% =====================================================================
%  Equality cases in the sharp one-dimensional matrix Lieb--Thirring
%  inequality: a complete, self-contained proof.
%
%  This file is written to be read straight through.  Each technical
%  object is introduced with a plain sentence saying what it is for,
%  then used.  The two analytic machines (the "trace engine" in
%  Section 4 and the "field/path engine" in Section 5) are proved in
%  full, so the main theorem never relies on an unproved black box.
%
%  Road map:
%    Section 1  the problem, the answer, and the idea of the proof
%    Section 2  notation and the four constants
%    Section 3  two elementary lemmas (spectrum, Young inequality)
%    Section 4  the trace engine: an exact integrated deficit identity
%    Section 5  the field and the connecting path (existence)
%    Section 6  the finite-interval action estimate
%    Section 7  the sharp inequality itself
%    Section 8  from equality to pointwise matrix relations
%    Section 9  fixed channels, the profile, complex W, attainability
% =====================================================================
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\title{Equality cases in the sharp one-dimensional\\[2pt]
matrix Lieb--Thirring inequality\\[6pt]\large a complete proof}
\author{}
\date{}

\begin{document}
\maketitle

\begin{abstract}
We classify all equality cases in the sharp one-dimensional Lieb--Thirring
inequality, for every finite matrix size and every exponent
$1/2<\gamma<3/2$.  For a measurable Hermitian positive semidefinite
potential $W$ with $\int_\R\tr(W^{\gamma+1/2})<\infty$, equality holds
exactly when, in one constant unitary basis, $W$ is a direct sum of scalar
one-bound-state solitons and zero channels, with independent scales and
centres.  Every step is proved here.
\end{abstract}

\tableofcontents

% =====================================================================
\section{The problem, the answer, and the idea}\label{sec:intro}
% =====================================================================

\subsection{What is being estimated}
Fix an integer $m\ge1$ (the number of internal ``channels'').  A
\emph{potential} is a measurable function $W:\R\to\C^{m\times m}$ whose
values are Hermitian and positive semidefinite.  Think of $-\Delta$ acting
on vector-valued functions $\psi:\R\to\C^m$, and of $W$ as a position
dependent multiplication operator.  The Schr\"odinger operator
\begin{equation}\label{eq:Ham}
 H_W=-\frac{d^2}{dx^2}-W
\end{equation}
is unbounded below when $W$ is large, but it has only finitely many
negative eigenvalues below any fixed level (Lemma~\ref{lem:spectral}).
Write $\lambda_j(H_W)$ for the eigenvalues, counted with multiplicity, and
for $1/2<\gamma<3/2$ set
\begin{equation}\label{eq:moment}
 \Tr(H_W)_-^\gamma
   =\sum_{\lambda_j(H_W)<0}\bigl|\lambda_j(H_W)\bigr|^\gamma .
\end{equation}
Here $\tr$ is the unnormalised trace on $\C^{m\times m}$, and $W^p$ is
defined by spectral calculus.

\subsection{The inequality and its constant}
The sharp Lieb--Thirring inequality in one dimension is
\begin{equation}\label{eq:sharp}
 \Tr(H_W)_-^\gamma
   \le C_\gamma\int_\R\tr\bigl(W(x)^p\bigr)\,dx,
 \qquad p=\gamma+\tfrac12,
\end{equation}
with the explicit constant
\begin{equation}\label{eq:Cgamma}
 C_\gamma
 =\Bigl(\frac{\gamma-1/2}{\gamma+1/2}\Bigr)^{\gamma-1/2}
   \frac{\Gamma(\gamma+1)}{\sqrt\pi\,\Gamma(\gamma+3/2)} .
\end{equation}
The inequality itself is due to Lieb and Thirring and, in this matrix
form, to the companion paper \cite{MLT}.  The new content here is the
\emph{equality} statement, proved together with the inequality so that
both rest on the same estimates.

\subsection{The answer}
Set
\begin{equation}\label{eq:r}
 r=\frac{1}{\gamma-1/2}>1 .
\end{equation}
The extremal potentials are built from the single scalar profile
\begin{equation}\label{eq:profile}
 w(x)=(r+1)a^2\,\sech^2\!\bigl(r\,a\,(x-x_0)\bigr),
 \qquad a>0,\ x_0\in\R ,
\end{equation}
which has exactly one negative eigenvalue, namely $-a^2$, and saturates
\eqref{eq:sharp} by itself.

\begin{theorem}[Equality classification]\label{thm:main}
Let $m\ge1$, $1/2<\gamma<3/2$, $p=\gamma+\tfrac12$, and let $W$ be a
measurable Hermitian positive semidefinite potential with
$\int_\R\tr(W^p)<\infty$.  Then \eqref{eq:sharp} holds.  Moreover,
equality holds if and only if there are an integer $k\in\{0,\dots,m\}$,
a constant unitary matrix $U$, positive numbers $a_1,\dots,a_k$, and real
centres $x_1,\dots,x_k$ such that, for almost every $x$,
\begin{equation}\label{eq:classification}
 W(x)=U\,\diag\bigl(w_1(x),\dots,w_k(x),0,\dots,0\bigr)\,U^* ,
 \qquad
 w_j(x)=(r+1)a_j^2\,\sech^2\!\bigl(r a_j (x-x_j)\bigr).
\end{equation}
The case $k=0$ means $W=0$.  Each nonzero channel carries exactly one
negative eigenvalue $-a_j^2$, so every extremal potential has at most $m$
negative eigenvalues counted with multiplicity.  The scales $a_j$ and
centres $x_j$ are independent, and coincident profiles are allowed.
\end{theorem}

\subsection{The idea of the proof}
The proof has six moves.  We state them now so the reader knows where
each later section is going.

\begin{enumerate}
\item[(1)] \emph{Linearise the spectrum.}  Approximate finitely many
negative eigenfunctions by smooth compactly supported trial functions and
arrange them as the rows of a matrix $U(x)$.  A lower triangular matrix
$C$ keeps the rows in successive spans; this is what makes the energies
$\ell_i^2$ appear in order.  This is the standard trial-function step of
the inequality.

\item[(2)] \emph{Build an action.}  For a diagonal matrix $K$ of trial
root-energies, define the energy
$\mathcal E_K(A)=\int(\|A'\|_{\HS}^2+\tr(K^2AA^T)-\tr((AA^T)^q))$.  It is
the trial-function energy written in matrix language.

\item[(3)] \emph{Insert a matrix path.}  A symmetric path $B(x)$ running
from $K$ at the left end of the interval to $-K$ at the right end turns
the action into a square plus a boundary term (complete the square):
\[
 \mathcal E_K(A)=\text{(boundary)}+\int\bigl(\|A'-BA\|_{\HS}^2-F(A,B)\bigr).
\]
The boundary term produces exactly $b_\sigma\sum\ell_i^{2\gamma}$.

\item[(4)] \emph{Keep the defect.}  Equality forces the square
$\|A'-BA\|_{\HS}^2$ and an algebraic defect
$\Phi_K(A,B)=\|[B,K^2]\|_{\op}^2+\|K^2-B^2-(AA^T)^r\|_{\op}^2$ to vanish.
The trace engine of Section~4 is what makes the coefficient of $\Phi_K$
uniform in the number of rows.  This uniformity is essential because the
number of negative eigenvalues is not known in advance to be finite.

\item[(5)] \emph{Pass to a limit.}  The number of rows may grow without
bound while the number $m$ of columns stays fixed.  The fixed column
count bounds the tail rows by the small energies, giving strong pointwise
convergence and exact algebraic relations in the limit.

\item[(6)] \emph{Solve the leftover ODE.}  The limit relations force the
channel subspaces to be constant in $x$.  On each channel a matrix
Riccati equation reduces to a scalar equation whose only bounded
solutions are the profiles \eqref{eq:profile}.  This yields
Theorem~\ref{thm:main}, and a direct factorization check proves that every
listed potential attains equality.
\end{enumerate}

\subsection{Brief history}
Lieb and Thirring introduced spectral moment inequalities for the kinetic
energy of fermions \cite{LiebThirring1975,LiebThirring1976}.  The sharp
scalar endpoint $\gamma=1/2$ is due to Hundertmark--Lieb--Thomas
\cite{HLT1998}, extended to operator-valued potentials by
Hundertmark--Laptev--Weidl \cite{HLW2000}.  The semiclassical constant is
sharp for $\gamma\ge3/2$; the relevant tools are the moment lifting of
Aizenman--Lieb \cite{AizenmanLieb1978} and the operator-valued inequalities
of Laptev--Weidl \cite{LaptevWeidl2000}.  Benguria--Loss \cite{BenguriaLoss2000}
gave a commutation proof at $\gamma=3/2$.  The one-state scalar profile
\eqref{eq:profile} goes back to Keller \cite{Keller1961}; its role in the
Lieb--Thirring problem is explained in \cite[Section~4]{LiebThirring1976}.
At the endpoint $\gamma=3/2$ the equality structure changes: Frank, Gontier
and Lewin \cite{FrankGontierLewin2025} show that several bound states can
share a single scalar channel (KdV multisolitons).  The strict inequality
$r>1$ used below is exactly what rules that out in the open interval.

% =====================================================================
\section{Notation and the four constants}\label{sec:notation}
% =====================================================================

Throughout,
\begin{equation}\label{eq:params}
 \sigma=\gamma-\tfrac12\in(0,1),\qquad
 p=1+\sigma=\gamma+\tfrac12,\qquad
 r=\frac1\sigma>1,\qquad
 q=r+1 .
\end{equation}
Note the identities
\begin{equation}\label{eq:identities}
 2\sigma+1=2\gamma,\qquad rp=q=1+r .
\end{equation}
Two constants recur:
\begin{equation}\label{eq:Dconst}
 D_\sigma=\frac{\sigma^\sigma}{(1+\sigma)^{1+\sigma}},
 \qquad
 b_\sigma=\int_{-1}^{1}(1-s^2)^\sigma\,ds .
\end{equation}
They are linked to \eqref{eq:Cgamma} by the beta integral:
\begin{equation}\label{eq:Crelation}
 C_\gamma=\frac{D_\sigma}{b_\sigma}.
\end{equation}
We use the operator norm $\|\cdot\|_{\op}$ and the Hilbert--Schmidt norm
$\|A\|_{\HS}^2=\tr(A^*A)$.  For real matrices $T$ also denotes transpose.
$\Sym_n$ is the space of real symmetric $n\times n$ matrices with inner
product $\tr(BH)$; the commutator is $[B,H]=BH-HB$.  The identity matrix
or operator has the size forced by context.

% =====================================================================
\section{Two elementary lemmas}\label{sec:prelim}
% =====================================================================

\subsection{The spectrum is discrete and the form is closed}
The first lemma says we may freely talk about ``the'' negative eigenvalues,
even when $W$ is only measurable.

\begin{lemma}[Spectral facts]\label{lem:spectral}
Let $p>1$ and let $W:\R\to\C^{m\times m}$ be measurable, Hermitian and
positive semidefinite, with $\int_\R\tr(W^p)<\infty$.  The quadratic form
\begin{equation}\label{eq:form}
 h_W[\psi]=\int_\R\|\psi'\|^2-\int_\R\langle\psi,W\psi\rangle,
 \qquad \psi\in H^1(\R;\C^m),
\end{equation}
is closed and bounded below.  Multiplication by $W^{1/2}$ is compact from
$H^1$ to $L^2$.  The negative spectrum of $H_W$ consists of isolated
eigenvalues of finite multiplicity, accumulating at most at $0$.  Every
eigenfunction lies in $H^1$ and solves $-\psi''-W\psi=\lambda\psi$ in
distributions.
\end{lemma}

\begin{proof}
Put $w=\|W\|_{\op}$.  Then $0\le w^p\le\tr(W^p)$, so $w\in L^p(\R)$.  The
one-dimensional Sobolev bound, applied to the vector $x\mapsto\|\psi(x)\|$,
reads $\|\psi\|_\infty^2\le 2\|\psi\|_2\|\psi'\|_2$.  Hence, by
H\"older and Young,
\begin{equation}\label{eq:formbound}
 \int_\R\langle\psi,W\psi\rangle
 \le 2^{1/p}\|w\|_p\|\psi'\|_2^{1/p}\|\psi\|_2^{2-1/p}
 \le \varepsilon\|\psi'\|_2^2
     +C_\varepsilon\|\psi\|_2^2 .
\end{equation}
So a suitably shifted form norm is equivalent to the $H^1$ norm: the form
is closed and bounded below, and by polarization the potential term is
continuous on $H^1$.

Let $T\psi=W^{1/2}\psi$.  For $R,L>0$ define the truncated multiplier
$T_{R,L}$ by keeping only $\{|x|\le R,\ w\le L\}$.  By the compact
embedding $H^1([-R,R])\hookrightarrow L^2([-R,R])$, each $T_{R,L}$ is
compact, and
\[
 \|(T-T_{R,L})\psi\|_2^2
 \le C\,\bigl\|w\,\mathbf 1_{\{|x|>R\}\cup\{w>L\}}\bigr\|_p
      \|\psi\|_{H^1}^2 .
\]
The prefactor tends to zero as $R,L\to\infty$, so $T$ is a norm limit of
compact operators, hence compact.

Now fix $a>0$.  If the spectral subspace of $H_W$ in $(-\infty,-a]$ were
infinite dimensional, it would contain an orthonormal sequence bounded in
$H^1$ by \eqref{eq:formbound}.  A subsequence converges weakly to zero in
$H^1$, so its $T$-images converge strongly to zero; but for unit vectors
in that subspace $h_W[\psi]\ge-a\ge-\|T\psi\|_2^2$, forcing
$\|T\psi\|_2^2\ge a$, a contradiction.  Thus every such subspace is finite
dimensional.

Finally test the weak form equation against smooth compactly supported
functions: this gives the stated distributional equation.  Its potential
term is locally integrable because $W\in L^p_{\mathrm{loc}}$ and $\psi$ is
locally bounded.
\end{proof}

\subsection{Young's inequality for matrices}
This is the scalar inequality $yz-z^q\le D_\sigma y^p$ promoted to
matrices, together with its equality condition.

\begin{lemma}[Trace Young inequality]\label{lem:young}
For positive semidefinite matrices $W,Z$ of the same finite size, put
\begin{equation}\label{eq:youngdefect}
 Y(W,Z)=D_\sigma\tr(W^p)-\tr(WZ)+\tr(Z^q).
\end{equation}
Then $Y(W,Z)\ge0$, with equality if and only if $W=qZ^r$.  Moreover
\begin{equation}\label{eq:youngcoercive}
 Y(W,Z)\ge\tfrac12\tr(Z^q)
       -(2^{p-1}-1)D_\sigma\tr(W^p).
\end{equation}
\end{lemma}

\begin{proof}
Maximise $z\mapsto yz-z^q$ over $z\ge0$ for fixed $y\ge0$.  The maximum
$D_\sigma y^p$ is attained exactly at $y=qz^r$; this is the scalar Young
inequality behind $D_\sigma$ in \eqref{eq:Dconst}.  Choose an orthonormal
eigenbasis $(v_j)$ of $Z$ with eigenvalues $z_j$, and set
$y_j=\langle v_j,Wv_j\rangle$.  By \eqref{eq:youngdefect} and the scalar
inequality,
\[
 \tr(WZ)-\tr(Z^q)
 =\sum_j (y_j z_j - z_j^q)
 \le D_\sigma\sum_j y_j^p .
\]
Because $t^p$ is strictly convex on $[0,\infty)$,
$\sum_j y_j^p\le\sum_j\langle v_j,W^p v_j\rangle=\tr(W^p)$.
Thus $Y(W,Z)\ge0$.  Equality in the convexity step forces every $v_j$ to
lie in an eigenspace of $W$; equality in the scalar step then forces
$Wv_j=qz_j^r v_j$ for every $j$, i.e.\ $W=qZ^r$.  This covers zero
eigenvalues as well.

For \eqref{eq:youngcoercive}, apply the already proven $Y\ge0$ to the pair
$(2W,Z)$ and expand:
$\tr(2WZ)\le\tfrac12\tr(Z^q)+2^{p-1}D_\sigma\tr(W^p)$, which rearranges
to \eqref{eq:youngcoercive}.
\end{proof}

% =====================================================================
\section{The trace engine: an exact integrated deficit}\label{sec:trace}
% =====================================================================

The action estimate compares a matrix $N$ against its power $N^q$.  The
classical inequality is $\tr(NV)\ge\tr(N^q)$ under a symbol condition, but
equality analysis needs the exact size of the gap.  The engine below gives
that gap as an integral of two manifestly nonnegative squares.  The point
is that it is an identity, not just an inequality, so a vanishing total
gap forces each square to vanish.

Throughout this section $0<\sigma<1$, $\beta=\sigma-\tfrac12$, and
$p=1+\sigma$, $r=1/\sigma$, $q=1+r$ as in \eqref{eq:params}.

\subsection{The normalisation constant}
Define $c_\sigma>0$ by
\begin{equation}\label{eq:trace:normalization}
 c_\sigma^{-1}
 =\int_0^\infty t^\beta
       \bigl(t^{-1/2}-(t+1)^{-1/2}\bigr)\,dt .
\end{equation}
The integrand is $O(t^{\sigma-1})$ near $0$ and $O(t^{\sigma-2})$ near
$\infty$, so the integral converges (this uses $0<\sigma<1$).  A change of
variable gives the scaling identity we will use constantly:
\begin{equation}\label{eq:trace:scaling}
 c_\sigma\int_0^\infty t^\beta
       \bigl(t^{-1/2}-(t+y)^{-1/2}\bigr)\,dt=y^\sigma,
 \qquad y\ge0 .
\end{equation}
Indeed, substitute $t=yu$; the left side becomes
$c_\sigma y^{\beta+1/2}\int_0^\infty u^\beta(u^{-1/2}-(u+1)^{-1/2})\,du
=y^\sigma$ by \eqref{eq:trace:normalization} and $\beta+\tfrac12=\sigma$.

\subsection{Statement of the engine}

\begin{theorem}[Exact integrated trace deficit]\label{thm:trace}
Let $n\ge1$ and let $B,V\in\C^{n\times n}$ be Hermitian.  Suppose
\begin{equation}\label{eq:trace:symbol}
 Y_s=(sI-B)^2,\qquad X_s=Y_s+V\ge bI\quad(s\in\R)
\end{equation}
for some $b>0$.  Define
\begin{equation}\label{eq:trace:resolvents}
 R(t)=\frac1\pi\int_\R(X_s+tI)^{-1}\,ds,
 \qquad
 P(t)=t^{-1/2}I-R(t),\qquad t>0,
\end{equation}
and
\begin{equation}\label{eq:trace:N}
 N=c_\sigma\int_0^\infty t^\beta P(t)\,dt .
\end{equation}
Then these integrals converge absolutely in matrix norm, $R(t)>0$, and
there are continuous Hermitian matrices $L(t)=R(t)^{-1}>0$ and $Z(t)$
with
\begin{equation}\label{eq:trace:LZ}
 V+tI=L^2+Z^2+i[B,Z],
 \qquad
 LZ+ZL+i[B,L]=0 .
\end{equation}
If $N\ge0$, then
\begin{equation}\label{eq:trace:deficit}
 \begin{aligned}
 \tr(NV)-\tr(N^q)
 =c_\sigma\int_0^\infty t^\beta\Bigl(
   &\tr(RZ^2)\\
   &+q\bigl\|R^{-1/2}-R^{1/2}(tI+N^r)^{1/2}\bigr\|_{\HS}^2
   \Bigr)\,dt .
 \end{aligned}
\end{equation}
In particular $\tr(NV)\ge\tr(N^q)$, and equality holds if and only if both
integrands vanish for every $t>0$.
\end{theorem}

The hypothesis concerns the full symbol $X_s$; the perturbation $V$ itself
need not be positive.  The proof has three parts: recover $L,Z$ from
half-line boundary values, evaluate a logarithmic determinant integral,
and complete a matrix square.

\subsection{Step 1: half-line boundary maps}
Quadratic growth of $X_s$ together with \eqref{eq:trace:symbol} gives an
integrable bound for $X_s^{-1}$, hence $R(t)>0$ and $R(t)=O(1)$ as
$t\downarrow0$.  Diagonalising $B$ gives the free identity
$\frac1\pi\int_\R(Y_s+tI)^{-1}\,ds=t^{-1/2}I$.  For $t\ge1$ both
$X_s+tI$ and $Y_s+tI$ dominate $c(s^2+t)I$ with $c>0$ independent of
$s,t$; the resolvent identity bounds the difference of inverses by
$C(s^2+t)^{-2}$.  Consequently
\begin{equation}\label{eq:trace:Pbounds}
 \|P(t)\|_{\op}=
 \begin{cases}
  O(t^{-1/2}), & t\downarrow0,\\
  O(t^{-3/2}), & t\to\infty,
 \end{cases}
\end{equation}
which proves convergence of $N$ in \eqref{eq:trace:N}.

Fix $t>0$ and write $\nabla=\partial_x-iB$.  Consider
\begin{equation}\label{eq:trace:ode}
 (-\nabla^2+V+tI)\psi=0 .
\end{equation}
For each $v\in\C^n$ there is a unique $H^1$ solution on either half-line
with $\psi(0)=v$.  The sesquilinear form
$a_t(\phi,\psi)=\int(\langle\nabla\phi,\nabla\psi\rangle
+\langle\phi,(V+tI)\psi\rangle)\,dx$ is coercive on $H^1(\R;\C^n)$, since
its Fourier symbol $X_s+tI$ dominates $c_t(1+s^2)I$; zero extension gives
coercivity on the trace-zero subspaces of each half-line, and the Riesz
representation theorem supplies the unique weak solution.  The equation
gives $\psi''\in L^2$, so $\psi\in H^2$, and both $\psi$ and $\psi'$
vanish at the infinite end.

Define the right and left boundary matrices $S,T$ by
\begin{equation}\label{eq:trace:boundary}
 \nabla\psi(0+)=-Sv,\qquad \nabla\psi(0-)=Tv .
\end{equation}
Integrating by parts gives
$a_t(\psi_v,\psi_w)=\langle v,Sw\rangle$ on the right and
$=\langle v,Tw\rangle$ on the left, so $S,T$ are Hermitian.  Translation
and uniqueness show $\nabla\psi=-S\psi$ throughout the right half-line
and $\nabla\psi=T\psi$ throughout the left half-line.  Substituting into
\eqref{eq:trace:ode} with arbitrary boundary data,
\begin{equation}\label{eq:trace:riccati}
 V+tI=S^2+i[B,S]=T^2-i[B,T] .
\end{equation}

Glue the two half-line solutions with the same boundary value $v$.  The
result is continuous with derivative jump $-(S+T)v$, so
$(-\nabla^2+V+tI)\psi=\delta_0(S+T)v$.  Fourier inversion with kernel
$e^{-isx}$ gives
\[
 v=\frac1{2\pi}\int_\R(X_s+tI)^{-1}\,ds\,(S+T)v
   =\tfrac12R(t)(S+T)v .
\]
The inverse symbol is $O(s^{-2})$, so its Fourier inverse is continuous
and evaluation at zero is legitimate.  Therefore
\begin{equation}\label{eq:trace:LZdef}
 L:=\tfrac12(S+T)=R^{-1}>0,
 \qquad Z:=\tfrac12(S-T)
\end{equation}
satisfy \eqref{eq:trace:LZ}: adding and subtracting the two halves of
\eqref{eq:trace:riccati} gives the first equation, and the second holds
because $LZ+ZL=\tfrac12(S^2-T^2)$ and $i[B,L]=\tfrac i2[B,S+T]$.  The
equation $LZ+ZL+i[B,L]=0$ determines $Z$ uniquely from $L>0$ (in an
eigenbasis of $L$ the map $Z\mapsto LZ+ZL$ multiplies the $(j,k)$ entry
by $L_{jj}+L_{kk}>0$), which also proves continuity of $Z(t)$.

Two consequences will be used.  First, the diagonal of the second
relation gives $\tr Z=0$.  Second, multiplying $i[R,B]=-(ZR+RZ)$ (the
$R$-conjugate of the second relation) into the trace of $R(V+tI)$ gives
the signed identity
\begin{equation}\label{eq:trace:RV}
 \tr\bigl(R(V+tI)\bigr)=\tr L-\tr(RZ^2),
\end{equation}
where the sign comes from $i\,\tr(R[B,Z])=-2\tr(RZ^2)$.

\subsection{Step 2: a logarithmic determinant integral}
Define
\begin{equation}\label{eq:trace:F}
 F(t)=\frac{\tr V}{\sqrt t}
       -\frac1\pi\int_\R
        \log\frac{\det(X_s+tI)}{\det(Y_s+tI)}\,ds .
\end{equation}
Both determinants are positive and the logarithmic ratio is
$O(|s|^{-2})$ at infinity; its $t$-derivative is the trace of a resolvent
difference with an integrable bound, locally uniform in $t>0$.  Hence $F$
is continuously differentiable.  The first relation in
\eqref{eq:trace:riccati} factors the symbol,
\[
 X_s+tI=(sI-B+iS)(sI-B-iS).
\]
If $u_j+iv_j$ are the eigenvalues of $B+iS$, counted with algebraic
multiplicity, then the square-integrability argument behind
\eqref{eq:trace:ode} forces $v_j>0$, and
$\det(X_s+tI)=\prod_j((s-u_j)^2+v_j^2)$.  The denominator has the same
form with centres the eigenvalues of $B$ and widths $\sqrt t$.
Translations of individual logarithms only change symmetric-cutoff
integrals by $o(1)$ (the derivative
$\frac{d}{da}\int_{-M}^M h(s-a)\,ds=h(M+a)-h(M-a)\to0$ uniformly for
bounded $a$).  Using the one-factor integral
$\int_\R\log\frac{s^2+v^2}{s^2+w^2}\,ds=2\pi(v-w)$ for $v,w>0$, and
$\sum_jv_j=\tr S=\tr L$, we obtain
\begin{equation}\label{eq:trace:Feval}
 F(t)=\frac{\tr V}{\sqrt t}-2\tr(L-\sqrt t I).
\end{equation}

Set
\begin{equation}\label{eq:trace:G}
 G(t)=\tr(L-2\sqrt t I+tR).
\end{equation}
Both $F,G\ge0$: for $G$ apply $\lambda-2\sqrt t+t/\lambda\ge0$ to the
positive eigenvalues $\lambda$ of $L$; for $F$ use
\eqref{eq:trace:Feval} with $\tr V=\tr(L^2+Z^2)-nt$ (the trace of the
first relation in \eqref{eq:trace:LZ}) to get
$F(t)=t^{-1/2}\tr((L-\sqrt t I)^2+Z^2)\ge0$.  Combining
\eqref{eq:trace:RV} and \eqref{eq:trace:Feval} yields the identity
\begin{equation}\label{eq:trace:FG}
 \tr(PV)=F+G+\tr(RZ^2),
\end{equation}
with every term on the right nonnegative.  By \eqref{eq:trace:Pbounds},
$F$, $G$, and $\tr(RZ^2)$ are integrable against $t^\beta dt$, and
$t^{\beta+1}F(t)$ vanishes at both ends.  Differentiating
\eqref{eq:trace:F} and using \eqref{eq:trace:Feval}--\eqref{eq:trace:G},
\[
 F'(t)=-\frac{\tr V}{2t^{3/2}}+\tr P(t),
 \qquad
 tF'(t)=-\tfrac12F(t)-G(t).
\]
Integration by parts with the vanishing endpoint limits gives
$\int_0^\infty t^\beta G\,dt=\sigma\int_0^\infty t^\beta F\,dt$; since
$\sigma=1/r$ and $\int F=r\int G$,
\begin{equation}\label{eq:trace:weighted}
 \tr(NV)=q\,c_\sigma\int_0^\infty t^\beta G(t)\,dt
       +c_\sigma\int_0^\infty t^\beta\tr(RZ^2)\,dt .
\end{equation}

\subsection{Step 3: the noncommutative square}
For a positive semidefinite matrix $D$, set
\[
 F_0(t,D)=\frac{\tr D}{\sqrt t}
       -2\tr\bigl((tI+D)^{1/2}-\sqrt t I\bigr).
\]
For a scalar eigenvalue $d\ge0$ the summand is
$\int_0^d(t^{-1/2}-(t+y)^{-1/2})\,dy$; Tonelli's theorem and
\eqref{eq:trace:scaling} give
\begin{equation}\label{eq:trace:F0}
 c_\sigma\int_0^\infty t^\beta F_0(t,D)\,dt=\frac1p\tr(D^p).
\end{equation}
Now expand the Hilbert--Schmidt square without any commutation (the cross
factors $R^{-1/2}R^{1/2}$ cancel, and cyclicity handles the rest):
\begin{align}
 \mathcal D_D(t)
  &=\bigl\|R^{-1/2}-R^{1/2}(tI+D)^{1/2}\bigr\|_{\HS}^2 \notag\\
  &=\tr\bigl(R^{-1}+R(tI+D)-2(tI+D)^{1/2}\bigr) \notag\\
  &=G(t)-\tr(P(t)D)+F_0(t,D).
  \label{eq:trace:square}
\end{align}
All three terms are integrable by \eqref{eq:trace:Pbounds},
\eqref{eq:trace:FG}, and \eqref{eq:trace:F0}.  Integrating
\eqref{eq:trace:square} with weight $t^\beta$ and using
$c_\sigma\int t^\beta P= N$ gives
\begin{equation}\label{eq:trace:duality}
 c_\sigma\int_0^\infty t^\beta G\,dt
 =\tr(ND)-\frac1p\tr(D^p)
  +c_\sigma\int_0^\infty t^\beta\mathcal D_D\,dt .
\end{equation}
When $N\ge0$ choose $D=N^r$.  Since $rp=q$, the first two terms equal
$\bigl(1-\tfrac1p\bigr)\tr(N^q)=\tr(N^q)/q$.  Substituting into
\eqref{eq:trace:weighted} proves \eqref{eq:trace:deficit}.  Both integrands
in \eqref{eq:trace:deficit} are nonnegative, so equality forces them to
vanish for almost every $t$, hence by continuity for every $t>0$.

Finally, if $B$ and $V$ commute the symbol condition implies $V\ge bI$;
simultaneous diagonalisation gives $R(t)=(tI+V)^{-1/2}$, $N=V^\sigma$,
$Z=0$, and both remainders vanish, as scalar equality requires.

% =====================================================================
\section{The field and the connecting path}\label{sec:field}
% =====================================================================

To run the complete-the-square step of the action estimate we need two
objects: a symmetric matrix field $M_\delta$ that mimics the power
$(K^2-B^2)^\sigma$ near values that matter, and a symmetric path $B(x)$
running from $K$ to $-K$ that solves an ODE of the form
$\varepsilon B'=M_\delta(B)-AA^T$.  This section constructs both.  Set
$\delta>0$ and recall $\beta=\sigma-\tfrac12$, and let $c_\sigma$ be the
constant of \eqref{eq:trace:normalization}.

\subsection{A scalar building block}
On $y\ge0$, define
\begin{equation}\label{eq:field:f}
 f_\delta(y)=\frac{c_\sigma}{\pi}\int_0^\infty t^\beta
   \Bigl(\frac1{t+\delta}-\frac1{t+\delta+y}\Bigr)\,dt,
\end{equation}
and extend to $y<0$ by the tangent line $f_\delta'(0)y$.  For $y\ge0$,
\begin{equation}\label{eq:field:fderivative}
 f_\delta'(y)=a_\sigma(y+\delta)^{\beta-1},
 \qquad
 a_\sigma=\frac{c_\sigma}{\pi}\int_0^\infty
   \frac{u^\beta}{(1+u)^2}\,du>0 .
\end{equation}
Both integrals converge because $-\tfrac12<\beta<\tfrac12$.  Thus $f_\delta$
is $C^1$, strictly increasing, and concave on $y\ge0$; the tangent
extension makes it $C^1$ on all of $\R$ and lets the functional calculus
below accept arguments with negative eigenvalues.

Set
\begin{equation}\label{eq:field:Q}
 Q_s(B)=s^2I-2sB+K^2,
\end{equation}
and
\begin{equation}\label{eq:field:M}
 M_\delta(B)=\lim_{R\to\infty}\int_{-R}^{R}
   \bigl(f_\delta(Q_s(B))-f_\delta(s^2)I\bigr)\,ds,
 \qquad B\in\Sym_n,
\end{equation}
where the symmetric cutoff cancels the leading odd term
$-2sf_\delta'(s^2)B$.  The scalar counterpart is
\begin{equation}\label{eq:field:mu}
 \mu_\delta(z)=\int_\R\bigl(f_\delta(s^2+z)-f_\delta(s^2)\bigr)\,ds,
 \qquad z\in\R .
\end{equation}

\begin{proposition}[Regularised field]\label{prop:field}
The limit \eqref{eq:field:M} exists locally uniformly, and
$M_\delta:\Sym_n\to\Sym_n$ is locally Lipschitz with symmetric values.
There is a $C^1$ function $J_\delta:\Sym_n\to\R$, $J_\delta(0)=0$, with
\begin{equation}\label{eq:field:primitive}
 dJ_\delta(B)[H]=-\tr\bigl(M_\delta(B)H\bigr).
\end{equation}
The scalar function $\mu_\delta$ is locally Lipschitz and strictly
increasing with $\mu_\delta(0)=0$, and for $z\ge0$
\begin{equation}\label{eq:field:mupositive}
 \mu_\delta(z)=(z+\delta)^\sigma-\delta^\sigma .
\end{equation}
For every real unit vector $e$ and every standard basis vector $e_i$,
\begin{align}
 e^{\mathsf T}M_\delta(B)e
  &\le\mu_\delta\bigl(e^{\mathsf T}K^2e-(e^{\mathsf T}Be)^2\bigr),
  \label{eq:field:direction}\\
 Be_i=ye_i\quad&\Longrightarrow\quad
 M_\delta(B)e_i=\mu_\delta(k_i^2-y^2)\,e_i .
  \label{eq:field:coordinate}
\end{align}
For every diagonal sign matrix $S$,
\begin{equation}\label{eq:field:equivariance}
 M_\delta(SBS)=S\,M_\delta(B)\,S .
\end{equation}
Finally,
\begin{equation}\label{eq:field:endpoints}
 J_\delta(-K)-J_\delta(K)
 =\sum_{i=1}^n\int_{-k_i}^{k_i}\mu_\delta(k_i^2-s^2)\,ds
 \;\longrightarrow\;
 b_\sigma\sum_{i=1}^n k_i^{2\sigma+1}
 \quad(\delta\downarrow0).
\end{equation}
\end{proposition}

\begin{proof}
\emph{Matrix calculus bound.}  If $h$ is Lipschitz on an interval
containing the spectra of Hermitian $X,Y$, then
\begin{equation}\label{eq:field:HS}
 \|h(X)-h(Y)\|_{\HS}\le\operatorname{Lip}(h)\,\|X-Y\|_{\HS}.
\end{equation}
Indeed, take eigenbases $e_i$ of $X$ and $v_j$ of $Y$; the mixed entries
of $h(X)-h(Y)$ and of $X-Y$ are $(h(\lambda_i)-h(\nu_j))\langle e_i,v_j\rangle$
and $(\lambda_i-\nu_j)\langle e_i,v_j\rangle$; square the scalar Lipschitz
inequality and sum.

\emph{Convergence of the field.}  On a bounded set of $B$'s the spectrum
of $E_s(B)=-2sB+K^2$ lies in $[-a|s|,a|s|]$ for all large $|s|$.  For
$h_s(y)=f_\delta(s^2+y)-f_\delta(s^2)-f_\delta'(s^2)y$,
\eqref{eq:field:fderivative} gives
$\sup_{|y|\le a|s|}|h_s'(y)|=O(|s|^{2\beta-3})$.  By \eqref{eq:field:HS},
\[
 f_\delta(Q_s(B))-f_\delta(s^2)I=-2s f_\delta'(s^2)B+T_s(B),
\]
where $\|T_s(B)\|_{\HS}$ and the local Lipschitz constant of $T_s$ are
$O(|s|^{2\beta-2})$ (include $f_\delta'(s^2)K^2$ in $T_s$ for the size
bound, and apply \eqref{eq:field:HS} to $h_s(E_s(B))$ using
$E_s(B_1)-E_s(B_2)=-2s(B_1-B_2)$).  Since $2\beta-2<-1$, these remainders
are integrable at infinity; the odd term cancels at each symmetric cutoff,
and on bounded $s$-intervals \eqref{eq:field:HS} applies directly to
$f_\delta$.  This gives the stated convergence and regularity.

\emph{Primitive.}  Let $g_\delta'=f_\delta$.  Trace differentiation gives
$d\,\tr(g_\delta(X))[H]=\tr(f_\delta(X)H)$ (true for polynomials by
cyclicity, then for $g_\delta$ by $C^1$ approximation on compact spectral
intervals).  For $s\ne0$ the function
\[
 B\mapsto\frac{\tr\bigl(g_\delta(Q_s(B))-g_\delta(K^2)\bigr)}{2s}
        +f_\delta(s^2)\tr B
\]
has differential $-\tr\bigl((f_\delta(Q_s(B))-f_\delta(s^2)I)H\bigr)$; at
$s=0$ the differential is constant in $B$ and also has a primitive.  Hence
the negative of each finite-cutoff field has vanishing integral around
every closed piecewise smooth curve in $\Sym_n$, i.e.\ it is a gradient.
Local uniform convergence passes this to $-M_\delta$, and path integration
from $0$ defines $J_\delta$, proving \eqref{eq:field:primitive}.

\emph{Scalar bounds and profile.}  For bounded $z$, the integrand in
\eqref{eq:field:mu} and its Lipschitz constant in $z$ are $O(|s|^{2\beta-2})$
at infinity, so $\mu_\delta$ is locally Lipschitz and absolutely
convergent; strict increase and $\mu_\delta(0)=0$ follow from $f_\delta$.
For $z\ge0$, Tonelli applied to \eqref{eq:field:f} gives
\[
 \mu_\delta(z)=c_\sigma\int_0^\infty t^\beta
   \bigl((t+\delta)^{-1/2}-(t+\delta+z)^{-1/2}\bigr)\,dt,
\]
which the scaling identity \eqref{eq:trace:scaling} evaluates as
\eqref{eq:field:mupositive}.

\emph{Direction and coordinate.}  For fixed $a,z$, translations under a
symmetric cutoff are legitimate:
\begin{equation}\label{eq:field:shift}
 \lim_{R\to\infty}\int_{-R}^{R}
   \bigl(f_\delta((s-a)^2+z)-f_\delta(s^2)\bigr)\,ds=\mu_\delta(z),
\end{equation}
because $F_z(s)=f_\delta(s^2+z)$ is even and $F_z'(s)=O(|s|^{2\beta-1})\to0$,
so the difference of shifted and unshifted cutoffs is bounded by
$|a|^2\sup_{R-|a|\le u\le R+|a|}|F_z'(u)|\to0$.  Concavity of $f_\delta$
in an eigenbasis of $Q_s(B)$ gives
$e^{\mathsf T}f_\delta(Q_s(B))e\le f_\delta(e^{\mathsf T}Q_s(B)e)$, whose
scalar argument is $(s-e^{\mathsf T}Be)^2+e^{\mathsf T}K^2e-(e^{\mathsf T}Be)^2$.
Integrating with \eqref{eq:field:shift} gives
\eqref{eq:field:direction}; if $Be_i=ye_i$, the functional calculus is
exact on $e_i$, giving \eqref{eq:field:coordinate}.  Since $SK^2S=K^2$,
conjugation inside the integral gives \eqref{eq:field:equivariance}.

\emph{Endpoints.}  On diagonal arguments, \eqref{eq:field:primitive} and
\eqref{eq:field:coordinate} read
$dJ_\delta(\diag(b_1,\dots,b_n))=-\sum_i\mu_\delta(k_i^2-b_i^2)\,db_i$.
Integrating from $K$ to $-K$ gives the first equality in
\eqref{eq:field:endpoints}.  For $z\ge0$, $0\le\mu_\delta(z)\le z^\sigma$
and $\mu_\delta(z)\to z^\sigma$; dominated convergence and the substitution
$s=k_iu$ give $\int_{-k_i}^{k_i}(k_i^2-s^2)^\sigma ds=b_\sigma k_i^{2\sigma+1}$,
hence the limit.
\end{proof}

\subsection{Positive values of the field}
The next proposition is the only place where the sign of an argument of
$M_\delta$ matters.  It says that if $M_\delta(B)$ is itself a positive
matrix (as it will be once it equals $AA^{\mathsf T}$), the whole symbol
$Q_s(B)$ is positive definite and the trace engine applies.

\begin{proposition}[Positive values of the field]\label{prop:field-pos}
If $M=M_\delta(B)\ge0$, then
\begin{equation}\label{eq:field:posbounds}
 Q_s(B)\ge0\ (s\in\R),
 \qquad 0\le M\le\kappa^{2\sigma}I .
\end{equation}
Set $V=K^2-B^2+\delta I$.  Then $(sI-B)^2+V=Q_s(B)+\delta I\ge\delta I$,
and with $R(t)$ and $N$ as in \eqref{eq:trace:resolvents}--\eqref{eq:trace:N}
one has $N=M+\delta^\sigma I$.  In particular the hypotheses of
Theorem~\ref{thm:trace} hold with this $B$ and $V$.
\end{proposition}

\begin{proof}
By \eqref{eq:field:direction} and strict increase of $\mu_\delta$,
positivity of $M$ forces $e^{\mathsf T}K^2e\ge(e^{\mathsf T}Be)^2$ for every
unit $e$.  Completing the scalar square,
\[
 e^{\mathsf T}Q_s(B)e=(s-e^{\mathsf T}Be)^2
   +e^{\mathsf T}K^2e-(e^{\mathsf T}Be)^2\ge0,
\]
so $Q_s(B)\ge0$ as a Hermitian matrix on $\C^n$.  Also
$M\le\mu_\delta(\kappa^2)I\le\kappa^{2\sigma}I$ by \eqref{eq:field:direction}
and \eqref{eq:field:mupositive}.

Write $Y_s=(sI-B)^2$.  By diagonalisation of $B$ and \eqref{eq:field:shift},
the scalar subtraction $f_\delta(s^2)I$ in \eqref{eq:field:M} may be
replaced by $f_\delta(Y_s)$.  Since $Q_s(B)$ and $Y_s$ are both positive,
\eqref{eq:field:f} gives their difference under $f_\delta$ as a Stieltjes
integral
\[
 \frac{c_\sigma}{\pi}\int_0^\infty t^\beta
 \Bigl((Y_s+(t+\delta)I)^{-1}-(Q_s(B)+(t+\delta)I)^{-1}\Bigr)\,dt .
\]
For fixed $B,\delta$ both inverted matrices are bounded below by
$c(1+t+s^2)I$, and their difference is $K^2-B^2$, so the resolvent
identity bounds the integrand by $Ct^\beta(1+t+s^2)^{-2}$; integrating in
$s$ gives $C't^\beta(1+t)^{-3/2}$, integrable on $(0,\infty)$.  Fubini
therefore applies and yields
\[
 M=c_\sigma\int_0^\infty t^\beta
   \bigl((t+\delta)^{-1/2}I-R(t)\bigr)\,dt .
\]
Subtracting this from \eqref{eq:trace:N} and using \eqref{eq:trace:scaling}
gives $N=M+\delta^\sigma I$.
\end{proof}

\subsection{The connecting path}
The final object of this section walks $B$ from $K$ to $-K$ while tracking a
prescribed density $AA^{\mathsf T}$.  This is where topology enters: a
nonlinear equation for $C$ must be solved, and the only obstruction would
be a parity count of zeros.

\begin{proposition}[Connecting path]\label{prop:path}
Let $\Omega=(x_-,x_+)$ be bounded and let $U\in H^1_0(\Omega;\R^{n\times d})$,
$n,d\ge1$, satisfy $\int_\Omega UU^{\mathsf T}\,dx=I_n$.  For every
$\delta,\varepsilon>0$ there are a real lower triangular matrix
$C_\varepsilon\in\R^{n\times n}$ and a path
$B_\varepsilon\in C^1(\overline\Omega;\Sym_n)$ such that, with
$A_\varepsilon=C_\varepsilon U$,
\begin{equation}\label{eq:path:ode}
 \varepsilon B_\varepsilon'
   =M_\delta(B_\varepsilon)-A_\varepsilon A_\varepsilon^{\mathsf T},
 \qquad B_\varepsilon(x_-)=K,\quad B_\varepsilon(x_+)=-K,
\end{equation}
and
\begin{equation}\label{eq:path:bounds}
 \sup_{x\in\overline\Omega}\|B_\varepsilon(x)\|_{\op}\le\kappa,
 \qquad
 \sup_{0<\varepsilon\le1}\|C_\varepsilon\|_{\HS}<\infty,
\end{equation}
the second bound for fixed $\delta,K,\Omega$.
\end{proposition}

\begin{proof}
\emph{A cutoff and a homotopy.}  Use the continuous representative of $U$.
Choose a smooth nonnegative compactly supported cutoff $\chi$ on $\Sym_n$
equal to $1$ on $\{B:\|B\|_{\op}\le\kappa\}$ and invariant under conjugation
by diagonal sign matrices (averaging over the finite sign group produces
one).  The truncated field $\widehat M=\chi M_\delta$ is bounded and
globally Lipschitz.  Fix $\varepsilon>0$.  For $C$ lower triangular and
$\theta\in[0,1]$, the equation
\begin{equation}\label{eq:path:homotopy}
 \varepsilon B'=\theta\widehat M(B)-(CU)(CU)^{\mathsf T},
 \qquad B(x_-)=K,
\end{equation}
has a unique $C^1$ solution on $\overline\Omega$; define the continuous
endpoint map $\Phi(C,\theta)=B(x_+)+K\in\Sym_n$.

\emph{Sign symmetry.}  Let $\mathcal G$ be the group of diagonal sign
matrices, acting by $C\mapsto SC$ on lower triangular matrices and by
conjugation on $\Sym_n$.  Equivariance of the field and uniqueness of the
ODE give
\begin{equation}\label{eq:path:equivariance}
 \Phi(SC,\theta)=S\,\Phi(C,\theta)\,S .
\end{equation}

\emph{Three facts about the zero set.}
(i) \emph{Compactness.}  At a zero, integrating \eqref{eq:path:homotopy}
gives $CC^{\mathsf T}=2\varepsilon K+\theta\int_\Omega\widehat M(B(x))\,dx$,
so the zeros lie in a bounded set and their joint set in
$\R^{n\times n}_{\text{low}}\times[0,1]$ is compact.  For
$0<\varepsilon\le1$ this also gives the uniform bound
\begin{equation}\label{eq:path:Cbound}
 \|C\|_{\HS}^2\le2\tr K
   +n|\Omega|\sup_{B\in\Sym_n}\|\widehat M(B)\|_{\op}.
\end{equation}
(ii) \emph{Freeness.}  No row of $C$ can vanish at a zero.  Otherwise the
reflection changing only row $i$ would fix $C$; by \eqref{eq:path:equivariance}
and ODE uniqueness $Be_i=ye_i$ throughout, and \eqref{eq:field:coordinate}
gives $\varepsilon y'=\theta\chi(B(x))\mu_\delta(k_i^2-y^2)$ with $y(x_-)=k_i$.
Uniqueness forces $y\equiv k_i$, contradicting $y(x_+)=-k_i$.  Thus
$\mathcal G$ acts freely on the zero set.
(iii) \emph{One regular orbit at $\theta=0$.}  There
$\Phi(C,0)=2K-\varepsilon^{-1}CC^{\mathsf T}$.  Its lower triangular zeros
are exactly $C=\diag(d_1,\dots,d_n)$ with $d_i=\pm\sqrt{2\varepsilon k_i}$:
the first row of $CC^{\mathsf T}=2\varepsilon K$ fixes $|d_1|$ and forces the
rest of the first column to vanish, and induction proceeds.  These $2^n$
zeros form one sign orbit.  At each, the derivative in a lower triangular
direction $H$ has diagonal entries $-2\varepsilon^{-1}d_iH_{ii}$ and lower
off-diagonal entries $-\varepsilon^{-1}d_jH_{ij}$ ($i>j$), an isomorphism
$\R^{n\times n}_{\text{low}}\to\Sym_n$.

\emph{Continuation by parity.}  Suppose there were no zero at $\theta=1$.
Compactness and freeness give a bounded invariant open set
$O\subset\R^{n\times n}_{\text{low}}$ containing the projection of every
zero, with closure where all rows are nonzero, and a positive lower bound
for $\|\Phi\|_{\HS}$ on $(\partial O\times[0,1])\cup(\overline O\times\{1\})$.
Reparametrise $\theta$ to agree with the initial map on an initial collar
and retain this gap; extend $F$ constantly past both ends, smooth by
convolution, splice to the initial map on a smaller collar, and average in
the form $\Psi(C,\theta)=\frac1{|\mathcal G|}\sum_S S\Psi_0(SC,\theta)S$ to
get a smooth equivariant approximation $\Psi$ still equal to the initial map
on the collar and close enough to preserve the gap.

Freeness allows equivariant perturbations with arbitrary target directions:
at $C_0\in\overline O$ choose a ball whose sign translates are disjoint and
a bump $b$ supported there with $b(C_0)=1$; then
$V_H(C)=\sum_S b(SC)SHS$ is equivariant with $V_H(C_0)=H$.  Finitely many
such maps $V_1,\dots,V_N$ span $\Sym_n$ at every point.  Let $h\ge0$ be
smooth, vanish near $0$, and positive off the collar, and set
$\Psi_\alpha=\Psi+h(\theta)\sum_j\alpha_jV_j$.  For small $\alpha$ the gap
persists; at a zero with $0<\theta<1$ the full differential in
$(C,\theta,\alpha)$ is onto (parameter directions span the target where
$h>0$; the $C$-differential is the above isomorphism where $h=0$).  Hence
the interior zero set is a smooth manifold, and Sard's theorem
\cite{Sard1942} applied to projection onto the parameter ball gives
arbitrarily small $\alpha$ for which $0$ is a regular value.

For such $\alpha$ the zero set in $O\times[0,1]$ is a compact smooth
one-dimensional manifold with boundary.  The gap excludes boundary zeros
except at $\theta=0$, where the initial collar gives exactly the $2^n$
regular zeros.  The free action has a quotient that is again a compact
one-dimensional manifold with boundary (a neighbourhood disjoint from its
nontrivial translates gives a local interval or half-interval chart); all
$2^n$ boundary points form one orbit, so the quotient has exactly one
boundary point.  A compact one-dimensional manifold is a finite union of
circles and closed intervals, which cannot have exactly one boundary point.
Contradiction.  Hence $\Phi(\cdot,1)$ has a zero.

\emph{Removing the cutoff.}  For the resulting path and any real unit $e$,
put $y=e^{\mathsf T}Be$.  Whenever $|y|>\kappa$, \eqref{eq:field:direction}
gives $e^{\mathsf T}M_\delta(B)e\le\mu_\delta(e^{\mathsf T}K^2e-y^2)<0$.
Since $\chi\ge0$ and $(CU)(CU)^{\mathsf T}\ge0$, \eqref{eq:path:ode} implies
$y'\le0$ there.  Both endpoints lie in $[-\kappa,\kappa]$; an excursion
above $\kappa$ cannot start from $\kappa$ while $y$ is nonincreasing, and an
excursion below $-\kappa$ cannot return to $-\kappa$ while $y$ is
nonincreasing.  So $\|B(x)\|_{\op}\le\kappa$ for all $x$, which forces
$\chi(B)=1$ and hence \eqref{eq:path:ode} with the untruncated field.
Equation~\eqref{eq:path:Cbound} supplies the remaining bound in
\eqref{eq:path:bounds}.
\end{proof}

% =====================================================================
\section{The finite-interval action estimate}\label{sec:action}
% =====================================================================

This is the central estimate.  It turns the trial functions into a
quantitative lower bound that measures, in operator norm, how far the
data are from the two algebraic equality relations.  The uniformity of the
coefficient $c_*$ in the row count is what will later allow an
a-priori-infinite eigenvalue list.

Fix positive integers $n,d$, a bounded interval $\Omega=(x_-,x_+)$, and a
diagonal $K=\diag(\ell_1,\dots,\ell_n)$ with $0<\ell_i\le\kappa$.  For
$A\in H^1_0(\Omega;\R^{n\times d})$ set $\rho=AA^{\mathsf T}$ and
\begin{equation}\label{eq:act:energy}
 \mathcal E_K(A)=\int_\Omega
  \bigl(\|A'\|_{\HS}^2+\tr(K^2\rho)-\tr(\rho^q)\bigr)\,dx .
\end{equation}
For pointwise values $A\in\R^{n\times d}$ and $B\in\Sym_n$, define the
algebraic defect
\begin{equation}\label{eq:act:defect}
 \Phi_K(A,B)=\min\Bigl\{1,\;
   \|[B,K^2]\|_{\op}^2+\bigl\|K^2-B^2-(AA^{\mathsf T})^r\bigr\|_{\op}^2
   \Bigr\}.
\end{equation}
Thus $\Phi_K=0$ encodes the two relations $[B,K^2]=0$ and
$\rho=(K^2-B^2)^\sigma$, while $A'-BA=0$ is the first-order relation.

\begin{theorem}[Quantitative action estimate]\label{thm:action}
Let $0<\sigma<1$ and $\kappa>0$.  There is $c_*=c_*(\sigma,\kappa)>0$ with
the following property.  For $n,d,\Omega,K$ as above, suppose
$U\in H^1_0(\Omega;\R^{n\times d})$ satisfies
$\int_\Omega UU^{\mathsf T}\,dx=I_n$.  For every $\eta>0$ there exist a
real lower triangular $C\in\R^{n\times n}$ and a path
$B\in C^1(\overline\Omega;\Sym_n)$ with
\begin{equation}\label{eq:act:endpointB}
 B(x_-)=K,\qquad B(x_+)=-K,\qquad
 \sup_{x\in\overline\Omega}\|B(x)\|_{\op}\le\kappa,
\end{equation}
such that, for $A=CU$,
\begin{equation}\label{eq:act:bound}
 \mathcal E_K(A)\ge b_\sigma\sum_{i=1}^n\ell_i^{2\gamma}
   +\int_\Omega\Bigl(\|A'-BA\|_{\HS}^2+c_*\Phi_K(A,B)\Bigr)\,dx-\eta .
\end{equation}
The constant $c_*$ is independent of $n,d,\Omega,U$ and of
$\min_i\ell_i$.
\end{theorem}

The proof has two ingredients.  The first is a constraint lemma, obtained
by feeding the trace engine (Proposition~\ref{prop:field-pos} plus
Theorem~\ref{thm:trace}) into a penalty.  The second completes the square
along the path of Proposition~\ref{prop:path}; the boundary term there is
exactly $J_\delta(-K)-J_\delta(K)$, which tends to
$b_\sigma\sum\ell_i^{2\gamma}$.  We prove the constraint lemma first.

\subsection{The nonlinear term is controlled by the defect}
For $0<\delta\le1$, recall the field $M_\delta$ of Section~\ref{sec:field}
and the function $F(A,B)=\tr(\rho^q)-\tr\bigl((K^2-B^2)\rho\bigr)$,
$\rho=AA^{\mathsf T}$.

\begin{lemma}[Constraint estimate]\label{lem:constraint}
The constant $c_*(\sigma,\kappa)>0$ can be chosen so that whenever
$0<\delta\le1$, $\|B\|_{\op}\le\kappa$, and $M_\delta(B)=AA^{\mathsf T}$,
\begin{equation}\label{eq:act:constraint-bound}
 F(A,B)+c_*\Phi_K(A,B)\le E_{n,\delta},
 \qquad
 E_{n,\delta}
 =n\bigl(\delta\kappa^{2\sigma}+\delta^\sigma(\kappa^2+\delta)\bigr)
   +\delta^{2\sigma}.
\end{equation}
\end{lemma}

\begin{proof}
Write $\rho=M_\delta(B)\ge0$.  By Proposition~\ref{prop:field-pos},
\begin{equation}\label{eq:act:positive-symbol}
 Q_s:=s^2I-2sB+K^2\ge0\ (s\in\R),
 \qquad 0\le\rho\le\kappa^{2\sigma}I .
\end{equation}
Put $V=K^2-B^2+\delta I$.  With $R(t)$, $N$, $c_\sigma$ from
\eqref{eq:trace:resolvents}--\eqref{eq:trace:N}, Proposition~\ref{prop:field-pos}
also gives $N=\rho+\delta^\sigma I$.  Since
$(sI-B)^2+V=Q_s+\delta I\ge\delta I$ and $N\ge0$,
Theorem~\ref{thm:trace} applies and supplies $L(t)=R(t)^{-1}>0$ and
Hermitian $Z(t)$ with
\begin{equation}\label{eq:act:LZ}
 V+tI=L^2+Z^2+i[B,Z],\qquad LZ+ZL+i[B,L]=0,
\end{equation}
and the exact identity
\begin{equation}\label{eq:act:trace-deficit}
 \Delta:=\tr(NV)-\tr(N^q)
 =c_\sigma\int_0^\infty t^\beta\Bigl(\tr(RZ^2)
   +q\bigl\|R^{-1/2}-R^{1/2}(tI+N^r)^{1/2}\bigr\|_{\HS}^2\Bigr)\,dt\ge0 .
\end{equation}
Here the matrices may be complex Hermitian although $A,B$ are real; $i$ is
the imaginary unit.

\emph{A dimension-free bound from $[1,2]$.}  On $1\le t\le2$ one has
\begin{equation}\label{eq:act:Rbounds}
 a_\kappa I\le R(t)\le A_\kappa I:
\end{equation}
for the upper bound use $Q_s\ge0$ on a bounded interval and
$Q_s\ge\tfrac12s^2I$ for $|s|\ge\max\{1,4\kappa\}$; for the lower bound
restrict to $|s|\le1$ where $Q_s+(\delta+t)I\le(4+2\kappa+\kappa^2)I$.
These constants depend on $\kappa$ alone.  Also
$\|N\|_{\op}\le\kappa^{2\sigma}+1$, so $H_t=(tI+N^r)^{1/2}$ and $L(t)$ are
bounded in operator norm by constants depending only on $\sigma,\kappa$.
The two integrands control $\|Z(t)\|_{\HS}^2$ and $\|L(t)-H_t\|_{\HS}^2$
(use $\tr(RZ^2)=\tr(ZRZ)$ and
$R^{-1/2}-R^{1/2}H_t=R^{1/2}(L-H_t)$).  Averaging over $[1,2]$ gives some
$t$ in that interval with
\begin{equation}\label{eq:act:smallsquares}
 \|Z(t)\|_{\op}+\|L(t)-H_t\|_{\op}\le C_1\sqrt\Delta,
\end{equation}
$C_1=C_1(\sigma,\kappa)$; when $\Delta=0$ take a point where both
nonnegative integrands vanish.

\emph{Reading off the defect.}  Suppose $\Delta\le1$ and use this $t$.  The
first relation of \eqref{eq:act:LZ} gives
$V-N^r=(L^2-H_t^2)+Z^2+i[B,Z]$, whose norm is $\le C_2\sqrt\Delta$ by the
bounds above and \eqref{eq:act:smallsquares}.  The second relation gives
$\|[B,L]\|_{\op}\le2\|L\|_{\op}\|Z\|_{\op}$, and
$[B,V]=[B,L^2]+[B,Z^2]+i[B,[B,Z]]$ has the same bound.  Thus
\begin{equation}\label{eq:act:Vcoercivity}
 \|V-N^r\|_{\op}+\|[B,V]\|_{\op}\le C_3\sqrt\Delta .
\end{equation}
Now $N=\rho+\delta^\sigma I$ commutes with $\rho$, and $y\mapsto y^r$ is
Lipschitz on $[0,\kappa^{2\sigma}+1]$ because $r>1$, so
$\|N^r-\rho^r\|_{\op}\le C_4\delta^\sigma$.  Since $[B,V]=[B,K^2]$ and
$\delta\le\delta^\sigma$, \eqref{eq:act:Vcoercivity} yields
\[
 \|[B,K^2]\|_{\op}^2+\|K^2-B^2-\rho^r\|_{\op}^2\le C_5(\Delta+\delta^{2\sigma}).
\]
Choosing $c_*>0$ small (also $c_*\le1$), for $\Delta\le1$
\begin{equation}\label{eq:act:coercivity}
 c_*\Phi_K(A,B)\le\Delta+\delta^{2\sigma},
\end{equation}
and for $\Delta>1$ this holds because $\Phi_K\le1$.

\emph{Relating $\Delta$ to $F$.}  Expand $N$ and $V$:
\[
 F(A,B)=-\Delta+\tr(\rho^q)-\tr(N^q)+\delta\tr\rho+\delta^\sigma\tr V .
\]
Since $N=\rho+\delta^\sigma I$ and $r>1$,
$\tr(\rho^q)\le\tr(N^q)$; also
$\tr\rho\le n\kappa^{2\sigma}$ and $V\le(\kappa^2+\delta)I$.  Adding
\eqref{eq:act:coercivity} proves \eqref{eq:act:constraint-bound}.
\end{proof}

\subsection{The penalty limit and the proof of the action estimate}
\begin{proof}[Proof of Theorem~\ref{thm:action}]
Fix $0<\delta\le1$.  Proposition~\ref{prop:path} gives, for every
$0<\varepsilon\le1$, a lower triangular $C_\varepsilon$ and a path
$B_\varepsilon$ with the endpoints and norm bound of
\eqref{eq:act:endpointB} such that, for $A_\varepsilon=C_\varepsilon U$,
\begin{equation}\label{eq:act:path}
 \varepsilon B_\varepsilon'
   =M_\delta(B_\varepsilon)-A_\varepsilon A_\varepsilon^{\mathsf T}.
\end{equation}
For fixed $\delta,K,\Omega,U$ the matrices $C_\varepsilon$ are bounded
uniformly in $0<\varepsilon\le1$.  Suppress the subscript and write
$u_\delta(A,B)=\|M_\delta(B)-AA^{\mathsf T}\|_{\HS}^2$.  Since
$A\in H^1_0$, $B\in C^1$, and $M_\delta$ has the primitive $J_\delta$ of
\eqref{eq:field:primitive}, the product and chain rules for absolutely
continuous functions and \eqref{eq:act:path} give, almost everywhere,
\begin{align*}
 \bigl(J_\delta(B)+\tr(A^{\mathsf T}BA)\bigr)'
 &=2\tr((A')^{\mathsf T}BA)
    -\tr\bigl((M_\delta(B)-AA^{\mathsf T})B'\bigr)\\
 &=2\tr((A')^{\mathsf T}BA)-\varepsilon^{-1}u_\delta(A,B).
\end{align*}
Completing the square in the action and integrating, using $A=0$ at both
endpoints, gives the exact formula
\begin{equation}\label{eq:act:identity}
 \mathcal E_K(A)
 =J_\delta(-K)-J_\delta(K)
  +\int_\Omega\Bigl(\|A'-BA\|_{\HS}^2-F(A,B)
       +\varepsilon^{-1}u_\delta(A,B)\Bigr)\,dx .
\end{equation}

\emph{Penalty limit.}  The continuous representative of $U$ is bounded on
$\overline\Omega$; the bounds on $C_\varepsilon$ and $\|B_\varepsilon\|_{\op}$
place all pairs $(A_\varepsilon(x),B_\varepsilon(x))$ in one compact set
$\mathcal K_\delta\subset\R^{n\times d}\times\Sym_n$ with
$\|B\|_{\op}\le\kappa$, on which $F,\Phi_K,u_\delta$ are continuous.
Lemma~\ref{lem:constraint} then implies
\begin{equation}\label{eq:act:penalty}
 \limsup_{\varepsilon\downarrow0}\sup_{(A,B)\in\mathcal K_\delta}
   \bigl(F(A,B)+c_*\Phi_K(A,B)-\varepsilon^{-1}u_\delta(A,B)\bigr)
   \le E_{n,\delta}.
\end{equation}
Otherwise some sequence $\varepsilon_j\downarrow0$ and pairs in the
compact set make the expression at least $E_{n,\delta}+\tau$, $\tau>0$;
boundedness of $F+c_*\Phi_K$ would force $u_\delta=O(\varepsilon_j)$, and a
convergent subsequence would give a limit pair with $u_\delta=0$ and
$F+c_*\Phi_K\ge E_{n,\delta}+\tau$, contradicting Lemma~\ref{lem:constraint}.

For any $\tau>0$ choose $\varepsilon$ so the supremum in
\eqref{eq:act:penalty} is at most $E_{n,\delta}+\tau$.  Substituting into
\eqref{eq:act:identity},
\begin{align*}
 \mathcal E_K(A_\varepsilon)
 \ge{}&J_\delta(-K)-J_\delta(K)-|\Omega|(E_{n,\delta}+\tau)\\
 &+\int_\Omega\Bigl(\|A_\varepsilon'-B_\varepsilon A_\varepsilon\|_{\HS}^2
     +c_*\Phi_K(A_\varepsilon,B_\varepsilon)\Bigr)\,dx .
\end{align*}
By \eqref{eq:field:endpoints} the endpoint term exceeds
$b_\sigma\sum\ell_i^{2\gamma}$ up to an error going to $0$ with $\delta$,
and $|\Omega|E_{n,\delta}\to0$ as $\delta\downarrow0$.  First choose
$\delta$ so that both are below $\eta/2$, then $\tau$ with
$|\Omega|\tau<\eta/2$, then the required $\varepsilon$.  This proves
\eqref{eq:act:bound}.  All compactness arguments are made after fixing
$\delta$; no bound uniform in $\delta$ is needed.
\end{proof}

% =====================================================================
\section{The sharp inequality itself}\label{sec:bound}
% =====================================================================

Before turning to equality we record that the action estimate already
gives the sharp bound \eqref{eq:sharp}; this is the inequality half of
Theorem~\ref{thm:main}.

\begin{proposition}[Sharp spectral bound]\label{prop:sharp}
Every potential in Theorem~\ref{thm:main} satisfies
\begin{equation}\label{eq:bound:sharp}
 \Tr(H_W)_-^\gamma\le\frac{D_\sigma}{b_\sigma}\int_\R\tr(W^p).
\end{equation}
\end{proposition}

\begin{proof}
First suppose $W$ is real symmetric.  Choose any finite list of negative
eigenvalues $-k_1^2,\dots,-k_n^2$ with $k_1\ge\dots\ge k_n>0$ and real
orthonormal eigenfunctions $\varphi_i$.  For $0<\xi<k_n^2$ put
$\ell_i=(k_i^2-\xi)^{1/2}$ and $K=\diag(\ell_1,\dots,\ell_n)$.  There are
real smooth compactly supported orthonormal $u_i$ with form matrix
\begin{equation}\label{eq:bound:trial}
 \bigl(h_W[u_i,u_j]\bigr)_{i,j=1}^n\le-K^2 .
\end{equation}
Indeed, approximate the $\varphi_i$ in $H^1$ and multiply the family by the
inverse square root of its Gram matrix: the Gram matrix tends to $I_n$ and
the form matrix tends to $-\diag(k_i^2)$ by Lemma~\ref{lem:spectral}, so an
operator-norm error below $\xi$ gives \eqref{eq:bound:trial}.

Let $U$ have rows $u_i$ and let $\Omega$ contain their supports.  For every
lower triangular $C$, the $i$th row $a_i$ of $A=CU$ satisfies
\[
 h_W[a_i]+\ell_i^2\|a_i\|_2^2
 \le\sum_{l\le i}C_{il}^2(\ell_i^2-\ell_l^2)\le0 .
\]
Set $Z=A^TA$.  Since $AA^T$ and $Z$ have the same nonzero eigenvalues with
multiplicity, summing the preceding bound and applying Lemma~\ref{lem:young}
gives
\[
 \mathcal E_K(A)\le\int_\R\{\tr(WZ)-\tr(Z^q)\}\le D_\sigma\int_\R\tr(W^p).
\]
Theorem~\ref{thm:action}, with its nonnegative defects discarded, gives
$b_\sigma\sum_i\ell_i^{2\gamma}\le D_\sigma\int\tr(W^p)+\eta$ for every
$\eta>0$.  Let $\eta\downarrow0$, then $\xi\downarrow0$, and finally exhaust
the negative eigenvalues.  The case of empty negative spectrum is immediate.

For complex Hermitian $W=E+iF$ with $E,F$ real, the realification
\begin{equation}\label{eq:bound:realification}
 \widetilde W=\begin{pmatrix}E&-F\\F&E\end{pmatrix}
\end{equation}
is real symmetric and positive semidefinite; the constant unitary matrix
$2^{-1/2}\bigl(\begin{smallmatrix}I&iI\\I&-iI\end{smallmatrix}\bigr)$
conjugates $\widetilde W$ to $W\oplus\overline W$.  Hence
$\tr(\widetilde W^p)=2\tr(W^p)$ and
$\Tr(H_{\widetilde W})_-^\gamma=2\Tr(H_W)_-^\gamma$ (conjugation identifies
the spectra of $H_W$ and $H_{\overline W}$).  Applying the real inequality
to $\widetilde W$ and dividing by two proves \eqref{eq:bound:sharp}.
\end{proof}

% =====================================================================
\section{From equality to pointwise matrix relations}\label{sec:compact}
% =====================================================================

We now assume equality holds and extract information.  The difficulty is
that the number of negative eigenfunctions may be infinite while the
potential has only $m$ channels.  The fixed column count $m$ is what keeps
the tails under control.

Throughout this section $W$ is real symmetric, $W\ge0$ almost everywhere,
and $0<\int_\R\tr(W^p)<\infty$.  Suppose equality holds in the form
\begin{equation}\label{eq:comp:equality}
 b_\sigma\sum_{i\in\mathcal I}k_i^{2\gamma}
   =D_\sigma\int_\R\tr(W^p)\,dx,
\end{equation}
where $-k_i^2$ are the negative eigenvalues of $H_W$, repeated according to
multiplicity, and $k_1\ge k_2\ge\cdots>0$.  By Lemma~\ref{lem:spectral}
the index set $\mathcal I$ is a nonempty finite initial segment of $\mathbb N$
or all of $\mathbb N$, each eigenvalue has finite multiplicity, and $k_i\to0$
in the infinite case.  Fix real orthonormal eigenfunctions
$\varphi_i\in H^1(\R;\R^m)$, viewed as row vectors.  Write
\begin{equation}\label{eq:comp:H}
 \mathscr H=\ell^2(\mathcal I;\R),\quad
 \Lambda=\diag(k_i:i\in\mathcal I),\quad
 \kappa=k_1 .
\end{equation}
For operators between real Hilbert spaces the superscript $T$ also denotes
the adjoint.

\begin{proposition}[Pointwise relations at equality]\label{prop:relations}
Under these assumptions there are rows $a_i\in H^1(\R;\R^m)$,
$i\in\mathcal I$, each a finite linear combination of eigenfunctions with
eigenvalue $-k_i^2$, such that
\begin{equation}\label{eq:comp:roweig}
 -a_i''-a_iW=-k_i^2a_i \quad\text{in distributions.}
\end{equation}
The rows $a_i$ may vanish.  On a common set of full measure they define a
Hilbert--Schmidt operator $A(x):\R^m\to\mathscr H$, and there is a bounded
self-adjoint operator $B$ on $\mathscr H$, $\|B\|_{\op}\le\kappa$, with
\begin{equation}\label{eq:comp:limit}
 [B,\Lambda^2]=0,\qquad
 \Lambda^2-B^2=(AA^T)^r,\qquad
 a_i'(x)=(BA)_{i,\cdot}\quad(i\in\mathcal I).
\end{equation}
Moreover
\begin{equation}\label{eq:comp:potential}
 W(x)=q\bigl(A(x)^TA(x)\bigr)^r \quad\text{almost everywhere.}
\end{equation}
The operator $B$ exists separately at each such point; no measurable choice
is required.
\end{proposition}

\subsection{Finite spectral approximations and vanishing errors}
Let $n_j=j$ when $\mathcal I=\mathbb N$ and $n_j=|\mathcal I|$ when
$\mathcal I$ is finite.  Choose $0<\xi_j<1/j$ with
\begin{equation}\label{eq:comp:shift}
 \ell_{j,i}:=(k_i^2-\xi_j)^{1/2}>0\ (1\le i\le n_j),
 \qquad
 \sum_{i\le n_j}\bigl(k_i^{2\gamma}-\ell_{j,i}^{2\gamma}\bigr)<1/j .
\end{equation}
The same shift $\xi_j$ is used for every eigenvalue, so the differences of
their squares are unchanged.  Set
$K_j=\diag(\ell_{j,1},\dots,\ell_{j,n_j})$.

By \eqref{eq:bound:trial} we can choose real smooth compactly supported
orthonormal $u_{j,1},\dots,u_{j,n_j}$ with $H^1$ errors below $1/j$ and
form-matrix error below $\xi_j$:
\begin{equation}\label{eq:comp:form}
 \|u_{j,i}-\varphi_i\|_{H^1}<1/j,
 \qquad
 \bigl(h_W[u_{j,i},u_{j,l}]\bigr)_{i,l=1}^{n_j}\le-K_j^2 .
\end{equation}
Let $U_j$ have rows $u_{j,i}$ and choose a bounded interval $\Omega_j$
containing their supports.  Apply Theorem~\ref{thm:action} with
$K=K_j$, $d=m$, $\kappa=k_1$, and $\eta=1/j$: it gives a lower triangular
$C_j$, a symmetric path $B_j$ with $\|B_j\|_{\op}\le\kappa$, and
$A_j=C_jU_j$.  Extend $A_j$ by zero outside $\Omega_j$ and $B_j$ by $K_j$
to the left and $-K_j$ to the right; all errors vanish outside $\Omega_j$.

The lower triangular form of $C_j$ gives an extra nonnegative error.  Put
$Z_j=A_j^TA_j$, an $m\times m$ positive semidefinite matrix.  By
\eqref{eq:comp:form},
\begin{equation}\label{eq:comp:triangular}
 \mathcal E_{K_j}(A_j)
 \le\int_\R\bigl(\tr(WZ_j)-\tr(Z_j^q)\bigr)\,dx-T_j,
 \quad
 T_j=\sum_{l\le i\le n_j}(C_j)_{il}^{\,2}(k_l^2-k_i^2)\ge0 .
\end{equation}
Indeed, the $i$th row of $A_j$ is $\sum_{l\le i}(C_j)_{il}u_{j,l}$, whose
quadratic form plus $\ell_{j,i}^2$ times its squared $L^2$ norm is at most
$\sum_{l\le i}(C_j)_{il}^{\,2}(\ell_{j,i}^2-\ell_{j,l}^2)
=-\sum_{l\le i}(C_j)_{il}^{\,2}(k_l^2-k_i^2)$; summing over $i$ and using
that $A_jA_j^T$ and $Z_j$ share nonzero eigenvalues gives
\eqref{eq:comp:triangular}.  The ordering of the $k_i$ makes each summand
of $T_j$ nonnegative.

Recall the Young deficit \eqref{eq:youngdefect}; by
Lemma~\ref{lem:young}, $Y(W,Z)=0$ iff $W=qZ^r$.  Combining
Theorem~\ref{thm:action}, \eqref{eq:comp:equality}, and
\eqref{eq:comp:triangular},
\begin{align}
 0\le T_j+\int_\R\Bigl(
   Y(W,Z_j)+\|A_j'-B_jA_j\|_{\HS}^2+c_*\Phi_{K_j}(A_j,B_j)\Bigr)\,dx
 &\le d_j,\label{eq:comp:vanish}\\
 d_j:=b_\sigma\Bigl(\sum_{i\in\mathcal I}k_i^{2\gamma}
       -\sum_{i\le n_j}\ell_{j,i}^{2\gamma}\Bigr)+1/j
 &\longrightarrow0 .\notag
\end{align}
Here $c_*>0$ is fixed; its independence of $n_j$ is essential.  Hence all
integral errors and $T_j$ tend to zero.

A uniform bound on $A_j$ will also be needed.  The coercive estimate
\eqref{eq:youngcoercive} gives
\begin{equation}\label{eq:comp:coercivity}
 Y(W,Z_j)\ge\tfrac12\tr(Z_j^q)-(2^{p-1}-1)D_\sigma\tr(W^p),
\end{equation}
and since $Z_j$ is $m\times m$, scalar H\"older gives
$\|A_j\|_{\HS}^{2q}=(\tr Z_j)^q\le m^{q-1}\tr(Z_j^q)$.  With
\eqref{eq:comp:vanish} this bounds $A_j$ uniformly in $L^{2q}(\R)$ in the
Hilbert--Schmidt norm, independently of its number of rows.

\subsection{Convergence of each fixed row}
For fixed $i\in\mathcal I$, the coefficients $(C_j)_{il}$, $l\le i$, are
bounded as $j\to\infty$.  Choose a bounded interval $J_i$ so large that
\[
 \Bigl(\int_{J_i}\varphi_l(x)\varphi_s(x)^T\,dx\Bigr)_{l,s=1}^{i}
       \ge\tfrac34 I_i,
\]
possible because the full Gram matrix is the identity.  By
\eqref{eq:comp:form} the corresponding Gram matrix of the $u_{j,l}$ is at
least $\tfrac12 I_i$ for large $j$, so
\[
 \frac12\sum_{l\le i}(C_j)_{il}^{\,2}
 \le\int_{J_i}|(A_j)_{i,\cdot}|^2\,dx
 \le|J_i|^{1-1/q}\Bigl(\int_{J_i}\|A_j\|_{\HS}^{2q}\,dx\Bigr)^{1/q},
\]
uniformly bounded.  Lower triangularity keeps the number of coefficients
in a fixed row finite.  A diagonal subsequence makes each $(C_j)_{il}$
converge to a limit $c_{il}$, $l\le i$.  From $T_j\to0$ and
\eqref{eq:comp:triangular},
\begin{equation}\label{eq:comp:gapsuppress}
 c_{il}=0\quad\text{whenever }k_l>k_i .
\end{equation}
Define $a_i=\sum_{l\le i}c_{il}\varphi_l$, a finite sum; then
\eqref{eq:comp:form} gives
\begin{equation}\label{eq:comp:rowconv}
 (A_j)_{i,\cdot}\longrightarrow a_i\quad\text{in }H^1(\R;\R^m)
 \quad\text{for every fixed }i.
\end{equation}
The ordering of the eigenvalues together with
\eqref{eq:comp:gapsuppress} shows only eigenfunctions with eigenvalue
$-k_i^2$ occur in $a_i$, proving \eqref{eq:comp:roweig} (the distributional
form equation follows by testing against smooth compactly supported
functions; the potential is locally integrable and the rows locally
bounded).

Using the continuous, locally absolutely continuous representatives and
$H^1(\R)\subset C_b(\R)$, \eqref{eq:comp:rowconv} also gives pointwise
convergence at every $x$.  Passing to a further subsequence we may arrange
that the integrals of the nonnegative integrand in \eqref{eq:comp:vanish}
are summable and that, for every fixed $i$, the squared $L^2$ errors of
the derivatives in \eqref{eq:comp:rowconv} are summable (a diagonal choice
successively takes the first $j$ available derivative errors and the
integral error below $2^{-j}$).  Consequently, on a common set of full
measure,
\begin{equation}\label{eq:comp:pointwise}
 Y(W,Z_j)\to0,\quad
 \|A_j'-B_jA_j\|_{\HS}\to0,\quad
 \Phi_{K_j}(A_j,B_j)\to0,\quad
 (A_j')_{i,\cdot}(x)\to a_i'(x)\ \forall i .
\end{equation}
We also exclude the null set where $W$ is not a finite positive semidefinite
matrix.

\subsection{Tail control in the row-index space}
Pad $A_j,B_j,K_j$ with zeros in the unused coordinates of $\mathscr H$, so
$A_j:\R^m\to\mathscr H$ and $B_j,K_j$ are self-adjoint finite-rank operators.
Since the omitted $k_i$ tend to zero,
\begin{equation}\label{eq:comp:diagonal}
 \|K_j^2-\Lambda^2\|_{\op}
 \le\max\{\xi_j,\sup_{i>n_j}k_i^2\}\longrightarrow0 .
\end{equation}
Fix $x$ in the full-measure set of \eqref{eq:comp:pointwise} and suppress
it.  Write $\rho_j=A_jA_j^T$.  Convergence of $\Phi$ to zero gives
\begin{equation}\label{eq:comp:algebraerr}
 \|[B_j,K_j^2]\|_{\op}\to0,
 \qquad e_j:=\|K_j^2-B_j^2-\rho_j^r\|_{\op}\to0 .
\end{equation}
For $h\ge1$ let $P_h$ project onto the first $h$ coordinates, $Q_h=I-P_h$,
and $s_h=\sup_{i>h}k_i^2\to0$.  For unit $v$ in the range of $Q_h$,
\begin{equation}\label{eq:comp:tailquad}
 \|B_jv\|^2+\langle v,\rho_j^rv\rangle\le s_h+e_j .
\end{equation}
By Jensen for the spectral measure of $\rho_j$ ($r>1$),
$\langle v,\rho_jv\rangle^r\le\langle v,\rho_j^rv\rangle$, so
$\|Q_h\rho_jQ_h\|_{\op}\le(s_h+e_j)^{1/r}$.  Although $\mathscr H$ may be
infinite dimensional, $Q_h\rho_jQ_h$ has rank at most $m$ because $A_j$ has
$m$ columns; its trace is at most $m$ times its operator norm.  Hence
\begin{equation}\label{eq:comp:tailbounds}
 \|Q_hA_j\|_{\HS}^2\le m(s_h+e_j)^{1/r},
 \qquad
 \|B_jQ_h\|_{\op}\le(s_h+e_j)^{1/2}.
\end{equation}
The first bound converts decay of the eigenvalue parameters into decay of
the total mass of all omitted rows, with factor the fixed physical
dimension $m$ rather than the number of retained eigenfunctions.

For fixed $h$ the first $h$ rows of $A_j$ converge by
\eqref{eq:comp:rowconv}; \eqref{eq:comp:tailbounds} makes the remaining rows
uniformly small by taking $h$ large then $j$ large.  Thus $A_j$ is Cauchy in
Hilbert--Schmidt norm and converges to an operator $A:\R^m\to\mathscr H$
whose rows are the $a_i(x)$:
\begin{equation}\label{eq:comp:HSlimit}
 A_j\longrightarrow A\quad\text{in Hilbert--Schmidt norm.}
\end{equation}
This holds for the whole already-chosen subsequence and needs no further
choice depending on $x$.

For $B_j$ only pointwise compactness is needed.  Finite compressions
$P_hB_jP_h$ are bounded in finite-dimensional spaces, so a diagonal
subsequence makes them converge for every $h$; self-adjointness and the
second bound in \eqref{eq:comp:tailbounds} give
$\|B_j-P_hB_jP_h\|_{\op}\le2\|B_jQ_h\|_{\op}\le2(s_h+e_j)^{1/2}$.  The
subsequence converges in operator norm to a self-adjoint $B$,
$\|B\|_{\op}\le\kappa$ (possibly depending on $x$).

Now pass to all products.  Hilbert--Schmidt convergence of $A_j$ implies
operator-norm convergence of $A_jA_j^T$ and $A_j^TA_j$; continuous
functional calculus for nonnegative operators gives convergence of their
$r$th powers.  With \eqref{eq:comp:diagonal} and
\eqref{eq:comp:algebraerr} this proves
$[B,\Lambda^2]=0$ and $\Lambda^2-B^2=(AA^T)^r$.  Also
$B_jA_j\to BA$ in Hilbert--Schmidt norm; taking the $i$th row and using the
derivative convergence and the first-order error in
\eqref{eq:comp:pointwise} gives $a_i'(x)=(BA)_{i,\cdot}$ for every $i$.
Finally continuity of the finite-dimensional Young deficit and
$Y(W,A_j^TA_j)\to0$ give $Y(W,A^TA)=0$, and Lemma~\ref{lem:young} gives
$W=q(A^TA)^r$.

All statements hold on one common set of full measure.  No function
$x\mapsto B(x)$ is selected; only the existence of a suitable operator at
each such point enters later.  This proves Proposition~\ref{prop:relations}.

% =====================================================================
\section{Fixed channels and the equality profiles}\label{sec:rigid}
% =====================================================================

The pointwise relations of Proposition~\ref{prop:relations} involve an
auxiliary operator $B$ that may depend on $x$.  We now show the channel
subspaces are constant, so that on each channel $B$ becomes a fixed matrix
obeying an ODE whose solutions are the profiles of \eqref{eq:profile}.

\subsection{The channel subspaces are constant}
Assume first that $W$ is real and is a nonzero equality potential.  Use the
rows $a_i$, the numbers $k_i$, and $\Lambda=\diag(k_i)$ from
Proposition~\ref{prop:relations}.  For each distinct value $\alpha>0$ among
the $k_i$, let $n_\alpha$ be its finite multiplicity and let $G_\alpha$ be
the $n_\alpha\times m$ matrix whose rows are the $a_i$ with $k_i=\alpha$;
each entry is in $H^1(\R)$ and we use its continuous representative.

At every point of the common full-measure set of Proposition~\ref{prop:relations}
choose one of the operators $B$ it supplies.  Since $B$ commutes with
$\Lambda^2$ it preserves each eigenspace, and
$AA^T=(\Lambda^2-B^2)^{1/r}$ commutes with $B$ and $\Lambda^2$.  Writing
$B_\alpha$ for the block on the $\alpha^2$-eigenspace,
\begin{align}
 G_\alpha G_\beta^T&=0\quad(\alpha\ne\beta),
 \label{eq:rig:orthogonal}\\
 G_\alpha'&=B_\alpha G_\alpha,
 & [B_\alpha,G_\alpha G_\alpha^T]&=0\nonumber\\
 \alpha^2I-B_\alpha^2&=(G_\alpha G_\alpha^T)^r,
 & \|B_\alpha\|_{\op}&\le\kappa .
 \label{eq:rig:blocksquare}
\end{align}
These hold almost everywhere with no compatibility between the choices of
$B$ at different points.

\begin{lemma}\label{lem:right}
For every $\alpha$ the subspace $\mathcal R_\alpha=(\ker G_\alpha(x))^\perp
\subset\R^m$ is independent of $x$.  The nonzero $\mathcal R_\alpha$ are
mutually orthogonal and
\begin{equation}\label{eq:rig:dimbound}
 \sum_\alpha\dim\mathcal R_\alpha\le m .
\end{equation}
In particular only finitely many $G_\alpha$ are nonzero.  The potential $W$
has a continuous representative, equal to zero on the orthogonal complement
of the $\mathcal R_\alpha$ and to $q(G_\alpha^TG_\alpha)^r$ on
$\mathcal R_\alpha$.  With this representative every $G_\alpha$ is $C^2$.
\end{lemma}

\begin{proof}
Fix $v\in\R^m$.  From \eqref{eq:rig:blocksquare},
$\|(G_\alpha v)'\|\le\kappa\|G_\alpha v\|$ almost everywhere, and $G_\alpha v$
is locally absolutely continuous.  If it vanishes at $x_0$, integration and
Gronwall's inequality, first right and then left of $x_0$, show it vanishes
everywhere; hence $\ker G_\alpha(x)$ is constant, including off the original
full-measure set.

The products in \eqref{eq:rig:orthogonal} are continuous, so their
almost-everywhere vanishing is everywhere; thus the constant row spaces
$\mathcal R_\alpha$ are mutually orthogonal, giving
\eqref{eq:rig:dimbound} and finiteness of the nonzero blocks.  On the
full-measure set, $A^TA=\sum_\alpha G_\alpha^TG_\alpha$ with only finitely
many nonzero summands, each acting on its own $\mathcal R_\alpha$, so
\eqref{eq:comp:potential} gives
\begin{equation}\label{eq:rig:contpot}
 W=q\sum_\alpha(G_\alpha^TG_\alpha)^r \quad\text{almost everywhere}.
\end{equation}
The right side is continuous; we use it as the representative of $W$.  The
row equation \eqref{eq:comp:roweig} then reads
$G_\alpha''=\alpha^2G_\alpha-G_\alpha W$ in distributions, whose right side
is continuous; integrating twice, $G_\alpha$ is $C^2$.
\end{proof}

The row spaces split $\R^m$ into finitely many constant channels.  To find
the potential on one channel we also need a constant range for $G_\alpha$ in
the eigenfunction index space.

\begin{lemma}\label{lem:left}
For $\alpha$ with $\ell:=\dim\mathcal R_\alpha>0$, the range of
$G_\alpha(x)$ in $\R^{n_\alpha}$ is independent of $x$.  In constant
orthonormal bases of this range and of $\mathcal R_\alpha$ the restriction
of $G_\alpha$ is an invertible $C^2$ matrix $D:\R\to\R^{\ell\times\ell}$, and
$B_0=D'D^{-1}$ is $C^1$ with, everywhere,
\begin{equation}\label{eq:rig:active}
 B_0=B_0^T,\qquad
 \alpha^2I-B_0^2=(DD^T)^r>0,\qquad
 B_0'=r(B_0^2-\alpha^2I).
\end{equation}
\end{lemma}

\begin{proof}
Choose a constant orthonormal basis of $\mathcal R_\alpha$ and let $E(x)$ be
the matrix of $G_\alpha(x)$ restricted to it; it has full column rank $\ell$
everywhere.  Its range projection is $P=E(E^TE)^{-1}E^T$.  The commutator in
\eqref{eq:rig:blocksquare} shows $B_\alpha$ preserves the range of
$G_\alpha G_\alpha^T$, which is the range of $E$; thus $E'=B_\alpha E$ has
columns in this range almost everywhere.  Differentiating $P$ gives
$P'=(I-P)E'(E^TE)^{-1}E^T+E(E^TE)^{-1}(E')^T(I-P)=0$ almost everywhere,
and since $P$ is $C^1$ it is constant.

With constant orthonormal bases of both spaces, $D$ is invertible $C^2$ and
$D'D^{-1}$ is, almost everywhere, the restriction of $B_\alpha$ to its
invariant range.  Then the symmetry and the square identity in
\eqref{eq:rig:active} hold almost everywhere and extend by continuity;
positivity is strict because $D$ is invertible.  On $\mathcal R_\alpha$,
\eqref{eq:rig:contpot} makes the potential $q(D^TD)^r$, so
\eqref{eq:comp:roweig} becomes $D''=\alpha^2D-q(DD^T)^rD$ (using a
singular-value decomposition for the second form).  Differentiating
$B_0=D'D^{-1}$ with $q=r+1$,
\[
 B_0'=D''D^{-1}-B_0^2
     =\alpha^2I-q(\alpha^2I-B_0^2)-B_0^2
     =r(B_0^2-\alpha^2I).
\]
\end{proof}

\subsection{The matrix Riccati equation separates the profiles}
Fix a nonzero block and the matrix $B_0$.  Orthogonally diagonalise the
symmetric $B_0(x_*)$ at one point.  The right side of the Riccati equation is
a polynomial in $B_0$, so uniqueness of the matrix ODE preserves the
diagonal form.  The strict inequality in \eqref{eq:rig:active} places each
initial diagonal entry $b_j(x_*)$ in $(-\alpha,\alpha)$, and the scalar
equation $b_j'=r(b_j^2-\alpha^2)$ has the unique solution with that value
\begin{equation}\label{eq:rig:tanh}
 b_j(x)=-\alpha\tanh\!\bigl(r\alpha(x-y_j)\bigr),\qquad 1\le j\le\ell,
\end{equation}
for a unique centre $y_j$.  The diagonal matrix with these entries solves the
same initial-value problem as $B_0$ on all of $\R$, so uniqueness gives
equality everywhere.

Now \eqref{eq:rig:active} shows $DD^T$ is diagonal in these coordinates.
Since $D'=\diag(b_j)D$, each row of $D$ is a positive scalar multiple of its
value at $x_*$; the rows are nonzero and mutually orthogonal (as $DD^T$ is
diagonal).  Taking their normalised directions as a constant orthonormal
basis of $\mathcal R_\alpha$ makes $D$ diagonal with positive entries.  The
potential in this fixed basis is diagonal, with entries
\begin{equation}\label{eq:rig:profiles}
 q\bigl(\alpha^2-b_j(x)^2\bigr)
 =(r+1)\alpha^2\,\sech^2\!\bigl(r\alpha(x-y_j)\bigr).
\end{equation}
Combining these bases over the finitely many nonzero $\mathcal R_\alpha$ and
adding a basis of their orthogonal complement gives, for real $W$, the form
\eqref{eq:classification} with at most $m$ nonzero profiles by
\eqref{eq:rig:dimbound}.  The zero potential corresponds to $k=0$.

\subsection{Complex Hermitian potentials}
For a complex Hermitian $W$, take the realification $\widetilde W$ from
\eqref{eq:bound:realification}, which is equivalent to $W\oplus\overline W$
by a constant unitary matrix, with both $\int\tr(W^p)$ and the negative
moment doubled.  Equality for $W$ implies equality for the real
$\widetilde W$, so the real result gives a constant unitary $V$ with
\[
 W(x)\oplus\overline W(x)
 =V\diag(f_1(x),\dots,f_{2m}(x))V^*
 \quad\text{almost everywhere,}
\]
each $f_j$ a profile \eqref{eq:rig:profiles} or zero.  Let $P$ be the
projection onto the first summand and $\widehat P=V^*PV$.  Since $P$
commutes with $W\oplus\overline W$, we have
$(f_i-f_j)\widehat P_{ij}=0$ almost everywhere, so $\widehat P_{ij}=0$
whenever $f_i,f_j$ are not equal almost everywhere.  Partition the indices
into groups of identical functions (including the zero function).  On each
group the diagonal potential is a scalar multiple of the identity, so any
constant orthonormal basis of the intersection of that coordinate group with
$\operatorname{ran}\widehat P$ diagonalises it.  Transporting these bases by
$V$ gives a constant unitary basis of $\C^m$ in which $W$ has the asserted
profiles.  This proves necessity for complex Hermitian $W$.

\subsection{Every listed potential attains equality}
It remains to verify attainment directly.  Fix $a>0$ and $x_0\in\R$ and set
\[
 V(x)=(r+1)a^2\sech^2\!\bigl(ra(x-x_0)\bigr),
 \qquad H=-\partial_x^2-V .
\]
Put $w(x)=a\tanh\bigl(ra(x-x_0)\bigr)$ and $Q=\partial_x+w$ on
$H^1(\R)$.  Since $w,w'$ are bounded, direct differentiation gives
\begin{align}
 Q^*Q&=H+a^2,
 \label{eq:rig:factor}\\
 QQ^*&=-\partial_x^2+a^2+(r-1)a^2\sech^2\!\bigl(ra(x-x_0)\bigr)\ge a^2,
 \label{eq:rig:partner}
\end{align}
the last inequality because $r>1$.  The equation $Q\psi=0$ has the
one-dimensional $L^2$ solution space
\[
 \ker Q=\operatorname{span}\bigl\{\sech^{1/r}(ra(x-x_0))\bigr\},
\]
so $-a^2$ is a simple eigenvalue of $H$, and \eqref{eq:rig:factor} excludes
eigenvalues below it.  If $H\phi=\lambda\phi$ with $-a^2<\lambda<0$, put
$\mu=\lambda+a^2>0$, $g=Q\phi$; then $\|g\|_2^2=\mu\|\phi\|_2^2>0$,
$Q^*Q\phi=\mu\phi\in\operatorname{Dom}Q$, so $g\in\operatorname{Dom}(QQ^*)$
and $QQ^*g=\mu g$, contradicting \eqref{eq:rig:partner} because $\mu<a^2$.
Hence $-a^2$ is the only negative eigenvalue.

Finally $2p-1=2\gamma$ and the substitution $s=\tanh y$ give
\begin{align*}
 \int_\R V(x)^p\,dx
 &=\frac{(r+1)^p}{r}a^{2\gamma}\int_\R\sech^{2p}y\,dy\\
 &=\frac{(r+1)^p}{r}a^{2\gamma}\int_{-1}^{1}(1-s^2)^{p-1}\,ds
  =\frac{b_\sigma}{D_\sigma}a^{2\gamma},
\end{align*}
using $(r+1)^p/r=D_\sigma^{-1}$ and $p-1=\sigma$.  Therefore
$\Tr(H)_-^\gamma=a^{2\gamma}=C_\gamma\int_\R V^p$.  The zero potential
contributes zero to both sides; finite orthogonal direct sums add both
spectral moments and trace integrals, and constant unitary conjugation
changes neither.  So every potential listed in Theorem~\ref{thm:main}
attains equality, completing the proof.

% =====================================================================
\begin{thebibliography}{99}
\raggedright
\bibitem{AizenmanLieb1978}
Michael Aizenman and Elliott H. Lieb,
\emph{On semi-classical bounds for eigenvalues of Schr\"odinger operators},
Physics Letters A \textbf{66} (1978), no.~6, 427--429.

\bibitem{BenguriaLoss2000}
Rafael Benguria and Michael Loss,
\emph{A simple proof of a theorem of Laptev and Weidl},
Mathematical Research Letters \textbf{7} (2000), no.~2--3, 195--203.

\bibitem{FrankGontierLewin2025}
Rupert L. Frank, David Gontier and Mathieu Lewin,
\emph{Optimizers for the finite-rank Lieb--Thirring inequality},
American Journal of Mathematics \textbf{147} (2025), no.~2, 503--560.

\bibitem{HLW2000}
Dirk Hundertmark, Ari Laptev and Timo Weidl,
\emph{New bounds on the Lieb--Thirring constants},
Inventiones Mathematicae \textbf{140} (2000), no.~3, 693--704.

\bibitem{HLT1998}
Dirk Hundertmark, Elliott H. Lieb and Lawrence E. Thomas,
\emph{A sharp bound for an eigenvalue moment of the one-dimensional
Schr\"odinger operator},
Advances in Theoretical and Mathematical Physics \textbf{2} (1998),
no.~4, 719--731.

\bibitem{Keller1961}
Joseph B. Keller,
\emph{Lower bounds and isoperimetric inequalities for eigenvalues of the
Schr\"odinger equation},
Journal of Mathematical Physics \textbf{2} (1961), no.~2, 262--266.

\bibitem{LaptevWeidl2000}
Ari Laptev and Timo Weidl,
\emph{Sharp Lieb--Thirring inequalities in high dimensions},
Acta Mathematica \textbf{184} (2000), no.~1, 87--111.

\bibitem{LiebThirring1975}
Elliott H. Lieb and Walter E. Thirring,
\emph{Bound for the kinetic energy of fermions which proves the stability
of matter},
Physical Review Letters \textbf{35} (1975), no.~11, 687--689.

\bibitem{LiebThirring1976}
Elliott H. Lieb and Walter E. Thirring,
\emph{Inequalities for the moments of the eigenvalues of the Schr\"odinger
Hamiltonian and their relation to Sobolev inequalities},
in \emph{Studies in Mathematical Physics: Essays in Honor of Valentine
Bargmann}, Princeton University Press, 1976, 269--303.

\bibitem{MLT}
\emph{Sharp one-dimensional Lieb--Thirring inequalities for matrix
potentials}, companion preprint, 2026.

\bibitem{Sard1942}
Arthur Sard,
\emph{The measure of the critical values of differentiable maps},
Bulletin of the American Mathematical Society \textbf{48} (1942),
883--890.

\end{thebibliography}

\end{document}