31 Arithmetic cancellation and the geometric remainder

31.1 A transfer estimate with both inputs explicit

Let $B\subset\mathbb R^2$ be a fixed convex body containing the origin in its interior, and normalize its gauge so that the shortest nonzero integer vector has gauge one. Put $$A_B(u)=\#\bigl(uB\cap(\mathbb Z^2\setminus\{0\})\bigr),\qquad \Delta_B(u)=A_B(u)-a_Bu^2,\qquad a_B=\operatorname{Area}(B),$$ and write $$P_B(L)=\#\bigl(LB\cap\mathbb Z^2_{\mathrm{prim}}\bigr),\qquad E_B(L)=P_B(L)-\frac{a_B}{\zeta(2)}L^2.$$ All primitive vectors are counted with their signs; an unoriented count for a centrally symmetric body is half this count. The count $A_B$ is $A_0$ of Section 30.1, and $\Delta_B$ is its all-point discrepancy $D_B$. The earlier torus count $A_X$ in Section 29.1 includes the origin and uses hyperbolic length units; the conversion to the unit-shortest-vector gauge is given at the end of this section.

There are two inputs in the primitive problem: the geometric discrepancy $\Delta_B$, and cancellation in the Möbius function. A quantitative bound for the second input does not by itself specify the first. The following elementary estimate keeps their roles separate.

31.2 An unconditional sublinear consequence of the Dini condition

Define the nonincreasing geometric modulus $$\omega_B(u)=\sup_{v\ge u}\frac{|\Delta_B(v)|}{v}.$$ For $L/(D+1)\ge1$, direct comparison on the intervals $[L/(d+1),L/d]$ gives $$\sum_{d\le D}|\Delta_B(L/d)| \le 2L\int_{L/(D+1)}^L\omega_B(u)\frac{du}{u}.$$(31.3) Consequently, $$\int_1^\infty\omega_B(u)\frac{du}{u}<\infty \quad\Longrightarrow\quad E_B(L)=o_B(L).$$(31.4) Here the arithmetic input is only the prime number theorem, in the equivalent form $M(x)=o(x)$, not RH. Indeed, use the nonincreasing envelope $\eta(x)=\sup_{y\ge x}|M(y)|/y$, fix $U\ge2$, and choose $D=\lfloor L/U\rfloor$. After division by $L$, the large-divisor bound tends to zero for this fixed $U$, whereas the small-divisor bound is at most $2\int_{U/2}^{\infty}\omega_B(u)\,du/u$ for all sufficiently large $L$. Let $L\to\infty$ and then $U\to\infty$. This strengthens the earlier absolute-value Dini argument, which gave only $O_B(L)$.

31.3 Fixed zero-free half-planes and the recent quasi-RH claims

For the Riemann zeta function and $1/2\le\theta_0<1$, the standard zero-free-half-plane formulation is equivalent to $$\zeta(s)\ne0\quad(\Re s>\theta_0) \quad\Longleftrightarrow\quad M(x)=O_\varepsilon(x^{\theta_0+\varepsilon}) \quad\text{for every }\varepsilon>0.$$(31.5) Here nonvanishing excludes the pole at $s=1$, where $1/\zeta$ instead has a zero. We give the analytic step because a fixed zero-free half-plane and a height-dependent zero-free region are different inputs.

Suppose first that the left side of (31.5) holds. Fix $0<\delta<(1-\theta_0)/2$. At sufficiently large $|t|$, choose the branch of $\log\zeta$ on the disk centered at $2+it$ of radius $R_2=2-\theta_0-\delta/4$ that agrees with the Euler-product logarithm at its center. This disk contains neither zeros nor the pole. Polynomial growth of $\zeta$ in a fixed vertical strip gives $\Re\log\zeta\ll_\delta\log(2+|t|)$ there. Borel–Carathéodory bounds $|\log\zeta|$ by the same order on the smaller disk of radius $R=2-\theta_0-\delta/2$. On the disk of radius $r_0=1/2$ the absolutely convergent Euler-product logarithm is bounded. The three-circles theorem, with $r=2-\theta_0-\delta<R$, now gives $$|\log\zeta(\sigma+it)|\ll_\delta(\log(2+|t|))^{\lambda_\delta}, \qquad \theta_0+\delta\le\sigma\le2,\qquad \lambda_\delta=\frac{\log(r/r_0)}{\log(R/r_0)}<1.$$ Consequently, for every $\varepsilon>0$, $1/\zeta(\sigma+it)\ll_{\delta,\varepsilon}(1+|t|)^\varepsilon$ on $\sigma\ge\theta_0+\delta$; bounded heights and $\sigma\ge2$ are handled by analyticity of the reciprocal and the Euler product.

For a half-integer $x$, truncated Perron inversion on $c=1+1/\log x$, at height $T=x^2$, has error $O(x\log^2x/T)$ because $|\mu(n)|\le1$. Move the contour to $\sigma_0=\theta_0+\delta$. There are no poles of the reciprocal in the intervening strip. The new vertical integral is $O_{\delta,\varepsilon}(x^{\sigma_0}T^\varepsilon)$ and the two horizontal integrals are $O_{\delta,\varepsilon}(xT^{\varepsilon-1})$. Choosing $\delta$ and $\varepsilon$ sufficiently small gives every exponent $\theta_0+\varepsilon'$ requested on the right side of (31.5). Half-integers suffice since $M$ is constant between consecutive integers.

Conversely, for each compact subset of $\Re s>\theta_0$, choose $\varepsilon>0$ smaller than its distance from the boundary. The bound for $M$ makes $s\int_1^\infty M(x)x^{-s-1}\,dx$ locally uniformly convergent and holomorphic there. Partial summation identifies it with $1/\zeta(s)$ on $\Re s>1$, so analytic continuation precludes a zero of $\zeta$ in $\Re s>\theta_0$. This equivalence uses the Euler product, meromorphic continuation and vertical growth of $\zeta$; zero-freeness of an arbitrary Dirichlet series would not alone suffice.

The OpenAI manuscript dated 30 September 2026 claims the half-plane $\Re s>7/8$ for all finite-order Hecke $L$-functions over $\mathbb Q(\sqrt{-3})$, and hence for Dirichlet $L$-functions, including $\zeta$ [116, Theorem 1.1]. The manuscript dated 5 October 2026 gives a separate proof claiming $\Re s>11/12$ and describes it as a simpler intermediate result toward the September theorem [117, Theorem 1.1 and the following paragraph]. For both papers we use the theorem statements in the TeX sources at repository revision adc7f1241b42 (the initial release), as pinned in the claim ledger and bibliography. We checked those statements, not either complete proof or its formalization. All applications here are conditional on the explicit arithmetic hypothesis in Proposition 31.1. The corresponding Mertens exponents would be $7/8+\varepsilon$, $11/12+\varepsilon$, and, under full RH, $1/2+\varepsilon$. The generalized assertions about other $L$-functions are unnecessary for this primitive planar count.

31.4 Logarithmic geometric savings

The next comparison is conditional on an additional estimate for the all-point discrepancy: $$|\Delta_B(u)|\ll_B\frac{u}{(\log u)^\beta},\qquad u\ge3, \qquad\beta>0.$$(31.6) No such estimate is asserted here for a McShane–Rivin norm ball.

First suppose $M(x)\ll x^\theta$ for some fixed $\theta<1$. Choose any fixed $d$ with $1/(2-\theta)<d<1$, and take $D=\lfloor L^d\rfloor$. The large-divisor contribution in (31.2) is $O(L^{2-d(2-\theta)})$, with exponent strictly smaller than one. The small-divisor contribution is $$\ll L\int_{L^{1-d}/2}^L\frac{du}{u(\log u)^\beta} \ll_{\beta,d} L(\log L)^{1-\beta}.$$ Thus $$M(x)\ll x^\theta\ (\theta<1),\quad \Delta_B(u)\ll u(\log u)^{-\beta} \quad\Longrightarrow\quad E_B(L)\ll L(\log L)^{1-\beta}.$$(31.7) In particular, the geometric threshold $\beta=1$ gives $O_B(L)$. Any fixed power saving for $M$ suffices for this conclusion: full RH does not improve the logarithmic exponent in this elementary argument.

For comparison, the classical unconditional Vinogradov–Korobov estimate, recorded for example in Zhai [114, eq. (1.5)], has the form $$M(x)\ll x\exp\!\left(-c(\log x)^{3/5} (\log\log x)^{-1/5}\right).$$ Write $F(L)=(\log L)^{3/5}(\log\log L)^{-1/5}$ and take $D=\lfloor L\exp(-aF(L))\rfloor$ with a sufficiently small fixed $a>0$. Since $F(D)\sim F(L)$, the large-divisor part is $O(L\exp(-c'F(L)))$. The small-divisor part is bounded by $$L\int_{\exp(aF(L))/2}^L\frac{du}{u(\log u)^\beta}.$$ The resulting elementary comparison is $$\begin{array}{c|c|c} \text{geometric exponent }\beta &\text{unconditional VK input} &M(x)\ll x^\theta,\ \theta<1\\ \hline 0<\beta<1 &O\!\left(L(\log L)^{1-\beta}\right) &O\!\left(L(\log L)^{1-\beta}\right)\\[2pt] \beta=1 &O(L\log\log L) &O(L)\\[2pt] \beta>1 &O\!\left(LF(L)^{1-\beta}\right) &O\!\left(L(\log L)^{1-\beta}\right). \end{array}$$ These are consequences of one elementary splitting argument, not assertions of optimality. The constants can depend on $B$, $\beta$, the arithmetic exponent, and the chosen cutoff exponent.

31.5 Power-saving geometric discrepancies and established benchmarks

If instead $|\Delta_B(u)|\ll u^\alpha$ with $0\le\alpha<1$, then $$\sum_{d\le D}|\Delta_B(L/d)|\ll L^\alpha D^{1-\alpha}.$$ Balancing this with $L^2D^{\theta-2}$ gives $$D\asymp L^{(2-\alpha)/(3-\alpha-\theta)},\qquad E_B(L)\ll L^{\kappa(\alpha,\theta)},\qquad \kappa(\alpha,\theta)=\frac{2-\alpha\theta}{3-\alpha-\theta}.$$(31.8) For example, $\alpha=2/3$ and a Mertens exponent arbitrarily close to $1/2$ give $O_\varepsilon(L^{10/11+\varepsilon})$. Exponents arbitrarily close to $7/8$ and $11/12$ give, respectively, $O_\varepsilon(L^{34/35+\varepsilon})$ and $O_\varepsilon(L^{50/51+\varepsilon})$ by this method. These examples concern a hypothetical or separately established geometric estimate with $\alpha=2/3$; they do not assign that exponent to a McShane–Rivin ball. Stronger results for smooth curved domains use additional structure and should not be confused with this general transfer bound.

With the same power-saving geometric input, the unconditional VK estimate already gives $$E_B(L)\ll L\exp(-c_B F(L))$$ for some $c_B>0$, by choosing $D=L\exp(-aF(L))$ and balancing the two exponential factors. Thus sublinear primitive errors in suitable geometries do not require RH.

This unconditional sublinear conclusion is a classical one for sufficiently smooth, positively curved domains: Zhai records it in equation (1.4) of [114]. His parameter is the area scale $x=L^2$, so his $O(x^{1/2}\exp(-cF(x)))$ becomes $O(L\exp(-c'F(L)))$ here. The hypotheses include finite nonzero curvature throughout and a $C^4$ Gauss map. Those hypotheses do not hold for a McShane–Rivin ball.

Under RH, Baker [115, Theorem 1] proves $O_\varepsilon(L^{10/13+\varepsilon})$ for a specified class of piecewise $C^4$ domains whose arcs have nonzero curvature, with additional $C^4$ reciprocal-curve regularity and rational endpoint tangents, as specified by hypotheses (H1)–(H4) on pages 267–268. This is an established comparison result, not a claim about the best current exponent or about the flat norm balls considered here. In the opposite direction, the unconditional polygon obstruction $\Omega_\pm(L\sqrt{\log\log L})$ recalled in Section 30.2 persists regardless of RH. It belongs to rational polygons and is not a lower bound for strictly convex bodies. These examples show why the shape hypothesis must accompany any arithmetic hypothesis.

31.6 The remaining geometric obstruction

Strict convexity supplies $\Delta_B(u)=o(u)$ and hence $\omega_B(u)\to0$, but no specified rate for that convergence. Under any fixed arithmetic exponent $\theta<1$, the preceding power cutoff yields a bound of the form $$|E_B(L)|\ll L\log L\,\omega_B(cL^{1-d}) +L^{2-d(2-\theta)}.$$ It is again $o(L\log L)$, without a universal logarithmic saving specified by this argument. This is a limitation of the stated inputs and this transfer estimate, not an impossibility theorem for more special convex bodies.

One cannot simply multiply a bound for $M$ by the supremum of the weights $\Delta_B(L/d)$. Abel summation also sees their variation. On a dyadic interval $d\asymp D$, the counting part of these weights can have total variation of order $(L/D)^2$, despite a much smaller pointwise discrepancy. The large-divisor estimate above pays for that variation explicitly. Further cancellation would require additional information about the geometric weights or their correlation with $\mu(d)$.

Equivalently, for $\Re s>2$ the gauge zeta functions satisfy $$Z_B(s)=\sum_{m\ne0}\|m\|_B^{-s},\qquad Z_B^{\mathrm{prim}}(s)=\frac{Z_B(s)}{\zeta(s)}.$$ The estimate $\Delta_B(u)=o(u)$ provides the representation $$Z_B(s)=\frac{a_Bs}{s-2} +s\int_1^\infty\Delta_B(u)u^{-s-1}\,du, \qquad \Re s>1.$$ A zero-free result for the denominator does not on its own continue the geometric numerator through the line $\Re s=1$, nor provide the vertical growth bounds needed for a sharp counting inversion. This is the analytic reason to keep the arithmetic and geometric inputs separate.

For a body satisfying the infinite-flatness and rational-normal hypotheses of Proposition 29.1, the prime-row construction gives $\Omega(L^{1-\varepsilon})$ along an unbounded sequence for every $\varepsilon>0$. This excludes a fixed sublinear power bound for that body even if RH holds. It does not exclude $O(L)$, or even $o(L)$: the quantitative lower bound is $L W(h/L^2)$ up to constants, and $W(h/L^2)\to0$ under strict convexity. A useful next geometric target is therefore a quantitative modulus. One possible target is $\Delta_B(u)=O(u/\log u)$, provided it is compatible with that body’s local widths. Under the same supporting-row, two-sided-graph and irrational-contact hypotheses, the row is empty at $L=n/h$ and contributes $w_n(L^{-1})+O(1)$ all points at $L+L^{-1}$. The all-point main term changes by only $a_B(2+L^{-2})$, so $$\max\{|\Delta_B(L)|,|\Delta_B(L+L^{-1})|\} \ge \frac L2W(h/L^2)-O_B(1).$$ There is no totient density factor in this all-point estimate, and it holds for every sufficiently large integer $n$, without primality. Consequently the proposed target requires $W(h/L^2)=O(1/\log L)$ along these contact scales. Strict convexity and infinite flatness alone do not give that condition; the model $f(s)=\exp[-\exp(|s|^{-\beta})]$ in Section 30.5 violates it. Establishing a compatible modulus for any specified McShane–Rivin ball remains a separate problem.

The all-point target has a numerical diagnostic independent of the area constant, just as in Section 30.3. Define $$\mathcal D_A(L)=A_B(2L)-4A_B(L).$$ The dyadic inverse identity proves the equivalence $$\Delta_B(L)=O(L/\log L) \quad\Longleftrightarrow\quad \mathcal D_A(L)=O(L/\log L).$$(31.9) Indeed, the reverse implication follows from $\Delta_B(L)=-\sum_{j\ge0}4^{-j-1}\mathcal D_A(2^jL)$ and $\log(2^jL)\ge\log L$. For a symmetric torus norm ball, let $\lambda_\gamma$ be the simple-geodesic length in the same normalization as the gauge of $B$. The length list already used in Section 30.7, rescaled when imposing the unit-shortest-vector convention, gives the exact identity $$A_B(L)=2\sum_{\substack{\gamma\text{ unoriented simple}\\\lambda_\gamma\le L}} \left\lfloor\frac L{\lambda_\gamma}\right\rfloor.$$ Thus the proposed geometric threshold can be investigated without estimating the norm-ball area or introducing a new enumeration of curve types. No finite numerical range proves the uniform bound in (31.9).

The full campaign experiments/torus_campaign/run.py implements this signed-multiple diagnostic at its dyadic cutoffs, with the all-point rows and differences in artifacts/torus_campaign/result.json. The floor comparisons use exact rational representations of the stored binary64 lengths. Membership for the true geodesic lengths remains numerical; the exact identity does not certify the required length or completeness assumptions. The campaign uses hyperbolic lengths $\ell_\gamma$ without rescaling. If $s_X$ is the shortest hyperbolic length, the unit-shortest-vector gauge uses $\lambda_\gamma=\ell_\gamma/s_X$ and hence $A_B(L)=A_{\mathrm{hyp}}(s_XL)$. The campaign’s hyperbolic dyadic rows must be interpreted with that change of units; they do not establish a uniform logarithmic bound. The separate finite companion experiments/counting_mixing/run.py checks rational examples of Möbius inversion, signed multiples, row selectors, finite dyadic identities and conditional balance exponents. These finite checks do not supply either the required geometric modulus or an arithmetic hypothesis. Source, independent-review and generation records are in docs/execution/counting-mixing-review-2026-11-01/.