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\title{A Checked Improvement of the Numerical Constant in Section 5}
\author{Ingo Althoefer and GPT-5.5 Thinking}
\date{May 28, 2026}

\begin{document}
\maketitle

\section*{Purpose}

This note checks a small modification of the explicit numerical argument in
Section 5 of the manuscript \emph{The sum-product conjecture is false for real
numbers}.  The original Section 5 states that the construction gives
\[
        \max(|A+A|,|AA|) \le |A|^{2-0.00000087}.
\]
The modification below keeps the same construction
\[
        A=GP, \qquad G=B^\times(Y), \qquad P=X+B^+(\varepsilon X),
\]
but improves the treatment of the product set and uses a sharper regulator
estimate.  The checked outcome is the following safe statement.

\begin{theorem}
Using the same tower of totally real number fields and the same construction
as in Section 5, but with the two changes described below, there are
arbitrarily large finite sets $A\subset \RR$ such that
\[
        \max(|A+A|,|AA|) \le |A|^{2-0.0007}.
\]
The optimized value suggested by the same calculation is about
$0.000719$, but $0.0007$ is the safer quoted constant.
\end{theorem}

\section{The improved product-set estimate}

Let $K$ be a totally real field of degree $d$, and let
\[
H=\{x\in \RR^d:x_1+\cdots+x_d=0\}.
\]
Modulo the signs $\{\pm 1\}$, the logarithmic image of the unit group is a
lattice $\Lambda\subset H$ of rank $d-1$.  Put
\[
        K_Y=H\cap[-Y,Y]^d,
        \qquad S_Y=\Lambda\cap K_Y.
\]
Then $B^\times(Y)$ maps exactly two-to-one onto $S_Y$.

\begin{lemma}[lattice doubling]
For every $Y>0$,
\[
        |S_{2Y}|\le 5^{d-1}|S_Y|.
\]
Consequently, if $G=B^\times(Y)$, then
\[
        |GG|\le 5^{d-1}|G|.
\]
\end{lemma}

\begin{proof}
Choose a maximal set $T\subset S_{2Y}$ such that no two distinct elements of
$T$ differ by an element of $S_Y$.  Equivalently, for distinct $t,t'\in T$ we
have $t-t'\notin K_Y\cap\Lambda$.

The translates
\[
        t+\frac12 K_Y, \qquad t\in T,
\]
have disjoint interiors inside the affine space $H$.  Indeed, if two such
translates had intersecting interiors, then $t-t'$ would lie in the interior of
$K_Y$, hence in $K_Y\cap\Lambda=S_Y$, a contradiction.  Since
$T\subset K_{2Y}$, all these translates lie in
\[
        K_{2Y}+\frac12K_Y=\frac52K_Y.
\]
Comparing $(d-1)$-dimensional volumes gives
\[
        |T|\le \frac{\operatorname{vol}_{d-1}((5/2)K_Y)}
                    {\operatorname{vol}_{d-1}((1/2)K_Y)}
              =5^{d-1}.
\]
By maximality of $T$, every element of $S_{2Y}$ differs from some element of
$T$ by an element of $S_Y$.  Therefore
\[
        S_{2Y}\subset T+S_Y,
\]
and hence $|S_{2Y}|\le |T||S_Y|\le 5^{d-1}|S_Y|$.

Finally, $GG$ maps at most two-to-one to $S_Y+S_Y\subset S_{2Y}$, whereas
$G$ maps exactly two-to-one to $S_Y$.  Hence
\[
        |GG|\le 2|S_{2Y}|\le 2\cdot 5^{d-1}|S_Y|=5^{d-1}|G|.
\]
\end{proof}

This replaces the cruder bound
\[
        |GG|\le |B^\times(2Y)|
\]
followed by a separation estimate.  It is the main source of the improved
constant.

\section{A sharper regulator input}

The proof quoted in the manuscript gives, for every $s>1$,
\[
2^d R_Kh_K
   \le
2s(s-1)\left(\pi^{-d/2}\Delta_K^{1/2}\right)^s
          \Gamma(s/2)^d\zeta(s)^d.
\]
Assume now that the fields in the tower satisfy
\[
        \Delta_K\le C_2^d, \qquad C_2=857.57.
\]
Since $h_K\ge 1$, we obtain, for every fixed $s>1$,
\[
        R_K \le (1+o(1))^d
        \left(
        \frac{\zeta(s)\Gamma(s/2)}{2\pi^{s/2}}C_2^{s/2}
        \right)^d.
\]
The expression in parentheses is minimized near
\[
        s=1.371966384\ldots,
\]
where it is
\[
        101.9560759\ldots .
\]
Thus, after passing to sufficiently large degrees in the tower, we may safely
use
\[
        R_K\le 103^d.
\]


\section{The construction and its estimates}

Let
\[
        \phi=\frac{1+\sqrt5}{2},
        \qquad
        \varepsilon=\frac{\phi-1}{\phi+1}=0.2360679774\ldots .
\]
This is the largest value allowed by the directness condition
\[
        \frac{1+\varepsilon}{1-\varepsilon}\le \phi.
\]
Let
\[
        G=B^\times(Y),
        \qquad
        P=X+B^+(\varepsilon X),
        \qquad
        A=GP.
\]
As in Section 5, the representation $a=up$ with $u\in G$ and $p\in P$ is
unique.  Indeed, if $u_1p_1=u_2p_2$, then
$u_1/u_2=p_2/p_1$ is a unit, and every real embedding of $p_2/p_1$ lies in
$(\phi^{-1},\phi)$.  The unit separation lemma therefore gives
$u_1/u_2\in\{\pm1\}$, and the sign $-1$ is impossible because all embeddings of
$p_2/p_1$ are positive.  Hence
\[
        |A|=|G||P|.
\]

Using $R_K\le 103^d$ and $\Delta_K\le C_2^d$, the lower bounds for the two
lattices give, up to harmless $(1+o(1))^d$ factors,
\[
        |G|\ge \left(\frac{Y}{103}\right)^d,
        \qquad
        |P|\ge \left(\frac{\varepsilon X}{\sqrt{C_2}}\right)^d.
\]
Therefore
\[
        |A|\ge (BXY)^d,
        \qquad
        B=\frac{\varepsilon}{103\sqrt{C_2}}
          =0.0000782645\ldots .
\]

For the product set, the improved lattice-doubling lemma gives
\[
\begin{aligned}
        |AA|&\le |GG||PP| \\
             &\le 5^{d-1}|G||P|^2 \\
             &\le \left(\frac{515}{Y}\right)^d |A|^2,
\end{aligned}
\]
again with only harmless $(1+o(1))^d$ losses.  Notice that here we keep the full
unit ball $G$; we do not need to discard elements of $G$.

For the sumset, every element of $A$ has every real embedding bounded in
absolute value by $(1+\varepsilon)Xe^Y$.  Hence
\[
        A+A\subset B^+(2(1+\varepsilon)Xe^Y),
\]
and the additive lattice upper bound gives
\[
        |A+A|
        \le (4(1+\varepsilon)Xe^Y+1)^d.
\]
Dividing by the lower bound for $|A|^2$ yields
\[
        |A+A|\le
        \left(
        \frac{M_1e^Y}{XY^2}+\frac{M_2}{X^2Y^2}
        \right)^d |A|^2,
\]
where
\[
        M_2=\frac{103^2C_2}{\varepsilon^2}=163256270.4\ldots,
\]
and
\[
        M_1=4(1+\varepsilon)M_2=807183391.9\ldots .
\]

We shall also need an upper bound for $|A|$ in order to translate a factor of
the form $q^d$ into a power saving in $|A|$.  The standard upper bounds give
\[
        |G|\le 10(5Y+1)^{d-1},
        \qquad
        |P|\le (2\varepsilon X+1)^d.
\]
Thus, for fixed $X,Y$ and sufficiently large $d$,
\[
        |A|\le \exp(Ld),
\]
where $L$ may be taken to be any number larger than
\[
        \log(5Y+1)+\log(2\varepsilon X+1).
\]

\section{Numerical choice}

Take
\[
        Y=1415,
        \qquad
        X=\lfloor e^{1423}\rfloor .
\]
Then the product estimate gives
\[
        \frac{515}{Y}=0.3639575971\ldots <0.364.
\]
The first additive factor satisfies
\[
        \frac{M_1e^Y}{XY^2}<0.136,
\]
and the second additive factor is negligible.  Thus both estimates are bounded
by
\[
        (0.364+o(1))^d|A|^2.
\]
On the other hand, the upper bound for $|A|$ gives, for all sufficiently large
$d$,
\[
        |A|\le e^{1432d}.
\]
Finally,
\[
        -\log(0.364)=1.010601\ldots
        \quad\textrm{and}\quad
        0.0007\cdot 1432=1.0024.
\]
Therefore, after absorbing the harmless $(1+o(1))^d$ factors by taking $d$
sufficiently large,
\[
        (0.364+o(1))^d \le |A|^{-0.0007}.
\]
Consequently
\[
        \max(|A+A|,|AA|)\le |A|^{2-0.0007}.
\]

\section*{Conclusion}

The recheck is positive.  The improvement rests on two valid changes:

\begin{enumerate}
\item use the lattice-doubling estimate $|GG|\le 5^{d-1}|G|$ instead of bounding
$GG$ directly by the full ball $B^\times(2Y)$; and
\item optimize the regulator estimate over $s>1$, which allows the safe bound
$R_K\le 103^d$ along the same tower for all sufficiently large degrees.
\end{enumerate}

Together these changes improve the explicit constant from
$0.00000087$ to the safe value $0.0007$.

\end{document}