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\hypersetup{pdftitle={Catalan's constant is irrational},pdfauthor={OpenAI}}
\newtheorem{theorem}{Theorem}[section]
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\title{Catalan's constant is irrational}
\author{OpenAI}
\date{September 24, 2026}
\begin{document}
\maketitle
\begin{abstract}
We prove that Catalan's constant
$G=\sum_{j\geq0}(-1)^j/(2j+1)^2$ is irrational. The proof builds explicit
determinants whose entries are rational linear combinations of $1$, $G$
and $\zeta(2)$, arranges a Taylor-contact cancellation that removes the
$\zeta(2)$ terms, and then compares a sharp lower bound on the prime
valuations of these determinants with a sharp upper bound on their real
size. The two bounds are incompatible at the scale of the square of the
matrix size, so $G$ cannot be rational. The argument is self-contained
apart from the ordinary prime number theorem; every finite numerical
statement is certified by exact rational arithmetic.
\end{abstract}
\section{Introduction}\label{sec:introduction}
\subsection{What we prove}
Catalan's constant is the real number
\[
G=\beta(2)=\sum_{j=0}^{\infty}\frac{(-1)^j}{(2j+1)^2}
=1-\frac19+\frac1{25}-\frac1{49}+\cdots
\approx 0.9159655941772190\ldots
\]
where $\beta$ is the Dirichlet beta function. Equivalently
$G=L(2,\chi_{-4})$ for the odd quadratic character of conductor $4$.
The same number is one quarter of the volume of the smallest orientable
complete finite-volume hyperbolic three-manifold with two cusps, and it
appears in lattice sums and in a large family of integrals. Its
arithmetic type is the question we settle.
\begin{theorem}\label{thm:main}
Catalan's constant is irrational.
\end{theorem}
The word ``irrational'' means that $G\ne p/q$ for all integers $p,q$.
Everything below is aimed at Theorem~\ref{thm:main}; every other result
in the paper is a lemma, proposition or estimate feeding it.
\subsection{Why a good approximation is not enough}
Because we can compute $G$ to billions of digits, one might think its
arithmetic is within reach. This is false, and the reason drives the
whole proof.
Suppose $G$ really is a fraction $p/q$. Then for any integers $A,B$ the
combination
\[
A\,G-B=\frac{A\,p-q\,B}{q}
\]
is a rational with denominator dividing $q$, so it is either exactly
zero or at least $1/q$ in absolute value: it cannot be ``small but
nonzero''. This gives the standard criterion.
\begin{proposition}[Rationality criterion]\label{prop:criterion}
A real number $\alpha$ is irrational if and only if there are integers
$A_m\ne0$ and $B_m$ with
\[
0<|A_m\,\alpha-B_m|\longrightarrow0\quad(m\to\infty).
\]
\end{proposition}
\begin{proof}
Take $\alpha=p/q$ rational. If $A_m\alpha-B_m\ne0$ then
$|A_m\alpha-B_m|\ge1/q$, independent of $m$, so the limit cannot be $0$.
Conversely, if $\alpha$ is irrational, the continued fraction of $\alpha$
produces convergents $P_m/Q_m$ with $|Q_m\alpha-P_m|<1/Q_m\to0$, and these
forms are nonzero (an exact equality would make $\alpha=P_m/Q_m$
rational).
\end{proof}
So to prove irrationality we must build nonzero integer combinations that
tend to zero. The hard part is the word \emph{nonzero}: approximations
$P_m/Q_m$ only give $|Q_mG-P_m|=Q_m|G-P_m/Q_m|$, and multiplying by the
denominator $Q_m$ can easily destroy the decay.
\subsection{The obstruction for Catalan's constant, and earlier work}
For odd positive values $\beta(1),\beta(3),\ldots$ the question is
trivial, because these are rational multiples of powers of $\pi$. The
even values are genuinely arithmetic. Rivoal and Zudilin proved that
infinitely many even beta values are irrational, and that at least one of
$\beta(2),\beta(4),\ldots,\beta(14)$ is irrational
\cite[Theorems~1 and~2]{RivoalZudilin2003}. Zudilin reduced this list to
the six values through $\beta(12)$ \cite[Theorem~1]{Zudilin2019}, and
Lai and Zhou to the five values through $\beta(10)$
\cite[Theorem~6.1]{LaiZhou2022}. Fischler obtained stronger quantitative
linear-independence results for families of Dirichlet $L$-values
\cite[Theorems~1 and~2]{Fischler2020}. All of these are ``at least one
of a family'' statements; none identifies $\beta(2)$ itself.
For a single value one wants explicit integer forms $A_mG-B_m\to0$.
Ap\'ery's recurrences for $\zeta(2)$ and $\zeta(3)$ \cite{Apery1979} and
Beukers's integral proofs \cite{Beukers1979} succeed because the analytic
decay survives the necessary denominator clearing.
Zudilin constructed Ap\'ery-like recurrences and integral
representations for $G$ \cite[Theorems~1--3]{Zudilin2003}, and the
discussion after his Theorem~1 makes the problem explicit: the rational
quotients converge, but after clearing denominators the integer forms
fail to tend to zero. The same obstruction appears in the Pad\'e
approximation work of Rivoal \cite[Theorems~1 and~2]{Rivoal2006Catalan}
and Krattenthaler--Rivoal \cite[Theorems~1 and~2]{KrattenthalerRivoal2008Catalan},
in the hypergeometric identity work of Krattenthaler and Zudilin
\cite[Section~2 and Theorem~2]{KrattenthalerZudilin2019}, in
Nesterenko's half-integer hypergeometric approximations
\cite{Nesterenko2016Catalan}, and in the Rhin--Viola exponent
$0.6293\ldots$ reported by Viola \cite{Viola2022Catalan}. Explicit
families with exponents below $1$, such as those of Eskandari
\cite[Theorem~1.1]{Eskandari2026Approximations}, still fall short of
Proposition~\ref{prop:criterion}: an exponent below $1$ does not by
itself make $A_mG-B_m\to0$.
At neighboring characters, Calegari, Dimitrov and Tang proved the
$\mathbb Q$-linear independence of $1,\pi^2,L(2,\chi_{-3})$
\cite[Theorem~A]{CalegariDimitrovTang2024}, and their Remark~11.1.17
again records the denominator cost as the obstruction for two Catalan
families. Calegari proved irrationality of a $2$-adic analogue of $G$
\cite[Theorem~4.2]{Calegari2005}; as Calegari, Dimitrov and Tang explain
\cite[Section~5.2, footnote~2]{CalegariDimitrovTang2025}, the real
identity carries an extra $\pi^2$ term that the $2$-adic identity does
not, which is exactly why the $2$-adic result does not transfer. In our
construction the cancellation of an extra $\zeta(2)$ term is likewise
essential. Sun has also announced the same qualitative statement
\cite[Theorem~1.1]{Sun2026}; nothing from that preprint is used here.
\subsection{The plan: one determinant, two incompatible estimates}
Instead of a single linear form we use a whole integer matrix. Its
determinant $\Delta_N$ is signed, but that is fine: we only need that
$\Delta_N$ is a nonzero rational number whose size we can bound from
below and from above.
\emph{Step 1 (build and cancel).} For each $N$ we build an $n\times n$
determinant $\Delta_N$ with $n=48N$. Its entries are moments of two
elementary kernels against polynomial rows. Each moment is a rational
combination of $1$, $G$ and $\zeta(2)$. The rows are chosen so that the
coefficients of $\zeta(2)$ cancel \emph{exactly}, by matching Taylor
series. Thus, under the single hypothesis $G\in\mathbb Q$, every entry
and hence $\Delta_N$ lies in $\mathbb Q$. This is
Section~\ref{sec:foundations}.
\emph{Step 2 (it is nonzero).} For a large prime $p$ take $N=p$ and prove
$\Delta_p\ne0$, indeed $v_p(\Delta_p)=-96p$ exactly. The proof reduces
the $48p\times48p$ matrix modulo $p$ to three fixed rational
$48\times48$ matrices, and an exact computation over $\mathbb F_{101}$
proves all three invertible. This is Section~\ref{sec:nonvanishing}.
\emph{Step 3 (a lower bound from valuations).} A nonzero rational number
is pinned down by its prime valuations,
$\log|\Delta_N|=\sum_p v_p(\Delta_N)\log p$. Careful $p$-adic estimates
of the moments bound how negative $v_p(\Delta_N)$ can be at each prime,
and summing $\log p$ against the prime number theorem gives a lower
bound. This is Section~\ref{sec:odd-primes}, with the prime $2$ handled
in Section~\ref{sec:foundations}.
\emph{Step 4 (an upper bound from an integral).} Independently of any
arithmetic hypothesis, $\Delta_N$ has an integral representation as a
product of Vandermonde determinants. Estimating that integral gives an
upper bound on $|\Delta_N|$. This is Sections~\ref{sec:real-place}
and~\ref{sec:energy}, with the numerical constants certified in
Section~\ref{sec:certificate}.
The payoff is a pair of bounds of opposite type on the same normalized
logarithm. Define
\begin{equation}\label{eq:normalized-logarithm}
\mathcal L_N=\frac{\log|\Delta_N|}{(48N)^2}-\frac12\log 2 .
\end{equation}
If $G\in\mathbb Q$, then along any sequence of $N$ with $\Delta_N\ne0$,
\[
\liminf_{N\to\infty}\mathcal L_N>-2.29084
\qquad\text{(Proposition~\ref{prop:finite-lower})},
\]
and unconditionally
\[
\limsup_{N\to\infty}\mathcal L_N\le-2.290939875<-2.2909
\qquad\text{(Proposition~\ref{prop:real-upper})}.
\]
If $G$ were rational we could take the sequence $N=p$, $p\to\infty$
prime, on which both bounds apply and $\Delta_p\ne0$. Then
$-2.29084<\liminf\mathcal L_p\le\limsup\mathcal L_p\le-2.290939875$, but
$-2.29084>-2.2909$. Contradiction. Hence $G$ is irrational.
The whole proof is therefore an exercise in making the two constants
$2.29084$ and $2.290939875$ genuinely correct, with a margin.
Zudilin's determinantal criterion \cite{Zudilin2017Determinantal} is a useful
precedent, but its determinant is a positive Hankel determinant, whereas ours
is mixed and signed, so nonvanishing and the real estimate need separate
arguments; the logarithmic interactions of the resulting Vandermonde products
are estimated as in logarithmic potential theory \cite{Saff2010}.
\subsection{What makes the comparison close enough}
Three design choices are responsible for the narrow positive gap.
\emph{Integral rows with two layers of divisibility.} The rows are
integral Chebyshev polynomials in disguise, which supplies both the
Taylor contact that cancels $\zeta(2)$ and useful arithmetic
divisibility. At the large odd primes that dominate the leading bound,
the entries have denominators of order $p^2$ and of order $p$; both
layers must be tracked separately, because cancelling only the first
layer would leave a denominator bound too weak to win.
\emph{Nonvanishing by arithmetic, not by positivity.} The determinant is
mixed and signed, so there is no positivity to fall back on. Instead we
reduce a growing matrix to a fixed finite one using Frobenius and two
polynomial bases, and check the fixed matrices by exact modular
arithmetic; here the certificate over $\mathbb F_{101}$ enters.
\emph{Respect for both branches of the parametrization.} The rows live on
two branches of a rational parametrization, and the real estimate treats
both. A bounded holomorphic interpolant controls the mixed evaluations
when the integration nodes are spread out; a Hadamard bound controls them
when the nodes bunch up. The interpolation is uniform even as nodes
collide, because the evaluation determinant cancels algebraically inside
a finite-dimensional Hardy space, so no inverse-Vandermonde estimate is
needed. The resulting Vandermonde products turn the many-variable
supremum into an explicit energy problem, whose answer is the rational
trial data of Section~\ref{sec:certificate}.
\subsection{How to read this paper}
The logical spine is short: Section~\ref{sec:foundations} builds $\Delta_N$
and proves it rational under $G\in\mathbb Q$; Section~\ref{sec:odd-primes}
and the prime-$2$ part of Section~\ref{sec:foundations} give the lower
bound; Section~\ref{sec:nonvanishing} supplies the nonzero determinants;
Sections~\ref{sec:real-place} and~\ref{sec:energy} give the upper bound;
Section~\ref{sec:certificate} certifies the finite constants; and
Section~\ref{sec:conclusion} assembles the contradiction. A reader who
wants only the shape of the argument can read this introduction and
Section~\ref{sec:conclusion}; the intermediate sections are the estimates
that make the two constants correct.
The only outside analytic input in the arithmetic half is the ordinary
prime number theorem, used in the form $\theta(y)=\sum_{p\le y}\log p\sim
y$ and proved locally in the form actually needed
\cite[Step~VI, p.~707]{Zagier1997}. The real half uses Andr\'eief's
integration identity \cite[Equation~(1.7) and Section~2.2]{Forrester2018Andreief},
Cauchy's double alternant \cite[Section~2.1, Equation~(2.7)]{Krattenthaler1999},
the classical bounded-interpolation picture \cite{Sarason1967}
and a Chebyshev expansion of the logarithmic kernel
\cite[Lemma~3.1]{GaroufalidisPopescu2013}, all restated and proved where
used. Section~\ref{sec:conclusion} also records the hyperbolic-volume
consequences.
\subsection{Conventions}
All logarithms are natural. We use $v_p(p)=1$ and $v_p(0)=+\infty$; for
real estimates we put $\log0=-\infty$. Constants in asymptotic estimates
may depend on the fixed rational value hypothetically assigned to $G$,
but never on $N$ or on the integration variables. Finite certificates are
specified by rational data and rational arithmetic, so they can be
checked without numerical logarithms or root finding.
\section{Polynomial rows and mixed moments}\label{sec:foundations}
This section builds the determinant $\Delta_N$ and proves the one
arithmetic fact we need from it: under the hypothesis $G\in\mathbb Q$,
every entry of $\Delta_N$ is rational. The cancellation of $\zeta(2)$ is
the whole reason to design the rows as we do, so we motivate it first and
then carry out the construction in full.
\subsection{The two kernels and why moments are rational}
Two elementary kernels appear: on $(-1,1)$ the weight $|t|\,dt/f(t)$, and
on $(0,1)$ the weight $ds$. Both integrate against $1/(1-ts)$, and the
geometric series $1/(1-ts)=\sum_{j\ge0}(ts)^j$ converts these moments into
simple sums. The moments are governed by exactly two classical constants,
$G$ and $\zeta(2)$, and this is what makes the construction arithmetic.
For each positive integer $N$ set
\begin{equation}\label{eq:parameters}
\begin{gathered}
n=48N,\qquad a=11N,\qquad b=7N,\qquad q=g=4N,\qquad h=2N,\\
L=n+a=n+b+q=59N,\qquad C=L+g=63N,\\
H=C+h=65N,\qquad A=a+2g=19N.
\end{gathered}
\end{equation}
The matrix has size $n=48N$; the integers $a,b,q,g,h$ and the resulting
$L,C,H,A$ are the bookkeeping that makes both the arithmetic and the real
estimates come out. The numbers $48,11,7,4,2$ are not natural, but the
outcome of optimizing the two final constants.
Write $f(t)=\sqrt{1-t^2}$ with the positive branch on $(-1,1)$, and let
$w=t/(1+f)$. Then $t=2w/(1+w^2)$ and $f=(1-w^2)/(1+w^2)$. Let the star
involution send $w$ to $w^{-1}$; it fixes $t$ and sends $f$ to $-f$. For
$0\le r<n$ define the row polynomials
\begin{equation}\label{eq:rows}
R_r=(1-t)^h t^{C-1}w^{r-g},\qquad
P_r=\frac{R_r+R_r^*}{2},\qquad
D_r=\frac{t(R_r^*-R_r)}{2f}.
\end{equation}
Near $t=0$ we read $f$ through its Taylor branch with $f(0)=1$, so these
are polynomial identities rather than statements about irrationalities.
\begin{lemma}\label{lem:rows-polynomials}
Let $T_d,U_d$ be the Chebyshev polynomials, normalized by $T_0=1$,
$T_1=x$, $U_{-1}=0$, $U_0=1$ and $S_{d+1}=2xS_d-S_{d-1}$. With
$d=|r-g|$,
\[
P_r=(1-t)^h t^{C-1}T_d(1/t),\qquad
D_r=\operatorname{sgn}(r-g)(1-t)^h t^{C-1}U_{d-1}(1/t),
\]
and these are integer polynomials, the second being zero when $r=g$.
Moreover
\begin{equation}\label{eq:row-support}
\operatorname{supp}P_r\subseteq\{A,\ldots,H-1\},\qquad
\operatorname{supp}D_r\subseteq\{A+1,\ldots,H-1\}.
\end{equation}
\end{lemma}
\begin{proof}
The identities $(w+w^{-1})/2=1/t$ and $(w^{-1}-w)/2=f/t$ turn
$R_r\pm R_r^*$ into the displayed Chebyshev expressions; these are the
standard Laurent formulas \cite[Equations~(18.5.1)--(18.5.2)]{DLMFChebyshev}.
The degree bound $d\le n-1-g$ gives the stated supports after multiplying
by $(1-t)^h t^{C-1}$.
\end{proof}
The key algebraic fact is the Taylor contact
\begin{equation}\label{eq:contact}
\frac{tP_r}{f}-D_r=\frac{tR_r}{f}=O(t^{C+r-g})=O(t^{L}).
\end{equation}
Since $w=t/2+O(t^3)$, the combination $tR_r/f$ vanishes to order at least
$L$, exactly the order of the highest column index $j$. This is the
cancellation mechanism, and it removes the $\zeta(2)$ coefficient below.
\subsection{The moments and the determinant}
For nonnegative integers $i,j$ define the two mixed moments
\begin{equation}\label{eq:mixed-moments}
M(i,j)=\int_{-1}^1\int_0^1\frac{|t|}{f(t)}\frac{t^i s^j}{1-ts}\,ds\,dt,
\qquad
Z(i,j)=\int_0^1\int_0^1\frac{t^i s^j}{1-ts}\,ds\,dt,
\end{equation}
extended bilinearly to polynomial first arguments. Both integrals converge
absolutely: integrating in $s$ leaves $(-\log(1-u))/\sqrt{1-u^2}$, which
is integrable on $(0,1)$, and bounded polynomial factors do not disturb
this. Hence every determinant expansion below is absolutely integrable and
Fubini applies.
For a raw column index $b\le j<L$ put
\[
F_j(r)=M(P_r,j)-\frac32 Z(D_r,j)\qquad(0\le r<n).
\]
The determinant used throughout the proof is
\begin{equation}\label{eq:determinant}
\Delta_N=\det_{0\le r,k<n}
\left(M\bigl(P_r,s^{b+k}(1-s)^q\bigr)
-\frac32Z\bigl(D_r,s^{b+k}(1-s)^q\bigr)\right).
\end{equation}
If $\mathcal F$ is the $n\times(n+q)$ matrix of raw columns $F_j$, then
$\Delta_N=\det(\mathcal F\mathcal T)$ where the integer filter matrix
$\mathcal T$ has entries
\begin{equation}\label{eq:integer-filter}
\mathcal T_{j,k}=(-1)^{j-b-k}\binom{q}{j-b-k},
\qquad b\le j<L,\quad 0\le k<n.
\end{equation}
This filter is a fourth finite difference in disguise; it is why the raw
column range is $j<L$ and why the arithmetic estimate can be layered
(Section~\ref{sec:odd-primes}).
\subsection{Moment evaluation and cancellation of \texorpdfstring{$\zeta(2)$}{zeta(2)}}
Set
\begin{equation}\label{eq:moment-values}
c_l=4^{-l}\binom{2l}{l},\qquad
m_i=\int_{-1}^1t^i\frac{|t|}{f(t)}\,dt
=\begin{cases}\dfrac{2}{(i+1)c_{i/2}},&i\text{ even},\\[2pt]0,&i\text{ odd},\end{cases}
\end{equation}
with $m_{-1}=0$. The moment formula follows from symmetry, $m_0=2$, and
integration by parts $(i+1)m_i=i m_{i-2}$. Termwise integration gives
\begin{equation}\label{eq:moment-series}
M(i,j)=\sum_{u\ge0}\frac{m_{i+u}}{j+u+1},\qquad
M(i,j)-M(i+1,j+1)=\frac{m_i}{j+1}.
\end{equation}
Put $K^-_d=M(d,0)$ and $K^+_d=M(0,d)$. Their boundary recurrences are
\begin{equation}\label{eq:boundary-recurrences}
dK^-_d=(d-1)K^-_{d-2}+m_{d-2}+m_{d-1},\qquad K^-_1=2,
\end{equation}
\begin{equation}\label{eq:boundary-recurrences-plus}
(d-1)K^+_d=(d-2)K^+_{d-2}+\frac{2}{d-1}\qquad(d\ge2),
\end{equation}
from the telescoping identity
$\frac{d m_{d+u}-(d-1)m_{d+u-2}}{u+1}=m_{d+u-2}-m_{d+u}$ (valid termwise
and summing since $m_v\to0$). The two independent starts are
\begin{equation}\label{eq:boundary-starts}
K^-_0=K^+_0=4G,\qquad K^+_1=\frac{\pi^2}{4}=\frac32\zeta(2).
\end{equation}
Indeed $K^-_0=\int_0^{\pi/2}\log\frac{1+\sin\theta}{1-\sin\theta}\,d\theta
=-4\int_0^{\pi/4}\log\tan v\,dv=4G$, and $K^+_1=\sum_{k\ge1}1/(2k^2c_k)$
together with $(\arcsin x)^2=\sum_{k\ge1}x^{2k}/(2k^2c_k)$ at $x=1$ gives
$\pi^2/4=\frac32\zeta(2)$.
Now define $M^0(i,j)$ by the same recurrences
\eqref{eq:moment-series}--\eqref{eq:boundary-recurrences-plus} but with
the starts $K^-_0=K^+_0$ and $K^+_1$ replaced by zero. Concretely, let
$k^-_d,k^+_d$ satisfy \eqref{eq:boundary-recurrences}--\eqref{eq:boundary-recurrences-plus}
with
\begin{equation}\label{eq:rational-starts}
k^-_0=k^+_0=k^+_1=0,\qquad k^-_1=2 .
\end{equation}
Then
\begin{equation}\label{eq:diagonal-reduction}
M^0(i,j)=
\begin{cases}
k^-_{i-j}-\sum_{k=1}^{j}\dfrac{m_{i-j+k-1}}{k},&i\ge j,\\[6pt]
k^+_{j-i}-\sum_{k=j-i+1}^{j}\dfrac{m_{k-(j-i)-1}}{k},&i<j,
\end{cases}
\end{equation}
with the algebraic convention $M^0(-1,j)=k^+_{j+1}$ (not an integral). With
$B_i^{(d)}=\sum_{k=1}^i k^{-d}$ and $B_0^{(d)}=0$, put
\begin{equation}\label{eq:rational-z}
Z^0(i,j)=\begin{cases}
-B_i^{(2)},&i=j,\\[2pt]
\dfrac{B_i^{(1)}-B_j^{(1)}}{i-j},&i\ne j.
\end{cases}
\end{equation}
Then the two kernels separate cleanly:
\begin{equation}\label{eq:rational-decomposition}
\begin{split}
M(i,j)&=M^0(i,j)+4G\,c_{(i-j)/2}+\frac32\zeta(2)\,c_{(j-i-1)/2},\\
Z(i,j)&=Z^0(i,j)+[i=j]\,\zeta(2).
\end{split}
\end{equation}
Here $c_z=0$ unless $z$ is a nonnegative integer, and $[i=j]$ is one if
$i=j$ and zero otherwise. The reason is that the homogeneous parts of the
boundary recurrences propagate the starts by exactly these two $c$-kernels,
while diagonal reduction preserves both index differences; the $Z$ formula
is partial fractions plus the diagonal term.
For the arithmetic estimate of Section~\ref{sec:odd-primes} we record the
explicit solution of the rational recurrences. Define
\begin{equation}\label{eq:h-star}
H_z^*=\begin{cases}
c_{z/2},&z\ge0\text{ even},\\[2pt]
\dfrac{1}{z\,c_{(z-1)/2}},&z\ge1\text{ odd},
\end{cases}
\qquad
\frac{H_{z+2}^*}{H_z^*}=\frac{z+1}{z+2}.
\end{equation}
Then
\begin{equation}\label{eq:explicit-boundary-arrays}
\begin{split}
k^-_u&=\begin{cases}
H_u^*\displaystyle\sum_{\substack{1\le z\le u\\z\text{ even}}}\dfrac{2}{z^2(H_z^*)^2},&u\text{ even},\\[8pt]
H_u^*\displaystyle\sum_{\substack{1\le z\le u\\z\text{ odd}}}\dfrac{2}{z},&u\text{ odd},
\end{cases}\\
k^+_{u+1}&=H_u^*\displaystyle\sum_{\substack{1\le z\le u\\z\equiv u\ (2)}}\dfrac{2}{z^2H_z^*}\qquad(u\ge0).
\end{split}
\end{equation}
Indeed, dividing each recurrence by the appropriate $H^*$ and using the
ratio in \eqref{eq:h-star} turns the increments into $2/(z^2(H_z^*)^2)$
and $2/z$ for the minus recurrence and $2/(z^2H_z^*)$ for the plus
recurrence, with the stated starts; this proves all the formulas,
including the empty-sum cases.
Now we can see the cancellation.
\begin{proposition}\label{prop:rationality}
Each raw entry $F_j(r)$ lies in $\mathbb Q+\mathbb Q G$. Consequently, if
$G\in\mathbb Q$, then $\Delta_N\in\mathbb Q$ for every positive integer
$N$.
\end{proposition}
\begin{proof}
By \eqref{eq:rational-decomposition} the coefficient of $\zeta(2)$ in
$F_j(r)$ is
\[
\frac32\Bigl(\sum_i[t^i]P_r\,c_{(j-i-1)/2}-[t^j]D_r\Bigr)
=\frac32[t^j]\Bigl(\frac{tP_r}{f}-D_r\Bigr)=0
\]
because $j<L$ and the contact \eqref{eq:contact} gives $tP_r/f-D_r=O(t^L)$.
Written out, only the $G$ term and the rational arrays survive:
\begin{equation}\label{eq:rational-raw-entry}
F_j(r)=\sum_i[t^i]P_r\bigl(M^0(i,j)+4G\,c_{(i-j)/2}\bigr)
-\frac32\sum_i[t^i]D_r\,Z^0(i,j).
\end{equation}
Both sums are finite and involve only rational data, so $F_j(r)\in
\mathbb Q+\mathbb Q G$; if $G\in\mathbb Q$ the filter $\mathcal T$ is
integer and $\Delta_N\in\mathbb Q$.
\end{proof}
The inevitable $\zeta(2)$ in the moments is removed by choosing $D_r$ to
match the Taylor series of $tP_r/f$ through order $L$.
\subsection{The estimate at the prime two}
Assume now $G\in\mathbb Q$ (also inside $\mathbb Q_2$) and bound the
$2$-adic valuation of $\Delta_N$. This is the first half of the lower
bound; the odd primes are Section~\ref{sec:odd-primes}. Constants that may
depend on the fixed rational value of $G$ carry a subscript $G$.
\begin{proposition}\label{prop:two-adic}
With $\delta=(q+g+h)/n=5/24$, for every positive integer $N$,
\begin{equation}\label{eq:two-adic-bound}
v_2(\Delta_N)\ge-\Bigl(\frac{\delta}{2}+\frac{\delta^2}{8}\Bigr)n^2
-O_G\bigl(n\log(n+2)\bigr)=-\frac{505}{4608}n^2-O_G\bigl(n\log(n+2)\bigr).
\end{equation}
The same bound holds for every full minor of the raw column matrix.
\end{proposition}
\begin{proof}
Introduce the $2$-adic smoothed moments
\begin{equation}\label{eq:two-adic-smoothing}
M_{\rm s}(i,j)=\sum_{k\ge0}\frac{m_{i+k}}{j+k+1}\quad\text{in }\mathbb Q_2 .
\end{equation}
For even $u$ the factorial formula gives
$v_2(m_u)=1+u-v_2\binom{u}{u/2}\ge u+1-\log_2(u+1)$, and odd moments
vanish. Hence for $0\le i,j<H$ every term of \eqref{eq:two-adic-smoothing}
has valuation at least $i-2\log_2H-1$, so the tail converges uniformly and
$v_2(M_{\rm s}(i,j))\ge i-2\log_2H-1$.
The diagonal and boundary recurrences hold for $M_{\rm s}$ as well, with
tails tending to zero $2$-adically, so there are exactly two homogeneous
discrepancies against $M^0(i,j)+4G\,c_{(i-j)/2}$:
\[
e_1=4G-M_{\rm s}(0,0),\qquad e_2=-M_{\rm s}(0,1).
\]
Both are fixed elements of $\mathbb Q_2$, independent of $N$. Fix $K_G$
with $v_2(e_1),v_2(e_2)\ge-K_G$. Split each raw column as
\begin{equation}\label{eq:two-adic-column-types}
\begin{split}
F_j(r)&=S_j(r)+E_j(r)+B_j(r),\\
S_j(r)&=M_{\rm s}(P_r,j),\\
E_j(r)&=e_1\sum_i[t^i]P_r\,c_{(i-j)/2},\\
B_j(r)&=\sum_i[t^i]D_r\Bigl(e_2[i=j]-\tfrac32Z^0(i,j)\Bigr).
\end{split}
\end{equation}
Here $S_j$ is smoothed, $E_j$ carries the $e_1$ kernel and $B_j$ the
$e_2$ kernel.
Coefficient bounds (Chebyshev valuations $v_2([x^u]T_d)\ge u-1$,
$v_2([x^u]U_d)\ge u$, and $v_2(c_l)\ge-2l$) give
$v_2(a_{u,j})\ge u+j-H+1-K_G$ and $v_2(b_{u,j})\ge-2L_2-1-K_G$ with
$L_2=\lfloor\log_2H\rfloor$.
Expand $\det(\mathcal F\mathcal T)$ by Cauchy--Binet and fix a full raw
minor. A term with $m$ columns of type $E$, $l$ of type $B$ and $n-m-l$
of type $S$ has valuation at least
\[
-O_G\bigl(n\log(n+2)\bigr)
+\min_{m+l\le n}\Bigl\{-(H-b)m+\tfrac32m^2+\tfrac12l^2+C(n-m-l)\Bigr\}.
\]
Minimizing over reals, using $H-b=n(1+\delta)$ and $C\ge n$, the minimum
is at $l=n-m$ and equals
\[
\tfrac{n^2}{2}-(2+\delta)nm+2m^2
=2\Bigl(m-\tfrac{(2+\delta)n}{4}\Bigr)^2-\Bigl(\tfrac{\delta}{2}+\tfrac{\delta^2}{8}\Bigr)n^2.
\]
Summing over minors proves \eqref{eq:two-adic-bound}; the same expansion
gives the bound for each raw minor.
\end{proof}
The number $505/4608$ is added to the odd-prime loss in
Section~\ref{sec:odd-primes} to obtain the constant $-2.29084$ of
Proposition~\ref{prop:finite-lower}.
\section{Odd primes and the finite-place lower bound}\label{sec:odd-primes}
Throughout this section assume $G\in\mathbb Q$, so the columns $F_j$ of
Section~\ref{sec:foundations} are rational. For an odd prime $p$,
reduction modulo $p$ always means reduction of a member of $\mathbb Z_p$;
every congruence below includes the assertion that both sides are locally
integral.
The raw matrix has $n+q$ columns while the filtered determinant has size
$n$. At the large primes treated first, each raw column has at most two
powers of $p$ in its denominator. We make invertible column changes over
$\mathbb Z_p$ and bound how many columns can still have denominator $p^2$
or $p$. Such bounds control every full minor of the raw matrix, and
Cauchy--Binet transfers them to the integer column filter. The first
reduction identifies residues after multiplication by $p^2$; for
$p>H/2$ a second reduction identifies the remaining residues after
multiplication by $p$.
\subsection{Digit reduction of the rational moments}
We first prove the reductions needed at primes $2\sqrt H<p\le H$. We may
suppose $p$ does not divide the denominator of $G$: as $N\to\infty$ every
fixed denominator prime eventually lies below $2\sqrt H$. Put
\[
E(t)=(1-t^2)^{(p-1)/2}=\sum_{d=0}^{p-1}E_dt^d\quad\text{in }\mathbb F_p[t],
\qquad \epsilon=(-1)^{(p-1)/2},
\]
and set $E_d=0$ outside $0\le d<p$. Within this range,
\[
E_d=\begin{cases}c_{d/2}\pmod p,&d\text{ even},\\0,&d\text{ odd},\end{cases}
\qquad E_{p-1-d}=\epsilon E_d,
\]
the first from $(-1)^k\binom{(p-1)/2}{k}\equiv4^{-k}\binom{2k}{k}$ and the
second by reversing coefficients.
\begin{lemma}[Digit reductions]\label{lem:odd-digit-reductions}
For $0\le i,j<H$ let
\[
\ell=j\bmod p,\quad d=(j-i-1)\bmod p,
\qquad
i'=\frac{i+1+d-\ell}{p}-1,\quad j'=\Bigl\lfloor\frac jp\Big\rfloor,
\]
with $0\le\ell,d<p$. Then $i'\ge-1$, and, using the convention
$M^0(-1,j')=k^+_{j'+1}$,
\begin{equation}\label{eq:digit-layers}
\begin{split}
p^2M^0(i,j)&\equiv E_dM^0(i',j')\pmod p,\\
p^2Z^0(i,j)&\equiv
\begin{cases}
Z^0(\lfloor i/p\rfloor,j'),&i\equiv j\pmod p,\\
0,&i\not\equiv j\pmod p.
\end{cases}
\end{split}
\end{equation}
\end{lemma}
\begin{proof}
We first give the parity and carry calculation for the factors $H_z^*$ of
\eqref{eq:h-star}. Write $z=Pp+r$, $0\le r<p$, $0\le z<H$. Since
$H<p^2/4$, all factors at reduced indices $P$ are $p$-adic units. If $z$
is even,
\begin{equation}\label{eq:odd-even-H}
\begin{cases}
H_z^*\equiv H_P^*c_{r/2}\pmod p,&r\text{ even},\\
v_p(H_z^*)=1,&r\text{ odd}.
\end{cases}
\end{equation}
When $r$ is even, $P$ is even and the base-$p$ digits of $z/2$ are
$P/2,r/2$, so $(1+x)^p=1+x^p$ gives the first line including $4^{-z/2}$.
When $r$ is odd, $P$ is odd and the low digit of $z/2$ is $(p+r)/2$; the
factorial formula for $\binom z{z/2}$ has exactly one carry,
$\lfloor z/p\rfloor-2\lfloor(z/2)/p\rfloor=1$, with no $p^2$ contribution.
For odd $z$ one has
\begin{equation}\label{eq:odd-odd-H}
\frac1{H_z^*}\in\mathbb Z_p,\qquad
pH_z^*\equiv
\begin{cases}
\epsilon c_{r/2}H_P^*,&r\text{ even},\\
0,&r\text{ odd}.
\end{cases}
\end{equation}
If $r$ is odd, both $z$ and $c_{(z-1)/2}$ are units; if $r$ is even, $P$ is
odd and digit expansion gives $c_{(Pp-1)/2}\equiv\epsilon c_{(P-1)/2}$ and
$pH_{Pp}^*\equiv\epsilon H_P^*$. The recurrence
$H_{z+2}^*/H_z^*=(z+1)/(z+2)$ through $r=0,2,\ldots,p-1$ proves the nonzero
line. Since $m_{z-1}=2H_z^*$ for odd $z$ and vanishes for even $z$,
\[
pm_{z-1}\equiv E_{p-1-r}m_{P-1}\pmod p,
\]
including $z=0$ with $m_{-1}=0$.
For $u=Pp+r$, $0\le u<H$, we claim
\begin{equation}\label{eq:odd-starting-digits}
p^2k^-_u\equiv E_{p-1-r}k^-_P,
\qquad
p^2k^+_{u+1}\equiv E_r k^+_{P+1}\pmod p.
\end{equation}
For odd $u$ write $p^2k^-_u=(pH_u^*)\sum_{z\le u,\,z\text{ odd}}2p/z$;
only $z=mp$ with $m$ odd survives, and if $r$ is even,
\eqref{eq:odd-odd-H} gives
$\epsilon E_rH_P^*\sum_{m\le P}2/m=E_{p-1-r}k^-_P$, while if $r$ is odd
the prefactor vanishes. For even $u$ the quantities $p/(zH_z^*)$ with even
$z\le u$ are integral, so $p^2k^-_u$ is integral and vanishes mod $p$ when
$r$ is odd; when $r$ is even the surviving summands form complete blocks
$z=mp-\lambda$, $m=2,4,\ldots,P$, $\lambda=0,2,\ldots,p-1$, and the
downward recurrence gives
$\frac{p}{(mp-\lambda)H_{mp-\lambda}^*}\equiv\frac{c_{\lambda/2}}{mH_m^*}$.
With $s=(p-1)/2$,
$\sum_{j=0}^s c_j^2\equiv\binom{2s}{s}\equiv\epsilon$, and multiplying by
$H_u^*\equiv E_rH_P^*$ proves the first congruence also for even $u$.
For the second congruence suppose first $u$ even. In
$p^2k^+_{u+1}=H_u^*\sum_{z\equiv u(2)}2p^2/(z^2H_z^*)$ every term with
$p\nmid z$ vanishes since $v_p(H_z^*)\le1$; at $z=mp$, $m$ even and
$H_{mp}^*\equiv H_m^*$, giving $E_rk^+_{P+1}$ when $r$ even and zero when
$r$ odd. For odd $u$ write $(pH_u^*)\sum_{z\text{ odd}}2p/(z^2H_z^*)$; only
$z=mp$ survives, there $pH_{mp}^*\equiv\epsilon H_m^*$ and the two
$\epsilon$ cancel. This proves the second congruence in every case and the
stated local integrality.
If $i\ge j$ put $u=i-j=Pp+r$, so $d=p-1-r$ and $i'=j'+P$, and use
$M^0(i,j)=k^-_u-\sum_{k=1}^j m_{u+k-1}/k$. After multiplication by $p^2$
terms with $p\nmid k$ vanish by $pm_{z-1}\equiv E_{p-1-r}m_{P-1}$, and at
$k=mp$ the surviving term is $E_{p-1-r}m_{P+m-1}/m$, giving
$p^2M^0(i,j)\equiv E_{p-1-r}(k^-_P-\sum_{m=1}^{j'}m_{P+m-1}/m)=E_dM^0(i',j')$.
If $i<j$ put $u=j-i-1=Pp+r$, so $d=r$ and $i'=j'-P-1$, and use
$M^0(i,j)=k^+_{u+1}-\sum_{k=u+2}^j m_{k-u-2}/k$. For a multiple $k=mp$,
$k-u-1=(m-P-1)p+(p-1-r)$, and the reduced summand is $E_rm_{m-P-2}/m$; a
possible first multiple with $m=P+1$ contributes zero, so the reduced sum
runs over $m=P+2,\ldots,j'$ as in $M^0(i',j')$. Also
$i'=\lfloor(i-\ell)/p\rfloor\ge-1$; when $i'=-1$ and $j'=P$ the sum is
empty and the starting term is exactly $k^+_{j'+1}=M^0(-1,j')$. The reduced
indices are below $H/p<p/4$, so the right side is integral at $p$.
For $Z^0$, write $I=\lfloor i/p\rfloor$ and $J=\lfloor j/p\rfloor$. Since
$i,j<p^2$, $pB_i^{(1)}\equiv B_I^{(1)}$ and $p^2B_i^{(2)}\equiv B_I^{(2)}$.
The diagonal formula follows at once. Off the diagonal, if
$i\not\equiv j\pmod p$ the denominator $i-j$ is a unit and $p^2Z^0(i,j)\equiv0$;
if the residues agree, $i-j=p(I-J)$ with $I-J$ a unit, and the first
harmonic congruence gives $Z^0(I,J)$. This also proves integrality of
$p^2Z^0(i,j)$.
\end{proof}
\subsection{The leading layer and paired raw columns}
We gather the moment reductions into column vectors. For a fixed residue
$0\le\ell<p$ define, over $\mathbb F_p$,
\[
P'_u=[t^{(u+1)p+\ell}]\,tP(t)E(t)\ (u\ge-1),\qquad
D'_u=[t^{up+\ell}]\,D(t)\ (u\ge0),
\]
where $P,D$ denote the vectors of all row polynomials, with the dependence
on $\ell$ understood. Their supports are finite.
Lemma~\ref{lem:odd-digit-reductions}, gathered by coefficient residue,
gives
\begin{equation}\label{eq:column-layer}
X_{kp+\ell}:=p^2F_{kp+\ell}\bmod p
=\sum_{u\ge-1}P'_uM^0(u,k)-\frac32\sum_{u\ge0}D'_uZ^0(u,k).
\end{equation}
For a term $[t^i]P$, its residue $d$ satisfies $i+1+d=(i'+1)p+\ell$, so it
contributes exactly to $P'_{i'}$ including $i'=-1$; the $Z^0$ reduction
survives only when $i=up+\ell$, which is the extraction defining $D'_u$.
The $G$ coefficient is a sum of integral multiples of $c_{(i-j)/2}$, hence
integral at every odd prime, so it disappears after multiplication by
$p^2$. Every raw column therefore lies in $p^{-2}\mathbb Z_p^n$.
If $\ell\ge H-2p$, the degree bounds $\deg(tPE)\le H+p-1$ and $\deg D<H$
show $P'_u,D'_u$ vanish for $u>1$. Put
\[
V_\ell=2P'_1-\tfrac32D'_1,\qquad U_\ell=2P'_{-1}-\tfrac32D'_0,
\]
and note the small reduced values
\[
\begin{array}{c|rrr}
u&-1&0&1\\\hline
M^0(u,0)&0&0&2\\
M^0(u,1)&2&0&-2
\end{array}
\qquad
\begin{array}{c|rr}
u&0&1\\\hline
Z^0(u,0)&0&1\\
Z^0(u,1)&1&-1
\end{array}.
\]
Consequently, whenever the indicated columns exist,
\begin{equation}\label{eq:odd-paired-layer}
X_\ell=V_\ell,\qquad X_{p+\ell}=U_\ell-V_\ell.
\end{equation}
Since $tP$ and $D$ have no terms below degree $A+1$, $U_\ell=0$ for
$\ell\le A$.
We record how changes in the full column pool are used. Let $T$ be the $n$
by $(n+q)$ matrix of all raw columns. If $Q\in\operatorname{GL}_{n+q}(\mathbb Z_p)$
and every full minor of $TQ$ has valuation at least $-D$, the same holds
for every full minor of $T=(TQ)Q^{-1}$ by Cauchy--Binet, the coefficients
being minors of the integral matrix $Q^{-1}$. Applying Cauchy--Binet once
more to the integer filter matrix gives $v_p(\Delta_N)\ge-D$. This
transfer applies even when the columns or their leading residues are
linearly dependent.
For $2\sqrt H<p\le H/2$ use each pair of columns with
$\max(b,H-2p)\le\ell<\min(p,L-p,A)$. Their number is exactly
\[
d_0=\bigl(\min(p,L-p,A)-\max(b,H-2p)\bigr)_+,
\qquad y_+=\max(y,0).
\]
Replacing its high column by the sum of the pair is an integral elementary
column operation with integral inverse, and the pair sum lies in
$p^{-1}\mathbb Z_p^n$ by \eqref{eq:odd-paired-layer}, while all other
columns still lie in $p^{-2}\mathbb Z_p^n$. The pairs are disjoint, since
$\ell<p$. Any selection of $n$ columns from the pool of $n+q$ contains at
least $(d_0-q)_+$ of the improved columns, so
\begin{equation}\label{eq:odd-lower-pair-range}
v_p(\Delta_N)\ge-2n+(d_0-q)_+.
\end{equation}
\subsection{The central layer and integral column elimination}
For $p>H/2$ the next lemma identifies the residues of central columns
after multiplication by $p$, in terms of the same vectors $V_\ell$. The
elimination retains noncentral columns and uses $p$ times those columns to
cancel the corresponding $V_\ell$ terms.
\begin{lemma}[Central layer]\label{lem:odd-central-layer}
Suppose $H/2<p\le H$ and write $K=L-p$, $J=H-p$. For every central column,
meaning
\[
b\le j<\min(p,L),\qquad j\ge J,
\]
one has
\begin{equation}\label{eq:central-layer}
pF_j\in\mathbb Z_p^n,\qquad
pF_j\equiv-\sum_{0\le\ell<J}\frac{V_\ell}{j-\ell}\pmod p.
\end{equation}
All denominators $j-\ell$ here are units at $p$.
\end{lemma}
\begin{proof}
For every contributing row degree $0\le i<H$, $i-j\le H-1-J=p-1$ and
$j-i<p$, so $|i-j|<p$ and the starting value $k^-_{i-j}$ or $k^+_{j-i}$ in
the diagonal reduction is integral at $p$. That reduction is the starting
value minus $\sum_{0\le\ell<\min(i,j)}m_{i-\ell-1}/(j-\ell)$ with
$1\le j-\ell<p$. By $pm_{z-1}\equiv E_{p-1-r}m_{P-1}$, a nonzero reduction
after multiplication by $p$ requires $p\le i-\ell<2p$, and then
\[
pm_{i-\ell-1}\equiv2E_{2p+\ell-1-i}.
\]
Such an index has $0\le\ell<J$, and conversely its coefficient can be
collected over all $i$ without truncation since $j\ge J>\ell$ and
$i\ge p+\ell>\ell$. Hence
$pM^0(P,j)\equiv-\sum_{0\le\ell<J}2P'_1/(j-\ell)$ with $P'_1$ the extracted
vector of the residue $\ell$. For $Z^0(i,j)$, the harmonic sums have a pole
only if $i=p+\ell$ with $0\le\ell<J$, where the denominator $i-j$ is a unit
(equality mod $p$ would need $j=\ell<J$); then
$pZ^0(p+\ell,j)\equiv1/(\ell-j)$, and all other $Z^0(i,j)$ are integral.
Combining these with the factor $-3/2$ proves \eqref{eq:central-layer}; the
$G$ term is integral here as before.
\end{proof}
We cancel in the full column pool, where $R$ counts retained columns with
possible denominator $p^2$ and $S$ bounds the other columns with possible
denominator $p$; every remaining column is integral.
\begin{lemma}[Elimination over the local integers]\label{lem:odd-local-elimination}
For $H/2<p\le H$ define
\begin{align*}
R&=K_++(K-A)_++(J-\max(b,K))_+,\\
S&=(\min(A,K)-b)_++(\min(b,J)-\max(0,K))_+.
\end{align*}
Then every full raw minor, and hence the filtered determinant, satisfies
\begin{equation}\label{eq:odd-two-layer-bound}
v_p(\Delta_N)\ge-\min(2n,\,n+R,\,2R+S).
\end{equation}
\end{lemma}
\begin{proof}
The noncentral columns are the high columns $F_{p+\ell}$ for $0\le\ell<K$
and the low columns $F_\ell$ for $b\le\ell<J$ (empty ranges allowed: $p>b$,
$J<p$, $K<J$). For $b\le\ell<\min(A,K)$ retain the low column and replace
the high column by $F_{p+\ell}+F_\ell$. Let
$B=(\min(A,K)-b)_+$ be the number of these pair sums, each in
$p^{-1}\mathbb Z_p^n$. Retain every other noncentral column as a column in
$p^{-2}\mathbb Z_p^n$. The number retained is
\begin{equation}\label{eq:odd-retained-count}
R_0=K_++(J-b)_+-B,
\end{equation}
which equals the stated $R$ (split $K$ into $K\le0$, $0<K\le b$,
$b<K\le A$, $K>A$, using $K<J$ in the last two cases).
These retained columns represent every $V_\ell$ with $0\le\ell<J$ except
possibly those in
\[
\mathcal E=\{\ell:\max(0,K)\le\ell<\min(b,J)\},\quad
e=|\mathcal E|=(\min(b,J)-\max(0,K))_+.
\]
For $b\le\ell<J$ the retained low column has leading residue
$p^2F_\ell\equiv V_\ell$; for $0\le\ell<\min(b,K)$ the unpaired high column
has leading residue $-V_\ell$ since $\ell<b<A$ makes $U_\ell=0$. For each
represented residue choose a retained column $Y_\ell$ and sign
$\sigma_\ell$ with $p^2Y_\ell\equiv\sigma_\ell V_\ell$, and for each central
$j$ choose $a_{\ell j}\in\mathbb Z_p$ reducing to $\sigma_\ell/(j-\ell)$ and
replace that column by
\[
C_j=F_j+p\sum_{\substack{0\le\ell<J\\\ell\notin\mathcal E}}a_{\ell j}Y_\ell.
\]
This is a shear of the whole pool with integral coefficients and integral
inverse, preserving every retained column and pair sum; the resulting
central columns lie in $p^{-1}\mathbb Z_p^n$ and
$pC_j\equiv-\sum_{\ell\in\mathcal E}V_\ell/(j-\ell)\pmod p$. No information
about the next coefficient of $Y_\ell$ is used: if
$Y_\ell=p^{-2}y_0+p^{-1}y_1+y_2$, then $pY_\ell=p^{-1}y_0+y_1+py_2$ has an
integral unknown second coefficient.
Let $c$ be the number of central columns and let their scaled residues
define $\varphi:\mathbb F_p^c\to\mathbb F_p^n$,
$(a_j)_j\mapsto\sum_j a_j(pC_j\bmod p)$. Its image lies in the span of the
$e$ exceptional vectors, so $d:=\operatorname{rank}\varphi\le\min(c,e)$.
Choose a basis of $\mathbb F_p^c$ whose last $c-d$ vectors span
$\ker\varphi$, lift its basis matrix to $W\in\operatorname{Mat}_c(\mathbb Z_p)$;
then $W\in\operatorname{GL}_c(\mathbb Z_p)$ by the adjugate formula. This
change leaves at most $d$ columns in $p^{-1}\mathbb Z_p^n$; the other
$c-d$ are integral since their scaled residues vanish and they already
lay in $p^{-1}\mathbb Z_p^n$. If $c=0$ this is empty.
The pool now has $R$ columns with possible denominator $p^2$, at most
$B+d\le B+e=S$ other columns with possible denominator $p$, and integral
remaining columns. In a full minor with $r$ columns from the first group
and $s$ from the second, the loss is at most $2r+s$, and
$2r+s\le2n$, $2r+s\le n+R$, $2r+s\le2R+S$ hold since $r+s\le n$, $r\le R$,
$s\le S$. All changes and inverses are integral, so Cauchy--Binet proves
\eqref{eq:odd-two-layer-bound} for all original full minors and for
$\Delta_N$.
\end{proof}
The omission of the optional pair at $\ell=A$ is harmless: its physical
column is included in $R$. The proof also covers $K\le0$ and $J\le b$; in
particular, if $L\le p\le H$, then $R=0$ and $S=H-p$, and at the formal
endpoint $p=H$ all columns are integral.
\subsection{Summing the prime losses}
For a nonzero rational determinant the weighted valuations sum to its
logarithmic absolute value; combining the odd-prime estimates with the
bound at $2$ gives the lower bound used in the final comparison.
\begin{proposition}[Finite-place bound]\label{prop:finite-lower}
Assume $G\in\mathbb Q$. Along any sequence of positive integers $N\to\infty$
for which $\Delta_N\ne0$, one has
\begin{equation}\label{eq:finite-lower-final}
\liminf_{N\to\infty}
\Bigl(\frac{\log|\Delta_N|}{n^2}-\frac12\log2\Bigr)
\ge -\frac{8609}{4608}-\Bigl(\frac12+\frac{505}{4608}\Bigr)\log2
>-2.29084.
\end{equation}
\end{proposition}
\begin{proof}
Put $x=p/N$. For $x\le65/2$, \eqref{eq:odd-lower-pair-range} gives
\[
\frac{d_0}{N}=\bigl(\min(x,59-x,19)-\max(7,65-2x)\bigr)_+,
\]
which is $0$ for $0\le x\le23$, $2x-46$ for $23\le x\le29$, and $12$ for
$29\le x\le65/2$. Only the part exceeding $q/N=4$ improves a full minor.
For $x>65/2$ the formulas of Lemma~\ref{lem:odd-local-elimination} give the
following values, retaining the intermediate breakpoint $52$:
\[
\begin{array}{c|cc}
x\text{ range}&R/N&S/N\\\hline
[65/2,40]&105-2x&12\\
{}[40,52]&65-x&52-x\\
{}[52,58]&117-2x&x-52\\
{}[58,59]&59-x&6\\
{}[59,65]&0&65-x
\end{array}.
\]
The entries agree at common endpoints. Taking
$\min(96,48+R/N,2R/N+S/N)$ yields the extra breakpoint $69/2$, so for every
relevant prime $p>2\sqrt H$ the valuation is at least $-Nd(p/N)$ where $d$
is continuous, vanishes for $x\ge65$, and has the exact pieces
\begin{equation}\label{eq:odd-loss-table}
\begin{array}{c|c|c|c}
x\text{ range}&d(x)&\text{endpoints}&\displaystyle\int_{\text{range}}d\,dx\\\hline
[0,25]&96&96,96&2400\\
{}[25,29]&146-2x&96,88&368\\
{}[29,65/2]&88&88,88&308\\
{}[65/2,69/2]&153-2x&88,84&172\\
{}[69/2,40]&222-4x&84,62&803/2\\
{}[40,58]&182-3x&62,8&630\\
{}[58,59]&124-2x&8,6&7\\
{}[59,65]&65-x&6,0&18
\end{array}.
\end{equation}
In particular
\begin{equation}\label{eq:odd-loss-area}
\int_0^{65}d(x)\,dx=\frac{8609}{2}.
\end{equation}
The omitted small odd primes cost only $o(n^2)$. At any odd prime the
factorial formula gives $0\le v_p\binom{2l}{l}\le1+\log H/\log p$ for
$2l\le H$, and the displayed formulas for $H_z^*,m_i,k_d^\pm,Z^0$ and the
diagonal reduction give, with an absolute constant $C_0$ and with
$\operatorname{den}G$ the reduced positive denominator of $G$,
\[
v_p(F_j(r))\ge-C_0\Bigl(1+\frac{\log H}{\log p}\Bigr)-v_p(\operatorname{den}G).
\]
Expanding a determinant multiplies this by at most $n$, and summing over
$p\le2\sqrt H$ with weights $\log p$ gives
$O(n\sqrt H\log H+n\log(\operatorname{den}G))=o(n^2)$. For large $N$ every
prime $p>H$ is integral throughout, so these primes contribute nothing.
We use the prime number theorem $\theta(y)=\sum_{p\le y}\log p\sim y$
\cite[Step~VI, p.~707]{Zagier1997}. Fix a partition
$0=x_0<\cdots<x_m=65$; the theorem gives
$(\theta(Nx_j)-\theta(Nx_{j-1}))/N\to x_j-x_{j-1}$, and step functions
from the supremum and infimum of $d$ on each interval bound the weighted
sum, whose limiting difference tends to zero as the mesh shrinks. Thus
\begin{equation}\label{eq:odd-weighted-primes}
\frac1N\sum_{p\le65N}d(p/N)\log p\longrightarrow\int_0^{65}d(x)\,dx.
\end{equation}
The prime $2$ and primes below $2\sqrt H$ may be removed at cost $o(1)$,
using $\|d\|_\infty=96$ and $\theta(2\sqrt H)\le2\sqrt H\log(2\sqrt H)$.
Combining with \eqref{eq:odd-loss-area} and $n^2=2304N^2$,
\[
\sum_{p\text{ odd}}v_p(\Delta_N)\log p\ge-\frac{8609}{4608}n^2-o(n^2).
\]
Proposition~\ref{prop:two-adic} with $\delta=10/48$ gives
\[
v_2(\Delta_N)\ge-\Bigl(\frac\delta2+\frac{\delta^2}{8}\Bigr)n^2-o(n^2)
=-\frac{505}{4608}n^2-o(n^2).
\]
For a nonzero rational, $\log|\Delta_N|=\sum_pv_p(\Delta_N)\log p$; dividing
by $n^2$ and subtracting $\tfrac12\log2$ proves \eqref{eq:finite-lower-final}
with $n=48N$, so $n^2=2304N^2$ and $N\to\infty$ along the same sequence.
The last comparison has a short rational check. The identity
$\log2=2\sum_{j\ge0}3^{-(2j+1)}/(2j+1)$ and its positive geometric tail give
\[
\log2\le2\sum_{j=0}^{5}\frac{3^{-(2j+1)}}{2j+1}
+\frac{2\cdot3^{-13}}{13(1-3^{-2})}<\frac{693149}{10^6}.
\]
So the lower constant exceeds
\[
-\frac{8609}{4608}-\frac{2809}{4608}\frac{693149}{10^6}
=-\frac{10556055541}{4608000000}>-2.29084,
\]
completing the proof.
\end{proof}
\section{Nonvanishing along the prime sequence}\label{sec:nonvanishing}
The next proposition supplies the nonzero determinants needed in
Proposition~\ref{prop:finite-lower}.
\begin{proposition}\label{prop:nonvanishing}
Assume $G\in\mathbb Q$. For every sufficiently large prime $p$, the
determinant with scale parameter $N=p$ satisfies
\begin{equation}\label{eq:nonvanishing-exact-valuation}
v_p(\Delta_p)=-96p.
\end{equation}
In particular $\Delta_p\ne0$ for every sufficiently large prime $p$.
\end{proposition}
We reduce a matrix of size $48p$ to three fixed rational matrices of size
$48$. Throughout, an underlined row polynomial is formed with $N=1$, so
\[
n_0=48,\quad b_0=7,\quad q_0=g_0=4,\quad h_0=2,\quad
L_0=59,\quad C_0=63,\quad H_0=65.
\]
In addition to the usual rows $0,\ldots,47$, define the auxiliary row $48$
by the same formulas as in \eqref{eq:rows}. For $0\le r\le48$ and
$0\le k<48$ put
\begin{equation}\label{eq:nonvanishing-base-matrix}
\mathcal B(r,k)=\sum_{j=0}^{4}(-1)^j\binom4j
\bigl\{M^0(\underline P_r,7+k+j)-\tfrac32 Z^0(\underline D_r,7+k+j)\bigr\},
\end{equation}
with $M^0,Z^0$ acting linearly on the first argument. Thus $\mathcal B$ is
a fixed rational $49\times48$ matrix. Let
\[
\mathcal B_0=(\mathcal B(r,k))_{0\le r,k<48},\qquad
\mathcal B_1=(\mathcal B(r+1,k))_{0\le r,k<48}.
\]
The certificate below proves
\begin{equation}\label{eq:nonvanishing-fixed-determinants}
\det\mathcal B_0\ne0,\qquad
\det(\mathcal B_0+\mathcal B_1)\ne0,\qquad
\det(\mathcal B_0-\mathcal B_1)\ne0.
\end{equation}
We first show why these three fixed facts imply the proposition.
\subsection{Two palindromic bases}
Let $p$ be an odd prime and work over $\mathbb F_p$. The space
\[
\mathcal P_p=\{Q\in\mathbb F_p[w]:\deg Q\le2p-2,\ w^{2p-2}Q(1/w)=Q(w)\}
\]
has dimension $p$. For $0\le\ell,i<p$ consider
\[
U_\ell(w)=(2w)^\ell(1+w^2)^{p-1-\ell},\qquad
Q_i(w)=w^i\sum_{j=0}^{p-i-1}w^{2j}.
\]
Both families lie in $\mathcal P_p$, with lowest terms $2^\ell w^\ell$ and
$w^i$. Triangularity in degrees $0,\ldots,p-1$ makes each a basis. Define
$a_{\ell i}\in\mathbb F_p$ by $Q_i=\sum_\ell a_{\ell i}U_\ell$ and
$\mathbf a=(a_{\ell i})$. Lowest-term comparison gives
\begin{equation}\label{eq:nonvanishing-triangular-unit}
a_{\ell i}=0\ (\ell<i),\qquad a_{ii}=2^{-i},\qquad
\det\mathbf a=2^{-p(p-1)/2}\ne0,
\end{equation}
so this varying-size matrix introduces no exceptional odd primes.
Let $Q'_0=0$ and $Q'_i=Q_{p-i}$ for $1\le i<p$, written
$Q'_i=\sum_\ell b_{\ell i}U_\ell$. Then
\begin{equation}\label{eq:nonvanishing-transition}
\mathbf b=\mathbf a\Pi,\qquad \Pi e_0=0,\qquad \Pi e_i=e_{p-i}\ (1\le i<p),
\end{equation}
with $e_i$ the coordinate vectors. Explicitly
$Q_i=w^i(1-w^{2(p-i)})/(1-w^2)$ and $Q'_i=w^{p-i}(1-w^{2i})/(1-w^2)$. Their
sum is
\[
Q_i+w^pQ'_i=w^i\sum_{j=0}^{p-1}w^{2j}=w^i(1-w^2)^{p-1},
\]
since $\binom{p-1}{j}=(-1)^j$ in $\mathbb F_p$. With $t=2w/(1+w^2)$,
$f=(1-w^2)/(1+w^2)$, $E(t)=(1-t^2)^{(p-1)/2}=f^{p-1}$, division by
$(1+w^2)^{p-1}$ gives the identity
\begin{equation}\label{eq:nonvanishing-palindromic-expansion}
E(t)w^i=\sum_{\ell=0}^{p-1}t^\ell\bigl(a_{\ell i}+b_{\ell i}w^p\bigr).
\end{equation}
\subsection{Frobenius and the two extraction shifts}
Set $N=p$, write each row index uniquely as $r=pr_0+i$ with $0\le r_0<48$
and $0\le i<p$, and put $t'=t^p$, $w'=w^p$, $f'=f^p$. Frobenius gives
$t'=2w'/(1+(w')^2)$, $f'=(1-(w')^2)/(1+(w')^2)$ and $(f')^2=1-(t')^2$.
With $C=63p$, $h=2p$, $g=4p$ in \eqref{eq:rows},
\begin{align}
tR_rE(t)
&=(1-t')^2(t')^{63}(w')^{r_0-4}E(t)w^i\notag\\
&=\sum_{\ell=0}^{p-1}t^\ell t'
\bigl(a_{\ell i}\underline R_{r_0}(t',w')+b_{\ell i}\underline R_{r_0+1}(t',w')\bigr).
\label{eq:nonvanishing-frobenius-row}
\end{align}
Here the single factor $w'$ in the second term creates one successor row of
index at most $48$. Since star fixes $t,t'$ and negates $f,f'$, and
$fE=f'$, the row identities give $tR_rE=tP_rE-f'D_r$ and
$t'\underline R_j=t'\underline P_j-f'\underline D_j$. Taking the
star-invariant part of \eqref{eq:nonvanishing-frobenius-row}, then its
anti-invariant part and cancelling $f'$, yields
\begin{align}
tP_r(t)E(t)&=\sum_{\ell=0}^{p-1}t^\ell t'
\bigl(a_{\ell i}\underline P_{r_0}(t')+b_{\ell i}\underline P_{r_0+1}(t')\bigr),
\label{eq:nonvanishing-P-identity}\\
D_r(t)&=\sum_{\ell=0}^{p-1}t^\ell
\bigl(a_{\ell i}\underline D_{r_0}(t')+b_{\ell i}\underline D_{r_0+1}(t')\bigr).
\label{eq:nonvanishing-D-identity}
\end{align}
These are polynomial identities, since both sides are polynomials and
$t=2w/(1+w^2)$ is injective on $\mathbb F_p[t]$.
In the notation of \eqref{eq:column-layer} the extractions for a fixed
residue $\ell$ are
$P'_u=[t^{(u+1)p+\ell}]\,tP_rE$ and $D'_u=[t^{up+\ell}]\,D_r$. Every
polynomial is uniquely $\sum_\ell t^\ell A_\ell(t^p)$, so the last two
identities give
\begin{equation}\label{eq:nonvanishing-extracted-rows}
(P'_u,D'_u)=[(t')^u]
\bigl\{a_{\ell i}(\underline P_{r_0},\underline D_{r_0})(t')
+b_{\ell i}(\underline P_{r_0+1},\underline D_{r_0+1})(t')\bigr\}.
\end{equation}
The $t'$ in the first identity is exactly consumed by the $u+1$ shift in
the $P$ extraction; the $D$ extraction has none, and $P'_{-1}=0$ here.
There is a degree issue at the last successor row. The auxiliary
polynomials have
\begin{equation}\label{eq:nonvanishing-auxiliary-degrees}
\min\deg\underline P_{48}=62-44=18,\qquad
\min\deg\underline D_{48}=63-44=19.
\end{equation}
For the last block $r=47p+i$, $i>0$, \eqref{eq:nonvanishing-transition}
gives $b_{\ell i}=0$ for $\ell<p-i$ and $b_{p-i,i}=2^{-(p-i)}\ne0$, so the
successor contribution starts at $19p+(p-i)=20p-i$, while the first-row
contribution starts no earlier than $20p+i$. Directly from the rows,
$|r-g|=43p+i$ makes $tP_r$ and $D_r$ start at $C-|r-g|=20p-i$, matching
the extraction since $E(0)=1$. The leading Chebyshev coefficients are
powers of $2$, nonzero in odd characteristic; when $i=0$ the successor term
vanishes and the first term starts at $20p$. Thus the lower degree of the
auxiliary row injects no forbidden coefficient, and its contact
$t\underline R_{48}/f=O(t^{107})$ holds since $63+48-4=107>59$.
\subsection{The residue blocks and their determinant}
Take $p>260$ and exclude the primes dividing the denominator of the
hypothetical rational $G$. Then $p>2\sqrt{65p}=2\sqrt H$, so
\eqref{eq:digit-layers} and \eqref{eq:column-layer} apply at this same
prime. The rational raw entry is
\[
F_j(r)=M^0(P_r,j)-\tfrac32Z^0(D_r,j)+4G\sum_v[t^v]P_r\,c_{(v-j)/2},
\]
whose last sum is $p$-integral (integer polynomial coefficients, and every
$c_d$ integral at odd $p$), so the $G$ term vanishes after multiplication by
$p^2$. All scaled entries are $p$-integral by the cited layer formulas.
Substituting \eqref{eq:nonvanishing-extracted-rows} into
\eqref{eq:column-layer} gives
\begin{align}
p^2F_{kp+\ell}(pr_0+i)
\equiv{}&a_{\ell i}\bigl\{M^0(\underline P_{r_0},k)-\tfrac32Z^0(\underline D_{r_0},k)\bigr\}
\notag\\
&+b_{\ell i}\bigl\{M^0(\underline P_{r_0+1},k)-\tfrac32Z^0(\underline D_{r_0+1},k)\bigr\}
\pmod p.
\label{eq:nonvanishing-raw-bridge}
\end{align}
The base entries have denominators with prime factors at most $65$, so
their reductions exist for $p>260$.
The filter respects residues: writing $k=\ell+pk_0$ with $0\le\ell<p$ and
$0\le k_0<48$,
\begin{equation}\label{eq:nonvanishing-residue-filter}
s^{7p+k}(1-s)^{4p}=s^\ell(s^p)^{7+k_0}(1-s^p)^4\quad\text{in }\mathbb F_p[s].
\end{equation}
Multiplying raw entries by $p^2$ before reduction makes this a valid column
operation (coefficient differences divisible by $p$ multiply integral
scaled entries). The fourth difference uses exactly the five base indices
$7+k_0,\ldots,11+k_0$, all in $7,\ldots,58$. Each residue group thus holds
exactly $48$ base columns, and no operation changes $\ell$.
Order rows by $(i,r_0)$ and columns by $(\ell,k_0)$. The reduction of the
scaled $48p\times48p$ matrix has $(i,\ell)$ block
$a_{\ell i}\mathcal B_0+b_{\ell i}\mathcal B_1$. Writing this matrix as
$\mathcal L_p$,
\begin{equation}\label{eq:nonvanishing-block-matrix}
\begin{split}
\mathcal L_p&=\mathbf a^T\otimes\mathcal B_0+\mathbf b^T\otimes\mathcal B_1,\\
\mathcal L_p\bigl((\mathbf a^T)^{-1}\otimes I_{48}\bigr)
&=I_p\otimes\mathcal B_0+\Pi^T\otimes\mathcal B_1,
\end{split}
\end{equation}
using $\mathbf b^T=\Pi^T\mathbf a^T$.
The nonzero indices $1,\ldots,p-1$ split into $(p-1)/2$ pairs $\{i,p-i\}$,
on which $e_i+e_{p-i}$ and $e_i-e_{p-i}$ are eigenvectors of $\Pi$ with
eigenvalues $1,-1$; with $e_0$ of eigenvalue $0$ they form a basis since $2$
is invertible. Thus $\Pi^T$ has eigenvalue multiplicities
$1,(p-1)/2,(p-1)/2$ for $0,1,-1$. A similarity on the $p$-dimensional
factor proves the exact block identity
\begin{align}
\det\mathcal L_p
={}&(\det\mathbf a)^{48}\det\mathcal B_0\notag\\
&\cdot\det(\mathcal B_0+\mathcal B_1)^{(p-1)/2}\det(\mathcal B_0-\mathcal B_1)^{(p-1)/2}.
\label{eq:nonvanishing-factorization}
\end{align}
There is no factor from the similarity; returning to the original orders
changes at most the sign.
\subsection{An exact certificate for the fixed matrices}
Every entry of \eqref{eq:nonvanishing-base-matrix} is constructed by the
following rational arithmetic, which also fixes its reduction mod $101$.
Form $m_0=2$, $m_i=0$ for odd $i$, and $m_i=i m_{i-2}/(i+1)$ for even
$2\le i\le64$. Set $k^-_0=0$, $k^-_1=2$, $k^+_0=k^+_1=0$ and
\begin{align*}
k^-_d&=\frac{(d-1)k^-_{d-2}+m_{d-2}+m_{d-1}}{d} &&(2\le d\le64),\\
k^+_d&=\frac{(d-2)k^+_{d-2}+2/(d-1)}{d-1} &&(2\le d\le58).
\end{align*}
For $0\le i\le64$, $0\le j\le58$ form
$Y_{i0}=k^-_i$, $Y_{0j}=k^+_j$, $Y_{ij}=Y_{i-1,j-1}-m_{i-1}/j$ ($i,j\ge1$),
so $Y_{ij}=M^0(i,j)$. With $B_i^{(d)}=\sum_{u=1}^i u^{-d}$ and
$B_0^{(d)}=0$, set
\[
J_{ij}=\tfrac32\begin{cases}-B_i^{(2)},&i=j,\\[2pt](B_i^{(1)}-B_j^{(1)})/(i-j),&i\ne j,\end{cases}
\]
so $J_{ij}=\tfrac32Z^0(i,j)$ with its diagonal sign included.
For the polynomials define $E_0=t^{62}$, $E_1=t^{61}$, $I_0=0$, $I_1=t^{62}$
and, for either family $S=E,I$, $S_m=2S_{m-1}/t-S_{m-2}$ ($2\le m\le44$).
These integer polynomials are $E_m=t^{62}T_m(1/t)$ and
$I_m=t^{62}U_{m-1}(1/t)$. For $0\le r\le48$ let $u=|r-4|$ and form, in
degrees $0,\ldots,64$,
\[
y_r(t)=(1-t)^2E_u(t),\qquad
z_r(t)=\operatorname{sign}(r-4)(1-t)^2I_u(t).
\]
They are exactly $\underline P_r$ and $\underline D_r$. Contract to the
length-$52$ vector
\[
v_r(j)=\sum_{i=0}^{64}\bigl([t^i]y_r\,Y_{ij}-[t^i]z_r\,J_{ij}\bigr),
\qquad j=7,\ldots,58.
\]
Replacing $v$ four times successively by $(v(j)-v(j+1))_j$ gives the
length-$48$ row $r$ of $\mathcal B$, including the auxiliary row's
degree-$18$ term.
Every scalar denominator here is a product of nonzero integers of absolute
value at most $65$, so each is a unit mod $101$; polynomial division by
$t$ only shifts exponents of polynomials divisible by $t$. Hence the whole
construction runs over $\mathbb F_{101}$ and agrees with reduction of the
rational matrices.
The elimination rule is: for $\sigma\in\{0,1,-1\}$ start with
$R_i=\mathcal B(i,\cdot)+\sigma\mathcal B(i+1,\cdot)$, $0\le i<48$, in
increasing order; at step $i$, if $R_i(i)=0$, swap row $i$ with the first
later row having a nonzero entry in column $i$; record $d_i=R_i(i)$ and
replace every row $j>i$ by $R_j-(R_j(i)/d_i)R_i$. There are no swaps for
$\sigma=0,-1$; for $\sigma=1$ the only swaps (0-based) are $(30,31)$ and
$(45,46)$. The pivots over $\mathbb F_{101}$ are, $16$ per line:
\begin{center}
\small
\begin{tabular}{cl}
\toprule
$\sigma$ & Successive pivots in $\mathbb F_{101}$\\
\midrule
$0$ & $60,68,62,79,47,32,69,57,30,72,35,66,43,20,85,48$\\
& $88,4,77,54,60,79,26,68,83,39,40,65,1,68,78,24$\\
& $15,98,32,22,94,9,99,10,15,75,4,2,25,53,90,79$\\[2pt]
$1$ & $38,51,90,70,5,21,88,55,45,20,35,41,77,10,18,25$\\
& $76,14,38,72,6,66,56,35,83,58,56,11,17,20,30,24$\\
& $28,11,25,70,79,99,66,38,4,41,63,91,63,17,90,98$\\[2pt]
$-1$ & $82,92,26,87,21,74,87,88,88,3,14,23,38,58,36,20$\\
& $26,33,94,74,78,45,93,86,73,66,45,30,61,3,88,27$\\
& $20,58,69,48,78,39,48,1,66,18,86,93,52,92,39,77$\\
\bottomrule
\end{tabular}
\end{center}
Every listed pivot is nonzero, so the swaps and row additions are
invertible and all three matrices are invertible mod $101$, hence have
nonzero determinants over $\mathbb Q$; this proves
\eqref{eq:nonvanishing-fixed-determinants}. The modulus $101$ serves only
this fixed certificate; the growing determinant uses arbitrary large
primes as follows.
\begin{proof}[Completion of the proof of Proposition~\ref{prop:nonvanishing}]
Let $\mathcal E$ consist of $2$, the primes dividing the denominator of
$G$, the primes dividing any denominator of an entry of $\mathcal B$, and
the primes dividing the numerator or denominator of the three nonzero
rational determinants in \eqref{eq:nonvanishing-fixed-determinants}; this is
a fixed finite set. For $p>260$ outside $\mathcal E$ the three fixed
matrices reduce to invertible matrices over $\mathbb F_p$, and
\eqref{eq:nonvanishing-triangular-unit} together with
\eqref{eq:nonvanishing-factorization} give $\det\mathcal L_p\ne0$.
Let $\mathcal F_p$ be the original filtered matrix with determinant
$\Delta_p$, of size $n=48p$; then $p^2\mathcal F_p$ is integral over
$\mathbb Z_{(p)}$, and its reduction after the stated row and column
permutations is $\mathcal L_p$. Hence $p^{2n}\Delta_p=\det(p^2\mathcal F_p)$
is a unit in $\mathbb Z_{(p)}$, so $v_p(\Delta_p)=-2n=-96p$. Every
sufficiently large prime lies outside $\mathcal E$ and exceeds $260$.
\end{proof}
\section{The real-place determinant estimates}\label{sec:real-place}
We estimate the determinant \eqref{eq:determinant} without any rationality
assumption. Write
\begin{equation}\label{eq:real-normalized-parameters}
\alpha=\frac an=\frac{11}{48},\quad
\beta=\frac bn=\frac7{48},\quad
\gamma=\frac gn=\frac qn=\frac4{48},\quad
\eta=\frac hn=\frac2{48},\quad
D_*=n-1-2g=40N-1,
\end{equation}
so $0<D_*<n$ and $n\ge48$. For a real list $y$ of length $n$ let
$V(y)=\prod_{i<j}(y_j-y_i)$ and $\mathcal V(y)=|V(y)|$. All constants below
are independent of the locations and separations of the integration
variables.
\subsection{The integral and its two sheets}
Put $d\mu(t)=|t|\,dt/f(t)$ on $(-1,1)$ and set
$A_r(t)=P_r(t)-\frac32\mathbf 1_{\{t>0\}}\frac{f(t)}tD_r(t)$ and
$\psi_k(s)=s^{b+k}(1-s)^q$. The value at $t=0$ is immaterial; the
combination has a finite limit there because $D_r(t)/t$ is a polynomial.
The determinant entries are exactly
$\int_{-1}^1\int_0^1A_r(t)\psi_k(s)/(1-ts)\,ds\,d\mu(t)$. The measure has
mass $\mu((-1,1))=2$, and the scalar majorant
$\int\!\int(1-|t|s)^{-1}ds\,d\mu(t)$ is finite since integration in $s$
gives at worst an integrable logarithmic singularity at $|t|=1$; the
functions $A_r,\psi_k$ are bounded for fixed $n$. Thus all permutation
expansions below are absolutely integrable.
Applying Andr\'eief's identity twice
\cite[Equation~(1.7) and Section~2.2]{Forrester2018Andreief}, expanding
determinants and relabelling,
\begin{equation}\label{eq:real-double-integral}
\Delta_N=\frac1{(n!)^2}\int_{(-1,1)^n}\int_{(0,1)^n}
\det[A_r(t_i)]\,\det\!\Bigl[\frac1{1-t_is_j}\Bigr]\,
\det[\psi_k(s_j)]\,ds\,d\mu^n(t).
\end{equation}
Fubini applies by the majorant. Cauchy's double alternant
\cite[Section~2.1, Equation~(2.7)]{Krattenthaler1999} is
\begin{equation}\label{eq:real-cauchy-determinant}
\det\!\Bigl[\frac1{1-t_is_j}\Bigr]=\frac{V(t)V(s)}{\prod_{i,j}(1-t_is_j)};
\end{equation}
multiplying by the denominator gives a polynomial alternating separately in
$t$ and $s$, and expanding at $s=0$ shows the constant is $1$. Also
$\det[\psi_k(s_j)]=V(s)\prod_js_j^b(1-s_j)^q$.
Use the real coordinate
\begin{equation}\label{eq:real-coordinate}
x_i=\frac{t_i}{1+\sqrt{1-t_i^2}},\qquad
t_i=\frac{2x_i}{1+x_i^2},\qquad
d\mu(t_i)=\frac{4|x_i|}{(1+x_i^2)^2}\,dx_i,
\end{equation}
which maps $(-1,1)$ bijectively to $(-1,1)$. Except on a null set we may
assume the $x_i$ are distinct and nonzero. Define
\[
(\xi_{\mathrm f}(x),\xi_{\mathrm n}(x))=\begin{cases}(1/2,1/2),&x<0,\\(-1/4,5/4),&x>0,\end{cases}
\qquad
\mathcal R_n(x)=\det[\xi_{\mathrm f}(x_i)x_i^{g-r}+\xi_{\mathrm n}(x_i)x_i^{r-g}].
\]
The subscripts refer to evaluation at $x_i$ (near) and $1/x_i$ (far),
inside and outside the unit circle. Since $fD_r/t=(R_r^*-R_r)/2$, the
negative half-interval has row $(R_r+R_r^*)/2$ and the positive half has
$(5R_r-R_r^*)/4$; factoring $(1-t_i)^ht_i^{C-1}$ from column $i$ gives the
exact identity
\begin{equation}\label{eq:real-sheet-integral}
\Delta_N=\frac1{(n!)^2}\int\!\int\mathcal R_n(x)\,
\frac{V(t)V(s)^2}{\prod_{i,j}(1-t_is_j)}
\prod_js_j^b(1-s_j)^q\prod_it_i^{C-1}(1-t_i)^h\,ds\,d\mu^n(t).
\end{equation}
There is one $t$-Vandermonde and two $s$-Vandermondes, and all denominators
are positive.
\subsection{A uniform interpolating function}
Put $c_i=\xi_{\mathrm n}(x_i)/\xi_{\mathrm f}(x_i)$, nonzero on both
halves. For $0\le r<n$,
\begin{equation}\label{eq:real-sheet-factorization}
\xi_{\mathrm f}(x_i)x_i^{g-r}+\xi_{\mathrm n}(x_i)x_i^{r-g}
=\xi_{\mathrm f}(x_i)x_i^{g-(n-1)}
\bigl(x_i^{n-1-r}+c_i x_i^{D_*}x_i^r\bigr).
\end{equation}
Thus a holomorphic $h_*$ with $h_*(x_i)=c_ix_i^{D_*}$ turns the bracket into
$x_i^{n-1-r}+h_*(x_i)x_i^r$, and we bound the resulting evaluation
determinant through the norm of $h_*$. The first estimate applies when
\begin{equation}\label{eq:case-condition}
\sum_{i=1}^n\frac{1-x_i^2}{1+x_i^2}\le D_*.
\end{equation}
The construction is uniform even when the nodes cluster.
\begin{lemma}\label{lem:real-uniform-interpolation}
Let $x_1,\ldots,x_n$ be distinct nonzero reals in $(-1,1)$ satisfying
\eqref{eq:case-condition}. There is a function $h_*$ holomorphic on a
neighborhood of the closed unit disk with
\[
h_*(x_i)=c_ix_i^{D_*},\qquad
c_i=\begin{cases}1,&x_i<0,\\-5,&x_i>0,\end{cases}
\qquad
\sup_{|z|\le1}|h_*(z)|\le K_0n,\quad K_0=10e^{12}.
\]
The neighborhood may depend on the nodes; the bound does not.
\end{lemma}
\begin{proof}
Set $B(z)=\prod_i(z-x_i)/(1-x_iz)$ and $F(z)=z^{D_*}/B(z)$. For
$0<u\le1$ and real $0<|x|<1$,
\[
\frac{d}{d\log u}\frac12\log\frac{u^2+x^2}{1+x^2u^2}
=\frac{(1-x^4)u^2}{(u^2+x^2)(1+x^2u^2)}\le\frac{1-x^2}{1+x^2},
\]
using $u^2+u^{-2}\ge2$. Integrating from $u$ to $1$ gives
$|B(iu)|\ge u^{\sum_i(1-x_i^2)/(1+x_i^2)}\ge u^{D_*}$; the same holds at
$-iu$, and $F(0)=0$. For $1\le u\le1+2/n$ each factor has modulus at least
one because $(u^2+x^2)-(1+x^2u^2)=(u^2-1)(1-x^2)\ge0$, so
\begin{equation}\label{eq:real-diameter-bound}
|F(iu)|\le e^2\qquad(|u|\le1+2/n).
\end{equation}
Orient $\Gamma=[-i(1+2/n),i(1+2/n)]$ upwards and define, off $\Gamma$,
$\mathcal C(z)=\frac1{2\pi i}\int_\Gamma6F(w)/(w-z)\,dw$. Write
$c_-=1$, $c_+=-5$ and
\begin{equation}\label{eq:real-glued-interpolant}
h_*(z)=c_\pm z^{D_*}-B(z)\mathcal C(z)\quad\text{for }\pm\operatorname{Re}z>0.
\end{equation}
The density is holomorphic near $\Gamma$ (its poles $x_i$ are nonzero
real). Local deformation shows the left minus right lateral value of
$\mathcal C$ equals $6F$: moving a short upward piece of $\Gamma$ to its
left gives a positively oriented contour with residue $6F(z)$. Hence the
two continuations agree, because
$(c_--c_+)z^{D_*}-6B(z)F(z)=0$; this glues $h_*$ holomorphically across
$\Gamma$, including its two crossings of the unit circle. For each fixed
node list the poles $1/x_i$ and the endpoints of $\Gamma$ lie strictly
outside the closed disk, so a small exterior neighborhood avoids them and
the same gluing applies; $h_*$ is holomorphic there, and since $B(x_i)=0$
and $\mathcal C$ is regular at the nonzero real $x_i$, the interpolation
follows.
For the bound, in either radius-$1/4$ disk about $\pm i$ the quantities
$|w|,|w-x_i|,|1-x_iw|$ are at least $3/4$, so
\begin{equation}\label{eq:real-logarithmic-derivative}
|F'(w)/F(w)|\le\tfrac43D_*+\tfrac83n<4n.
\end{equation}
As $|F(\pm i)|=1$, integration within those disks gives $|F(w)|\le e^{12}$
whenever $|w\mp i|\le3/n$. If $|z|=1$ and
$|\operatorname{Re}z|\ge1/(10n)$, the segment is at distance at least
$1/(10n)$ and length at most $3$, so \eqref{eq:real-diameter-bound} gives
$|\mathcal C(z)|\le30e^2n$. If $0<|\operatorname{Re}z|<1/(10n)$, $z$ is
near $\pm i$; replacing the relevant portion of $\Gamma$ by three sides of
a rectangle whose far vertical side has real part
$-\operatorname{sign}(\operatorname{Re}z)/n$ (in the radius-$1/4$ disk,
containing neither $z$ nor a pole) leaves $\mathcal C(z)$ unchanged. Each
new side is within $3/n$ of $\pm i$; the contour has length at most $3$ and
distance at least $1/(2n)$ from $z$ (the vertical side by $\ge1/n$, the
outer horizontal side by $\ge2/n$, the inner horizontal side and remaining
diameter by $1/n-(1-\sqrt{1-(10n)^{-2}})>1/(2n)$). Thus
$|\mathcal C(z)|\le6e^{12}n$ here. Finally $|B(z)|=1$ on $|z|=1$, so
\eqref{eq:real-glued-interpolant} gives $|h_*(z)|\le K_0n$ there, and at
$z=\pm i$ by continuity; the maximum modulus principle extends the bound.
\end{proof}
\subsection{Evaluation in a finite-dimensional Hilbert space}
The next balance uses the kernel and compression viewpoint of bounded
analytic interpolation \cite{Sarason1967}; the finite-dimensional facts are
proved directly.
\begin{proposition}[Interpolation estimate]\label{prop:first-sheet-bound}
Under \eqref{eq:case-condition},
\begin{equation}\label{eq:real-first-sheet-bound}
|\mathcal R_n(x)|\le(1+K_0n)^n\,\mathcal V(1/x)\prod_i|x_i|^g.
\end{equation}
The prefactor is $\exp(O(n\log n))$, uniformly in the nodes.
\end{proposition}
\begin{proof}
Let $Q(z)=\prod_i(1-x_iz)$ and give
$\mathcal H_Q=\{v/Q:v\in\mathbb C[z],\ \deg v<n\}$ the $H^2$ norm
$\|v/Q\|^2=\frac1{2\pi}\int_0^{2\pi}|v(e^{i\theta})|^2/|Q(e^{i\theta})|^2\,d\theta$.
The zeros of $Q$ lie outside the closed disk, so this is a genuine norm.
Define $\widetilde v(z)=z^{n-1}v(1/z)$ and $R(v/Q)=\widetilde v/Q$; this is
a complex-linear isometry, since $|\widetilde v(e^{i\theta})|=|v(e^{-i\theta})|$
and $|Q(e^{i\theta})|=|Q(e^{-i\theta})|$.
The functions $k_i(z)=(1-x_iz)^{-1}$ lie in $\mathcal H_Q$, are linearly
independent (distinct poles $1/x_i$), and form a basis; for $u\in H^2$ the
Taylor inner product gives $\langle u,k_i\rangle=u(x_i)$, so the orthogonal
projection $\Pi_Q$ onto $\mathcal H_Q$ satisfies
$(\Pi_Qu)(x_i)=u(x_i)$. Multiplication by $h_*$ has $H^2$ operator norm at
most $\|h_*\|_\infty$, so $J=\Pi_QM_{h_*}|_{\mathcal H_Q}$ and $L=I+JR$
satisfy $\|L\|\le1+K_0n$.
Let $E:\mathcal H_Q\to\mathbb C^n$ send $v/Q$ to $(v(x_i))_i$; it is
invertible since a degree-$<n$ polynomial cannot vanish at all distinct
nodes, and $E(u)_i=Q(x_i)u(x_i)$. Because $LR=R+J$ and projection preserves
values, applying this to $LR(v/Q)$ cancels the factors $Q(x_i)$ and gives
\begin{equation}\label{eq:real-evaluation-cancellation}
ELR(v/Q)=\bigl(\widetilde v(x_i)+h_*(x_i)v(x_i)\bigr)_i.
\end{equation}
In the basis $1/Q,\ldots,z^{n-1}/Q$, $\det E=V(x)$ and $|\det R|=1$, so the
determinant of the mixed evaluations divided by $V(x)$ has absolute value
$|\det L|\le\|L\|^n\le(1+K_0n)^n$. This is an algebraic cancellation of the
evaluation determinant; no bound for the inverse evaluation matrix enters.
As $|\xi_{\mathrm f}(x_i)|\le1/2\le1$ and
$\mathcal V(1/x)=\mathcal V(x)\prod_i|x_i|^{-(n-1)}$,
\eqref{eq:real-first-sheet-bound} follows from
\eqref{eq:real-sheet-factorization}.
\end{proof}
\begin{proposition}[Hadamard estimate]\label{prop:second-sheet-bound}
For every distinct nonzero real list in $(-1,1)$, in particular when
\eqref{eq:case-condition} fails,
\begin{equation}\label{eq:real-second-sheet-bound}
|\mathcal R_n(x)|\le(3/2)^n n^{n/2}2^{-\binom n2}
\prod_i(1+x_i^2)^{(n-1)/2}|x_i|^{g-(n-1)}.
\end{equation}
\end{proposition}
\begin{proof}
Expand the determinant by choosing one sheet in each column. For
$z_i\in\{x_i,1/x_i\}$ the absolute monomial determinant is
$\mathcal V(z)\prod_i|z_i|^{-g}$. Write $z_i=\tan\phi_i$,
$-\pi/2<\phi_i<\pi/2$; then
\[
\mathcal V(z)=\prod_i(1+z_i^2)^{(n-1)/2}\prod_{i<j}|\sin(\phi_j-\phi_i)|.
\]
The last product is $2^{-\binom n2}$ times the Vandermonde on the
unit-circle points $e^{2i\phi_i}$, whose monomial matrix has column norms
$\sqrt n$, so Hadamard gives $2^{-\binom n2}n^{n/2}$. The far-to-near weight
ratio is
$|1/x|^{-g}(1+x^{-2})^{(n-1)/2}/(|x|^{-g}(1+x^2)^{(n-1)/2})=|x|^{-D_*}\ge1$,
so all weights are bounded by their far-sheet values; and the sum of
absolute sheet coefficients is at most
$\prod_i(|\xi_{\mathrm f}(x_i)|+|\xi_{\mathrm n}(x_i)|)\le(3/2)^n$.
\end{proof}
\subsection{The two principal integrands}
Put $t_i=2x_i/(1+x_i^2)$ and
\[
\mathcal J_n(x,s)=\frac{\mathcal V(t)\mathcal V(s)^2}{\prod_{i,j}(1-t_is_j)}
\prod_js_j^b(1-s_j)^q\prod_i|t_i|^{C-1}(1-t_i)^h.
\]
Define
\begin{equation}\label{eq:real-integrand-majorants}
\begin{split}
\mathcal I_{2,n}(x,s)&=\mathcal J_n(x,s)\mathcal V(1/x)\prod_i|x_i|^g,\\
\mathcal I_{1,n}(x,s)&=\mathcal J_n(x,s)2^{-\binom n2}
\prod_i(1+x_i^2)^{(n-1)/2}|x_i|^{g-(n-1)}.
\end{split}
\end{equation}
Let $\Omega_{2,n}$ be the interior configurations with distinct nonzero
$x_i$ satisfying \eqref{eq:case-condition}, and $\Omega_{1,n}$ the
complementary case. We use the interpolation estimate on $\Omega_{2,n}$ and
the Hadamard estimate on $\Omega_{1,n}$ (valid in both cases). After
$t=2x/(1+x^2)$, the index $\kappa$ is the exponent of the absolute
$x$-Vandermonde in the majorant: $\kappa=2$ for interpolation, $\kappa=1$
for Hadamard; the $s_j$ range over $(0,1)$.
The propositions and \eqref{eq:real-sheet-integral} give
\[
|\Delta_N|\le\frac{2^nK_n}{(n!)^2}\max_{\kappa\in\{1,2\}}\sup_{(x,s)\in\Omega_{\kappa,n}}\mathcal I_{\kappa,n}(x,s),
\quad K_n=\max\{(1+K_0n)^n,(3/2)^nn^{n/2}\},
\]
since $d\mu^n(t)ds$ has total mass $2^n$ and the two case sets partition the
domain up to a null set. In logarithmic form, with $\log0=-\infty$,
\begin{equation}\label{eq:real-supremum-reduction}
\frac{\log|\Delta_N|}{n^2}-\frac12\log2
\le\max_{\kappa\in\{1,2\}}\sup_{(x,s)\in\Omega_{\kappa,n}}
\Bigl(\frac{\log\mathcal I_{\kappa,n}(x,s)}{n^2}-\frac12\log2\Bigr)
+O\Bigl(\frac{\log(n+2)}n\Bigr).
\end{equation}
The error is independent of the integration lists; factorials, measure
masses and sheet sums are all accounted for.
Repeated nonzero $x_i$ give a zero row determinant and are handled by
continuity in the first estimate; repeated $s_j$ give a zero $s$
Vandermonde. Sets with a zero or boundary node have measure zero, and near
them the original expression \eqref{eq:real-double-integral} supplies the
integrable definition. No separation condition is needed in any bound, so
all such configurations remain in the suprema of
\eqref{eq:real-supremum-reduction}.
\section{A uniform energy bound}\label{sec:energy}
This section replaces the two many-variable majorants by one-variable
suprema plus explicit quadratic terms. The bound must be uniform before
taking those suprema; matched damping and retained endpoint corrections
provide this uniformity.
We retain $\alpha,\beta,\gamma,\eta$, the cases $\kappa=2,1$, and the
nonnegative majorants $\mathcal I_{\kappa,n}$ of
\eqref{eq:real-integrand-majorants}. Put
\begin{align}
W_\kappa(x)&=(\alpha+2\gamma)\log|x|+2\eta\log(1-x)
-(\kappa/2+\alpha+\eta+\gamma)\log(1+x^2),\notag\\
W_s(s)&=\beta\log s+\gamma\log(1-s),\notag\\
D(x)&=2\gamma-\frac{2x^2}{1+x^2}.
\label{eq:energy-fields}
\end{align}
Here $D(x)$ is a scalar function, distinct from the row polynomials $D_r$.
For a real sequence $u=(u_k)_{k\ge1}$ with $|u_k|\le Kr^k$ for some
$K<\infty$ and $0<r<1$, define
\begin{equation}\label{eq:energy-trial-notation}
\|u\|_*^2=\sum_{k\ge1}\frac{u_k^2}{k},\qquad
T(u,x)=\sum_{k\ge1}\frac{u_kT_k(x)}{k},\qquad
S(u,x)=\sum_{k\ge1}\frac{u_kx^k}{k},
\end{equation}
uniformly convergent for $-1\le x\le1$. The sequences $p,v$ below are freely
chosen trial coefficients; the inequality $-a^2\le b^2-2ab$ replaces two
negative quadratic sums by affine upper bounds, leaving two one-variable
suprema. Each admissible choice gives a valid bound, and the certificate
supplies one successful choice per case.
\begin{proposition}\label{prop:energy-reduction}
Let $p,v$ be two such sequences and take $\lambda=0$ in case $\kappa=2$ or
any fixed $\lambda\ge0$ in case $\kappa=1$. Uniformly over configurations in
the indicated case,
\begin{align}
&\limsup_{N\to\infty}\sup_{\text{case }\kappa}
\Bigl(\frac{\log\mathcal I_{\kappa,n}}{n^2}-\frac12\log2\Bigr)\notag\\
&\qquad\le(-1+\alpha+\gamma)\log2+\kappa\|p\|_*^2+\frac12\|v\|_*^2\notag\\
&\qquad\quad+\sup_{\substack{-1\le x<1\\x\ne0}}\{W_\kappa(x)+\lambda D(x)-2\kappa T(p,x)-S(v,x)\}\notag\\
&\qquad\quad+\sup_{0<s<1}\{W_s(s)+2T(v,s)\}.
\label{eq:energy-dual}
\end{align}
Consequently the maximum of the right sides for the two cases bounds
$\limsup_{N\to\infty}(n^{-2}\log|\Delta_N|-\tfrac12\log2)$. We use
$\log0=-\infty$.
\end{proposition}
\begin{proof}
Write $\langle F(x)\rangle=n^{-1}\sum_iF(x_i)$, and similarly for $s$; in a
double average $\langle K(x,x')\rangle_{\ne}$ sum over ordered pairs $i\ne
j$ and divide by $n^2$. The substitution $t=2x/(1+x^2)$ gives
\begin{equation}\label{eq:energy-transform-identities}
|t-t'|=\frac{2|x-x'|(1-xx')}{(1+x^2)(1+x'^2)},\quad
1-ts=\frac{1-2xs+x^2}{1+x^2},\quad
1-t=\frac{(1-x)^2}{1+x^2}.
\end{equation}
The exponent of $|x_i|$ in either majorant is exactly
$(C-1)+g-(n-1)=C+g-n=a+2g$, so there is no singular finite-size correction
at $x=0$. Keeping the other powers gives the exact identity
\begin{align}
\frac{\log\mathcal I_{\kappa,n}}{n^2}-\frac12\log2
&=c_{\kappa,n}\log2
+\Bigl\langle W_\kappa(x)+\frac{\kappa+2}{2n}\log(1+x^2)\Bigr\rangle
+\langle W_s(s)\rangle\notag\\
&\quad+\frac\kappa2\langle\log|x-x'|\rangle_{\ne}
+\frac12\langle\log(1-xx')\rangle_{\ne}
+\langle\log|s-s'|\rangle_{\ne}\notag\\
&\quad-\langle\log(1-2xs+x^2)\rangle,
\label{eq:energy-exact-finite}\\
c_{\kappa,n}&=\frac Cn+\frac{\kappa-2}{2}-\frac{\kappa+1}{2n}.\notag
\end{align}
The correction $(\kappa+2)/(2n)\langle\log(1+x^2)\rangle$ is bounded by
$2\log2/n$.
We use the classical Chebyshev expansion of the logarithmic kernel, in the
form of Haagerup's identity \cite[Lemma~3.1 and its proof]{GaroufalidisPopescu2013}.
On $[-1,1]$, with $u=\cos\theta$, $u'=\cos\theta'$, $0\le r<1$,
\begin{align}
L_r(u,u')&=-\log2+\log|1-re^{i(\theta+\theta')}|+\log|1-re^{i(\theta-\theta')}|\notag\\
&=-\log2-2\sum_{k\ge1}\frac{r^kT_k(u)T_k(u')}{k}.
\label{eq:energy-cosine-kernel}
\end{align}
Letting $r\uparrow1$ at distinct $u,u'$ gives $L_1(u,u')=\log|u-u'|$, since
the product of the two chord lengths is $2|u-u'|$. For $|z|,|z'|<1$,
\begin{equation}\label{eq:energy-power-cross-kernels}
\log(1-zz')=-\sum_{k\ge1}\frac{z^kz'^k}{k},\qquad
-\log(1-2zs+z^2)=2\sum_{k\ge1}\frac{z^kT_k(s)}{k},
\end{equation}
the last being the log of $|1-ze^{i\arccos s}|^2$. The singular kernels
cannot be summed at diagonals, so we regularize all appearances of the same
moment by the same factor, preserving completion of squares. Near
$(x,s)=(1,1)$, $|1-xe^{i\arccos s}|^2=(1-x)^2+2x(1-s)$ identifies the
scales $1-x$ and $\sqrt{1-s}$.
Fix $0<\varepsilon<1/8$ and define
\begin{equation}\label{eq:energy-damping}
\tau=1-\varepsilon,\qquad
\rho(x)=1-\varepsilon(1-x),\qquad
\sigma(s)=1-\varepsilon\sqrt{1-s}.
\end{equation}
Replace the $x$ cosine kernel by $L_{\tau^2}(x,x')$, the $s$ cosine kernel by
$L_{\sigma(s)\sigma(s')}(s,s')$, the power kernel by
$\log(1-\rho(x)\rho(x')xx')$, and the cross kernel by
$-\log|1-\rho(x)\sigma(s)xe^{i\arccos s}|^2$. Each original kernel is at
most its replacement plus $O(\varepsilon)$, with an absolute constant.
For $|\zeta|=1$ and $0<r\le1$,
$|1-r\zeta|^2-r|1-\zeta|^2=(1-r)^2\ge0$, so each cosine kernel costs at
most $-\log r\le O(\varepsilon)$. For the power kernel put $u=xx'$ and
$r=\rho(x)\rho(x')$. If $u\ge0$, monotonicity gives
$\log(1-u)\le\log(1-ru)$. If $u<0$, write $q=-u\le1$; then
\begin{equation}\label{eq:energy-opposite-sign}
\log(1+q)-\log(1+rq)\le\frac{(1-r)q}{1+rq}\le1-r\le4\varepsilon.
\end{equation}
For the cross kernel write $a=1-x$, $b=\sqrt{1-s}$, $\theta=\arccos s$,
$z=1-xe^{i\theta}$, $z_\varepsilon=1-\rho(x)\sigma(s)xe^{i\theta}$; then
$|z_\varepsilon-z|\le\varepsilon|x|(a+b)$. On $x\le0$, $|z|\ge1$ and
$|x|(a+b)\le3$; on $0\le x\le1/2$, $|z|\ge1/2$ and $|x|(a+b)\le1$; on
$1/2\le x<1$, $|z|^2=a^2+2xb^2\ge a^2+b^2$ and $|x|(a+b)\le\sqrt2|z|$.
Hence $|z_\varepsilon-z|\le3\varepsilon|z|$ everywhere and, in the
direction needed,
\begin{equation}\label{eq:energy-cross-comparison}
-\log|z|^2\le-\log|z_\varepsilon|^2+2\log(1+3\varepsilon)\le-\log|z_\varepsilon|^2+6\varepsilon.
\end{equation}
For each fixed finite interior configuration the damped series converge
absolutely. Put
\[
A_k=\langle\tau^kT_k(x)\rangle,\quad
B_k=\langle\rho(x)^kx^k\rangle,\quad
C_k=\langle\sigma(s)^kT_k(s)\rangle.
\]
Inserting diagonals gives two exact negative-square identities,
\begin{align}
\frac\kappa2\langle L_{\tau^2}(x,x')\rangle
&=-\frac\kappa2\log2-\kappa\sum_{k\ge1}\frac{A_k^2}{k},
\label{eq:energy-first-square}
\end{align}
\begin{align}
&\frac12\langle\log(1-\rho(x)\rho(x')xx')\rangle+\langle L_{\sigma(s)\sigma(s')}(s,s')\rangle\notag\\
&\qquad-\langle\log|1-\rho(x)\sigma(s)xe^{i\arccos s}|^2\rangle\notag\\
&\quad=-\log2-\frac12\sum_{k\ge1}\frac{(B_k-2C_k)^2}{k}.
\label{eq:energy-second-square}
\end{align}
The $B_k,C_k$ in the cross term are exactly those of the pure terms. Adding
the constants to $c_{\kappa,n}\log2$ gives
$\bigl(-1+\alpha+\gamma-\frac{\kappa+1}{2n}\bigr)\log2$; the omitted-diagonal
constants enter the next $O_\varepsilon(1/n)$ correction.
Control this correction before taking any supremum. A damped cosine diagonal
is at least $-\log2+2\log(1-r)$, so the $x$ cosine diagonals cost
$O_\varepsilon(1/n)$. For the power diagonal,
$1-\rho(x)^2x^2\ge1-x$ ($x\ge0$) and
$1-\rho(x)^2x^2\ge1-(1-\varepsilon)^2\ge\varepsilon$ ($x\le0$); as
$1\le1-x\le2$ for $x\le0$, the upward correction is at most
$O_\varepsilon(1/n)-\frac1{2n}\langle\log(1-x)\rangle$. Finally
$1-\sigma(s)^2=\varepsilon\sqrt{1-s}(2-\varepsilon\sqrt{1-s})\ge\varepsilon\sqrt{1-s}$,
so the $s$ cosine diagonals cost at most
$O_\varepsilon(1/n)-n^{-1}\langle\log(1-s)\rangle$. The cross term already
contains all pairs, and there is no loss involving $\log|x|$ or
$\log(1+x)$.
For real $a,b$, $-a^2\le b^2-2ab$. Apply this to the first square with
$b=p_k$ and the second with $b=v_k$; dividing by $k$ and summing,
\begin{align*}
-\kappa\sum_kA_k^2/k&\le\kappa\|p\|_*^2-2\kappa\sum_kp_kA_k/k,\\
-\tfrac12\sum_k(B_k-2C_k)^2/k&\le\tfrac12\|v\|_*^2-\sum_kv_kB_k/k+2\sum_kv_kC_k/k.
\end{align*}
For $\kappa=1$, failure of \eqref{eq:case-condition} says
$1-2\langle x^2/(1+x^2)\rangle>1-1/n-2\gamma$, hence
$\langle D(x)\rangle>-1/n$; adding $\lambda\langle D(x)\rangle$ then costs at
most $\lambda/n$ for an upper bound, explaining $\lambda\ge0$ and its sign in
\eqref{eq:energy-dual}.
Let $T_\varepsilon(p,x),S_\varepsilon(v,x),T_\varepsilon(v,s)$ denote the
three series of \eqref{eq:energy-trial-notation} with factors
$\tau^k,\rho(x)^k,\sigma(s)^k$. The preceding bounds reduce the estimate to
two one-variable suprema of
\begin{align*}
X_{n,\varepsilon}(x)&=\frac{19}{48}\log|x|
+\Bigl(\frac1{12}-\frac1{2n}\Bigr)\log(1-x)\\
&\quad-(\kappa/2+\alpha+\eta+\gamma)\log(1+x^2)+\lambda D(x)\\
&\quad-2\kappa T_\varepsilon(p,x)-S_\varepsilon(v,x),\\
Y_{n,\varepsilon}(s)&=\frac7{48}\log s
+\Bigl(\frac1{12}-\frac1n\Bigr)\log(1-s)+2T_\varepsilon(v,s),
\end{align*}
plus the constant in \eqref{eq:energy-dual} and
$O(\varepsilon)+O_\varepsilon(1/n)$, all bounded finite-size terms of
\eqref{eq:energy-exact-finite} included in this error. For $n\ge24$ both
weakened endpoint coefficients are at least $1/24$.
Set $P_1=\sum_k|p_k|$ and $V_1=\sum_k|v_k|$, finite. From
$1-(1-d)^k\le kd$,
\[
|T_\varepsilon(p,x)-T(p,x)|\le\varepsilon P_1,\quad
|S_\varepsilon(v,x)-S(v,x)|\le2\varepsilon V_1,\quad
|T_\varepsilon(v,s)-T(v,s)|\le\varepsilon V_1.
\]
The changes in the two tangent functions are at most
$2\kappa\varepsilon P_1+2\varepsilon V_1$ and $2\varepsilon V_1$, common
summable bounds; no uniform convergence of the raw square series is needed.
All remaining nonsingular terms are uniformly bounded for $n\ge24$ and
$0<\varepsilon<1/8$. The positive coefficients $19/48,7/48$ and the weakened
coefficients at least $1/24$ force uniform decay to $-\infty$ near
$x=0,x=1,s=0,s=1$; the values at $x=-1/2$, $s=1/2$ have a common finite
lower bound, and the regular endpoint $x=-1$ is included by continuous
extension. Hence compact sets independent of $n\ge24$ and small
$\varepsilon$ contain maximizers, on which the functions converge uniformly
as $n\to\infty$ with $\varepsilon$ fixed and then as $\varepsilon\downarrow0$.
Taking limits in this order removes $O_\varepsilon(1/n)$ first and proves
\eqref{eq:energy-dual}; applying \eqref{eq:real-supremum-reduction} finishes.
\end{proof}
\section{An exact certificate for the two barriers}\label{sec:certificate}
All finite decimals in this Section are exact rationals. We specify trial
sequences for \eqref{eq:energy-dual}, isolate every stationary point of its
two one-variable functions, and bound their values by rational arithmetic.
\begin{proposition}\label{prop:real-upper}
For the determinants of \eqref{eq:determinant},
\begin{equation}\label{eq:certified-real-upper}
\limsup_{N\to\infty}\Bigl(\frac{\log|\Delta_N|}{n^2}-\frac12\log2\Bigr)
\le-2.290939875<-2.2909.
\end{equation}
The separate bounds for the majorants in cases $\kappa=2$ and $\kappa=1$ are
$-2.290939875$ and $-2.296789875$.
\end{proposition}
\subsection{Exact trial sequences}
For $\kappa=2$ take $d=10$, $\lambda_2=0$; for $\kappa=1$ take $d=8$,
$\lambda_1=2.47405979$. Write each sequence as
\begin{equation}\label{eq:certificate-sequences}
u_k=l_k+\sum_zr_zz^k,\qquad l_k=0\ (k>d).
\end{equation}
The tables give all coefficients; every integer in them is multiplied by
$10^{-8}$. Finite parts, in increasing $k$:
\begin{center}
\begin{tabular}{ccr r r r r}
\toprule
$\kappa$&$u$&\multicolumn{5}{c}{finite coefficients}\\
\midrule
2&$p$&45559127&-50750856&-6578767&13970217&4786184\\
&&-4292433&-576704&1311615&671564&-346453\\
2&$v$&-23910158&21152432&-2885110&-11558199&6485289\\
&&1456821&-1912176&-2263524&2742210&-1162454\\
\bottomrule
\end{tabular}
\medskip
\begin{tabular}{ccr r r r}
\toprule
$\kappa$&$u$&\multicolumn{4}{c}{finite coefficients}\\
\midrule
1&$p$&11913521&-79993701&-21956443&37903579\\
&&10744022&-2073193&570246&-24939103\\
1&$v$&-89913025&52874280&45168341&-30708629\\
&&-19269841&33922112&9819141&-16992389\\
\bottomrule
\end{tabular}
\end{center}
Case $\kappa=1$ has no exponential terms. For $\kappa=2$, a real base $z$
has coefficient $a$ of $z^k$; a nonreal base has $a,b$ the coefficients of
$\Re(z^k),\Im(z^k)$, represented in \eqref{eq:certificate-sequences} by
$r_z=(a-ib)/2$, $r_{\bar z}=\overline{r_z}$ (again $\times10^{-8}$), each
nonreal base paired with its conjugate:
\begin{center}
\begin{tabular}{ccrr}
\toprule
$u$&base $z$&$a$&$b$\\
\midrule
$p$&$.85$&-9338452&---\\
&$.94$&-2141509&---\\
&$.7i$&-66277922&-31907569\\
&$.85i$&-1231651&6002645\\
&$.092+.92i$&-3225918&8928234\\
&$-.092+.92i$&-2105536&-9091287\\
\midrule
$v$&$-.8$&15199211&---\\
&$-.96$&4451662&---\\
&$.88$&2545398&---\\
&$.95$&-4932634&---\\
&$.984$&11618157&---\\
&$.78i$&27238714&-38447936\\
&$.9i$&-31341084&-30188786\\
&$.955i$&-6693542&11912254\\
&$.984i$&2055213&-21715849\\
\bottomrule
\end{tabular}
\end{center}
These are real exponentially decaying sequences: ten $p$ tail terms and
thirteen $v$ tail terms counting conjugates, with base moduli at most $.94$
and $.984$, and all $|l_k|,|r_z|<1$. The convergent logarithm series give
\begin{align}
\|u\|_*^2&=\sum_{k=1}^d\frac{l_k^2+2l_k\sum_zr_zz^k}{k}
-\sum_{z,z'}r_zr_{z'}\operatorname{Log}(1-zz'),\notag\\
T(u,x)&=\sum_{k=1}^d\frac{l_kT_k(x)}{k}
-\frac12\sum_zr_z\operatorname{Log}(1-2xz+z^2),\notag\\
S(u,x)&=\sum_{k=1}^d\frac{l_kx^k}{k}
-\sum_zr_z\operatorname{Log}(1-xz).
\label{eq:certificate-log-formulas}
\end{align}
Here $\operatorname{Log}$ is the principal logarithm, all sums real by
conjugation, and there is no branch ambiguity. For the norm and power terms
the arguments have positive real part; for the cosine term,
$1-2xz+z^2=(1-ze^{i\theta})(1-ze^{-i\theta})$ with $x=\cos\theta$ has both
factors of positive real part and principal arguments summing strictly inside
$(-\pi,\pi)$, so the sum of their logarithms is exactly the principal
logarithm of the product, even when the product is negative real.
For these choices put
\begin{align}
X_\kappa(x)&=W_\kappa(x)+\lambda_\kappa D(x)-2\kappa T(p,x)-S(v,x),\notag\\
Y_\kappa(x)&=\beta\log x+\gamma\log(1-x)+2T(v,x).
\label{eq:certificate-barriers}
\end{align}
The domain of $X_\kappa$ is $[-1,1)\setminus\{0\}$ and that of $Y_\kappa$ is
$(0,1)$.
\subsection{Derivative numerators and exhaustive root brackets}
Define
\begin{align*}
t_u(x)&=\sum_{k=1}^dl_kU_{k-1}(x)+\sum_z\frac{r_zz}{1-2xz+z^2},\\
h_u(x)&=\sum_{k=1}^dl_kx^{k-1}+\sum_z\frac{r_zz}{1-xz}.
\end{align*}
Differentiating \eqref{eq:certificate-log-formulas},
\begin{align}
X_\kappa'(x)&=\frac{\alpha+2\gamma}{x}-\frac{2\eta}{1-x}
-(\kappa+2\alpha+2\eta+2\gamma)\frac{x}{1+x^2}\notag\\
&\quad-\frac{4\lambda_\kappa x}{(1+x^2)^2}-2\kappa t_p(x)-h_v(x),\notag\\
Y_\kappa'(x)&=\frac\beta x-\frac\gamma{1-x}+2t_v(x).
\label{eq:certificate-derivatives}
\end{align}
For completely specified rational numerators use
\begin{align}
Q_X(x)&=x(1-x)(1+x^2)^{3-\kappa}
\prod_{z\text{ in }p}(1-2xz+z^2)\prod_{z\text{ in }v}(1-xz),\notag\\
Q_Y(x)&=x(1-x)\prod_{z\text{ in }v}(1-2xz+z^2),\notag\\
A_X(x)&=Q_X(x)X_\kappa'(x),\qquad A_Y(x)=Q_Y(x)Y_\kappa'(x).
\label{eq:certificate-numerators}
\end{align}
Empty products equal $1$. These equations, the coefficient tables, and
$T_0=1,T_1=x$, $U_0=1,U_1=2x$, $V_{k+1}=2xV_k-V_{k-1}$, specify four
polynomials over $\mathbb Q$ without numerical root calculation. The
denominators do not vanish on the stated domains: $|1-xz|\ge1-|z|>0$ and
$|1-2xz+z^2|\ge(1-|z|)^2>0$ for real $|x|\le1$; conjugate factors in
$Q_X,Q_Y$ have positive products, so $Q_X$ has the sign of $x$ on its domain
and $Q_Y>0$ on $(0,1)$.
Here is an exact root-count certificate. If
$A(x)=\sum_{j=0}^{d_0}A_jx^j$, on a subinterval $(b,c)$ form
\begin{equation}\label{eq:certificate-descartes-transform}
(1+t)^{d_0}A\Bigl(\frac{b+ct}{1+t}\Bigr)=\sum_{h=0}^{d_0}a_ht^h,\qquad
a_h=\sum_{j=0}^{d_0}\sum_{k=0}^jA_j\binom jkb^{j-k}c^k\binom{d_0-j}{h-k},
\end{equation}
out-of-range binomials zero. Delete zero coefficients and count consecutive
sign changes. Degrees and counts, per consecutive interval:
\begin{center}
\begin{tabular}{cccll}
\toprule
$\kappa$&function&$d_0$&division points&sign variations\\
\midrule
2&$X$&36&$-1,0,1$&$9,9$\\
2&$Y$&24&$0,.25,.5,.75,1$&$5,2,2,6$\\
1&$X$&13&$-1,-.5,0,.5,1$&$1,1,2,1$\\
1&$Y$&9&$0,1$&$5$\\
\bottomrule
\end{tabular}
\end{center}
The bound is Descartes' rule of signs \cite[Book~III, p.~373]{Descartes1637}.
For completeness it follows by factoring positive roots: multiplication by
$t-r$, $r>0$, raises the variation count by at least one. Rescale to $r=1$
and drop zero leading coefficients; if $a_0,\ldots,a_d$ are the old and
$b_0,\ldots,b_{d+1}$ the new coefficients, then
$\sum_{j=0}^hb_j=-a_h$ ($0\le h\le d$) and $b_{d+1}=a_d$. Each sign change
of the nonzero partial sums forces an intervening $b_j$ of the later sign;
starting from $b_0=-a_0$ this gives an ordered subsequence with all
variations of the $a_h$, and $b_{d+1}$ has the opposite sign to the last
partial sum, adding one more. So the count bounds the positive roots with
multiplicity. The substitution sends $t>0$ bijectively to $(b,c)$ and
preserves multiplicities.
The integers $m$ below specify disjoint open brackets $(m,m+2)/10^{10}$ for
roots of the corresponding derivative:
\begin{center}
\begin{tabular}{ccrrr}
\toprule
$\kappa$&function&\multicolumn{3}{c}{$m$}\\
\midrule
2&$X$&-9601109148&-8942317572&-7608305633\\
&&-6503394794&-5185864065&-4015634158\\
&&-3108806646&-2067921826&-1589849496\\
&&1531948062&2072448179&3208186484\\
&&4381119427&5851354199&7269030693\\
&&8390277402&9332614564&9709786219\\
\midrule
2&$Y$&176402802&330649406&764952882\\
&&1227250753&2149465998&3048189112\\
&&4322699096&5564757994&6801929373\\
&&8031988371&8877037851&9577761832\\
&&9838463999&9972727815&9992037196\\
\midrule
1&$X$&-9917299785&-2259572153&2543808026\\
&&4437270259&6348298970&\\
\midrule
1&$Y$&532669786&2504239325&5701738806\\
&&7966939383&9454490138&\\
\bottomrule
\end{tabular}
\end{center}
For each integer the exact sign check is
\begin{equation}\label{eq:certificate-bracket-sign}
N_A(m)N_A(m+2)<0,\qquad
N_A(s)=\sum_{j=0}^{d_0}A_j(10^{10})^{d_0-j}s^j.
\end{equation}
Equations \eqref{eq:certificate-numerators}, \eqref{eq:certificate-descartes-transform}
and \eqref{eq:certificate-bracket-sign} give rational recipes for every entry.
The bracket count in each interval equals the displayed variation count, so
by the intermediate value theorem and the variation bound there is exactly
one simple root per bracket and no other roots in those intervals. Roots at
division points are harmless because all finite division points are evaluated
separately below.
\subsection{Rational logarithms and all candidate values}
For a positive rational $s=2^my$ with $1\le y\le2$,
\begin{equation}\label{eq:certificate-H}
H(y)=2\sum_{j=0}^{17}\frac1{2j+1}\Bigl(\frac{y-1}{y+1}\Bigr)^{2j+1},
\end{equation}
with $\log s$ replaced by $mH(2)+H(y)$. If $q=(y-1)/(y+1)$, the positive
remainder obeys
$0\le\log y-H(y)\le2q^{37}/(37(1-q^2))$. For a nonzero rational complex
$a+ib$, the real log part is half the real log of $a^2+b^2$; for the
principal argument choose an octant: rotate by $-k\pi/4$, $-4\le k\le4$, so
the rotated real part is positive and the ratio $t$ satisfies $|t|\le1/2$,
and select the rotation with $k\pi/4+\arctan t$ principal (value $\pi$ on the
negative real axis). The ratio $t$ is rational, since the rotation is, up to
positive scale, a product of maps $(a,b)\mapsto(a+b,b-a)$ or $(a-b,a+b)$.
With
\begin{equation}\label{eq:certificate-J}
J(t)=\sum_{j=0}^{23}\frac{(-1)^jt^{2j+1}}{2j+1},
\end{equation}
substitute $k(J(1/2)+J(1/3))+J(t)$ for the argument; the identity
$\arctan(1/2)+\arctan(1/3)=\pi/4$ follows from the tangent addition formula
with both summands positive summing below $\pi/2$, and
$|\arctan t-J(t)|\le|t|^{49}/49$.
All positive inputs lie between $2^{-100}$ and $2^{100}$: real arguments at
evaluation points are between $.0007$ and $2$, and squared moduli of complex
arguments between $(.016)^4$ and $(1+.984)^4$ by the factor bounds after
\eqref{eq:certificate-numerators}; at most $100$ scaling steps suffice. With
$h_0=2(1/3)^{37}/(37\cdot8/9)$ and $j_0=(1/2)^{49}/49$, a real log costs at
most $101h_0$ and an argument at most $9j_0$, and
$2\cdot101h_0+9j_0<7\cdot10^{-16}$, so $10^{-15}$ bounds the absolute error
of a complex log (hence $10^{-11}$ per log). Rational polynomial terms are
exact. The total log coefficient mass is at most $2\cdot10^2+13^2/2=284.5$
in $\kappa\|p\|_*^2+\|v\|_*^2/2$, less than $2\cdot10+13+2=35$ in $X_\kappa$,
and less than $13+1=14$ in $Y_\kappa$, giving substitution errors below
$2.85\cdot10^{-13}$, $3.5\cdot10^{-14}$, $1.4\cdot10^{-14}$, all under
$10^{-8}$. Enclosing each series term between multiples of $10^{-40}$, adding
remainder bounds with outward signs and propagating intervals linearly adds
under $10^{-32}$; for complex weight $c$ use
$\Re(c\operatorname{Log}z)=\Re(c)\Re\operatorname{Log}z-\Im(c)\Im\operatorname{Log}z$.
This is a fully rational interval calculation with no numerical logarithms.
Table~\ref{tab:certificate-values} gives all fifty point-value upper bounds;
its argument column is $10^{10}x$, a row marked $B$ is a left bracket
endpoint, a row marked $P$ is a finite division point, and each bound is a
rational upper bound rounded up at the twelfth decimal after the remainder is
included. The largest unrounded interval width is below
$2.16146856249281\cdot10^{-16}$.
\begin{longtable}{ccrrr}
\caption{Certified rational upper bounds at every candidate evaluation point.}\label{tab:certificate-values}\\
\toprule
$\kappa$&function&$10^{10}x$&type&upper bound\\
\midrule
\endfirsthead
\multicolumn{5}{c}{\tablename~\thetable\ (continued)}\\
\toprule
$\kappa$&function&$10^{10}x$&type&upper bound\\
\midrule
\endhead
\bottomrule
\endfoot
2&$X$&-9601109148&B&-0.984034048775\\
2&$X$&-8942317572&B&-0.984375363807\\
2&$X$&-7608305633&B&-0.984034052640\\
2&$X$&-6503394794&B&-0.984105522615\\
2&$X$&-5185864065&B&-0.984034053414\\
2&$X$&-4015634158&B&-0.984092061375\\
2&$X$&-3108806646&B&-0.984034037901\\
2&$X$&-2067921826&B&-0.984364031075\\
2&$X$&-1589849496&B&-0.984034038450\\
2&$X$&1531948062&B&-0.984033385315\\
2&$X$&2072448179&B&-0.984776982913\\
2&$X$&3208186484&B&-0.984034026898\\
2&$X$&4381119427&B&-0.984264010107\\
2&$X$&5851354199&B&-0.984034029391\\
2&$X$&7269030693&B&-0.984245044505\\
2&$X$&8390277402&B&-0.984034016294\\
2&$X$&9332614564&B&-0.984567630027\\
2&$X$&9709786219&B&-0.984033926884\\
2&$X$&-10000000000&P&-0.986727371546\\
2&$Y$&176402802&B&-1.608946411646\\
2&$Y$&330649406&B&-1.609949505120\\
2&$Y$&764952882&B&-1.608960295828\\
2&$Y$&1227250753&B&-1.609094597854\\
2&$Y$&2149465998&B&-1.608960426500\\
2&$Y$&3048189112&B&-1.608992193672\\
2&$Y$&4322699096&B&-1.608960428486\\
2&$Y$&5564757994&B&-1.608979509517\\
2&$Y$&6801929373&B&-1.608960430478\\
2&$Y$&8031988371&B&-1.608990548812\\
2&$Y$&8877037851&B&-1.608960429811\\
2&$Y$&9577761832&B&-1.609055802895\\
2&$Y$&9838463999&B&-1.608960412835\\
2&$Y$&9972727815&B&-1.609694899071\\
2&$Y$&9992037196&B&-1.608958123669\\
2&$Y$&2500000000&P&-1.608973617030\\
2&$Y$&5000000000&P&-1.608971701898\\
2&$Y$&7500000000&P&-1.608976908052\\
1&$X$&-9917299785&B&-2.778491531574\\
1&$X$&-2259572153&B&-1.324666731948\\
1&$X$&2543808026&B&-1.324655807046\\
1&$X$&4437270259&B&-1.352236629957\\
1&$X$&6348298970&B&-1.324666329425\\
1&$X$&-10000000000&P&-2.775818077526\\
1&$X$&-5000000000&P&-1.544057819905\\
1&$X$&5000000000&P&-1.347557680876\\
1&$Y$&532669786&B&-1.428286151250\\
1&$Y$&2504239325&B&-1.515602647362\\
1&$Y$&5701738806&B&-1.428335732372\\
1&$Y$&7966939383&B&-1.465686164672\\
1&$Y$&9454490138&B&-1.428335358167\\
\end{longtable}
The same calculation gives upper bounds
$.778415976284$ ($\kappa=2$) and $.931985203901$ ($\kappa=1$) for
$\kappa\|p\|_*^2+\tfrac12\|v\|_*^2$, so the substituted expressions, before
error allowances, obey the stronger cutoffs
\begin{center}
\begin{tabular}{cccc}
\toprule
$\kappa$&$\kappa\|p\|_*^2+\|v\|_*^2/2$&$X$&$Y$\\
\midrule
2&$.77843$&$-.98400$&$-1.60891$\\
1&$.93205$&$-1.32442$&$-1.42805$\\
\bottomrule
\end{tabular}
\end{center}
each a strict upper bound. With the $10^{-8}$ allowance the coarser cutoffs we
use are
\begin{equation}\label{eq:certificate-coarse-cutoffs}
\begin{array}{c|ccc}
\kappa&\kappa\|p\|_*^2+\tfrac12\|v\|_*^2&X&Y\\\hline
2&.77844&-.98399&-1.60890\\
1&.9321&-1.3244&-1.4280.
\end{array}
\end{equation}
The procedure also gives $.693146<\log2<.693149$.
\subsection{From bracket values to global suprema}
A single derivative bound suffices on every whole closed bracket. All its
points are at distance greater than $.0007$ from both $0$ and $1$; the bracket
nearest $s=1$ has right endpoint $9992037198/10^{10}$, distance exactly
$.0007962802$ from $1$. For $|x|\le1$,
$U_{k-1}(\cos\theta)=\sin(k\theta)/\sin\theta$ gives $|U_{k-1}(x)|\le k$,
including endpoint limits. The polynomial part of $X_\kappa'$ is at most
$4\sum_{k=1}^{10}k+10=230$; its endpoint terms are bounded by
$(19/48+1/12)/.0007<685$; its $x/(1+x^2)$ term is below $3$; its multiplier
term below $4\lambda_1<10$; and its two log-tail derivatives by
$4\cdot10/(.06)^2+13/.016<11924$. Hence $|X_\kappa'|<12852<120000$
throughout every bracket. For $Y_\kappa'$ the polynomial part is at most
$110$, the endpoint terms below $(7/48+1/12)/.0007<328$, and the log tails
below $26/(.016)^2=101562.5$, so $|Y_\kappa'|<102001<120000$ on every whole
bracket. These hold in both cases, with empty tails in case $1$.
By the mean value theorem, moving from a left bracket endpoint to its
stationary point raises either value by less than
$120000\cdot2/10^{10}=.000024$. Every interior stationary point lies in one
of these brackets or is a division point; all finite division points are in
Table~\ref{tab:certificate-values} ($-1$ for $X_2$; $1/4,1/2,3/4$ for $Y_2$;
$-1,-1/2,1/2$ for $X_1$). The remaining endpoints are singular with limit
$-\infty$: $x\to0^\pm$ and $x\to1^-$ for $X_\kappa$, $x\to0^+$ and
$x\to1^-$ for $Y_\kappa$, unaffected by the bounded tangent and multiplier
terms. So the root exhaustion, the finite endpoint values and the derivative
estimate certify the global suprema, not just the sampled values.
Finally $-1+\alpha+\gamma=-11/16$. Apply Proposition~\ref{prop:energy-reduction},
use \eqref{eq:certificate-coarse-cutoffs}, and add $.000024$ per supremum.
Since $\log2>.693146$, the right side is strictly less than
\begin{align*}
-\frac{11}{16}(.693146)+.77844-.98399-1.60890+.000048&=-2.290939875,\\
-\frac{11}{16}(.693146)+.9321-1.3244-1.4280+.000048&=-2.296789875.
\end{align*}
Taking the larger proves Proposition~\ref{prop:real-upper}.
\section{Conclusion}\label{sec:conclusion}
\begin{proof}[Proof of Theorem~\ref{thm:main}]
Suppose $G$ is rational. Proposition~\ref{prop:rationality} makes $\Delta_N$
rational for every $N$, and Proposition~\ref{prop:nonvanishing} gives
$\Delta_p\ne0$ for every sufficiently large prime $p$. Along this prime
sequence, Proposition~\ref{prop:finite-lower} gives
\[
\liminf_{\substack{p\to\infty\\p\ {\rm prime}}}
\Bigl(\frac{\log|\Delta_p|}{(48p)^2}-\frac12\log2\Bigr)>-2.29084,
\]
while Proposition~\ref{prop:real-upper} gives, along the same sequence, an
upper limit strictly below $-2.2909$. Since $-2.29084>-2.2909$, this is a
contradiction. Hence $G$ is irrational.
\end{proof}
\paragraph{Hyperbolic volumes.}
Normalize sectional curvature to $-1$. Agol's theorem identifies $4G$ as the
minimum volume of an orientable complete finite-volume hyperbolic
three-manifold with exactly two cusps, attained by the Whitehead-link and
$(-2,3,8)$-pretzel-link complements
\cite[Introduction and Theorem~3.6]{Agol2010TwoCusps}. Theorem~\ref{thm:main}
therefore makes this minimum and both link volumes irrational. More
generally, every orientable arithmetic hyperbolic three-orbifold defined over
$\mathbb Q(i)$ has irrational volume. Its lattice
$\Gamma\le\operatorname{PSL}_2(\mathbb C)$ is, up to conjugacy, commensurable
with the projective norm-one group $\Gamma^1(\mathcal O)=\mathcal O^1/\{\pm1\}$
of a maximal order $\mathcal O$ in a quaternion algebra $B/\mathbb Q(i)$
\cite[Definition~38.3.4]{Voight2021Quaternion}. For each such algebra with
finite reduced discriminant $\mathfrak D$, the volume formula gives
\[
\operatorname{vol}\bigl(\Gamma^1(\mathcal O)\backslash\mathbb H^3\bigr)
=\frac{G}{3}\prod_{\mathfrak p\mid\mathfrak D}(N\mathfrak p-1),
\qquad N\mathfrak p=|\mathbb Z[i]/\mathfrak p|,
\]
obtained by inserting
$\zeta_{\mathbb Q(i)}(2)=\zeta(2)L(2,\chi_{-4})=(\pi^2/6)G$ into
\cite[Theorem~39.1.13]{Voight2021Quaternion}. After conjugating, a common
finite-index subgroup $H$ gives
\[
\operatorname{vol}(\Gamma\backslash\mathbb H^3)
=\frac{[\Gamma^1(\mathcal O):H]}{[\Gamma:H]}
\operatorname{vol}\bigl(\Gamma^1(\mathcal O)\backslash\mathbb H^3\bigr),
\]
a positive rational multiple of $G$, hence irrational. In the split case
$B=M_2(\mathbb Q(i))$, $\mathcal O=M_2(\mathbb Z[i])$, the empty product
yields
$\operatorname{vol}(\operatorname{PSL}_2(\mathbb Z[i])\backslash\mathbb H^3)=G/3$
\cite[Example~39.1.16]{Voight2021Quaternion}.
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\end{document}