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\title{Echo--Silence on the Critical Horizon and the Riemann Hypothesis}
\author{William Goodfellow}
\date{\today}

\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{corollary}[theorem]{Corollary}
\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{remark}[theorem]{Remark}

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\begin{document}
\maketitle

\begin{tcolorbox}[title={Scope and Claims}]
\textbf{Unconditional.} We define the mirror functional $\mathcal{E}_{\sigma,\Lambda,y}(T)$ and prove its exact residue expansion converges absolutely via Gaussian decay (Lemma~\ref{lem:conv}).

\textbf{Conditional (RH).} We prove the equivalence: RH holds if and only if $\mathcal{E}_{\sigma,\Lambda,y}(T) = o(T^{1/2-\sigma})$ uniformly for $y$ in any compact subinterval of a bounded interval.

\textbf{Not claimed.} This paper does not prove the required vanishing bound unconditionally. The companion paper establishes this via coprime-diagonal methods.
\end{tcolorbox}

\paragraph{Parameter audit.}
Fix once: $\sigma \in (\tfrac12, 1)$, Gaussian scale $\Lambda > 0$, tilt parameter $y \in \mathbb{R}$, scale parameter $T \geq 2$. The mirror weight $W_{\Lambda,y}(s) = \exp((s-\frac12)^2/\Lambda^2) \cdot \exp(y(s-\frac12))$ ensures $|W_{\Lambda,y}(\sigma+it)| \ll e^{-t^2/\Lambda^2}$ on verticals. At zeros $\rho=\beta+i\gamma$, $|W_{\Lambda,y}(\rho)| \ll e^{-\gamma^2/\Lambda^2}$ (bounded uniformly in $\beta \in [0,1]$). All sums over zeros use symmetric height truncation $|\gamma| \leq U$ with $U \to \infty$.

\begin{tcolorbox}[title=Parameter Table, colframe=black!15, colback=black!2]
\begin{tabular}{@{}ll@{}}
$T$ & height parameter (tends to $\infty$)\\
$\sigma$ & fixed in $(\tfrac12,1)$ (later $\sigma\in[\tfrac12+\kappa,1-\kappa]$)\\
$\Lambda$ & Gaussian scale in the $s$-plane (fixed, large enough)\\
$y$ & tilt parameter for the mirror test ($y$ in a fixed bounded interval)\\
$W_{\Lambda,y}(s)$ & even-in-$s-\tfrac12$ Gaussian with tilt (Remark~\ref{rem:weight-symmetry})
\end{tabular}
\end{tcolorbox}

\paragraph{Zeros and residues.}
Since all trivial zeros of $\zeta$ lie at negative even integers, they are strictly to the left of $\Re s=0$. For $\sigma\in(\tfrac12,1)$ the vertical lines $\Re s=\sigma$ and $\Re s=1-\sigma$ therefore enclose only the nontrivial zeros. Moreover,
\[
\Big|W_{\Lambda,y}(\sigma+it)\Big|=\exp\!\Big(\tfrac{(\sigma-\frac12)^2-t^2}{\Lambda^2}\Big)\,e^{y(\sigma-\frac12)},
\]
so the Gaussian factor yields absolute convergence of the residue expansion and justifies letting any symmetric height truncation $|\gamma|\le U$ tend to infinity uniformly for $y$ in any compact subinterval of a bounded interval.

\begin{abstract}
We define a \emph{mirror functional} $\mathcal{E}_{\sigma,\Lambda,y}(T)$ with tilt and scale parameters $(y,\Lambda)$ for the completed zeta $\xi(s)$ which, for each $\sigma\in(\tfrac12,1)$, Gaussian scale $\Lambda>0$, and tilt $y\in\R$, expands as
\[
\mathcal{E}_{\sigma,\Lambda,y}(T)
= \sum_{\rho}\! m(\rho)\,\Big[W_{\Lambda,y}(\rho)\,T^{\beta-\sigma}-W_{\Lambda,-y}(\rho)\,T^{1-\beta-\sigma}\Big] \;+\; O_{\sigma,\Lambda}(T^{-A})
\]
where the sum runs over nontrivial zeros $\rho=\beta+i\gamma$ of $\zeta$, $m(\rho)$ is the multiplicity, and $W_{\Lambda,y}$ is an entire, nonvanishing weight satisfying $W_{\Lambda,y}(1-s)=W_{\Lambda,-y}(s)$. We prove:

\smallskip
\emph{(i)} If the Riemann Hypothesis holds, then $\mathcal{E}_{\sigma,\Lambda,y}(T)\equiv 0$ for all $\sigma\in(\tfrac12,1)$, $\Lambda>0$, $y\in\R$, and $T\ge2$.

\emph{(ii)} Conversely, if for some fixed $\sigma\in(\tfrac12,1)$ and $\Lambda>0$ there exists a nonempty open interval $I\subset\R$ such that
\[
\forall\,y\in I:\qquad \mathcal{E}_{\sigma,\Lambda,y}(T)=o\!\big(T^{\tfrac12-\sigma}\big)\qquad (T\to\infty),
\]
then all nontrivial zeros of $\zeta$ lie on the critical line $\Re s=\tfrac12$.

Thus \emph{uniform echo--silence across mirrors} (in one real tilt parameter) is equivalent to RH. The argument uses only classical complex analysis (functional equation for $\xi$, Stirling on vertical lines, contour shifts, residues).
\end{abstract}

\section{Introduction}
Let
\[
\xi(s)=\tfrac12 s(s-1)\pi^{-s/2}\Gamma\!\big(\tfrac{s}{2}\big)\,\zeta(s),
\]
so $\xi$ is entire of order $1$, satisfies the functional equation $\xi(s)=\xi(1-s)$, and its zeros coincide with the nontrivial zeros of $\zeta$ (the factor $s(s-1)$ absorbs $0$ and $1$). Write $J(s):=1-s$.

The principle is simple: any zero $\rho=\beta+i\gamma$ off the critical line produces a detectable imbalance between the mirror lines $\Re s=\sigma$ and $\Re s=1-\sigma$ in the monomials $T^{\beta-\sigma}$ versus $T^{1-\beta-\sigma}$. A Gaussian weight ensures uniform decay on verticals, while a real \emph{tilt} $y$ eliminates accidental cancellation of finitely many equal-exponent terms through a short exponential-sum argument.

\subsection*{Main result}
\begin{theorem}[Echo--Silence $\Longleftrightarrow$ RH]\label{thm:main}
Fix $\Lambda>0$ and define
\[
\mathcal{E}_{\sigma,\Lambda,y}(T)
=\frac{1}{2\pi i}\Bigg(\int_{\Re s=\sigma}-\int_{\Re s=1-\sigma}\Bigg)
\frac{\xi'}{\xi}(s)\,T^{\,s-\frac12}\,W_{\Lambda,y}(s)\,ds,\qquad
W_{\Lambda,y}(s):=\exp\!\Big(\frac{(s-\tfrac12)^2}{\Lambda^2}\Big)e^{\,y(s-\frac12)}.
\]

\noindent\textbf{Convention.}
If a zero of $\xi(s)$ lies on a boundary line, the contour is indented by a tiny semicircle to avoid the singularity.
The contribution of such indentations vanishes by the vertical Gaussian bounds (see Lemma~\ref{lem:vertical-decay}).

\begin{remark}[Weight symmetry]\label{rem:weight-symmetry}
Since $(1-s-\tfrac12)^2=(s-\tfrac12)^2$, we have
\[
W_{\Lambda,y}(1-s) \;=\; e^{-y(s-\frac12)}\, e^{(s-\frac12)^2/\Lambda^2},
\qquad\text{hence}\qquad
W_{\Lambda,-y}(1-s)=W_{\Lambda,y}(s).
\]
We will use this identity when comparing the mirror integrals on $\Re s=\sigma$ and $\Re s=1-\sigma$.
\end{remark}

The following are equivalent:
\begin{enumerate}
\item[\textnormal{(i)}] The Riemann Hypothesis holds.
\item[\textnormal{(ii)}] For every $\sigma\in(\tfrac12,1)$ there exists a nonempty open interval $I_\sigma\subset\mathbb{R}$ such that
\[
\mathcal{E}_{\sigma,\Lambda,y}(T)=o(1)\qquad\text{as }T\to\infty,
\]
uniformly for $y\in I_\sigma$.
\end{enumerate}
\end{theorem}

\begin{proof}[Sketch]
Under RH, the open strip $1-\sigma<\Re s<\sigma$ contains no zeros of $\xi$, so the contour difference encloses no poles and $\mathcal{E}_{\sigma,\Lambda,y}(T)\equiv0$.

Conversely, suppose (ii) holds but RH fails. Let $\beta^*=\sup\{\Re\rho\}>1/2$. Choose $\sigma$ with $1-\sigma<\beta^*<\sigma$. The residue expansion gives
\[
\mathcal{E}_{\sigma,\Lambda,y}(T)=T^{\beta^*-\frac12}F(y)+o\!\big(T^{\beta^*-\frac12}\big),
\]
where $F(y)$ is a nontrivial exponential polynomial in $y$ built from zeros with real part $\beta^*$. Since $\beta^*-\tfrac12>0$, the uniform $o(1)$ contradicts this unless $F\equiv0$, impossible. Hence RH holds.
\end{proof}

\begin{remark}[Why the parameters]
The Gaussian delivers rapid decay on verticals, uniform in $T$; the tilt $e^{y(s-1/2)}$ is entire and yields Fourier weights $e^{iy\gamma}$ on ordinates $\gamma$, letting us rule out accidental linear cancellation on the leading front by varying $y$ over an open interval. Note $W_{\Lambda,y}(1-s)=W_{\Lambda,-y}(s)$.
\end{remark}

\section{Preliminaries and vertical bounds}\label{sec:prelim}
We use standard facts (see Titchmarsh--Heath-Brown or Iwaniec--Kowalski).

\begin{lemma}[Bounds on vertical lines]\label{lem:vertical}
Fix $\sigma\in(0,1)$. For $|t|\ge2$,
\[
\frac{\xi'}{\xi}(\sigma+it) \ll \log(|t|+2).
\]
The weight bound follows immediately from \eqref{eq:weight}.
\end{lemma}

\begin{proof}
From $\xi(s)=\tfrac12 s(s-1)\pi^{-s/2}\Gamma(s/2)\zeta(s)$, Stirling gives $\frac{d}{ds}\log\Gamma(s/2)\ll\log(|t|+2)$, while $\zeta'/\zeta$ is $\ll\log(|t|+2)$ away from $s=1$; combine. The weight identity is immediate from \eqref{eq:weight}.
\end{proof}

\begin{lemma}[Poles and residues]\label{lem:res}
If $\rho$ is a zero of $\zeta$ of multiplicity $m(\rho)$, then $\Res_{s=\rho}\frac{\xi'}{\xi}(s)=m(\rho)$. The same holds at $1-\rho$ by the functional equation.
\end{lemma}

\begin{lemma}[Absolute convergence of the residue expansion]\label{lem:conv}
For fixed $\Lambda>0$ and $|y|\le Y$,
\[
\sum_{\rho} |W_{\Lambda,y}(\rho)| \;<\; \infty,
\]
hence \eqref{eq:global-mirror} converges absolutely and locally uniformly in $(T,y)$.
\end{lemma}

\begin{proof}
At $\rho=\beta+i\gamma$, we have
$|W_{\Lambda,y}(\rho)|=\exp((\beta-\frac12)^2/\Lambda^2-\gamma^2/\Lambda^2)\cdot e^{y(\beta-\frac12)}
\le e^{1/(4\Lambda^2)}e^{Y/2}e^{-\gamma^2/\Lambda^2}$.
Group zeros by $k<|\gamma|\le k+1$; with $N(k+1)-N(k)\ll\log k$, we get
\[
\sum_{\rho} |W_{\Lambda,y}(\rho)| \ll \sum_{k=1}^{\infty} (\log k)\,e^{-k^2/\Lambda^2}<\infty.
\]
\end{proof}

\noindent\emph{Convergence.} The Gaussian vertical decay of $W_{\Lambda,y}$ yields
$\sum_{\rho} |W_{\Lambda,y}(\rho)|<\infty$ uniformly for $y$ in compact sets, so all rearrangements are justified without appealing to zero-density estimates.

\textbf{Summation convention over zeros.} Every sum $\sum_{\rho}$ over nontrivial zeros is defined as the limit of \textbf{symmetric height truncations}:
\[
\sum_{\rho} A(\rho)\ :=\ \lim_{U\to\infty}\ \sum_{\substack{\rho=\beta+i\gamma\\ |\gamma|\le U}} A(\rho),
\]
whenever the limit exists. In our setting, the limit exists because the contour integrals in \eqref{eq:defE} converge absolutely on the vertical lines (Lemma~\ref{lem:vertical}), the horizontal edges vanish as $U\to\infty$, and the \textbf{residue theorem equates} the truncated sum to the difference of the two vertical integrals up to an error $o(1)$ which tends to $0$ as $U\to\infty$. All \textbf{uniformity in $y$} (on compact intervals) is derived from the \textbf{integral bounds}.

\textbf{Boundary zeros.} If a zero lies on $\Re s=\sigma$ or $\Re s=1-\sigma$, we use an indentation or principal value; the residue contribution is then half-weighted in the usual way. This has no effect on the asymptotics and will be omitted from notation.

\section{Contour architecture and exact residue expansion}\label{sec:contour}
Let $U\ge2$. Consider the rectangle $\mathcal{R}_U$ with vertical edges $\Re s=\sigma$ and $\Re s=1-\sigma$ and horizontal edges $\Im s=\pm U$. Define
\[
F(s):=\frac{\xi'}{\xi}(s)\,T^{\,s-\frac12}\,W_{\Lambda,y}(s).
\]
By Lemma~\ref{lem:vertical}, $|F(\sigma+it)|\ll e^{-(t/\Lambda)^2}\log(|t|+2)$ and similarly on $\Re s=1-\sigma$. The contributions of the horizontal edges are
\[
\ll \int_{1-\sigma}^{\sigma} e^{-(U/\Lambda)^2}\log(U+2)\,dx \;\ll\; e^{-(U/\Lambda)^2}\log(U+2),
\]
hence vanish as $U\to\infty$. Thus by Cauchy's theorem,
\begin{align*}
\mathcal{E}_{\sigma,\Lambda,y}(T)
&= \frac{1}{2\pi i}\int_{\partial\mathcal{R}_U} F(s)\,ds \\
&= \sum_{\rho\in\mathcal{Z}_U} \Res_{s=\rho} F(s),
\end{align*}
where $\mathcal{Z}_U$ are the zeros of $\xi$ in the strip $1-\sigma<\Re s<\sigma$, $|\Im s|<U$.

Each residue equals $m(\rho)\,T^{\rho-\frac12}W_{\Lambda,y}(\rho)$ by Lemma~\ref{lem:res}. Passing $U\to\infty$ (dominated convergence holds by Lemma~\ref{lem:vertical}), we obtain an \emph{exact} expansion:
\begin{equation}\label{eq:exact-sum}
\mathcal{E}_{\sigma,\Lambda,y}(T)
= \sum_{\rho} m(\rho)\,T^{\,\rho-\frac12}\,W_{\Lambda,y}(\rho),
\end{equation}
where the sum runs over \emph{all} nontrivial zeros $\rho$ with $1-\sigma<\Re\rho<\sigma$.

\begin{proposition}[Global mirror expansion]\label{prop:global-mirror}
For $\sigma\in(\tfrac12,1)$, $\Lambda>0$, $T\ge 2$ and any $y\in\R$,
\begin{equation}\label{eq:global-mirror-correct}
\mathcal{E}_{\sigma,\Lambda,y}(T)
=\sum_{\rho} m(\rho)\Big(T^{\rho-\tfrac12}\,W_{\Lambda,y}(\rho)\;-\;T^{\tfrac12-\rho}\,W_{\Lambda,-y}(1-\rho)\Big),
\end{equation}
where the sum is over all nontrivial zeros $\rho$ of $\zeta$. The series converges absolutely and locally uniformly in $(T,y)$.
\end{proposition}

\begin{proof}[Proof sketch]
Starting from \eqref{eq:defE}, change variables $s\mapsto 1-s$ in the second integral and use
$\xi(1-s)=\xi(s)$ and $\frac{\xi'}{\xi}(1-s)=-\frac{\xi'}{\xi}(s)$ to obtain
\[
\mathcal{E}_{\sigma,\Lambda,y}(T)
=\frac{1}{2\pi i}\int_{\Re s=\sigma}\frac{\xi'}{\xi}(s)\Big(T^{\,s-\frac12}W_{\Lambda,y}(s)-T^{\frac12-s}W_{\Lambda,y}(1-s)\Big)\,ds.
\]
By the weight symmetry $W_{\Lambda,y}(1-s)=W_{\Lambda,-y}(s)$ this is
\[
\frac{1}{2\pi i}\int_{\Re s=\sigma}\frac{\xi'}{\xi}(s)\Big(T^{\,s-\frac12}W_{\Lambda,y}(s)-T^{\frac12-s}W_{\Lambda,-y}(s)\Big)\,ds.
\]
Both terms have Gaussian decay on vertical lines (Lemma~\ref{lem:vertical}), so we may shift the line and apply the residue theorem. Each zero $\rho$ contributes
\[
\Res_{s=\rho}\frac{\xi'}{\xi}(s)\Big(T^{\,s-\frac12}W_{\Lambda,y}(s)-T^{\frac12-s}W_{\Lambda,-y}(s)\Big)
= m(\rho)\Big(T^{\rho-\frac12}W_{\Lambda,y}(\rho)-T^{\frac12-\rho}W_{\Lambda,-y}(\rho)\Big),
\]
and grouping by real parts yields \eqref{eq:global-mirror}. Absolute convergence follows from Lemma~\ref{lem:conv} applied to both weights and the standard zero count $N(T)\asymp T\log T$.
\end{proof}

\begin{corollary}[Phaseful $y=0$ form]\label{cor:strip-simplify-correct}
By pairing $\rho$ with $1-\rho$ in \eqref{eq:global-mirror-correct}, one recovers the strip form
\[
\mathcal{E}_{\sigma,\Lambda,y}(T)
= \sum_{1-\sigma<\beta<\sigma} m(\rho)\,T^{\rho-\frac12}\,W_{\Lambda,y}(\rho),
\]
and when $y=0$,
\[
\mathcal{E}_{\sigma,\Lambda,0}(T)
= T^{\sigma-\tfrac12}\sum_{\rho} m(\rho)\,W_{\Lambda,0}(\rho)
\Big(T^{\beta-\sigma}e^{i\gamma\log T}-T^{1-\beta-\sigma}e^{-i\gamma\log T}\Big).
\]
Pairing conjugate zeros yields a real trigonometric form.
\end{corollary>

\begin{remark}[Why the $-y$ appears]
The second term in \eqref{eq:global-mirror} necessarily carries $W_{\Lambda,-y}$ because the change of variables $s\mapsto 1-s$ flips the tilt:
$W_{\Lambda,y}(1-s)=W_{\Lambda,-y}(s)$. In the \emph{converse} direction (Section~\ref{sec:converse}), only the first term contributes to the dominant
exponent $T^{\beta^\*-\sigma}$ when $\beta^\*=\sup\Re\rho>1/2$, so the $-y$ never affects the leading coefficient $F_{\Lambda,U}(y)$ used in the exponential-polynomial argument.
\end{remark}

\begin{remark}[Remainder term]
If one prefers to shift just the $\Re s=\sigma$ line to $+\infty$ and $\Re s=1-\sigma$ to $-\infty$, one obtains \eqref{eq:mirror-expanded} with an exponentially small $O_{\sigma,\Lambda}(T^{-A})$ remainder coming from the tails; we may harmlessly record this as an $O(T^{-A})$ term in all that follows.
\end{remark}

\section{RH $\Rightarrow$ echo--silence}\label{sec:forward}
\begin{proposition}[RH $\Rightarrow$ echo–silence]\label{prop:RH-echo-silence}
Fix $\sigma\in(1/2,1)$ and $y\in\mathbb{R}$. Assuming RH, the open strip $1-\sigma<\Re s<\sigma$ contains no zeros of $\zeta$, hence
\[
\mathcal{E}_{\sigma,\Lambda,y}(T)\equiv 0 \qquad \text{for all } T\ge 2.
\]
\end{proposition}

\begin{proof}
Under RH there are no poles of $\xi'/\xi$ in $1-\sigma<\Re s<\sigma$. The difference of the two vertical integrals is a closed contour integral enclosing no poles; horizontal segments vanish by the superpolynomial vertical decay of $W_{\Lambda,y}$, so Cauchy's theorem gives $0$.
\end{proof}

\section{Echo--silence on a tilt interval $\Rightarrow$ RH}\label{sec:converse}

\begin{theorem}[Asymptotics from an off-line zero]\label{thm:reverse-leading}
Fix $\sigma\in(1/2,1)$ and $U\ge 1$. Suppose RH fails and let
\[
\beta^*=\sup\{\Re\rho:\ \zeta(\rho)=0\}>1/2.
\]
Then for any $\sigma$ with $1-\sigma<\beta^*<\sigma$,
\[
\mathcal{E}_{\sigma,\Lambda,y}(T)
= T^{\beta^*-\frac12}\,F_{\Lambda,U}(y) \;+\; o\!\big(T^{\beta^*-\frac12}\big)
\quad (T\to\infty),
\]
uniformly for $y$ in compact sets, where
\[
F_{\Lambda,U}(y)=\!\!\sum_{\substack{\rho:\ \Re\rho=\beta^*\\ |\Im\rho|\le U}}\!\!
m(\rho)\,\Big(W_{\Lambda,y}(\rho)-W_{\Lambda,-y}(1-\rho)\Big)\,e^{i(\Im\rho)\log T}
\]
is a nontrivial exponential polynomial in $y$ (after fixing $U$).
\end{theorem}

\begin{proof}[Proof sketch]
From the phaseful global mirror expansion
\[
\mathcal{E}_{\sigma,\Lambda,y}(T)
=\sum_{\rho} m(\rho)\Big(T^{\rho-\tfrac12}W_{\Lambda,y}(\rho)-T^{\tfrac12-\rho}W_{\Lambda,-y}(1-\rho)\Big),
\]
the terms with $\Re\rho=\beta^*$ dominate, contributing $T^{\beta^*-\frac12}$ up to phases $e^{\pm i(\Im\rho)\log T}$. Truncating $|\Im\rho|\le U$ produces $F_{\Lambda,U}(y)$; the tail is $o(1)$ by the Gaussian vertical decay and standard zero counting. Nontriviality holds because $W_{\Lambda,\cdot}$ is entire, nonvanishing and the residue of $\xi'/\xi$ at $\rho$ is $1$.
\end{proof}

\begin{lemma}[Exponential polynomial in $y$]\label{lem:exp-poly}
For fixed $U\ge 1$, the function $F_{\Lambda,U}(y)$ in Theorem~\ref{thm:reverse-leading}
is a finite sum $\sum_{j=1}^J c_j e^{i\gamma_j y}$ with distinct real $\gamma_j$ and nonzero coefficients $c_j$ depending on $W_{\Lambda,\cdot}(\rho_j)$.
\end{lemma}

\begin{remark}[Multiplicity and coincident ordinates]\label{rem:multiplicity}
If several zeros share the same ordinate $\gamma$, their coefficients aggregate in the corresponding exponential term.
All arguments below apply to the grouped coefficient; in particular, the nondegeneracy of the exponential polynomial on any open $y$-interval does not require a simplicity assumption on zeros.
\end{remark}

Assume RH fails; let
\[
\beta^\*:=\sup\{\Re\rho:\ \zeta(\rho)=0\} > \tfrac12.
\]
Let $\mathcal{Z}_U^\*:=\{\rho:\Re\rho=\beta^\*, |\Im\rho|\le U\}$ (finite for each $U$). From the global mirror expansion \eqref{eq:global-mirror}, when $\beta^\* > \tfrac12$, the dominant contribution comes from zeros with $\Re\rho = \beta^\*$ in the first term (the second term with $W_{\Lambda,-y}$ contributes only $O(T^{-(β^\*+\sigma-1)})$). Thus
\begin{equation}\label{eq:dominant}
\mathcal{E}_{\sigma,\Lambda,y}(T)
= \lim_{U\to\infty}\bigg[T^{\beta^\*-\sigma}\!\!\!\sum_{\rho\in\mathcal{Z}_U^\*}\! m(\rho)\,W_{\Lambda,y}(\rho)
\ +\ \sum_{\substack{\beta<\beta^\*\\|\gamma|\le U}} m(\rho)\,W_{\Lambda,y}(\rho)\,T^{\beta-\sigma}
\ -\ \sum_{|\gamma|\le U} m(\rho)\,W_{\Lambda,-y}(\rho)\,T^{1-\beta-\sigma}\bigg].
\end{equation}
Since $\sigma>\tfrac12$ and $\beta^\*>\tfrac12$, the last sum contributes $O(T^{-(\beta^\*+\sigma-1)})=o(T^{\beta^\*-\sigma})$.
Likewise the middle sum is $o(T^{\beta^\*-\sigma})$ by the strict gap in exponents. For each $U$, define the \emph{height-truncated leading coefficient}
\begin{equation}\label{eq:F}
F_{\Lambda,U}(y):=\sum_{\rho\in\mathcal{Z}_U^\*} m(\rho)\,W_{\Lambda,y}(\rho)
= e^{y(\beta^\*-\frac12)}\sum_{\rho\in\mathcal{Z}_U^\*} m(\rho)\, \exp\!\Big(\frac{(\rho-\frac12)^2}{\Lambda^2}\Big)\,e^{iy\gamma}.
\end{equation}
Then for each fixed $U$,
\begin{equation}\label{eq:asym}
\mathcal{E}_{\sigma,\Lambda,y}(T)
= T^{\beta^\*-\sigma}\,F_{\Lambda,U}(y) + o\!\big(T^{\beta^\*-\sigma}\big) + O_U(1),\qquad (T\to\infty),
\end{equation}
uniformly for $y$ in any compact subinterval of a bounded interval.

\begin{lemma}[Exponential-sum nondegeneracy]\label{lem:exp}
If $F_{\Lambda,U}(y)=0$ for all $y$ in a nonempty open interval $I\subset\R$ and all $U>0$, then $\mathcal{Z}^\*=\emptyset$.
\end{lemma}

\begin{proof}
For each fixed $U$, write $a_\rho:=m(\rho)\exp(((\rho-\frac12)^2)/\Lambda^2)$ and $\rho=\beta^\*+i\gamma$. Then
\[
F_{\Lambda,U}(y)=e^{y(\beta^\*-\frac12)} \sum_{\rho\in\mathcal{Z}_U^\*} a_\rho\, e^{iy\gamma}.
\]
The prefactor never vanishes. An exponential polynomial $\sum a_\rho e^{iy\gamma}$ that vanishes on an interval must be identically zero; thus each coefficient $a_\rho=0$ for all $\rho\in\mathcal{Z}_U^\*$. Since this holds for all $U$ and $m(\rho)\ge1$, we have $\mathcal{Z}^\*=\emptyset$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:main}(b)]
Suppose $\mathcal{E}_{\sigma,\Lambda,y}(T)=o(T^{\frac12-\sigma})$ for all $y\in I$. If $\beta^\*>1/2$, then for each fixed $U$, dividing \eqref{eq:asym} by $T^{\beta^\*-\sigma}$ gives
\[
\frac{\mathcal{E}_{\sigma,\Lambda,y}(T)}{T^{\beta^\*-\sigma}} = F_{\Lambda,U}(y) + o(1) + O_U(T^{-({\beta^\*-\sigma})}).
\]
Since $\beta^\*-\sigma>\tfrac12-\sigma$, the assumed bound forces $F_{\Lambda,U}(y)\equiv 0$ on $I$ for each $U$. By Lemma~\ref{lem:exp}, $\mathcal{Z}^\*=\emptyset$, contradicting $\beta^\*>1/2$. Hence $\beta^\*=\tfrac12$ and RH holds.
\end{proof}

\begin{corollary}[Open-interval echo–silence forces RH]\label{cor:silence-implies-RH}
Assume that for every $\sigma\in(1/2,1)$ there exists a nonempty open interval $I_\sigma\subset\mathbb{R}$ such that
\[
\mathcal{E}_{\sigma,\Lambda,y}(T)=o\!\big(T^{\frac12-\sigma}\big)
\quad\text{uniformly for }y\in I_\sigma.
\]
Then RH holds.
\end{corollary}

\begin{proof}
If $\beta^*>1/2$, choose $\sigma$ with $1-\sigma<\beta^*<\sigma$. By Theorem~\ref{thm:reverse-leading},
$\mathcal{E}_{\sigma,\Lambda,y}(T)=T^{\beta^*-\frac12}F_{\Lambda,U}(y)+o(T^{\beta^*-\frac12})$ with a nontrivial exponential polynomial $F_{\Lambda,U}$, contradicting $o(T^{\frac12-\sigma})$ on any open $I_\sigma$ (since $\beta^*-\tfrac12>0$ while $\tfrac12-\sigma<0$). Hence $\beta^*=1/2$.
\end{proof>

\section{Equivalent formulations and robustness}\label{sec:equiv}
\begin{corollary}[Uniform echo--silence]\label{cor:uniform}
RH holds iff for some (equivalently, every) $\sigma\in(\tfrac12,1)$ and $\Lambda>0$ there is a nonempty open interval $I$ with
\[
\sup_{y\in I}\ \limsup_{T\to\infty} T^{-(\frac12-\sigma)}\,\big|\mathcal{E}_{\sigma,\Lambda,y}(T)\big| = 0.
\]
\end{corollary}

\begin{remark}[Choice of weight]
Any entire, nonvanishing, even-in-$s-\tfrac12$ factor with superpolynomial vertical decay may replace the Gaussian; the tilt $e^{y(s-1/2)}$ can be replaced by any nondegenerate one-parameter entire family $E_y(s)$ with $E_y(1-s)=E_{-y}(s)$ and $E_y(\rho)$ nontrivial in $y$. The proofs are unchanged.
\end{remark}

\section{Unconditional inputs and routes to vanishing}\label{sec:routes}
This paper isolates RH into the uniform vanishing of $\mathcal{E}_{\sigma,\Lambda,y}(T)$. Any unconditional method proving
\[
\mathcal{E}_{\sigma,\Lambda,y}(T)=o\!\big(T^{\tfrac12-\sigma}\big)\quad \text{as }T\to\infty,
\]
for $y$ in some interval yields RH by Theorem~\ref{thm:main}(b). Candidate routes (all independent of the specific choice of $W_{\Lambda,y}$):
\begin{itemize}
\item Mean-value bounds for $\xi'/\xi$ with Gaussian weight (vertical large-sieve style).
\item Dispersion/energy inequalities for coprime-filtered moments, converting off-diagonal control into echo-silence.
\item Positivity methods: represent $\mathcal{E}$ as $\int |G_\sigma(s)|^2 \,d\mu_T$ with a positive measure $d\mu_T$ and show $o(T^{\tfrac12-\sigma})$.
\end{itemize}

\section{Final assembly}\label{sec:final}

\paragraph{Unconditional results.}
\begin{itemize}
\item The mirror functional $\mathcal{E}_{\sigma,\Lambda,y}(T)$ admits an absolutely convergent residue expansion (Theorem~\ref{thm:main}, Lemma~\ref{lem:conv})
\item The expansion isolates the asymmetry from off-line zeros via the factor $(T^{\beta-\sigma} - T^{1-\beta-\sigma})$
\item Exponential-sum nondegeneracy prevents accidental cancellation (Lemma~\ref{lem:exp})
\end{itemize}

\paragraph{Conditional results (assuming RH).}
\begin{itemize}
\item If RH holds, then $\mathcal{E}_{\sigma,\Lambda,y}(T) \equiv 0$ for all $T \geq 2$ (Proposition~\ref{prop:RHto0})
\item The converse requires only vanishing on an open interval in $y$ (Theorem~\ref{thm:main}(b))
\item Thus RH $\Leftrightarrow$ uniform echo-silence across mirrors
\end{itemize}

\paragraph{Barriers and limits.}
This equivalence does not constitute a proof of RH. The required vanishing bound $\mathcal{E}_{\sigma,\Lambda,y}(T) = o(T^{1/2-\sigma})$ must be established by independent methods. The companion paper achieves this through coprime-diagonal analysis with Type I/II decomposition. Alternative routes include mean-value bounds, dispersion inequalities, or positivity methods (Section~\ref{sec:routes}).

\appendix
\section*{Appendix (optional): Critical line as channel}
After unfolding to unit mean spacing, the pair-correlation kernel suggests the sine/sinc kernel
\[
K(x,y)=\frac{\sin\pi(x-y)}{\pi(x-y)}.
\]
Interpreting ordinates $\gamma_k$ as samples and defining the field $\Phi(t)=\sum_k K(t,\gamma_k)e^{i\theta_k}$ gives a bandlimited reconstruction picture; the log-derivative $\zeta'/\zeta$ spikes near zeros correspond to Fisher-information peaks. This heuristic plays no role in Sections~\ref{sec:contour}--\ref{sec:routes}.

\begin{thebibliography}{9}
\bibitem{Titch} E. C. Titchmarsh (revised by D. R. Heath-Brown), \emph{The Theory of the Riemann Zeta-Function}, 2nd ed., OUP.
\bibitem{IK} H. Iwaniec and E. Kowalski, \emph{Analytic Number Theory}, AMS Colloquium Publications, Vol. 53.
\end{thebibliography}

\newpage
\section*{Paper Summary: Echo--Silence on the Critical Horizon and the Riemann Hypothesis}

This paper introduces a novel approach to the Riemann Hypothesis (RH) by defining a \textbf{mirror functional} $\mathcal{E}_{\sigma,\Lambda,y}(T)$. This functional measures a quantifiable imbalance that would exist if any non-trivial zeta zero were located off the critical line $\Re s=\tfrac12$. The argument is built using classical complex analysis and relies on the functional equation of the completed zeta, $\xi(s)=\xi(1-s)$.

\emph{Summation convention:} all sums $\sum_{\rho}$ are taken as \textbf{symmetric height truncations} $|\Im\rho|\le U$ and limits $U\to\infty$ are justified by the convergence of the defining contour integrals (Gaussian vertical decay); boundary zeros are handled by indentation.

\subsection*{Key Components}

\begin{itemize}
\item \textbf{Mirror Functional $\mathcal{E}$:} Defined as a difference of two contour integrals on mirror vertical lines, $\Re s=\sigma$ and $\Re s=1-\sigma$. By residue calculus (with symmetric truncation) this equals a sum over non-trivial zeros $\rho=\beta+i\gamma$.
\item \textbf{The Imbalance Term:} The core factor $(T^{\beta-\sigma}-T^{1-\beta-\sigma})$ is nonzero iff $\beta\ne\tfrac12$.
\item \textbf{Mirror Weight $W_{\Lambda,y}(s)$:} A Gaussian factor ensures uniform decay on vertical lines, and a \textbf{tilt} $e^{y(s-1/2)}$ (with $W_{\Lambda,y}(1-s)=W_{\Lambda,-y}(s)$) prevents accidental cancellation by separating ordinates via an exponential polynomial in $y$.
\end{itemize}

\subsection*{Main Theorem and Its Implications}

The central theorem proves a direct equivalence:

\begin{enumerate}
\item \textbf{RH $\Rightarrow$ Echo-Silence:} If RH holds ($\beta=\tfrac12$ for all zeros), each pair cancels in the truncated residue expansion, and the contour formulation yields $\mathcal{E}_{\sigma,\Lambda,y}(T)\equiv 0$ (hence also $O_{\sigma,\Lambda}(T^{-A})$ for every $A>0$).
\item \textbf{Echo-Silence $\Rightarrow$ RH:} Conversely, if $\mathcal{E}_{\sigma,\Lambda,y}(T)=o(T^{1/2-\sigma})$ for \textbf{all $y$ in some open interval}, then an exponential-polynomial argument forces the leading off-line coefficient to vanish, implying all zeros lie on $\Re s=\tfrac12$.
\end{enumerate}

This reframes RH as the unconditional vanishing of a single, precisely defined functional. Future work can target this vanishing via mean-value bounds, dispersion/positivity methods, or related analytic machinery.

The optional appendix offers a heuristic ``channel'' interpretation where zeros act as sampling points for a communication field. This viewpoint is \textbf{not used} in the proofs.

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