, and the prime two contributes $O(N)$. The split primes form one half of the reduced classes modulo four or eight. The fixed-modulus prime number theorem and partial summation (equivalently \cite{NIST}, equation (27.11.11)) give \[ \sum_{\substack{q\le x\\(d/q)=1}}\frac{\log q}{q-1} =\left(\frac12+o(1)\right)\log x. \] Consequently the logarithmic contribution of primes at most $\tau N$ is at most \[ (1+o_\tau(1))N\log N. \] Here the rounding contribution is $2\pi(\tau N)\log V_N=O_\tau(N)$, not an additional term of order $N\log N$. At least $(1-o_\tau(1))N\log N$ remains above the cutoff, and each value contributes at most $\log V_N=(2+o(1))\log N$. At least $N/2-o_\tau(N)$ indices have a prime above the cutoff. Removing the first $R$ indices proves \eqref{eq:private-count}. For large $N$, $\tau N>2(N+1)$ and $(\tau N)^2>V_N$. The latter inequalities give uniqueness and valuation one. If the same prime divided $Q_d(n)$ and $Q_d(m)$ at distinct indices in the prefix, it would divide $(n-m)(n+m)$, whose two non-zero factors have absolute value below the prime. This is impossible. \end{proof} Let $\mathcal R$ denote the set in \eqref{eq:private-count}, and let $q_n$ be its private prime. For any source $h$, changing only the signs at these primes can set all $h(Q_d(n))$, $n\in\mathcal R$, to $+1$, or all to $-1$. Every other value of this trace in the prefix is unchanged. The two new means differ by exactly $2|\mathcal R|/N$. Given an earlier mean $a$, at least one branch differs from $a$ by at least $|\mathcal R|/N$. The harmonic cost is less than \begin{equation}\label{eq:private-cost} \sum_{n\in\mathcal R}\frac1{q_n}<\frac{|\mathcal R|}{\tau N}\le\frac1\tau. \end{equation} No privacy between different traces is used. These primes construct the base $f$; the separate exclusive reservoir will protect the other norms among its completions. \subsection{A calendar fixed before the input} Take the depth-one reservoir $S=E_1$ of Section~\ref{sec:sparse}. It is decidable, has the counting bound in \eqref{eq:bad-reservoir-size}, and has infinite reciprocal sum. Choose an increasing computable sequence of integers $\tau_\ell\ge4$ with \[ \sum_{\ell\ge1}\tau_\ell^{-1}<\varepsilon. \] For example $\tau_\ell=\max(4,\lceil2^{\ell+2}/\varepsilon\rceil)$ suffices. Start with $R_0\ge\max(B,2)$ and $V_0>(R_0+1)^2+2$. At stage $j\ge1$, first take a finite block $F_j$ of successive primes of $S$ above $V_{j-1}$ until its reciprocal sum is at least one. Let \[ E_j=F_1\cup\cdots\cup F_j,\qquad D_j=E_j\cup\{q:q\text{ prime},\ q\le j\}. \] Enumerate all sign configurations on the finite set $D_j$. For each configuration, with a new global task number $\ell$, choose by finite search an integer $N_\ell$ larger than the currently protected number $R$ of origins, large enough that \begin{equation}\label{eq:calendar-choice} \tau_\ell N_\ell>\max(E_j\cup\{B,j\}),\qquad |\mathcal R_\ell|\ge\left(\frac12-\frac1{j+4}\right)N_\ell. \end{equation} Here $\mathcal R_\ell$ is the private-prime set of Lemma~\ref{lem:private} after the $R$ previously protected origins. The search may additionally require $N_\ell\ge\tau_\ell$ and the elementary size conditions in that lemma. It terminates by \eqref{eq:private-count}. Replace $R$ by $N_\ell$ after each task. When all configurations have been treated, set $V_j=(R+1)^2+2$ and start the next stage. This calendar uses prime factors, finite counts and the \emph{number} of configurations, but not the source values or any chosen signs. It defines \[ E=\bigcup_{j\ge1}F_j \] before $f_\star$ is supplied. The blocks each have reciprocal mass at least one, while $E\subseteq S$, proving \eqref{eq:bad-reservoir-size}. To decide membership of a prime, run the finite calendar until a cutoff $V_j$ exceeds it. The long gaps between free blocks are essential: they allow every configuration already exposed to receive a discrepancy witness before any further free prime can affect that witness. \paragraph{Calendar procedure (no source queries).} The following finite-stage procedure summarises the construction. It emits instructions that can subsequently be replayed for any $f_\star$; it need not store an infinite calendar. \begin{quote} \small \textbf{Initialise} $R\gets R_0$, $V\gets V_0$, $E^{\rm cur}\gets\varnothing$, $\ell\gets0$.\\[2pt] \textbf{For} $j=1,2,\ldots$: \begin{enumerate}[label=\arabic*.,leftmargin=1.7em,itemsep=3pt,topsep=3pt] \item Take successive primes of $S$ above $V$ until their reciprocal sum is at least one; call this finite block $F_j$. Set $E^{\rm cur}\gets E^{\rm cur}\cup F_j$ and $D_j\gets E^{\rm cur}\cup\{q\text{ prime}:q\le j\}$. \item \textbf{For each} configuration $\eta\in\{-1,1\}^{D_j}$ in a fixed order: \begin{enumerate}[label=\alph*.,leftmargin=1.5em,itemsep=2pt,topsep=2pt] \item Set $\ell\gets\ell+1$. Search for $N_\ell>R$ satisfying \eqref{eq:calendar-choice}, $N_\ell\ge\tau_\ell$, $\tau_\ell N_\ell>2(N_\ell+1)$ and $(\tau_\ell N_\ell)^2>(N_\ell+1)^2+2$. \item Compute the private indices $\mathcal R_\ell$ and their primes $q_n$. Emit the instruction \[ (j,\ell,D_j,\eta,R,N_\ell,(q_n)_{n\in\mathcal R_\ell}). \] No value of $\eta$ or of a source is used in the search for $N_\ell$. \item Set $R\gets N_\ell$. \end{enumerate} \item After all configurations are emitted, set $V\gets(R+1)^2+2$. This is $V_j$. Only then begin the next free block. \end{enumerate} \textbf{Reservoir:} $E=\bigcup_{j\ge1}F_j$. \end{quote} \subsection{One base treats every completion} Now begin from $f^{(0)}=f_\star$ and follow the fixed calendar. At the task for a configuration $\eta$ on $D_j$, extend $\eta$ by one at all other primes and put $h=f^{(\ell-1)}\eta$. Its mean at the previously protected size $R$ is the reference $a=M_{d_0,h}(R)$. Apply the two private-prime prescriptions to $h$ on $\mathcal R_\ell$ and retain the branch farther from $a$, with a fixed rule for ties. Make the corresponding changes to $f^{(\ell-1)}$, obtaining $f^{(\ell)}$. The old reference mean is unchanged, and \begin{equation}\label{eq:persistent-discrepancy} \bigabs{M_{d_0,f^{(\ell)}\eta}(N_\ell) -M_{d_0,f^{(\ell)}\eta}(R)} \ge\frac12-\frac1{j+4}. \end{equation} Three invariants make the limiting source legitimate. A prime used at an earlier task divides an already protected value, so it cannot later be private to a new index in a larger prefix. No sign is changed twice. A used prime is larger than every prime in $E_j$, while every future free block lies beyond all values of the current stage; hence a used prime never belongs to $E$. Finally, later private modifications preserve all old values of the target trace, and later free primes are too large to divide them. Thus every discrepancy witness remains valid for all extensions of the current configuration on $E$. \paragraph{Source-dependent replay.} For an emitted instruction, let $h=f^{(\ell-1)}\eta$ and compute $a=M_{d_0,h}(R)$. For each $\sigma\in\{-1,1\}$, compute the finite set of flips at the recorded private primes that makes $h(Q_{d_0}(n))=\sigma$ for all $n\in\mathcal R_\ell$. Compare the two resulting means at $N_\ell$ to $a$, choose the farther one (choose $\sigma=+1$ in a tie), and apply its flips to $f^{(\ell-1)}$. These are the source queries and sign decisions; they do not change the emitted calendar. \begin{center} \fbox{\begin{minipage}{.94\linewidth} \small \textbf{Invariants of the calendar and replay.} \begin{enumerate}[label=\arabic*.,leftmargin=1.7em,itemsep=2pt,topsep=4pt] \item Every size and prime block is chosen independently of the source. \item Each recorded pivot is private in the entire new target-trace prefix, not only among the new indices. \item Used pivots lie outside the current free set and every future free block; hence they lie outside $E$ and are never used twice. \item Later base changes preserve all earlier target values. Later free primes also miss these values, so each witness survives every extension of its exposed signs. \item All configurations on $D_j$ receive their witnesses before the next free block is allowed to enter. \end{enumerate} \end{minipage}} \end{center} Define $f$ by the stabilised prime signs. The total cost is strictly below $\sum_\ell1/\tau_\ell<\varepsilon$. Primes at most $B$ and all primes of $E$ were unchanged, proving \eqref{eq:bad-source-close}. To compute $f(m)$, advance until the increasing thresholds $\tau_\ell N_\ell$ exceed $m$; no later change affects a prime factor of $m$. All computations use finitely many values of $f_\star$. This also proves continuity and relative computability. Fix, \emph{after the construction}, any $A\subseteq E$ and any finite prime set $F$. For every sufficiently large $j$, all primes of $F$ lie at most $j$. The restriction of $\beta_A\beta_F$ to $D_j$ is one of the configurations treated at that stage. Free primes outside $E_j$ lie beyond its two observation prefixes. The persistent witness \eqref{eq:persistent-discrepancy} therefore applies to the final source $f\beta_A\beta_F$. The two sizes go to infinity and the lower bound tends to $1/2$, proving \eqref{eq:bad-oscillation}. The argument covers all subsets, including non-computable ones, without taking an uncountable intersection of probability-one events. Lemma~\ref{lem:fibre} gives the protected norms for $F=\varnothing$. This completes the proof of Theorem~\ref{thm:bad-reservoir}. \subsection{Resistance to summable auxiliary changes and the exact boundary} The finite-set robustness has a useful extension. \begin{corollary}\label{cor:bad-robust} For the source in Theorem~\ref{thm:bad-reservoir}, \eqref{eq:bad-oscillation} remains true when $F$ is replaced by any prime set $C$ with $\sum_{q\in C}q^{-1/2}<\infty$. If $C\subseteq\Pd$, the two protected norms remain unchanged as well. \end{corollary} \begin{proof} A prime dividing some $Q_{d_0}(n)$ in the prefix is at most $(N+1)^2+2\le3N^2$ for $N\ge2$. It meets at most $2N/q+2$ indices, so the proportion affected is at most $6q^{-1/2}$. For $C_Y=C\cap[2,Y]$, the union bound gives, uniformly in $A$ and $N\ge2$, \[ |M_{d_0,f\beta_A\beta_C}(N)-M_{d_0,f\beta_A\beta_{C_Y}}(N)| \le12\sum_{\substack{q\in C\\q>Y}}q^{-1/2}. \] Oscillations of two bounded sequences at uniform distance at most $\delta$ differ by at most $2\delta$. Apply the finite-set theorem and let $Y\to\infty$. \end{proof} Proposition~\ref{prop:probability-only} applies to the very reservoir and base of Theorem~\ref{thm:bad-reservoir}. Thus their fixed word laws converge in probability and in every finite moment, while \[ \{\omega:M_{d_0,f\omega}(N)\text{ converges as }N\to\infty\}=\varnothing. \] There is no contradiction with almost-sure convergence along a suitable subsequence extracted from convergence in probability. The obstruction concerns the full sequence of prefixes; its discrepancy scales depend on the sign assignment. Nor is this a negative statement about logarithmic means, which the construction does not control. The reservoir violates \eqref{eq:A}, by the positive availability theorem, but we have not proved that every reservoir violating \eqref{eq:A} supports an obstructive source. The exact polynomial classification of Theorem~\ref{thm:classification} and the incomplete classification of arbitrary subreservoirs are different results. % Original source: sections/09_collective.tex \section{Collective rigidity and certificates for numerical values}\label{sec:collective} Two protected quadratic fields create no additional odd product class beyond themselves. With three independent classes, a fourth norm can become compulsory. This section concerns the different triple \begin{equation}\label{eq:collective-triple} \D_0=\{-2,2,3\},\qquad n^2+2,\quad n^2-2,\quad n^2-3. \end{equation} The third trace is $n^2-3$, not $n^2+1$. No prefix hypothesis is needed for the exact statement about these three traces. \subsection{The fourth norm is forced by all three traces} \begin{proposition}[A collective norm constraint]\label{prop:collective-fibre} A quotient $u\in\M$ is one on all positive values of the traces in \eqref{eq:collective-triple} if and only if its prime signs can differ from one only on \[ S=\{q:q\text{ prime},\ q\equiv5\pmod{24}\}. \] Every such quotient also satisfies \begin{equation}\label{eq:fourth-norm} u(a^2+3b^2)=1\qquad(a,b\in\Q,\ (a,b)\ne(0,0)). \end{equation} \end{proposition} \begin{proof} For primes $q>3$, all three symbols $(-2/q),(2/q),(3/q)$ are $-1$ exactly in the class $5\pmod{24}$. Such a prime also has $(-3/q)=-1$, because \[ [-3]=[-2][2][3]\quad\text{in }\Q^\times/\Q^{\times2}. \] The inert valuation formula \eqref{eq:inert-valuation} proves sufficiency for the three norms and for \eqref{eq:fourth-norm}. For necessity, the traces give $2=Q_2(2)$ and $3=Q_{-2}(1)$. A prime $p\ge5$ outside $S$ has a centred root $1\le r\le(p-1)/2$ in some $Q_e$, $e\in\D_0$, with \begin{equation}\label{eq:collective-descent} 00$, such that \begin{equation}\label{eq:certificate-general} m\prod_{(r,e)\in\mathcal C(m)}(r^2-e)=z_m^2, \qquad z_m\in\N_{\ge1}, \end{equation} and \begin{equation}\label{eq:certificate-bounds} |\mathcal C(m)|\le\lfloor\log_2m\rfloor,\qquad r\le\max\{2,(m-1)/2\}\quad((r,e)\in\mathcal C(m)). \end{equation} Here $\log_2$ is the base-two logarithm. For $m=1$ the set is empty. No such certificate exists when $m$ has an odd valuation at a prime $5\pmod{24}$. \end{theorem} \begin{proof} Set $\mathcal C(1)=\varnothing$, $\mathcal C(2)=\{(2,2)\}$ and $\mathcal C(3)=\{(1,-2)\}$. For a prime $p\ge5$ outside $S$, choose the first form in the order $-2,2,3$ having a root modulo $p$, and then its smallest centred positive root $r$. Put $c=(r^2-e)/p

0}^\times$ generated by non-zero absolute values of the three rational norms. The descent and inert valuation formula give \begin{equation}\label{eq:H0} \mathcal H_0=\{t\in\Q_{>0}:v_q(t)\text{ is even for every }q\in S\}. \end{equation} For rational $t$, apply the integer statement to its coprime numerator and denominator. A square is itself a norm; inverses and products of norms are norms. Grouping the factors of \eqref{eq:certificate-general} by field therefore represents every permitted positive integer, and then every member of \eqref{eq:H0}, as a product of three positive rational norms, one from each field. The positivity follows by grouping the positive trace factors and their inverses. In particular $t^2+3\in\mathcal H_0$ for every $t\in\Q$. For $m=n^2+3$, however, the certificate permits indices $O(n^2)$, not indices bounded by the original $n$. Its logarithmic number of readings is not a statement about their statistical distribution or their combined loads across different values. \subsection{Every specialisation, but no rational norm identity} Over the rational function field, define \begin{equation}\label{eq:HT} \mathcal H_T=(\Q(T)^\times)^2 \prod_{e\in\D_0}\Norm_{\Q(\sqrt e)(T)/\Q(T)} (\Q(\sqrt e)(T)^\times). \end{equation} Products of these norm groups mean finite products; repeated factors from one field can be grouped. \begin{theorem}[A functional obstruction]\label{thm:functional} The polynomial $T^2+3$ does not belong to $\mathcal H_T$. In fact there is no identity \begin{equation}\label{eq:no-recipe} T^2+3=c\,C(T)^2\prod_{j=1}^s\bigl(A_j(T)^2-e_jB_j(T)^2\bigr) \end{equation} with $c\in\Q^\times$, $C,A_j,B_j\in\Q(T)$, $e_j\in\D_0$, and non-zero multiplicative factors. \end{theorem} \begin{proof} Use the discrete valuation $v_P$ at the irreducible polynomial $P=T^2+3$. Its residue field is $\Q(\sqrt{-3})$. None of $-2,2,3$ is a square in that field: if $(a+b\sqrt{-3})^2$ is rational, then $ab=0$, so a rational square class becoming square there is either $1$ or $-3$. For $e\in\D_0$ and $A,B\in\Q(T)$ with non-zero norm, it follows that \begin{equation}\label{eq:functional-even} v_P(A^2-eB^2)=2\min\{v_P(A),v_P(B)\}. \end{equation} If the two valuations agree, cancellation of the leading residues would make $e$ a square in the residue field; if they differ, the term of smaller valuation cannot cancel. Set $v_P(0)=+\infty$ for a zero coordinate. This argument includes poles. Every factor on the right of \eqref{eq:no-recipe} has even $v_P$, and the rational constant has valuation zero. The left side has valuation one, a contradiction. \end{proof} A fixed rational recipe of the form \eqref{eq:no-recipe} cannot be valid at infinitely many integers where it is defined: that would make its difference from $T^2+3$ a rational function with infinitely many zeros, hence an identity. Nor can finitely many such recipes cover infinitely many integers. This is compatible with the computable descent, whose choices adapt to the factorisation of each input. It is not a lower bound on all possible algorithms. \begin{corollary}[Infinitely many independent functional classes]\label{cor:functional-independent} The classes of $(T+j)^2+3$, $j\ge0$, are linearly independent over $\F_2$ in $\Q(T)^\times/\mathcal H_T$. Every rational specialisation of each polynomial nevertheless belongs to $\mathcal H_0$. \end{corollary} \begin{proof} The polynomials are distinct and irreducible. A non-empty product has valuation one at any factor occurring in it, whose residue field is again $\Q(\sqrt{-3})$. Formula~\eqref{eq:functional-even} rules out membership in $\mathcal H_T$. The specialisation statement follows from \eqref{eq:H0} and \eqref{eq:fourth-norm}. \end{proof} The contrast is between a norm group over $\Q(T)$ and the groups of its numerical specialisations, not a failure of a local-global principle over a fixed number field. Results on multinorm principles, such as \cite{PollioRapinchuk}, concern a different comparison. The valuation proof above is self-contained and does not invoke such a principle. % Original source: sections/10_families.tex \section{Odd-product closure for a finite protected family}\label{sec:families} The collective example admits a general description. It also identifies exactly which part of the two-field criterion must change when more traces are protected. In this section let \[ \mathcal D'=\{d_1,\ldots,d_s\} \] be a non-empty finite family of distinct squarefree integers, none a rational square. Protect all positive trace values $n^2-d_i$, and protect the prime signs up to \begin{equation}\label{eq:general-prefix} B\ge B_0:=\max\left(3,4\max_i|d_i|\right). \end{equation} This finite prefix is an assumption of the general result. It was not needed for the concrete triple of Section~\ref{sec:collective}. \subsection{The free primes and the consistent sign pattern} \begin{lemma}[Exact general fibre]\label{lem:general-fibre} The quotient of two sources with the protected data can be chosen freely exactly at the primes \begin{equation}\label{eq:general-free} S_{\mathcal D',B}=\{q>B:(d_i/q)=-1\text{ for all }i\}. \end{equation} It is one at all other primes. Every such quotient preserves all non-zero absolute rational norms $|a^2-d_i b^2|$. \end{lemma} \begin{proof} Inert valuations prove sufficiency. For $p>B$ outside \eqref{eq:general-free}, one of the forms has a centred root $r$. Since $|d_i|