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\title{A lower bound for the sixth moment\\of the Riemann zeta function}
\author{Niklas Fraccaro}
\date{}
\begin{document}
\maketitle
\begin{abstract}
We prove that the lower limit of the sixth moment of the Riemann zeta
function, normalized by $(a_3/9!)T(\log T)^9$, is greater than $41.007$.
Our approach extends the two-piece amplifier construction used by Durkan and Page by allowing independent profiles, centered derivatives, and finite real linear combinations of two projected test families. An asymmetric twisted fourth-moment estimate and their amplified moment formula yield a common numerator kernel. Explicit balanced and unbalanced
kernels determine the projected Gram form. A rational certificate with
250 tests, followed by approximation by admissible profiles, gives the
stated bound.
\end{abstract}
\smallskip
\noindent\textit{2020 Mathematics Subject Classification.} Primary 11M06.\par
\noindent\textit{Keywords.} Riemann zeta function, sixth moment, amplified moments,
twisted fourth moment.

\section*{Introduction}
Write
\[
 M_3(T)=\int_0^T\left|\zeta\!\left(\tfrac12+it\right)\right|^6\dd t,
 \qquad
 a_3=\prod_{\ell\ \mathrm{prime}}
       (1-\ell^{-1})^9\sum_{j=0}^{\infty}\binom{j+2}{2}^{\!2}\ell^{-j}.
\]
Equivalently,
\[
 a_3=\prod_{\ell\ \mathrm{prime}}(1-\ell^{-1})^4
       (1+4\ell^{-1}+\ell^{-2}).
\]
This product converges to a positive number. The conjecture of Conrey
and Ghosh~\cite{CG}, also consistent with the random matrix prediction
of Keating and Snaith~\cite{KS}, is
\[
 M_3(T)\sim 42\,\frac{a_3}{9!}\,T(\log T)^9.
\]
Durkan and Page~\cite[Theorem~1]{DP} prove the lower bound with constant
$34.4$ in this normalization, using two-piece amplifiers. We obtain the
following improvement.

\begin{maintheorem}\label{thm:main}
We have
\[
 \liminf_{T\to\infty}\frac{M_3(T)}{(a_3/9!)T(\log T)^9}>41.007.
\]
\end{maintheorem}

The construction is a finite-dimensional approximation to $\zeta^3$.
We use two families of tests and project each onto the real line in
$\C$ selected by the functional equation for $\zeta^3$. The projection
preserves its real correlation with $\zeta^3$ without increasing its norm.
Allowing the profiles and centered derivatives to vary independently
provides additional directions beyond the two-piece test. Once the
mixed moments and the Gram matrix have been evaluated, Cauchy--Schwarz
reduces the lower bound to a quotient of two explicit quadratic
expressions. The certificate in Section~\ref{sec:certificate} makes this
quotient greater than $41.007$ with a strict margin.

There are two distinct analytic inputs for the numerator. For the first
family we retain the two polynomial lengths separately in the argument
of Bettin, Bui, Li and Radziwi\l\l~\cite{BBLR}; the resulting range is
$3a+5b<2$ for $a\leq b$. For the second family we use
Durkan--Page~\cite[Theorem~16, $k=1$]{DP}, with common cutoff
$\eta<1/4$. The denominator requires only two-zeta moment calculations.
The essential compatibility is that any pair of admissible tests has
total support exponent less than one.

The rational endpoint data are accompanied by an exact Lean 4
certificate and a formalization of the sixth-moment bound~\cite{Code}.
The Lean development establishes the result conditional on a single
explicit analytic hypothesis, \texttt{DivisorProof.Watt41}, formulated
on the basis of Proposition~4.1 of Watt~\cite{Watt}.
This hypothesis is stated but not proved in Lean; the repository
documents the regularity and uniformity conventions needed to relate
it to Watt's published estimate. The formalization uses some different
technical formulations and intermediate arguments and is not intended
as a line-by-line translation of the present proof. 

Section~\ref{sec:tests} sets up the tests and the quotient. Section~\ref{sec:twisted}
establishes the asymmetric fourth-moment input, and
Sections~\ref{sec:mixed} and~\ref{sec:gram} derive the numerator and
denominator kernels. Section~\ref{sec:profiles} passes from admissible
profiles to endpoint values. The certificate and the passage to the
unweighted moment follow in Sections~\ref{sec:certificate}
and~\ref{sec:passage}. The contour pairing, reflection identity and
rational integration formulas are recorded in the appendices.

\section{The projected test space}\label{sec:tests}
\subsection{Notation and test functions}
Let $T>1$, $L=\log T$ and $s=\tfrac12+it$. We use the functional
equation in the convention
\[
 \zeta(s)=\chi(s)\zeta(1-s),\qquad
 \chi(s)=\pi^{s-1/2}\frac{\Gamma((1-s)/2)}{\Gamma(s/2)}.
\]
On the critical line, $1-s=\bar s$ and $|\chi(s)|=1$, so that
$\zeta(s)^3=\chi(s)^3\overline{\zeta(s)^3}$.
For a real compactly supported profile $f$, define
\[
 A_{f,T}(w)=\sum_{n\geq1}f\!\left(\frac{\log n}{L}\right)n^{-w}.
\]
If $\supp f\subseteq[0,a]$, this is a Dirichlet polynomial of length
at most $T^a$. For $p\in\R[x]$, put
\begin{align*}
 C_{p,T}(w)
 &=p\!\left(\frac12+\frac1L\frac{\mathrm d}{\mathrm dw}\right)\zeta(w)\\
 &=\left.p(\partial_z)\bigl[e^{z/2}\zeta(w+z/L)\bigr]\right|_{z=0},
 &p^-(x)&=p(-x).
\end{align*}
At fixed $T$ we suppress the subscript $T$. Since the profiles and
polynomial coefficients are real,
\[
 A_f(1-s)=\overline{A_f(s)},\qquad C_p(1-s)=\overline{C_p(s)}.
\]
The certificate uses test polynomials $p$ of degree at most four; this
cutoff describes the chosen finite certificate, rather than an
intrinsic restriction of the construction. All profiles and test
polynomials remain fixed as $T\to\infty$. The degrees of polynomial
profiles need not be at most four.

\subsection{Two families and the two-piece construction}
The two raw test families are
\[
 U_{p;f,g}(t)=\chi(s)C_p(s)A_f(s)A_g(1-s),\qquad
 V_{p;f,g}(t)=C_p(s)A_f(s)A_g(s).
\]
We project them by
\[
 \Pi H(t)=\frac{H(t)+\chi(s)^3\overline{H(t)}}2.
\]
Pointwise, this is the orthogonal projection for the real inner product
$\Re(z\bar w)$ onto the line
\[
 \{w\in\C:w=\chi(s)^3\bar w\},
\]
which also contains $\zeta(s)^3$. For $p=1$, $f=g$ and $A=A_f$,
\begin{equation}\label{eq:two-piece}
 2\Pi\bigl(U_{1;f,f}+V_{1;f,f}\bigr)
 =\zeta(s)\bigl(A(s)+\chi(s)A(1-s)\bigr)^2.
\end{equation}
The expression on the right is the two-piece amplifier test used by
Durkan and Page~\cite[Section~1.3 and equation~(1.9)]{DP}.
Thus their test is recovered algebraically. The identity alone
does not assert that the same profiles are analytically admissible for
both families.

For the moment calculations we use
\begin{equation}\label{eq:projection}
 \begin{aligned}
 \zeta(s)^3\overline{\Pi H}
   &=\Re\bigl(\zeta(s)^3\overline H\bigr),\\
 (\Pi H)\overline{\Pi J}
   &=\frac12\Re\bigl(H\overline J+\overline{\chi(s)^3}HJ\bigr).
 \end{aligned}
\end{equation}
Also $|\Pi H|\leq|H|$. The second term in the Gram identity explains
why the ordinary moments $H\overline J$ alone do not suffice.

\subsection{Centering and shift conventions}
The centering is adapted to the functional equation. Exactly,
\[
 e^{z/2}\zeta(s+z/L)
 =\chi(s)H_T(t,z)e^{-z/2}\zeta(1-s-z/L),\qquad
 H_T(t,z)=e^z\frac{\chi(s+z/L)}{\chi(s)}.
\]
For $T\leq t\leq2T$ and $z$ in a fixed complex disc,
$H_T(t,z)=1+O(L^{-1})$ uniformly. This produces the polynomial
reflection $p\mapsto p^-$ below.
The replacement is justified inside normalized moments, not used as an
exact identity between centered derivatives. We first prove the moment
limits locally uniformly in the scaled complex shifts, including
coincident shifts after their apparent poles have been removed. Fixed
differential operators may then be applied by Cauchy's integral
formula.

\subsection{Admissible profiles and the order of limits}
The numerator formulas use the following classes:
\begin{align*}
 U:\quad &f\in C_c^\infty((0,a)),\quad g\in C_c^\infty((0,b)),
       \quad 0<a\leq b,\quad 3a+5b<2;\\
 V:\quad &f,g\text{ are real polynomials on }[0,\eta],
       \text{ extended by zero},\quad 0<\eta<\tfrac14.
\end{align*}
For a finite $V$-family, $\eta$ is chosen in common. To apply the Gram
formulas we additionally take the $V$-profiles to vanish at $0$ and
$\eta$, so their zero extensions are continuous. The beta profiles in
Section~\ref{sec:profiles} have this property at all sufficiently large
indices.

For each test, the sum of its two support exponents is less than $1/2$.
For $U$, this follows from $4(a+b)\leq3a+5b<2$; for $V$, from
$2\eta<1/2$. Consequently every pair of tests, including a mixed pair,
has total support exponent less than one. This is the common length
condition for the Gram asymptotics.

All profiles, polynomials and linear combinations remain fixed as
$T\to\infty$. Point masses enter only as limits of admissible profiles:
a sufficiently large profile index is fixed first, and then
$T\to\infty$ is taken. No estimate uniform in a profile index growing
with $T$ is required.

\subsection{The quotient to be evaluated}
For real $\Phi\in C_c^\infty((1,2))$, write
$\widehat\Phi(0)=\int_\R\Phi(x)\dd x$. The moment formulas will also
be used for signed $\Phi$. In Cauchy--Schwarz we choose $0\leq\Phi\leq1$
and $\widehat\Phi(0)>0$.
Choose finitely many raw tests $B_1,\ldots,B_m$ from $U,V$, and let
$c\in\R^m$. Set
\[
 F_c=\Pi\!\left(\sum_{j=1}^m c_jB_j\right).
\]
We shall determine a real vector $k$ and a real symmetric matrix $G$,
independent of $T$ and $\Phi$, such that
\begin{equation}\label{eq:moment-limits}
 \begin{aligned}
 \frac1{a_3TL^9}\int_\R\Phi(t/T)\zeta(s)^3
          \overline{\Pi B_j(t)}\dd t&\longrightarrow\widehat\Phi(0)k_j,\\
 \frac1{a_3TL^9}\int_\R\Phi(t/T)(\Pi B_j(t))
          \overline{\Pi B_k(t)}\dd t&\longrightarrow\widehat\Phi(0)G_{jk}.
 \end{aligned}
\end{equation}
Thus $c^\top k$ and $c^\top Gc$ describe the correlation of the test
with $\zeta^3$ and its squared norm. The off-diagonal entries of $G$
record the interaction between its components.
When $c^\top Gc>0$, the denominator below is positive for all
sufficiently large $T$, and Cauchy--Schwarz gives
\[
 \int_\R\Phi(t/T)|\zeta(s)|^6\dd t
 \geq
 \frac{\left|\int_\R\Phi(t/T)\zeta(s)^3\overline{F_c(t)}\dd t\right|^2}
      {\int_\R\Phi(t/T)|F_c(t)|^2\dd t}.
\]
Together with~\eqref{eq:moment-limits}, this yields
\begin{equation}\label{eq:quotient-bound}
 \liminf_{T\to\infty}
 \frac{\int_\R\Phi(t/T)|\zeta(s)|^6\dd t}{(a_3/9!)TL^9}
 \geq\widehat\Phi(0)\frac{9!(c^\top k)^2}{c^\top Gc}.
\end{equation}
It remains to establish the two moment limits and certify an
admissible choice with positive denominator and quotient greater than
$41.007$. A finite dyadic decomposition will then recover $M_3(T)$.

\section{An asymmetric twisted fourth moment}\label{sec:twisted}

The $U$-numerator requires a twisted fourth moment in which the two Dirichlet
polynomials may have different lengths.  We obtain the needed range by retaining
this distinction in the proof of Bettin--Bui--Li--Radziwi\l\l\ \cite{BBLR}.
We first specify the complete main term, so that the assertion also has an
unambiguous meaning when shifts coincide.

For $\lambda=(\alpha,\beta,\gamma,\delta)$, put
\[
 \tau_{u,v}(n)=\sum_{de=n}d^{-u}e^{-v},
 \qquad \lambda^*=(\gamma,\delta,\alpha,\beta),
\]
and define
\begin{align*}
 \mathcal A_\lambda
 &=\frac{\zeta(1+\alpha+\gamma)\zeta(1+\alpha+\delta)
          \zeta(1+\beta+\gamma)\zeta(1+\beta+\delta)}
         {\zeta(2+\alpha+\beta+\gamma+\delta)},\\
 \mathcal B_\lambda(h)
 &=\prod_{\ell^\nu\parallel h}
   \frac{\displaystyle\sum_{r\geq0}
       \tau_{\alpha,\beta}(\ell^r)
       \tau_{\gamma,\delta}(\ell^{r+\nu})\ell^{-r}}
        {\displaystyle\sum_{r\geq0}
       \tau_{\alpha,\beta}(\ell^r)
       \tau_{\gamma,\delta}(\ell^r)\ell^{-r}},\\
 Z_\lambda(h,k)
 &=\mathcal A_\lambda\mathcal B_\lambda(h)
                        \mathcal B_{\lambda^*}(k)
 \qquad ((h,k)=1).
\end{align*}
Here $\ell$ denotes a prime.  With $d=(h,k)$, let
\begin{equation}\label{eq:twisted-coefficient}
 \mathcal M_\lambda(T;f,g)
 =\sum_{h,k\geq1}
 f\!\left(\frac{\log h}{L}\right)
 g\!\left(\frac{\log k}{L}\right)
 \frac{d}{hk}Z_\lambda(h/d,k/d).
\end{equation}
This is a finite coefficient sum and does not depend on $t$.
The six shifts and their phase exponents are
\begin{equation}\label{eq:twisted-six-swaps}
\begin{array}{c|c|c}
 j&\lambda^{(j)}&e_j(\lambda)\\ \hline
 0&(\alpha,\beta,\gamma,\delta)&0\\
 1&(-\gamma,-\delta,-\alpha,-\beta)&-\alpha-\beta-\gamma-\delta\\
 2&(-\gamma,\beta,-\alpha,\delta)&-\alpha-\gamma\\
 3&(-\delta,\beta,\gamma,-\alpha)&-\alpha-\delta\\
 4&(\alpha,-\gamma,-\beta,\delta)&-\beta-\gamma\\
 5&(\alpha,-\delta,\gamma,-\beta)&-\beta-\delta.
\end{array}
\end{equation}
Define
\begin{align}
 I_\lambda(T;f,g)
 &=\int_{\R}\Phi(t/T)
   \zeta(s+\alpha)\zeta(s+\beta)
   \zeta(\bar s+\gamma)\zeta(\bar s+\delta)
   A_f(s)A_g(\bar s)\,\dd t,
   \label{eq:twisted-moment}\\
 \mathfrak M_\lambda(T;f,g)
 &=\int_{\R}\Phi(t/T)
   \sum_{j=0}^{5}\left(\frac{t}{2\pi}\right)^{e_j(\lambda)}
   \mathcal M_{\lambda^{(j)}}(T;f,g)\,\dd t.
   \label{eq:twisted-main}
\end{align}
At collisions, the sum in \eqref{eq:twisted-main} is interpreted by its
removable continuation; no individual singular term is evaluated separately.

\begin{theorem}[Asymmetric twisted fourth moment]\label{thm:twisted}
Let $0<a\leq b$, $3a+5b<2$, $f\in C_c^\infty((0,a))$,
$g\in C_c^\infty((0,b))$, and let $\Phi\in C_c^\infty((1,2))$ be real.
For every compact $K\subset\C^4$, with $\lambda=\boldsymbol\sigma/L$,
\begin{equation}\label{eq:twisted-asymptotic}
 I_\lambda(T;f,g)=\mathfrak M_\lambda(T;f,g)+o(T)
\end{equation}
uniformly for $\boldsymbol\sigma\in K$.  More precisely, for every
$\varepsilon>0$, the error is
\[
 O_{K,f,g,\Phi,\varepsilon}\!\left(
 T^{1/2+a+b+\varepsilon}
 +T^{3/4+(3a+5b)/8+\varepsilon}\right).
\]
The same assertion holds with $f$ and $g$ interchanged in the moment and
the main term.  Thus the shorter polynomial may occupy either side of the
critical line.
\end{theorem}

\begin{proof}
We use the sharper intermediate estimates in the proof of
\cite[Proposition~3.1]{BBLR}; the main-term calculation is that of
\cite[Sections~2 and~4]{BBLR}.  In the proof, we suppress $f,g$ in
$I_\lambda$, $\mathcal M_\lambda$ and $\mathfrak M_\lambda$.
All profiles and the weight remain fixed.

Write $\lambda=\boldsymbol\sigma/L$.  We first work on compact sets on which
\[
 -\sigma_1,\quad -\sigma_2,\quad \sigma_3,\quad \sigma_4
\]
are pairwise distinct, and remove this restriction at the end.

\medskip\noindent\emph{1. The quadratic-divisor estimate.}
After the approximate functional equation and a smooth dyadic decomposition,
the off-diagonal terms involve
\[
 hm_1m_2-kn_1n_2=\pm\ell,
 \qquad h\asymp A,\quad k\asymp B,\quad
 m_i\asymp M_i,\quad n_i\asymp N_i,\quad \ell\asymp H.
\]
Put $M=M_1M_2$, $N=N_1N_2$ and $P=ABMN$.
The relevant blocks satisfy $AM\asymp BN$.
Order the factors on each side and, if necessary, interchange the sides so that
\[
 M_1\ll M_2P^\eta,\qquad N_1\ll N_2P^\eta,
 \qquad BN_1\leq AM_1.
\]
These are the ordering and orientation conditions in
\cite[Proposition~3.1, equation~(13)]{BBLR}.
Interchanging the sides does not affect the later bounds in
$Q=AB$ and $R=\max\{A,B\}$.  The smooth dyadic weights satisfy the
derivative bounds required there.

For $H\ll(AB)^{1/2+\eta}$, with arbitrarily small fixed $\eta>0$, the two
ranges in the proof of \cite[Proposition~3.1, pp.~16--17]{BBLR} give
\begin{align}
 E_{\mathrm I}
 &\ll_\varepsilon T^\varepsilon(PH^2)^{1/4}
   \left(AB+H^{1/4}A^{3/8}B^{1/8}P^{1/8}\right),
   \label{eq:twisted-large-gcd}\\
 E_{\mathrm {II}}
 &\ll_\varepsilon T^\varepsilon(PH^2)^{1/4}
   \left(AB+(ABH)^{1/4}P^{1/8}\right).
   \label{eq:twisted-small-gcd}
\end{align}
Here range~I is the large-$d$ range for $d=(hm_1,kn_1)$, treated by
equation~(15) of that paper, and range~II is its complement, treated by
equation~(16).  We have retained the displayed bounds before their dependence
on $A,B$ is combined into $A+B$.  The remaining Poisson and weight-replacement
error is $O_\varepsilon(T^\varepsilon H^2)$.
All dyadic parameters are bounded by fixed powers of $T$, so the small-power
losses in those estimates have been absorbed into $T^\varepsilon$.

Set
\[
 Q=AB,\qquad R=\max\{A,B\},\qquad X=(MN)^{1/2}.
\]
Since $A^{3/8}B^{1/8}\leq Q^{1/8}R^{1/4}$,
$Q^{1/4}\leq Q^{1/8}R^{1/4}$ and $P=QX^2$, we obtain
\begin{equation}\label{eq:twisted-discrepancy}
 E\ll_\varepsilon T^\varepsilon
 \left(H^{1/2}Q^{5/4}X^{1/2}
       +H^{3/4}Q^{1/2}R^{1/4}X^{3/4}+H^2\right),
\end{equation}
where $E$ is the discrepancy between the dyadic quadratic-divisor sum and
its zero-frequency contribution.

\medskip\noindent\emph{2. Conversion to the moment estimate.}
For arbitrarily small positive parameters $\rho$ and $u$, the approximate
functional equation and decay of the oscillatory integral restrict the
relevant blocks to
\begin{equation}\label{eq:twisted-block-ranges}
 X\ll T^{1+u/2},\qquad
 H\ll T^\rho\frac{\sqrt Q\,X}{T}.
\end{equation}
This is the dyadic form of the truncations in \cite[Section~2.4]{BBLR}:
the approximate functional equation restricts $MN$, while integration by
parts restricts the frequency
$\log(hm_1m_2/(kn_1n_2))\asymp H/(\sqrt Q\,X)$.
The discarded ranges are negligible.

The compact supports of $f,g$ inside $(0,a),(0,b)$ bound $A,B$ below by
fixed positive powers of $T$.  Choosing $\rho,u$ sufficiently small in
\eqref{eq:twisted-block-ranges} therefore gives $H\ll Q^{1/2+\eta}$.
Moreover, a nonempty off-diagonal block has $H\gg1$, so the same inequalities
give the required lower bound on $MN$ relative to a sufficiently small fixed
power of $AB$.  Thus the hypotheses used in
\eqref{eq:twisted-discrepancy} hold.

The coefficient normalization is $(ABMN)^{-1/2}=(\sqrt Q\,X)^{-1}$,
and the $t$-integral has length $O(T)$.  Hence the conversion factor from
the quadratic-divisor discrepancy to the moment is $T/(\sqrt Q\,X)$.
Writing $\Lambda=T^\rho$, multiplication of
\eqref{eq:twisted-discrepancy} by this factor gives
\begin{equation}\label{eq:twisted-block-error}
 E_{\mathrm{block}}\ll_\varepsilon T^\varepsilon
 \left(\Lambda^{1/2}T^{1/2}Q
 +\Lambda^{3/4}T^{1/4}Q^{3/8}R^{1/4}X^{1/2}
 +\Lambda^2\frac{\sqrt Q\,X}{T}\right).
\end{equation}
Using $Q\ll T^{a+b}$, $R\ll T^b$ and $X\ll T^{1+u/2}$, the three
terms are bounded, respectively, by
\[
 T^{1/2+a+b+\rho/2+\varepsilon},\qquad
 T^{3/4+(3a+5b)/8+3\rho/4+u/4+\varepsilon},\qquad
 T^{(a+b)/2+2\rho+u/2+\varepsilon}.
\]
The third is smaller than the first.  There are only logarithmically many
dyadic blocks.  Taking $\rho,u$ sufficiently small and absorbing this
summation into an arbitrarily small power of $T$, we obtain
\begin{equation}\label{eq:twisted-off-diagonal-error}
 E_{\mathrm{off}}\ll_\varepsilon
 T^{1/2+a+b+\varepsilon}
 +T^{3/4+(3a+5b)/8+\varepsilon}.
\end{equation}
This is where the asymmetric condition $3a+5b<2$ enters.

\medskip\noindent\emph{3. Evaluation of the main terms.}
The diagonal and zero-frequency terms are evaluated by the Mellin and
contour argument of \cite[Sections~2 and~4]{BBLR}.
The two coefficient sequences in that argument are now
$f(\log h/L)$ and $g(\log k/L)$; keeping their lengths separate changes
the error bookkeeping, but not the arithmetic identities.

Choose an even auxiliary function $G$, rapidly decreasing in vertical
strips, normalized by $G(0)=1$, and with zeros at the shifted poles crossed
in the contour shifts.  The residues at zero in the two diagonal integrals
give the unswapped and fully swapped terms in $\mathfrak M_\lambda(T)$.
Pairing each original zero-frequency integral with its dual leaves a contour
difference whose residue at zero is the corresponding one-swap term.
The other crossed poles are cancelled by $G$.
For the $(\beta,\delta)$ swap the residue is
\[
 \left(\frac{t}{2\pi}\right)^{-\beta-\delta}
 \mathcal M_{(\alpha,-\delta,\gamma,-\beta)}(T),
\]
and permutation gives the other three terms.
Appendix~\ref{app:crossed} records the incoming BBLR expression and the
contour pairing in our notation.

The gamma-factor and dual-weight replacements contribute
$O_\varepsilon(T^{1/2+a+b+\varepsilon})$.
Moving the diagonal contours to $\Re z=-1/8$ contributes
$O(T^{3/4+(a+b)/2}(\log T)^{48})$, which is absorbed in
\eqref{eq:twisted-off-diagonal-error}, since
$(a+b)/2\leq(3a+5b)/8$.
Consequently,
\begin{equation}\label{eq:twisted-separated-bound}
 I_\lambda(T)-\mathfrak M_\lambda(T)
 \ll_\varepsilon T^{1/2+a+b+\varepsilon}
          +T^{3/4+(3a+5b)/8+\varepsilon}
\end{equation}
uniformly on compact sets of separated normalized shifts.

\medskip\noindent\emph{4. The exponent gap and colliding shifts.}
Let $\Delta=2-3a-5b>0$.  The exponents in
\eqref{eq:twisted-separated-bound} satisfy
\[
 \frac34+\frac{3a+5b}{8}=1-\frac{\Delta}{8},
 \qquad
 \frac12+a+b=1-\frac{\Delta+b-a}{4}
 \leq1-\frac{\Delta}{4}.
\]
Thus $0<\varepsilon<\Delta/16$ gives $o(T)$ on compact separated shift sets.

Although individual terms in $\mathfrak M_\lambda(T)$ have apparent poles
at collisions, their sum is holomorphic.  On each collision hyperplane the
paired terms have equal remaining factors and opposite residues; for
example, on $\alpha+\gamma=0$ the residues of
$\zeta(1+\alpha+\gamma)$ and $\zeta(1-\alpha-\gamma)$ cancel.
Multiplying by the product of the six differences among
$-\alpha,-\beta,\gamma,\delta$ clears the possible poles.  The resulting
holomorphic numerator is divisible by every factor, hence by their product,
so intersections of collision hyperplanes are removable as well.

Fix a compact $K\subset\C^4$ of normalized shifts, and choose a polydisc
centred at the origin containing $K$, with pairwise distinct coordinate
radii.  On its distinguished boundary, the four signed coordinates
$-\sigma_1,-\sigma_2,\sigma_3,\sigma_4$ are uniformly separated: the
absolute value of each difference is at least the difference of the
corresponding radii.  Both the moment and the complete six-term expression
are holomorphic in this polydisc for sufficiently large $T$.
Applying the maximum-modulus principle successively in the four variables
therefore extends \eqref{eq:twisted-separated-bound}, and in particular its
$o(T)$ consequence, from that boundary to $K$.
This proves \eqref{eq:twisted-asymptotic} with the asserted uniform error.
Finally, complex conjugation and interchange of the two shift pairs give
the version with $f,g$ reversed.
\end{proof}

\section{Mixed moments and a common numerator kernel}\label{sec:mixed}

\subsection{The kernel and the two moment limits}
The six swapped main terms admit a determinant representation. When three
shift parameters coincide, the resulting divided difference becomes an
average over a three-dimensional simplex. This gives the kernel below.
The $V$-calculation produces the same kernel with its second endpoint
reflected, although its analytic input is different. In this section
$a,b$ denote logarithmic positions averaged against the profiles, rather
than fixed support bounds.

Let $X=(X_0,X_1,X_2,X_3)$ be uniform on the simplex
$X_j\geq0$, $\sum_jX_j=1$. We normalize this probability measure by
\[
 \E F(X)=6\int_{\substack{x_1,x_2,x_3\geq0\\x_1+x_2+x_3\leq1}}
 F(1-x_1-x_2-x_3,x_1,x_2,x_3)\,\dd x_1\dd x_2\dd x_3,
 \qquad \E1=1.
\]
For $a,b\geq0$, put
\begin{align}
 P(u,v)&=\frac{uv(1-u)(1-v)(1-u-v)}{12},\notag\\
 q(a,b,t)&=\left(-\frac12,a-t-\frac12,\frac12-b+t,\frac12\right),\notag\\
 K_z(a,b)&=\int_0^{\min(a,b)}
 P(b-t,a-t)\E e^{zq(a,b,t)\cdot X}\,\dd t.
 \label{eq:num-kernel}
\end{align}
For a real polynomial $p$, define
\begin{equation}\label{eq:num-poly-kernel}
 K_p(a,b)=p(\partial_z)K_z(a,b)\big|_{z=0}
 =\int_0^{\min(a,b)}P(b-t,a-t)
       \E p\bigl(q(a,b,t)\cdot X\bigr)\,\dd t.
\end{equation}
The generator $K_z$ is entire in $z$, including when its simplex nodes
coincide.

\begin{proposition}[Numerator limits]\label{prop:numerator}
For an admissible $U$- or $V$-test and a real polynomial $p$ of degree at
most four, the first limit in \eqref{eq:moment-limits} holds with
\begin{align}
 k_U(p;f,g)&=\int_0^\infty\int_0^\infty
              f(a)g(b)K_p(a,b)\,\dd a\dd b,
              \label{eq:num-u-limit}\\
 k_V(p;f,g)&=\int_0^\eta\int_0^\eta
              f(a)g(b)K_p(a,1-b)\,\dd a\dd b.
              \label{eq:num-v-limit}
\end{align}
There is no extra projection factor in these formulas.
\end{proposition}

We prove the corresponding locally uniform limits for
\begin{align*}
 \mathcal U_T(z)&=\int_{\R}\Phi(t/T)\zeta(s)^3\overline{\chi(s)}
 A_f(\bar s)A_g(s)e^{z/2}\zeta(\bar s+z/L)\,\dd t,\\
 \mathcal V_T(z)&=\int_{\R}\Phi(t/T)\zeta(s)^3
 A_f(\bar s)A_g(\bar s)e^{z/2}\zeta(\bar s+z/L)\,\dd t.
\end{align*}
Differentiation at zero gives the unprojected correlations. The projection
identity \eqref{eq:projection} then takes their real parts, and the limiting
kernels are real.

\subsection{From the six \texorpdfstring{$U$}{U}-main terms to the simplex}
Since $\overline{\chi(s)}\zeta(s)=\zeta(\bar s)$,
\begin{equation}\label{eq:num-u-fourth}
 \mathcal U_T(z)=e^{z/2}I_{(0,0,z/L,0)}(T;g,f).
\end{equation}
The order $g,f$ is essential. Theorem~\ref{thm:twisted} permits this reversal,
so it remains to extract the leading $L^9$ coefficient of its main term.

First take separated scaled shifts $\sigma=(\sigma_1,\ldots,\sigma_4)$.
Write
\[
 D(\sigma)=\prod_{i=1}^2\prod_{j=3}^4(\sigma_i+\sigma_j),
 \qquad E_{r,s}(x)=\frac{e^{-rx}-e^{-sx}}{s-r},
 \qquad E_{r,r}(x)=xe^{-rx}.
\]
Use the transformations $\sigma^{(j)}$ and the linear forms $e_j(\sigma)$
from \eqref{eq:twisted-six-swaps}. If $A,B$ are the logarithmic positions
of the first and second profiles in $I_\lambda$, the leading kernel is
\begin{equation}\label{eq:num-six-kernel}
 \int_0^{\min(A,B)}\sum_{j=0}^5
 \frac{e^{e_j(\sigma)}}{D(\sigma^{(j)})}
 E_{\sigma^{(j)}_3,\sigma^{(j)}_4}(A-t)
 E_{\sigma^{(j)}_1,\sigma^{(j)}_2}(B-t)\,\dd t.
\end{equation}
The main term is $a_3TL^9\widehat\Phi(0)$ times the profile average of
\eqref{eq:num-six-kernel}, up to $o(TL^9)$.

The normalization and profile positions can be read directly from Mellin
inversion. The principal factor of the unswapped term is
\begin{equation}\label{eq:num-nine-factors}
 \frac{1}{D(\sigma)(u+v)(u+\sigma_3)(u+\sigma_4)
                         (v+\sigma_1)(v+\sigma_2)}.
\end{equation}
The nine denominator factors are precisely the pairwise sums of the row
shifts $(\sigma_1,\sigma_2,u)$ and the column shifts
$(\sigma_3,\sigma_4,v)$. Each corresponds to a polar zeta factor; the
remaining Euler factor tends to $a_3$, as in
Lemma~\ref{lem:gram-nine-edge}. With
$\widetilde f(u)=\int f(x)e^{ux}\,\dd x$, our inversion convention is
\[
 f(\log n/L)=\frac{1}{2\pi i}\int_{(c)}\widetilde f(u)n^{-u/L}\,\dd u.
\]
Replacing $L^{-1}\zeta(1+w/L)$ by $1/w$ therefore gives
\eqref{eq:num-nine-factors}. Finally,
\[
 \frac1{u+v}=\int_0^\infty e^{-t(u+v)}\,\dd t,
 \qquad
 \mathcal L^{-1}\!\left(\frac1{(u+r)(u+s)}\right)(x)=E_{r,s}(x),
\]
which yields \eqref{eq:num-six-kernel} term by term. The phase
$(t/2\pi)^{e_j(\sigma)/L}$ tends uniformly to $e^{e_j(\sigma)}$ on the
support of $\Phi$.

For completeness, the following bound justifies the limiting Mellin
calculation, including its tails. On a compact separated shift set $K$,
choose $R$ with $|\sigma_i|\leq R$ and take $c>R+1$. On the fixed scaled
contours $u=c+i\xi$, $v=c+i\upsilon$, the five arguments involving $u,v$
are
\[
 \mathcal W=\{u+v,u+\sigma_3,u+\sigma_4,v+\sigma_1,v+\sigma_2\}.
\]
For $w\in\mathcal W$, $\Re w\geq1$, and absolute convergence gives, for
$L\geq1$,
\[
 |L^{-1}\zeta(1+w/L)|\leq L^{-1}\zeta(1+\Re w/L)
 \leq L^{-1}+(\Re w)^{-1}\leq2.
\]
The four factors depending only on the shifts are uniformly bounded after
division by $L$, since their scaled arguments stay away from zero. The
regular Euler factor is bounded on these contours: its local factors are
$1+O(\ell^{-2+\delta})$, uniformly in the imaginary parts, for some fixed
$0<\delta<1$ and all sufficiently large $T$. Smooth compact support also
gives $|\widetilde f(c+i\xi)|\ll_f(1+|\xi|)^{-2}$, and similarly for $g$.
Thus the normalized integrand is bounded by
\begin{equation}\label{eq:num-mellin-majorant}
 \frac{C_{K,f,g}}{(1+|\xi|)^2(1+|\upsilon|)^2}.
\end{equation}
On bounded contour segments the nine normalized zeta factors converge
uniformly to their reciprocal arguments and the regular factor converges
to $a_3$. The majorant controls the tails uniformly, giving the locally
uniform $o(TL^9)$ remainder. No further power saving in $T$ is needed here.
Holomorphic continuation and the distinguished-boundary argument of
Section~\ref{sec:twisted} then include coincident shifts.

There is a useful closed form for the six summands. Put
\[
 x=(-\sigma_1,-\sigma_2,\sigma_3,\sigma_4),\qquad
 \rho(u,v)=(-1,v-1,-u,0),\qquad
 \Delta(x)=\prod_{0\leq i<j\leq3}(x_j-x_i).
\]
Then
\begin{equation}\label{eq:num-determinant}
 \begin{split}
 &\sum_{j=0}^5\frac{e^{e_j(\sigma)}}{D(\sigma^{(j)})}
 E_{\sigma^{(j)}_3,\sigma^{(j)}_4}(A-t)
 E_{\sigma^{(j)}_1,\sigma^{(j)}_2}(B-t)\\
 &\hspace{18mm}=e^{x_0+x_1}
 \frac{\det\bigl(e^{x_i\rho_j(A-t,B-t)}\bigr)_{i,j=0}^3}{\Delta(x)}.
 \end{split}
\end{equation}
Expand the determinant along its first two columns: the six choices of
rows give the six swaps. Since the numerator is alternating in $x$, the
quotient extends jointly holomorphically across $\Delta(x)=0$.

In \eqref{eq:num-u-fourth}, $A=b$, $B=a$ and $x=(0,0,z,0)$. The confluent
form of the Hermite--Genocchi divided-difference identity gives
\begin{equation}\label{eq:num-confluent}
 e^{z/2}\left.\frac{\det(e^{x_i\rho_j})_{i,j=0}^3}{\Delta(x)}
 \right|_{x=(0,0,z,0)}
 =\frac{\Delta(\rho)}{12}\E e^{z(\rho+\frac12\mathbf1)\cdot X}.
\end{equation}
One can check the normalization without invoking a general formula:
for distinct $\rho_j$ and $z\ne0$,
\[
 \E e^{z\rho\cdot X}=
 \frac6{z^3}\sum_{j=0}^3
 \frac{e^{z\rho_j}}{\prod_{k\ne j}(\rho_j-\rho_k)},
\]
while the three coincident rows on the left of \eqref{eq:num-confluent}
supply the factor $1/(2z^3)$. Continuity removes the restrictions. For
the general divided-difference identity, see \cite[formula~(83)]{deBoor}.
Since
\[
 \Delta(\rho(u,v))=12P(u,v),\qquad
 \rho(b-t,a-t)+\tfrac12\mathbf1=q(a,b,t),
\]
we obtain, locally uniformly in $z$,
\[
 \frac{\mathcal U_T(z)}{a_3TL^9}\longrightarrow
 \widehat\Phi(0)\int_0^\infty\int_0^\infty
 f(a)g(b)K_z(a,b)\,\dd a\dd b.
\]
Cauchy's formula proves \eqref{eq:num-u-limit}.

\subsection{The \texorpdfstring{$V$}{V}-moment and the reflection identity}
The $V$-moment uses the short-polynomial, single-$\chi$ fourth moment of
Durkan--Page \cite[Theorem~16, $k=1$, and Section~5.2]{DP}.
We spell out the orientation and the residue formula, since both matter
for the signs in the reflected kernel.

For scaled shifts $\sigma$, define the conjugate orientation
\[
 \mathcal J^-_\sigma(T;f,g)=\int_{\R}\Phi(t/T)\chi(s)
 \prod_{i=1}^2\zeta(s+\sigma_i/L)
 \prod_{j=3}^4\zeta(\bar s+\sigma_j/L)
 A_f(\bar s)A_g(\bar s)\,\dd t.
\]
Relative to the displayed definition of $J_{1,1}$ in \cite{DP}, this
conjugates the integral and interchanges the two shift pairs; it is the
orientation used after reversal of conjugation in their Section~5.2.
We read the formula below from that reversed-orientation calculation,
with the shifts labelled as in $\mathcal J^-_\sigma$, rather than
identifying $\mathcal J^-_\sigma$ with their displayed $J_{1,1}$.
Our generator satisfies the exact identity
\begin{equation}\label{eq:num-v-dp-orientation}
 \mathcal V_T(z)=e^{z/2}\mathcal J^-_{(0,0,0,z)}(T;f,g),
\end{equation}
because $\chi(s)\zeta(\bar s)=\zeta(s)$.
The usual polynomial convention
$P(\log(T^\eta/n)/\log T^\eta)$ corresponds to
$f(a)=P(1-a/\eta)$. A formula for one squared profile extends to $f,g$
by $A_fA_g=(A_{f+g}^2-A_{f-g}^2)/4$, using their common cutoff $\eta$.

Here is the normalized residue expression from that calculation in our
variables. Set
\[
 Q_\sigma(w)=(w+\sigma_1)(w+\sigma_2)(w-\sigma_3)(w-\sigma_4),
 \qquad
 F_\sigma(a,\omega)=\frac1{2\pi i}\oint
       \frac{e^{au}(u-\omega)}{Q_\sigma(u)}\,\dd u,
\]
and
\begin{equation}\label{eq:num-dp-residue-kernel}
 R_\sigma(a,b)=\frac{e^{-\sigma_3-\sigma_4}}{2\pi i}\oint
 \frac{e^\omega F_\sigma(a,\omega)F_\sigma(b,\omega)}
      {Q_\sigma(\omega)}\,\dd\omega.
\end{equation}
All contours are positively oriented circles containing the four roots
$-\sigma_1,-\sigma_2,\sigma_3,\sigma_4$, with radii fixed on the compact
shift set under consideration. The $k=1$ instance of the four-residue
calculation gives
\begin{equation}\label{eq:num-dp-incoming}
 \frac{\mathcal J^-_\sigma(T;f,g)}{a_3TL^9}
 =\widehat\Phi(0)\int_0^\eta\int_0^\eta
        f(a)g(b)R_\sigma(a,b)\,\dd a\dd b+o(1).
\end{equation}
To match \eqref{eq:num-dp-residue-kernel} to the four terms of the source,
take, for example, the outer residue at $\omega=-\sigma_1$.
The factor $u+\sigma_1$ cancels in each inner contour, leaving the three
factors $u+\sigma_2,u-\sigma_3,u-\sigma_4$. Its outer coefficient is
\[
 \frac{e^{-\sigma_1-\sigma_3-\sigma_4}}
 { (\sigma_2-\sigma_1)(\sigma_1+\sigma_3)(\sigma_1+\sigma_4)}.
\]
This is the first residue in \cite[Theorem~16]{DP}; the other three roots
give the remaining terms. Each profile inversion has scale $L^3$,
and the outer residue contributes $L^3$, explaining $L^9$. Equivalently,
the two factors $(\log T^\eta)^3$ in the source lose their $\eta^3$
factors when its simplex variables are rescaled to our logarithmic
positions. The arithmetic factor is $a_3$.
The source error is $O(TL^8)$ for fixed polynomial profiles in the range
$\eta<1/4$. Its uniformity on separated compact shift sets, followed by
holomorphic continuation as in Section~\ref{sec:twisted}, gives the locally
uniform limit in \eqref{eq:num-dp-incoming}. The contour representation
also makes removability at colliding shifts explicit.

Define the entire functions
\begin{align*}
 B_a(z)&=\int_0^a(a-u)e^{-zu}\,\dd u,\qquad
 C_a(z)=-\frac12\int_0^a(a-u)^2e^{-zu}\,\dd u,\\
 H_m(z)&=\frac1{m!}\int_0^1(1-u)^me^{-zu}\,\dd u
 \qquad(m=0,1,2),
\end{align*}
and put
\begin{equation}\label{eq:num-j-kernel}
 J(a,b;z)=B_aB_bH_2+(B_aC_b+C_aB_b)H_1+C_aC_bH_0,
\end{equation}
with every factor on the right evaluated at $z$. The elementary inner
inverse transform is
\[
 \frac1{2\pi i}\oint e^{au}\frac{u-\omega}{u^3(u+z)}\,\dd u
 =B_a(z)+\omega C_a(z).
\]
Thus
\begin{equation}\label{eq:num-j-contour}
 J(a,b;z)=\frac1{2\pi i}\oint
 \frac{e^\omega}{\omega^3(\omega+z)}
 (B_a(z)+\omega C_a(z))(B_b(z)+\omega C_b(z))\,\dd\omega,
\end{equation}
where the contours enclose $0,-z$. Expanding the numerator gives
$H_2,H_1,H_0$, respectively, and proves \eqref{eq:num-j-kernel}.
At $\sigma=(0,0,0,z)$, \eqref{eq:num-dp-residue-kernel} becomes
$R_\sigma(a,b)=e^{-z}J(a,b;-z)$. In view of
\eqref{eq:num-v-dp-orientation}, we have therefore proved
\begin{equation}\label{eq:num-v-generator-limit}
 \frac{\mathcal V_T(z)}{a_3TL^9}\longrightarrow
 \widehat\Phi(0)\int_0^\eta\int_0^\eta
 f(a)g(b)e^{-z/2}J(a,b;-z)\,\dd a\dd b,
\end{equation}
locally uniformly near zero.

The connection with the $U$-kernel is now an identity of entire functions.
\begin{lemma}[Reflection]\label{lem:num-reflection}
If $a,b>0$ and $a+b<1$, then
\begin{equation}\label{eq:num-reflection}
 e^{-z/2}J(a,b;-z)=K_z(a,1-b)\qquad(z\in\C).
\end{equation}
\end{lemma}
\begin{proof}
For $z\ne0$, the translated simplex nodes of $K_z(a,1-b)$ are
$0,a-t,b+t,1$, with $0\leq t\leq a$. The divided-difference formula
reduces this kernel to four elementary exponential integrals. Their sum
agrees with the result of integrating the definitions of $B,C,H$ in
$e^{-z/2}J(a,b;-z)$; Appendix~\ref{app:reflection} gives the common
numerator. Entirety includes $z=0$.
\end{proof}
Combining \eqref{eq:num-v-generator-limit} and \eqref{eq:num-reflection},
then applying $p(\partial_z)$ by Cauchy's formula, proves
\eqref{eq:num-v-limit} and completes Proposition~\ref{prop:numerator}.
The reflection is an identity of limiting kernels: the analytic
$V$-asymptotic still uses its own short-length range $\eta<1/4$.

\section{The denominator kernels and the projected Gram form}
\label{sec:gram}

The projection identity \eqref{eq:projection} splits each Gram entry into an
ordinary product and a reflected product. After applying the functional
equation, these reduce to the balanced and unbalanced moments below; the
remaining reflected $VV$ product is oscillatory and contributes no main term.

Throughout this section the real profiles $f,g,h,k$ are continuous and supported
in $[0,A_i]$, for $i=1,2,3,4$, respectively, with
\[
 S_*:=A_1+A_2+A_3+A_4<1.
\]
This includes every pair of tests from Section~\ref{sec:tests}. The endpoint
variables $a,b,c,d$ below are integration variables, not support bounds. The
polynomials $p,q$ have degree at most four.

\subsection{The common arithmetic diagonal}

\begin{lemma}[Nine-edge diagonal limit]\label{lem:gram-diagonal}\label{lem:gram-nine-edge}
Let $F$ be bounded, measurable and compactly supported on
$\R_{\geq0}^3\times\R_{\geq0}^3$, and suppose that its discontinuity set has
zero measure for the limiting measure below. Writing
$\log\mathbf n/L=(\log n_1/L,\ldots,\log n_3/L)$, we have
\begin{equation}\label{eq:gram-diagonal}
 \frac1{L^9}
 \sum_{n_1n_2n_3=m_1m_2m_3}
 \frac{F(\log\mathbf n/L,\log\mathbf m/L)}{n_1n_2n_3}
 \longrightarrow
 a_3\int_{x_{ij}\geq0}
 F\left(\left(\sum_jx_{ij}\right)_i,
         \left(\sum_ix_{ij}\right)_j\right)
 \prod_{i,j=1}^3\dd x_{ij}.
\end{equation}
\end{lemma}

\begin{proof}
The Dirichlet series of the diagonal factors as
\[
 \sum_{\prod_i n_i=\prod_jm_j}
 \frac{\prod_i n_i^{-u_i}\prod_jm_j^{-v_j}}{\prod_i n_i}
 =E(\mathbf u,\mathbf v)\prod_{i,j=1}^3\zeta(1+u_i+v_j).
\]
At each prime the nine terms of first degree agree with the nine zeta factors.
Their quotient is therefore $1+O(\ell^{-2+\varepsilon})$ uniformly near the
origin, and $E$ is regular there. At zero its local factor is
\[
 (1-\ell^{-1})^9\sum_{r\geq0}\binom{r+2}{2}^{\!2}\ell^{-r}
 =(1-\ell^{-1})^4(1+4\ell^{-1}+\ell^{-2}),
\]
so $E(\mathbf0,\mathbf0)=a_3$. Substituting $u_i=\alpha_i/L$ and
$v_j=\beta_j/L$, with positive real parts, gives the limiting Laplace transform
\[
 a_3\prod_{i,j}(\alpha_i+\beta_j)^{-1}.
\]
This is the transform of the pushforward of Lebesgue measure on the nine
nonnegative variables $x_{ij}$ in \eqref{eq:gram-diagonal}. Exponentially
tilting makes the measures finite, and the continuity theorem for Laplace
transforms gives weak convergence. Removing the tilt on a compact set proves
the assertion, including bounded tests whose discontinuities have limiting
measure zero.
\end{proof}

\subsection{The balanced kernel}

For $a+b+c+d<1$, let $\mathcal R(a,b;c,d)$ consist of
$r=(r_{11},r_{12},r_{21},r_{22})\geq0$ with row sums at most $a,b$ and column
sums at most $c,d$. Put
\[
 t_r=\sum_{i,j}r_{ij},\qquad
 \Delta=\frac{a+b-c-d}{2},\qquad
 w_r=\frac{1-a-b-c-d+2t_r}{2},
\]
and define
\begin{equation}\label{eq:gram-balanced-kernel}
 B_{p,q}(a,b;c,d)=
 \int_{\mathcal R(a,b;c,d)}\int_{-w_r}^{w_r}
 p(\Delta-x)q(-\Delta-x)\,\dd x\,\dd r.
\end{equation}

\begin{proposition}[Balanced moment]\label{prop:gram-balanced}
The normalized integral
\[
 \frac1{a_3TL^9}\int_{\R}\Phi(t/T)
 C_p(s)C_q(\bar s)A_f(s)A_g(s)A_h(\bar s)A_k(\bar s)\,\dd t
\]
converges to
\[
 \widehat\Phi(0)\int f(a)g(b)h(c)k(d)
 B_{p,q}(a,b;c,d)\,\dd a\,\dd b\,\dd c\,\dd d.
\]
\end{proposition}

\begin{proof}
Apply the smoothed two-zeta approximate functional equation, first with two
scaled shifts; compare \cite[Lemmas~4 and~5]{Young}. Write $H=h_1h_2$ and
$K=k_1k_2$ for the two amplifier products. The relevant length condition in
that argument is $HK\leq T^{S_*}$, so its parameter $2\theta$ is $S_*<1$.
The direct and dual sums are effectively restricted to $mn\leq T^{1+\delta}$,
where $0<\delta<1-S_*$. If $Hm\ne Kn$ and their ratio is near one, integrality
gives
\[
 T\left|\log\frac{Hm}{Kn}\right|
 \gg T^{(1-S_*-\delta)/2}.
\]
Outside that range the frequency is larger. Repeated integration by parts
against the smooth time weight therefore makes the off-diagonal negligible.
The tails and gamma-factor replacements cost
$O_\varepsilon(T^{1/2+S_*/2+2\varepsilon})$, with logarithmic losses absorbed
into $T^\varepsilon$; choose $\varepsilon<(1-S_*)/4$.

To specify the smoothing, one may start with the pole-corrected Gaussian
\[
 G_{s_1,s_2}(u)=e^{u^2}\prod_{\nu=1}^2
       \left(1-\frac{u^2}{s_\nu^2}\right)
       \left(1-\frac{u^2}{(1-s_\nu)^2}\right),
 \qquad s_1=s+z/L,\quad s_2=\bar s+w/L.
\]
It is even, equals one at zero, and cancels the poles crossed in the
approximate functional equation. Removing its correction polynomial and
replacing the gamma ratios are included in the preceding error bound;
the estimates are uniform for $z,w$ in fixed complex discs.

We give the limiting-cutoff step explicitly, since the smoothing still depends
on $T$. After these replacements the weight is
\[
 V_0(x)=\frac1{2\pi i}\int_{(1)}e^{u^2}x^{-u}\frac{\dd u}{u}
       =\frac1{\sqrt\pi}\int_{\frac12\log x}^{\infty}e^{-v^2}\,\dd v.
\]
For $t=T\tau$, $1\leq\tau\leq2$, and $Q=\log(mn)/L$, this is
\[
 V_0(2\pi mn/t)
 =\frac1{\sqrt\pi}\int_{\{L(Q-1)+\log(2\pi/\tau)\}/2}^{\infty}
 e^{-v^2}\,\dd v.
\]
It lies between zero and one and converges uniformly to
$\mathbf1_{Q\leq1}$ outside every fixed strip $|Q-1|<\epsilon$, uniformly
in $\tau$. On a fixed compact set, bound the strip by a continuous function
which is one on $|Q-1|\leq\epsilon$ and zero on $|Q-1|\geq2\epsilon$.
Lemma~\ref{lem:gram-diagonal} applies to this majorant; its limiting integral
tends to zero as $\epsilon\downarrow0$, because the preimage of $Q=1$ is a
null hyperplane in the nine edge variables. Thus the smoothing can be replaced
by the sharp cutoff on compact sets. For completeness, no mass is lost at
infinity: under the exponential tilt used in the lemma, the shifted profile
weight grows at most exponentially in $Q$, and the Gaussian decay bounds
its product with the smoothing by $C e^{-Q^2}$ for $Q\geq2$ and all
sufficiently large $L$.
The tilted measures have bounded total mass, so truncating at $Q\leq R$
has error $O(e^{-R})$, uniformly in the fixed shift discs. Letting first
$T\to\infty$, then $\epsilon\downarrow0$ and $R\to\infty$, justifies the
cutoff $\log(mn)/L\leq1$.

Apply Lemma~\ref{lem:gram-diagonal} and disintegrate its nine variables by the
four profile positions. The remaining matrix of edge variables is
\[
 \begin{pmatrix}
 u&c-r_{11}-r_{21}&d-r_{12}-r_{22}\\
 a-r_{11}-r_{12}&r_{11}&r_{12}\\
 b-r_{21}-r_{22}&r_{21}&r_{22}
 \end{pmatrix},
 \qquad r\in\mathcal R(a,b;c,d).
\]
All changes of coordinates have Jacobian one. The two zeta lengths are the
first row and first column sums,
\[
 R_0=u+c+d-t_r,\qquad C_0=u+a+b-t_r.
\]
The cutoff $R_0+C_0\leq1$ is exactly $0\leq u\leq w_r$. For the direct term
the polynomial weight is $p(1/2-R_0)q(1/2-C_0)$; setting $x=u-w_r$ gives the
integrand of \eqref{eq:gram-balanced-kernel} on $[-w_r,0]$. For the dual term
the weight is $p(C_0-1/2)q(R_0-1/2)$; setting $x=w_r-u$ gives the same
integrand on $[0,w_r]$. Adding the two halves proves the kernel formula.

The shifted estimates are uniform on fixed complex discs. Cauchy's formula
therefore justifies the fixed polynomial derivatives used above.
\end{proof}

\subsection{The unbalanced kernel}

Put $H=a+b+c-d$ and $\tau=(1-H)/2$. If $H\leq0$, define
$M_{p,q}(a,b,c;d)=0$. For $H>0$, let
\[
 \mathcal S(a,b,c;d)=
 \{(r_1,r_2):0\leq r_1\leq a,\ 0\leq r_2\leq b,
       \ 0\leq r_3:=H-r_1-r_2\leq c\},
\]
and set
\begin{equation}\label{eq:gram-unbalanced-kernel}
\begin{split}
 M_{p,q}(a,b,c;d)
 ={}&\int_{\mathcal S(a,b,c;d)}
 \int_{\prod_{i=1}^3[-r_i/2,r_i/2]}
 p(y_1+y_2+y_3-\tau)\\[-2pt]
 &\hspace{39mm}\cdot q(y_1+y_2+y_3+\tau)
 \,\dd\mathbf y\,\dd r_1\,\dd r_2.
\end{split}
\end{equation}

\begin{proposition}[Unbalanced moment]\label{prop:gram-unbalanced}
The normalized integral
\[
 \frac1{a_3TL^9}\int_{\R}\Phi(t/T)\chi(s)^{-1}
 C_p(s)C_q(\bar s)A_f(s)A_g(s)A_h(s)A_k(\bar s)\,\dd t
\]
converges to
\[
 \widehat\Phi(0)\int f(a)g(b)h(c)k(d)
 M_{p,q}(a,b,c;d)\,\dd a\,\dd b\,\dd c\,\dd d.
\]
\end{proposition}

\begin{proof}
For the shifted generator, the functional equation moves the first zeta
factor to the negative side. Retain its centering factor during the estimates:
\[
 H_T(t,z)=\left(\frac{t}{2\pi T}\right)^{-z/L}+O(T^{-1})
\]
uniformly on fixed $z$-discs. Expand the two negative-side zeta factors by
the one-sided smoothed series
\[
 \sum_{m,n\geq1}m^{-w_1}n^{-w_2}e^{-mn/T^3},
 \qquad w_1=\bar s-z/L,\quad w_2=\bar s+w/L.
\]
To justify the replacement, Mellin inversion gives
\[
 \sum_{m,n\geq1}m^{-w_1}n^{-w_2}e^{-mn/T^3}
 =\frac1{2\pi i}\int_{(2)}
 \Gamma(u)T^{3u}\zeta(w_1+u)\zeta(w_2+u)\,\dd u.
\]
Truncate at $|\Im u|\leq T/4$ and shift to $\Re u=-1/2-\varepsilon$, with
$0<\varepsilon<1/4$. Only the pole of $\Gamma(u)$ at zero is crossed: the
zeta poles have imaginary part $t+O(L^{-1})$, outside this rectangle.
Stirling's formula makes the horizontal edges and the discarded right-hand
tails exponentially small. On the new vertical edge, the functional equation
and the absolutely convergent zeta series on $\Re s>1$ give
\[
 |\zeta(w_1+u)\zeta(w_2+u)|\ll_\varepsilon T^{1+2\varepsilon}.
\]
Since $|T^{3u}|=T^{-3/2-3\varepsilon}$ and $\Gamma(u)$ is integrable along
that line, the pointwise replacement error is
$O_\varepsilon(T^{-1/2+\varepsilon})$, uniformly on fixed shift discs.
Multiplication by the four amplifiers costs at most $O(T^{S_*/2})$, and
integration over $t\asymp T$ gives
$O_\varepsilon(T^{1/2+S_*/2+2\varepsilon})=o(T)$ after choosing
$\varepsilon<(1-S_*)/4$. The factor $H_T(t,z)$ is retained throughout:
its pure-power part has the derivative bounds needed below and tends to one
only in the diagonal limit. Its $O(T^{-1})$ remainder is absorbed in the same
error budget.

The diagonal equation is $h_1h_2h_3=kmn$. Away from it, the near-diagonal
logarithmic frequency is at least a constant times $T^{-(A_1+A_2+A_3)}$.
Its product with $T$ therefore tends to infinity by a fixed power. The
exponentially smoothed series has polynomially bounded total coefficient
mass, so sufficiently many integrations by parts discard the off-diagonal
uniformly in the shifts. On the diagonal $mn\leq T^{A_1+A_2+A_3}$, and hence
$e^{-mn/T^3}\to1$ uniformly. There is no additional condition analogous to
$R_0+C_0\leq1$ in the balanced case.

Apply Lemma~\ref{lem:gram-diagonal}, now with the three positive profiles
fixing the row sums $a,b,c$ and the fourth profile fixing the last column
sum $d$. The matrix can be written
\[
 \begin{pmatrix}
 v_1&r_1-v_1&a-r_1\\
 v_2&r_2-v_2&b-r_2\\
 v_3&r_3-v_3&c-r_3
 \end{pmatrix},
 \qquad r_1+r_2+r_3=H,\quad 0\leq v_i\leq r_i.
\]
Again the Jacobian is one. After reflection the polynomial weight is
\[
 p(v_1+v_2+v_3-1/2)\,q(1/2-H+v_1+v_2+v_3).
\]
Putting $y_i=v_i-r_i/2$ gives \eqref{eq:gram-unbalanced-kernel}.
Uniform holomorphic convergence and Cauchy's formula justify the
differentiation.
\end{proof}

\subsection{Reflection errors and all projected Gram blocks}

We now assemble the projected Gram form. First, we control the error in the
centered functional equation. Cauchy's formula applied to $H_T(t,z)-1$ gives,
for fixed $q$,
\begin{equation}\label{eq:gram-reflection-error}
 \left|\chi(s)^{-1}C_q(s)-C_{q^-}(\bar s)\right|
 \ll_q L^{-1}\sum_{j=0}^4|C_{x^j}(\bar s)|.
\end{equation}
After multiplication by the two profile polynomials belonging to one test,
the squared weighted $L^2$ norm of this error is $O(TL^7)$ by
Proposition~\ref{prop:gram-balanced}. The other factor has squared norm
$O(TL^9)$. Cauchy--Schwarz therefore makes each resulting Gram error
$O(TL^8)=o(TL^9)$. A nonnegative smooth majorant of $|\Phi|$ suffices.
This argument uses only second moments of the test functions.

Second, the reflected $VV$ term is negligible. After one reflection it has
the form
\[
 \int_{\R}\Phi(t/T)\chi(s)^{-2}C_p(s)C_q(\bar s)
 A_f(s)A_g(s)A_h(s)A_k(s)\,\dd t.
\]
Move the positive-side zeta to the negative side and use the same one-sided
expansion as above. The remaining $\chi^{-1}$ has phase derivative
$\log(t/2\pi)+O(t^{-1})$. The full phase derivative is
\[
 \log(t/2\pi)+\log\frac{mn}{h_1h_2h_3h_4}+O(t^{-1})
 \geq(1-S_*)\log T+O(1).
\]
There is no stationary point. Integration by parts, with the replacement
bound from Proposition~\ref{prop:gram-unbalanced}, gives $o(TL^9)$.

For two tests with polynomial/profile data $(p;a,b)$ and $(q;c,d)$, define
the point kernels
\begin{equation}\label{eq:gram-blocks}
\begin{aligned}
 G_{UU}(p,q;a,b,c,d)
   &=\tfrac12\{B_{p,q}(a,d;b,c)+B_{p,q^-}(a,c;b,d)\},\\
 G_{UV}(p,q;a,b,c,d)
   &=\tfrac12\{M_{q,p}(b,c,d;a)+M_{p,q^-}(a,c,d;b)\},\\
 G_{VV}(p,q;a,b,c,d)&=\tfrac12B_{p,q}(a,b;c,d),\\
 G_{VU}(p,q;a,b,c,d)&=G_{UV}(q,p;c,d,a,b).
\end{aligned}
\end{equation}
For profiles $f,g,h,k$, the Gram entry is their fourfold average against
the appropriate kernel in \eqref{eq:gram-blocks}. This proves the second
limit in \eqref{eq:moment-limits}.

For example, $U_{p;f,g}\overline{U_{q;h,k}}$ has positive profiles $f,k$
and negative profiles $g,h$, giving the first $UU$ term. The real part of
$\overline{\chi(s)^3}U_{p;f,g}U_{q;h,k}$, followed by
\eqref{eq:gram-reflection-error}, gives the second term with $q^-$.
For $UV$, conjugating its ordinary product gives the unbalanced ordering
$(g,h,k;f)$ with polynomials $(q,p)$; its reflected product gives
$(f,h,k;g)$ with $(p,q^-)$. The reflected $VV$ product is the nonstationary
term just treated. These calculations also account for every factor $1/2$
in \eqref{eq:gram-blocks}.

All formulas hold for real signed $\Phi$. Wherever positivity was used
in a bound, dominate $|\Phi|$ by a smooth nonnegative weight, or write $\Phi$
as a difference of two such weights. Symmetry and nonnegativity of the
resulting Gram form follow directly from its origin as a limit of squared
norms.

\section{From admissible profiles to endpoint kernels}\label{sec:profiles}
The rational certificate uses point evaluations, whereas the moment
theorems require actual profiles. We therefore take two successive
limits, in the order specified in Section~\ref{sec:tests}.

For a $U$-endpoint $a>0$, choose a nonnegative
$\psi\in C_c^\infty((0,1))$ with integral one and put
\[
 f_{a,n}(x)=\frac1{h_n}\psi\!\left(\frac{a-x}{h_n}\right),
 \qquad h_n\downarrow0,\quad h_n<a.
\]
This is a probability density supported strictly inside $(0,a)$, and
$f_{a,n}(x)\dd x$ converges weakly to $\delta_a$. For a pair
$0<a\leq b$ with $3a+5b<2$, use these profiles at $a,b$. The strict
support bounds remain valid at every index.

For the $V$-family, fix the common cutoff
\[
 \eta=\frac{19999}{80000}<\frac14.
\]
For each endpoint $0<a<\eta$, set $k_n=\lfloor na/\eta\rfloor$ and
use the polynomial probability density on $[0,\eta]$
\begin{equation}\label{eq:beta-profile}
 b_{a,n}(x)=\frac{n+1}{\eta}\binom n{k_n}
       \left(\frac x\eta\right)^{k_n}
       \left(1-\frac x\eta\right)^{n-k_n},
\end{equation}
extended by zero. Its mean and variance are respectively
\[
 \frac{\eta(k_n+1)}{n+2}\longrightarrow a,
 \qquad
 \frac{\eta^2(k_n+1)(n-k_n+1)}{(n+2)^2(n+3)}
 \leq\frac{\eta^2}{4(n+3)}.
\]
Thus $b_{a,n}(x)\dd x\Rightarrow\delta_a$. For sufficiently large $n$,
$1\leq k_n\leq n-1$, so its zero extension is continuous at both
ends. Its support bound is $\eta$, irrespective of $a$. Its degree
may grow with $n$, but $n$ stays fixed during every $T\to\infty$ limit.

All the required kernels are continuous in the endpoint variables on
their admissible domains. For $K_p$ this follows from its compact
integral representation. For $B_{p,q}$ and
$M_{p,q}$, write the integrals on a fixed surrounding box,
using indicators for their defining inequalities. As the endpoints
vary, the indicators converge away from finitely many affine
hyperplanes, and the integrands have a common compact bound.
Dominated convergence applies; when the unbalanced height $H$
approaches zero, its integration region shrinks to a null set. The
same argument applies to all the projected Gram kernels.

Weak convergence of the product profile measures now gives, for any
fixed finite collection of endpoint tests,
\begin{equation}\label{eq:profile-limits}
 \begin{gathered}
 k_{j,n}\longrightarrow k_j,\qquad G_{jk,n}\longrightarrow G_{jk},\\
 N_n:=\sum_jc_jk_{j,n}\longrightarrow N:=\sum_jc_jk_j,\qquad
 D_n:=\sum_{j,k}c_jc_kG_{jk,n}\longrightarrow D:=\sum_{j,k}c_jc_kG_{jk}.
 \end{gathered}
\end{equation}
Here $k_j=K_{p_j}(a_j,b_j)$ for a $U$-row and
$k_j=K_{p_j}(a_j,1-b_j)$ for a $V$-row, while $G_{jk}$ is the
corresponding projected point kernel from Section~\ref{sec:gram}.
Since there are finitely many entries, one common sufficiently large
profile index works for all of them. Only the denominator $D_n$ of
the chosen vector must eventually be positive; invertibility of the
whole Gram matrix is unnecessary.

\section{The explicit certificate}\label{sec:certificate}
At fixed endpoints and polynomial degrees, the kernels determine $k$
and $G$, and coefficient selection is governed by
$(c^\top k)^2/(c^\top Gc)$. On a subspace where $G$ is positive
definite and $k\ne0$, Cauchy--Schwarz in the $G$-inner product gives
the maximizing direction $G^{-1}k$. This describes the finite
optimization problem; the proof uses only the rational vector
specified below. Neither its 250 tests nor the degree-four cutoff
assert optimality.

The certificate consists of the 250 rows
$(\text{type},A,B,r,c)$ in \texttt{certificate\_250.tsv} at the fixed
repository version~\cite{Code}. A row denotes endpoints
$a=A/10^8$, $b=B/10^8$, coefficient $c$, and test polynomial
$p_r(x)=T_r(2x)$, where $T_r$ is the Chebyshev polynomial:
\[
 \begin{array}{c|ccccc}
 r&0&1&2&3&4\\ \hline
 p_r(x)&1&2x&8x^2-1&32x^3-6x&128x^4-32x^2+1.
 \end{array}
\]
There are 220 $U$-rows and 30 $V$-rows. The data satisfy
\[
 0<a\leq b,\qquad
 \min_{U\text{-rows}}(2-3a-5b)=\frac{19997}{10^8}>0,
 \qquad
 \max_{V\text{-rows}}\{a,b\}=0.249975<\eta.
\]
Thus Section~\ref{sec:profiles} applies. Rows of the same type and
with the same endpoints can be combined into one polynomial, leaving
178 groups without changing $N$ or $D$.

\subsection{Exact evaluation}
Rational endpoints and polynomial tests make every kernel entry
rational. For the numerator, simplex moments reduce the calculation
to polynomial integration of degree at most nine. For the balanced
kernel, slicing by the sum of its four edge variables gives a
piecewise cubic density, and the resulting integrand has degree at
most twelve. For the unbalanced kernel, box integration followed by
inclusion--exclusion leaves polynomial simplex integrals.
Appendix~\ref{app:exact} records the integration formulas, breakpoints
and exact interpolation rule used in this finite computation.

\subsection{The certified inequalities}
Let $N,D$ be the endpoint values in~\eqref{eq:profile-limits}.
Exact rational evaluation of the explicit kernels gives
\begin{equation}\label{eq:certificate-bounds}
 N>46159,\qquad 0<D<18854000000000.
\end{equation}
Consequently,
\begin{equation}\label{eq:certificate-ratio}
 R:=\frac{9!N^2}{D}
 >\frac{9!\cdot46159^2}{18854000000000}
 =\frac{1208080410327}{29459375000}
 =41.00835168\ldots>41.007.
\end{equation}
The strict margin allows for profile approximation and smoothing.
At the snapshot~\cite{Code}, the exact numerator and denominator
bounds are proved in
\path{SixthMoment/Certificate/Exact/NumeratorBound.lean} and
\path{SixthMoment/Certificate/Exact/DenominatorBound.lean};
\path{SixthMoment/Certificate/Bounds.lean} records the final
rational comparison. Together with the endpoint table and the
formulas in Appendix~\ref{app:exact}, these specify the numerical
part of the proof reproducibly.

\section{Passage to the sixth moment}\label{sec:passage}
By~\eqref{eq:profile-limits}, $N_n\to N$ and $D_n\to D>0$, so
$R_n:=9!N_n^2/D_n\to R$. Fix $q<R$. Choose a sufficiently large
fixed $n$, a weight $0\leq\Phi\leq1$ in $C_c^\infty((1,2))$, and a
fixed integer $J\geq1$ such that
\[
 mR_n(1-2^{-J})>q,\qquad m=\int_\R\Phi.
\]
This is possible because $m$ may be arbitrarily close to one. The
moment limits and~\eqref{eq:quotient-bound} give
\[
 \liminf_{Y\to\infty}
 \frac{\int_\R\Phi(t/Y)|\zeta(\tfrac12+it)|^6\dd t}
      {(a_3/9!)Y(\log Y)^9}\geq mR_n.
\]
For $Y_j=T/2^{j+1}$, $0\leq j<J$, the weights $\Phi(t/Y_j)$ have
disjoint supports inside $[0,T]$. Summing these finitely many lower
bounds and using $\log Y_j/\log T\to1$ yields
\[
 \liminf_{T\to\infty}\frac{M_3(T)}{(a_3/9!)T(\log T)^9}
 \geq mR_n\sum_{j=0}^{J-1}2^{-j-1}
 =mR_n(1-2^{-J})>q.
\]
Letting $q\uparrow R$, we conclude that
\[
 \liminf_{T\to\infty}\frac{M_3(T)}{(a_3/9!)T(\log T)^9}
 \geq R>41.007,
\]
which proves Theorem~\ref{thm:main}.

\appendix
\section{The crossed main terms}\label{app:crossed}

We make explicit the source-to-notation step in the contour calculation of
Theorem~\ref{thm:twisted}.  Write $f_h=f(\log h/L)$ and
$g_k=g(\log k/L)$, and put
\[
 C_\lambda(z)=(\alpha+z,-\delta-z,\gamma+z,-\beta-z).
\]
In \cite[Section~2.4, pp.~11--12]{BBLR}, use the coefficient sequences
$f_h,g_k$.  After the change of variable called
$s\mapsto\beta+\delta+2s$ there, the arithmetic part of the unnumbered
formula for $M^{**}_{\alpha,\beta,\gamma,\delta}$ is
\begin{align}
 &\frac{\zeta(1+\alpha-\beta)\zeta(1+\gamma-\delta)}
             {\zeta(2+\alpha-\beta+\gamma-\delta)}
   \zeta(1+\alpha+\gamma+2z)\zeta(1-\beta-\delta-2z)
   \notag\\[-2pt]
 &\hspace{8mm}\times
 \sum_{d\geq1}\sum_{(h_0,k_0)=1}
 \frac{f_{dh_0}g_{dk_0}}
 {d\,h_0^{1-\beta-z}k_0^{1-\delta-z}}
 \eta_{\lambda,h_0,k_0}(0,0,\beta+\delta+2z)
 =\mathcal M_{C_\lambda(z)}(T).
 \label{eq:twisted-incoming-crossed}
\end{align}
Here $\eta$ is the finite Euler factor defined in
\cite[equation~(23)]{BBLR}.
The equality follows from the prime-power identity
\cite[equation~(9)]{BBLR}, with all four shifts increased by $z$:
the factors $h_0^{\beta+z}$ and $k_0^{\delta+z}$ convert the two
$\eta$ factors into $\mathcal B_{C_\lambda(z)}(h_0)$ and
$\mathcal B_{C_\lambda(z)^*}(k_0)$.  This also identifies the coefficient
normalization $1/(dh_0k_0)$ with $d/(hk)$ in
\eqref{eq:twisted-coefficient}.

Before integration against $\Phi(t/T)\,\dd t$, the original and dual
zero-frequency terms corresponding to the $(\beta,\delta)$ swap therefore
combine into
\begin{equation}\label{eq:twisted-contour-pair}
 \left(\frac{t}{2\pi}\right)^{-\beta-\delta}
 \frac{1}{2\pi i}
 \left(\int_{(c)}-\int_{(-c)}\right)
 \frac{G(z)}{z}\mathcal M_{C_\lambda(z)}(T)\,\dd z,
\end{equation}
where $c>0$ is fixed and sufficiently small.  The gamma-factor replacement
has already been included in the error in Theorem~\ref{thm:twisted}.

Indeed, if $R_\lambda=(-\delta,-\gamma,-\beta,-\alpha)$, then
$C_{R_\lambda}(z)=C_\lambda(-z)$ up to permutations within the two shift
pairs.  The arithmetic factor is invariant under these permutations.
Changing $z$ to $-z$ in the dual integral and using the evenness of $G$
gives \eqref{eq:twisted-contour-pair}.
Besides $z=0$, the possible poles in this strip are
\[
 z=-\frac{\alpha+\gamma}{2},\qquad
 z=-\frac{\beta+\delta}{2}.
\]
They are cancelled by the prescribed zeros of $G$.
The residue theorem thus evaluates \eqref{eq:twisted-contour-pair} as
\[
 \left(\frac{t}{2\pi}\right)^{-\beta-\delta}
 \mathcal M_{(\alpha,-\delta,\gamma,-\beta)}(T).
\]
Permuting $\alpha,\beta$ and $\gamma,\delta$ gives the other three crossed
terms.  Together with the two diagonal residues, they are exactly the six
terms in \eqref{eq:twisted-main}.

\section{The reflection calculation}\label{app:reflection}
\begin{proof}[Proof of Lemma~\ref{lem:num-reflection}]
It suffices to check $z\ne0$. Write $E(a,z)=e^{az}-1-az$ and
\begin{align*}
 N(a,b,z)={}&z(1-z/2)E(a,z)E(b,z)\\
 &-\frac{z^3}{2}\bigl(a^2E(b,z)+b^2E(a,z)\bigr)
 +\frac{a^2b^2z^4}{4}(e^z-1).
\end{align*}
Elementary integration in the definitions of $B,C,H$ gives
$z^7J(a,b;-z)=N(a,b,z)$. For the other side,
$\min(a,1-b)=a$ and the translated simplex nodes are
$0,a-t,b+t,1$. The divided-difference formula gives
\begin{align*}
 2z^3e^{z/2}K_z(a,1-b)=\int_0^a\bigl[
 &-(b-a+2t)(1-a+t)(1-b-t)\\
 &+e^{z(a-t)}(b+t)(1-b-t)\\
 &-e^{z(b+t)}(a-t)(1-a+t)\\
 &+e^z(a-t)(b+t)(b-a+2t)\bigr]\,\dd t.
\end{align*}
Integrating the polynomial and exponential-quadratic terms yields
$2N(a,b,z)/z^4$. Thus both sides of \eqref{eq:num-reflection} equal
$N(a,b,z)/(z^7e^{z/2})$. Their integral definitions are entire, so the
identity also holds at zero.
\end{proof}

\section{Exact rational evaluation}\label{app:exact}
For the numerator, the simplex moment formula is
\begin{equation}\label{eq:exact-simplex}
 \E(q\cdot X)^r=\frac{6r!}{(r+3)!}
       \sum_{i_0+i_1+i_2+i_3=r}q_0^{i_0}q_1^{i_1}q_2^{i_2}q_3^{i_3}.
\end{equation}
For $r\leq4$, this reduces the numerator kernel to integration of a
polynomial in $t$ of degree at most nine over a rational interval.

For the balanced kernel, the dependence on the four edge variables
is only through $t=\sum r_{ij}$. Set
\[
 e=\min(a+b,c+d),\quad \ell=(t-b)_+,\quad u=\min(a,t),
 \quad m=(t-d)_+,\quad v=\min(c,t),
\]
where $x_+=\max(x,0)$. The density of this slice is
\begin{equation}\label{eq:exact-slice}
 \begin{aligned}
 W(t)&=\int_\ell^u\int_m^v
        \bigl(\min(r,s)-\max(0,r+s-t)\bigr)\dd s\dd r,\\
 6W(t)&=F_t(u,v)-F_t(\ell,v)-F_t(u,m)+F_t(\ell,m),\\
 F_t(r,s)&=3r^2s+(r-s)_+^3-(r+s-t)_+^3.
 \end{aligned}
\end{equation}
With $\Delta=(a+b-c-d)/2$, it follows that
\[
 B_{p,q}(a,b;c,d)
 =\int_0^e W(t)
   \int_{-(1-a-b-c-d+2t)/2}^{(1-a-b-c-d+2t)/2}
     p(\Delta-x)q(-\Delta-x)\dd x\dd t.
\]
Split $[0,e]$ at the points of
$\{a,b,c,d,a+c,a+d,b+c,b+d\}$ that lie in the interval. On each
resulting interval, $W$ is cubic and the inner integral has degree
at most nine, so the product has degree at most twelve. Rational
antiderivatives therefore suffice. Equivalently, the implementation
uses the exact 13-node interpolation rule
\[
 \int_\ell^u Q(t)\dd t
 =(u-\ell)\sum_{j=0}^{12}w_j
    Q\!\left(\ell+\frac j{12}(u-\ell)\right),\qquad
 w_j=\int_0^1\prod_{\substack{0\leq k\leq12\\k\ne j}}
                   \frac{12x-k}{j-k}\dd x.
\]
The weights are rational, and the identity is exact for $\deg Q\leq12$.

For the unbalanced kernel, first expand the integrand and use
\[
 \int_{-r/2}^{r/2}y^{2j}\dd y
   =\frac{r^{2j+1}}{2^{2j}(2j+1)},\qquad
 \int_{-r/2}^{r/2}y^{2j+1}\dd y=0.
\]
Evaluate the remaining capped-simplex polynomial integral by
inclusion--exclusion over $r_1\leq a$, $r_2\leq b$, $r_3\leq c$.
After translating each violated cap, expand the polynomial and use,
for $H\geq0$,
\begin{equation}\label{eq:exact-capped-simplex}
 \int_{\substack{r_1,r_2\geq0\\r_1+r_2\leq H}}
     r_1^ir_2^j(H-r_1-r_2)^k\dd r_1\dd r_2
 =\frac{i!j!k!}{(i+j+k+2)!}H^{i+j+k+2}.
\end{equation}
Terms with negative translated $H$ are zero. These rules, the
projected Gram formulas and the rational input rows completely
specify the finite computation.

\begin{thebibliography}{9}

\bibitem{BBLR}
S.~Bettin, H.~M.~Bui, X.~Li and M.~Radziwi\l\l,
\emph{A quadratic divisor problem and moments of the Riemann zeta-function},
J. Eur. Math. Soc. \textbf{22} (2020), 3953--3980.
\href{https://doi.org/10.4171/JEMS/999}{doi:10.4171/JEMS/999}.
The equation, proposition and page numbers used here refer to
\href{https://arxiv.org/abs/1609.02539v1}{arXiv:1609.02539v1}.

\bibitem{CG}
J.~B.~Conrey and A.~Ghosh,
\emph{A conjecture for the sixth power moment of the Riemann zeta-function},
Int. Math. Res. Not. \textbf{1998}, no.~15, 775--780.
\href{https://doi.org/10.1155/S1073792898000476}{doi:10.1155/S1073792898000476}.
\href{https://arxiv.org/abs/math/9807187}{arXiv:math/9807187}.

\bibitem{deBoor}
C.~de~Boor,
\emph{Math/CS 887: Approximation Theory}, class notes, Spring 2003,
version of 21 November 2009, pp.~52--53, formula~(83).
\url{https://pages.cs.wisc.edu/~deboor/887/notes.pdf}.

\bibitem{DP}
B.~Durkan and T.~Page,
\emph{Amplified moments of the Riemann zeta function},
\href{https://arxiv.org/abs/2606.27323v2}{arXiv:2606.27323v2} (2026).

\bibitem{Code}
N.~Fraccaro,
\emph{Sixth moment: Lean formalization and exact certificate},
repository snapshot \texttt{950bfaff8724d4adb1c37535634a139b70228df0}.
\url{https://github.com/Niklas-Fraccaro/sixth-moment-lean/tree/950bfaff8724d4adb1c37535634a139b70228df0}.
The endpoint table is \texttt{certificate\_250.tsv}; the analytic limits
and exact evaluations are in \texttt{SixthMoment/Numerator},
\texttt{SixthMoment/Denominator} and \texttt{SixthMoment/Certificate}.

\bibitem{KS}
J.~P.~Keating and N.~C.~Snaith,
\emph{Random matrix theory and $\zeta(1/2+it)$},
Comm. Math. Phys. \textbf{214} (2000), 57--89.
\href{https://doi.org/10.1007/s002200000261}{doi:10.1007/s002200000261}.

\bibitem{Watt}
N.~Watt,
\emph{Kloosterman sums and a mean value for Dirichlet polynomials},
J. Number Theory \textbf{53} (1995), 179--210.
\href{https://doi.org/10.1006/jnth.1995.1086}{doi:10.1006/jnth.1995.1086}.

\bibitem{Young}
M.~P.~Young,
\emph{A short proof of Levinson's theorem},
Arch. Math. \textbf{95} (2010), 539--548.
\href{https://doi.org/10.1007/s00013-010-0199-9}{doi:10.1007/s00013-010-0199-9}.
The lemma numbers used here refer to
\href{https://arxiv.org/abs/1002.4403v1}{arXiv:1002.4403v1}.

\end{thebibliography}
\end{document}