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\title{The periodic spin-one Haldane gap}
\author{OpenAI}
\date{September 24, 2026}
\begin{document}
\maketitle
\begin{abstract}
We prove a uniform positive spectral gap for the antiferromagnetic
spin-one Heisenberg chain on even periodic rings, in the
$\liminf$ (even-periodic) formulation of the spin-one Haldane
conjecture. The proof combines a spatial transfer operator, a
purity bootstrap that doubles length and inverse temperature
together, and two finite, exactly verified initial data sets.
\end{abstract}
\section{Introduction}\label{sec:introduction}
Haldane predicted that antiferromagnetic Heisenberg chains behave
differently for integer and half-odd-integer spin
\cite{Haldane1981,Haldane1983PRL,Haldane1983PLA}. For spin one the
prediction is a positive gap above the ground energy. We prove this
for the pure bilinear model on even periodic rings, with bounds that
are uniform in the ring length.
\begin{theorem}\label{thm:main}
Let $S^x,S^y,S^z\in M_3(\C)$ be self-adjoint spin-one matrices with
\[
[S^\alpha,S^\beta]=i\textstyle\sum_\eta\varepsilon_{\alpha\beta\eta}S^\eta,
\qquad \sum_\alpha(S^\alpha)^2=2I_3 .
\]
For even $L\ge4$ put $\cH_L=(\C^3)^{\otimes L}$, and on it
\begin{equation}\label{eq:main-H}
H_L=\sum_{j=0}^{L-1}\sum_{\alpha=x,y,z}S_j^\alpha S_{j+1}^\alpha,
\qquad S_L^\alpha=S_0^\alpha .
\end{equation}
Let $E_0(L)$ be the least energy, $E_1(L)$ the next distinct energy,
$\gamma_L=E_1(L)-E_0(L)$, and $\Delta_1=\liminf_{L\to\infty,
L\text{ even}}\gamma_L$. Then the ground state of $H_L$ is unique
for every even $L\ge60$, and
\begin{align}
\gamma_L&>\frac4{105}\log\frac{80}{79} && (L\ge60\text{ even}),
\label{eq:gap60}\\
\gamma_L&>\frac{\log 20}{784} && (L\ge2304\text{ even}),
\label{eq:gap2304}
\end{align}
so $\Delta_1\ge\log(20)/784>0$. The same conclusions hold for any
simultaneously similar spin-one realization satisfying the same
commutator and Casimir normalizations.
\end{theorem}
The next distinct energy always exists: in the $S^z$ basis the
all-zero and all-$+1$ product vectors have energy expectations $0$
and $L$, so $H_L$ is not scalar. Because $H_L$ is self-adjoint and
non-scalar in finite dimension, its full spectrum has a unique least
eigenvalue and a next distinct one; $\gamma_L$ is the difference.
\subsection*{How the proof works}
A shifted partition function $Z_L(b)=\Tr e^{-b(H_L+aLI)}$ can be
read in two ways at once. Physically it is a sum of positive Gibbs
weights, one per physical eigenvector. Spatially
(Section~\ref{sec:transfer}) it is a moment $\Tr(X_b^L)$ of a
self-adjoint, Hilbert--Schmidt operator $X_b$; for even $L$ this is a
sum of nonnegative weights, one per spatial eigenvalue. One object,
two probability distributions.
For a probability distribution $(p_i)$ call $\sum_i p_i^2$ its
\emph{purity}; purity near one means one weight carries almost all
the mass. Squaring and renormalizing the weights turns purity defect
$1-P$ into roughly its square (Lemma~\ref{lem:purity}). Squaring
physical weights doubles the inverse temperature, while squaring
spatial weights doubles the length. Exact cancellation identities
couple the two improvements (Lemma~\ref{lem:rounded-update}), so
starting from two finite purity bounds both defects decay doubly
exponentially as length and inverse temperature double together
(Proposition~\ref{prop:bootstrap}). Moment inequalities fill in
every intermediate even length. Finally, purity of the physical
Gibbs distribution bounds each excited-to-ground weight ratio, hence
the energy gap.
Two finite initializations supply the seeds. One begins at length
$60$ and inverse temperature $105/4$, the other at length $2304$ and
inverse temperature $784$ with a smaller defect, giving the stronger
bound. Both rest on short-chain thermal traces
(Appendices~\ref{sec:thermal120}--\ref{sec:thermal72}) and one
explicit family of periodic matrix-product trial vectors
(Appendix~\ref{sec:trials}). Every finite step is an exact integer or
rational computation with stated truncation errors; no floating-point
eigenvalue estimate is a premise.
\subsection*{Historical context}
The half-odd-integer side has a rigorous obstruction in the
low-energy bound of Lieb, Schultz and Mattis
\cite[Appendix~B, Theorem~2]{LiebSchultzMattis1961}, extended by
Affleck and Lieb \cite{AffleckLieb1986}; possible ground-state
degeneracy is part of that picture. Integer spin admits a unique
gapped state, but this distinction alone does not prove a gap for a
given interaction.
The Affleck--Kennedy--Lieb--Tasaki construction gave a rigorous
gapped spin-one chain with biquadratic interactions
\cite{AKLT1987,AKLT1988}. Its bond interaction is $h+\tfrac13h^2$ up
to a scalar, with $h=\mathbf S_j\cdot\mathbf S_{j+1}$, whereas
\eqref{eq:main-H} uses $h$ itself. Yarotsky proved that a unique
ground state and a uniform
periodic gap persist under sufficiently small translation-invariant
finite-range perturbations of the AKLT point
\cite[arXiv v1, Theorems~1 and~3]{Yarotsky2006}; this controls a
neighborhood, not a gapped path to the pure bilinear point.
For the pure chain, White and Huse's DMRG study estimated a bulk gap
of $0.41050(2)$ and effective spin-one-half end modes
\cite{WhiteHuse1993}; neutron scattering found experimental evidence
in the nearly one-dimensional spin-one antiferromagnet
$\mathrm{CsNiCl}_3$ \cite{Buyers1986}. Batista and Ortiz proposed a
spin--fermion proof \cite{BatistaOrtiz2001}; its gap step uses a
first-order perturbative reduction without a uniform remainder, so it
does not close the argument. Tasaki proved a nontrivial
infinite-chain topological index for the model under a gap assumption
for odd open chains with endpoint fields
\cite[arXiv v4, Assumption~2 and Theorem~3]{Tasaki2025}; his boundary
hypotheses require an argument separate from the even-periodic
estimates here.
Finite-volume gap criteria include those of Knabe and Nachtergaele
for frustration-free chains \cite{Knabe1988,Nachtergaele1996}, and the
sharper thresholds of Gosset and Mozgunov \cite{GossetMozgunov2016}.
Banerjee, Guo and Chowdhury use semidefinite methods to bound string
correlators
\cite[Eqs.~(2),~(13), and Figure~4]{BanerjeeGuoChowdhury2026}, a
different quantity from the uniform spectral gap studied here.
Our finite inputs are thermal traces of small twisted chains and
energies of periodic matrix-product vectors
\cite{FNW1992,PerezGarcia2007}. The traces are rigorous enclosures of
matrix-exponential traces; validated matrix exponentiation is
classical \cite{BochevMarkov1989,GoldsztejnNeumaier2014}, and the
conditioned Poisson polynomial uses the Fox--Glynn estimate
\cite[Proposition~1]{FoxGlynn1988}.
The transfer-matrix idea goes back to Suzuki \cite{Suzuki1985}; our
ordered bond expansion follows Aizenman and Nachtergaele
\cite[arXiv v1, Section~2.2]{AizenmanNachtergaele1994}, followed by a
direct operator construction on bond histories.
\section{The model and its twisted partition functions}\label{sec:model}
A single object will carry the whole proof: a shifted partition
function that is simultaneously a physical Gibbs trace and a spatial
moment. This section fixes coordinates, twists, and an elementary
energy bound used to control the finite exponential computations.
We use self-adjoint spin-one matrices throughout; the non-self-adjoint
case reduces to this by the next lemma.
\begin{lemma}[Representation]\label{lem:representation}
Let $S^x,S^y,S^z\in M_3(\C)$ satisfy
\[
[S^\alpha,S^\beta]=i\textstyle\sum_\gamma\varepsilon_{\alpha\beta\gamma}S^\gamma,
\qquad \sum_\alpha(S^\alpha)^2=2I .
\]
They are simultaneously similar to the standard self-adjoint spin-one
matrices; if they are already self-adjoint, the similarity is unitary.
The tensor-power similarity preserves the full spectrum of every
periodic $H_n$.
\end{lemma}
\begin{proof}
Put $S^\pm=S^x\pm iS^y$. The assumptions give
\begin{equation}\label{eq:ladder-identities}
[S^z,S^\pm]=\pm S^\pm,\qquad
S^-S^+=2I-(S^z)^2-S^z,\qquad S^+S^-=2I-(S^z)^2+S^z .
\end{equation}
Take an $S^z$-eigenvector $v$ of eigenvalue $m$ with maximal real part.
Then $S^+v=0$, so $m=1$ or $m=-2$. The case $m=-2$ would permit
indefinite lowering: a last nonzero lowered vector $w$ has weight
$-2-k$, while $S^+S^-w=0$ forces weight $-1$ or $2$, contradicting
finite dimension. Hence $m=1$.
The vectors $v$, $S^-v/\sqrt2$, $(S^-)^2v/2$ are nonzero by
\eqref{eq:ladder-identities} and have distinct weights $1,0,-1$; they
form a basis, and there is no weight $-2$, so they are the whole
standard spin-one module with ladder coefficients $\sqrt2$. When the
matrices are self-adjoint and $\norm v=1$ the vectors are orthonormal.
Conjugating each tensor factor conjugates every term of $H_n$, so the
spectrum is unchanged.
\end{proof}
In the weight basis $|s\rangle$, $s\in\{-1,0,1\}$, the generator is
$S^z|s\rangle=s|s\rangle$ with ladder entries $\sqrt2$. In the
Cartesian basis indexed by axes, the real matrices
\[
(S^\alpha)_{ij}=-i\varepsilon_{\alpha i j},\qquad G_\alpha=iS^\alpha
\]
obey the same relations; each $G_\alpha$ is real of operator norm one.
For $h=\sum_\alpha S^\alpha\otimes S^\alpha$ contraction of the epsilon
tensors gives
\begin{equation}\label{eq:cartesian-bond}
-h=\sum_\alpha G_\alpha\otimes G_\alpha,\qquad
h=F-|\Omega\rangle\langle\Omega|,\quad
\Omega=\sum_{i=1}^3|ii\rangle,
\end{equation}
where $F(u\otimes v)=v\otimes u$. Hence $h$ has eigenvalues
$-2,-1,1$ and $h\le I$.
\subsection{Twists and partition functions}
Number the sites $0,\dots,n-1$ with $n\ge4$. A proper rotation
$g\in SO(3)$ is unitary on $\C^3$. Define
\begin{equation}\label{eq:twisted-hamiltonian}
H_n(g)=\sum_{j=0}^{n-2}h_{j,j+1}+h_{n-1,0}(g),\qquad
h_{n-1,0}(g)=\sum_\alpha S_{n-1}^\alpha\,(gS^\alpha g^{-1})_0 ,
\end{equation}
so only the last bond is conjugated, on its site-$0$ factor; $H_n(g)$
is self-adjoint and $H_n(I)=H_n$. With the shift
\begin{equation}\label{eq:shift}
a=\frac{700741}{500000},\qquad
Z_n(b,g)=\Tr\exp[-b(H_n(g)+anI)],\qquad Z_n(b)=Z_n(b,I),
\end{equation}
all $Z_n(b,g)$ are strictly positive, and the scalar shift $an$ will
cancel from every purity ratio and from the final gap.
We need the diagonal rotations
\[
P_x=\operatorname{diag}(1,-1,-1),\quad
P_y=\operatorname{diag}(-1,1,-1),\quad
P_z=\operatorname{diag}(-1,-1,1),
\]
$D_2=\{I,P_x,P_y,P_z\}$, and the cyclic rotation
$C:e_x\mapsto e_y\mapsto e_z\mapsto e_x$; write $P=P_x$. By a global
cyclic rotation the three nonidentity elements of $D_2$ have equal
partition functions. After a global rotation and a change to the
weight basis, $P$ and $C$ act as $\exp(-i\theta S^z)$ with
$\theta=\pi$ and $\theta=2\pi/3$.
For later use with explicit matrices, a word $x=(x_0,\dots,x_{n-1})$
in the weight basis has diagonal entry $\sum_j x_{j-1}x_j$ (indices
mod $n$); each allowed step changing two neighboring letters by
opposite units contributes $1$, and a seam step from column $y$ to row
$x$ contributes the phase
\begin{equation}\label{eq:seam-phase}
\exp\!\bigl(i\theta(y_0-x_0)\bigr),
\end{equation}
since $gS^\pm g^{-1}=e^{\mp i\theta}S^\pm$. These formulas tie the
finite matrices below to the physical $H_n(g)$ in
\eqref{eq:twisted-hamiltonian}.
\subsection{A uniform energy bound}
\begin{lemma}[Energy bounds]\label{lem:energy}
For every $n\ge4$ and $g\in SO(3)$, $-\tfrac{3n}{2}I\le H_n(g)\le nI$.
\end{lemma}
\begin{proof}
The upper bound is $h\le I$ plus unitary conjugation of the seam. For
the lower bound consider $T=-h_{12}-h_{23}$ on three open sites in
Cartesian coordinates. By \eqref{eq:cartesian-bond} an unequal pair
swaps with coefficient $-1$, an equal pair changes to either other equal
pair with coefficient $1$, and there are no diagonal terms. Give each
three-letter word a positive weight $q$ according to its equality
pattern:
\begin{center}
\begin{tabular}{c|c|c}
word type & $q$ & absolute weighted row sum\\ \hline
$aaa$ & $4$ & $3+3+3+3=12$\\
$aab$ or $abb$ & $3$ & $4+3+2=9$\\
$aba$ & $2$ & $3+3=6$\\
$abc$ & $2$ & $2+2=4$
\end{tabular}
\end{center}
Every row sum is at most $3q$. With $D$ the diagonal of these weights,
$D^{-1}TD$ has absolute row sums at most three, hence spectral radius
at most three; since $T$ is self-adjoint and similar to this matrix,
$h_{12}+h_{23}\ge-3I$.
An open triad containing the seam is unitarily equivalent to an
untwisted triad (delete the seam bond and rotate all sites on one side,
using rotation invariance of the internal interaction), so the bound
holds for every consecutive pair of bonds of $H_n(g)$. Summing the $n$
triad inequalities counts each bond twice and gives
$H_n(g)\ge-\tfrac{3n}{2}I$.
\end{proof}
The trial inputs of Appendix~\ref{sec:trials} give $E_0(n)+an<0$ at
$n=72,120$, hence $Z_n(b)>1$ for every $b>0$; those lower bounds are
paired with finite thermal upper bounds in Section~\ref{sec:initialization}.
\section{Partition functions as spatial moments}\label{sec:transfer}
We now construct a self-adjoint spatial operator whose $L$th moment is
the physical partition function \eqref{eq:shift}. This is what turns
even spatial powers into positive weights. The operator itself need
not be positive.
Let $\mathcal G\subset SO(3)$ be the group of proper signed permutation
matrices (it contains $D_2$ and $C$). For $g\in\mathcal G$ write
\[
g e_\alpha=\epsilon_g(\alpha)e_{\sigma_g(\alpha)},
\qquad \epsilon_g(\alpha)\in\{-1,1\}.
\]
Rotation covariance of the Cartesian matrices gives
\begin{equation}\label{eq:rotation-covariance}
gG_\alpha g^{-1}=\epsilon_g(\alpha)G_{\sigma_g(\alpha)} .
\end{equation}
\begin{proposition}[Spatial transfer]\label{prop:transfer}
For each $b>0$ there are a Hilbert space $\mathcal K_b$, a compact
self-adjoint operator $X_b$, and a unitary representation
$g\mapsto U_g$ of $\mathcal G$ commuting with $X_b$, such that
\begin{equation}\label{eq:spatial-trace}
Z_n(b,g)=\Tr(X_b^nU_g)\qquad(n\ge4,\ g\in\mathcal G).
\end{equation}
Both $X_b$ and the representation commute with complex conjugation, and
$X_b$ is Hilbert--Schmidt, so $X_b^k$ is trace class for every integer
$k\ge2$. If $\lambda_i(b)$ are the nonzero eigenvalues, counted with
multiplicity, then
\begin{equation}\label{eq:spatial-moments}
\sum_i|\lambda_i(b)|^k<\infty\ (k\ge2),\qquad
Z_n(b)=\sum_i\lambda_i(b)^n\ (n\ge4).
\end{equation}
\end{proposition}
\begin{proof}
\emph{Histories and kernel.} A bond history is a finite ordered list
\[
x=((t_1,\alpha_1),\dots,(t_k,\alpha_k)),\quad
0<t_1<\dots<t_k<b,\ \alpha_j\in\{x,y,z\},
\]
including the empty list. Let $\Omega_b$ be the disjoint union of
these labeled simplexes, with Lebesgue measure on each and unit mass on
the empty history. The measure is finite:
\begin{equation}\label{eq:history-mass}
\mu(\Omega_b)=\sum_{k\ge0}\frac{3^kb^k}{k!}=e^{3b},
\qquad \mathcal K_b=L^2(\Omega_b,\mu;\C).
\end{equation}
For two histories $x,y$, merge their labeled times increasingly and let
$k_b(x,y)$ be the trace of the resulting time-ordered product of the
$G_\alpha$ (empty pair: the identity, trace three; time collisions: a
null event, set kernel zero). Since $\norm{G_\alpha}=1$ the kernel is
real with $|k_b(x,y)|\le3$, and exchanging $x,y$ keeps the merged
product unchanged, so $k_b(y,x)=k_b(x,y)$. The integral operator
$K_b$ is thus self-adjoint and Hilbert--Schmidt with
$\norm{K_b}_{\mathrm{HS}}^2\le9e^{6b}$. Set $X_b=e^{-ab}K_b$; compact
self-adjoint spectral theory gives real square-summable eigenvalues,
and boundedness gives $\sum_i|\lambda_i|^k<\infty$ for $k\ge2$.
\emph{Rotations.} The permutation $\sigma_g$ acts on labels preserving
times and measure. Put $s_g(x)=\prod_{\alpha\in x}\epsilon_g(\alpha)$
and $(U_g f)(x)=s_g(\sigma_g^{-1}x)f(\sigma_g^{-1}x)$. The cocycle rule
$s_{gh}(x)=s_g(\sigma_hx)s_h(x)$ shows these real unitaries form a
representation. Rotating every matrix inside a site trace, via
\eqref{eq:rotation-covariance}, gives
$k_b(\sigma_gx,\sigma_gy)=s_g(x)s_g(y)k_b(x,y)$, so $U_g$ commutes with
$K_b$, $X_b$; the kernel of $K_bU_g$ is
\begin{equation}\label{eq:twisted-kernel}
(K_bU_g)(x,y)=k_b(x,\sigma_gy)s_g(y).
\end{equation}
\emph{The cycle trace.} For $n\ge2$ both $K_b^{n-1}$ and $K_bU_g$ are
Hilbert--Schmidt (the iterated kernel of $K_b^{n-1}$ is bounded because
$k_b$ is bounded and $\mu$ finite). The Hilbert--Schmidt product trace
identity $\Tr(AB)=\int a(x,y)b(y,x)\,d\mu x\,d\mu y$ and
\eqref{eq:twisted-kernel} give
\begin{equation}\label{eq:cycle-integral}
\Tr(K_b^nU_g)=\int_{\Omega_b^n}\Bigl(\prod_{j=0}^{n-2}k_b(x_j,x_{j+1})\Bigr)
k_b(x_{n-1},\sigma_gx_0)s_g(x_0)\,d\mu^n ,
\end{equation}
with all iterated integrals absolutely integrable, so no diagonal
formula for $K_b$ is needed.
\emph{Identification.} Expand $\exp[-bH_n(g)]$ in its ordered-time
series. By \eqref{eq:cartesian-bond} each event picks a bond and one of
three labels and inserts the corresponding two $G$ factors; at the seam
the site-$0$ factor is rotated. At total degree $k$ there are $(3n)^k$
choices each of tensor norm at most one, so absolute convergence holds:
\begin{equation}\label{eq:history-absolute}
3^n\sum_{k\ge0}\frac{(3nb)^k}{k!}=3^ne^{3nb}.
\end{equation}
We may regroup by bond; interleavings of bond histories partition the
global time simplex up to null collisions, and at each site the tensor
trace is the kernel on the two incident histories in their original
chronological order. For an explicit seam check, call its history
$e_0$ and the others $e_1,\dots,e_{n-1}$ around the ring. The physical
integrand is
\[
s_g(e_0)k_b(\sigma_ge_0,e_1)k_b(e_1,e_2)\cdots k_b(e_{n-1},e_0),
\]
which equals the right side of \eqref{eq:cycle-integral} after setting
$x_0=e_0,x_1=e_{n-1},\dots,x_{n-1}=e_1$ (spatial enumeration reversed,
chronological products untouched). Thus
$\Tr e^{-bH_n(g)}=\Tr(K_b^nU_g)$, and multiplying by $e^{-abn}$ gives
\eqref{eq:spatial-trace}; the compact trace formula gives
\eqref{eq:spatial-moments}.
\end{proof}
For even $n\ge4$, \eqref{eq:spatial-trace} supplies the probabilities
\begin{equation}\label{eq:spatial-probabilities}
p_i=\frac{|\lambda_i(b)|^n}{Z_n(b)},\qquad
\sum_i p_i=1,\qquad
\sum_i p_i^2=\frac{Z_{2n}(b)}{Z_n(b)^2}.
\end{equation}
Odd moments are also useful for polynomial estimates, but evenness is
essential here. The spatial eigenvalues are distinct from the physical
energies of $H_n$.
\subsection{Rotation sectors}
Small twisted partition functions split the spatial spectrum into a few
lists with known multiplicities; this later isolates a dominant weight.
\begin{lemma}[Sector moments]\label{lem:sectors}
Fix $b>0$ and let $X_b,U_g$ be as in Proposition~\ref{prop:transfer}.
There are real eigenvalue lists (with multiplicity) whose integer
moments $I_k,O_k,N_k$ converge absolutely for $k\ge2$, with
\begin{align}
Z_n(b)&=I_n+2O_n+3N_n,\label{eq:sector-total}\\
Z_n(b,P)&=I_n+2O_n-N_n,\label{eq:sector-pi}\\
Z_n(b,C)&=I_n-O_n\qquad(n\ge4).\label{eq:sector-cyclic}
\end{align}
The $I$ list is the restriction to the joint invariant subspace of
$D_2$ and $C$; the $O$ list occurs twice and the $N$ list three times
in the full spatial list. All three moments are nonnegative at even
order, and $J_n=I_n+2O_n$ is the moment on the $D_2$-invariant subspace.
\end{lemma}
\begin{proof}
The four $D_2$ characters define orthogonal projections
$\Pi_\chi=\frac14\sum_{g\in D_2}\chi(g)U_g$ commuting with $X_b$ and
summing to the identity. Conjugation by $U_C$ permutes the three
nontrivial character spaces cyclically, so their restrictions are
unitarily equivalent; write $N_k$ for the common moment. In the
trivial-character space, split by the eigenvalues $1,\omega,\bar\omega$
of $U_C$ ($\omega=e^{2\pi i/3}$); these reduce $X_b$, complex
conjugation exchanges the last two spaces and preserves each real
eigenvalue of $X_b$ with multiplicity, so their common moment is $O_k$
and the first space gives $I_k$. Absolute convergence is
\eqref{eq:spatial-moments}.
The decomposition gives \eqref{eq:sector-total}. On the three
nontrivial spaces a fixed $U_P$ has signs $+1,-1,-1$ and is the identity
on the trivial space, giving \eqref{eq:sector-pi}. The operator $U_C$
permutes the nontrivial spaces (zero contribution) and has coefficients
$1,\omega,\bar\omega$ on the remaining three with
$\omega+\bar\omega=-1$, giving \eqref{eq:sector-cyclic}. The $J_n$
identity is the decomposition of the trivial space, with nonnegativity
at even order.
\end{proof}
The invariant $I$ space need not be one-dimensional; the point is that
every weight outside it occurs at least twice in the full list, which
identifies a unique largest weight without discarding any sector.
\section{From two purities to a uniform gap}\label{sec:bootstrap}
The spatial and physical descriptions of the \emph{same} partition
function give two ways to concentrate its weights. This section turns
two initial concentration estimates into a gap at every larger even
length; the initial estimates are supplied in
Section~\ref{sec:initialization}.
By Proposition~\ref{prop:transfer}, each partition function is both a
physical Gibbs trace (counting every physical eigenvector with
multiplicity) and, at even lengths, a nonnegative spatial moment
(counting absolute powers of every spatial eigenvalue). Define, for
even $n\ge4$,
\begin{equation}\label{eq:purities}
S(n,b)=\frac{Z_{2n}(b)}{Z_n(b)^2},\qquad
T(n,b)=\frac{Z_n(2b)}{Z_n(b)^2}.
\end{equation}
The shift $aL$ cancels from both ratios. First, the inequality behind
their improvement under squaring.
\begin{lemma}[Squaring probabilities]\label{lem:purity}
Let $(w_i)$ be a finite or countable family of nonnegative numbers
with $\sum_iw_i=1$ and purity $P=\sum_iw_i^2\in(0,1]$. The normalized
squares $w_i^2/P$ have purity $P'$ with
\begin{equation}\label{eq:purity-square}
1-P'\le f(1-P),\qquad f(u)=\frac{u^2}{2(1-u)^2}\ \ (0\le u<1),
\end{equation}
and $f$ is increasing. Consequently
\begin{equation}\label{eq:two-squarings}
0<S(n,b),T(n,b)\le1,\quad
1-S(2n,b)\le f(1-S(n,b)),\quad
1-T(n,2b)\le f(1-T(n,b)).
\end{equation}
\end{lemma}
\begin{proof}
Some weight is positive, so $P>0$, and $w_i^2\le w_i$ gives $P\le1$.
Nonnegativity allows rearrangement of all countable sums, so
\[
1-P'=\frac{2\sum_{i<j}w_i^2w_j^2}{P^2}
\le\frac{2\bigl(\sum_{i<j}w_iw_j\bigr)^2}{P^2}
=\frac{(1-P)^2}{2P^2},
\]
and $f'(u)=u/(1-u)^3\ge0$. For $S(n,b)$ use the spatial
probabilities $|\lambda_i(b)|^n/Z_n(b)$ of
\eqref{eq:spatial-probabilities}: squaring them changes $n$ to $2n$.
For $T(n,b)$ list all physical eigenvalues $\varepsilon_k(n)$ with
multiplicity and use
\begin{equation}\label{eq:gibbs-weights}
w_k(n,b)=\frac{e^{-b(\varepsilon_k(n)+an)}}{Z_n(b)} .
\end{equation}
Their purity is $T(n,b)$, and squaring them changes $b$ to $2b$.
\end{proof}
\subsection{The coupled update}
The two purities constrain one another because they share partition
functions; direct cancellation gives
\begin{align}
S(n,2b)&=S(n,b)^2\frac{T(2n,b)}{T(n,b)^2},
\label{eq:space-time-cancel}\\
T(4n,b)&=T(2n,b)^2\frac{S(2n,2b)}{S(2n,b)^2}.
\label{eq:time-space-cancel}
\end{align}
Both denominators are at most one, and the following form lets the two
initial bounds differ and be rounded upward.
\begin{lemma}[Coupled update]\label{lem:rounded-update}
Suppose $n\ge4$ even, $b>0$, and $p,q\in[0,1)$ with
$1-S(n,b)\le p$, $1-T(2n,b)\le q$. If $p_+,q_+\in[0,1)$ obey
\begin{equation}\label{eq:rounded-update}
p_+\ge f\bigl(1-(1-p)^2(1-q)\bigr),\qquad
q_+\ge f\bigl(1-(1-q)^2(1-p_+)\bigr),
\end{equation}
then $1-S(2n,2b)\le p_+$ and $1-T(4n,2b)\le q_+$.
\end{lemma}
\begin{proof}
Equation \eqref{eq:space-time-cancel} gives
$S(n,2b)\ge(1-p)^2(1-q)$; spatial squaring in Lemma~\ref{lem:purity}
and monotonicity of $f$ prove the first bound. Next
\eqref{eq:time-space-cancel} gives $T(4n,b)\ge(1-q)^2(1-p_+)$, and
temporal squaring proves the second. Every input of $f$ lies in $[0,1)$
because the product subtracted from one is positive and at most one.
\end{proof}
In words: round the first output up, feed it into the second, round
that up. The pair based at $(n,b)$ becomes a pair based at $(2n,2b)$
whose physical member sits at length $4n$; neither bound has to
decrease.
\subsection{Quadratic decay and intermediate lengths}
Once both defects are small one common bound suffices. Set
\begin{equation}\label{eq:bootstrap-coefficient}
G(u)=\frac12\Bigl((1-u)^{-1}+(1-u)^{-2}+(1-u)^{-3}\Bigr)^2,
\end{equation}
increasing on $[0,1)$, with
\begin{equation}\label{eq:bootstrap-factorization}
f\bigl(1-(1-u)^3\bigr)=G(u)u^2 .
\end{equation}
\begin{proposition}[Uniform gap from two initial purities]
\label{prop:bootstrap}
For every even $L\ge4$ let $H_L$ be a non-scalar finite-dimensional
self-adjoint Hamiltonian, and suppose that $a\in\R$ and, for each
$b>0$, a real finite or countable list $(\lambda_i(b))$ satisfy
\[
\sum_i|\lambda_i(b)|^4<\infty,\qquad
Z_L(b):=\Tr e^{-b(H_L+aLI)}=\sum_i|\lambda_i(b)|^L
\quad(L\ge4\text{ even}).
\]
Define $S,T$ by \eqref{eq:purities} and $G$ by
\eqref{eq:bootstrap-coefficient}. Let $n_0\ge4$ even, $b_0>0$, and let
$u_0,C$ satisfy
\begin{equation}\label{eq:bootstrap-hypotheses}
0<u_0<\tfrac16,\quad G(u_0)\le C,\quad r:=Cu_0<1,\quad C(1-3u_0)>3,
\end{equation}
with seeds
\begin{equation}\label{eq:bootstrap-seeds}
S(n_0,b_0)\ge1-u_0,\qquad T(2n_0,b_0)\ge1-u_0 .
\end{equation}
For $j\ge0$ set
\begin{equation}\label{eq:dyadic-parameters}
n_j=2^jn_0,\quad b_j=2^jb_0,\quad u_j=C^{-1}r^{2^j}.
\end{equation}
Then
\begin{equation}\label{eq:dyadic-purities}
S(n_j,b_j)\ge1-u_j,\qquad T(2n_j,b_j)\ge1-u_j,
\end{equation}
and for every even $L\ge n_0$ the physical ground state is unique with
gap
\begin{equation}\label{eq:bootstrap-gap}
\gamma_L>\frac{\log(1/r)}{b_0}.
\end{equation}
\end{proposition}
\begin{proof}
\emph{Doubling.} By definition $u_{j+1}=Cu_j^2$, $0<u_{j+1}<u_j\le
u_0$. Suppose both defects at $(n_j,b_j)$ are at most $u=u_j$. The
first bound in Lemma~\ref{lem:rounded-update} is at most
\[
f\bigl(1-(1-u)^3\bigr)=G(u)u^2\le Cu^2=u_{j+1}\le u .
\]
Use $p_+=u_{j+1}$ in the second update; its input is at most
$1-(1-u)^3$, so its output is also at most $Cu^2=u_{j+1}$. Induction
gives \eqref{eq:dyadic-purities}.
\emph{Intermediate even lengths.} For a nonnegative list $(a_i)$ with
finite positive $p$th-power sum and $q\ge p>0$, normalization gives
$\sum a_i^q\le(\sum a_i^p)^{q/p}$ (the normalized $p$th powers are
probabilities, whose $q/p$ powers sum to at most one). Applied at the
two temperatures, whenever $n\le L\le2n$ with both even,
\[
Z_L(b)\le Z_n(b)^{L/n},\qquad
Z_L(2b)\ge Z_{2n}(2b)^{L/(2n)},
\]
which combine to
\begin{equation}\label{eq:purity-interpolation}
T(L,b)\ge\bigl[T(2n,b)S(n,b)^2\bigr]^{L/(2n)} .
\end{equation}
The bracket lies in $(0,1]$ and the exponent in $[1/2,1]$. For
$n=n_j$, $b=b_j$:
\begin{equation}\label{eq:intermediate-purity}
T(L,b_j)\ge(1-u_j)^3\ge1-3u_j>\tfrac12
\qquad(n_j\le L\le2n_j,\ L\text{ even}).
\end{equation}
\emph{The physical spectrum.} The largest weight in
\eqref{eq:gibbs-weights} is at least its purity, since
$\sum_kw_k^2\le(\max_kw_k)\sum_kw_k$, hence $>1/2$. These weights count
all physical eigenvectors, so two ground-state eigenvectors would give
two equal maximal weights, impossible: the ground state is unique.
Since $H_L$ is non-scalar and self-adjoint in finite dimension, a next
distinct energy exists. Let $w_0$ be the ground weight and $w_1$ the
weight of an eigenvector at that next energy. Then
\begin{align}
e^{-b_j\gamma_L}
=\frac{w_1}{w_0}
\le\frac{1-w_0}{w_0}
\le\frac{3u_j}{1-3u_j}
\le\frac{3}{C(1-3u_0)}r^{2^j}
<r^{2^j},
\label{eq:gap-ratio}
\end{align}
using the shift cancellation and \eqref{eq:bootstrap-hypotheses}.
Taking logarithms and $b_j=2^jb_0$ gives \eqref{eq:bootstrap-gap} on
$n_j\le L\le2n_j$; the dyadic intervals cover every even $L\ge n_0$.
\end{proof}
\subsection{The two parameter choices}
We will prove the seeds \eqref{eq:bootstrap-seeds} for
\begin{equation}\label{eq:bootstrap-parameters}
\begin{array}{c|c|c|c|c}
n_0&b_0&u_0&C&r=Cu_0\\ \hline
60&105/4&1/8&79/10&79/80\\
2304&784&1/100&5&1/20
\end{array}
\end{equation}
The scalar hypotheses are checked directly. For the first row,
$G(1/8)=913952/117649<79/10$ and
$3/[C(1-3u_0)]=48/79<1$. For the second, $1-(1-u)^3\le3u$ gives for
$0<u\le1/100$
\[
G(u)\le\frac9{2(1-3u)^2}\le\frac{45000}{9409}<5,\qquad
\frac3{C(1-3u_0)}=\frac{60}{97}<1 .
\]
So the two rows give the bounds $(4/105)\log(80/79)$ and $(\log20)/784$
of \eqref{eq:bootstrap-gap}, once their seeds hold. The second row will
come from six applications of Lemma~\ref{lem:rounded-update} to bounds
based at $(36,49/4)$. Everything left is finite: establish the two seed
inequalities without losing the margins these updates need.
\section{Two finite initializations}\label{sec:initialization}
The bootstrap needs a spatial purity at length $n$ and a physical
purity at length $2n$, at the same inverse temperature. We build two
such seed pairs. The first starts at length $60$; the second reaches
length $2304$ and gives the stronger eventual gap. Each seed rests on
a few short-chain partition functions and one periodic trial vector.
\subsection{Thermal data and moment bounds}
Write $J_n=I_n+2O_n$ for the $D_2$-invariant moment, $I_n$ for the
moment also fixed by $C$, and $N_n$ for one nontrivial $D_2$ sector
(Lemma~\ref{lem:sectors}). Since each $O_n$ occurs twice and each
$N_n$ three times,
\begin{equation}\label{eq:sector-inversion}
\begin{aligned}
J_n&=\frac{Z_n+3Z_n(P)}4,& N_n&=\frac{Z_n-Z_n(P)}4,\\
I_n&=\frac{Z_n+3Z_n(P)+8Z_n(C)}{12},&
O_n&=\frac{Z_n+3Z_n(P)-4Z_n(C)}{12},
\end{aligned}
\end{equation}
all at one common inverse temperature.
\begin{lemma}[Finite thermal inputs]\label{lem:thermal-inputs}
For $a=700741/500000$ the tables below give centers of the shifted
$Z_n(b,g)$ in units $10^{-6}$, with absolute error below $10^{-5}$ at
$b=21/2$ and below $30\cdot10^{-6}$ at $b=49/4$:
\[
\begin{array}{c|rrr}
&\multicolumn{3}{c}{b=21/2}\\
n&g=1&g=P&g=C\\ \hline
6&8945303&346734&939987\\
8&3741281&567167&984108\\
10&2327547&721090&1002613
\end{array}
\qquad
\begin{array}{c|rr}
&\multicolumn{2}{c}{b=49/4}\\
n&g=1&g=P\\ \hline
4&124885379&65757\\
5&2925&2960742\\
6&12870614&286599\\
7&54069&1920414\\
8&4639283&510140\\
9&195455&1514846\\
10&2654823&675419\\
11&377629&1314544\\
12&1900547&787405
\end{array}
\]
\end{lemma}
Appendices~\ref{sec:thermal120} and~\ref{sec:thermal72} prove these
tables by rational enclosures of the physical matrix exponentials with
explicit error bounds; they are rigorous, not numerical estimates. The
other finite input is Lemma~\ref{lem:trial-inputs}:
\begin{equation}\label{eq:trial-thermal-consequence}
E_0(72)<-72a,\qquad E_0(120)<-120a,
\end{equation}
so $Z_{72}(b)>1$ and $Z_{120}(b)>1$ for every $b>0$ (one Boltzmann
factor already exceeds one).
Two elementary lemmas convert the tables into spectral bounds.
\begin{lemma}[Capped masses]\label{lem:cappedmass}
Let $(t_i)$ be finite or countable with $0\le t_i\le h$ and
$\sum_it_i\le m$, where $h>0, m\ge0$. For $r\ge1$ put
\begin{equation}\label{eq:capped-function}
\mathcal C(m,h,r)=qh^r+(m-qh)^r,\qquad q=\lfloor m/h\rfloor.
\end{equation}
Then $\sum_it_i^r\le\mathcal C(m,h,r)$; without the cap,
$\sum_it_i^r\le m^r$.
\end{lemma}
\begin{proof}
For a finite family at fixed total mass, move mass from the smaller of
two coordinates in $(0,h)$ to the larger until one reaches $0$ or $h$;
convexity of $x\mapsto x^r$ shows the sum of $r$th powers does not
decrease. Repetition leaves at most one coordinate in $(0,h)$, giving
\eqref{eq:capped-function} at the actual mass; the right side is
continuous and nondecreasing in the mass, so bounded by its value at
$m$. Take increasing unions of finite subfamilies for the countable
case. The uncapped bound follows from $t_i^r\le t_i(\sum_jt_j)^{r-1}$.
\end{proof}
\begin{lemma}[Polynomial filter]\label{lem:polynomial-filter}
Let $\epsilon\ge0$, $v>0$, and let $(\lambda_i)$ be a real eigenvalue
list with moments $M_j=\sum_i\lambda_i^j$ converging absolutely, and
$|M_j-c_j|\le\epsilon$ for the exponents occurring in a polynomial
$F(x)=\sum_jf_jx^j\ge0$ on $\R$. If
\[
B=\sum_jf_jc_j+\epsilon\sum_j|f_j|
\quad\text{and}\quad F(x)>B\ \text{for }|x|\ge v,
\]
then every $|\lambda_i|<v$.
\end{lemma}
\begin{proof}
Absolute convergence gives $\sum_iF(\lambda_i)\le B$; nonnegativity
means a single $|\lambda_i|\ge v$ would contradict this.
\end{proof}
All moments converge by Proposition~\ref{prop:transfer}, and the
coefficient sums in \eqref{eq:sector-inversion} are at most one in
absolute value, so the thermal error bound also bounds each sector
moment error (odd moments included, with signs kept in the filter).
\subsection{The initialization at length sixty}
\begin{proposition}\label{prop:seed60}
At inverse temperature $b=105/4$, $S(60,b)>7/8$ and $T(120,b)>7/8$.
\end{proposition}
\begin{proof}
Work at $b_0=21/2$. From the $b_0=21/2$ table and
\eqref{eq:sector-inversion} the tenth moments obey
\[
m_I=\frac{1042655}{10^6},\quad m_O=\frac{40041}{10^6},\quad
m_N=\frac{401625}{10^6},
\]
with margins $19/(12\cdot10^6)$, $7/(12\cdot10^6)$,
$3/(4\cdot10^6)$ over centers plus $10^{-5}$. Apply
Lemma~\ref{lem:polynomial-filter} to
\[
F_I(x)=x^6(-2+10x^2)^2,\qquad F_N(x)=x^6(-25+100x^2)^2 .
\]
For a filter $x^6(c_0+c_1x^2)^2$ the summed upper bound is the centered
combination of moments $6,8,10$ with coefficients $c_0^2,2c_0c_1,c_1^2$
plus $10^{-5}(|c_0|+|c_1|)^2$. Rational evaluation gives
\[
|\lambda|<U:=\frac{100245}{100000}\ (I\text{ sector}),\qquad
|\lambda|<V:=\frac{88643}{100000}\ (N\text{ sector}),
\]
since the filter at the endpoint exceeds its bound by more than
$0.0301$ ($I$) and $0.3463$ ($N$), and each filter increases with $|x|$
beyond its endpoint ($c_1>0$, $c_0+c_1x^2>0$ there).
The inequalities $U^{10}<m_I<2U^{10}$ and $V^{10}<m_N<2V^{10}$ allow the
two-piece \eqref{eq:capped-function}. For $k=30,120$ set
\[
A_k=U^k+(m_I-U^{10})^{k/10},\quad B_k=m_O^{k/10},\quad
C_k=V^k+(m_N-V^{10})^{k/10},
\]
bounding $I_k,O_k,N_k$. These are even moments, hence nonnegative, so
\begin{equation}\label{eq:short-moment-upper}
Z_k(b_0)\le A_k+2B_k+3C_k,\qquad
Z_k(b_0,P),Z_k(b_0,C)\le A_k+2B_k .
\end{equation}
Put $m=A_{30}+2B_{30}+3C_{30}$, $p=A_{30}+2B_{30}$,
$q=A_{120}+2B_{120}+3C_{120}$, and
\[
D=\frac{1450}{1000},\quad E=\frac{1202}{1000},\quad
\tau=\frac{1225}{1000},\quad\tau_0=\frac{1264}{1000},\quad
R_q=\frac{684}{10000}.
\]
The scalar estimates are
\begin{equation}\label{eq:seed60-scalars}
m^5<D^2,\quad p^5<E^2,\quad q\ge1,\quad (q-1)^5<R_q^2,\quad
\frac{D+11E}{12}\le\tau,\quad \frac{D+3E}{4}\le\tau_0,
\end{equation}
with positive margins exceeding $0.001068$, $0.000433$,
$0.0000472$ in the first, second and fourth strict inequalities; all
numbers are rational and stated above.
Raise the inverse temperature to $b=(5/2)b_0$. For positive Boltzmann
factors $z_i$, $\sum_i z_i^{5/2}\le(\sum_iz_i)^{5/2}$, so
\eqref{eq:short-moment-upper}--\eqref{eq:seed60-scalars} give
\[
Z_{30}(b)\le D,\qquad Z_{30}(b,P),Z_{30}(b,C)\le E ;
\]
by \eqref{eq:sector-inversion} the $I$ and $J$ portions of the thirtieth
spatial moment are at most $\tau$ and $\tau_0$.
Let $w_i=|\lambda_i(X_b)|^{30}$, so $\sum_iw_i\le D$, while
\eqref{eq:trial-thermal-consequence} gives $\sum_iw_i^4=Z_{120}(b)>1$.
Since $D(0.88)^3<1$, some $w_0>0.88$ exists, and it is unique because
$2(0.88)>D$. It lies in $I$: an $O$ weight occurs at least twice and an
$N$ weight at least three times (counting multiplicities, including
equal moduli of opposite-sign spatial eigenvalues).
To sharpen $w_0$, suppose $0\le t\le0.998$ and $w_0\ge t$. Let $x$ be
the remaining $I$ mass, $y$ the full $O$-pair mass, $z$ the full $N$-triple
mass; their squared weights sum to at most $x^2+y^2/2+z^2/3$ with
\[
0\le x\le\tau-t,\quad 0\le y,\quad
x+y\le\tau_0-t,\quad z\le D-t-x-y .
\]
The divisors use identical repeated lists. Since $D>\tau_0$ the last
bound is nonnegative on the whole polygon, and the convex quadratic
$x^2+y^2/2+(D-t-x-y)^2/3$ attains its maximum at a vertex:
\begin{equation}\label{eq:residual-R}
R(t)=\max_{(x,y)\in\mathcal V_t}
\Bigl(x^2+\tfrac{y^2}{2}+\tfrac{(D-t-x-y)^2}{3}\Bigr),\quad
\mathcal V_t=\{(0,0),(\tau-t,0),(\tau-t,\tau_0-\tau),(0,\tau_0-t)\}.
\end{equation}
Thus the remaining squared weights sum to at most $R(t)$, their fourth
powers to at most $R(t)^2$. Exact evaluation gives
\[
\begin{array}{c|ccc}
t&0.88&0.99&0.998\\ \hline
R(t)&0.1359&0.0721&0.068404
\end{array}
\]
and
\[
1-0.99^4-R(0.88)^2=0.02093518,\qquad
1-0.998^4-R(0.99)^2=0.002777621984.
\]
If $w_0\le0.99$ the first would contradict $\sum_iw_i^4>1$, and the
second then excludes $w_0\le0.998$. Keeping the dominant contribution
in the fourth moment,
\[
S(60,b)=\frac{\sum_iw_i^4}{(\sum_iw_i^2)^2}
\ge\Bigl(1+\frac{R(0.998)}{0.998^2}\Bigr)^{-2}>0.8756>\frac78 .
\]
For the physical purity at length $120$, at temperature $b_0$: the
largest Boltzmann factor $z_0$ exceeds one (trial bound) while the sum
of all factors is at most $q$. After raising to the power $5/2$ the
ratio of remaining mass to $z_0^{5/2}$ is at most $(q-1)^{5/2}<R_q$,
so
\[
T(120,b)\ge(1+R_q)^{-2}>0.8760>\frac78 .
\]
This last step concerns physical Boltzmann factors, separate from the
spatial weights used for $S$.
\end{proof}
\subsection{The initialization giving the stronger gap}
\begin{proposition}\label{prop:seed2304}
At inverse temperature $b=784$,
\[
1-S(2304,b)\le\frac{84513}{10^7}<\frac1{100},\qquad
1-T(4608,b)\le\frac{27493}{5\cdot10^6}<\frac1{100}.
\]
\end{proposition}
\begin{proof}
We isolate one large $J$ eigenvalue, bound the rest, get seeds at
length $36$, then apply six rounded updates to reach $2304$.
Start at $b_0=49/4$; the $b_0=49/4$ table gives centers for $J_n,N_n$
via \eqref{eq:sector-inversion} with error at most $\epsilon=30\cdot10^{-6}$.
Use
\[
F_J(x)=x^4(7+17x-280x^2-185x^3+1000x^4)^2,\quad
F_N(x)=x^4(12-394x^2+1000x^4)^2,
\]
of degree $12$, so the table supplies every needed moment. The summed
upper bounds of Lemma~\ref{lem:polynomial-filter} are
\[
B_J=318604.54848175,\qquad B_N=48169.880699 .
\]
Set $v_J=100139/100000$, $v_N=872/1000$. Direct comparison gives
\[
F_J(v_J)-B_J>180,\quad F_J(-v_J)-B_J>496888,\quad
F_N(\pm v_N)-B_N>654 .
\]
To cover the whole exterior rays, write $Q(x)=7+17x-280x^2-185x^3+1000x^4$;
for $y\ge0$,
\[
\begin{aligned}
Q(1+y)&=559+2902y+5165y^2+3815y^3+1000y^4,\\
Q(-1-y)&=895+3978y+6275y^2+4185y^3+1000y^4.
\end{aligned}
\]
both positive and increasing. Likewise $12-394r^2+1000r^4$ is positive
and increasing for $r\ge v_N$ (value at $v_N$ positive, derivative
$4r(1000r^2-197)>0$). Hence every $J$ eigenvalue has modulus below
$v_J$ and every $N$ eigenvalue below $v_N$.
The twelfth-moment upper bounds are $m_J=1.0657205$, $m_N=0.2783155$.
Put $\ell=999/1000$ and, for $k=36,72$,
\begin{equation}\label{eq:seed72-tail}
B(k)=3\mathcal C(m_N,v_N^{12},k/12),\quad
R_A(k)=B(k)+(m_J-\ell^{12})^{k/12}
\end{equation}
(with $m_J>\ell^{12}$). If all $J$ eigenvalues had modulus at most
$\ell$, Lemma~\ref{lem:cappedmass} would give
$Z_{72}(b_0)\le\mathcal C(m_J,\ell^{12},6)+B(72)<1$, with strict margin
above $0.0693$; the trial inequality contradicts this. So some
$\lambda_0$ in $J$ has $|\lambda_0|>\ell$; removing it leaves twelfth
moment at most $m_J-\ell^{12}$ in $J$, and with the three $N$ sectors
$R_A(k)$ bounds the remaining full $k$th moment. Thus
\begin{equation}\label{eq:seed72-pq}
\begin{aligned}
1-S(36,b_0)&\le p_0:=1-\Bigl(1+\frac{R_A(36)}{\ell^{36}}\Bigr)^{-2},\\
1-T(72,b_0)&\le q_0:=1-\bigl(v_J^{72}+R_A(72)\bigr)^{-2}.
\end{aligned}
\end{equation}
The first retains $\lambda_0^{72}$; the second uses $Z_{72}(2b_0)>1$ and
$Z_{72}(b_0)\le v_J^{72}+R_A(72)$. Numerically $p_0<0.047916$,
$q_0<0.181521$, but the updates use their exact rational definitions.
Apply Lemma~\ref{lem:rounded-update} six times, rounding each new defect
up to a multiple of $10^{-7}$ and using the rounded spatial defect in
the physical update. With $f(u)=u^2/[2(1-u)^2]$ and
$\lceil x\rceil_7=10^{-7}\lceil10^7x\rceil$:
\[
p_{j+1}=\bigl\lceil f\bigl(1-(1-p_j)^2(1-q_j)\bigr)\bigr\rceil_7,
\quad
q_{j+1}=\bigl\lceil f\bigl(1-(1-q_j)^2(1-p_{j+1})\bigr)\bigr\rceil_7 .
\]
Exact rational evaluation gives
\[
\begin{array}{c|cc}
j&p_j&q_j\\ \hline
1&151249/2500000&433453/2500000\\
2&343303/5000000&1632381/10000000\\
3&44601/625000&361773/2500000\\
4&632939/10000000&1055161/10000000\\
5&375793/10000000&445941/10000000\\
6&84513/10000000&27493/5000000
\end{array}
\]
Each update lies in $[0,1)$, where $f$ increases, so upward rounding
preserves the bounds. Length and inverse temperature have multiplied by
$2^6$: $36\cdot64=2304$ and $(49/4)\cdot64=784$. The last row proves
the proposition.
\end{proof}
\subsection{A periodic interface for boundary estimates}
\begin{corollary}[Periodic inputs for boundary estimates]
\label{cor:boundary-inputs}
Let $X_b$ be the transfer operator of Proposition~\ref{prop:transfer}
with $a=700741/500000$, and put
\[
\begin{gathered}
b_0=49/4,\quad b_1=49/2,\quad \ell=999/1000,\\
U=100139/100000,\quad V=872/1000,\quad m_J=1.0657205,\quad m_N=.2783155 .
\end{gathered}
\]
At $b=b_0$ the full spatial list is the $D_2$-invariant list $J$ plus
three identical lists $N$. Every $J$ eigenvalue has modulus below $U$
and the sum of its twelfth powers is at most $m_J$; every $N$ eigenvalue
has modulus below $V$ and each $N$ list's twelfth powers sum to at most
$m_N$; some $J$ eigenvalue has modulus above $\ell$. Moreover
$1-S(72,b_1)\le151249/2500000$. For $n=72,120$, $E_0(n)+na<0$, so a
physical ground Boltzmann factor of $Z_n(b)$ exceeds one for every
$b>0$.
\end{corollary}
\begin{proof}
The decomposition is Lemma~\ref{lem:sectors} with $J=I+2O$; the filters,
twelfth-moment bounds and dominant eigenvalue are proved at the start of
Proposition~\ref{prop:seed2304}. The first row of the rounded update
table gives the display at $(72,49/2)$; the variational assertion is
Lemma~\ref{lem:trial-inputs}.
\end{proof}
\section{Proof of the gap theorem}\label{sec:assembly}
The two initializations feed the same purity theorem. The finite inputs
they use are proved in Appendices~\ref{sec:thermal120}--\ref{sec:trials};
those proofs are part of the argument.
\begin{proof}[Proof of Theorem~\ref{thm:main}]
Proposition~\ref{prop:transfer} identifies each partition function with
a moment of a self-adjoint spatial transfer operator, so $S$ and $T$ of
Section~\ref{sec:bootstrap} are well defined. The $H_L$ are self-adjoint
and non-scalar, so the bootstrap criterion applies.
Proposition~\ref{prop:seed60} gives
$S(60,105/4)>7/8$ and $T(120,105/4)>7/8$. Apply
Proposition~\ref{prop:bootstrap} with
$(n_0,b_0,u_0,C)=(60,105/4,1/8,79/10)$; its parameter inequalities hold
by Section~\ref{sec:bootstrap}, and $Cu_0=79/80$. Hence the ground
state is unique and $\gamma_L>\frac4{105}\log(80/79)$ for every even
$L\ge60$, proving \eqref{eq:gap60}.
Proposition~\ref{prop:seed2304} gives both defects at most $1/100$ at
the pair with $(n_0,b_0)=(2304,784)$. Apply the same theorem with
$u_0=1/100$, $C=5$; since $Cu_0=1/20$,
\[
\gamma_L>\frac{\log20}{784}\qquad(L\ge2304\text{ even}),
\]
proving \eqref{eq:gap2304}. The limit inferior over even lengths gives
$\Delta_1\ge\log(20)/784>0$. Finally, Lemma~\ref{lem:representation}
carries every spectral conclusion to any simultaneously similar spin-one
realization, via the tensor-power similarity.
\end{proof}
The bounds become uniform on \emph{all} even lengths as follows: the
second bound covers $L\ge2304$, and the finitely many even $L$ with
$4\le L\le2302$ are finite-dimensional self-adjoint Hamiltonians, each
with a positive gap above its full ground sector. This yields
$\inf_{L\ge4,\,L\text{ even}}\gamma_L>0$, though it gives neither a
numerical value nor simplicity at those small lengths.
\subsection{An infinite-volume consequence}
The finite-volume estimate also controls local excitations in a
thermodynamic limit, using the limiting argument of
\cite[arXiv v4, End Matter, Definition~11 and Theorem~12]{Tasaki2025}.
A local operator on the spin chain $\Z$ acts on finitely many sites; a
state $\omega$ is a linear functional on local operators with
$\omega(I)=1$, $\omega(A^*A)\ge0$. Put
$[H,A]=\sum_{j\in\Z}[\mathbf S_j\cdot\mathbf S_{j+1},A]$, only finitely
many terms nonzero.
\begin{corollary}\label{cor:local-limit}
For the self-adjoint Hamiltonians \eqref{eq:main-H}, the normalized
ground states of the even periodic rings have a subsequence, as
$L\to\infty$, converging on every local operator to a state $\omega$;
each such limit obeys
\begin{equation}\label{eq:local-limit-gap}
\omega(A^*[H,A])\ge\frac{\log20}{784}\bigl(\omega(A^*A)-|\omega(A)|^2\bigr)
\end{equation}
for every local operator $A$.
\end{corollary}
\begin{proof}
Place the even ring of length $L$ on $\{-L/2,\dots,L/2-1\}$ with its
closing bond at the endpoints. Compactness of the density matrices on
each finite interval and a diagonal subsequence give convergence on all
local operators; the compatible limits form a positive normalized
functional $\omega$. For $L\ge2304$, Theorem~\ref{thm:main} gives a
unique ground state $\psi_L$ and gap above $\delta=\log(20)/784$.
Writing $\omega_L(A)=\langle\psi_L,A\psi_L\rangle$, the spectral
inequality on the orthogonal complement of $\psi_L$ gives
\[
\omega_L(A^*[H_L,A])\ge\delta\bigl(\omega_L(A^*A)-|\omega_L(A)|^2\bigr).
\]
For fixed local $A$ and large $L$, the closing bond commutes with $A$ and
$[H_L,A]=[H,A]$; passing to the local limit proves the claim.
\end{proof}
In particular $\omega$ is a ground state (no local perturbation lowers
its energy). For $\omega(A)=0$, \eqref{eq:local-limit-gap} is the
locally unique gapped property of
\cite[arXiv v4, Definition~11]{Tasaki2025}, with gap at least
$\log(20)/784$. The argument does not identify limits of different
subsequences, nor uniqueness among all infinite-volume ground states.
\appendix
\section{Thermal calculation at inverse temperature $21/2$}\label{sec:thermal120}
We prove the $b=21/2$ table of Lemma~\ref{lem:thermal-inputs}. The
calculation uses the physical Hamiltonians at $n=6,8,10$ with twists
$1,P,C$; all arithmetic is integral or rational, complex numbers being
written in $\Z[\omega]$ with $\omega=e^{2\pi i/3}$.
\subsection{Finite matrices and their columns}
Fix $b=21/2$ and one length, write $H=H_n(g)$. The three twists have
angles $\theta=0,\pi,2\pi/3$ about the weight-basis axis. Set
\begin{equation}\label{eq:thermal120-matrices}
W=I-\frac{H_n(g)+(3n/2)I}{16},\qquad B=16W .
\end{equation}
Lemma~\ref{lem:energy} gives $\norm W\le1$, $\norm B\le16$. In a fixed
total-weight sector the vectors are words
$x\in\{-1,0,1\}^n$ of fixed $\sum_jx_j$, the diagonal of $B$ is
\[
B_{x,x}=16-\frac{3n}{2}-\sum_{j=0}^{n-1}x_{j-1}x_j ,
\]
and for each $j$ the move $y_j=x_j-\delta$, $y_k=x_k+\delta$
($k=j+1\bmod n$, $\delta=\pm1$, letters in $\{-1,0,1\}$) has entry $-1$,
except at the seam $j=n-1,k=0$ where it is $-e^{i\theta\delta}$ (the
phase \eqref{eq:seam-phase}). On a coordinate $u+v\omega$ the cyclic
twist acts by
\begin{equation}\label{eq:thermal120-phases}
-\omega(u+v\omega)=v+(v-u)\omega,\qquad
-\omega^{-1}(u+v\omega)=(u-v)+u\omega,
\end{equation}
so every matrix-vector product is integer arithmetic in two coordinates.
Only some columns are needed. Let $s(x)=(x_{n-1},x_0,\dots,x_{n-2})$.
The seam formula gives $H_{s(x),s(y)}=e^{i\theta(y_{n-1}-x_{n-1})}H_{x,y}$,
so $|x\rangle\mapsto e^{-i\theta x_{n-1}}|s(x)\rangle$ commutes with $H$.
Reversal of the word and global negation of letters both send $H$ to its
complex conjugate, and composing either permutation with conjugation
gives an antiunitary symmetry. Each symmetry preserves the norm of the
corresponding column of any real exponential of $H$. In every sector of
nonnegative total weight take the lexicographically least word in each
orbit under rotations and reversal, give it the orbit size as
multiplicity, and double for strictly positive total weight to include
the opposite sector. These multiplicities sum to $3^n$ and reproduce
the full Frobenius norm of the exact exponential. No equivariance of the
rounded columns is needed: each computed representative is compared with
its own exact column at its own multiplicity.
\subsection{A rational polynomial and its error}
Write
\[
E=e^{84(W-I)}=e^{-(b/2)(H_n(g)+(3n/2)I)},\qquad x_n=(3/2-a)bn,
\]
so $Z_n(b,g)=e^{x_n}\norm{E}_F^2$ (Frobenius norm). For $0\le j\le256$
let
\begin{equation}\label{eq:thermal120-probabilities}
p_j=\frac{84^j/j!}{\sum_{l=0}^{256}84^l/l!},\qquad
P(W)=\sum_{j=0}^{256}p_jW^j .
\end{equation}
These are mean-$84$ Poisson probabilities conditioned to $\le256$; for
a mean-$84$ Poisson variable $V$,
\[
\Pr(V>256)\le\frac{\mathbb E(2^V)}{2^{257}}=\frac{e^{84}}{2^{257}}
<\frac{27^{28}}{2^{257}}<2^{-117},
\]
and since $\norm{W^j}\le1$ conditioning changes the operator average by
at most twice this probability (Fox--Glynn coefficient argument
\cite[Proposition~1]{FoxGlynn1988}):
\begin{equation}\label{eq:thermal120-poisson-error}
\norm{P(W)-E}<2^{-116}.
\end{equation}
Put $Q=2^{56}$, $q_j=\lfloor Qp_j\rfloor$. From $z=0$, for
$j=256,\dots,0$:
\begin{equation}\label{eq:thermal120-horner}
z\leftarrow\operatorname{fl}(Bz/16)+q_je_r ,
\end{equation}
with $\operatorname{fl}$ the floor on each real coordinate (on $u,v$
separately for $u+v\omega$). The ideal suffix at step $j$ is
$Q\sum_{l=j}^{256}p_lW^{l-j}e_r$ of norm at most $Q$, and one
coordinate floor adds less than $2\sqrt{d_0}$ in a sector of dimension
$d_0$ while the coefficient floor adds at most one; contraction of $W$
bounds the error at every stage by
\begin{equation}\label{eq:thermal120-column-error}
257(2\sqrt{3^n}+1)\le125159.
\end{equation}
The zero coefficients above $q_{172}$ may be skipped without changing
this count (all $257$ steps are still charged).
The repeated computed-column list, divided by $Q$ and multiplied by
$e^{x_n/2}$, differs from $\sqrt{Z_n(b,g)}$ by at most
\begin{equation}\label{eq:thermal120-scaled-error}
e^{x_n/2}\sqrt{3^n}\Bigl(\frac{257(2\sqrt{3^n}+1)}Q+2^{-116}\Bigr)
<4\cdot10^{-7},
\end{equation}
because $\sqrt{3^n}\le243$, $x_n/2<6$, $e^{x_n/2}<3^6$, and
$3^6\cdot243(125159/2^{56}+2^{-116})<4\cdot10^{-7}$.
\subsection{Exact evaluation by packed integers}
For efficiency the recurrence runs on batches of at most $192$ columns,
storing a digit list $y_0,\dots,y_{w-1}$ as
$\operatorname{pack}(y)=\sum_{t=0}^{w-1}y_t2^{64t}$ (arbitrary precision).
Integer linear arithmetic preserves this identity before the floor even
with carries. For either coefficient, $|u|,|v|\le2|u+v\omega|$, so by
\eqref{eq:thermal120-column-error} every coordinate of $Bz$ is at most
\begin{equation}\label{eq:thermal120-lane-bound}
32(Q+125159)=2305843009217699040<2^{63}.
\end{equation}
Let $s_w=\sum_{t=0}^{w-1}2^{64t}$ and
$\mathrm{bias}=2^{63}s_w$, $\mathrm{mask}=(2^{60}-1)s_w$,
$\mathrm{bias16}=2^{59}s_w$. For $Y=\operatorname{pack}(y)$ obeying
\eqref{eq:thermal120-lane-bound}, the operation
\[
\texttt{(((Y+bias)>{}>4)\&mask)-bias16}
\]
equals $\operatorname{pack}((\lfloor y_t/16\rfloor)_t)$. Adding the
bias makes every base-$2^{64}$ digit lie in $[0,2^{64})$; the global
shift moves four low bits of each higher lane into the four high bits of
the previous lane; the mask deletes exactly those crossing bits, keeping
$\lfloor(y_t+2^{63})/16\rfloor$ in lane $t$; subtracting $2^{59}$ gives
$\lfloor y_t/16\rfloor$, also for negative $y_t$. Hence the packed
recurrence equals \eqref{eq:thermal120-horner}. Final coordinates obey
$2(Q+125159)<2^{63}$, so biasing and unpacking recovers every integer
exactly; for $u+v\omega$ the squared modulus is $u^2-uv+v^2$, and summing
these integers with representative multiplicities gives the integer
$t_{n,g}$.
\subsection{Exact totals and trace enclosures}
The integer recurrence gives:
\begin{center}
\begin{tabular}{ccl}
\toprule
$n$&$g$&$t_{n,g}$\\
\midrule
6&$1$&93637037510293158533967385845216\\
6&$P$&3629519998654831111446492683663\\
6&$C$&9839536903588146388838689926305\\
8&$1$&4947327904487991279560670127283\\
8&$P$&749999795795292705866802820621\\
8&$C$&1301346729026243689610427021457\\
10&$1$&388818111664728958208623998490\\
10&$P$&120458568770462253326886628639\\
10&$C$&167487042402286114495695779656\\
\bottomrule
\end{tabular}
\end{center}
Enclose the scalar factor by
\[
L_n=\sum_{j=0}^{100}\frac{x_n^j}{j!},\qquad
U_n=L_n+2\frac{x_n^{101}}{101!}.
\]
Since $0<x_n<12$ and later terms have ratio at most $x_n/102<1/2$,
$L_n<e^{x_n}<U_n$. For each center $c_{n,g}$ of the $b=21/2$ table,
exact rational comparison gives
\begin{equation}\label{eq:thermal120-rational-enclosure}
c_{n,g}-2\cdot10^{-6}
<\frac{L_nt_{n,g}}{Q^2}
<\frac{U_nt_{n,g}}{Q^2}
<c_{n,g}+2\cdot10^{-6}.
\end{equation}
These put every scaled norm below $4$, so by
\eqref{eq:thermal120-scaled-error} replacing its square by the true trace
costs less than $(8+4\cdot10^{-7})4\cdot10^{-7}$. Together with
\eqref{eq:thermal120-rational-enclosure},
\[
|Z_n(21/2,g)-c_{n,g}|
<2\cdot10^{-6}+(8+4\cdot10^{-7})4\cdot10^{-7}<10^{-5}.
\]
This proves the $b=21/2$ table.
\section{Thermal calculation at inverse temperature $49/4$}\label{sec:thermal72}
We prove the $b=49/4$ table, using integer approximations to columns of a
physical matrix exponential. Throughout
\[
b=\frac{49}{4},\qquad J=280,\qquad Q=2^{55},\qquad 4\le n\le12,
\]
with twists the identity and rotation by $\pi$ about $z$ (the latter has
the same partition function as $P$ by a global rotation).
\subsection{The physical matrix and its rational approximation}
In the weight basis a word $w=(w_0,\dots,w_{n-1})$, $w_i\in\{-1,0,1\}$,
has conserved total weight. On each weight block set
\begin{equation}\label{eq:thermal72-matrix}
R=I-\frac{\frac32nI+H}{16},\qquad D=32R ,
\end{equation}
an integer matrix with diagonal
$D_{w,w}=32-3n-2\sum_{i=0}^{n-1}w_{i-1}w_i$, an entry $-2$ for an allowed
opposite unit step at letters $i-1,i$, and $+2$ on the twisted seam
$i=0$ (indices mod $n$). Lemma~\ref{lem:energy} gives
\[
0\le\tfrac32nI+H\le\tfrac52nI,\qquad
\spec(R)\subseteq[1-5n/32,1]\subseteq[-\tfrac78,1],
\]
so $R$ is a self-adjoint contraction and
\[
F=\exp\!\bigl[-\tfrac b2(\tfrac32nI+H)\bigr]=\exp[98(R-I)] .
\]
On the full physical space (direct sum over weight blocks),
\begin{equation}\label{eq:thermal72-trace}
Z_n(b,g)=e^{x_n}\norm{F}_F^2,\qquad x_n=bn\bigl(\tfrac32-a\bigr).
\end{equation}
Define the conditioned Poisson probabilities
\begin{equation}\label{eq:thermal72-coefficients}
p_j=\frac{v_j}{\sum_{\ell=0}^Jv_\ell},\quad
v_j=98^j\frac{J!}{j!},\quad 0\le j\le J .
\end{equation}
Since $\norm R\le1$, discarding the Poisson tail and renormalizing
changes the exponential by at most twice the tail probability
(Fox--Glynn):
\[
\norm{F-\sum_{j=0}^Jp_jR^j}
\le2e^{-98}\sum_{j=281}^\infty\frac{98^j}{j!}
\le2^{-280}e^{98}<2^{-84}<Q^{-1},
\]
using $2^j\ge2^{281}$ in the tail and $e^{-98}\sum_{j\ge0}196^j/j!=e^{98}$,
then $e<4$.
\subsection{Integer columns and their error}
For a unit basis column $e_r$, start at $X=0$ and for $j=J,\dots,0$ do
\begin{equation}\label{eq:thermal72-horner}
X\leftarrow\Bigl\lfloor\frac{DX}{32}\Bigr\rfloor+c_je_r,\qquad
c_j=\lfloor Qp_j\rfloor,
\end{equation}
with coordinatewise floor; all quantities are integers, and since
$c_j=0$ for $j>191$ the initial zero steps may be omitted. Compare with
the ideal suffix $Y_j=Q\sum_{\ell=j}^Jp_\ell R^{\ell-j}e_r$,
$\norm{Y_j}_2\le Q$. In a block of dimension $d_0$ a vector floor adds
less than $\sqrt{d_0}$ and a coefficient floor less than one, so
contraction bounds the accumulated error by $281(\sqrt{d_0}+1)$. With
\begin{equation}\label{eq:thermal72-errors}
d_n=\lfloor\sqrt{3^n}\rfloor+1,\quad
E_n=281(d_n+1)+1,\quad \mathcal E_n=d_nE_n,
\end{equation}
every computed column obeys
\begin{equation}\label{eq:thermal72-column-error}
\norm{X-QFe_r}_2\le E_n .
\end{equation}
The bound charges all $281$ ideal steps (including the omitted ones).
The recursion is exactly signed-$64$-bit arithmetic: every coordinate is
at most $Q+281(d_n+1)<2Q$ with $d_n\le730$; any row of $D$ has absolute
sum at most $|32-3n|+2n+4n\le76$. Hence every product and partial sum is
below $76\cdot2Q=5476377146882523136<2^{63}-1$, and division by $32$ plus
$c_j\le Q$ stays below $207165582859042816<2^{63}-1$. Starting at $X=0$
the bounds establish exactness of the next integer operation, and
contraction establishes the next coordinate bound; the induction does not
assume the overflow-freeness it proves. Final squares and sums use
arbitrary-precision integers.
\subsection{Representative columns}
One column per orbit suffices. For positive total weight use the orbit
under cyclic shifts and reversal and give the representative twice the
orbit size (this includes the uncomputed negative-weight block); in the
zero-weight block include digit negation in the orbit and use its size
without a factor of two. Take the lexicographically least word,
handle periodic words by set cardinalities; the multiplicities $\nu_r$
satisfy $\sum_r\nu_r=3^n$.
Symmetries: reversal fixes the seam as an unordered bond and global
negation preserves every entry. For the twisted chain a bare shift moves
the seam; left rotation by $q$ moves it to $(k-1,k)$, $k=-q\bmod n$, and
the diagonal unitary
\[
G_k|w\rangle=(-1)^{\sum_{j=0}^{k-1}w_j}|w\rangle,\quad G_0=I,
\]
flips the hopping signs exactly at the two boundary bonds of that arc,
restoring the seam. The resulting signed shift commutes with $H$, so
exact exponential columns at rotated words have equal norms; the
untwisted chain needs no correction.
Let $X_r$ be the output of \eqref{eq:thermal72-horner} at representative
$r$ and
\begin{equation}\label{eq:thermal72-total}
W_{n,g}=\sum_r\nu_r\sum_wX_r(w)^2 .
\end{equation}
Rounding need not preserve the signed symmetries, so form abstract direct
sums repeating each exact (resp.\ computed) representative column $\nu_r$
times: the first has squared norm $Q^2\norm{F}_F^2$, the second $W_{n,g}$.
By \eqref{eq:thermal72-column-error},
\begin{equation}\label{eq:thermal72-frobenius}
\bigl|\sqrt{W_{n,g}}-Q\norm{F}_F\bigr|\le\sqrt{3^n}E_n<\mathcal E_n,
\end{equation}
which proves the full trace estimate without equivariance of the rounded
columns.
\subsection{Exact totals and rational enclosures}
Equations \eqref{eq:thermal72-coefficients},
\eqref{eq:thermal72-horner}, \eqref{eq:thermal72-total} give (second
column: no twist; third: $\pi$ twist):
\begin{center}
\begingroup
\footnotesize
\setlength{\tabcolsep}{4pt}
\begin{tabular}{r|rr}
\(n\)&\(W_{n,I}\)&\(W_{n,P}\)\\ \hline
4&1298092848946206529211955779616040&683496640248050713756440676071\\
5&9095379218684502088743241789&9205948219074287936685289758037\\
6&11971284263768914176431179720622&266573596119930013833919176645\\
7&15044150708248976451784593468&534330762492192297990602166370\\
8&386135473475360677201628357188&42459883501386986922766507327\\
9&4866417737994347551816846903&37716485750353424806876104310\\
10&19772960374215472827775892486&5030484777222468978572235564\\
11&841347687722481981930750335&2928767918985354021900689432\\
12&1266666171463993322695232916&524785463625387865619045258
\end{tabular}
\endgroup
\end{center}
Turn these integers into the $b=49/4$ table. Let
\[
r_{n,g}=\lfloor\sqrt{W_{n,g}}\rfloor,\quad
s_n=\sum_{j=0}^{100}\frac{x_n^j}{j!},\quad
t_n=\frac{x_n^{100}}{100!}.
\]
Since $0\le x_n<15$, later omitted terms have ratio at most $15/101$ and
the tail is at most $15t_n/86<t_n$, so $s_n\le e^{x_n}\le s_n+t_n$.
Equations \eqref{eq:thermal72-trace} and \eqref{eq:thermal72-frobenius}
give
\begin{equation}\label{eq:thermal72-enclosure}
L_{n,g}=\frac{s_n\max(0,r_{n,g}-\mathcal E_n)^2}{Q^2},\quad
U_{n,g}=\frac{(s_n+t_n)(r_{n,g}+1+\mathcal E_n)^2}{Q^2},\quad
L_{n,g}\le Z_n(b,g)\le U_{n,g},
\end{equation}
all rational/integer with the integer square root fixed by
$r^2\le W<(r+1)^2$. For the integer centers $C_{n,g}$ of the $b=49/4$
table, direct evaluation yields
\[
L_{n,g}>\frac{C_{n,g}-30}{10^6},\qquad
U_{n,g}<\frac{C_{n,g}+30}{10^6}\qquad(4\le n\le12,\ g=I,P),
\]
each strict inequality with margin above $13/10^6$. This gives a
reproducible finite verification of all eighteen enclosures and proves the
$b=49/4$ table.
\section{Two variational inputs}\label{sec:trials}
The initializations need a lower bound on the untwisted partition function
at lengths $72$ and $120$. One fixed triple of integer matrices defines
periodic matrix-product vectors at both lengths with energy per bond below
$-a=-700741/500000$. A block decomposition reduces the computation to
powers of integer matrices of order at most $162$.
\subsection{Norm and energy contractions}
For $N\ge4$ and real $D\times D$ matrices $A_s$ ($s=-1,0,1$) define, in
the orthonormal spin-one weight basis,
\begin{equation}\label{eq:trial-vector}
\psi_N(s_0,\dots,s_{N-1})=\Tr(A_{s_0}\cdots A_{s_{N-1}}),
\end{equation}
the periodic matrix-product representation of
\cite[Section~2.1, Equations~(2)--(3)]{PerezGarcia2007}. Put
\begin{align}
B_{st}&=A_sA_t,\quad T=\sum_{s=-1}^1A_s\otimes A_s,
\label{eq:trial-transfer}\\
O&=\sum_{s,t=-1}^1st\,B_{st}\otimes B_{st}
+\sum_{\substack{s=-1,0\\t=0,1}}
\bigl(B_{st}\otimes B_{s+1,t-1}
+B_{s+1,t-1}\otimes B_{st}\bigr).
\label{eq:trial-insertion}
\end{align}
All traces are ordinary, unnormalized.
\begin{lemma}\label{lem:trial-contraction}
Let $N\ge4$, $D\ge1$ and $A_{-1},A_0,A_1\in M_D(\R)$; define
$\psi_N,T,O$ by \eqref{eq:trial-vector}--\eqref{eq:trial-insertion} and,
for the periodic spin-one Hamiltonian $H_N$, set
$V_N=\Tr(T^N), U_N=\Tr(T^{N-2}O)$. Then
$\norm{\psi_N}^2=V_N$ and
$\langle\psi_N,H_N\psi_N\rangle=NU_N$. In particular, if $V_N>0$ and
$U_N<-aV_N$, then $E_0(N)<-Na$ and $Z_N(b)>1$ for every $b>0$.
\end{lemma}
\begin{proof}
Expanding $T^N$ and using $\Tr(X\otimes Y)=\Tr(X)\Tr(Y)$, the sum of
squares of the (real) amplitudes \eqref{eq:trial-vector} is $V_N$; no
self-adjointness or positivity of $T$ is assumed. In the weight basis a
bond has diagonal $st$ on $(s,t)$ and off-diagonal $1$ from $(s,t)$ to
$(s+1,t-1)$ for $s=-1,0$, $t=0,1$ (from
$h=S^z\otimes S^z+(S^+\otimes S^-+S^-\otimes S^+)/2$ and ladder entries
$\sqrt2$). Inserting this bond into the two amplitude layers gives
exactly $O$; summing the other indices gives $U_N$ for one bond, and
cyclicity of the amplitude trace makes $\psi_N$ translation invariant, so
every bond including the seam contributes equally. For $V_N>0$ the
variational principle gives $E_0(N)\le NU_N/V_N<-Na$, so the shifted
ground energy is negative and its term in $\Tr e^{-b(H_N+NaI)}$ exceeds
one for every $b>0$.
\end{proof}
\subsection{One integer matrix triple}
Set $D=26$, index eight types $\alpha=1,\dots,8$ by
$(\ell_1,\dots,\ell_8)=(1,1,1,1,3,3,3,5)$, and order virtual labels
$(\alpha,d)$ first by $\alpha$ then by
$d=-\ell_\alpha,-\ell_\alpha+2,\dots,\ell_\alpha$. The matrices $A_s$
use the diagonal parameters $D_\alpha$ of Table~\ref{tab:trial-diagonal}
and the parameters $P_{\alpha\beta}$ of Table~\ref{tab:trial-pairs},
defined only when $\ell_\beta=\ell_\alpha+2$.
\begin{table}[ht]
\centering
\begin{tabular}{c|rrrrrrrr}
$\alpha$&1&2&3&4&5&6&7&8\\ \hline
$D_\alpha$&$-289$&$-84$&$-9$&33&$-22$&27&71&$-11$
\end{tabular}
\caption{Diagonal parameters, common to both lengths.}
\label{tab:trial-diagonal}
\end{table}
\begin{table}[ht]
\centering
\begin{tabular}{c|r@{\qquad}c|r@{\qquad}c|r}
$(\alpha,\beta)$&$P_{\alpha\beta}$&
$(\alpha,\beta)$&$P_{\alpha\beta}$&
$(\alpha,\beta)$&$P_{\alpha\beta}$\\ \hline
$(1,5)$&$-11$&$(2,7)$&$-49$&$(4,6)$&9\\
$(1,6)$&$-35$&$(3,5)$&$-6$&$(4,7)$&$-32$\\
$(1,7)$&$-28$&$(3,6)$&$-25$&$(5,8)$&1\\
$(2,5)$&$-23$&$(3,7)$&$-20$&$(6,8)$&2\\
$(2,6)$&16&$(4,5)$&$-34$&$(7,8)$&$-16$
\end{tabular}
\caption{Parameters joining virtual types whose $\ell$ differ by $2$.}
\label{tab:trial-pairs}
\end{table}
Entry rules. Row $(\alpha,d)$, column $(\beta,e)$, with $\ell=\ell_\alpha$,
$k=\ell_\beta$, $s\in\{-1,0,1\}$. The entry is zero unless
\begin{equation}\label{eq:trial-support}
e=d+2s,\qquad |k-\ell|\le2,\qquad (k=\ell\Longrightarrow\alpha=\beta).
\end{equation}
For an allowed entry set
\[
p=\begin{cases}D_\alpha,& k=\ell,\\ P_{\alpha\beta},& k=\ell+2,\\
-P_{\beta\alpha},& k=\ell-2,\end{cases}
\]
$r=(d+\ell)/2$, and the entry $pf$ where
\begin{equation}\label{eq:trial-entry}
\begin{array}{c|ccc}
&s=-1&s=0&s=1\\ \hline
k=\ell+2&2(\ell+1-r)(\ell+2-r)&2(r+1)(\ell-r+1)&(r+1)(r+2)\\
k=\ell&2(\ell-r+1)&d&-(r+1)\\
k=\ell-2&2&-2&1
\end{array}
\end{equation}
lists the values of $f$. These rules do not depend on $N$.
\subsection{Finite contractions and variational bounds}
Evaluate the contractions at $N=72$ and $N=120$. The support rule
\eqref{eq:trial-support} makes this a short block calculation: on paired
virtual labels $((\alpha,d),(\beta,e))$ both $T$ and $O$ preserve $d-e$
(the two layers shift by the same $2s$ for $T$, and by the same total
physical weight $s+t$ for $O$, also in each off-diagonal term). The
difference blocks are exhaustive with dimensions
\begin{equation}\label{eq:trial-blocks}
\begin{array}{c|rrrrrrrrrrr}
d-e&-10&-8&-6&-4&-2&0&2&4&6&8&10\\ \hline
\dim&1&8&32&80&136&162&136&80&32&8&1
\end{array},
\end{equation}
summing to $676=26^2$, each block once.
The whole integer computation: form $A_s$, $B_{st}$, $T$, $O$ from the
tables and \eqref{eq:trial-transfer}, \eqref{eq:trial-insertion},
\eqref{eq:trial-support}, \eqref{eq:trial-entry}; restrict to each
difference block giving $T_\delta,O_\delta$; compute powers by
\[
P_0=I,\quad P_{2j}=P_j^2\ (j\ge1),\quad P_{2j+1}=P_j^2T_\delta\ (j\ge0),
\]
then accumulate
\begin{equation}\label{eq:trial-integer-algorithm}
U_N=\sum_\delta\sum_{i,j}(P_{N-2})_{ij}(O_\delta)_{ji},\quad
V_N=\sum_\delta\sum_{i,j}(P_{N-2})_{ij}(T_\delta^2)_{ji},
\end{equation}
using only arbitrary-precision integers. For $N=72$ the power is $70$;
for $N=120$ use $T^{118}=T^{64}T^{32}T^{16}T^4T^2$ and
$T^{120}=T^{64}T^{32}T^{16}T^8$ within each block. The resulting exact
integer comparisons are
\begin{equation}\label{eq:trial-integer-results}
V_N>0,\qquad -c_NV_N\le10^{12}U_N<(-c_N+1)V_N,
\qquad
\begin{array}{c|r}
N&c_N\\ \hline
72&1401482112601\\
120&1401482105174
\end{array}.
\end{equation}
\begin{lemma}\label{lem:trial-inputs}
For the untwisted periodic spin-one Heisenberg Hamiltonian and
$a=700741/500000$, $E_0(72)<-72a$ and $E_0(120)<-120a$; consequently
$Z_{72}(b)>1$ and $Z_{120}(b)>1$ for every $b>0$.
\end{lemma}
\begin{proof}
For each row of \eqref{eq:trial-integer-results},
$(-c_N+1)/10^{12}<-700741/500000$, so the corresponding explicit matrices
have $V_N>0$ and $U_N/V_N<-a$; apply Lemma~\ref{lem:trial-contraction}.
\end{proof}
The support rule places these trial vectors in total weight zero, but the
variational bound is for the minimum of the full physical spectrum; the
only symmetry used in the contraction is cyclic translation. The inputs
hold at every positive inverse temperature, so each initialization uses
them at its own temperature.
\begin{thebibliography}{99}
\bibitem[Haldane(1981)]{Haldane1981} F.~D.~M. Haldane, Ground state properties of antiferromagnetic chains with unrestricted spin, ILL preprint SP-81/95 (1981); reproduced as arXiv:1612.00076.
\bibitem[Haldane(1983a)]{Haldane1983PLA} F.~D.~M. Haldane, Continuum dynamics of the 1-D Heisenberg antiferromagnet, Phys.\ Lett.\ A \textbf{93}, 464--468 (1983).
\bibitem[Haldane(1983b)]{Haldane1983PRL} F.~D.~M. Haldane, Nonlinear field theory of large-spin Heisenberg antiferromagnets, Phys.\ Rev.\ Lett.\ \textbf{50}, 1153--1156 (1983).
\bibitem[Lieb, Schultz and Mattis(1961)]{LiebSchultzMattis1961} E. Lieb, T. Schultz and D. Mattis, Two soluble models of an antiferromagnetic chain, Ann.\ Phys.\ \textbf{16}, 407--466 (1961).
\bibitem[Affleck and Lieb(1986)]{AffleckLieb1986} I. Affleck and E.~H. Lieb, A proof of part of Haldane's conjecture on spin chains, Lett.\ Math.\ Phys.\ \textbf{12}, 57--69 (1986).
\bibitem[Affleck, Kennedy, Lieb and Tasaki(1987)]{AKLT1987} I. Affleck, T. Kennedy, E.~H. Lieb and H. Tasaki, Rigorous results on valence-bond ground states in antiferromagnets, Phys.\ Rev.\ Lett.\ \textbf{59}, 799--802 (1987).
\bibitem[Affleck, Kennedy, Lieb and Tasaki(1988)]{AKLT1988} I. Affleck, T. Kennedy, E.~H. Lieb and H. Tasaki, Valence bond ground states in isotropic quantum antiferromagnets, Comm.\ Math.\ Phys.\ \textbf{115}, 477--528 (1988).
\bibitem[Yarotsky(2006)]{Yarotsky2006} D.~A. Yarotsky, Ground states in relatively bounded quantum perturbations of classical lattice systems, Comm.\ Math.\ Phys.\ \textbf{261}, 799--819 (2006); arXiv:math-ph/0412040.
\bibitem[White and Huse(1993)]{WhiteHuse1993} S.~R. White and D.~A. Huse, Numerical renormalization-group study of low-lying eigenstates of the antiferromagnetic $S=1$ Heisenberg chain, Phys.\ Rev.\ B \textbf{48}, 3844--3852 (1993).
\bibitem[Buyers et al.(1986)]{Buyers1986} W.~J.~L. Buyers, R.~M. Morra, R.~L. Armstrong, M.~J. Hogan, P. Gerlach and K. Hirakawa, Experimental evidence for the Haldane gap in a spin-1 nearly isotropic antiferromagnetic chain, Phys.\ Rev.\ Lett.\ \textbf{56}, 371--374 (1986).
\bibitem[Batista and Ortiz(2001)]{BatistaOrtiz2001} C.~D. Batista and G. Ortiz, Generalized Jordan--Wigner transformations, Phys.\ Rev.\ Lett.\ \textbf{86}, 1082--1085 (2001); arXiv:cond-mat/0008374.
\bibitem[Tasaki(2025)]{Tasaki2025} H. Tasaki, Ground state of the $S=1$ antiferromagnetic Heisenberg chain is topologically nontrivial if gapped, Phys.\ Rev.\ Lett.\ \textbf{134}, 076602 (2025); arXiv:2407.17041 (v4).
\bibitem[Fannes, Nachtergaele and Werner(1992)]{FNW1992} M. Fannes, B. Nachtergaele and R.~F. Werner, Finitely correlated states on quantum spin chains, Comm.\ Math.\ Phys.\ \textbf{144}, 443--490 (1992).
\bibitem[P\'erez-Garc\'ia et al.(2007)]{PerezGarcia2007} D. P\'erez-Garc\'ia, F. Verstraete, M.~M. Wolf and J.~I. Cirac, Matrix product state representations, Quantum Inf.\ Comput.\ \textbf{7}, 401--430 (2007); arXiv:quant-ph/0608197.
\bibitem[Knabe(1988)]{Knabe1988} S. Knabe, Energy gaps and elementary excitations for certain VBS-quantum antiferromagnets, J.\ Stat.\ Phys.\ \textbf{52}, 627--638 (1988).
\bibitem[Nachtergaele(1996)]{Nachtergaele1996} B. Nachtergaele, The spectral gap for some spin chains with discrete symmetry breaking, Comm.\ Math.\ Phys.\ \textbf{175}, 565--606 (1996); arXiv:cond-mat/9410110.
\bibitem[Gosset and Mozgunov(2016)]{GossetMozgunov2016} D. Gosset and E. Mozgunov, Local gap threshold for frustration-free spin systems, J.\ Math.\ Phys.\ \textbf{57}, 091901 (2016); arXiv:1512.00088.
\bibitem[Banerjee, Guo and Chowdhury(2026)]{BanerjeeGuoChowdhury2026} S. Banerjee, H. Guo and D. Chowdhury, Bootstrap certification of string order in quantum spin chains, arXiv:2609.22420 (2026).
\bibitem[Suzuki(1985)]{Suzuki1985} M. Suzuki, Transfer-matrix method and Monte Carlo simulation in quantum spin systems, Phys.\ Rev.\ B \textbf{31}, 2957--2965 (1985).
\bibitem[Aizenman and Nachtergaele(1994)]{AizenmanNachtergaele1994} M. Aizenman and B. Nachtergaele, Geometric aspects of quantum spin states, Comm.\ Math.\ Phys.\ \textbf{164}, 17--63 (1994); arXiv:cond-mat/9310009.
\bibitem[Fox and Glynn(1988)]{FoxGlynn1988} B.~L. Fox and P.~W. Glynn, Computing Poisson probabilities, Comm.\ ACM \textbf{31}, 440--445 (1988).
\bibitem[Bochev and Markov(1989)]{BochevMarkov1989} P. Bochev and S. Markov, A self-validating numerical method for the matrix exponential, Computing \textbf{43}, 59--72 (1989).
\bibitem[Goldsztejn and Neumaier(2014)]{GoldsztejnNeumaier2014} A. Goldsztejn and A. Neumaier, On the exponentiation of interval matrices, Reliable Computing \textbf{20}, 53--72 (2014).
\end{thebibliography}
\end{document}