\section{Positive kernels and the reduction to half-integer radii}
\label{sec:kernels}

By Theorem~\ref{thm:classification}, every limit of an admissible sequence
of gap $g$ is a measure $\sigma(R,t_0)$ in which at least one radius, which
we call $R$, is a half-integer. In this section we prove that the other
radius is a half-integer as well (Theorem~\ref{thm:half-integer-limits}),
and that near the remaining limits four scalar states at one fixed energy
must move more slowly than every power of $a$
(Theorem~\ref{thm:four-scalar-slow}). Both results are obtained from
Gaussian kernels that are nonnegative on the physical spectrum and whose
Lorentzian Fourier transforms are finite sums of narrow Gaussians. The
exact Gaussian identity of Lemma~\ref{lem:gaussian-identity} turns
information on finitely many low states into a bound for the sum of the
kernel over all states; the bound localizes every state of a logarithmic
energy window near finitely many circle lattices, with its exact spin, and
this is excluded by Theorems~\ref{thm:fixed-localization}
and~\ref{thm:moving-localization}.

\subsection{Conventions}\label{subsec:ker-conventions}

We write $[u,v]$ for the point of $\RR^2$ whose light-cone coordinates
are $u$ and $v$, that is, the point $(P,Q)=\bigl(\frac{u+v}2,
\frac{u-v}2\bigr)$. All vectors entering kernels are given in this form.
Recall that $|X|^2=P^2+Q^2=\frac12(u^2+v^2)$, that the spin of a real point
is $uv=P^2-Q^2$, and that $B(X,Y)=uv_Y+vu_Y=2(PP_Y-QQ_Y)$ for
$Y=[u_Y,v_Y]=(P_Y,Q_Y)$. For such $Y$ put
\[
 Y^\sharp=[v_Y,u_Y]=(P_Y,-Q_Y),\qquad
 B(X,Y^\sharp)=uu_Y+vv_Y=2(PP_Y+QQ_Y).
\]
The map $Y\mapsto Y^\sharp$ is linear and maps $\Lambda_r$ onto
$\Lambda_{1/r}$. A real point is \emph{off both axes} if $PQ\neq0$,
equivalently $u^2\neq v^2$. The open disc of radius $\varrho$ about
$x\in\RR^2$, for the Euclidean norm of $(P,Q)$, is denoted
$D(x,\varrho)$. The \emph{vacuum support} is
\[
 V=\supp\nu_{\vac}=\bigl\{(P,Q):P,Q\in\{0,\pm1\}\bigr\}
 =\bigl\{[0,0],\ \pm[1,1],\ \pm[1,-1],\ \pm[2,0],\ \pm[0,2]\bigr\}
 \subset\Lambda_1 ,
\]
and by \eqref{eq:physical-measures}
\begin{equation}\label{eq:ker-vacuum-masses}
 \nu_{\vac}(\{(0,0)\})=1,\qquad
 \nu_{\vac}(\{(\pm1,0)\})=\nu_{\vac}(\{(0,\pm1)\})=-\tfrac12,\qquad
 \nu_{\vac}(\{(\pm1,\pm1)\})=\tfrac14 .
\end{equation}

Let $Z$ be admissible at $c=1+12a$. For a state $j$ and
$\sigma\in\{\pm1\}^2$ we write
\[
 X_j^\sigma=\bigl(\sigma_1\sqrt{h_j},\,\sigma_2\sqrt{\bar h_j}\bigr),
 \qquad
 X_{j,a}^\sigma=\bigl(\sigma_1\sqrt{h_j-\tfrac a2},\,
 \sigma_2\sqrt{\bar h_j-\tfrac a2}\bigr)
\]
for its \emph{physical points} and its \emph{shifted points}, with one
determination of each square root fixed. Thus
$\rho_a=\sum_j\frac14\sum_\sigma\delta_{X_j^\sigma}$, and for every entire
$f$
\begin{equation}\label{eq:ker-point-evaluations}
 \ev f(h_j,\bar h_j)=\tfrac14\sum_\sigma f(X_j^\sigma),\qquad
 \ev f\bigl(h_j-\tfrac a2,\bar h_j-\tfrac a2\bigr)
 =\tfrac14\sum_\sigma f(X_{j,a}^\sigma).
\end{equation}
Each coordinate of $X^\sigma_{j,a}$ is real or purely imaginary, and
\begin{equation}\label{eq:ker-shifted-points}
 \bigl|\operatorname{Re}X^\sigma_{j,a}-X^\sigma_j\bigr|\leq\sqrt a,
 \qquad
 \bigl|\operatorname{Im}X^\sigma_{j,a}\bigr|^2\leq a ,
\end{equation}
because $|\sqrt{h-a/2}-\sqrt h|\leq\sqrt{a/2}$ for $h\geq a/2$, while for
$0\leq h<a/2$ the shifted coordinate is imaginary of modulus at most
$\sqrt{a/2}$ and $\sqrt h<\sqrt{a/2}$. The same notation and the same
bounds are used for the four vacuum weights $(k,l)\in\{0,1\}^2$, whose
physical points form $V$. For a function $f$ that is not even, the
physical pairing is by definition the pairing with the even measure
$\mu_a$,
\begin{equation}\label{eq:ker-physical-pairing}
 \langle\mu_a,f\rangle=\sum_{k,l\in\{0,1\}}(-1)^{k+l}\ev f(k,l)
 +\sum_j\ev f(h_j,\bar h_j),
\end{equation}
so that every state enters through all four of its physical points, each
with weight $\frac14$. The analytic evaluation $\mathcal L_a(f)$ of
\eqref{eq:analytic-evaluation} is the same expression with all weights
shifted by $-\frac a2$.

\subsection{Trigonometric kernels and their Fourier transforms}

A \emph{trigonometric polynomial} is a function
\begin{equation}\label{eq:ker-trig-polynomial}
 \Phi(X)=\sum_{i\in\Xi}c_i\,e^{2\pi iB(X,\xi_i)},
\end{equation}
where $\Xi$ is a finite index set, $\xi_i\in\RR^2$ and $c_i\in\RR$. We
always assume that the family $(\xi_i,c_i)_{i\in\Xi}$ is invariant under
$\xi\mapsto-\xi$, so that $\Phi$ is real on $\RR^2$. The points $\xi_i$
are called the \emph{centres} of $\Phi$; they need not be distinct, and
each index is treated separately. For $\epsilon>0$ and $\xi\in\RR^2$ put
\begin{equation}\label{eq:ker-broad-narrow}
 \Phi_\epsilon(X)=e^{-2\pi\epsilon^2|X|^2}\,\Phi(X)
 =e^{-\pi\epsilon^2(u^2+v^2)}\,\Phi(X),
 \qquad
 \gamma_{\epsilon,\xi}(X)=\epsilon^{-2}e^{-2\pi|X-\xi|^2/\epsilon^2}.
\end{equation}
We call $\Phi_\epsilon$ a \emph{kernel}; its first factor is the broad
Gaussian and $\gamma_{\epsilon,\xi}$ is a narrow Gaussian. As in
Section~\ref{sec:setup}, $|X-\xi|^2$ denotes the polynomial
$(P-P_\xi)^2+(Q-Q_\xi)^2$ when $X$ is complex, so that
\begin{equation}\label{eq:ker-narrow-modulus}
 |\gamma_{\epsilon,\xi}(X)|
 =\epsilon^{-2}\exp\!\Bigl(-\frac{2\pi}{\epsilon^2}
 \bigl(|\operatorname{Re}X-\xi|^2-|\operatorname{Im}X|^2\bigr)\Bigr).
\end{equation}

The building blocks are, for real $Y$ and an integer $D\geq1$,
\begin{equation}\label{eq:ker-sine-square}
 S_Y(X)=\sin^2\bigl(\pi B(X,Y)\bigr),\qquad
 \Omega_D(X)=\sin^2(\pi Du)+\sin^2(\pi Dv)=S_{[0,D]}(X)+S_{[D,0]}(X),
\end{equation}
and we write
\[
 C=[2,0],\qquad C^\sharp=[0,2],\qquad
 \Omega=\Omega_2=S_C+S_{C^\sharp}=\sin^2(2\pi u)+\sin^2(2\pi v).
\]
A \emph{product kernel} is built from a product of four sine-square
factors $S_{Y_1}S_{Y_1^\sharp}S_{Y_2}S_{Y_2^\sharp}$, and a \emph{sum
kernel} from $\Omega_DS_YS_{Y^\sharp}$ or from $\Omega_D$ alone. If
$Y\in\Lambda_r$, then $B(\cdot,Y)$ is integral on $\Lambda_r$ and
$B(\cdot,Y^\sharp)$ is integral on $\Lambda_{1/r}$; hence
\begin{equation}\label{eq:ker-vanishing}
 S_YS_{Y^\sharp}=0\ \text{ on }\Lambda_r\cup\Lambda_{1/r}
 \quad(Y\in\Lambda_r),
 \qquad
 \Omega_D=0\ \text{ on }\{[u,v]:Du,Dv\in\ZZ\}.
\end{equation}

\begin{lemma}[Fourier expansion of kernels]\label{lem:kernel-fourier}
Let $\Phi$ be a trigonometric polynomial as in
\eqref{eq:ker-trig-polynomial} and let $\epsilon>0$. Then
$\Phi_\epsilon\in\mathcal G_{2\pi\epsilon^2}$,
\begin{equation}\label{eq:ker-fourier}
 \mathbb S\Phi_\epsilon=\sum_{i\in\Xi}c_i\,\gamma_{\epsilon,\xi_i}
 ,\qquad\text{that is,}\qquad
 \mathbb S\Phi_\epsilon(X)=\epsilon^{-2}\sum_{i\in\Xi}c_i
 \exp\!\Bigl(-\frac{\pi}{\epsilon^2}
 \bigl((u-u_{\xi_i})^2+(v-v_{\xi_i})^2\bigr)\Bigr),
\end{equation}
and for every admissible partition function at $c=1+12a$
\begin{equation}\label{eq:ker-sum-rule}
 \mathcal L_a(\Phi_\epsilon)
 =\sum_{i\in\Xi}c_i\,\mathcal L_a(\gamma_{\epsilon,\xi_i}),
\end{equation}
all evaluations being absolutely convergent. The centres and
coefficients of the building blocks are as follows.
\begin{enumerate}[(i)]
\item For $\Phi=S_{Y_1}S_{Y_2}S_{Y_3}S_{Y_4}$: $\Xi=\{-1,0,1\}^4$,
 $\xi_\eta=\sum_{i=1}^4\eta_iY_i$ and
 \[
 c_\eta=\frac1{16}\Bigl(-\frac12\Bigr)^{|\eta|},\qquad
 |\eta|=\#\{i:\eta_i\neq0\}.
 \]
 The coefficient is negative exactly at the centres with $|\eta|\in\{1,3\}$.
\item For $\Phi=\Omega_DS_YS_{Y^\sharp}$: the $45$ centres are
 $\zeta+\eta_1Y+\eta_2Y^\sharp$ with
 $\zeta\in\{0,\pm[D,0],\pm[0,D]\}$ and $\eta\in\{-1,0,1\}^2$, and
 \[
 \begin{array}{c|ccc}
 & \eta=0 & |\eta|=1 & |\eta|=2\\\hline
 \zeta=0 & \tfrac14 & -\tfrac18 & \tfrac1{16}\\[2pt]
 \zeta\neq0 & -\tfrac1{16} & \tfrac1{32} & -\tfrac1{64}
 \end{array}
 \]
 are the coefficients. There are $1$, $4$, $4$ centres in the first row
 and $4$, $16$, $16$ in the second.
\item For $\Phi=\Omega_D$: the centres are $0$ with coefficient $1$ and
 $\pm[D,0]$, $\pm[0,D]$ with coefficient $-\frac14$.
\end{enumerate}
\end{lemma}
\begin{proof}
In momentum coordinates $e^{2\pi iB(X,\xi)}=e^{z\cdot X}$ with
$z=4\pi i(P_\xi,-Q_\xi)$, so
$\Phi_\epsilon$ is a finite sum of functions $\phi_{b,z}$ with
$b=2\pi\epsilon^2$ and belongs to $\mathcal G_b$. By
\eqref{eq:S-gaussian}, with $J=\operatorname{diag}(1,-1)$ and
$z=4\pi iJ\xi$,
\[
 \mathbb S\bigl(e^{-2\pi\epsilon^2|X|^2}e^{2\pi iB(X,\xi)}\bigr)
 =\frac1{\epsilon^2}\exp\!\Bigl(
 \frac{(4\pi i)^2\,J(\xi+X)\cdot J(\xi+X)}{8\pi\epsilon^2}\Bigr)
 =\gamma_{\epsilon,-\xi}(X).
\]
Summing over $i$ and using the invariance of the family under
$\xi\mapsto-\xi$ gives \eqref{eq:ker-fourier}; the second form is the
same formula in light-cone coordinates, since
$2|X-\xi|^2=(u-u_\xi)^2+(v-v_\xi)^2$. Lemma~\ref{lem:gaussian-identity}
gives $\mathcal L_a(\Phi_\epsilon)=\mathcal L_a(\mathbb S\Phi_\epsilon)$
with absolute convergence, and each $\gamma_{\epsilon,\xi}$ lies in
$\mathcal G_{2\pi/\epsilon^2}$, which gives \eqref{eq:ker-sum-rule}.

The tables follow from
\[
 S_Y(X)=\tfrac12-\tfrac14e^{2\pi iB(X,Y)}-\tfrac14e^{-2\pi iB(X,Y)},
 \qquad
 \Omega_D(X)=1-\tfrac14\sum_{\zeta\in\{\pm[D,0],\pm[0,D]\}}
 e^{2\pi iB(X,\zeta)},
\]
the second because $B(X,[0,D])=Du$ and $B(X,[D,0])=Dv$. Multiplying four
factors of the first kind gives the coefficient
$(\frac12)^{4-|\eta|}(-\frac14)^{|\eta|}$ of (i). In (ii) the coefficient
is the product of $1$ or $-\frac14$, according as $\zeta=0$ or not, with
$(\frac12)^{2-|\eta|}(-\frac14)^{|\eta|}\in\{\frac14,-\frac18,\frac1{16}\}$.
In each case the family is invariant under $\xi\mapsto-\xi$.
\end{proof}

\begin{lemma}[Evenness and positivity]\label{lem:kernel-positivity}
Let $Y\in\RR^2$ and let $D\geq1$ be an integer.
\begin{enumerate}[(i)]
\item As entire functions on $\CC^2$,
 \begin{equation}\label{eq:ker-pair-square}
 \begin{aligned}
 S_Y(X)S_{Y^\sharp}(X)&=\tfrac14\bigl(\cos(4\pi Q_YQ)-\cos(4\pi P_YP)
 \bigr)^2,\\
 \Omega_D(X)&=1-\cos(2\pi DP)\cos(2\pi DQ).
 \end{aligned}
 \end{equation}
 In particular $S_YS_{Y^\sharp}$ and $\Omega_D$ are even.
\item If $\Phi$ is a product of functions $S_YS_{Y^\sharp}$, then
 $\ev{\Phi_\epsilon}(h,\bar h)\geq0$ for all real $h,\bar h$ of either
 sign. Consequently, for an admissible partition function at $c=1+12a$,
 the physical and the shifted evaluations of a product kernel at every
 state are nonnegative, also when one shifted momentum is imaginary.
\item If $\Phi$ is $\Omega_D$ times a product of functions
 $S_YS_{Y^\sharp}$, then $\Phi_\epsilon\geq0$ on $\RR^2$, so the physical
 evaluations $\ev{\Phi_\epsilon}(h_j,\bar h_j)=\Phi_\epsilon(X_j^\sigma)$
 are nonnegative. The shifted evaluation of $\Omega_D$ at a point with
 $P$ real and $Q=i\tau$ imaginary equals
 $1-\cos(2\pi DP)\cosh(2\pi D\tau)$ and may be negative.
\end{enumerate}
\end{lemma}
\begin{proof}
With $\alpha=2\pi P_YP$ and $\beta=2\pi Q_YQ$ one has
$\pi B(X,Y)=\alpha-\beta$ and $\pi B(X,Y^\sharp)=\alpha+\beta$, and
$\sin(\alpha-\beta)\sin(\alpha+\beta)=\frac12(\cos2\beta-\cos2\alpha)$.
Similarly $\sin^2(\pi Du)+\sin^2(\pi Dv)
=1-\frac12\bigl(\cos(2\pi Du)+\cos(2\pi Dv)\bigr)$ and $u=P+Q$, $v=P-Q$.
This is (i). The function $x\mapsto\cos(4\pi P_Y\sqrt x)$ is entire with
real Taylor coefficients, hence real for real $x$ of either sign.
Therefore $\ev{(S_YS_{Y^\sharp})}(h,\bar h)$ is the square of a real
number for real $h,\bar h$; the broad Gaussian contributes the positive
factor $e^{-2\pi\epsilon^2(h+\bar h)}$, and (ii) follows. Part (iii) is
immediate from \eqref{eq:ker-sine-square} and
\eqref{eq:ker-pair-square}.
\end{proof}

Only physical evaluations enter the budgets below; part (ii) is
recorded because it shows that product kernels give positive sum rules
directly on the analytic evaluations.

\subsection{Central-shift and tail estimates}

\begin{lemma}[Central shift for kernels]\label{lem:kernel-shift}
Let $0<g\leq\frac12$, and let $\Phi$ be a trigonometric polynomial with
$|\xi_i|\leq\varrho_*$ for all $i$ and $\sum_i|c_i|\leq c_*$. There is
$C=C(g,\varrho_*,c_*)$ such that, for $0<\epsilon\leq\frac12$, every
admissible partition function of gap $g$ at $c=1+12a$ with
$0<a\leq\min\{g^2/4,\epsilon^2/4\}$ satisfies
\begin{align}
 \bigl|\ev{\Phi_\epsilon}\bigl(h-\tfrac a2,\bar h-\tfrac a2\bigr)
 -\ev{\Phi_\epsilon}(h,\bar h)\bigr|
 &\leq C\,a\,e^{-\pi\epsilon^2(h+\bar h)}
 \qquad(h,\bar h\geq0),\label{eq:ker-shift-single}\\
 \sum_j\bigl|\ev{\Phi_\epsilon}\bigl(h_j-\tfrac a2,\bar h_j-\tfrac a2\bigr)
 -\ev{\Phi_\epsilon}(h_j,\bar h_j)\bigr|
 &\leq C\,a\,\epsilon^{-2},\label{eq:ker-shift-sum}\\
 \Bigl|\sum_{k,l\in\{0,1\}}(-1)^{k+l}
 \ev{\Phi_\epsilon}\bigl(k-\tfrac a2,l-\tfrac a2\bigr)
 -\langle\nu_{\vac},\Phi_\epsilon\rangle\Bigr|
 &\leq C\,a .\label{eq:ker-shift-vacuum}
\end{align}
In particular, if $\Phi$ vanishes on $V$, the shifted vacuum value of
$\Phi_\epsilon$ is at most $Ca$ in absolute value.
\end{lemma}
\begin{proof}
Let $|\operatorname{Im}P|,|\operatorname{Im}Q|\leq1$ and $|\alpha|\leq2$.
Since $|\operatorname{Im}B(X,\xi)|\leq2|\operatorname{Im}X||\xi|$, each
exponential $e^{2\pi iB(X,\xi_i)}$ and its derivatives of order at most
two are bounded by $(1+4\pi\varrho_*)^2e^{4\sqrt2\pi\varrho_*}$. The
broad Gaussian has modulus
$e^{-2\pi\epsilon^2(|\operatorname{Re}X|^2-|\operatorname{Im}X|^2)}
\leq e^{4\pi}e^{-2\pi\epsilon^2|\operatorname{Re}X|^2}$, and its
derivatives of order at most two are this function times polynomials of
degree at most two in $\epsilon^2P$, $\epsilon^2Q$ and $\epsilon^2$, hence
bounded by a constant times
$(1+\epsilon^2|\operatorname{Re}X|)^2e^{-2\pi\epsilon^2|\operatorname{Re}X|^2}
\leq C'e^{-\pi\epsilon^2|\operatorname{Re}X|^2}$. Therefore
\begin{equation}\label{eq:ker-kernel-derivatives}
 |\partial^\alpha\Phi_\epsilon(X)|\leq M\,e^{-\pi\epsilon^2|\operatorname{Re}X|^2},
 \qquad M=C''(\varrho_*)\,c_* ,
\end{equation}
on this strip. Corollary~\ref{cor:central-shift} with $\beta=\pi\epsilon^2
\leq1$ gives \eqref{eq:ker-shift-single}, and applying it to the four
vacuum weights gives \eqref{eq:ker-shift-vacuum}. Since
$a\leq\beta/(4\pi)$, summing \eqref{eq:ker-shift-single} over the states
with \eqref{eq:thermal-moment} gives \eqref{eq:ker-shift-sum}.
\end{proof}

Narrow test functions are not uniformly bounded on a strip of fixed
width. For them we use the following variant of
Corollary~\ref{cor:central-shift}, in which only the strip actually
visited by the central shift enters.

\begin{lemma}[Central shift on the thin strip]\label{lem:thin-shift}
Let $0<a\leq1$, let $f$ be entire on $\CC^2$, and suppose that for some
$M<\infty$ and $\beta>0$
\[
 |\partial^\alpha f(X)|\leq M\,e^{-\beta|\operatorname{Re}X|^2}
 \qquad\text{for }|\alpha|\leq2,\ \
 |\operatorname{Im}P|,|\operatorname{Im}Q|\leq\sqrt{a/2}.
\]
Then for all $h,\bar h\geq0$
\[
 \bigl|\ev f\bigl(h-\tfrac a2,\bar h-\tfrac a2\bigr)-\ev f(h,\bar h)\bigr|
 \leq a\,M\,e^{3\beta}\,e^{-\beta(h+\bar h)} .
\]
\end{lemma}
\begin{proof}
It suffices to show that
$|\partial_h\ev f(h',\bar h')|\leq\frac12Me^{2\beta}e^{-\beta(h'+\bar h')}$
for real $h',\bar h'\geq-\frac a2$, and the same for $\partial_{\bar h}$;
integrating $\frac{d}{ds}\ev f(h-\frac s2,\bar h-\frac s2)$ over
$0\leq s\leq a$ then gives the claim, with $e^{\beta a}\leq e^{\beta}$.
Fix $\bar h'$ and a sign of $\sqrt{\bar h'}$, and let $\phi$ be the
function $P\mapsto f(P,\pm\sqrt{\bar h'})$. If $h'\leq2$,
Lemma~\ref{lem:sq-momentum-derivative} with $k=1$ expresses the
derivative as $\frac14\int_{-1}^1\phi''(t\sqrt{h'})\dd t$; the points
$(t\sqrt{h'},\pm\sqrt{\bar h'})$ lie in the thin strip and the square of
their real part is at least $\max\{\bar h',0\}\geq h'+\bar h'-2$. If
$h'>2$, the derivative is
$\bigl(\phi'(\sqrt{h'})-\phi'(-\sqrt{h'})\bigr)/(4\sqrt{h'})$, the
points are in the thin strip, and the square of their real part is at
least $h'+\bar h'$. Averaging over the sign of $\sqrt{\bar h'}$ gives the
bound in both cases.
\end{proof}

\begin{lemma}[Tails of narrow Gaussians]\label{lem:narrow-tail}
Let $0<g\leq\frac12$ and $\varrho_*\geq1$. There is
$C_1=C_1(g,\varrho_*)$ such that for every admissible partition function
of gap $g$ at $c=1+12a$ with $0<a\leq g^2/4$, every $x\in\RR^2$ with
$|x|\leq\varrho_*$, every $0<\varrho\leq\frac14$ and every
$0<\epsilon\leq\frac12$,
\begin{equation}\label{eq:ker-narrow-tail}
 \sum_j\frac14\sum_{\sigma:\,|X_j^\sigma-x|\geq\varrho}
 \exp\!\Bigl(-\frac{\pi|X_j^\sigma-x|^2}{2\epsilon^2}\Bigr)
 \leq C_1\,e^{-\pi\varrho^2/(2\epsilon^2)} .
\end{equation}
If moreover $a\leq\varrho^2/4$, then for every state and every $\sigma$
with $|X_j^\sigma-x|\geq\varrho$,
\begin{equation}\label{eq:ker-far-point}
 |\gamma_{\epsilon,x}(X^\sigma_{j,a})|\leq\epsilon^{-2}e^{2\pi a/\epsilon^2}
 \exp\!\Bigl(-\frac{\pi|X_j^\sigma-x|^2}{2\epsilon^2}\Bigr),
\end{equation}
and the same holds for the shifted points of the four vacuum weights.
\end{lemma}
\begin{proof}
Put $E_1=4\varrho_*^2\geq4$. By \eqref{eq:count-bound} there are at most
$A_gE_1e^{\pi E_1/8}$ states with $\Delta_j\leq E_1$, and each of their
terms is at most $e^{-\pi\varrho^2/(2\epsilon^2)}$. If $\Delta_j>E_1$,
then $|X_j^\sigma|=\sqrt{\Delta_j}>2|x|$, so
$|X_j^\sigma-x|\geq\frac12|X_j^\sigma|$ and the term is at most
$e^{-\beta\Delta_j}$ with $\beta=\pi/(8\epsilon^2)\geq\pi/2$. Since
$2\pi a\leq\pi/8\leq\beta/2$, \eqref{eq:thermal-tail} bounds the sum of
these terms by
\[
 A_g\beta\Bigl(\frac{2E_1}{\beta}+\frac4{\beta^2}\Bigr)e^{-\beta E_1/2}
 \leq A_g\bigl(2E_1+\tfrac8\pi\bigr)e^{-\pi/(4\epsilon^2)}
 \leq A_g\bigl(2E_1+\tfrac8\pi\bigr)e^{-\pi\varrho^2/(2\epsilon^2)} .
\]
This proves \eqref{eq:ker-narrow-tail}. If $a\leq\varrho^2/4$ and
$|X_j^\sigma-x|\geq\varrho$, then by \eqref{eq:ker-shifted-points}
$|\operatorname{Re}X^\sigma_{j,a}-x|\geq|X_j^\sigma-x|-\sqrt a
\geq\frac12|X_j^\sigma-x|$ and $|\operatorname{Im}X^\sigma_{j,a}|^2\leq a$,
and \eqref{eq:ker-far-point} follows from \eqref{eq:ker-narrow-modulus}.
\end{proof}

\subsection{The peak budget}

By \eqref{eq:ker-sum-rule}, the analytic evaluation of a kernel is a
signed sum of narrow Gaussian peaks. The next lemma bounds this sum from
above. Peaks with positive coefficient need only an upper bound for the
number of states near the centre; peaks with negative coefficient need a
lower bound, which is available when enough states are known to sit close
to the centre, away from the momentum axes. The leading terms cancel by
a relation among the limiting masses. The lemma is stated for a single
partition function; all its hypotheses are explicit, and the limiting
measure enters only through the numbers $m_i$.

\begin{lemma}[Peak budget]\label{lem:peak-budget}
Let $0<g\leq\frac12$, $\varrho_*\geq1$, $c_*,m_*<\infty$ and
$0<\varrho\leq\min\{\frac14,\frac12\sqrt g\}$. Let $\Phi$ be a
trigonometric polynomial \eqref{eq:ker-trig-polynomial} with
$|\xi_i|\leq\varrho_*$ and $\sum_i|c_i|\leq c_*$, and let
$(m_i)_{i\in\Xi}$ be real numbers with $0\leq m_i\leq m_*$ and
\begin{equation}\label{eq:ker-mass-balance}
 \sum_{i\in\Xi}c_i\,m_i=0 .
\end{equation}
Let $Z$ be admissible of gap $g$ at $c=1+12a$, let $\varepsilon\geq0$,
and assume the following.
\begin{enumerate}[(B1)]
\item[(B0)] $\Phi$ vanishes on the vacuum support $V$.
\item[(B1)] \emph{Origin.} If $\xi_i=0$, then $c_i>0$ and $m_i=1$.
\item[(B2)] \emph{Positive centres.} If $c_i>0$ and $\xi_i\neq0$, then
 $\dist(\xi_i,V)\geq\varrho$ and
 $\rho_a\bigl(D(\xi_i,\varrho)\bigr)\leq m_i$.
\item[(B3)] \emph{Negative centres.} If $c_i<0$, then one of the
 following holds.
 \begin{enumerate}[(a)]
 \item \emph{Anchored centre.} $|P_{\xi_i}|\geq2\varrho$ and
 $|Q_{\xi_i}|\geq2\varrho$; the disc $D(\xi_i,\varrho)$ contains no
 point of $V$ other than possibly $\xi_i$ itself, and in that case
 $\xi_i\in\{(\pm1,\pm1)\}$; and, with
 $w_i=\nu_{\vac}(\{\xi_i\})\in\{0,\frac14\}$, there are $n_i\geq0$
 distinct states $j$ with
 \[
 s_j=P_{\xi_i}^2-Q_{\xi_i}^2,\qquad
 \bigl|\Delta_j-|\xi_i|^2\bigr|\leq\varepsilon ,
 \qquad\text{and}\qquad w_i+\tfrac14n_i\geq m_i .
 \]
 \item \emph{Empty centre.} $m_i=0$, $\dist(\xi_i,V)\geq\varrho$ and
 $\rho_a\bigl(D(\xi_i,\varrho)\bigr)=0$.
 \end{enumerate}
\end{enumerate}
Then there is $C=C(g,\varrho_*,c_*,m_*)$, independent of $\varrho$, $Z$,
$a$, $\epsilon$ and $\varepsilon$, such that
\begin{equation}\label{eq:ker-peak-budget}
 \sum_j\ev{\Phi_\epsilon}(h_j,\bar h_j)
 \leq\frac{C}{\varrho^{2}}\Bigl(\bigl(a+\varepsilon^2\bigr)\epsilon^{-4}
 +e^{-\pi\varrho^2/(4\epsilon^2)}\Bigr)
\end{equation}
whenever
\begin{equation}\label{eq:ker-budget-range}
 0<\epsilon\leq\tfrac12,\qquad
 0<a\leq\min\bigl\{\tfrac14g^2,\ \tfrac14\epsilon^2,\
 \tfrac14\varrho^2\bigr\},\qquad
 \varepsilon\leq\varrho^2 .
\end{equation}
In particular, for fixed $\varrho$ the right side of
\eqref{eq:ker-peak-budget} is at most
$C'\bigl((a+\varepsilon^2)\epsilon^{-4}+e^{-\kappa/\epsilon^2}\bigr)$
with $\kappa=\pi\varrho^2/4$ and $C'=C\varrho^{-2}$.
\end{lemma}

No lower bound on the mass near a positive centre is assumed, positive
centres are allowed to carry limiting mass, and the states near a
positive centre need not be close to it on the scale $\epsilon$. No
bound of this kind could hold at a positive centre placed at one of the
negative atoms $(\pm1,0)$, $(0,\pm1)$ of $\nu_{\vac}$, which is why (B2)
and (B3) keep all centres other than the origin and $(\pm1,\pm1)$ away
from $V$.

\begin{proof}
Write $\gamma_i=\gamma_{\epsilon,\xi_i}$ and
$\mathsf N=\epsilon^{-2}e^{-\pi\varrho^2/(2\epsilon^2)}$. Since $\Phi$ is
real on $\RR^2$, the entire function $\ev{\Phi_\epsilon}$ is real at real
arguments and $\mathcal L_a(\Phi_\epsilon)$ is real, so by
\eqref{eq:ker-sum-rule}
\begin{equation}\label{eq:ker-real-parts}
 \mathcal L_a(\Phi_\epsilon)
 =\sum_{i\in\Xi}c_i\operatorname{Re}\mathcal L_a(\gamma_i).
\end{equation}
Fix $i$ and write $\xi=\xi_i$. By \eqref{eq:ker-point-evaluations},
$\mathcal L_a(\gamma_i)$ is a sum over the pairs $(j,\sigma)$ and over
the shifted points of the four vacuum weights. Call a pair, or a vacuum
point, \emph{far} if the corresponding physical point is at distance at
least $\varrho$ from $\xi$, and \emph{near} otherwise. By
\eqref{eq:ker-far-point}, \eqref{eq:ker-narrow-tail} and
$e^{2\pi a/\epsilon^2}\leq e^{\pi/2}$, the far terms satisfy
\begin{equation}\label{eq:ker-far-total}
 \sum_{(j,\sigma)\ \mathrm{far}}\tfrac14\bigl|\gamma_i(X^\sigma_{j,a})\bigr|
 +\sum_{\mathrm{far\ vacuum\ points}}\tfrac14\bigl|\gamma_i\bigr|
 \leq e^{\pi/2}(C_1+4)\,\mathsf N=:\mathsf F .
\end{equation}
We also use $e^{x}\leq1+e^{\pi/2}x$ for $0\leq x\leq\frac\pi2$, with
$x=2\pi a/\epsilon^2$.

\emph{The origin.} Let $\xi=0$. Every state is far, because
$|X_j^\sigma|=\sqrt{\Delta_j}\geq\sqrt g\geq2\varrho$, and so are the
vacuum weights other than $(0,0)$. The shifted point of the weight
$(0,0)$ is $(i\sqrt{a/2},i\sqrt{a/2})$ up to signs, where $\gamma_i$
equals $\epsilon^{-2}e^{2\pi a/\epsilon^2}$. Hence
\[
 \operatorname{Re}\mathcal L_a(\gamma_i)
 \leq\epsilon^{-2}+2\pi e^{\pi/2}a\,\epsilon^{-4}+\mathsf F
 =m_i\epsilon^{-2}+2\pi e^{\pi/2}a\,\epsilon^{-4}+\mathsf F .
\]

\emph{Positive centres.} Let $c_i>0$ and $\xi\neq0$. All vacuum points
are far. At a near pair, \eqref{eq:ker-narrow-modulus} and
\eqref{eq:ker-shifted-points} give
$|\gamma_i(X^\sigma_{j,a})|\leq\epsilon^{-2}e^{2\pi a/\epsilon^2}$, and
the near pairs have total weight
$\sum\frac14=\rho_a(D(\xi,\varrho))\leq m_i$. Hence
\[
 \operatorname{Re}\mathcal L_a(\gamma_i)
 \leq m_i\epsilon^{-2}+2\pi e^{\pi/2}m_*a\,\epsilon^{-4}+\mathsf F .
\]

\emph{Empty centres.} In case (B3)(b) every pair and every vacuum point
is far, so $|\mathcal L_a(\gamma_i)|\leq\mathsf F$, and $m_i=0$.

\emph{Anchored centres.} Assume (B3)(a). If a state $j$ has a near
physical point, then $\sqrt{h_j}\geq|P_\xi|-\varrho\geq\varrho$ and
likewise $\sqrt{\bar h_j}\geq\varrho$, so
$h_j-\frac a2$ and $\bar h_j-\frac a2$ are positive, all four shifted
points of $j$ are real, and $\gamma_i$ is positive at each of them. If
$j$ has no near point, all its pairs are far. Consequently the real part
of the sum over the states is at least the sum of
$\frac14\gamma_i(X^\sigma_{j,a})$ over any chosen set of pairs with
near physical points, minus the far terms. Let $j$ be one of the $n_i$
states of (B3)(a), and let $\sigma$ be the sign pattern of $\xi$. Since
$s_j=P_\xi^2-Q_\xi^2$, we have
$|h_j-P_\xi^2|=|\bar h_j-Q_\xi^2|=\frac12|\Delta_j-|\xi|^2|
\leq\frac12\varepsilon$. As $P_\xi^2\geq4\varrho^2$,
$\varepsilon\leq\varrho^2$ and $a\leq\varrho^2$, the number
$h_j-\frac a2$ is positive, and
\[
 \Bigl|\sqrt{h_j-\tfrac a2}-|P_\xi|\Bigr|
 \leq\frac{|h_j-\frac a2-P_\xi^2|}{|P_\xi|}
 \leq\frac{\varepsilon+a}{4\varrho} ,
\]
and the same for the second coordinate. The same computation with $a=0$
shows that $X_j^\sigma$ is near. Therefore
$|X^\sigma_{j,a}-\xi|^2\leq(\varepsilon+a)^2/(8\varrho^2)$ and
\[
 \gamma_i(X^\sigma_{j,a})\geq\epsilon^{-2}\max\{0,1-x\},\qquad
 x=\frac{\pi(\varepsilon+a)^2}{4\varrho^2\epsilon^2}.
\]
If $\xi\in\{(\pm1,\pm1)\}$, the vacuum weight $(1,1)$ enters
$\mathcal L_a$ with sign $+1$; its four shifted points
$\sqrt{1-a/2}\,(\pm1,\pm1)$ are real, the one with the signs of $\xi$ is
at squared distance $2(1-\sqrt{1-a/2})^2\leq a^2/2$ from $\xi$, and
$\gamma_i$ is at least $\epsilon^{-2}(1-\pi a^2/\epsilon^2)
\geq\epsilon^{-2}(1-x)$ there. The physical points of the other vacuum
weights are at distance at least one from $\xi$ and are far. If
$\xi\notin V$, all vacuum points are far. Altogether
\[
 \operatorname{Re}\mathcal L_a(\gamma_i)
 \geq\bigl(w_i+\tfrac14n_i\bigr)\epsilon^{-2}\max\{0,1-x\}-\mathsf F
 \geq m_i\epsilon^{-2}
 -\frac{\pi m_*(\varepsilon^2+a)}{2\varrho^2\epsilon^{4}}-\mathsf F ,
\]
where we used $w_i+\frac14n_i\geq m_i\geq0$ and
$(\varepsilon+a)^2\leq2\varepsilon^2+2a$.

\emph{Conclusion.} Multiplying each of the four bounds by $c_i$, with the
appropriate sign, and inserting them into \eqref{eq:ker-real-parts} we
obtain, by \eqref{eq:ker-mass-balance},
\[
 \mathcal L_a(\Phi_\epsilon)
 \leq\epsilon^{-2}\sum_ic_im_i
 +c_*\Bigl(2\pi e^{\pi/2}m_*\frac{a}{\epsilon^4}
 +\frac{\pi m_*(\varepsilon^2+a)}{2\varrho^2\epsilon^4}+\mathsf F\Bigr)
 \leq\frac{C}{\varrho^2}\Bigl(\frac{a+\varepsilon^2}{\epsilon^4}
 +e^{-\pi\varrho^2/(4\epsilon^2)}\Bigr),
\]
because $\mathsf N\leq4(\pi e\varrho^2)^{-1}e^{-\pi\varrho^2/(4\epsilon^2)}$.
Finally, by (B0) and Lemma~\ref{lem:kernel-shift},
\[
 \sum_j\ev{\Phi_\epsilon}(h_j,\bar h_j)
 \leq\mathcal L_a(\Phi_\epsilon)+Ca+Ca\epsilon^{-2},
\]
and the last two terms are bounded by $Ca\epsilon^{-4}\varrho^{-2}$.
\end{proof}

Along a convergent sequence, all hypotheses of
Lemma~\ref{lem:peak-budget} except the displacement bound in (B3)(a)
follow from vague convergence.

\begin{lemma}[The hypotheses along a convergent sequence]
\label{lem:budget-hypotheses}
Let an admissible sequence of gap $g$ converge to $\rho_0$, and put
$\mu_0=\nu_{\vac}+\rho_0$. For each $a$ in the sequence let
$\Phi^{(a)}=\sum_{i\in\Xi}c_ie^{2\pi iB(\cdot,\xi_i(a))}$ be a
trigonometric polynomial with a fixed index set and fixed coefficients,
such that $\xi_i(a)\to\xi_i^0$ as $a\to0$ for every $i$, and
$\xi_i(a)=\xi_i^0$ for all $a$ whenever $\xi_i^0\in V$. Let $\Phi^{(0)}$
be the trigonometric polynomial with centres $\xi_i^0$, assume that
$\Phi^{(0)}$ vanishes on $\supp\mu_0$, and put
$m_i=\mu_0(\{\xi_i^0\})$.
\begin{enumerate}[(i)]
\item The masses satisfy \eqref{eq:ker-mass-balance}:
 $\sum_ic_i\,\mu_0(\{\xi_i^0\})=0$.
\item There is $\varrho_0>0$ such that for every
 $0<\varrho\leq\varrho_0$ and all sufficiently small $a$ in the
 sequence: condition \textup{(B2)} holds for $\Phi^{(a)}$ at every index
 with $\xi_i^0\notin V$, and condition \textup{(B3)(b)} holds at every
 index with $\xi_i^0\notin V\cup\supp\rho_0$.
\item Let $\xi_i^0$ be off both axes, with
 $\xi_i^0\notin V\setminus\{(\pm1,\pm1)\}$, and put
 $n_i=4\rho_0(\{\xi_i^0\})$. For $0<\varrho\leq\varrho_0$ and all
 sufficiently small $a$, exactly $n_i$ states of $\rho_a$ have a
 physical point in $D(\xi_i^0,\varrho)$; they have the spin
 $s(\xi_i^0)$ of $\xi_i^0$, and their dimensions converge to
 $|\xi_i^0|^2$. If $s(\xi_i(a))=s(\xi_i^0)$ and these dimensions are
 within $\varepsilon$ of $|\xi_i(a)|^2$, then \textup{(B3)(a)} holds
 for $\Phi^{(a)}$ at the index $i$ with these $n_i$ states, and
 $w_i+\frac14n_i=m_i$.
\end{enumerate}
The thresholds in $a$ depend on the sequence and are not effective.
\end{lemma}
\begin{proof}
(i) For $a=0$ the evaluation $\mathcal L_0(f)$ is the pairing
$\langle\mu_0,f\rangle$, and Lemma~\ref{lem:gaussian-identity} (in the
case $a=0$) together with \eqref{eq:ker-fourier} gives
$\langle\mu_0,\Phi^{(0)}_\epsilon\rangle
=\sum_ic_i\langle\mu_0,\gamma_{\epsilon,\xi_i^0}\rangle$. Since $\Phi^{(0)}$
vanishes on $\supp\mu_0$,
\[
 0=\epsilon^2\langle\mu_0,\Phi^{(0)}_\epsilon\rangle
 =\sum_ic_i\int e^{-2\pi|X-\xi_i^0|^2/\epsilon^2}\dd\mu_0(X)
 \qquad(0<\epsilon\leq1).
\]
The integrands are dominated by $e^{-2\pi|X-\xi_i^0|^2}$, which is
integrable for $|\nu_{\vac}|+\rho_0$ by \eqref{eq:count-bound} with
$a=0$, and converge pointwise to the indicator of $\{\xi_i^0\}$ as
$\epsilon\to0$. Dominated convergence gives (i).

(ii) The set $V\cup\supp\rho_0$ is locally finite. Choose
$\varrho_0\leq\min\{\frac14,\frac12\sqrt g\}$ so small that, for every $i$, the closed disc of radius
$2\varrho_0$ about $\xi_i^0$ contains no point of $V\cup\supp\rho_0$
other than possibly $\xi_i^0$, and $|P_{\xi_i^0}|,|Q_{\xi_i^0}|\geq
3\varrho_0$ whenever $\xi_i^0$ is off both axes. Let
$0<\varrho\leq\varrho_0$ and let $a$ be so small that
$|\xi_i(a)-\xi_i^0|<\frac12\varrho$ for all $i$. Then
$D(\xi_i(a),\varrho)\subset D'=D(\xi_i^0,\frac32\varrho)$. Since
$\rho_0(\partial D')=0$, vague convergence gives
$\rho_a(D')\to\rho_0(D')=\rho_0(\{\xi_i^0\})$, and since all these
numbers lie in $\frac14\ZZ$, equality holds for small $a$. If
$\xi_i^0\notin V$, then $\rho_0(\{\xi_i^0\})=m_i$ and
$\dist(\xi_i(a),V)\geq2\varrho_0-\frac12\varrho\geq\varrho$, which is
(B2); if moreover $\xi_i^0\notin\supp\rho_0$, then $m_i=0$ and
$\rho_a(D(\xi_i(a),\varrho))=0$, which is (B3)(b).

(iii) Let $U'$ be the union of the four discs of radius
$\frac32\varrho$ about the points $(\pm P_{\xi_i^0},\pm Q_{\xi_i^0})$.
It is bounded, open and invariant under the two reflections, and its
boundary does not meet $\supp\rho_0$ because $\rho_0$ is even. A state
of $\rho_0$ supported in $U'$ has weights
$(P_{\xi_i^0}^2,Q_{\xi_i^0}^2)$, both positive, and gives mass $\frac14$
to $\xi_i^0$; so there are $n_i$ such states. By
Corollary~\ref{cor:cluster-stabilization}, for small $a$ exactly $n_i$
states of $\rho_a$ are supported in $U'$, their spins (which are
integers and converge) equal $s(\xi_i^0)$, and their dimensions converge
to $|\xi_i^0|^2$. A state of $\rho_a$ with a physical point in
$D(\xi_i^0,\varrho)$ is supported in $U'$; conversely, by the
convergence of dimensions and spins, each of the $n_i$ states supported
in $U'$ has, for small $a$, a physical point in $D(\xi_i^0,\varrho)$.
For small $a$ the geometric requirements of (B3)(a) hold at
$\xi_i(a)$ by the choice of $\varrho_0$, and
$m_i=\nu_{\vac}(\{\xi_i^0\})+\rho_0(\{\xi_i^0\})=w_i+\frac14n_i$,
because $\xi_i(a)=\xi_i^0$ when $\xi_i^0\in V$.
\end{proof}

\subsection{Local zeros and the three-projection lemma}

A budget \eqref{eq:ker-peak-budget} bounds the value of the
trigonometric polynomial at every state of a logarithmic window, since
all terms are nonnegative. Two deterministic lemmas convert such a
bound into information on the state. The first concerns a state close
to a known zero of the kernel; the second concerns an arbitrary state
and several kernels.

Let $X_0=[u_0,v_0]=(P_0,Q_0)$ be a real point off both axes. A state
with the spin $u_0v_0$ of $X_0$ and with dimension $|X_0|^2+\delta$ has
weights $(P_0^2+\frac\delta2,Q_0^2+\frac\delta2)$; for
$|\delta|<2\min\{P_0^2,Q_0^2\}$ we denote by
\begin{equation}\label{eq:ker-fixed-spin-curve}
 X(\delta)=\Bigl(\operatorname{sgn}(P_0)\sqrt{P_0^2+\tfrac\delta2},\
 \operatorname{sgn}(Q_0)\sqrt{Q_0^2+\tfrac\delta2}\Bigr)
\end{equation}
its physical point with the signs of $X_0$. This is the
\emph{fixed-spin curve} through $X_0$, parametrized by the dimension. For
$Y=[u_Y,v_Y]$ put
\begin{equation}\label{eq:ker-rate}
 \lambda(X_0,Y)=\frac{|u_0v_Y-v_0u_Y|}{|u_0^2-v_0^2|}.
\end{equation}

\begin{lemma}[Local zeros along a fixed-spin curve]\label{lem:local-zero}
Let $\varrho_*\geq1$ and $\varpi>0$, let $X_0=(P_0,Q_0)$ be a real point
with $|P_0|,|Q_0|\geq\varpi$ and $|X_0|\leq\varrho_*$, and let
$|\delta|\leq\varpi^2$. Let $Y\in\RR^2$ with $|Y|\leq\varrho_*$.
\begin{enumerate}[(i)]
\item $\bigl|S_Y(X(\delta))-S_Y(X_0)\bigr|\leq\pi\varrho_*|\delta|/\varpi$.
\item If $B(X_0,Y)\in\ZZ$ and $\lambda=\lambda(X_0,Y)>0$, then
 $S_Y(X(\delta))\geq\lambda^2\delta^2$ whenever
 $|\delta|\leq\min\{2\lambda\varpi^3/\varrho_*,\ 1/(3\lambda)\}$.
\item Let $\Phi$ be a function on $\RR^2$ with
 $\Phi\geq\prod_{i=1}^kS_{Y_i}\prod_{i=1}^{k'}S_{Y_i'}$, where
 $k\geq1$, all $|Y_i|,|Y_i'|\leq\varrho_*$, and, for some $\lambda>0$
 and $\varkappa>0$,
 \[
 B(X_0,Y_i)\in\ZZ,\quad\lambda(X_0,Y_i)\geq\lambda\quad(1\leq i\leq k),
 \qquad S_{Y_i'}(X_0)\geq\varkappa\quad(1\leq i\leq k').
 \]
 Put
 $\delta_1=\min\bigl\{\varpi^2,\ 2\lambda\varpi^3/\varrho_*,\
 \varpi/(3\varrho_*),\ \varkappa\varpi/(2\pi\varrho_*)\bigr\}$.
 Then for $|\delta|\leq\delta_1$
 \[
 \Phi(X(\delta))\geq\Bigl(\frac\varkappa2\Bigr)^{k'}\lambda^{2k}\delta^{2k};
 \qquad\text{so}\qquad
 |\delta|\leq\frac1\lambda\Bigl(\Bigl(\frac2\varkappa\Bigr)^{k'}
 \Gamma\Bigr)^{1/(2k)}\quad\text{if }\ \Phi(X(\delta))\leq\Gamma .
 \]
\end{enumerate}
\end{lemma}
\begin{proof}
Write $X(\delta)=(P,Q)$. For $|\delta|\leq\varpi^2$ we have
$|P|,|Q|\geq\varpi/\sqrt2$, and $dP/d\delta=1/(4P)$,
$d^2P/d\delta^2=-1/(16P^3)$, and likewise for $Q$. The phase
$\beta(\delta)=B(X(\delta),Y)=2(PP_Y-QQ_Y)$ therefore satisfies
\[
 \beta'=\frac12\Bigl(\frac{P_Y}{P}-\frac{Q_Y}{Q}\Bigr),\qquad
 |\beta'|\leq\frac{\varrho_*}{\varpi},\qquad
 |\beta''|\leq\frac{|P_Y|+|Q_Y|}{8(\varpi/\sqrt2)^3}
 \leq\frac{\varrho_*}{2\varpi^3},
\]
and $\beta'(0)=(P_YQ_0-Q_YP_0)/(2P_0Q_0)
=(u_0v_Y-v_0u_Y)/(u_0^2-v_0^2)$, so that $|\beta'(0)|=\lambda(X_0,Y)
\leq\varrho_*/\varpi$. Since $x\mapsto\sin^2(\pi x)$ has derivative
bounded by $\pi$, (i) follows. In (ii),
$|\beta(\delta)-\beta(0)-\beta'(0)\delta|\leq\varrho_*\delta^2/(4\varpi^3)
\leq\frac12\lambda|\delta|$, so the distance from $\beta(\delta)$ to the
integer $\beta(0)$ lies between $\frac12\lambda|\delta|$ and
$\frac32\lambda|\delta|\leq\frac12$, and
$\sin^2(\pi x)\geq4x^2$ for $|x|\leq\frac12$ gives the claim. In (iii),
each factor $S_{Y_i}$ is at least $\lambda(X_0,Y_i)^2\delta^2\geq
\lambda^2\delta^2$ by (ii), whose conditions hold because
$2\lambda(X_0,Y_i)\varpi^3/\varrho_*\geq2\lambda\varpi^3/\varrho_*$ and
$1/(3\lambda(X_0,Y_i))\geq\varpi/(3\varrho_*)$; each factor $S_{Y'_i}$ is
at least $\frac12\varkappa$ by (i).
\end{proof}

We use the elementary inequality
\begin{equation}\label{eq:ker-sine-distance}
 \sin^2(\pi x)\geq4\dist(x,\ZZ)^2\qquad(x\in\RR).
\end{equation}

\begin{lemma}[Three projections]\label{lem:three-projection}
Let $\ell\geq1$ and $p\geq1$ be integers with $\gcd(p,2\ell)=1$, put
$p_k=p+k\ell$, let $0<t_-\leq t\leq t_+$ and
$\tau_*=\max\{t_+,1/t_-\}$, and let
\[
 B_k(t)=[t,\,p_k/t]\in\Lambda_t,\qquad k=0,1,2 .
\]
Let $U,V_0,\Delta V\geq1$ be integers, $V_j=V_0+j\Delta V$,
$A_j=[U,V_j]$ for $j=0,1,2$, and $D=2U\Delta V$. Let $X=[u,v]$ be a real
point with $uv=s\in\ZZ$ and $|u|,|v|\leq\Upsilon$, where
$\Upsilon\geq1$, and let $0<\zeta\leq1$.
\begin{enumerate}[(a)]
\item \emph{Free vectors.} Suppose that for every $k\in\{0,1,2\}$
 \begin{equation}\label{eq:ker-free-alternative}
 \dist\bigl(B(X,B_k(t)),\ZZ\bigr)\leq\zeta\quad\text{or}\quad
 \dist\bigl(B(X,B_k(t)^\sharp),\ZZ\bigr)\leq\zeta ,
 \end{equation}
 and that $(11+6p)\,\tau_*\Upsilon\zeta<1/(4\ell^2)$. Then there is
 $\lambda=[u_\lambda,v_\lambda]\in\Lambda_t\cup\Lambda_{1/t}$ with
 \[
 u_\lambda v_\lambda=s,\qquad
 |u-u_\lambda|,\ |v-v_\lambda|\leq(3+2p)\tau_*\zeta,\qquad
 \bigl||X|^2-|\lambda|^2\bigr|\leq3(3+2p)\tau_*\zeta\Upsilon .
 \]
\item \emph{Fixed vectors.} Put $C_A=3+2V_2$. Suppose that for every
 $j\in\{0,1,2\}$
 \begin{equation}\label{eq:ker-fixed-alternative}
 \dist\bigl(B(X,A_j),\ZZ\bigr)\leq\zeta\quad\text{or}\quad
 \dist\bigl(B(X,A_j^\sharp),\ZZ\bigr)\leq\zeta ,
 \end{equation}
 and that $3C_A\zeta\Upsilon<D^{-2}$. Then there are integers $m,n$ with
 \[
 \frac{mn}{D^2}=s,\qquad
 \Bigl|u-\frac mD\Bigr|,\ \Bigl|v-\frac nD\Bigr|\leq C_A\zeta,\qquad
 \Bigl||X|^2-\frac{m^2+n^2}{2D^2}\Bigr|\leq3C_A\zeta\Upsilon .
 \]
\item \emph{Product kernels.} If\/ $0\leq\Gamma\leq1$ and
 \[
 S_{A_j}(X)S_{A_j^\sharp}(X)\,S_{B_k(t)}(X)S_{B_k(t)^\sharp}(X)\leq\Gamma
 \qquad\text{for all }0\leq j,k\leq2,
 \]
 then \eqref{eq:ker-free-alternative} holds for all $k$ or
 \eqref{eq:ker-fixed-alternative} holds for all $j$, with
 $\zeta=\frac12\Gamma^{1/8}$.
\item \emph{Sum kernels.} If\/ $0\leq\Gamma\leq1$ and
 $\Omega(X)\,S_{B_k(t)}(X)S_{B_k(t)^\sharp}(X)\leq\Gamma$ for
 $k=0,1,2$, put $\zeta=\frac12\Gamma^{1/6}$. Then
 \eqref{eq:ker-free-alternative} holds for all $k$, or there are
 integers $m,n$ with $|u-\frac m2|,|v-\frac n2|\leq\frac12\zeta$. In the
 second case, if $6\zeta\Upsilon<1$, then $\frac{mn}4=s$ and
 $\bigl||X|^2-\frac18(m^2+n^2)\bigr|\leq\frac32\zeta\Upsilon$.
\end{enumerate}
\end{lemma}
\begin{proof}
(a) Call $k$ of the first kind if the first alternative in
\eqref{eq:ker-free-alternative} holds, and of the second kind otherwise.
Two of the three indices, $k_1<k_2$, are of the same kind. If they are
of the first kind put $(\hat u,\hat v)=(u/t,tv)$, otherwise
$(\hat u,\hat v)=(v/t,tu)$. In both cases
$\hat u\hat v=s$, $|\hat u|,|\hat v|\leq\tau_*\Upsilon$, and, since
$B(X,B_k(t))=p_ku/t+tv$ and $B(X,B_k(t)^\sharp)=tu+p_kv/t$,
\[
 p_{k_i}\hat u+\hat v=n_i+e_i,\qquad n_i\in\ZZ,\quad|e_i|\leq\zeta
 \qquad(i=1,2).
\]
Let $D'=(k_2-k_1)\ell\in\{\ell,2\ell\}$, $m'=n_2-n_1$ and
$n'=D'n_1-p_{k_1}m'$. Subtracting the two relations and substituting back,
\[
 \hat u=\frac{m'}{D'}+e_{\hat u},\quad|e_{\hat u}|\leq\frac{2\zeta}{D'}
 \leq2\zeta;\qquad
 \hat v=\frac{n'}{D'}+e_{\hat v},\quad
 |e_{\hat v}|\leq\zeta+\frac{2p_{k_1}\zeta}{D'}\leq(3+2p)\zeta ,
\]
because $p_{k_1}/D'\leq p+1$: if $k_1=1$ then $k_2=2$ and $D'=\ell$.
Hence
\[
 \Bigl|s-\frac{m'n'}{D'^2}\Bigr|
 =\bigl|e_{\hat u}\hat v+(\hat u-e_{\hat u})e_{\hat v}\bigr|
 \leq(5+2p)\tau_*\Upsilon\zeta+2(3+2p)\zeta^2
 \leq(11+6p)\tau_*\Upsilon\zeta<\frac1{D'^2}.
\]
Both $s$ and $m'n'/D'^2$ lie in $D'^{-2}\ZZ$, so $m'n'=sD'^2$ and
$D'^2\mid m'n'$. We claim that $D'\mid m'$. The integer $p_{k_1}$ is coprime
to $D'$: for $k_1=0$ because $\gcd(p,2\ell)=1$, and for $k_1=1$ because
$D'=\ell$ and $\gcd(p+\ell,\ell)=1$. Suppose that a prime has exponent
$e\geq1$ in $D'$ and exponent $r<e$ in $m'$. It does not divide
$p_{k_1}$, so its exponent in $n'=D'n_1-p_{k_1}m'$ is exactly $r$, and its
exponent in $m'n'$ is $2r<2e$, contradicting $D'^2\mid m'n'$. Hence
$D'\mid m'$, and then $D'\mid n'$. With $m=m'/D'$ and $n=n'/D'$ we get
$mn=s$, $|\hat u-m|\leq2\zeta$ and $|\hat v-n|\leq(3+2p)\zeta$. In the
first kind, $\lambda=[mt,n/t]\in\Lambda_t$ satisfies
$|u-mt|\leq2\tau_*\zeta$ and $|v-n/t|\leq(3+2p)\tau_*\zeta$; in the
second kind, $\lambda=[n/t,mt]\in\Lambda_{1/t}$ satisfies the same bounds
with $u$ and $v$ exchanged. In both cases $u_\lambda v_\lambda=mn=s$.
Finally, with $c=(3+2p)\tau_*\zeta\leq1\leq\Upsilon$,
\[
 \bigl||X|^2-|\lambda|^2\bigr|
 \leq\tfrac12|u-u_\lambda|\bigl(2|u|+|u-u_\lambda|\bigr)
 +\tfrac12|v-v_\lambda|\bigl(2|v|+|v-v_\lambda|\bigr)
 \leq c(2\Upsilon+c)\leq3c\Upsilon .
\]

(b) Since $B(X,A_j)=V_ju+Uv$ and $B(X,A_j^\sharp)=Uu+V_jv$, two indices
$j_1<j_2$ satisfy, with $(x_1,x_2)=(u,v)$ or $(x_1,x_2)=(v,u)$,
\[
 V_{j_i}x_1+Ux_2=n_i+e_i,\qquad n_i\in\ZZ,\quad|e_i|\leq\zeta\qquad(i=1,2).
\]
Let $D''=(j_2-j_1)\Delta V\in\{\Delta V,2\Delta V\}$. Subtracting,
$x_1=(n_2-n_1)/D''+e'$ with $|e'|\leq2\zeta$, and then
\[
 x_2=\frac{D''n_1-V_{j_1}(n_2-n_1)}{D''U}+e'',\qquad
 |e''|\leq\frac{\zeta+2V_{j_1}\zeta}{U}\leq(1+2V_2)\zeta .
\]
Both $D''$ and $D''U$ divide $D=2U\Delta V$, so $x_1$ and $x_2$, hence
$u$ and $v$, are within $C_A\zeta$ of $D^{-1}\ZZ$; let $m/D$ and $n/D$ be
the corresponding grid values. Then
$|s-mn/D^2|\leq C_A\zeta(2\Upsilon+C_A\zeta)\leq3C_A\zeta\Upsilon<D^{-2}$,
and both numbers lie in $D^{-2}\ZZ$, so they are equal. The bound for
$|X|^2$ is obtained as in (a).

(c) If $S_{A_j}S_{A_j^\sharp}(X)>\Gamma^{1/2}$ for some $j$, then
$S_{B_k}S_{B_k^\sharp}(X)\leq\Gamma^{1/2}$ for every $k$; otherwise
$S_{A_j}S_{A_j^\sharp}(X)\leq\Gamma^{1/2}$ for every $j$. If a product of
two numbers of $[0,1]$ is at most $\Gamma^{1/2}$, one of them is at most
$\Gamma^{1/4}$, and by \eqref{eq:ker-sine-distance} the corresponding
phase is within $\frac12\Gamma^{1/8}$ of an integer.

(d) If $\Omega(X)>\Gamma^{1/3}$, then
$S_{B_k}S_{B_k^\sharp}(X)\leq\Gamma^{2/3}$ for every $k$, one factor is at
most $\Gamma^{1/3}$, and its phase is within $\frac12\Gamma^{1/6}$ of an
integer. Otherwise $\sin^2(2\pi u)$ and $\sin^2(2\pi v)$ are at most
$\Gamma^{1/3}$, so $2u$ and $2v$ are within $\zeta$ of integers $m,n$.
Then $|s-\frac{mn}4|\leq\frac12\zeta(2\Upsilon+\frac12\zeta)
\leq\frac32\zeta\Upsilon<\frac14$, and both numbers lie in
$\frac14\ZZ$; the bound for $|X|^2$ follows as in (a).
\end{proof}

In (b) and (d) the associated point $[m/D,n/D]$, respectively
$[m/2,n/2]$, has integral spin, so by Lemma~\ref{lem:grid-circles} it
lies in the finite union of circle lattices
\begin{equation}\label{eq:ker-grid-circles}
 \Sigma_D=\bigcup_{k\mid D^2}\Lambda_{k/D},
\end{equation}
respectively in $\Sigma_2=\Lambda_{1/2}\cup\Lambda_1\cup\Lambda_2$. The
set $\Sigma_D$ contains $\Lambda_1$ and is invariant under the
reflection $[u,v]\mapsto[v,u]$, because $k\mapsto D^2/k$ permutes the
divisors of $D^2$.

\subsection{Two kernel families and their finite hypotheses}
\label{subsec:ker-families}

Throughout, $I\subset(0,\infty)$ is a compact interval, $\ell\geq1$ is an
integer, and $p$ is an integer with
\begin{equation}\label{eq:ker-p-conditions}
 \gcd(p,2\ell)=1,\qquad p>\sup_{t\in I}t^2+2 .
\end{equation}
For an integer $n\geq1$ and $t>0$ put
\begin{gather*}
 Y_n(t)=[t,\,n/t]\in\Lambda_t,\qquad Y_n(t)^\sharp=[n/t,\,t]\in\Lambda_{1/t},\\
 p_k=p+k\ell,\qquad B_k(t)=Y_{p_k}(t)\quad(k=0,1,2).
\end{gather*}
The point $Y_n(t)$ has spin $n$ and squared norm
$E_n(t)=\frac12(t^2+n^2/t^2)$. For $t\in I$ the points $B_k(t)$ are off
both axes, because $p_k>t^2$, and lie outside $V$, because
$p_k/t>t+2/t>2$; and $E_p$ is strictly decreasing on $I$, because
$E_p'(t)=t-p^2/t^3<0$. For a half-integer $R$ we put
\begin{equation}\label{eq:ker-Sigma-t}
 \Sigma(t)=V\cup\Lambda_R\cup\Lambda_{1/R}\cup\Lambda_t\cup\Lambda_{1/t}.
\end{equation}
If $\rho_0=\sigma(R,t_0)$ and $\mu_0=\nu_{\vac}+\rho_0=\frac12\rhoF_R
+\frac12\rhoF_{t_0}$, then by \eqref{eq:circle-measures}
\begin{equation}\label{eq:ker-limit-masses}
 \mu_0(\{x\})=\tfrac14\bigl(e_R+e_{1/R}+e_{t_0}+e_{1/t_0}\bigr)(\{x\}),
 \qquad\supp\mu_0\subset\Sigma(t_0),\qquad\supp\rho_0\subset\Sigma(t_0).
\end{equation}

\subsubsection*{The product family}
\emph{Product data} consist of $I$, $\ell$, $p$ as above, a half-integer
$R=K/2$ with $K\geq3$ and $K\neq4$, and integers $U,V_0,\Delta V\geq1$
such that, with $V_j=V_0+j\Delta V$ and $A_j=[U,V_j]$ for $j=0,1,2$,
\begin{equation}\label{eq:ker-A-conditions}
 A_j\in\Lambda_R,\qquad U\geq3,\qquad V_j\geq3,\qquad|U-V_j|\geq3 .
\end{equation}
Thus $A_j\in\Lambda_R\cap\Lambda_1$ is off both axes, and
$A_j^\sharp\in\Lambda_{1/R}\cap\Lambda_1$. The ten trigonometric
polynomials of the family are
\begin{equation}\label{eq:ker-product-family}
 \Phi^{*}_t=S_CS_{C^\sharp}\,S_{B_0(t)}S_{B_0(t)^\sharp},\qquad
 \Phi^{j,k}_t=S_{A_j}S_{A_j^\sharp}\,S_{B_k(t)}S_{B_k(t)^\sharp}
 \quad(0\leq j,k\leq2).
\end{equation}
Since $C^\sharp=[0,2]=[0\cdot R,\,K/R]\in\Lambda_R$ and
$C,A_j\in\Lambda_1$, \eqref{eq:ker-vanishing} shows that each of them
vanishes on $\Sigma(t)$. By Lemma~\ref{lem:kernel-fourier}(i) their
centres are
\begin{equation}\label{eq:ker-product-centres}
 \xi_\eta(t)=\eta_1Y+\eta_2Y^\sharp+\eta_3B_k(t)+\eta_4B_k(t)^\sharp,
 \qquad\eta\in\{-1,0,1\}^4,
\end{equation}
with $Y=C$, $k=0$ for $\Phi^*_t$ and $Y=A_j$ for $\Phi^{j,k}_t$. The
\emph{product hypotheses} at $t_0\in I$ are:
\begin{enumerate}[(H1)]
\item Neither $2t_0$ nor $2/t_0$ is an integer, and
 $B_k(t_0)\notin\Lambda_{1/t_0}\cup\Lambda_R\cup\Lambda_{1/R}$ for
 $k=0,1,2$.
\item For each of the ten polynomials, every centre $\xi_\eta(t_0)$ with
 $|\eta|=3$ lies outside $\Sigma(t_0)$.
\item For each of the ten polynomials, every centre $\xi_\eta(t_0)$ with
 $|\eta|\in\{2,4\}$ lies outside $V$.
\item $S_{B_0(t_0)}(A_j)\,S_{B_0(t_0)^\sharp}(A_j)\neq0$ for $j=0,1,2$,
 and $S_C\,S_{C^\sharp}\,S_{B_0(t_0)^\sharp}$ does not vanish at
 $B_k(t_0)$ for $k=1,2$.
\end{enumerate}
The centres with $|\eta|=1$ are $\pm C,\pm C^\sharp\in V$ for $\Phi^*_t$,
and $\pm A_j$, $\pm A_j^\sharp$, $\pm B_k(t)$, $\pm B_k(t)^\sharp$, which
lie outside $V$ by \eqref{eq:ker-A-conditions} and the remark above.
No condition is imposed on coincidences among the centres.

\subsubsection*{The sum family}
\emph{Sum data} consist of $I$, $\ell$, $p$ as above, a radius
$R\in\{1,2\}$, and an integer $w$ with $1\leq w<p$. We write
\[
 W(t)=Y_w(t)=[t,\,w/t],\qquad\mathcal N=\{p_0,p_1,p_2,w\},\qquad
 \Phi^{(n)}_t=\Omega\,S_{Y_n(t)}S_{Y_n(t)^\sharp}\quad(n\in\mathcal N).
\]
Since $\Lambda_1\cup\Lambda_2\cup\Lambda_{1/2}\subset\frac12\ZZ^2$ in
light-cone coordinates, \eqref{eq:ker-vanishing} shows that
$\Phi^{(n)}_t$ vanishes on $\Sigma(t)$. By
Lemma~\ref{lem:kernel-fourier}(ii) its centres are
$\zeta+\eta_1Y_n(t)+\eta_2Y_n(t)^\sharp$ with
$\zeta\in\{0,\pm C,\pm C^\sharp\}$ and $\eta\in\{-1,0,1\}^2$. We also
need the $27$ points
\begin{equation}\label{eq:ker-row-preimages}
 x_{l,k,j}(t)=[2,2]-lC^\sharp-kW(t)-jW(t)^\sharp
 =\bigl[\,2-kt-jw/t,\ 2-2l-kw/t-jt\,\bigr],\qquad0\leq l,k,j\leq2 .
\end{equation}
The \emph{sum hypotheses} at $t_0\in I$ are:
\begin{enumerate}[(S1)]
\item Neither $2t_0$ nor $2/t_0$ is an integer, and
 $Y_n(t_0)\notin\Lambda_{1/t_0}$ for $n\in\mathcal N$.
\item For $n\in\mathcal N$, $\zeta\in\{\pm C,\pm C^\sharp\}$ and
 $\eta\in\{\pm1\}^2$, the centre
 $\zeta+\eta_1Y_n(t_0)+\eta_2Y_n(t_0)^\sharp$ lies outside $\Sigma(t_0)$.
\item For $n\in\mathcal N$, $\zeta\in\{\pm C,\pm C^\sharp\}$ and
 $\eta\in\{\pm1\}^2$, the centres
 $\eta_1Y_n(t_0)+\eta_2Y_n(t_0)^\sharp$, $\zeta+\eta_1Y_n(t_0)$ and
 $\zeta+\eta_1Y_n(t_0)^\sharp$ lie outside $V$.
\item $S_{B_0(t_0)^\sharp}\bigl(Y_n(t_0)\bigr)\neq0$ for
 $n\in\{p_1,p_2,w\}$, and neither $2(t_0+p/t_0)$ nor $2(t_0-p/t_0)$ is an
 integer.
\item $x_{l,k,j}(t_0)\notin\Sigma(t_0)$ whenever $(k,j)\neq(0,0)$.
\end{enumerate}
Under (S1) each $Y_n(t_0)$, $n\in\mathcal N$, is off both axes (if
$t_0^2=n$ then $Y_n(t_0)=[n/t_0,t_0]\in\Lambda_{1/t_0}$), lies outside
$\frac12\ZZ^2\supset V\cup\Lambda_R\cup\Lambda_{1/R}$, and satisfies
$\Omega(Y_n(t_0))\geq\sin^2(2\pi t_0)>0$.

\begin{lemma}[The exceptional set is finite]\label{lem:exceptional-set}
For given product data, the set of $t_0\in I$ at which one of
\textup{(H1)--(H4)} fails is finite. For given sum data, the set of
$t_0\in I$ at which one of \textup{(S1)--(S5)} fails is finite.
\end{lemma}
\begin{proof}
All points considered below have the form
\begin{equation}\label{eq:ker-general-centre}
 \xi(t)=\bigl[a_1+b_1t+b_2n/t,\ a_2+b_2t+b_1n/t\bigr],
 \qquad a_1,a_2,b_1,b_2\in\ZZ,\quad n\geq1,
\end{equation}
with data from a finite list: for \eqref{eq:ker-product-centres},
$[a_1,a_2]=\eta_1Y+\eta_2Y^\sharp$, $(b_1,b_2)=(\eta_3,\eta_4)$ and
$n=p_k$; for the centres of $\Phi^{(n)}_t$, $[a_1,a_2]=\zeta$ and
$(b_1,b_2)=\eta$; for \eqref{eq:ker-row-preimages},
$[a_1,a_2]=[2,2-2l]$, $(b_1,b_2)=(-k,-j)$ and $n=w$. For $t\in I$ these
points stay in a fixed ball, and a point of $\Lambda_t$ or
$\Lambda_{1/t}$ in that ball has charges in a fixed finite set, as has a
point of $V\cup\Lambda_R\cup\Lambda_{1/R}$. Thus $\xi(t)\in\Sigma(t)$ for
some $t\in I$ means that one of finitely many systems
\[
 \xi(t)=[mt,\,n'/t],\qquad\xi(t)=[m/t,\,n't],\qquad\xi(t)=[x_1,x_2]
\]
holds, with integers $m,n'$ and a fixed point $[x_1,x_2]$. After
multiplication by $t$, each coordinate equation is a polynomial equation
of degree at most two in $t$, so a system has finitely many solutions
unless both of its polynomials vanish identically. Comparing
coefficients, and using $n\neq0$, this happens only as follows:
\begin{equation}\label{eq:ker-identities}
\begin{aligned}
 &\xi(t)\in\Lambda_t\ \text{identically}&&\Longrightarrow\quad
 a_1=a_2=0,\ b_2=0;\\
 &\xi(t)\in\Lambda_{1/t}\ \text{identically}&&\Longrightarrow\quad
 a_1=a_2=0,\ b_1=0;\\
 &\xi(t)=[x_1,x_2]\ \text{identically}&&\Longrightarrow\quad
 b_1=b_2=0 .
\end{aligned}
\end{equation}
Indeed, the first system reads $a_1t+(b_1-m)t^2+b_2n=0$ and
$a_2t+b_2t^2+b_1n-n'=0$, the second reads $a_1t+b_1t^2+b_2n-m=0$ and
$a_2t+(b_2-n')t^2+b_1n=0$, and the third has first equation
$(a_1-x_1)t+b_1t^2+b_2n=0$.

\emph{Product data.} For $Y=[u_Y,v_Y]\in\{C,A_0,A_1,A_2\}$ we have
$u_Y^2\neq v_Y^2$, so
$\eta_1Y+\eta_2Y^\sharp=[\eta_1u_Y+\eta_2v_Y,\eta_1v_Y+\eta_2u_Y]$
vanishes only for $\eta_1=\eta_2=0$. If $|\eta|=3$, none of the three
cases of \eqref{eq:ker-identities} can occur, since they would force
$|\eta|\leq1$, $|\eta|\leq1$ and $|\eta|\leq2$ respectively; hence (H2)
fails only on a finite set. If $|\eta|\in\{2,4\}$ and
$(\eta_3,\eta_4)\neq(0,0)$, the third case cannot occur, so
$\xi_\eta(t)\in V$ for finitely many $t$ only; if $\eta_3=\eta_4=0$ the
centre is one of $\pm[u_Y+v_Y,u_Y+v_Y]$, $\pm[u_Y-v_Y,v_Y-u_Y]$, which is
never in $V$: for $Y=C$ these are $\pm[2,2]$, $\pm[2,-2]$, and for
$Y=A_j$ both coordinates have modulus at least $3$ by
\eqref{eq:ker-A-conditions}. Hence (H3) fails only on a finite set. In
(H1), $2t_0\in\ZZ$ or $2/t_0\in\ZZ$ holds at finitely many points of
$I$;
$B_k(t)\in\Lambda_{1/t}$ requires $t^2=m$ with a bounded integer $m$; and
$B_k(t)\in\Lambda_R\cup\Lambda_{1/R}$ requires $t$ to be the first
coordinate of one of finitely many fixed points. In (H4) the phases are
\begin{gather*}
 B(A_j,B_0(t))=\frac{Up}t+V_jt,\qquad
 B(A_j,B_0(t)^\sharp)=Ut+\frac{V_jp}t,\\
 B(B_k(t),C)=\frac{2p_k}t,\qquad
 B(B_k(t),C^\sharp)=2t,
\end{gather*}
and $B(B_k(t),B_0(t)^\sharp)=t^2+pp_k/t^2$. Each is a nonconstant
rational function of $t$, bounded on $I$, and therefore integer-valued at
finitely many points of $I$.

\emph{Sum data.} In (S2) we have $[a_1,a_2]=\zeta\neq0$ and
$b_1b_2\neq0$, and in (S3) we have $(b_1,b_2)\neq(0,0)$ and the target
is a fixed point; no case of \eqref{eq:ker-identities} occurs. In (S5),
$a_1=2\neq0$ excludes the first two cases and $(k,j)\neq(0,0)$ the third.
(S1) is treated like (H1), and the phases in (S4) are
$B(Y_n(t),B_0(t)^\sharp)=t^2+pn/t^2$ and $2(t\pm p/t)$, again nonconstant
rational functions.
\end{proof}

\subsection{The case $R\notin\{1,2\}$}

\begin{proposition}[Localization and exclusion for product data]
\label{prop:generic-localization}
Let $0<g\leq\frac12$, let product data be given, and let $t_0$ be an
interior point of $I$ at which \textup{(H1)--(H4)} hold. Then no
admissible sequence of gap $g$ converges to $\sigma(R,t_0)$.
\end{proposition}
\begin{proof}
Suppose that an admissible sequence of gap $g$ converges to
$\rho_0=\sigma(R,t_0)$, and let $\mu_0=\nu_{\vac}+\rho_0$. All constants
below depend on the data, on $g$ and on $t_0$; statements hold for all
sufficiently small $a$ in the sequence.

\emph{Step 0: limiting masses.} By \eqref{eq:ker-limit-masses}, a point
$[2,0]=(1,1)$ lies in $\Lambda_r$ if and only if $2/r\in\ZZ$, and in
$\Lambda_{1/r}$ if and only if $2r\in\ZZ$. Since $2R=K\in\ZZ$,
$2/R=4/K\notin\ZZ$, and neither $2t_0$ nor $2/t_0$ is an integer, we
get
$\mu_0(\{C\})=\frac14=\nu_{\vac}(\{C\})$, and the same at $-C$,
$\pm C^\sharp$. By (H1), $\mu_0(\{B_k(t_0)\})=\frac14$. By (H4),
$A_j\notin\Lambda_{t_0}\cup\Lambda_{1/t_0}$: otherwise $B(A_j,B_0(t_0))$
or $B(A_j,B_0(t_0)^\sharp)$ would be an integer. Hence
$\mu_0(\{A_j\})=\frac14n_j^0$ with $n_j^0\in\{1,2\}$, according as
$A_j\notin\Lambda_{1/R}$ or $A_j\in\Lambda_{1/R}$. These points are
outside $V$, except $\pm C,\pm C^\sharp$, so the same values hold for
$\rho_0$ at $B_k(t_0)$ and $A_j$, and $\rho_0(\{\pm C\})=
\rho_0(\{\pm C^\sharp\})=0$.

\emph{Step 1: the moving radius.} By
Lemma~\ref{lem:budget-hypotheses}(iii), applied to the constant family
$\Phi^*_{t_0}$ and its centre $B_0(t_0)$, exactly one state of $\rho_a$
has a physical point near $B_0(t_0)$; it has spin $p$, and its dimension
$\Delta_{0,a}$ converges to $E_p(t_0)$. Since $E_p$ is strictly
decreasing on $I$ and $t_0$ is interior, there is a unique $t_a\in I$
with
\begin{equation}\label{eq:ker-t-a}
 E_p(t_a)=\Delta_{0,a},\qquad t_a\to t_0 .
\end{equation}
In the same way, using the constant families $\Phi^{j,k}_{t_0}$ and
their centres $B_k(t_0)$ and $A_j$: for $k=1,2$ exactly one state has a
physical point near $B_k(t_0)$; it has spin $p_k$, and we write its dimension as
$E_{p_k}(t_a)+\delta_{k,a}$, where $\delta_{k,a}\to0$. Exactly $n_j^0$
states have a physical point near $A_j$; they have spin $UV_j$, and
their dimensions $|A_j|^2+\delta$ satisfy $\delta\to0$.

\emph{Step 2: the budget of\/ $\Phi^*$.} We apply
Lemmas~\ref{lem:budget-hypotheses} and~\ref{lem:peak-budget} to
$\Phi^{(a)}=\Phi^*_{t_a}$, whose centres $\xi_\eta(t_a)$ converge to
$\xi_\eta(t_0)$, with $m_\eta=\mu_0(\{\xi_\eta(t_0)\})$. By (H2), (H3)
and the remark after (H4), the only centres with $\xi_\eta(t_0)\in V$ are
the origin, for $\eta=0$, and $\pm C$, $\pm C^\sharp$; these do not
depend on $a$. The polynomial $\Phi^*_{t_0}$ vanishes on $\supp\mu_0$,
so \eqref{eq:ker-mass-balance} holds. Condition (B0) holds, and (B1)
holds because the coefficient at $\eta=0$ is $\frac1{16}$ and
$\mu_0(\{0\})=1$. The centres with
$|\eta|\in\{2,4\}$ have positive coefficients and satisfy (B2) by (H3).
The centres with $|\eta|=3$ satisfy (B3)(b) by (H2) and
\eqref{eq:ker-limit-masses}. The remaining centres have $|\eta|=1$ and
negative coefficients. At $\pm C$, $\pm C^\sharp$ condition (B3)(a)
holds with $n_i=0$ and $w_i=\frac14=m_i$. At $\pm B_0(t_a)$ and
$\pm B_0(t_a)^\sharp$, which are the four physical points of a state of
spin $p$ and dimension $E_p(t_a)$, condition (B3)(a) holds with
$n_i=1$, the state of Step~1, and with $\varepsilon=0$, by
\eqref{eq:ker-t-a}. Fix $0<\varrho\leq\frac12\sqrt g$ as
in Lemma~\ref{lem:budget-hypotheses}, for $\Phi^*_{t_a}$ and for the nine
families $\Phi^{j,k}_{t_a}$, put $\kappa=\pi\varrho^2/4$ and choose
\begin{equation}\label{eq:ker-width-L}
 \epsilon^2=\kappa/L,\qquad\text{so that}\qquad e^{-\kappa/\epsilon^2}=a .
\end{equation}
The range \eqref{eq:ker-budget-range} holds for small $a$, and
\eqref{eq:ker-peak-budget} gives
\begin{equation}\label{eq:ker-initial-budget}
 \sum_j\Phi^*_{t_a,\epsilon}(X_j^\sigma)\leq C_0\,a\,L^2 .
\end{equation}
Here the kernel is even, so its value at a state does not depend on
$\sigma$, and all terms are nonnegative by
Lemma~\ref{lem:kernel-positivity}. For a state with $\Delta_j\leq E_*$,
where $E_*$ is any fixed bound, the broad Gaussian is at least $\frac12$
for small $a$, and therefore
\begin{equation}\label{eq:ker-initial-single}
 \Phi^*_{t_a}(X_j^\sigma)\leq2C_0\,a\,L^2\qquad(\Delta_j\leq E_*).
\end{equation}

\emph{Step 3: anchoring.} Consider one of the $n_j^0$ states near
$A_j$, with dimension $|A_j|^2+\delta$. Its physical point with the signs of
$A_j$ is $X(\delta)$ in the notation
\eqref{eq:ker-fixed-spin-curve}, with $X_0=A_j$. We apply
Lemma~\ref{lem:local-zero}(iii) to $\Phi=\Phi^*_{t_a}$ with the two
vanishing factors $S_C$, $S_{C^\sharp}$ and the two nonvanishing factors
$S_{B_0(t_a)}$, $S_{B_0(t_a)^\sharp}$. Indeed $B(A_j,C)=2V_j$ and
$B(A_j,C^\sharp)=2U$ are integers, and by \eqref{eq:ker-rate}
\[
 \lambda(A_j,C)=\frac{2V_j}{|U^2-V_j^2|}>0,\qquad
 \lambda(A_j,C^\sharp)=\frac{2U}{|U^2-V_j^2|}>0 ,
\]
while $S_{B_0(t_a)}(A_j)$ and $S_{B_0(t_a)^\sharp}(A_j)$ converge to the
positive numbers of (H4). Since $\delta\to0$, the lemma applies for
small $a$, and with \eqref{eq:ker-initial-single} and $k=2$ it gives
\begin{equation}\label{eq:ker-anchor-A}
 |\delta|\leq C\,(aL^2)^{1/4}=C\,a^{1/4}L^{1/2}
 \qquad\text{for every state near }A_j .
\end{equation}
Next let $k\in\{1,2\}$ and consider the state near $B_k(t_0)$. Take
$X_0=B_k(t_a)$, which stays uniformly off both axes, the vanishing
factor $S_{B_0(t_a)}$ and the nonvanishing factors $S_C$, $S_{C^\sharp}$,
$S_{B_0(t_a)^\sharp}$. Here $B(B_k(t),B_0(t))=p+p_k$ is an integer, and
\[
 \lambda\bigl(B_k(t),B_0(t)\bigr)
 =\frac{|t\cdot p/t-(p_k/t)\cdot t|}{|t^2-p_k^2/t^2|}
 =\frac{k\ell}{p_k^2/t^2-t^2}
\]
is bounded below on $I$; the three nonvanishing factors at $B_k(t_a)$
converge to the positive numbers of (H4). Lemma~\ref{lem:local-zero}(iii)
with $k=1$ gives
\begin{equation}\label{eq:ker-anchor-B}
 |\delta_{k,a}|\leq C\,(aL^2)^{1/2}=C\,a^{1/2}L\qquad(k=1,2),
\end{equation}
and $\delta_{0,a}=0$ by \eqref{eq:ker-t-a}.

\emph{Step 4: the nine budgets.} Fix $j,k$ and apply
Lemmas~\ref{lem:budget-hypotheses} and~\ref{lem:peak-budget} to
$\Phi^{(a)}=\Phi^{j,k}_{t_a}$. The only centre with $\xi_\eta(t_0)\in V$
is the origin. Conditions (B0)--(B2) and (B3)(b) for $|\eta|=3$ hold as
in Step~2. The centres with $|\eta|=1$ are $\pm A_j$, $\pm A_j^\sharp$,
which are the physical points of a state of spin $UV_j$ and dimension
$|A_j|^2$, and $\pm B_k(t_a)$, $\pm B_k(t_a)^\sharp$. By Step~1 and
Lemma~\ref{lem:budget-hypotheses}(iii), condition (B3)(a) holds at the
former with the $n_j^0$ states near $A_j$ and at the latter with the
single state near $B_k(t_0)$, and by \eqref{eq:ker-anchor-A} and
\eqref{eq:ker-anchor-B} with
\[
 \varepsilon=\varepsilon_a=C\,a^{1/4}L^{1/2} .
\]
With the width \eqref{eq:ker-width-L}, \eqref{eq:ker-peak-budget} gives
\begin{equation}\label{eq:ker-nine-budgets}
 \sum_j\Phi^{j,k}_{t_a,\epsilon}(X_j^\sigma)
 \leq C\bigl((a+a^{1/2}L)L^2+a\bigr)\leq C\,a^{1/2}L^3,
 \qquad0\leq j,k\leq2 .
\end{equation}

\emph{Step 5: localization.} Let $j$ be a state with $\Delta_j\leq L$,
and let $X=[u,v]$ be one of its physical points. The broad Gaussian at
$X$ is at least $e^{-2\pi\kappa}$, all terms of
\eqref{eq:ker-nine-budgets} are nonnegative, and so
\[
 \Phi^{j',k}_{t_a}(X)\leq\Gamma_a:=C\,a^{1/2}L^3
 \qquad\text{for all }0\leq j',k\leq2 .
\]
Moreover $uv=s_j\in\ZZ$ and $|u|,|v|\leq\sqrt{2\Delta_j}\leq
\Upsilon:=\sqrt{2L}$. We apply Lemma~\ref{lem:three-projection}(c) with
$t=t_a$, $t_\pm$ the endpoints of $I$, and
\[
 \zeta=\tfrac12\Gamma_a^{1/8}=C\,a^{1/16}L^{3/8} .
\]
Since $\zeta\Upsilon\to0$, the smallness conditions of parts (a) and (b)
of that lemma hold for small $a$. Consequently there is a real point
$X_j'$, equal either to a point of $\Lambda_{t_a}\cup\Lambda_{1/t_a}$ or
to a point $[m/D,n/D]$ with $D=2U\Delta V$, whose spin is exactly $s_j$,
and
\begin{equation}\label{eq:ker-generic-localization}
 \bigl|\Delta_j-|X_j'|^2\bigr|\leq C\,\zeta\,\Upsilon
 \leq C\,a^{1/16}L^{7/8}.
\end{equation}
In the second case $X_j'\in\Sigma_D$ by \eqref{eq:ker-grid-circles}.

\emph{Step 6: contradiction.} The set $\Sigma_D$ is a fixed finite union
of circle lattices, invariant under $[u,v]\mapsto[v,u]$ and containing
$\Lambda_1$; the radii $t_a$ lie in the compact interval $I$; and by
Step~5 every state with $\Delta_j\leq u_*L$, $u_*=1>1/(2\pi)$, has a real
point $X_j'\in\Sigma_D\cup\Lambda_{t_a}\cup\Lambda_{1/t_a}$ whose spin is
exactly $s_j$ and which satisfies
$|\Delta_j-|X_j'|^2|\leq Ca^{1/16}L^{7/8}$, for all sufficiently small
$a$ in the sequence. Discarding finitely many terms leaves an admissible
sequence of gap $g$, and for it this contradicts
Theorem~\ref{thm:moving-localization}.
\end{proof}

\subsection{The case $R\in\{1,2\}$: the scalar row}

For $R\in\{1,2\}$ the limit $\sigma(R,t_0)$ has a long scalar state of
dimension two, at the points $(\pm1,\pm1)$ where $\nu_{\vac}$ has
positive atoms. Its displacement enters every budget through the
centres $\pm C$, $\pm C^\sharp$, and must be bounded first. This is done
with one explicit combination of translates of a narrow test function,
whose Fourier transform is dominated by a sum kernel.

For $x\in\RR^2$ and $\epsilon>0$ put
\begin{equation}\label{eq:ker-psi}
\begin{gathered}
 \psi_{\epsilon,x}(X)=(P-P_x)\,e^{-2\pi|X-x|^2/\epsilon^2},\qquad
 \psi_\epsilon=\psi_{\epsilon,[2,2]},\\
 \text{that is,}\quad
 \psi_\epsilon(P,Q)=(P-2)\,e^{-2\pi((P-2)^2+Q^2)/\epsilon^2}.
\end{gathered}
\end{equation}
These functions are not even. For sum data and $t\in I$ we define the
\emph{row} as the following third-order iterated second difference,
acting on functions:
\begin{equation}\label{eq:ker-row}
 (\mathsf D_tf)(X)=\sum_{l,k,j=0}^{2}b_lb_kb_j\,
 f\bigl(X+lC^\sharp+kW(t)+jW(t)^\sharp\bigr),\qquad
 (b_0,b_1,b_2)=(1,-2,1).
\end{equation}
Pairing $\mathsf D_tf$ with $\mu_a$ is the same as pairing $f$ with the
measure obtained from $\mu_a$ by the operator
$(\tau_{C^\sharp}-1)^2(\tau_{W(t)}-1)^2(\tau_{W(t)^\sharp}-1)^2$, where
$\tau_Y\delta_X=\delta_{X+Y}$; we shall only use the function side.

\begin{lemma}[Preimages of the row]\label{lem:row-preimages}
Let $t>0$, let $w\geq1$ be an integer and $0<\epsilon\leq1$.
\begin{enumerate}[(i)]
\item With the points $x_{l,k,j}(t)$ of \eqref{eq:ker-row-preimages},
 \[
 \mathsf D_t\psi_\epsilon
 =\sum_{l,k,j=0}^2b_lb_kb_j\,\psi_{\epsilon,x_{l,k,j}(t)} .
 \]
 The three points with $k=j=0$ do not depend on $t$: they are
 $x_{0,0,0}=[2,2]=(2,0)$, $x_{1,0,0}=[2,0]=(1,1)$ and
 $x_{2,0,0}=[2,-2]=(0,2)$, with coefficients $1,-2,1$.
\item For real $X$ and $x$,
 $|\psi_{\epsilon,x}(X)|\leq e^{-\pi|X-x|^2/(2\epsilon^2)}$.
\item \emph{The middle preimage.} $\psi_{\epsilon,(1,1)}$ vanishes at
 $(1,1)$. If $|z|\leq1$ and $h=1+\frac z2$, then
 \[
 \Bigl|\tfrac14\psi_{\epsilon,(1,1)}\bigl(\sqrt h,\sqrt h\bigr)
 -\frac z{16}\Bigr|\leq\frac{z^2}{32}+\frac{|z|^3}{2\epsilon^2}.
 \]
\item \emph{The axis preimages.} If $0\leq\delta\leq1$,
 $h=4+\frac\delta2$ and $\bar h=\frac\delta2$, then
 $|\psi_{\epsilon,(2,0)}(\sqrt h,\pm\sqrt{\bar h})|\leq\frac\delta8$. If
 $h=\frac\delta2$ and $\bar h=4+\frac\delta2$, then
 $\psi_{\epsilon,(0,2)}(\sqrt h,\sqrt{\bar h})
 +\psi_{\epsilon,(0,2)}(-\sqrt h,\sqrt{\bar h})=0$.
\end{enumerate}
\end{lemma}
\begin{proof}
(i) For $y\in\RR^2$,
$\psi_\epsilon(X+y)=(P+P_y-2)e^{-2\pi|X+y-(2,0)|^2/\epsilon^2}
=\psi_{\epsilon,(2,0)-y}(X)$, and $(2,0)-y=[2,2]-y$ equals $x_{l,k,j}(t)$
for $y=lC^\sharp+kW(t)+jW(t)^\sharp$.
(ii) With $\nu=|X-x|$ we have
$|\psi_{\epsilon,x}(X)|\leq\nu e^{-\pi\nu^2/\epsilon^2}
e^{-\pi\nu^2/\epsilon^2}$ and
$\nu e^{-\pi\nu^2/\epsilon^2}\leq\epsilon(2\pi e)^{-1/2}\leq1$.
(iii) Let $q=\sqrt{1+z/2}-1$. Then $|q|\leq\frac12|z|$ and
$q-\frac z4=-\frac12q^2$, so $|q-\frac z4|\leq\frac18z^2$. The value in
question is $\frac14q\,e^{-4\pi q^2/\epsilon^2}$, and
$|q|(1-e^{-4\pi q^2/\epsilon^2})\leq4\pi|q|^3/\epsilon^2
\leq\pi|z|^3/(2\epsilon^2)$.
(iv) In the first case $0\leq\sqrt h-2=\frac\delta2/(\sqrt h+2)
\leq\frac\delta8$ and the Gaussian is at most one. In the second case
$\psi_{\epsilon,(0,2)}(P,Q)=Pe^{-2\pi(P^2+(Q-2)^2)/\epsilon^2}$ is odd in
$P$.
\end{proof}

\begin{lemma}[The multiplier of the row]\label{lem:row-multiplier}
Let $0<g\leq\frac12$, let $I\subset(0,\infty)$ be a compact interval,
$w\geq1$ an integer, $t\in I$, $W(t)=[t,w/t]$ and
$0<\epsilon\leq\frac12$.
\begin{enumerate}[(i)]
\item For $x\in\RR^2$,
 \[
 \mathbb S\psi_{\epsilon,x}(X)
 =i\epsilon^4P\,e^{-2\pi\epsilon^2|X|^2}e^{2\pi iB(X,x)};
 \qquad\text{in particular}\quad
 \mathbb S\psi_\epsilon(X)=i\epsilon^4P\,e^{-2\pi\epsilon^2|X|^2}e^{8\pi iP}.
 \]
\item $\mathbb S(\mathsf D_t\psi_\epsilon)=\omega_t\cdot\mathbb S\psi_\epsilon$
 with the multiplier
 \[
 \omega_t(X)=\prod_{Y\in\{C^\sharp,\,W(t),\,W(t)^\sharp\}}
 \bigl(e^{-2\pi iB(X,Y)}-1\bigr)^2 ,
 \]
 which vanishes on $\{[u,v]:2u\in\ZZ\}\supset V$ and satisfies, on
 $\RR^2$,
 \begin{equation}\label{eq:ker-multiplier-bound}
 \begin{aligned}
 |\omega_t|&=64\,S_{C^\sharp}S_{W(t)}S_{W(t)^\sharp}
 \leq64\,\Omega\,S_{W(t)}S_{W(t)^\sharp},\\
 \bigl|\mathbb S(\mathsf D_t\psi_\epsilon)(X)\bigr|
 &\leq16\,\epsilon^3\,e^{-\pi\epsilon^2|X|^2}\,
 \Omega(X)S_{W(t)}(X)S_{W(t)^\sharp}(X).
 \end{aligned}
 \end{equation}
\item There is $C=C(g,I,w)$ such that for every admissible partition
 function of gap $g$ at $c=1+12a$ with
 $0<a\leq\min\{g^2/4,\epsilon^2/4\}$,
 \begin{equation}\label{eq:ker-row-shifts}
 \bigl|\mathcal L_a(\mathsf D_t\psi_\epsilon)
 -\langle\mu_a,\mathsf D_t\psi_\epsilon\rangle\bigr|\leq Ca\epsilon^{-1},
 \qquad
 \bigl|\mathcal L_a\bigl(\mathbb S\mathsf D_t\psi_\epsilon\bigr)
 -\langle\mu_a,\mathbb S\mathsf D_t\psi_\epsilon\rangle\bigr|
 \leq Ca\epsilon ,
 \end{equation}
 and consequently, with $\Phi^{(w)}_t=\Omega S_{W(t)}S_{W(t)^\sharp}$ and
 the kernel $\Phi^{(w)}_{t,\epsilon/\sqrt2}$ of width $\epsilon/\sqrt2$,
 \begin{equation}\label{eq:ker-row-inequality}
 \bigl|\langle\mu_a,\mathsf D_t\psi_\epsilon\rangle\bigr|
 \leq16\,\epsilon^3\sum_j\Phi^{(w)}_{t,\epsilon/\sqrt2}(X_j^\sigma)
 +2Ca\epsilon^{-1}.
 \end{equation}
\end{enumerate}
Here the pairings with $\mu_a$ are the physical pairings
\eqref{eq:ker-physical-pairing}: the row is evaluated on the unshifted
weights, and the whole effect of the central shift is contained in
\eqref{eq:ker-row-shifts}.
\end{lemma}
\begin{proof}
(i) By the computation in the proof of Lemma~\ref{lem:kernel-fourier},
$\gamma_{\epsilon,x}=\mathbb Sf_{-x}$ with
$f_\xi(X)=e^{-2\pi\epsilon^2|X|^2}e^{2\pi iB(X,\xi)}$, and
$\mathbb S^2f(X)=f(-X)$; hence $\mathbb S\gamma_{\epsilon,x}(X)=f_{-x}(-X)
=e^{-2\pi\epsilon^2|X|^2}e^{2\pi iB(X,x)}$. Since
$\psi_{\epsilon,x}=\frac{\epsilon^4}{4\pi}\partial_{P_x}\gamma_{\epsilon,x}$
and the difference quotients in $P_x$ converge in the Schwartz space, on
which $\mathbb S$ is continuous, we get
$\mathbb S\psi_{\epsilon,x}=\frac{\epsilon^4}{4\pi}\partial_{P_x}
\bigl(e^{-2\pi\epsilon^2|X|^2}e^{2\pi iB(X,x)}\bigr)$, and
$\partial_{P_x}B(X,x)=2P$. For $x=(2,0)$, $B(X,x)=4P$.

(ii) A change of variables in \eqref{eq:S-transform} gives
$\mathbb S\bigl(f(\cdot+Y)\bigr)=e^{-2\pi iB(\cdot,Y)}\,\mathbb Sf$.
Hence $\mathbb S\mathsf D_tf$ is $\mathbb Sf$ times the product over
$Y\in\{C^\sharp,W(t),W(t)^\sharp\}$ of
$\sum_lb_le^{-2\pi ilB(X,Y)}=(e^{-2\pi iB(X,Y)}-1)^2$. On $\RR^2$,
$|e^{-2\pi i\theta}-1|^2=4\sin^2(\pi\theta)$, and
$S_{C^\sharp}=\sin^2(2\pi u)\leq\Omega$. Finally
$|\mathbb S\psi_\epsilon(X)|=\epsilon^4|P|e^{-2\pi\epsilon^2|X|^2}
\leq(2\pi e)^{-1/2}\epsilon^3e^{-\pi\epsilon^2|X|^2}$ and
$64(2\pi e)^{-1/2}\leq16$.

(iii) By (i) of Lemma~\ref{lem:row-preimages},
$\mathsf D_t\psi_\epsilon$ is a combination, with coefficients of total
modulus $64$, of functions $\psi_{\epsilon,x}$ with $|x|\leq\varrho_1$,
where $\varrho_1$ depends on $I$ and $w$. Write
$\psi_{\epsilon,x}(X)=\epsilon\,\varphi((X-x)/\epsilon)$ with
$\varphi(Y)=Y_1e^{-2\pi Y\cdot Y}$. For $|\alpha|\leq2$ we have
$|\partial^\alpha\varphi(Y)|\leq c(1+|Y|)^3
e^{-2\pi(|\operatorname{Re}Y|^2-|\operatorname{Im}Y|^2)}$. If
$|\operatorname{Im}P|,|\operatorname{Im}Q|\leq\sqrt{a/2}$, then
$|\operatorname{Im}(X-x)/\epsilon|^2\leq a/\epsilon^2\leq\frac14$, and
therefore
\[
 |\partial^\alpha\psi_{\epsilon,x}(X)|
 \leq c'\epsilon^{-1}e^{-\pi|\operatorname{Re}X-x|^2/\epsilon^2}
 \leq c'e^{\pi\varrho_1^2}\,\epsilon^{-1}\,e^{-|\operatorname{Re}X|^2},
\]
using $\epsilon\leq1$ and $|X-x|^2\geq\frac12|X|^2-|x|^2$.
Lemma~\ref{lem:thin-shift} with $\beta=1$ and $M=C\epsilon^{-1}$ bounds
the change of each evaluation under the central shift by
$Ca\epsilon^{-1}e^{-(h+\bar h)}$; summing over the states with
\eqref{eq:thermal-moment}, and over the four vacuum weights, gives the
first bound of \eqref{eq:ker-row-shifts}. For the second, by (i) and
(ii) the function $\mathbb S\mathsf D_t\psi_\epsilon$ is
$i\epsilon^4P\,e^{-2\pi\epsilon^2|X|^2}$ times a combination, with
coefficients of total modulus $64$, of exponentials
$e^{2\pi iB(X,\xi)}$ with $|\xi|\leq\varrho_1$. Arguing as for
\eqref{eq:ker-kernel-derivatives}, on the strip
$|\operatorname{Im}P|,|\operatorname{Im}Q|\leq1$ its derivatives of
order at most two are bounded by a constant times
\[
 \epsilon^4(1+|\operatorname{Re}X|)(1+\epsilon^2|\operatorname{Re}X|)^2
 e^{-2\pi\epsilon^2|\operatorname{Re}X|^2}
 \leq C\epsilon^3e^{-\pi\epsilon^2|\operatorname{Re}X|^2},
\]
and Corollary~\ref{cor:central-shift} with $\beta=\pi\epsilon^2$,
followed by \eqref{eq:thermal-moment}, gives the bound $Ca\epsilon$.

The function $\mathsf D_t\psi_\epsilon$ belongs to
$\mathcal G_{2\pi/\epsilon^2}$, so
$\mathcal L_a(\mathsf D_t\psi_\epsilon)
=\mathcal L_a(\mathbb S\mathsf D_t\psi_\epsilon)$ by
Lemma~\ref{lem:gaussian-identity}. In
$\langle\mu_a,\mathbb S\mathsf D_t\psi_\epsilon\rangle$ the vacuum terms
vanish because $\omega_t=0$ on $V$, and by
\eqref{eq:ker-point-evaluations} and \eqref{eq:ker-multiplier-bound} the
term of a state is at most
$16\epsilon^3e^{-\pi\epsilon^2\Delta_j}\Phi^{(w)}_t(X_j^\sigma)$ in
modulus, the function $\Phi^{(w)}_t$ being even. Combining this with
\eqref{eq:ker-row-shifts} gives \eqref{eq:ker-row-inequality}.
\end{proof}

\begin{proposition}[The scalar near dimension two]\label{prop:scalar-row}
Let $0<g\leq\frac12$, let sum data be given, and let $t_0$ be an
interior point of $I$ at which \textup{(S1)--(S5)} hold. Suppose that
an admissible sequence of gap $g$ converges to $\sigma(R,t_0)$. Then
there is $\varrho>0$ such that, for all sufficiently small $a$ in the
sequence, exactly one state has a physical point in $D((1,1),\varrho)$;
it has spin zero, and its dimension $2+z_a$ satisfies $z_a\to0$ and
\begin{equation}\label{eq:ker-scalar-bound}
 |z_a|\leq C\,a\,L^{5/2};\qquad\text{in particular}\quad
 |z_a|<a^{1/3}
\end{equation}
for all sufficiently small $a$ in the sequence. The constant $C$ depends
on $g$, the data and $t_0$.
\end{proposition}
\begin{proof}
Let $\rho_0=\sigma(R,t_0)$, $\mu_0=\nu_{\vac}+\rho_0$, and
$\varsigma_a=|z_a|$.

\emph{Step 0: the low clusters.} Since $R\in\{1,2\}$, both $2R$ and
$2/R$ are integers, so $C\in\Lambda_R\cap\Lambda_{1/R}$, while
$C\notin\Lambda_{t_0}\cup\Lambda_{1/t_0}$ by (S1). By
\eqref{eq:ker-limit-masses}, $\mu_0(\{C\})=\frac12$ and
$\rho_0(\{C\})=\frac14$, and the same holds at $-C$, $\pm C^\sharp$.
Lemma~\ref{lem:budget-hypotheses}(iii), applied to $\Phi^{(p)}_{t_0}$ and
its centre $C=(1,1)$, gives the first assertion, with spin zero and
$z_a\to0$. By (S1) and the remark after (S5),
$\mu_0(\{Y_n(t_0)\})=\rho_0(\{Y_n(t_0)\})=\frac14$ for $n\in\mathcal N$.
As in Step~1 of the proof of Proposition~\ref{prop:generic-localization},
exactly one state has a physical point near $B_0(t_0)$, and its
dimension $\Delta_{0,a}$ defines $t_a\in I$ by \eqref{eq:ker-t-a}, with
$t_a\to t_0$; for $n\in\{p_1,p_2,w\}$ exactly one state has a physical
point near $Y_n(t_0)$, it has spin $n$, and its dimension is
$E_n(t_a)+\delta_{n,a}$ with $\delta_{n,a}\to0$. Finally, the point
$[2,2]=(2,0)$ lies in $\Lambda_r$, and in $\Lambda_{1/r}$, if and only if
$2r$ and $2/r$ are both integers. Hence
$\rho_0(\{(\pm2,0)\})=\mu_0(\{(\pm2,0)\})=\frac12$, which is the measure
of one state of weights $(4,0)$, and likewise at $(0,\pm2)$. By
Corollary~\ref{cor:cluster-stabilization}, exactly one state of $\rho_a$
has physical points near $(\pm2,0)$; it has spin $4$ and dimension
$4+\delta_a^+$ with $\delta_a^+\to0$. Exactly one state has physical
points near $(0,\pm2)$; it has spin $-4$ and dimension $4+\delta_a^-$
with $\delta_a^-\to0$. Here $\delta^\pm_a>0$ because the weights of a
state are positive.

\emph{Step 1: the scale.} Fix $0<\varrho\leq\frac12\sqrt g$ as in
Lemma~\ref{lem:budget-hypotheses} for the four families
$\Phi^{(n)}_{t_a}$, $n\in\mathcal N$, and so small that the closed discs
of radius $2\varrho$ about the points $x_{l,k,j}(t_0)$ contain no point
of $V\cup\supp\rho_0$ other than possibly their centres; by (S5) the
centre itself is not in $V\cup\supp\rho_0$ when $(k,j)\neq(0,0)$. Put
\begin{equation}\label{eq:ker-scale-H}
 b_a=\varsigma_a+a,\qquad H=\log(1/b_a)\leq L,\qquad
 \beta_a=a+\varsigma_a^2,\qquad\epsilon^2=\frac{\pi\varrho^2}{16H}.
\end{equation}
Then $b_a\to0$, $e^{-\pi\varrho^2/(4\epsilon^2)}=b_a^4\leq8\beta_a$, and
$aH\leq b_a\log(1/b_a)\to0$, so that the range
\eqref{eq:ker-budget-range} holds for the widths $\epsilon$ and
$\epsilon/\sqrt2$ when $a$ is small.

\emph{Step 2: the budget of\/ $\Phi^{(p)}$.} We verify the hypotheses of
Lemma~\ref{lem:peak-budget} for $\Phi^{(a)}=\Phi^{(p)}_{t_a}$, using
Lemma~\ref{lem:budget-hypotheses} with
$m_i=\mu_0(\{\xi_i^0\})$ and the table of
Lemma~\ref{lem:kernel-fourier}(ii). The centres lying in $V$ at $t_0$
are the origin and $\pm C,\pm C^\sharp$, by (S2), (S3) and the remark
after (S5); they are fixed. The origin has coefficient $\frac14$ and
$\mu_0(\{0\})=1$. The centres $\pm B_0\pm B_0^\sharp$, $\zeta\pm B_0$
and $\zeta\pm B_0^\sharp$ with $\zeta\neq0$ have positive coefficients
and satisfy (B2) by (S3). The centres $\zeta\pm B_0\pm B_0^\sharp$ with
$\zeta\neq0$ satisfy (B3)(b) by (S2). At $\pm C$, $\pm C^\sharp$
condition (B3)(a) holds with $w_i=\frac14$, $n_i=1$, the scalar state of
Step~0, whose spin is zero and whose dimension differs from $|C|^2=2$ by
$\varsigma_a$; here $m_i=\frac12$. At $\pm B_0(t_a)$,
$\pm B_0(t_a)^\sharp$ it holds with $n_i=1$ and no displacement. With
$\varepsilon=\varsigma_a$, \eqref{eq:ker-peak-budget} gives
\begin{equation}\label{eq:ker-one-two-initial}
 \sum_j\Phi^{(p)}_{t_a,\epsilon}(X_j^\sigma)
 \leq C\bigl((a+\varsigma_a^2)H^2+b_a^4\bigr)\leq C\beta_aH^2 ,
\end{equation}
and, as in \eqref{eq:ker-initial-single},
$\Phi^{(p)}_{t_a}(X_j^\sigma)\leq C\beta_aH^2$ for every state with
$\Delta_j\leq E_*$.

\emph{Step 3: the state near $W$.} Its physical point with the signs of
$W(t_a)$ is $X(\delta_{w,a})$ for $X_0=W(t_a)$. We apply
Lemma~\ref{lem:local-zero}(iii) to
$\Phi^{(p)}_{t_a}\geq S_{C^\sharp}S_{B_0(t_a)}S_{B_0(t_a)^\sharp}$, with
the vanishing factor $S_{B_0(t_a)}$ and the nonvanishing factors
$S_{C^\sharp}$ and $S_{B_0(t_a)^\sharp}$. Indeed
$B(W(t),B_0(t))=p+w\in\ZZ$,
\[
 \lambda\bigl(W(t),B_0(t)\bigr)=\frac{p-w}{|t^2-w^2/t^2|}
\]
is bounded below near $t_0$ because $w<p$,
$S_{C^\sharp}(W(t_a))=\sin^2(2\pi t_a)\to\sin^2(2\pi t_0)>0$, and
$S_{B_0(t_a)^\sharp}(W(t_a))$ converges to the positive number of (S4).
With $k=1$ the lemma gives
\begin{equation}\label{eq:ker-anchor-W}
 |\delta_{w,a}|\leq C\,(\beta_aH^2)^{1/2}=C\beta_a^{1/2}H .
\end{equation}

\emph{Step 4: the axis states.} Let $X=[u,v]$, with $u,v>0$, be a
physical point of the state near $(\pm2,0)$. Then $uv=4$, $u,v\to2$, and
\[
 \delta_a^+=\tfrac12(u^2+v^2)-uv=\tfrac12(u-v)^2\leq(u-2)^2+(v-2)^2
 \leq\tfrac1{16}\,\Omega(X),
\]
by \eqref{eq:ker-sine-distance}, since $\dist(2u,\ZZ)=2|u-2|$ and
$\dist(2v,\ZZ)=2|v-2|$ for small $a$. The phases
$B([2,2],B_0(t))$ and $B([2,2],B_0(t)^\sharp)$ both equal $2(t+p/t)$, so
by (S4) the product $S_{B_0(t_a)}S_{B_0(t_a)^\sharp}$ at $X$ is bounded
below for small $a$. With Step~2 this gives
$\delta_a^+\leq C\beta_aH^2$. For the state near $(0,\pm2)$ take
$X=[u,v]$ with $u>0>v$; then $uv=-4$,
$\delta_a^-=\frac12(u+v)^2\leq(u-2)^2+(v+2)^2\leq\frac1{16}\Omega(X)$,
the phases at $[2,-2]$ are $\pm2(t-p/t)$, and (S4) gives in the same way
\begin{equation}\label{eq:ker-axis-bounds}
 0<\delta_a^\pm\leq C\beta_aH^2 .
\end{equation}

\emph{Step 5: the budget of\/ $\Phi^{(w)}$ at width
$\epsilon/\sqrt2$.} The hypotheses of Lemma~\ref{lem:peak-budget} for
$\Phi^{(w)}_{t_a}$ are verified as in Step~2, with $W$ in place of
$B_0$. The only difference is that the state at the centres
$\pm W(t_a)$, $\pm W(t_a)^\sharp$ is displaced by $\delta_{w,a}$. By
\eqref{eq:ker-anchor-W} we may take $\varepsilon=C\beta_a^{1/2}H$, which
also bounds $\varsigma_a$ and tends to zero. Applying
\eqref{eq:ker-peak-budget} with the width $\epsilon/\sqrt2$,
\begin{equation}\label{eq:ker-W-budget}
 \sum_j\Phi^{(w)}_{t_a,\epsilon/\sqrt2}(X_j^\sigma)
 \leq C\bigl((a+\beta_aH^2)H^2+b_a^8\bigr)\leq C\beta_aH^4 .
\end{equation}

\emph{Step 6: the row.} By \eqref{eq:ker-row-inequality} with $t=t_a$,
\eqref{eq:ker-W-budget} and $\epsilon^2=\pi\varrho^2/(16H)$,
\begin{equation}\label{eq:ker-row-upper}
 \bigl|\langle\mu_a,\mathsf D_{t_a}\psi_\epsilon\rangle\bigr|
 \leq C\epsilon^3\beta_aH^4+Ca\epsilon^{-1}\leq C\beta_aH^{5/2}.
\end{equation}
We now evaluate the left side by Lemma~\ref{lem:row-preimages}(i), one
preimage $x=x_{l,k,j}(t_a)$ at a time. Recall that
$\langle\mu_a,\psi_{\epsilon,x}\rangle$ is a sum over the physical
points of all states, with weights $\frac14$, and over the points of
$V$ with the weights \eqref{eq:ker-vacuum-masses}. By
Lemma~\ref{lem:row-preimages}(ii) and \eqref{eq:ker-narrow-tail}, the
points at distance at least $\varrho$ from $x$ contribute at most
$(C_1+4)e^{-\pi\varrho^2/(2\epsilon^2)}=(C_1+4)b_a^8$ in total. It
remains to find the points within $\varrho$ of $x$.

If $(k,j)\neq(0,0)$, there are none for small $a$: the disc
$D(x_{l,k,j}(t_a),\varrho)$ is contained in
$D'=D(x_{l,k,j}(t_0),\frac32\varrho)$, which contains no point of $V$,
and $\rho_a(D')\to\rho_0(D')=0$ by vague convergence, hence
$\rho_a(D')=0$.

For $x_{1,0,0}=(1,1)$ the points within $\varrho$ are the vacuum atom at
$(1,1)$, where $\psi_{\epsilon,(1,1)}$ vanishes, and the point
$(\sqrt h,\sqrt h)$, $h=1+\frac12z_a$, of the scalar state of Step~0,
which has weight $\frac14$. By Lemma~\ref{lem:row-preimages}(iii),
\[
 \langle\mu_a,\psi_{\epsilon,(1,1)}\rangle
 =\frac{z_a}{16}+O\bigl(\varsigma_a^2+\varsigma_a^3H+b_a^8\bigr).
\]
For $x_{0,0,0}=(2,0)$ the points within $\varrho$ are the two points
$(\sqrt h,\pm\sqrt{\bar h})$ of the state of dimension $4+\delta_a^+$,
each of weight $\frac14$; by Lemma~\ref{lem:row-preimages}(iv) and
\eqref{eq:ker-axis-bounds} they contribute at most
$\frac1{16}\delta_a^+\leq C\beta_aH^2$ in modulus. For
$x_{2,0,0}=(0,2)$ the points within $\varrho$ are the two points
$(\pm\sqrt h,\sqrt{\bar h})$ of the state of dimension $4+\delta_a^-$,
and their contributions cancel by
Lemma~\ref{lem:row-preimages}(iv). Since the coefficient of the middle
preimage is $b_1=-2$,
\begin{equation}\label{eq:ker-row-value}
 \langle\mu_a,\mathsf D_{t_a}\psi_\epsilon\rangle
 =-\frac{z_a}8+O\bigl(\beta_aH^2\bigr).
\end{equation}

\emph{Conclusion.} By \eqref{eq:ker-row-upper} and
\eqref{eq:ker-row-value},
$\frac18\varsigma_a\leq C(a+\varsigma_a^2)H^{5/2}$. Since
$\varsigma_aH^{5/2}\leq b_a\log^{5/2}(1/b_a)\to0$, the term
$C\varsigma_a^2H^{5/2}$ is at most $\frac1{16}\varsigma_a$ for small $a$,
and then $\varsigma_a\leq16CaH^{5/2}\leq16CaL^{5/2}$.
\end{proof}

\begin{proposition}[Localization and exclusion for sum data]
\label{prop:one-two-localization}
Let $0<g\leq\frac12$, let sum data be given, and let $t_0$ be an
interior point of $I$ at which \textup{(S1)--(S5)} hold. Then no
admissible sequence of gap $g$ converges to $\sigma(R,t_0)$.
\end{proposition}
\begin{proof}
Suppose that such a sequence exists, and keep the notation of the proof
of Proposition~\ref{prop:scalar-row}. We use only the bound
$\varsigma_a<a^{1/3}$ of \eqref{eq:ker-scalar-bound}. Choose the width
as in \eqref{eq:ker-width-L}, $\epsilon^2=\kappa/L$ with
$\kappa=\pi\varrho^2/4$.

\emph{Step 1.} The hypotheses of Lemma~\ref{lem:peak-budget} for
$\Phi^{(p)}_{t_a}$ were verified in Step~2 of the preceding proof, with
$\varepsilon=\varsigma_a<a^{1/3}$. Hence
\[
 \sum_j\Phi^{(p)}_{t_a,\epsilon}(X_j^\sigma)
 \leq C\bigl((a+a^{2/3})L^2+a\bigr)\leq C\,a^{2/3}L^2 ,
\]
and $\Phi^{(p)}_{t_a}(X_j^\sigma)\leq Ca^{2/3}L^2$ for states with
$\Delta_j\leq E_*$.

\emph{Step 2.} Let $k\in\{1,2\}$. For the state near $B_k(t_0)$ we apply
Lemma~\ref{lem:local-zero}(iii) at $X_0=B_k(t_a)$ to
$\Phi^{(p)}_{t_a}\geq S_{C^\sharp}S_{B_0(t_a)}S_{B_0(t_a)^\sharp}$. As in
Step~3 of the proof of Proposition~\ref{prop:generic-localization}, the
factor $S_{B_0(t_a)}$ vanishes at $B_k(t_a)$ with rate
$k\ell/(p_k^2/t_a^2-t_a^2)$; the factors $S_{C^\sharp}$ and
$S_{B_0(t_a)^\sharp}$ at $B_k(t_a)$ converge to $\sin^2(2\pi t_0)>0$ and
to the positive number of (S4). Therefore
\[
 |\delta_{p_k,a}|\leq C\,(a^{2/3}L^2)^{1/2}=C\,a^{1/3}L\qquad(k=1,2),
\]
and $\delta_{p_0,a}=0$.

\emph{Step 3.} For $k=0,1,2$ the hypotheses of
Lemma~\ref{lem:peak-budget} hold for $\Phi^{(p_k)}_{t_a}$, as in Step~2
of the preceding proof, with $\varepsilon=Ca^{1/3}L$, which bounds both
$\varsigma_a$ and $|\delta_{p_k,a}|$. Hence
\[
 \sum_j\Phi^{(p_k)}_{t_a,\epsilon}(X_j^\sigma)
 \leq C\bigl((a+a^{2/3}L^2)L^2+a\bigr)\leq C\,a^{2/3}L^4,\qquad k=0,1,2 .
\]

\emph{Step 4.} Let $j$ be a state with $\Delta_j\leq L$ and $X=[u,v]$ one
of its physical points. As in Step~5 of the proof of
Proposition~\ref{prop:generic-localization},
\[
 \Omega(X)\,S_{B_k(t_a)}(X)S_{B_k(t_a)^\sharp}(X)\leq\Gamma_a:=C\,a^{2/3}L^4
 \qquad(k=0,1,2),
\]
$uv=s_j\in\ZZ$ and $|u|,|v|\leq\Upsilon=\sqrt{2L}$.
Lemma~\ref{lem:three-projection}(d) and~(a), with
$\zeta=\frac12\Gamma_a^{1/6}=Ca^{1/9}L^{2/3}$, give a real point $X_j'$,
either in $\Lambda_{t_a}\cup\Lambda_{1/t_a}$ or of the form $[m/2,n/2]$,
whose spin is exactly $s_j$ and which satisfies
\begin{equation}\label{eq:ker-one-two-localization}
 \bigl|\Delta_j-|X_j'|^2\bigr|\leq C\,\zeta\,\Upsilon\leq C\,a^{1/9}L^{7/6}.
\end{equation}
In the second case
$X_j'\in\Sigma_2=\Lambda_{1/2}\cup\Lambda_1\cup\Lambda_2$ by
\eqref{eq:ker-grid-circles}.

\emph{Step 5.} As in Step~6 of the proof of
Proposition~\ref{prop:generic-localization}, after discarding finitely
many terms of the sequence this contradicts
Theorem~\ref{thm:moving-localization}, applied with the fixed union
$\Sigma_2$, the interval $I$, $u_*=1$, and the error bound
\eqref{eq:ker-one-two-localization}.
\end{proof}

\begin{remark}\label{rem:one-two-exponent}
If the full bound $|z_a|\leq CaL^{5/2}$ of \eqref{eq:ker-scalar-bound} is
used in Step~1, the same steps give budgets $O(aL^2)$ and $O(aL^4)$ and
the localization error $O(a^{1/6}L^{7/6})$ in
\eqref{eq:ker-one-two-localization}. Any positive power of $a$ suffices
for Theorem~\ref{thm:moving-localization}.
\end{remark}

\subsection{Choice of the integer data}

By Lemma~\ref{lem:exceptional-set}, fixed data leave finitely many
exceptional radii. We now show that for every radius $t_0>1$ which is
not a half-integer the data can be chosen, depending on $t_0$, so that
the finite hypotheses hold at $t_0$. All points involved have the form
\eqref{eq:ker-general-centre},
\[
 \xi=\bigl[a_1+b_1t_0+b_2n/t_0,\ a_2+b_2t_0+b_1n/t_0\bigr],
 \qquad a_1,a_2,b_1,b_2\in\ZZ,\quad n\geq1 ,
\]
and we recall that $\Lambda_{t_0}=\{[mt_0,n'/t_0]\}$ and
$\Lambda_{1/t_0}=\{[m/t_0,n't_0]\}$ with $m,n'\in\ZZ$.

\begin{lemma}[Incidences at an irrational radius]
\label{lem:irrational-incidences}
Let $t_0>0$ be irrational and let $\xi$ be as above.
\begin{enumerate}[(i)]
\item If $(b_1,b_2)\neq(0,0)$ and $\xi\in\QQ^2$, then $b_2=\mathsf s b_1$
 with a sign $\mathsf s\in\{\pm1\}$, and $t_0+\mathsf s n/t_0\in\QQ$.
 For each sign there is at most one integer $n\geq1$ with
 $t_0+\mathsf s n/t_0\in\QQ$; if $t_0^2\in\QQ$, this happens only for
 $\mathsf s=-1$ and $n=t_0^2$.
\item If $(a_1,a_2)\neq(0,0)$ and $\xi\in\Lambda_{t_0}\cup\Lambda_{1/t_0}$,
 then $t_0^2\notin\QQ$, $a_1a_2\neq0$, and $t_0$ satisfies a relation
 $t_0^2=e_1t_0+e_0$ with $e_0,e_1\in\QQ\setminus\{0\}$. Moreover
 $b_2\neq0$ and $n=a_1e_0/(e_1b_2)$ if $\xi\in\Lambda_{t_0}$, and
 $b_1\neq0$ and $n=a_2e_0/(e_1b_1)$ if $\xi\in\Lambda_{1/t_0}$. In
 particular, for given $t_0$, $a_1,a_2,b_1,b_2$ there are at most two
 such integers $n$.
\end{enumerate}
\end{lemma}
\begin{proof}
(i) Both $b_1t_0+b_2n/t_0$ and $b_2t_0+b_1n/t_0$ are rational, hence so
is $(b_1^2-b_2^2)t_0$, and $b_1^2=b_2^2\neq0$. Then
$b_1(t_0+\mathsf s n/t_0)\in\QQ$. If this holds for $n\neq n'$ with
the same sign, then $(n-n')/t_0\in\QQ$, which is impossible. If
$t_0^2\in\QQ$, then $(t_0^2+\mathsf s n)/t_0\in\QQ$ forces
$t_0^2+\mathsf s n=0$.

(ii) Since $\xi\mapsto\xi^\sharp$ maps $\Lambda_{1/t_0}$ onto
$\Lambda_{t_0}$ and exchanges $(a_1,b_1)$ with $(a_2,b_2)$, it suffices
to treat $\xi\in\Lambda_{t_0}$. Multiplying the two coordinate equations
$\xi=[mt_0,n'/t_0]$ by $t_0$ gives
\begin{equation}\label{eq:ker-charge-equations}
 a_1t_0+(b_1-m)t_0^2+b_2n=0,\qquad a_2t_0+b_2t_0^2+b_1n-n'=0 .
\end{equation}
If $t_0^2\in\QQ$, these give $a_1t_0\in\QQ$ and $a_2t_0\in\QQ$, so
$a_1=a_2=0$, contrary to the hypothesis. If $1,t_0,t_0^2$ are linearly
independent over $\QQ$, the first equation gives $a_1=0$ and $b_2n=0$,
so $b_2=0$, and the second gives $a_2=0$; again a contradiction. Hence
$t_0^2=e_1t_0+e_0$ with unique rational $e_0,e_1$, and $e_1\neq0$.
Substituting and comparing the coefficients of $1$ and $t_0$,
\[
 a_1=-(b_1-m)e_1,\qquad b_2n=-(b_1-m)e_0,\qquad a_2=-b_2e_1 .
\]
If $b_2=0$, then $a_2=0$, hence $a_1\neq0$, $b_1\neq m$ and $e_0=0$,
which would make $t_0=e_1$ rational. So $b_2\neq0$, $a_2\neq0$, and
$n=a_1e_0/(e_1b_2)$; since $n\neq0$, also $a_1\neq0$ and $e_0\neq0$.
\end{proof}

\begin{lemma}[Data for an irrational radius]\label{lem:data-irrational}
Let $I\subset(0,\infty)$ be a compact interval and $t_0\in I$
irrational. Put $\ell=t_0^2$ if $t_0^2$ is an integer and $\ell=1$
otherwise.
\begin{enumerate}[(a)]
\item Let $R=K/2$ with $K\geq3$, $K\neq4$, and put $U=2K$,
 $V_j=2(2K+3+j)$ for $j=0,1,2$. These are product data for every $p$
 satisfying \eqref{eq:ker-p-conditions}, and for all but finitely many
 such $p$ the hypotheses \textup{(H1)--(H4)} hold at $t_0$.
\item Let $R\in\{1,2\}$. There is an integer $w\geq1$ such that, for all
 but finitely many $p>w$ satisfying \eqref{eq:ker-p-conditions}, the
 hypotheses \textup{(S1)--(S5)} hold at $t_0$.
\end{enumerate}
\end{lemma}
\begin{proof}
The integers $p$ with $\gcd(p,2\ell)=1$ contain the progression
$1+2\ell\NN$, so infinitely many $p$ satisfy
\eqref{eq:ker-p-conditions}. If $t_0^2=\ell$ is an integer, then
$\ell\geq2$ and $\gcd(p_k,\ell)=\gcd(p,\ell)=1$ for $k=0,1,2$. Below,
``excluding $p$'' means discarding finitely many values of $p$. By
Lemma~\ref{lem:irrational-incidences}(i) we exclude the at most six $p$
for which
\begin{equation}\label{eq:ker-rational-sum}
 t_0+\mathsf s p_k/t_0\in\QQ\qquad\text{for some }\mathsf s\in\{\pm1\},
 \ k\in\{0,1,2\}.
\end{equation}
Three facts are used in both parts. First, neither $2t_0$ nor $2/t_0$ is
an integer. Secondly, $Y_n(t_0)=[t_0,n/t_0]\in\Lambda_{1/t_0}$ means
$t_0^2=m$ and $n=n't_0^2$ with integers $m,n'$, so it requires $t_0^2$
to be an integer dividing $n$; this fails for $n=p_k$. Thirdly, the
phase $t_0^2+pp_k/t_0^2$ of $S_{B_0(t_0)^\sharp}$ at $B_k(t_0)$ is, for
each $k$, an integer for at most one $p$: if $t_0^2\notin\QQ$ and two values
$p<p'$ gave integers, then, with $p'_k=p'+k\ell$, the number
$(p'p'_k-pp_k)/t_0^2$ would be a nonzero integer and $t_0^2$ rational; if $t_0^2=A/B$ in lowest terms with $B>1$,
integrality of $A/B+pp_kB/A$ gives $B\mid A^2$ after multiplication by
$AB$, which is impossible; and if $t_0^2=\ell$ is an integer,
integrality gives $\ell\mid pp_k$, contradicting
$\gcd(pp_k,\ell)=1$.

(a) The conditions \eqref{eq:ker-A-conditions} hold:
$A_j=[4R,\,K(2K+3+j)/R]\in\Lambda_R$, $U=2K\geq6$, and
$V_j-U=2K+6+2j\geq3$.

(H1): $B_k(t_0)$ has the irrational coordinate $t_0$, so it is not in
$\Lambda_R\cup\Lambda_{1/R}\subset\QQ^2$; the rest was shown above.

(H2): Let $|\eta|=3$. In the notation \eqref{eq:ker-general-centre},
$[a_1,a_2]=\eta_1Y+\eta_2Y^\sharp\neq0$ and
$(b_1,b_2)=(\eta_3,\eta_4)\neq(0,0)$. By
Lemma~\ref{lem:irrational-incidences}(i) and the exclusion
\eqref{eq:ker-rational-sum}, $\xi_\eta(t_0)\notin\QQ^2\supset
V\cup\Lambda_R\cup\Lambda_{1/R}$. By
Lemma~\ref{lem:irrational-incidences}(ii),
$\xi_\eta(t_0)\in\Lambda_{t_0}\cup\Lambda_{1/t_0}$ is possible only if
$p_k$ is one of at most two numbers determined by $t_0$ and
$(a_1,a_2,b_1,b_2)$. The latter range over a finite set which does not
depend on $p$, so we exclude finitely many $p$. In particular, if
$t_0^2\in\QQ$ no such incidence occurs: for the fixed vector $C$ or
$C^\sharp$, for instance, the two equations
\eqref{eq:ker-charge-equations} force $a_1t_0\in\QQ$ and
$a_2t_0\in\QQ$, while one of $a_1,a_2$ is $\pm2$.

(H3): Let $|\eta|\in\{2,4\}$. If $(\eta_3,\eta_4)=(0,0)$, the centre is
$\pm Y\pm Y^\sharp$ and is not in $V$, as in the proof of
Lemma~\ref{lem:exceptional-set}. Otherwise
$\xi_\eta(t_0)\notin\QQ^2\supset V$ by
Lemma~\ref{lem:irrational-incidences}(i) and \eqref{eq:ker-rational-sum}.

(H4): The phases at $A_j$ are $Up/t_0+V_jt_0$ and $Ut_0+V_jp/t_0$. If
one of them were an integer for two values $p\neq p'$, then
$U(p-p')/t_0$ or $V_j(p-p')/t_0$ would be an integer and $t_0$ rational;
so we exclude at most six $p$. At $B_k(t_0)$ the phases of $S_C$ and
$S_{C^\sharp}$ are $2p_k/t_0$ and $2t_0$, which are irrational, and the
phase of $S_{B_0(t_0)^\sharp}$ was treated above.

(b) We first choose $w$. Let $\mathfrak W$ be the finite set of integers
$n$ given by Lemma~\ref{lem:irrational-incidences}(ii) for
$[a_1,a_2]=[2,2-2l]$ and $(b_1,b_2)=(-k,-j)$ with $0\leq l,k,j\leq2$,
$(k,j)\neq(0,0)$. We require
\begin{equation}\label{eq:ker-w-conditions}
 t_0+w/t_0\notin\QQ,\qquad t_0-w/t_0\notin\QQ,\qquad
 w\notin\mathfrak W,\qquad\gcd(w,t_0^2)=1\ \text{ if }t_0^2\in\ZZ .
\end{equation}
By Lemma~\ref{lem:irrational-incidences}(i) the first two conditions
exclude at most two integers, and none other than $w=t_0^2$ when
$t_0^2\in\QQ$; so an integer $w\geq1$ with \eqref{eq:ker-w-conditions}
exists, for instance in the progression $1+t_0^2\NN$ when $t_0^2$ is an
integer. Fix it, and let $p>w$.

(S1): shown above for $n=p_k$; for $n=w$, $t_0^2$ would be an integer
dividing $w$, contrary to \eqref{eq:ker-w-conditions}.

(S2): Here $[a_1,a_2]=\zeta\in\{\pm[2,0],\pm[0,2]\}$, so $a_1a_2=0$ and
$(a_1,a_2)\neq0$, and Lemma~\ref{lem:irrational-incidences}(ii) shows
that the centre is not in $\Lambda_{t_0}\cup\Lambda_{1/t_0}$. It is not
in $\QQ^2\supset V\cup\Lambda_R\cup\Lambda_{1/R}$ by
Lemma~\ref{lem:irrational-incidences}(i), using
\eqref{eq:ker-w-conditions} for $n=w$ and \eqref{eq:ker-rational-sum}
for $n=p_k$.

(S3): The centres have $(b_1,b_2)\neq(0,0)$ and are not in $\QQ^2$, for
the same reason.

(S4): For $n=p_1,p_2$ the phase $t_0^2+pn/t_0^2$ was treated above. For
$n=w$ the same three cases apply: if $t_0^2\notin\QQ$, at most one $p$
gives an integer, since otherwise $(p'-p)w/t_0^2$ would be a nonzero
integer; if $t_0^2=A/B$ with $B>1$, integrality gives $B\mid A^2$; if
$t_0^2=\ell$, it gives $\ell\mid pw$, contradicting
$\gcd(pw,\ell)=1$. Finally $2(t_0\pm p/t_0)\in\ZZ$ holds for at most one
$p$ for each sign, because two such values would make $2(p-p')/t_0$ an
integer.

(S5): Let $(k,j)\neq(0,0)$. If $x_{l,k,j}(t_0)\in\QQ^2$, then by
Lemma~\ref{lem:irrational-incidences}(i) $j=\mathsf s k$, which for
$k,j\geq0$ means $j=k>0$ and $\mathsf s=1$, and
$t_0+w/t_0\in\QQ$, contrary to \eqref{eq:ker-w-conditions}. Since
$a_1=2\neq0$, Lemma~\ref{lem:irrational-incidences}(ii) shows that
$x_{l,k,j}(t_0)\in\Lambda_{t_0}\cup\Lambda_{1/t_0}$ would force
$w\in\mathfrak W$.
\end{proof}

For rational radii the incidences are decided by denominators. Let
$\alpha,\beta\geq1$ be coprime integers. For integers $c_1,c_2$,
\begin{equation}\label{eq:ker-denominators}
 \alpha\Bigl(c_1\frac\alpha\beta+c_2\frac\beta\alpha\Bigr)\in\ZZ
 \iff\beta\mid c_1,\qquad
 \beta\Bigl(c_1\frac\alpha\beta+c_2\frac\beta\alpha\Bigr)\in\ZZ
 \iff\alpha\mid c_2 ,
\end{equation}
because the two numbers are $c_1\alpha^2/\beta+c_2\beta$ and
$c_1\alpha+c_2\beta^2/\alpha$.

\begin{lemma}[Incidences at a rational radius]
\label{lem:rational-incidences}
Let $t_0=\alpha/\beta$ in lowest terms with $\beta\geq3$ and
$\alpha\geq4$, let $n\geq1$ be an integer coprime to $\alpha$, and let
$\xi$ be as in \eqref{eq:ker-general-centre} with $t=t_0$ and
$|b_1|,|b_2|\leq2$.
\begin{enumerate}[(i)]
\item If $|b_1|,|b_2|\leq1$ and one of $2\xi_1$, $2\xi_2$ is an integer,
 then $b_1=b_2=0$. Here $\xi=[\xi_1,\xi_2]$.
\item If $2\xi_1$ and $2\xi_2$ are both integers, then $b_1=b_2=0$.
\item If $\xi\in\Lambda_{t_0}$, then $b_2=0$ and $\alpha\mid a_1$. If
 $\xi\in\Lambda_{1/t_0}$, then $b_1=0$ and $\beta\mid a_1$,
 $\alpha\mid a_2$.
\end{enumerate}
\end{lemma}
\begin{proof}
We have $\xi_1=a_1+b_1\alpha/\beta+b_2n\beta/\alpha$ and
$\xi_2=a_2+b_2\alpha/\beta+b_1n\beta/\alpha$, and $\alpha\mid cn$ is
equivalent to $\alpha\mid c$. By \eqref{eq:ker-denominators}, if
$2\xi_1\in\ZZ$ then $\beta\mid2b_1$ and $\alpha\mid2b_2$, and if
$2\xi_2\in\ZZ$ then $\beta\mid2b_2$ and $\alpha\mid2b_1$. For
$|b_i|\leq1$ we have $|2b_i|\leq2<\beta,\alpha$, which gives (i). In
(ii), $\alpha\beta$ divides $2b_1$ and $2b_2$, and
$|2b_i|\leq4<12\leq\alpha\beta$. For (iii), let $\xi=[m\alpha/\beta,
n'\beta/\alpha]\in\Lambda_{t_0}$. The first coordinate gives
$(b_1-m)\alpha/\beta+b_2n\beta/\alpha=-a_1\in\ZZ$, so $\alpha\mid b_2$ and
$\beta\mid(b_1-m)$ by \eqref{eq:ker-denominators}; hence $b_2=0$,
because $|b_2|\leq2<\alpha$, and then $a_1=(m-b_1)\alpha/\beta$ is a
multiple of $\alpha$. If $\xi=[m\beta/\alpha,n'\alpha/\beta]
\in\Lambda_{1/t_0}$, the first coordinate gives
$b_1\alpha/\beta+(b_2n-m)\beta/\alpha=-a_1$, so $\beta\mid b_1$, hence
$b_1=0$, and $\alpha\mid(b_2n-m)$, hence $\beta\mid a_1$; the second
coordinate gives $(b_2-n')\alpha/\beta=-a_2$ with $\beta\mid(b_2-n')$,
hence $\alpha\mid a_2$.
\end{proof}

\begin{lemma}[Data for a rational radius]\label{lem:data-rational}
Let $t_0=\alpha/\beta>1$ in lowest terms with $\beta\geq3$, let
$I\subset(0,\infty)$ be a compact interval containing $t_0$, and put
$\ell=\alpha$. Let $p$ satisfy $p\equiv1\pmod{2\alpha}$ and
$p>\sup_It^2+2$; then \eqref{eq:ker-p-conditions} holds.
\begin{enumerate}[(a)]
\item Let $R=K/2$ with $K\geq3$, $K\neq4$, and put
 \[
 U=2K\alpha\beta,\qquad V_j=2+2\alpha\beta(K+2+j)\quad(j=0,1,2).
 \]
 These are product data, and \textup{(H1)--(H4)} hold at $t_0$.
\item Let $R\in\{1,2\}$ and $w=1$. Then \textup{(S1)--(S5)} hold at $t_0$.
\end{enumerate}
\end{lemma}
\begin{proof}
Since $\alpha>\beta\geq3$ we have $\alpha\geq4$, and the integers $p_k$
and $w=1$ are congruent to $1$ modulo $\alpha$, hence coprime to
$\alpha$; so Lemma~\ref{lem:rational-incidences} applies to all points
considered. Neither $2t_0=2\alpha/\beta$ nor $2/t_0=2\beta/\alpha$ is an
integer. By Lemma~\ref{lem:rational-incidences}(iii) with $a_1=a_2=0$
and $(b_1,b_2)=(1,0)$, $Y_n(t_0)\notin\Lambda_{1/t_0}$ for
$n\in\{p_0,p_1,p_2,1\}$. The phase $t_0^2+pn/t_0^2
=\alpha^2/\beta^2+pn\beta^2/\alpha^2$ is not an integer: otherwise
multiplication by $\alpha^2\beta^2$ would give $\beta^2\mid\alpha^4$.

(a) We have $A_j=[4\alpha\beta R,\,K(1+\alpha\beta(K+2+j))/R]
\in\Lambda_R$, $U,V_j\geq3$, and
$V_j-U=2+2\alpha\beta(2+j)\geq3$; so
\eqref{eq:ker-A-conditions} holds. Note that
\begin{equation}\label{eq:ker-rational-congruences}
 U\equiv0,\qquad V_j\equiv2\pmod{\alpha\beta}.
\end{equation}

(H1): By Lemma~\ref{lem:rational-incidences}(i) with $(b_1,b_2)=(1,0)$,
neither $2t_0$ nor $2p_k/t_0$ is an integer, whereas a point of
$\Lambda_R=\{[mK/2,2n'/K]\}$ has $2\xi_1\in\ZZ$ and a point of
$\Lambda_{1/R}$ has $2\xi_2\in\ZZ$. The rest was shown above.

(H2): Let $|\eta|=3$, so $(b_1,b_2)=(\eta_3,\eta_4)\neq(0,0)$. By
Lemma~\ref{lem:rational-incidences}(i), neither $2\xi_1$ nor $2\xi_2$
is an integer, so $\xi_\eta(t_0)\notin V\cup\Lambda_R\cup\Lambda_{1/R}$.
If $\xi_\eta(t_0)\in\Lambda_{t_0}$, then $\eta_4=0$ by
Lemma~\ref{lem:rational-incidences}(iii), hence $\eta_1\eta_2\neq0$, and
$\alpha$ divides $a_1=\eta_1u_Y+\eta_2v_Y$. For $Y=A_j$ this number is
congruent to $2\eta_2=\pm2$ modulo $\alpha$ by
\eqref{eq:ker-rational-congruences}, and for $Y=C$ it equals
$2\eta_1=\pm2$; both are impossible since $\alpha\geq4$. If
$\xi_\eta(t_0)\in\Lambda_{1/t_0}$, then $\eta_3=0$,
$\eta_1\eta_2\neq0$, and $\alpha$ divides $a_2=\eta_1v_Y+\eta_2u_Y$,
which is congruent to $2\eta_1$ for $Y=A_j$ and equals $2\eta_2$ for
$Y=C$; again impossible.

(H3): If $(\eta_3,\eta_4)\neq(0,0)$, then $\xi_\eta(t_0)\notin V$ by
Lemma~\ref{lem:rational-incidences}(i). Otherwise the centre is
$\pm Y\pm Y^\sharp\notin V$, as in the proof of
Lemma~\ref{lem:exceptional-set}.

(H4): By \eqref{eq:ker-rational-congruences},
\[
 B(A_j,B_0(t_0))=\frac{Up\beta}\alpha+\frac{V_j\alpha}\beta
 \equiv\frac{2\alpha}\beta,\qquad
 B(A_j,B_0(t_0)^\sharp)=\frac{U\alpha}\beta+\frac{V_jp\beta}\alpha
 \equiv\frac{2p\beta}\alpha\pmod{\ZZ},
\]
and neither is an integer, because $\beta\nmid2$ and $\alpha\nmid2$. At
$B_k(t_0)$ the phases of $S_C$ and $S_{C^\sharp}$ are
$2p_k\beta/\alpha$ and $2\alpha/\beta$, not integers, and the phase of
$S_{B_0(t_0)^\sharp}$ was treated above.

(b) (S1) and the first part of (S4) were shown above. If
$2(t_0\pm p/t_0)=2\alpha/\beta\pm2p\beta/\alpha$ were an integer,
\eqref{eq:ker-denominators} would give $\beta\mid2$. In (S2) and (S3)
the centres have $|b_1|,|b_2|\leq1$ and $(b_1,b_2)\neq(0,0)$, so by
Lemma~\ref{lem:rational-incidences}(i) they are not in $\frac12\ZZ^2
\supset V\cup\Lambda_R\cup\Lambda_{1/R}$; in (S2) moreover
$b_1b_2\neq0$, which excludes $\Lambda_{t_0}\cup\Lambda_{1/t_0}$ by
Lemma~\ref{lem:rational-incidences}(iii). In (S5),
$[a_1,a_2]=[2,2-2l]$ and $(b_1,b_2)=(-k,-j)\neq(0,0)$. By
Lemma~\ref{lem:rational-incidences}(ii) the point is not in
$\frac12\ZZ^2$. If it were in $\Lambda_{t_0}$, then $\alpha\mid a_1=2$;
if it were in $\Lambda_{1/t_0}$, then $\beta\mid a_1=2$. Both are
impossible.
\end{proof}

\begin{theorem}[Both limiting radii are half-integers]
\label{thm:half-integer-limits}
Let $0<g\leq\frac12$. Every limit of an admissible sequence of gap $g$
is of the form $\sigma(R,T)$ with
\[
 2R\in\ZZ,\qquad2T\in\ZZ,\qquad1\leq R,T\leq(2g)^{-1/2}.
\]
\end{theorem}
\begin{proof}
By Theorem~\ref{thm:classification} the limit is $\sigma(R,t_0)$ with
$1\leq R,t_0\leq R_{\max}=(2g)^{-1/2}$ and at least one half-integer
radius; since $\sigma(R,t_0)=\sigma(t_0,R)$ we may assume $2R\in\ZZ$.
Suppose that $2t_0\notin\ZZ$. Then $t_0>1$ is an interior point of
$I=[\frac12,R_{\max}+1]$. If $R\notin\{1,2\}$, then $R=K/2$ with
$K\geq3$ and $K\neq4$, and Lemma~\ref{lem:data-irrational}(a) or
Lemma~\ref{lem:data-rational}(a), according as $t_0$ is irrational or
rational (in which case its denominator is at least $3$), provides
product data for which (H1)--(H4) hold at $t_0$;
Proposition~\ref{prop:generic-localization} then shows that no
admissible sequence of gap $g$ converges to $\sigma(R,t_0)$. If
$R\in\{1,2\}$, parts (b) of the same lemmas provide sum data for which
(S1)--(S5) hold at $t_0$, and
Proposition~\ref{prop:one-two-localization} gives the same
contradiction.
\end{proof}

\subsection{The four-scalar anchor}

By Theorem~\ref{thm:half-integer-limits} only finitely many limits
remain for each gap, and all of them are supported on one rational
grid. Put
\begin{equation}\label{eq:ker-grid-denominator}
 N_g=\bigl\lfloor2(2g)^{-1/2}\bigr\rfloor,\qquad d=2\,N_g!,\qquad
 E_d=\tfrac12d^2,\qquad\eta_d=\frac1{8d^2}.
\end{equation}
Since $g\leq\frac12$ we have $N_g\geq2$ and $d\geq4$. If $R=K/2$ is a
half-integer with $1\leq R\leq(2g)^{-1/2}$, then $2\leq K\leq N_g$, so
$dR=N_g!\,K$ and $d/R=4N_g!/K$ are integers. Consequently, in light-cone
coordinates,
\begin{equation}\label{eq:ker-common-grid}
 \Lambda_R\cup\Lambda_{1/R}\subset d^{-1}\ZZ^2,\qquad
 [\pm d,0],\ [0,\pm d]\in\Lambda_R\cap\Lambda_{1/R}
\end{equation}
for every such $R$. The points $[\pm d,0]$, $[0,\pm d]$ are the four
points $(\pm\frac d2,\pm\frac d2)$; they have spin zero and squared norm
$E_d$.

\begin{lemma}[The common four-scalar cluster]
\label{lem:four-scalar-cluster}
Let $0<g\leq\frac12$. There is $a_1=a_1(g)>0$ such that every admissible
partition function of gap $g$ at $c=1+12a$ with $0<a<a_1$ has exactly
four states of spin zero with dimension in
$(E_d-\eta_d,E_d+\eta_d)$. Writing their dimensions as $E_d+z_{i,a}$,
$1\leq i\leq4$, and
\[
 \delta_a=\max_{1\leq i\leq4}|z_{i,a}| ,
\]
we have $\delta_a\to0$ uniformly: for every $\eta>0$ there is
$a(\eta)>0$ such that $\delta_a<\eta$ for every admissible partition
function of gap $g$ with $0<a<a(\eta)$.
\end{lemma}

The quantity $\delta_a$ depends on the partition function and not only
on $a$; we suppress this in the notation.

\begin{proof}
Let $\rho_0=\sigma(R,T)$ with half-integers $1\leq R,T\leq(2g)^{-1/2}$,
and $\mu_0=\nu_{\vac}+\rho_0$. By \eqref{eq:ker-common-grid} and
\eqref{eq:ker-limit-masses}, $\mu_0$ has mass
$\frac14\cdot4=1$ at each of the points $(\pm\frac d2,\pm\frac d2)$, and
these are outside $V$ because $\frac d2\geq2$. Hence $\rho_0$ has total
mass four on these points, that is, exactly four states of weights
$(\frac14d^2,\frac14d^2)$. Every point of $\supp\rho_0$ of spin zero is
of the form $[m/d,0]$ or $[0,n/d]$ and has dimension in
$\frac1{2d^2}\ZZ$; since $\eta_d<\frac1{2d^2}$, no state of $\rho_0$ of
spin zero other than these four has dimension in
$[E_d-\eta_d,E_d+\eta_d]$. Let
\[
 U_d=\bigl\{X\in\RR^2:\ |P^2-Q^2|<\tfrac12,\ \
 E_d-\eta_d<|X|^2<E_d+\eta_d\bigr\}.
\]
This set is bounded, open and invariant under the two reflections; its
boundary does not meet $\supp\rho_0$, because the points of
$\supp\rho_0$ have integral spin and those of spin zero have dimension
in $\frac1{2d^2}\ZZ$, which does not contain $E_d\pm\eta_d$;
and the states of $\rho_0$ supported in $U_d$ are the four states
above. A state of an admissible partition function has spin zero and
dimension in $(E_d-\eta_d,E_d+\eta_d)$ if and only if it is supported in
$U_d$.

Suppose that the lemma fails. Then there are $\eta>0$ and admissible
partition functions of gap $g$ at $c=1+12a_n$, $a_n\to0$, for each of
which the number of states supported in $U_d$ is not four, or is four
with $\delta_{a_n}\geq\eta$. By Theorem~\ref{thm:compactness} a
subsequence converges to some $\rho_0$, which is of the above form by
Theorem~\ref{thm:half-integer-limits}. By
Corollary~\ref{cor:cluster-stabilization}, along this subsequence the
number of states supported in $U_d$ is eventually four and their
dimensions converge to $E_d$, a contradiction.
\end{proof}

The kernel adapted to the grid is built from $\Omega_d$ alone,
\[
 \Omega_{d,\epsilon}(X)=e^{-2\pi\epsilon^2|X|^2}
 \bigl(\sin^2(\pi du)+\sin^2(\pi dv)\bigr).
\]

\begin{proposition}[The four-scalar budget]
\label{prop:four-scalar-budget}
Let $0<g\leq\frac12$. There are constants $C$, $\kappa>0$ and $a_0>0$,
depending only on $g$, such that for every admissible partition function
of gap $g$ at $c=1+12a$ with $0<a<a_0$, and every
$0<\epsilon\leq\frac12$ with $a\leq\frac14\epsilon^2$,
\begin{equation}\label{eq:ker-four-scalar-budget}
 \sum_je^{-2\pi\epsilon^2\Delta_j}
 \bigl(\sin^2(\pi du_j)+\sin^2(\pi dv_j)\bigr)
 \leq C\Bigl(\bigl(a+\delta_a^2\bigr)\epsilon^{-4}
 +e^{-\kappa/\epsilon^2}\Bigr),
\end{equation}
where $[u_j,v_j]$ is any physical point of the state $j$.
\end{proposition}
\begin{proof}
We apply Lemma~\ref{lem:peak-budget} to $\Phi=\Omega_d$, with
$\varrho=\min\{\frac14,\frac12\sqrt g\}$, $\varrho_*=d/\sqrt2$,
$c_*=2$ and $m_*=1$. By Lemma~\ref{lem:kernel-fourier}(iii) the centres
are the origin, with coefficient $1$, and the four points
$[\pm d,0]$, $[0,\pm d]$, with coefficient $-\frac14$. We take $m_i=1$
at all five centres, so that \eqref{eq:ker-mass-balance} reads
$1-4\cdot\frac14=0$. Condition (B0) holds because the points of $V$ have
integral light-cone coordinates, (B1) holds, and (B2) is empty. Let
$\xi=(\pm\frac d2,\pm\frac d2)$ be one of the negative centres. Its
coordinates have modulus $\frac d2\geq2\geq2\varrho$, the disc
$D(\xi,\varrho)$ contains no point of $V$, and $w_i=0$. For $a<a_1(g)$,
Lemma~\ref{lem:four-scalar-cluster} provides $n_i=4$ states of spin
$0=P_\xi^2-Q_\xi^2$ with $|\Delta_j-|\xi|^2|=|z_{i,a}|\leq\delta_a$, and
$w_i+\frac14n_i=1=m_i$. Thus (B3)(a) holds with
$\varepsilon=\delta_a$. By the uniform convergence in
Lemma~\ref{lem:four-scalar-cluster} there is $a_0\leq a_1$, depending
only on $g$, such that $\delta_a\leq\varrho^2$ and
$a\leq\min\{\frac14g^2,\frac14\varrho^2\}$ whenever $a<a_0$. Then
\eqref{eq:ker-budget-range} holds, and \eqref{eq:ker-peak-budget} is the
assertion, with $\kappa=\pi\varrho^2/4$ and a constant depending only on
$g$ and $d=d(g)$. The left side of \eqref{eq:ker-peak-budget} is the
left side of \eqref{eq:ker-four-scalar-budget} because
$\Omega_{d,\epsilon}$ is even.
\end{proof}

The constants in Proposition~\ref{prop:four-scalar-budget} do not depend
on a limiting measure: Lemma~\ref{lem:peak-budget} is applied to each
partition function separately, and the only input from compactness is
the threshold $a_0(g)$ of Lemma~\ref{lem:four-scalar-cluster}.

\begin{corollary}[Localization on the scale of the four scalars]
\label{cor:four-scalar-localization}
Let $0<g\leq\frac12$ and put
\[
 b_a=\sqrt a+\delta_a,\qquad H_a=\log(1/b_a).
\]
For every $M>0$ there are $C_M<\infty$ and $a_M>0$, depending only on
$g$ and $M$, such that for every admissible partition function of gap
$g$ with $0<a<a_M$ and every state with $\Delta_j\leq MH_a$, with
physical point $[u_j,v_j]$, there are unique integers $m,n$ with
\begin{equation}\label{eq:ker-four-scalar-association}
 \frac{mn}{d^2}=s_j,\qquad(m,n)\neq(0,0),\qquad
 \Bigl|u_j-\frac md\Bigr|+\Bigl|v_j-\frac nd\Bigr|\leq C_M\,b_aH_a .
\end{equation}
Moreover
$\bigl|\Delta_j-\frac{m^2+n^2}{2d^2}\bigr|\leq C_M\,b_aH_a^{3/2}$, and if
$|m|=|n|$, that is, if the associated point lies on a momentum axis, then
$0\leq\Delta_j-\frac{m^2+n^2}{2d^2}\leq C_M\,b_a^2H_a^2$.
\end{corollary}
\begin{proof}
By Lemma~\ref{lem:four-scalar-cluster}, $b_a\to0$ uniformly over the
class as $a\to0$. Take $\epsilon^2=\kappa/(2H_a)$ in
Proposition~\ref{prop:four-scalar-budget}; this is admissible for small
$a$ because $aH_a\leq b_a^2\log(1/b_a)\to0$. Then
$e^{-\kappa/\epsilon^2}=b_a^2$ and $a+\delta_a^2\leq b_a^2$, so the right
side of \eqref{eq:ker-four-scalar-budget} is at most $Cb_a^2H_a^2$. For a
state with $\Delta_j\leq MH_a$ the broad Gaussian is at least
$e^{-\pi\kappa M}$, and all terms are nonnegative; hence
$\sin^2(\pi du_j)+\sin^2(\pi dv_j)\leq C_Mb_a^2H_a^2$, and by
\eqref{eq:ker-sine-distance} there are integers $m,n$ with
$|u_j-\frac md|,|v_j-\frac nd|\leq c:=C_Mb_aH_a$. For small $a$ we have
$c<1/(2d)$, so $m,n$ are unique. Since
$|u_j|,|v_j|\leq\sqrt{2MH_a}$,
\[
 \Bigl|s_j-\frac{mn}{d^2}\Bigr|\leq c\bigl(|u_j|+|v_j|\bigr)+c^2
 \leq C_M\,b_aH_a^{3/2},
\]
which is smaller than $d^{-2}$ for small $a$; as both numbers lie in
$d^{-2}\ZZ$, they are equal. The same computation bounds
$|\Delta_j-\frac{m^2+n^2}{2d^2}|$. If $(m,n)=(0,0)$, then
$\Delta_j=\frac12(u_j^2+v_j^2)\leq c^2<g$, which is impossible. If
$|m|=|n|$, then $|s_j|=m^2/d^2=\frac{m^2+n^2}{2d^2}$ and
\[
 0\leq\Delta_j-|s_j|=\tfrac12\bigl(|u_j|-|v_j|\bigr)^2
 \leq\tfrac12\Bigl(\Bigl||u_j|-\frac{|m|}d\Bigr|
 +\Bigl||v_j|-\frac{|n|}d\Bigr|\Bigr)^2\leq2c^2 .
\]
All thresholds depend only on $g$ and $M$, because the convergence
$b_a\to0$ is uniform.
\end{proof}

\begin{theorem}[The four scalars move slowly]\label{thm:four-scalar-slow}
Let $0<g\leq\frac12$ and $p>0$. For every $k\in\NN$ there is
$a(p,k)>0$ such that
\[
 \delta_a\geq k\,a^p
\]
for every admissible partition function of gap $g$ at $c=1+12a$ with
$0<a<a(p,k)$. In other words, $\delta_a/a^p\to\infty$ as $a\to0$,
uniformly over the admissible partition functions of gap $g$.
\end{theorem}
\begin{proof}
It suffices to treat $p\leq\frac12$. If the statement fails, there are
$k\in\NN$ and an admissible sequence of gap $g$ with $a\to0$ and
$\delta_a<ka^p$. Apply Proposition~\ref{prop:four-scalar-budget} with
$\epsilon^2=\kappa/(2L)$, so that $e^{-\kappa/\epsilon^2}=a^2$; the
condition $a\leq\frac14\epsilon^2$ holds for small $a$. The right side
of \eqref{eq:ker-four-scalar-budget} is then at most
$C(a+k^2a^{2p})L^2+Ca^2\leq C'a^{2p}L^2$. For a state with
$\Delta_j\leq L$ the broad Gaussian is at least $e^{-\pi\kappa}$, so
$\sin^2(\pi du_j)+\sin^2(\pi dv_j)\leq Ca^{2p}L^2$. As in the proof of
Corollary~\ref{cor:four-scalar-localization}, for small $a$ there are
integers $m,n$ with
\[
 \Bigl|u_j-\frac md\Bigr|,\ \Bigl|v_j-\frac nd\Bigr|\leq C\,a^pL,\qquad
 \frac{mn}{d^2}=s_j,\qquad
 \Bigl|\Delta_j-\frac{m^2+n^2}{2d^2}\Bigr|\leq C\,a^pL^{3/2},
\]
where the exactness of the spin uses $|u_j|,|v_j|\leq\sqrt{2L}$ and
$a^pL^{3/2}\to0$. The point $X_j'=[m/d,n/d]$ has integral spin, so by
Lemma~\ref{lem:grid-circles} it lies in the fixed finite union
$\Sigma_d$ of \eqref{eq:ker-grid-circles}, which contains $\Lambda_1$
and is invariant under $[u,v]\mapsto[v,u]$. Thus every state with
$\Delta_j\leq u_*L$, $u_*=1>1/(2\pi)$, has a real point
$X_j'\in\Sigma_d$ with spin exactly $s_j$ and
$|\Delta_j-|X_j'|^2|\leq Ca^pL^{3/2}$, for all sufficiently small $a$ in
the sequence. Discarding finitely many terms leaves an admissible
sequence of gap $g$, and for it this contradicts
Theorem~\ref{thm:fixed-localization}.
\end{proof}

Theorem~\ref{thm:four-scalar-slow} gives no upper bound for $\delta_a$
by a power of $a$; it is used below only in the form
$\delta_a^{-1}\leq a^{-p}$ for every fixed $p>0$ and small $a$.