64c3814
==== TABLE OF CONTENTS ====
Positively-curved-Einstein-four-manifolds//figures/proof-map.tex
Positively-curved-Einstein-four-manifolds//figures/spectral-domains.tex
Positively-curved-Einstein-four-manifolds//paper.tex
Positively-curved-Einstein-four-manifolds//references.tex
Positively-curved-Einstein-four-manifolds//sections/01-introduction.tex
Positively-curved-Einstein-four-manifolds//sections/02-preliminaries.tex
Positively-curved-Einstein-four-manifolds//sections/03-analytic.tex
Positively-curved-Einstein-four-manifolds//sections/04-volume.tex
Positively-curved-Einstein-four-manifolds//sections/05-hessian.tex
Positively-curved-Einstein-four-manifolds//sections/06-coupling.tex
Positively-curved-Einstein-four-manifolds//sections/07-certificates.tex
Positively-curved-Einstein-four-manifolds//sections/08-classification.tex
Positively-curved-Einstein-four-manifolds//sections/appendix-a-analytic.tex
Positively-curved-Einstein-four-manifolds//sections/appendix-b-topology.tex
===========================
==== BEGIN Positively-curved-Einstein-four-manifolds//figures/proof-map.tex ====
\begin{figure}[htbp]
\centering
\begin{tikzpicture}[
box/.style={draw=blue!40!black,rounded corners=2pt,fill=blue!3,
text width=4.15cm,align=center,inner sep=5pt,font=\footnotesize},
arr/.style={-{Latex[length=2mm]},draw=blue!55!black,thick},
note/.style={font=\scriptsize,fill=white,inner sep=2pt,align=center}]
\node[box] (scalar) at (0,0) {Scalar and upper volume estimates\\Sections~\ref{sec:analytic}--\ref{sec:volume}};
\node[box] (hessian) at (6.2,0) {Weighted Hessian inequality\\Section~\ref{sec:hessian}};
\node[box] (balance) at (0,-1.55) {Equal second moments\\Section~\ref{sec:volume}};
\node[box] (integral) at (6.2,-1.55) {$\mathbf E\mathcal I\ge0$ for the two blocks\\Section~\ref{sec:coupling}};
\node[box] (signs) at (0,-3.1) {Exact pointwise bounds\\Sections~\ref{sec:coupling}--\ref{sec:certificates}};
\node[box] (vanish) at (6.2,-3.1) {Half-conformal flatness\\Strictness in Section~\ref{sec:classification}};
\node[box] (lower) at (0,-4.65) {Conditional\\lower volume bound\\Section~\ref{sec:classification}};
\node[box] (parallel) at (6.2,-4.65) {Parallel curvature and model tensor\\Section~\ref{sec:classification}};
\node[box] (simply) at (0,-6.2) {Simple connectivity\\Appendix~\ref{app:topology}};
\node[box] (global) at (6.2,-6.2) {Global model isometry\\Section~\ref{sec:classification}};
\draw[arr] (scalar)--(balance);
\draw[arr] (hessian)--(integral);
\draw[arr] (balance)--node[note,above]{variance\\term}(integral);
\draw[arr] (integral)--(vanish);
\draw[arr] (signs)--node[note,above]{pointwise\\signs}(vanish);
\draw[arr] (vanish)--(parallel);
\draw[arr] (lower)--(parallel);
\draw[arr] (parallel)--(global);
\draw[arr] (simply)--(global);
\end{tikzpicture}
\caption{Assembly of the proof. Moment balance enters the nonnegative
integral; the exact pointwise bounds and active-gradient strictness force
one-block vanishing. The conditional lower bound identifies the remaining
parallel tensor, and simple connectivity permits global continuation.}
\label{fig:proof-map}
\end{figure}
==== END Positively-curved-Einstein-four-manifolds//figures/proof-map.tex ====
==== BEGIN Positively-curved-Einstein-four-manifolds//figures/spectral-domains.tex ====
\begin{figure}[htbp]
\centering
\begin{tikzpicture}[scale=1.7, every node/.style={font=\small}]
\begin{scope}
\fill[blue!4] (0,0)--(2,0)--(0,2)--cycle;
\draw[thick] (0,0)--(2,0)--(0,2)--cycle;
\draw (0,0)--(1,1)--(1,0);
\node[below left] at (0,0) {$O$};
\node[below] at (1,0) {$S$};
\node[below right] at (2,0) {$U$};
\node[above right] at (1,1) {$C$};
\node[above left] at (0,2) {$W_0$};
\node at (1,-0.48) {$\mathcal D$: certificate $\mathcal A$};
\end{scope}
\begin{scope}[xshift=3.6cm]
\fill[blue!4] (0,0)--(2,0)--(1,1)--cycle;
\draw[thick] (0,0)--(2,0)--(1,1)--cycle;
\draw (0.5,0.5)--(1,0)--(1.5,0.5);
\draw (1,0)--(1,1);
\draw (0.5,0.5)--(1.5,0.5);
\node[below left] at (0,0) {$O$};
\node[below] at (1,0) {$S$};
\node[below right] at (2,0) {$U$};
\node[above] at (1,1) {$C$};
\node[above left] at (0.5,0.5) {$A$};
\node[above right] at (1.5,0.5) {$B$};
\node[above right,inner sep=2pt] at (1,0.5) {$E$};
\node at (1,-0.48) {$\mathcal D_+$: certificate $\mathcal B$};
\end{scope}
\end{tikzpicture}
\caption{The full and ordered curvature domains, with the subdivisions
used in the triangular coefficient table.}
\end{figure}
==== END Positively-curved-Einstein-four-manifolds//figures/spectral-domains.tex ====
==== BEGIN Positively-curved-Einstein-four-manifolds//paper.tex ====
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\hypersetup{pdftitle={Positively curved Einstein four-manifolds},pdfauthor={OpenAI}}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{lemma}[theorem]{Lemma}
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\title{Positively curved Einstein four-manifolds}
\author{OpenAI}
\date{September 23, 2026}
\begin{document}
\maketitle
\begin{abstract}
We prove the classification conjecture for connected smooth closed Einstein four-manifolds with strictly positive sectional curvature. Up to positive scaling and isometry, every such manifold is the round four-sphere, the complex projective plane with its Fubini--Study metric, or real projective four-space with its round metric. No orientability assumption is needed.
\end{abstract}
\input{sections/01-introduction.tex}
\clearpage
\begingroup
\small
\tableofcontents
\endgroup
\clearpage
\input{sections/02-preliminaries.tex}
\input{sections/03-analytic.tex}
\input{sections/04-volume.tex}
\input{sections/05-hessian.tex}
\input{sections/06-coupling.tex}
\input{sections/07-certificates.tex}
\input{sections/08-classification.tex}
\appendix
\input{sections/appendix-a-analytic.tex}
\input{sections/appendix-b-topology.tex}
\clearpage
\input{references.tex}
\end{document}
==== END Positively-curved-Einstein-four-manifolds//paper.tex ====
==== BEGIN Positively-curved-Einstein-four-manifolds//references.tex ====
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==== END Positively-curved-Einstein-four-manifolds//references.tex ====
==== BEGIN Positively-curved-Einstein-four-manifolds//sections/01-introduction.tex ====
\section{Introduction}\label{sec:introduction}
A Riemannian metric is Einstein if its Ricci tensor is a constant
multiple of the metric. This fixes the trace of the sectional curvatures
while allowing individual curvatures to vary. The classification
conjecture in dimension four asks whether the round $S^4$, the round
$\mathbb{RP}^4$, and Fubini--Study $\mathbb{CP}^2$ exhaust the connected
closed Einstein four-manifolds with strictly positive sectional curvature,
up to positive scaling and isometry. Yang explicitly lists these three
models in Conjecture~1 of~\cite{Yang2000}. We prove the classification
in the closed setting.
\begin{theorem}\label{thm:classification}
Let $(M,g)$ be a connected smooth closed Einstein four-manifold with
strictly positive sectional curvature, without an orientability
assumption. There is a constant $a>0$ such that $(M,ag)$ is isometric to
the standard round $S^4$, to $\mathbb{CP}^2$ with its Fubini--Study
metric, or to $\mathbb{RP}^4$ with its standard round metric.
\end{theorem}
Strict sectional positivity makes the Ricci tensor positive definite,
so the Einstein constant is positive. The oriented classification is the
principal task. Until the final subsection of
Section~\ref{sec:classification}, we therefore work on a connected
oriented smooth closed four-manifold, choose an orientation, and rescale
the metric so that $\operatorname{Ric}=3g$. The final orientation-cover
argument uses the oriented conclusion proved there and determines the
possible nonorientable quotient.
\paragraph{Earlier results.}
The problem combines rigidity of the Einstein equation with the global
restrictions of positive sectional curvature. Berger's early rigidity
work under strict quarter pinching~\cite{Berger1961} was followed by
quantitative sectional-curvature bounds of Yang, Costa, and
Cao--Tran~\cite{Yang2000,Costa2004,CaoTran2018}. These bounds impose
additional quantitative hypotheses, some expressed as upper bounds
relative to the Einstein constant rather than as pinching ratios.
Gursky--LeBrun prove the Fubini--Study conclusion for compact oriented
Einstein manifolds with nonnegative sectional curvature and a nonzero
positive-definite intersection form~\cite[Theorem~A]{GurskyLeBrun1999}.
Koca proves the Fubini--Study conclusion for a complete positively sectionally
curved Einstein metric Hermitian for an integrable complex
structure~\cite[Theorem~2]{Koca2014}. Under the different condition of
nonnegative isotropic curvature, the four-dimensional local-symmetry
result is due to Micallef--Wang~\cite[Theorem~4.4]{MicallefWang1993};
Brendle extends the Einstein rigidity result to higher dimensions
\cite[Theorem~1]{Brendle2010}.
Recent results impose other geometric or topological conditions.
Dameno obtains half-conformal flatness from fibrewise constancy of the
scalar curvature of an associated twistor metric, and obtains the
standard models in the compact positive-scalar case
\cite[Theorem~1.1 and Corollary~1.2]{Dameno2026}.
Catino--Dameno use separation of the largest and middle eigenvalues
of one Weyl block to obtain conformal K\"ahler geometry after at most a double cover
\cite[Theorems~1.1 and~1.4]{CatinoDameno2026}.
Di Cerbo treats the closed oriented nonflat case with nonnegative
sectional curvature and equality everywhere in a Weyl-norm bound,
obtaining a locally symmetric $S^2\times S^2$ universal cover
\cite[Proposition~3.2]{DiCerbo2026}.
Write $\chi$ for the Euler characteristic and $\tau$ for the signature.
Cheng proves the oriented positive-sectional classification under
$\chi\le3$, and consequently in the presence of a nontrivial Killing
field~\cite[Theorem~1.2 and Corollary~1.4]{Cheng2026}. His argument
combines radial Riccati comparison with Weyl-norm gaps and
characteristic-number identities. Gursky--Malchiodi prove the classification for closed oriented Einstein
four-manifolds with positive sectional curvature under
$2\chi-3|\tau|\le4$, using a Green-function comparison and ADM mass
\cite[Theorem~1.1]{GurskyMalchiodi2026}. These cited results retain
additional curvature, geometric, topological, or symmetry hypotheses.
The argument below addresses the interaction of the two Weyl blocks
without such an additional restriction.
\paragraph{The two-block problem.}
The Hodge star on two-forms has eigenvalues $+1$ and $-1$; its two
rank-three eigenbundles are denoted by $\Lambda^+$ and $\Lambda^-$.
In the normalization $\operatorname{Ric}=3g$, the curvature operator
restricts to $\operatorname{Id}+T$ and $\operatorname{Id}+U$ on these
bundles, where $T,U$ are symmetric trace-free endomorphisms. They are
the two Weyl curvature blocks. The Einstein equation removes the
off-diagonal Ricci block but does not force either Weyl block to vanish.
Our central intermediate conclusion is \emph{half-conformal flatness}:
\[
T\equiv0\qquad\text{or}\qquad U\equiv0.
\]
Use Hilbert--Schmidt norms and put
\[
v=|T|/\sqrt6,\qquad b=|U|/\sqrt6,\qquad
x=-\lambda_{\min}(T),\qquad y=-\lambda_{\min}(U).
\]
The exact sectional-positivity condition is $x+y<2$. The inequalities
$v\le x\le2v$ and $b\le y\le2b$ imply $v+b<2$, but this weaker norm
condition alone is insufficient for the tensor estimate below.
Gursky--LeBrun use the corresponding nonstrict cone and norm inequalities
and analyze the case of two nonzero Weyl blocks
\cite[Lemma~1 and Theorem~B]{GurskyLeBrun1999}.
For a function $f$, write $\mathbf E f$ for its integral divided by total
Riemannian volume. The gap and its parallel equality case are established
in Gursky--LeBrun~\cite[Theorem~1]{GurskyLeBrun1999}: a nonzero block
satisfies $\mathbf E v^2\ge1$, or the exchanged inequality for $b$.
We derive this statement in our normalization, including zeros of a block.
Scalar coupling of the two blocks also has a clear antecedent. Yang
chooses a constant $t$ for which $|W^+|-t|W^-|$ has mean zero and combines
the Weyl equation, refined Kato inequality, and Poincar\'e inequality
\cite[\S6]{Yang2000}. Cao--Tran similarly balance regularized powers of
the norms~\cite[Proposition~2.2]{CaoTran2018}. Here a volume and moment
argument forces equality of the two second moments, and a weighted
Hessian square supplies the additional tensor estimate.
\paragraph{From moment balance to half-conformal flatness.}
The characteristic identities relate the two second moments to volume
and to the dimensions of harmonic self-dual and anti-self-dual two-form
spaces. An upper polar-volume deficit, the two gaps, and a strengthened
scalar variance inequality force
$\mathbf E v^2=\mathbf E b^2$ when both blocks are nonzero
(Proposition~\ref{prop:balanced-moments}). This balance controls the mean
of $v-b$, enabling the variance inequality to contribute the sign needed
in the final integral argument.
The main differential estimate uses the Bianchi identity twice. It first
places the gradient of a Weyl block in one irreducible tensor type.
The curvature commutator then determines all but one component of its
full, generally nonsymmetric Hessian. A curvature-weighted quadratic
form is positive on the remaining component because its lower bound is
$4-2x-2y>0$. Completing its square with different scalar weights for the
two blocks and integrating by parts produces inequalities involving only
first derivatives (Proposition~\ref{prop:weighted-hessian}). The proof
retains both the derivatives of the scalar weights and the derivative of
the curvature operator acting on the matrix slot. This local estimate
does not use moment balance.
The two weighted inequalities, a scalar identity, and the variance bound
produce an integrand $\mathcal I$ with nonnegative mean. Explicit rational
comparison functions give $c\ge1/10$, nonnegative quantities $r,j$ built
from the norm gradients, and a scalar $\Theta$ such that
\[
\mathbf E\mathcal I\ge0,\qquad
\mathcal I\le-\frac{(\Theta-cr)^2}{c}-c(j-r^2),
\qquad r^2\le j.
\]
The pointwise bounds follow from the weighted Hessian calculation and
exact finite polynomial signs. Moreover, $r^2<j$ almost everywhere that
either norm gradient is nonzero. Integration therefore makes both norms
constant. If both blocks were nonzero, their separate gaps would then
contradict $v+b<2$. This proves half-conformal flatness.
The final geometric argument has a different role. Once one block
vanishes, the characteristic identities leave two harmonic counts. The
exceptional count forces $s_0=\mathbf E(v^2+b^2)=3$ and volume
$4\pi^2/3$; a conditional lower polar-volume estimate excludes precisely
that case. Equality in the surviving Weyl gap makes the curvature
parallel. Matching its tensor with a standard model and using simple
connectivity then gives a global isometry. Figure~\ref{fig:proof-map}
records the assembly.
\input{figures/proof-map.tex}
The radial estimates use discrete Jacobi determinants and retain the
order of curvature matrices along each geodesic. The lower refinement is
conditional on the moment value just described. The article gives the
geometric and analytic arguments, together with the finite polynomial
formulas and sign proofs. The accompanying checker reproduces the saved
finite arithmetic within the precise scope stated in its coverage record.
\paragraph{Organization.}
Section~\ref{sec:preliminaries} fixes curvature conventions and derives
the differential identities, including their behavior at zeros.
Section~\ref{sec:analytic} proves the weighted variance and Weyl-gap
estimates and states the characteristic identities.
Section~\ref{sec:volume} proves the upper deficit and moment balance.
Section~\ref{sec:hessian} derives the weighted Hessian inequality, and
Section~\ref{sec:coupling} combines it with the scalar estimates.
Section~\ref{sec:certificates} proves the resulting finite signs.
Section~\ref{sec:classification} proves one-block vanishing, the
conditional lower volume bound, and the full metric classification.
Appendix~\ref{app:analytic} contains the deferred analytic and heat
proofs, while Appendix~\ref{app:topology} proves the required topology.
==== END Positively-curved-Einstein-four-manifolds//sections/01-introduction.tex ====
==== BEGIN Positively-curved-Einstein-four-manifolds//sections/02-preliminaries.tex ====
\section{Curvature blocks and their covariant derivatives}
\label{sec:preliminaries}
Our first task is to express sectional positivity and the differential
Bianchi identity in a common set of variables. The curvature decomposition
is classical; see Derdzi\'nski~\cite[Section~2]{Derdzinski1983}.
We give the algebra and fix its normalizations, since the later Hessian
estimate depends on the precise representation constants.
Rescale the metric so that $\operatorname{Ric}=3g$, and write
\[
d\nu=\frac{d\operatorname{vol}_g}{\operatorname{Vol}(M)},
\qquad \mathbf E f=\int_M f\,d\nu.
\]
The Laplacian $\Delta$ is the trace of the second covariant derivative.
Our curvature convention is
\[
[\nabla_i,\nabla_j]\xi_k=R_{ijka}\xi_a
\]
for a covector, with $R_{ijij}>0$ on positively curved orthonormal
two-planes. All tensor connections are induced by the Levi--Civita
connection. We take covariant derivatives before choosing any
pointwise eigenbasis.
\subsection{The two Weyl blocks}
The Hodge-star eigenspaces $\Lambda^+$ and $\Lambda^-$ have eigenvalues
$+1$ and $-1$, respectively. On an oriented Euclidean four-space, choose
triples of skew orthogonal
complex structures $I_1,I_2,I_3$ and $J_1,J_2,J_3$ satisfying
\[
I_1I_2=I_3=-I_2I_1,\qquad
J_1J_2=J_3=-J_2J_1,
\]
and their cyclic analogues, with every $I_s$ commuting with every $J_t$.
The two triples can be constructed from left and right quaternionic
multiplication. As two-forms, $I_s/\sqrt2$ and $J_s/\sqrt2$ are
orthonormal bases of $\Lambda^+$ and $\Lambda^-$, respectively.
Vector and covector indices are identified by the metric.
\begin{lemma}[Curvature blocks and sectional positivity]
\label{lem:curvature-blocks}
There are symmetric trace-free endomorphisms
$T\in\operatorname{Sym}_0(\Lambda^+)$ and
$U\in\operatorname{Sym}_0(\Lambda^-)$ such that the curvature operator is
\[
\mathcal R_{\Lambda^2}
=\begin{pmatrix}\operatorname{Id}+T&0\\
0&\operatorname{Id}+U\end{pmatrix}.
\]
At a point where these endomorphisms are diagonal, with eigenvalues
$t_s,u_s$, the curvature tensor is
\[
R_{ikjl}
=\frac12\sum_s(1+t_s)I_{s,ik}I_{s,jl}
+\frac12\sum_s(1+u_s)J_{s,ik}J_{s,jl}.
\]
Put
\[
x=-\min_s t_s,\qquad y=-\min_s u_s,\qquad
v=\frac{|T|}{\sqrt6},\qquad b=\frac{|U|}{\sqrt6},
\]
where the matrix norms are Hilbert--Schmidt norms. Then
\[
\sec_g>0\quad\Longleftrightarrow\quad x+y<2,
\qquad
v\le x\le2v,\qquad b\le y\le2b.
\]
Both Weyl blocks, viewed as curvature four-tensors, have zero
divergence. In particular, for $Z=\nabla T$, whose matrix entries are
one-forms, the differential Bianchi constraint is
\[
\sum_s I_s Z_{sr}=0\qquad(r=1,2,3).
\]
\end{lemma}
\begin{proof}
The first Bianchi identity says that the traces of the two diagonal
blocks of the curvature operator agree. Their sum is half the scalar
curvature, hence is $6$. Contraction identifies the off-diagonal block
with the trace-free Ricci tensor: its nine entries give the coefficients
in the independent symmetric trace-free matrices $I_sJ_t$.
Thus the Einstein condition makes this block zero, and each diagonal
block has trace $3$. Expanding in the orthonormal two-form bases gives
the displayed curvature tensor.
A unit simple two-form has self-dual and anti-self-dual parts of norm
$1/\sqrt2$. Conversely, a unit two-form whose two parts have these
norms has exterior square zero, and is simple: in its orthogonal
normal form $\alpha=a\,e^1\wedge e^2+b\,e^3\wedge e^4$, the condition
$\alpha\wedge\alpha=0$ is $ab=0$. Therefore all pairs of plus and minus
parts of norm $1/\sqrt2$ occur. The smallest sectional curvature is
$1-(x+y)/2$, proving the equivalence.
For the norm comparison, write the eigenvalues of $T$ as
$-x,a,x-a$. Minimality of $-x$ gives $-x\le a\le2x$, and hence
\[
\frac32x^2
\le x^2+a^2+(x-a)^2
\le6x^2.
\]
Division by $6$ gives $v\le x\le2v$. The proof for $U$ is identical.
Contracting the differential Bianchi identity shows that the curvature
tensor is divergence-free when Ricci is parallel. Project its last
pair of indices onto $\Lambda^+$ or $\Lambda^-$; these projections
are parallel. Subtracting the parallel scalar-curvature contribution
proves divergence freedom of each Weyl block. In a quaternionic basis,
the resulting constraint on $\nabla T$ is exactly the stated identity.
\end{proof}
\subsection{Algebra of the derivative representations}
Let $\Sigma_s$ denote the infinitesimal action of $I_s$ on the plus
indices of a tensor representation. On $\Lambda^+$ it is rotation
about axis $s$ with speed $2$. A complexified irreducible representation
has \emph{spin $j$} when
\[
\sum_{s=1}^3\Sigma_s^2=-4j(j+1)\operatorname{Id}.
\]
Pairs $(j,\ell)$ specify the spins of the two commuting rotation
algebras. This notation describes associated tensor bundles of the
oriented orthonormal frame bundle; it requires no spin structure.
\begin{lemma}[Representation rules]
\label{lem:spin-algebra}
A spin-$j$ representation has weights
$-j,-j+1,\ldots,j$ for $\Sigma_3/(2i)$, each once, and
\[
H_j\otimes H_\ell
\cong H_{j+\ell}\oplus H_{j+\ell-1}\oplus\cdots\oplus H_{|j-\ell|}.
\]
In particular, the tangent, plus two-form, and plus trace-free
symmetric-matrix representations have types
\[
\left(\frac12,\frac12\right),\qquad (1,0),\qquad (2,0).
\]
The Casimir operators and the orthogonal projections onto these
irreducible types are parallel bundle endomorphisms.
\end{lemma}
\begin{proof}
The generators are skew-Hermitian and satisfy
$[\Sigma_1,\Sigma_2]=2\Sigma_3$, cyclically. Diagonalize
$\Sigma_3/(2i)$ and take the usual conjugate raising and lowering
operators, which shift its weight by one. If a unit highest vector
has weight $j$, commuting the raising and lowering operators gives
the squared lowering coefficient at weight $j-r$ as
\[
(r+1)(2j-r).
\]
Nonnegativity and termination of the chain imply that $2j$ is a
nonnegative integer and give the listed weights and Casimir.
Taking invariant orthogonal complements decomposes an arbitrary
representation. Multiplying the weight lists, and successively
removing the chain of highest remaining weight, proves the tensor
product rule with multiplicity one. Commuting actions operate on the
corresponding multiplicity spaces. The same chain description shows
that an intertwiner between single irreducible copies is a scalar.
The actions on the tangent space and on $\Lambda^+$ give the first
two types. The symmetric square of the spin-one representation is
spin two plus the scalar trace, giving the third. Finally, these
splittings are defined by invariant algebraic operators. The
connection induced by the orthonormal frame bundle preserves every
such operator. Thus the projections are parallel even though a
chosen local frame, or a curvature eigenbasis, need not be parallel.
\end{proof}
Write $F=\operatorname{Sym}_0(\Lambda^+)$. For a symmetric trace-free
matrix $A$, define the self-adjoint operator on $F$
\[
B_F(A)S=\frac{AS+SA}{2}
-\frac{\operatorname{tr}(AS)}3\operatorname{Id}.
\]
It is characterized by
$\langle B_F(A)S,S\rangle=\operatorname{tr}(AS^2)$.
On any plus spin-$j$ representation with $j\ge1$, define
\[
Q_j(A)=\frac{\sum_{r,s}A_{rs}\Sigma_r\Sigma_s}{4j(2j-1)}.
\]
For diagonal $A$ the numerator is $\sum_s a_s\Sigma_s^2$.
\begin{lemma}[Curvature-weighted operators]
\label{lem:quadratic-curvature-action}
If $A$ has smallest eigenvalue $-x$, then
\[
Q_j(A)\le \frac{x(j+1)}{2j-1}\operatorname{Id}.
\]
Moreover, $Q_2(A)=B_F(A)$. The compression of $B_F(A)$, acting on an
$F$ factor, to a highest-spin coupling of total plus spin $j$ is
$Q_j(A)$, provided the operator acts trivially on any additional
irreducible minus factor.
\end{lemma}
\begin{proof}
In a diagonal basis, $a_s+x\ge0$ and every $\Sigma_s^2$ is
nonpositive. Thus
\[
\sum_s a_s\Sigma_s^2
\le -x\sum_s\Sigma_s^2
=4xj(j+1)\operatorname{Id},
\]
which proves the bound.
To identify $Q_2$, diagonalize $S=\operatorname{diag}(d_1,d_2,d_3)$.
The speed-two matrix action gives
$|\Sigma_sS|^2=8(d_j-d_k)^2$, with $s,j,k$ cyclic; these three
derivatives are mutually orthogonal. Consequently, allowing $A$
to be nondiagonal in this basis,
\[
\begin{aligned}
\langle Q_2(A)S,S\rangle
&=-\frac13\sum_s A_{ss}(d_j-d_k)^2\\
&=-\frac13\sum_s A_{ss}
\bigl(2\operatorname{tr}(S^2)-3d_s^2\bigr)
=\operatorname{tr}(AS^2).
\end{aligned}
\]
Here trace-freeness of both matrices gives the middle identity and
removes the scalar term. Polarization proves the operator equality.
By Lemma~\ref{lem:spin-algebra}, spin two occurs once in
$\operatorname{End}(H_j)\cong H_j\otimes H_j$. Hence an equivariant
operator family linear in $A$ is unique up to scalar. The same is
true with an irreducible minus factor when the operators commute
with its action. Test the scalar at
$A=\operatorname{diag}(-1,-1,2)$ on a highest vector. At weight $m$,
\[
Q_j(A)=\frac{j(j+1)-3m^2}{j(2j-1)},
\]
so its highest-weight eigenvalue is $-1$. The $F$ highest vector
also has $B_F(A)$ eigenvalue $-1$. A product of highest vectors is
the highest vector of the stretched coupling, proving the claimed
compression. For the same reason, compression of a single generator
from a spin-$\ell$ factor to a stretched spin-$j$ product is
$\ell/j$ times the total generator.
\end{proof}
\begin{proposition}[Differential equations for the Weyl blocks]
\label{prop:weyl-derivative-equations}
The tensor $Z=\nabla T$ belongs to the parallel subbundle
$G$ of type $(5/2,1/2)$. On this bundle, put
$B_{\mathrm{op}}=Q_{5/2}(T)$. Then
\[
\Delta T=6(T-N),\qquad
N=(T^2)_0=T^2-2v^2\operatorname{Id},
\]
and
\begin{equation}
\Delta Z=(13-20B_{\mathrm{op}})Z.
\tag{1}\label{eq:1}
\end{equation}
Every statement has an exchanged counterpart for $U$.
\end{proposition}
\begin{proof}
The derivative space $T^*M\otimes F$ has types
$(5/2,1/2)$ and $(3/2,1/2)$. The divergence takes values in a space
whose plus spins are at most $3/2$. It is nonzero: a nonzero covector
tensored with $\operatorname{diag}(1,-1,0)$ has nonzero divergence.
It therefore vanishes on the first summand and is injective on the
second, by irreducibility. Lemma~\ref{lem:curvature-blocks} puts $Z$
in the first summand.
Let
\[
C=\sum_s I_s\otimes\Sigma_s
\]
on $T^*M\otimes F$. Expansion of the combined Casimir gives
$C=-4$ on $G$ and $C=6$ on its complement. In particular
$P_G=(6-C)/10$ is a parallel orthogonal projector. The bundle
curvature on $F$ is
\[
F_{ik}=\frac12\sum_s(1+t_s)I_{s,ik}\Sigma_s
\]
at a point of diagonalization. Taking the covariant divergence of
$CZ=-4Z$, using skewness of $I_s$ to take a commutator and
$\sum_{ik}I_{s,ik}I_{r,ik}=4\delta_{sr}$, gives
\[
-4\Delta T=\sum_s(1+t_s)\Sigma_s^2T
=-24T+24B_F(T)T=-24(T-N).
\]
To differentiate this equation, note first that
$\nabla N=2B_F(T)Z$. For each derivative index $j$, the covariant
commutation formula reads
\[
\Delta Z_j
=\nabla_j\Delta T+\nabla^i(F_{ij}T)
+F_{ij}\nabla^iT
+\operatorname{Ric}_{ja}Z_a.
\]
The divergence of $F$ vanishes because it is the induced action of
the divergence-free Riemann curvature. Thus the differentiated
equation contributes $6Z-12B_F(T)Z$, the Ricci term contributes
$3Z$, and the remaining bundle term is
$-\sum_s(1+t_s)I_s\Sigma_sZ$. Its constant part is $-CZ=4Z$.
The left side already lies in $G$, since $P_G$ is parallel and
$P_GZ=Z$. We may therefore compress the right side to $G$,
although its individual summands need not preserve $G$.
Lemma~\ref{lem:quadratic-curvature-action} gives
$P_GB_F(T)P_G=Q_{5/2}(T)$. On $G$, the total plus generators are
$I_s+\Sigma_s$, so
\[
40Q_{5/2}(T)
=P_G\sum_s t_s(I_s+\Sigma_s)^2P_G.
\]
The $I_s^2$ term vanishes by $\sum_s t_s=0$, and the $\Sigma_s^2$
term compresses to $24Q_{5/2}(T)$. Hence
\[
P_G\sum_s t_s I_s\Sigma_sP_G=8Q_{5/2}(T).
\]
The scalar contributions sum to $13$, while the two curvature
contributions sum to $-20Q_{5/2}(T)$, proving \eqref{eq:1}.
For a nondiagonal $T$ the same calculation uses the symmetric
contraction $\sum_{r,s}T_{rs}(I_r+\Sigma_r)(I_s+\Sigma_s)$:
its $I_rI_s$ part is zero by symmetry and trace-freeness.
Thus the computation is tensorial and entails no differentiation
of a moving eigenframe.
\end{proof}
\subsection{The gradient parametrization and the norm equation}
Write
\[
p=\nabla v,\qquad q=\nabla b,\qquad
e=\frac{|\nabla T|^2}{6},\qquad
\bar e=\frac{|\nabla U|^2}{6}.
\]
The norms $v,b$ are Lipschitz. Their gradients are understood almost
everywhere and vanish almost everywhere on the respective zero sets.
Where $v>0$, diagonalize $T$ and put, cyclically in $i,j,k$,
\[
t_i=vs_i,\qquad
\widetilde s_i=\frac{s_j-s_k}{\sqrt3},\qquad
\sigma=\frac16\sum_i s_i^3,\qquad
\rho=\frac16\sum_i s_i^2\widetilde s_i.
\]
Here $\sum s_i=0$ and $\sum s_i^2=6$. On the plane of triples with
zero sum, the map $s\mapsto\widetilde s$ is a skew-adjoint orthogonal
quarter-turn. Thus
\[
\sum_i\widetilde s_i=0,\qquad
\sum_i\widetilde s_i^2=6,\qquad
\sum_i s_i\widetilde s_i=0,\qquad
s_i^2+\widetilde s_i^2=4.
\]
For later use, these identities imply
\[
\sigma^2+\rho^2=1,\qquad
\frac{N}{v^2}
=\operatorname{diag}(\sigma s_i+\rho\widetilde s_i),
\qquad
\frac16\sum_i\widetilde s_i^3=-\rho.
\]
Indeed, the trace-free triple $n_i=s_i^2-2$ has squared norm $6$:
use $\sum_i s_i^4=(\sum_i s_i^2)^2/2=18$.
Its coefficients in the orthogonal basis $s,\widetilde s$ are
$\sigma,\rho$, proving the first two assertions. The last follows
by factoring the alternating cubic
$\sum_i s_i^2(s_j-s_k)$. If $r=x/v$, the other two eigenvalues
besides $-r$ have sum $r$ and product $r^2-3$, so
\[
1\le r\le2,\qquad
\sigma=\frac{3r-r^3}{2}.
\]
Normalized quantities are used only where the corresponding block
is nonzero. Polynomial products such as
$\sigma v^3=\operatorname{tr}(T^3)/6$ have their natural extensions
across its zero set. Bars denote the analogous quantities for $U$.
The next formula gives, in particular, the refined Kato inequality
$|\nabla T|^2\ge(5/3)|\nabla|T||^2$ for a half-Weyl block.
Gursky--LeBrun derive this estimate from the same irreducible Bianchi
gradient constraint~\cite[Lemma~4, equations~(10)--(12)]{GurskyLeBrun1999};
Calderbank--Gauduchon--Herzlich give its general representation-theoretic
setting~\cite[equation~(7.1)]{CalderbankGauduchonHerzlich2000}.
We retain all three vector parameters: the excess over the norm-gradient
energy will be needed in the coupled estimate.
\begin{lemma}[Parametrization of the Bianchi gradient space]
\label{lem:gradient-parametrization}
There are unique vectors $k,l$ such that
\begin{equation}
\begin{gathered}
Z_{ii}=s_i p-\sqrt3\,\widetilde s_i k,\qquad
Z_{jk}=-I_i\left(l+s_i k+\frac{\widetilde s_i}{\sqrt3}p\right),
\qquad (i,j,k)\ \text{cyclic},\\
e=\frac53|p|^2+5|k|^2+|l|^2.
\end{gathered}
\tag{2}\label{eq:2}
\end{equation}
More generally, for any fixed normalized trace-free diagonal triple
$s$, the same formulas parametrize the entire algebraic space $G$
by three arbitrary real four-vectors.
\end{lemma}
\begin{proof}
Write $D_i=Z_{ii}$ and $a_i=I_iZ_{jk}$. Quaternion multiplication
turns the three divergence equations into
$D_i+a_j-a_k=0$. Trace-freeness and the orthogonal basis
$s,\widetilde s$ give uniquely
\[
p=\frac16\sum_i s_iD_i,\qquad
k=-\frac1{6\sqrt3}\sum_i\widetilde s_iD_i,\qquad
D_i=s_ip-\sqrt3\,\widetilde s_i k.
\]
Since
$s_j-s_k=\sqrt3\,\widetilde s_i$ and
$\widetilde s_j-\widetilde s_k=-\sqrt3\,s_i$,
the vectors $s_i k+\widetilde s_i p/\sqrt3$ have exactly the
differences required of the $a_i$. Their difference from $a_i$ is
therefore a common vector $l$, proving the formulas and their
converse. This also establishes their full algebraic scope.
Orthogonality of $s,\widetilde s$ gives
\[
\sum_i|Z_{ii}|^2=6|p|^2+18|k|^2,\qquad
2\sum_{j<k}|Z_{jk}|^2=4|p|^2+12|k|^2+6|l|^2.
\]
The factor two in the second sum is the Hilbert--Schmidt multiplicity
of off-diagonal entries. Division by six proves the norm identity.
For $Z=\nabla T$,
$\nabla|T|^2=2v\sum_i s_iZ_{ii}=12vp$; comparison with
$|T|^2=6v^2$ identifies the parameter $p$ with $\nabla v$.
\end{proof}
The pointwise norm equation alone is insufficient for integration:
a Weyl block may vanish. We next prove that its regularized Laplacians
converge to an actual integrable density, so no measure supported on the
zero set is lost.
\begin{proposition}[Weak norm equation, including the zero set]
\label{prop:weak-norm-equation}
The function $e/v$, defined to be zero on $\{v=0\}$, belongs to
$L^1(M)$, and
\begin{equation}
\Delta v=6v(1-\sigma v)+\frac{e-|p|^2}{v}
\tag{3}\label{eq:3}
\end{equation}
holds as an equality of distributions with an $L^1$ density.
The quotient is zero almost everywhere on the zero set.
This identity may be tested against bounded Lipschitz functions.
The corresponding assertions hold for $b$.
\end{proposition}
\begin{proof}
On $\{v>0\}$, pair the equation for $\Delta T$ with $T$ to obtain
\[
\Delta(v^2)=12(v^2-\sigma v^3)+2e.
\]
This is globally an identity of smooth functions. Put
$v_\varepsilon=\sqrt{v^2+\varepsilon^2}$. Direct differentiation gives
\[
\Delta v_\varepsilon
=\frac{6(v^2-\sigma v^3)}{v_\varepsilon}
+q_\varepsilon,\qquad
q_\varepsilon=\frac e{v_\varepsilon}
-\frac{v^2|p|^2}{v_\varepsilon^3}.
\]
The refined derivative estimate in \eqref{eq:2} yields
\[
\frac25\frac e{v_\varepsilon}
\le q_\varepsilon\le\frac e{v_\varepsilon}.
\]
These assertions hold almost everywhere, including the zero set:
every smooth component of $T$ has derivative zero almost everywhere
on its zero set, hence $Z=0$ almost everywhere on $\{T=0\}$.
One can see this elementary fact at density points of a level set;
a nonzero derivative would instead make that scalar level set a
local hypersurface.
The first term in $\Delta v_\varepsilon$ is uniformly bounded in
absolute value by $6(v+v^2)$. Its integral is the negative of the
integral of $q_\varepsilon$. Consequently
$\mathbf E(e/v_\varepsilon)$ is uniformly bounded.
Monotone convergence proves the asserted $L^1$ bound for $e/v$.
Off the zero set, $q_\varepsilon$ is dominated by $e/v$ and tends
to $(e-|p|^2)/v$; on the zero set both are zero almost everywhere.
Dominated convergence therefore gives an $L^1$ limit, rather than
only a weak limit of measures. The zeroth-order term converges in
$L^1$ as well. Since $v_\varepsilon\to v$ uniformly, their
Laplacians converge distributionally, proving \eqref{eq:3} without
a residual singular measure.
Finally, approximate a bounded Lipschitz test function uniformly and
in $H^1$ by smooth functions, using local mollification and a
partition of unity. Uniform convergence controls its pairing with
the $L^1$ right side, and $H^1$ convergence controls its gradient
pairing with $\nabla v$. This proves the asserted testing rule.
\end{proof}
==== END Positively-curved-Einstein-four-manifolds//sections/02-preliminaries.tex ====
==== BEGIN Positively-curved-Einstein-four-manifolds//sections/03-analytic.tex ====
\section{Scalar estimates and the characteristic identities}
\label{sec:analytic}
This section supplies two kinds of global information: analytic bounds
for the sizes of the Weyl blocks, and integer identities relating their
second moments to volume. The former will control derivatives in the
coupled inequality; the latter will restrict the possible imbalance
between the two blocks.
The standing setting is the closed, connected, oriented Einstein
four-manifold with $\operatorname{Ric}=3g$ fixed in the introduction.
Appendix~\ref{app:analytic} supplies the analytic foundations used here;
its proofs depend only on this setting and the curvature conventions of
Section~\ref{sec:preliminaries}. We retain the weighted variance argument
and the complete Weyl-gap proof in the main text because their precise
weights and behavior at zeros are used in the two-block argument.
\subsection{A strengthened scalar variance estimate}
\begin{proposition}[Weighted variance]
\label{prop:weighted-variance}
For every real $a\in H^1(M)$,
\begin{equation}
\operatorname{Var}(a)
:=\mathbf E\bigl[(a-\mathbf E a)^2\bigr]
\le\mathbf E\frac{|\nabla a|^2}{K},
\qquad K=4+v^2+b^2.
\tag{4}\label{eq:4}
\end{equation}
\end{proposition}
\begin{proof}
First let $\phi$ be smooth and write
$S=\operatorname{Hess}_0\phi$. The integrated Bochner identity
and $\operatorname{Ric}=3g$ give
\[
\mathbf E\bigl[(\Delta\phi)^2\bigr]
=\mathbf E\left(4|\nabla\phi|^2+\frac43|S|^2\right).
\]
Define operators on covariant two-tensors by
\[
(T_D)_{ikjl}=\frac12\sum_s t_s I_{s,ik}I_{s,jl},
\qquad
(U_D)_{ikjl}=\frac12\sum_s u_s J_{s,ik}J_{s,jl},
\]
where $ij$ is the row index and $kl$ is the column index.
These tensorial definitions may be evaluated after pointwise
diagonalization. They annihilate metric trace. On the symmetric
trace-free tensors, the nine matrices $I_iJ_j$ form an orthogonal
basis, with $T_D$ and $U_D$ eigenvalues $t_i$ and $u_j$.
For example, $T_D$ acts by
$-\frac12\sum_s t_s I_s(\,\cdot\,)I_s$; the quaternion relations
give the eigenvalue $t_i$ on $I_iJ_j$.
Thus
\[
T_D+U_D\le 2(x+y)\operatorname{Id}<4\operatorname{Id}
\]
on this space.
The integrated curvature contraction has a useful sign and a precise
coefficient. Since $T_D$ annihilates trace, replace $S$ by
$\operatorname{Hess}\phi$ in its quadratic form. Integrate the
derivative $k$, use $\nabla_k(T_D)_{ikjl}=0$, and use skewness in
$i,k$ to obtain
\[
\begin{aligned}
\mathbf E\langle T_DS,S\rangle
&=-\mathbf E (T_D)_{ikjl}\phi_l\nabla_k\nabla_i\phi_j\\
&=\frac12\mathbf E (T_D)_{ikjl}R_{ikja}\phi_a\phi_l.
\end{aligned}
\]
In this contraction the minus curvature block is orthogonal to the
plus block. Substituting their expansions gives
\[
\frac12\sum_{ikj}(T_D)_{ikjl}R_{ikja}
=\frac12\sum_s t_s(1+t_s)
\sum_j I_{s,jl}I_{s,ja}
=3v^2\delta_{la}.
\]
The final equality uses $\sum t_s=0$ and $\sum t_s^2=6v^2$.
Consequently
\[
\mathbf E\langle(T_D+U_D)S,S\rangle
=3\mathbf E\bigl[(v^2+b^2)|\nabla\phi|^2\bigr].
\]
The pointwise upper operator bound, inserted into the positive
$|S|^2$ term of the Bochner identity, now yields
\[
\mathbf E\bigl[(\Delta\phi)^2\bigr]\ge\mathbf E K|\nabla\phi|^2.
\]
For smooth $a$, the closed-manifold Poisson statement in
Appendix~\ref{app:analytic-preliminaries} gives a unique smooth
zero-mean solution of $\Delta\phi=a-\mathbf E a$. Integration
by parts and Cauchy--Schwarz with the pointwise weight $K$ give
\[
\begin{aligned}
\operatorname{Var}(a)
&=-\mathbf E\langle\nabla a,\nabla\phi\rangle\\
&\le
\left(\mathbf E\frac{|\nabla a|^2}{K}\right)^{1/2}
\left(\mathbf E K|\nabla\phi|^2\right)^{1/2}\\
&\le
\left(\mathbf E\frac{|\nabla a|^2}{K}\right)^{1/2}
\operatorname{Var}(a)^{1/2}.
\end{aligned}
\]
Division proves the assertion when the variance is nonzero; the
zero case is immediate. Smooth functions are dense in $H^1$ by the
local mollification argument recalled in
Appendix~\ref{app:analytic-preliminaries}. Since $K$ is bounded above
and below by positive constants, this approximation also passes the
weighted energy to the limit and proves the assertion on $H^1$.
\end{proof}
\subsection{The sharp scalar inequality and the Weyl gap}
The following is the four-dimensional Einstein case of Ilias's sharp
positive-Ricci Sobolev inequality~\cite[Theorem~3]{Ilias}. Its proof in
the present normalization is given in Appendix~\ref{app:sharp-sobolev}.
The nonlinear Hessian argument
belongs to the Bochner approach to elliptic rigidity; compare the
systematic treatment in~\cite{BidautVeronVeron}.
\begin{proposition}[Normalized Sobolev inequality]
\label{prop:sharp-sobolev}
For every real $f\in H^1(M)$,
\[
\bigl(\mathbf E|f|^4\bigr)^{1/2}
\le\mathbf E\left(f^2+\frac12|\nabla f|^2\right).
\]
\end{proposition}
The next proposition is the Weyl-norm gap established in
Gursky--LeBrun~\cite[Theorem~1]{GurskyLeBrun1999}; we give the derivation in
our normalization, including the equality case and the possible
zeros of the Weyl block.
\begin{proposition}[Weyl-norm gap]
\label{prop:weyl-gap}
For either Weyl block,
\begin{equation}
v\equiv0\qquad\text{or}\qquad \mathbf E v^2\ge1.
\tag{5}\label{eq:5}
\end{equation}
If $T\not\equiv0$ and $\mathbf E v^2=1$, then
$\nabla T=0$ and $v=\sigma=1$. The corresponding statements hold
for $U$.
\end{proposition}
\begin{proof}
Put $A_\varepsilon=v^2+\varepsilon^2$ and
$f_\varepsilon=A_\varepsilon^{1/6}$. These are smooth functions.
The equation for $\Delta(v^2)$ and \eqref{eq:2} give
\[
\begin{aligned}
\Delta f_\varepsilon
=A_\varepsilon^{-5/6}
\left[
2(v^2-\sigma v^3)+\frac13e
-\frac59\frac{v^2}{A_\varepsilon}|p|^2
\right]
\ge 2A_\varepsilon^{-5/6}(v^2-v^3).
\end{aligned}
\]
Indeed, $\sigma\le1$, $v^2/A_\varepsilon\le1$, and
$e\ge5|p|^2/3$. Multiply by $f_\varepsilon$ and integrate:
\[
\mathbf E|\nabla f_\varepsilon|^2
\le 2\mathbf E\frac{v^3-v^2}{A_\varepsilon^{2/3}}
\le 2\mathbf E v^{5/3}.
\]
Their $L^2$ norms are uniformly bounded as well.
Thus $f_\varepsilon$ is bounded in $H^1$, and its uniform limit
$f=v^{1/3}$ belongs to $H^1$ by weak compactness. This bound is
obtained directly from the regularized equation; it requires no
positive lower bound on $v$.
Apply Proposition~\ref{prop:sharp-sobolev} to $f_\varepsilon$
and use the preceding energy bound:
\[
\begin{aligned}
\bigl(\mathbf E f_\varepsilon^4\bigr)^{1/2}
&\le \mathbf E\left[
A_\varepsilon^{1/3}
+\frac{v^3-v^2}{A_\varepsilon^{2/3}}\right]\\
&=\mathbf E\frac{\varepsilon^2+v^3}{A_\varepsilon^{2/3}}.
\end{aligned}
\]
The term $\varepsilon^2/A_\varepsilon^{2/3}$ is at most
$\varepsilon^{2/3}$, while
$v^3/A_\varepsilon^{2/3}\le v^{5/3}$.
Dominated convergence therefore gives
\[
\bigl(\mathbf E f^4\bigr)^{1/2}
\le\mathbf E(f^2v)
\le\bigl(\mathbf E f^4\bigr)^{1/2}
\bigl(\mathbf E v^2\bigr)^{1/2}.
\]
When $T\not\equiv0$, the common factor is positive, proving
\eqref{eq:5}.
If $\mathbf E v^2=1$, equality holds in Cauchy--Schwarz.
Thus $v=\alpha f^2$ almost everywhere for a constant $\alpha>0$.
It follows that $v$ takes only the values $0$ and $\alpha^3$
almost everywhere, and hence everywhere by continuity.
Connectedness and $T\not\equiv0$ force the positive constant value
on all of $M$. The moment condition gives $v=1$.
Equation~\eqref{eq:3} then becomes
$0=6(1-\sigma)+e$. Both terms are nonnegative, so
$\sigma=1$ and $e=0$. Smoothness gives $\nabla T=0$ everywhere.
\end{proof}
\subsection{The Euler and signature formulas}
\begin{proposition}[Characteristic identities]
\label{prop:characteristic-identities}
Let $n_+$ and $n_-$ be the finite nonnegative integer dimensions of
the spaces of harmonic self-dual and anti-self-dual two-forms. The Euler
characteristic and signature of $M$ are, respectively,
\[
\chi=2+n_++n_-,\qquad \tau=n_+-n_-.
\]
Then
\begin{equation}
\begin{aligned}
8\pi^2\chi
&=6\operatorname{Vol}(M)
\bigl(1+\mathbf E(v^2+b^2)\bigr),\\
12\pi^2\tau
&=6\operatorname{Vol}(M)\,\mathbf E(v^2-b^2).
\end{aligned}
\tag{6}\label{eq:6}
\end{equation}
If $U=0$, then $n_-=0$; likewise $T=0$ implies $n_+=0$.
\end{proposition}
These are the Euler and signature formulas in the present curvature
normalization. Appendix~\ref{app:characteristic-identities} identifies
the topological integers through Hodge theory and proves the formulas
by the form-complex heat-supertrace method of
McKean--Singer~\cite{McKeanSinger1967}; compare the general formulations
and local coefficients in Gilkey~\cite[Theorems~1.7.6 and
4.8.16]{Gilkey1984}. That appendix also proves the zero-block kernel
vanishing statements, which will be needed after one-block vanishing.
==== END Positively-curved-Einstein-four-manifolds//sections/03-analytic.tex ====
==== BEGIN Positively-curved-Einstein-four-manifolds//sections/04-volume.tex ====
\section{Upper volume deficit and balanced Weyl moments}
\label{sec:volume}
Write $V_M=\operatorname{Vol}(M,g)$ and $s_0=\mathbf E(v^2+b^2)$.
Our goal in this section is to prove that the two second moments agree
whenever neither Weyl block vanishes. The characteristic identities
turn an upper polar-volume deficit into an integer restriction; a
weighted scalar variance estimate then excludes the sole remaining
unequal-moment case.
Radial comparison also plays a central role in Cheng's classification
under the additional assumption $\chi(M)\le3$
\cite[Proposition~3.1]{Cheng2026}. His Riccati argument does not require
curvature matrices at different times to commute. We use a discrete
Jacobi determinant to retain their order and obtain the following
moment-dependent deficit.
\begin{proposition}[Upper volume deficit]\label{prop:upper-volume-deficit}
One has
\[
\frac{V_M}{8\pi^2/3}\le 1-\frac{3}{40}s_0.
\tag{7}\label{eq:7}
\]
\end{proposition}
Here is how the estimate will be used. If both blocks are nonzero, their
gaps and $v+b<2$ give $2\le s_0<4$. Equations~\eqref{eq:6} and
\eqref{eq:7} then imply $\chi<7$ and $|\tau|/\chi<4/15$.
The parity and nonnegativity constraints from $n_\pm$ leave only
$(\chi,|\tau|)=(5,1)$ when $\tau\ne0$. The last part of this section
rules out that case and proves Proposition~\ref{prop:balanced-moments}.
We first establish the upper deficit and the common radial machinery;
the conditional lower comparison needed after one-block vanishing is
proved in Section~\ref{sec:classification}.
Along a unit-speed geodesic, choose a parallel orthonormal normal frame
and let $\mathcal R(t)$ be the normal Jacobi curvature matrix. Set
\[
A(t)=\mathcal R(t)-\mathrm{Id},\qquad
\mathcal J''+\mathcal R\mathcal J=0,\qquad
\mathcal J(0)=0,\quad \mathcal J'(0)=\mathrm{Id},\qquad
D(t)=\det\mathcal J(t).
\]
Positive sectional curvature and $\operatorname{Ric}=3g$ give
$0<\mathcal R<3\mathrm{Id}$ and $\operatorname{tr}A=0$.
Let $\mathcal E$ denote integration over the unit tangent bundle against
normalized Riemannian measure on the base and normalized spherical
measure in the fibers.
\begin{lemma}[Geodesic averages]\label{lem:geodesic-averages}
At every time along the geodesic flow,
\[
\mathcal E|A|^2=\frac32s_0,\qquad
\mathcal E|A'|^2\le\frac32\,\mathcal E|A|^2.
\tag{8}\label{eq:8}
\]
\end{lemma}
\begin{proof}
The indicated probability measure is invariant under geodesic flow.
Indeed, in cotangent coordinates the Hamiltonian flow of kinetic energy
preserves canonical volume: its divergence vanishes by equality of the
mixed second derivatives of the Hamiltonian. Applying this observation
to energy shells gives invariance of the measure on the unit-energy
level. In orthonormal fiber coordinates this is precisely Riemannian
base volume times spherical measure.
The plus and minus contributions to $A$ are orthogonal conjugates of
$T/2$ and $U/2$, using the normal bases $I_s\dot\gamma$ and
$J_s\dot\gamma$. The same description applies to their derivatives in
parallel frames. The mixed contractions have zero spherical average:
they are invariant bilinear pairings between the inequivalent types
$(2,0)$ and $(0,2)$, or between $(5/2,1/2)$ and $(1/2,5/2)$.
Consequently,
\[
\mathcal E|A|^2=\frac14\mathbf E(|T|^2+|U|^2),\qquad
\mathcal E|A'|^2
=\frac1{16}\mathbf E(|\nabla T|^2+|\nabla U|^2).
\]
Here the second factor $1/4$ in the derivative identity is the spherical
average of a squared directional derivative in four dimensions.
Integrating the equation for $\Delta T$ yields
\[
\mathbf E e=6\mathbf E(\sigma v^3-v^2)\le6\mathbf E v^2,
\]
since $\sigma\le1$ and $v<2$. The exchanged identity gives the same
bound for $\bar e$. The stated formulas follow.
\end{proof}
\subsection{Minimizing geodesics and upper polar integration}
The global input is the classical index-form and closed-geodesic argument
of Myers, Synge, and Klingenberg
\cite{Myers1941,Synge1936,Klingenberg1959}; see also
\cite[Theorem~1.5 and Remark~1.2(ii)]{AbreschMeyer1997}.
Its full proof is in Appendix~\ref{app:topology}, under all the standing
oriented, closed, normalized positivity hypotheses.
\begin{lemma}[Diameter, injectivity, and topology]\label{lem:volume-topology}
The diameter of $M$ is at most $\pi$, its injectivity radius is at least
$\pi/\sqrt3$, and $M$ is simply connected. Every minimizing geodesic
segment has no point conjugate to its initial point in its interior.
\end{lemma}
The last assertion allows a conjugate endpoint. It is needed in the
upper polar argument even for segment lengths between $\pi/\sqrt3$ and
$\pi$.
Put $D_c(t)=D(t)$ before the first conjugate time and $D_c(t)=0$ at and
after it.
\begin{lemma}[Upper polar bound]\label{lem:upper-polar}
One has
\[
V_M\le2\pi^2\int_0^\pi\mathcal E D_c(t)\,dt.
\]
\end{lemma}
\begin{proof}
Fix a center $p$ and take all vectors of length
less than $\pi$ whose radial geodesics have no conjugate point at or
before their endpoints. This is an open set: the Jacobi determinant
divided by $t^3$, continued as $1$ at zero, has a positive minimum on
each such radial interval. Every point is the exponential image of a
minimizing vector of length at most $\pi$, by
Lemma~\ref{lem:volume-topology}. A minimizing vector is in this open
set unless it is critical for $\exp_p$, has length $\pi$, or is zero.
The exponential map is smooth with bounded derivative on the closed
radius-$\pi$ ball. The area formula with multiplicity states that the
integral of its absolute Jacobian over a measurable set equals the
integral of the number of preimages over its image. In particular,
critical vectors and the boundary sphere have images of zero
four-dimensional volume: their respective Jacobian integral and domain
volume are zero. On the open preconjugate set the determinant is
positive. The area formula therefore bounds $V_M$ by the integral of
$D$ over this set, with multiplicities only increasing that integral.
Nonconjugate cut vectors are included in the set; no identification of
cut points with conjugate points is used. Polar coordinates give the
claimed bound at each center, since the area of the unit
three-sphere is $2\pi^2$. Average over centers and use Tonelli's theorem.
\end{proof}
\subsection{Discrete determinant comparisons}
The upper polar bound reduces the volume problem to estimates for a
matrix Jacobi determinant. We approximate the Jacobi equation by a finite
Dirichlet matrix. Its Schur complements and sine-mode compressions
retain the order of matrix multiplication, without requiring simultaneous diagonalization along the geodesic.
The same discrete determinant will be used again for the conditional
lower comparison in Section~\ref{sec:classification}.
\begin{lemma}[The discrete Jacobi determinant]\label{lem:discrete-jacobi}
Fix a smooth symmetric normal curvature matrix on $[0,t]$, put $a=t/N$,
and let $\mathcal L_N$ be the block tridiagonal matrix on nodes
$1,\ldots,N-1$ with diagonal blocks $2\mathrm{Id}-a^2\mathcal R(ia)$
and off-diagonal blocks $-\mathrm{Id}$. Then
\[
a^3\det\mathcal L_N\longrightarrow D(t).
\]
If $t$ is before the first conjugate time, $\mathcal L_N$ is positive
definite for all sufficiently large $N$. The Jacobi convergence is
uniform over families with uniformly bounded coefficients and first
derivatives.
\end{lemma}
\begin{proof}
Define matrices without using inverses by
\[
Y_0=0,\quad Y_1=\mathrm{Id},\qquad
Y_{i+1}=(2\mathrm{Id}-a^2\mathcal R(ia))Y_i-Y_{i-1}.
\]
The position $aY_i$ and backward velocity $Y_i-Y_{i-1}$ evolve by
first changing velocity by $-a\mathcal R(ia)(aY_i)$ and then changing
position by $a$ times the new velocity. On the exact Jacobi solution
one step has error $O(a^2)$, while the update amplifies errors by at most
$1+Ca$. Iteration proves convergence of position and velocity, uniformly
on the fixed interval. The constants are uniform under the stated
coefficient bounds.
Suppose first that $t$ is preconjugate. Away from zero the exact Jacobi
matrix is invertible with uniformly bounded inverse on compact time
intervals, so the converging $aY_i$ are invertible there. Near zero use
the identity, still without inverses,
\[
Y_i=i\mathrm{Id}-a^2\sum_{r<i}(i-r)\mathcal R(ra)Y_r.
\]
For sufficiently small fixed $ia$ it gives $\max_{r\le i}|Y_r|\le Ci$
by absorbing the error, and then
$Y_i/i=\mathrm{Id}+O((ia)^2)$. Thus all the early matrices are invertible
as well. All inverses needed for elimination have now been obtained
independently of that elimination.
The Schur blocks are
\[
S_i=Y_{i+1}Y_i^{-1}
=2\mathrm{Id}-a^2\mathcal R(ia)-S_{i-1}^{-1}.
\]
They are symmetric, starting with
$S_1=2\mathrm{Id}-a^2\mathcal R(a)$. Near zero they differ from
$((i+1)/i)\mathrm{Id}$ by $O((ia)^2+a^2)$; away from zero, convergence of
position and velocity gives $S_i=\mathrm{Id}+O(a)$. Choose the small
initial interval first, and then $N$ large. All Schur blocks are
positive definite. Block elimination proves positivity of
$\mathcal L_N$ and gives
\[
\det\mathcal L_N=\prod_{i=1}^{N-1}\det S_i=\det Y_N.
\]
The desired limit follows. The determinant identity holds for arbitrary
coefficients by continuity, since invertible Schur blocks form a dense
set; convergence therefore gives the limit without the preconjugacy
assumption as well.
\end{proof}
\begin{lemma}[Upper radial comparison]\label{lem:upper-radial}
For $0<t<\pi$, set
\[
\beta=\frac{t^2}{\pi^2-t^2},\qquad
B=\frac2t\int_0^t A(s)\sin^2(\pi s/t)\,ds.
\]
Before conjugacy,
\[
D(t)\le\sin^3t\,\det(\mathrm{Id}-\beta B),
\qquad \mathrm{Id}-\beta B\succeq0.
\]
At every time $0<t<\pi$,
\[
1-\frac{D_c(t)}{\sin^3t}
\ge\frac16\min\{1,\beta^2\max(1,3-\beta)\}\,|B|^2.
\]
\end{lemma}
\begin{proof}
Compare $\mathcal L_N$ with the matrix for $\mathcal R=\mathrm{Id}$,
whose scalar sine-mode eigenvalues are
$2-2\cos(k\pi/N)-a^2$. They are positive for large $N$ because $t<\pi$.
Normalize $\mathcal L_N$ by the inverse square root of this comparison
matrix. The resulting positive matrix has trace $3$ on every diagonal
sine-mode block, since $\operatorname{tr}A(ia)=0$.
Split off its first three-dimensional mode and write the normalized
matrix as
$\left(\begin{smallmatrix}P&C\\C^*&Q\end{smallmatrix}\right)$.
Its Schur complement is positive and bounded above by $Q$. Thus
\[
\det\begin{pmatrix}P&C\\C^*&Q\end{pmatrix}
=\det P\,\det(Q-C^*P^{-1}C)
\le\det P\,\det Q\le\det P.
\]
The last inequality is arithmetic--geometric mean, since
$\operatorname{tr}Q=\dim Q$. The first-mode compression tends to
$\mathrm{Id}-\beta B$, because its normalized sine vector has entries
$\sqrt{2/N}\sin(\pi i/N)$ and
\[
\frac{a^2}{2-2\cos(\pi/N)-a^2}\longrightarrow\beta.
\]
Lemma~\ref{lem:discrete-jacobi} gives the determinant ratio limit
$D(t)/\sin^3t$. Positive compressions have a positive semidefinite
limit, and no inverse of that limiting compression is used.
This proves the first comparison.
The eigenvalues of $B$ lie in $[-1,2]$ and sum to zero, so $|B|^2\le6$.
Before conjugacy the determinant loss is
\[
1-\det(\mathrm{Id}-\beta B)
=\frac{\beta^2}{2}|B|^2+\beta^3\det B.
\]
Only $\det B<0$ is unfavorable. Write the eigenvalues then as
$-r,a_1,a_2$, where $a_1,a_2\ge0$ and $a_1+a_2=r$.
One has
\[
|\det B|\le\frac r6|B|^2,\qquad r\le1,\qquad\beta r\le2.
\]
The first inequality follows from $(a_1-a_2)^2\ge0$; the last follows
from $\beta a_i\le1$. These inequalities give the displayed loss bound.
If $\det B\ge0$ the same bound is immediate. After conjugacy its left
side is $1$, so $|B|^2\le6$ gives it directly.
\end{proof}
\subsection{An exact constant for the upper deficit}
We use one rational sine minorant for both the upper calculation here
and the conditional lower calculation in Section~\ref{sec:classification}.
\begin{lemma}[A rational sine minorant]\label{lem:sine-minorant}
Put
\[
a_0=\frac{157}{50},\qquad b_0=\frac{22}{7},\qquad
S(z)=a_0z-\frac{(b_0z)^3}{6}
+\frac{(a_0z)^5}{120}-\frac{(b_0z)^7}{5040}.
\]
The constants satisfy $a_0<\pi<b_0$, and
$0\le S(z)\le\sin(\pi z)$ for $0\le z\le1/2$. Its coefficients are
\[
S(z)=\frac{157}{50}z-\frac{5324}{1029}z^3
+\frac{95388992557}{37500000000}z^5
-\frac{155897368}{259416045}z^7.
\]
\end{lemma}
\begin{proof}
The elementary bounds $a_0<\pi<b_0$ and the alternating sine expansion
through the negative seventh-degree term give $S(z)\le\sin(\pi z)$ on
this interval. Also
\[
S(z)\ge z\left(a_0-\frac{b_0^3}{24}
-\frac{b_0^7}{322560}\right)\ge\frac95z\ge0.
\]
Expanding the four powers gives the displayed rational coefficients.
\end{proof}
Define
\[
f(u)=\begin{cases}
u^4(3+5u^2+6u^4),&0\le u\le5/8,\\
15/16,&5/8<u\le1.
\end{cases}
\]
The following strict rational lower bounds include the prefactor. The
rows use the indicated polynomial representatives up to their shared
endpoints, which do not affect the integrals.
\begin{center}
\begin{tabular}{@{}llll@{}}
\toprule
Integrand and prefactor & Interval & Degree & Lower bound\\
\midrule
$\dfrac{9a_0}{48}f(u)(1-u^2/2)S(\min(u,1-u))^3$
&$[0,1/2]$&31&$11869/10^6$\\
same &$[1/2,5/8]$&31&$30175/10^6$\\
same &$[5/8,1]$&23&$39518/10^6$\\
\bottomrule
\end{tabular}
\end{center}
For a rational polynomial $p(u)=\sum c_ju^j$ on $[r,s]$, these values
are obtained from
$\int_r^s p(u)\,du=\sum c_j(s^{j+1}-r^{j+1})/(j+1)$.
The three rows sum to more than $163/2000>2/25$; the cubic power $S^3$
is essential in each row.
\begin{proof}[Proof of Proposition~\ref{prop:upper-volume-deficit}]
For the sine-squared probability weight defining $B$, its time variance
is $t^2(1/12-1/(2\pi^2))$. The variance identity and \eqref{eq:8} give
\[
\mathcal E|B|^2\ge\mathcal E|A|^2
\left(1-\frac32t^2\left(\frac1{12}-\frac1{2\pi^2}\right)\right).
\]
Indeed, parallel identification along a geodesic and Cauchy--Schwarz
show
$\mathcal E|A(s)-A(r)|^2\le(s-r)^2\mathcal E|A'|^2$.
With $u=t/\pi$ the factor in parentheses is at least $1-u^2/2$.
For $0\le u<1$, furthermore,
\[
f(u)\le\min\{1,\beta^2\max(1,3-\beta)\},\qquad
\beta=\frac{u^2}{1-u^2}.
\]
At $u=1$ we give the capped comparison function its continuous value $1$.
For $u\le5/8$, comparison with $\beta^2(3-\beta)$ reduces to
$6-13u^2+6u^4\ge0$. On this interval $f$ is increasing and
$f(5/8)=7511875/8388608<1$, which also verifies the cap by $1$. The comparison function $\min\{1,\beta^2\max(1,3-\beta)\}$, as a function of $u$, is nondecreasing and remains $1$ after it first reaches $1$; at $u=5/8$ its value is
$57500/59319>15/16$, which handles the rest of the interval.
Lemmas~\ref{lem:upper-polar} and \ref{lem:upper-radial} therefore show
that the coefficient of $s_0$ saved from the round volume is at least
\[
\eta_+=\frac{9\pi}{48}\int_0^1
f(u)(1-u^2/2)\sin^3(\pi u)\,du.
\]
The first three rational bounds above give $\eta_+>2/25>3/40$, proving
\eqref{eq:7}.
\end{proof}
\subsection{A moment obstruction}
The upper volume deficit is proved. Together with the characteristic
identities and the two norm gaps, it leaves a single
possible unequal-moment integer case when both blocks are nonzero, up to
exchanging the blocks. The next scalar lemma
will exclude that case. Its proof reduces a constrained probability
measure to two support points; all numerical signs are established by
explicit rational coefficients.
For $0\le z\le2$, define
\[
H(z)=\sqrt z\left(1+\frac{z^2}{40}-\frac{z^4}{1152}\right),\qquad
\Phi(z)=16H(z)^2+9z(1-z).
\]
\begin{lemma}[Two constrained moments]\label{lem:moment-obstruction}
Every probability measure $\mu$ on $[0,2]$ satisfying
\[
\int z^2\,d\mu\ge1,\qquad
\int(28z+6z^2)\,d\mu\le25
\]
satisfies
\[
16\left(\int H\,d\mu\right)^2-\int\Phi\,d\mu<0.
\]
\end{lemma}
\begin{proof}
The feasible measures form a nonempty weakly compact convex set.
Suppose the continuous functional
$J(\mu)=16(\int H\,d\mu)^2-\int\Phi\,d\mu$ has a nonnegative maximum,
attained at $\mu$, and set $d=\int H\,d\mu$. First $d\ge11/20$.
Indeed,
\[
1+z^2/40-z^4/1152\ge1+(31/1440)z^2,
\]
and
\[
25+\frac{31}{45}z^2-\frac{346}{25}z
\ge\frac{17}{225}>0\qquad(0\le z\le2).
\]
The last quadratic is decreasing on this interval and has the stated
value at $2$. It follows that
$\Phi(z)\ge16(11/20)^2z^2$. The moment constraint and $J(\mu)\ge0$
now give the claim about $d$.
For another feasible measure $\nu$, direct expansion with
$\ell=32dH-\Phi$ gives
\[
J((1-t)\mu+t\nu)-J(\mu)
=t\left(\int\ell\,d\nu-\int\ell\,d\mu\right)
+16t^2\left(\int H\,d\nu-d\right)^2.
\]
For $t\downarrow0$, maximality shows that $\mu$ maximizes the linear
functional $\int\ell$. This conclusion does not require concavity of
$J$.
Put $g_1(z)=28z+6z^2-25$ and $g_2(z)=1-z^2$.
The measure with mass $9/32$ at $2$ and the rest at $0$ has respective
constraint expectations $-5/2$ and $-1/8$, so both constraints are
strictly feasible. The compact convex set of triples
\[
\left\{\left(\int g_1\,d\nu,\int g_2\,d\nu,\int\ell\,d\nu\right):
\nu\text{ a probability measure on }[0,2]\right\}
\]
is disjoint from the open set of triples with first two coordinates
negative and third coordinate greater than $\int\ell\,d\mu$.
Separation gives coefficients $-\alpha,-\gamma,\beta$, with
$\alpha,\gamma,\beta\ge0$. The last coefficient cannot be zero:
otherwise the strictly feasible triple would violate separation.
Normalize $\beta=1$. Evaluating on $\mu$ and on point masses shows
that $\mu$ is supported on the global maxima of
\[
F_d(z)=32dH(z)-\Phi(z)
-\alpha(28z+6z^2)-\gamma(1-z^2),
\qquad \alpha,\gamma\ge0.
\]
For completeness, separation bounds
$\int(\ell-\alpha g_1-\gamma g_2)\,d\nu$ by
$\int\ell\,d\mu$ for every $\nu$; for $\mu$ the reverse inequality
follows from its nonpositive constraints. Equality and then point
masses give precisely the asserted support condition.
We next verify that $F_d'''(z)>0$ on $(0,2]$. Set
\[
L(z)=\frac38+\frac{3z^2}{64}-\frac{35z^4}{1024},\qquad
V(z)=\frac{24}{5}-\frac{16z^2}{15}-\frac{7z^4}{48}
+\frac{7z^6}{1152}.
\]
Direct differentiation gives $H'''(z)=L(z)z^{-5/2}$ and
$\Phi'''(z)=V(z)$. For $x=\sqrt{z/2}$, the polynomial inequalities
\[
L(2x^2)>0,\qquad
D_*(x):=\frac{88}{5}L(2x^2)-\frac{17}{3}x^5V(2x^2)>0
\qquad(0\le x\le1)
\]
are certified below. Since $32d\ge88/5$ and $4\sqrt2<17/3$, they imply
$32dL(z)-z^{5/2}V(z)>0$ when $V(z)\ge0$; when $V(z)<0$ it is immediate.
The constraint polynomials have zero third derivative, proving the
claim.
Also $F_d'(0+)=+\infty$. Since $F_d''$ is strictly increasing, $F_d'$
can cross from positive to negative at most once. There is at most
one interior local maximum; an interval of maxima is impossible.
The endpoint $0$ is not a maximum, so the support contains at most
one interior point and the endpoint $2$. A point mass is infeasible:
its first moment condition would give $z\ge1$, and then
$28z+6z^2\ge34$. Thus the support is $\{w,2\}$, with weight $t$ at $2$.
Necessarily $0<w<1$, and the moment constraints become
\[
t_-(w):=\frac{1-w^2}{4-w^2}\le t\le
t_+(w):=\frac{25-28w-6w^2}{80-28w-6w^2}.
\]
The denominators are positive. Nonemptiness reduces to
$(2-w)(10-37w)\ge0$, hence $0<w\le10/37$.
For fixed $w$, $J$ is a convex quadratic in $t$, so it is bounded above
by the larger of its two endpoint values. Those values are negative:
\[
16t(1-t)\bigl(H(2)-H(w)\bigr)^2
>9\bigl((1-t)(w^2-w)+2t\bigr)
\qquad(t=t_-(w),\ t_+(w)).
\]
To verify these inequalities exactly, put
\[
x=\sqrt{w/2}\le\frac{37}{100},\qquad
R_*(x)=1+\frac{x^4}{10}-\frac{x^8}{72}.
\]
For each of
\[
(A_-,B_-)=(1-w^2,4-w^2),\qquad
(A_+,B_+)=(25-28w-6w^2,80-28w-6w^2),\quad w=2x^2,
\]
the required inequality, after multiplication by its positive
$B_\pm^2$, is $P_\pm(x)>0$, where
\[
P_\pm(x)=32A_\pm(B_\pm-A_\pm)
\bigl(R_*(1)-xR_*(x)\bigr)^2
-9\bigl((B_\pm-A_\pm)(w^2-w)+2A_\pm\bigr)B_\pm.
\]
This uses $H(2)-H(w)=\sqrt2(R_*(1)-xR_*(x))$.
The endpoint $w=10/37$, at which the weight interval can collapse, is
included in the following strict certificates.
Here are all the coefficient bounds used in this proof. If
$p(r+(s-r)y)=\sum_{j=0}^n a_jy^j$, define
\[
h_i=\sum_{j=0}^i a_j\frac{\binom{i}{j}}{\binom nj}
\quad(0\le i\le n).
\]
Binomial expansion gives
$p(r+(s-r)y)=\sum_i h_i\binom ni y^i(1-y)^{n-i}$; thus positive $h_i$
give positivity on $[r,s]$. If the original coefficients are
$p(x)=\sum p_kx^k$, they are obtained explicitly from
\[
a_j=(s-r)^j\sum_{k=j}^n p_k\binom{k}{j}r^{k-j}.
\]
The degrees and strict lower bounds for the resulting coefficients are
\begin{center}
\begin{tabular}{@{}llll@{}}
\toprule
Polynomial & Degree & Interval & Lower bound\\
\midrule
$L(2x^2)$ &8&$[0,1]$&$1/100$\\
$D_*(x)$ &17&$[j/4,(j+1)/4]$, $j=0,1,2,3$&$2/5$\\
$P_-(x)$ &22&$[0,37/100]$&$1$\\
$P_+(x)$ &22&$[0,37/100]$&$1$\\
\bottomrule
\end{tabular}
\end{center}
For direct reproduction, the first two polynomial inputs are
\[
L(2x^2)=\frac38+\frac3{16}x^4-\frac{35}{64}x^8,
\qquad
V(2x^2)=\frac{24}{5}-\frac{64}{15}x^4-\frac73x^8+\frac7{18}x^{12}.
\]
For example the least coefficient of $L(2x^2)$ is $1/64$; the least
coefficient of $D_*$ over its four intervals is
$2459668689/5793382400>2/5$. For $P_-$ and $P_+$ the respective least
coefficients are
\[
\begin{aligned}
\frac{58454914732902747377800095783827783501884053}
{50000000000000000000000000000000000000000000}&>1,\\
\frac{132428032291164570287168821949356607977047}
{770000000000000000000000000000000000000}&>1.
\end{aligned}
\]
All inputs and endpoints in these coefficient calculations are rational.
This proves the derivative and endpoint signs. The strict endpoint
inequalities say exactly that $J(t_-(w)),J(t_+(w))<0$, contradicting its
assumed nonnegative maximum.
\end{proof}
\begin{proposition}[Balanced Weyl moments]\label{prop:balanced-moments}
If neither Weyl block is identically zero, then
$\tau=0$ and $\mathbf E v^2=\mathbf E b^2$.
\end{proposition}
\begin{proof}
By \eqref{eq:5} and $v+b<2$,
$2\le s_0<4$. Equations \eqref{eq:6} and \eqref{eq:7} give
\[
\chi\le2(1-3s_0/40)(1+s_0)<7,
\qquad
\frac{|\tau|}{\chi}
=\frac23\frac{|\mathbf E v^2-\mathbf E b^2|}{1+s_0}
\le\frac23\frac{s_0-2}{1+s_0}<\frac4{15}.
\]
Here the first polynomial is increasing on $[2,4]$ and equals $7$ at
$4$. The integers $\chi=2+n_++n_-$ and $\tau=n_+-n_-$ have the same
parity. If $\tau\ne0$, the only possibility is $\chi=5$, $|\tau|=1$.
Exchange the blocks so that $P=\mathbf E b^2$ is the smaller moment.
The ratio in \eqref{eq:6} then gives
\[
\mathbf E v^2=\frac{13P+3}{7}.
\]
Combining this with $v^2<(2-b)^2$ yields
\[
P\ge1,\qquad28\mathbf E b+6P\le25.
\]
Integrating \eqref{eq:3} for $b$, using $\bar\sigma\le1$ and
$\bar e\ge5|\nabla b|^2/3$, gives
\[
P-\mathbf E b\ge\frac49\mathbf E|\nabla\sqrt b|^2.
\]
The integrability of $\bar e/b$ proved with \eqref{eq:3} implies
$c=\sqrt b\in H^1$ by regularization. Put
\[
G(c)=c+\frac{c^5}{40}-\frac{c^9}{1152},\qquad H(b)=G(c).
\]
The range of $c$ is bounded, and $G$ has bounded derivative there, so the
Sobolev chain rule applies, including at $c=0$. On $0\le b\le2$,
\[
G'(c)=1+\frac{b^2}{8}-\frac{b^4}{128}
\in\left[0,\frac{\sqrt{4+b^2}}2\right]
\subseteq\left[0,\frac{\sqrt K}2\right].
\]
For the first upper bound, put $z=b^2/4\in[0,1]$ and square the positive
quantity $1+z/2-z^2/8$; its square is
$1+z-z^3/8+z^4/64\le1+z$.
The variance inequality \eqref{eq:4}, extended by Sobolev approximation,
therefore gives
\[
\operatorname{Var}(H(b))
\le\frac14\mathbf E|\nabla\sqrt b|^2
\le\frac9{16}(P-\mathbf E b).
\]
Equivalently,
\[
J:=16(\mathbf E H(b))^2-\mathbf E\Phi(b)\ge0,
\qquad \Phi(b)=16H(b)^2+9b(1-b).
\tag{9}\label{eq:9}
\]
The distribution of $b$ is a probability measure on $[0,2]$ satisfying
the two moment constraints. Lemma~\ref{lem:moment-obstruction}
contradicts \eqref{eq:9}. Thus $\tau=0$, and \eqref{eq:6} gives equality
of the two second moments.
\end{proof}
==== END Positively-curved-Einstein-four-manifolds//sections/04-volume.tex ====
==== BEGIN Positively-curved-Einstein-four-manifolds//sections/05-hessian.tex ====
\section{A weighted Hessian inequality for the Weyl blocks}
\label{sec:hessian}
We continue the case in which neither Weyl block vanishes identically.
The preceding argument gives
\(\mathbf E v^2=\mathbf E b^2\). This equality will enter the coupling
in Section~\ref{sec:coupling}; the weighted Hessian inequality proved here
does not itself require it. Our task is to extract a first-derivative
inequality from the Bianchi-constrained second derivatives. Unequal
weights on the two blocks will create a useful interaction, but their
derivatives have costs that must be retained exactly.
Set \(\delta=2-v-b\), and retain the notation
\(Z=\nabla T\), \(p=\nabla v\), \(q=\nabla b\), and
\(e=|Z|^2/6\).
The argument is stated for $T$; exchanging the two orientations gives
its companion for $U$. All expectations use normalized Riemannian volume.
The local square will first be proved for an arbitrary global positive
Lipschitz weight. We will choose the two fixed weights only when their
derivative costs enter.
\subsection{The constrained Hessian and its curvature operator}
Retain $F=\operatorname{Sym}_0(\Lambda^+)$ and the Bianchi gradient
bundle $G\subset T^*M\otimes F$ from Section~\ref{sec:preliminaries}.
Their types are $(2,0)$ and $(5/2,1/2)$. The irreducible gradient
constraint and the refined Kato coefficient used below are established
in Gursky--LeBrun~\cite[Lemma~4, equations~(10)--(12)]{GurskyLeBrun1999};
Calderbank--Gauduchon--Herzlich~\cite{CalderbankGauduchonHerzlich2000}
place them in the general generalized-gradient framework. We use the
parallel projector $P_G=(6-C)/10$ and the compression identities already
proved in Proposition~\ref{prop:weyl-derivative-equations}. They are
pointwise tensor identities, including at repeated eigenvalues, and do
not require a smoothly chosen eigenbasis.
\begin{lemma}[The two Hessian components]
\label{lem:hessian-components}
Let \(\mathcal H=T^*M\otimes G\), and write
\(X=\nabla Z\), so \(X_{ij}=\nabla_i\nabla_jT\).
Then
\[
X=Y+D_0,
\qquad
Y\text{ has type }(3,1),\qquad
D_0\text{ has type }(2,0).
\]
If \(\operatorname{Tr}:\mathcal H\to F\) is contraction of the
two derivative slots and \(\operatorname{Tr}^*\) its adjoint, then
\[
\operatorname{Tr}\operatorname{Tr}^*=\frac{12}{5}\mathrm{Id},
\qquad
D_0=\frac5{12}\operatorname{Tr}^*\Delta T,
\qquad
|D_0|^2=\frac5{12}|\Delta T|^2.
\]
All the type projections in this statement are parallel.
\end{lemma}
\begin{proof}
The product rule gives four inequivalent, multiplicity-one summands
\[
\mathcal H=(3,1)\oplus(3,0)\oplus(2,1)\oplus(2,0).
\]
Skewing the derivative slots takes values in
\[
\Lambda^2T^*M\otimes F=(3,0)\oplus(2,0)\oplus(1,0)\oplus(2,1).
\]
For the actual Hessian $X=\nabla\nabla T$, its skew difference is
$[\nabla_i,\nabla_j]T=F_{ij}T$, where $F_{ij}$ is the curvature of the
bundle $F$. It is essential that the tensor on which this curvature
acts is the same Weyl block $T$ that occurs in $F_{ij}$. The minus
rotation algebra acts trivially on $F$, so there is no minus component.
The dependence on $T$ is an equivariant linear term and an equivariant
term in its symmetric square. The latter has plus spins $4,2,0$;
its only common type with $\Lambda^+\otimes F=(3,0)\oplus(2,0)\oplus(1,0)$
is $(2,0)$. The linear term has that type as well. Thus the skew
commutator has neither a $(3,0)$ nor a $(2,1)$ component. This conclusion
is specific to the Hessian of $T$, not to the derivative of an arbitrary
section of $G$.
Skewing is injective on each of those two domain summands. The
\((3,0)\) summand is entirely skew: its stretched plus coupling makes the
two derivative plus spin halves symmetric, while their minus coupling
to spin zero is alternating. To treat \((2,1)\), suppose that a tensor
of this type is symmetric in its full derivative slots. Its derivative
minus pair has spin \(1\) and is symmetric, so its derivative plus pair
must also be symmetric and have spin \(1\). The inner Bianchi constraint
makes the coupling operator \(\sum I_s^{(j)}\Sigma_s\) equal to \(-4\).
Symmetry in the derivative slots gives the same value for
\(\sum I_s^{(i)}\Sigma_s\). The total plus Casimir would therefore be
\[
-8-24+2(-4-4)=-48,
\]
whereas total spin \(2\) requires \(-24\). This contradiction proves
injectivity. Moreover, the two images lie in distinct target isotypic
components, so they cannot cancel one another. The corresponding
components of \(X\) must vanish separately.
For the trace calculation, let
\(K:F\to T^*M\otimes T^*M\otimes F\) be
\(K(S)_{ij}=\delta_{ij}S\). The adjoint of the restricted trace is
\(\operatorname{Tr}^*=P_{G}^{(j)}K\), where the projector acts
on the inner derivative and matrix slots. Now
\[
\operatorname{Tr}K=4\mathrm{Id},
\qquad
\operatorname{Tr}\bigl(C^{(j)}K(S)\bigr)
=\sum_{i,s}(I_s)_{ii}\Sigma_sS=0.
\]
Thus \(\operatorname{Tr}\operatorname{Tr}^*=12\mathrm{Id}/5\).
By equivariance, trace vanishes on \((3,1)\) and is supported on the
unique \((2,0)\) summand. The displayed nonzero composition makes its
restriction to that five-dimensional summand an isomorphism onto
\(F\). It therefore determines the whole of \(D_0\), including
its skew part, and gives both the formula and the norm identity.
One can also see explicitly that the trace component retains the
curvature commutator. At a point, its adjoint is
\[
(\operatorname{Tr}^*S)_{ij}
=\frac35\delta_{ij}S+\frac1{10}\sum_s I_{s,ij}\Sigma_sS.
\]
Hence
\[
(D_0)_{ij}-(D_0)_{ji}
=\frac1{12}\sum_s I_{s,ij}\Sigma_s\Delta T
=\frac12\sum_s(1+t_s)I_{s,ij}\Sigma_sT.
\]
For the last equality use $\Delta T=6(T-N)$ and, in a pointwise
diagonal frame, $\Sigma_sN=-t_s\Sigma_sT$. The invariant version is
$\Sigma_rN=-\sum_sT_{rs}\Sigma_sT$; no eigenframe has been
differentiated. Thus the displayed skew difference is exactly the
bundle-curvature action.
Finally, the type projections are invariant expressions in the combined
Casimirs, and trace is a metric contraction. They are parallel and do
not depend on any curvature eigenframe. The spin labels describe tensor
representations of the oriented frame group; no spin structure is used.
\end{proof}
The derivative-slot operators $T_D,U_D$ were defined in the weighted
variance proof, and $B_F(T)$ acts on the matrix slot. Compress all three
orthogonally to $\mathcal H=T^*M\otimes G$. The useful combination is
visible before the quadratic form is chosen. Let
$\Sigma_s^{\mathrm{tot}}$ be the total plus generator on $\mathcal H$ and put
\[
\widetilde Q=\frac1{60}
\sum_{r,s}T_{rs}\Sigma_r^{\mathrm{tot}}\Sigma_s^{\mathrm{tot}}.
\]
Compressing the inner derivative generator to $G$ gives
$I_s^{(j)}=\Sigma_s^G/5$: equivariance fixes the spin-one family up to a
scalar, and the highest weight gives $(1/2)/(5/2)=1/5$. At a point where
$T$ is diagonal, expansion of the total square therefore gives
\[
T_D=\frac1{10}\sum_s t_s I_s^{(i)}\Sigma_s^G
=3\widetilde Q-2Q_{5/2}(T),\qquad
B_F(T)=Q_{5/2}(T)
\]
on the compressed inner factor. The inner square contributes
$40Q_{5/2}(T)$, while the trace-free outer half-spin square is zero.
Consequently
\[
T_D+2B_F(T)=3\widetilde Q.
\]
This combined identity cancels the inner curvature action. On the free
$(3,1)$ type, $\widetilde Q=Q_3(T)\le4x/5$ and $U_D\le2y$. Thus the choice
\[
\mathcal P=4-\frac56\bigl(T_D+2B_F(T)\bigr)-U_D
=4-\frac52\widetilde Q-U_D
\]
has its plus and minus losses bounded by $2x$ and $2y$, respectively.
The scalar $4$ then meets the sectional boundary $x+y=2$. This explains
the useful coefficients; no uniqueness of the choice is asserted.
\begin{lemma}[Positivity on the highest Hessian type]
\label{lem:hessian-operator}
The combined compressed operator $\mathcal P$ preserves all four types
in $\mathcal H$. It is positive definite on $(3,1)$, and on the trace
type it satisfies
\[
\mathcal P\operatorname{Tr}^*
=\operatorname{Tr}^*(4-B_F(T)).
\]
The individual derivative-slot and matrix-slot summands need not preserve
the four types.
\end{lemma}
\begin{proof}
The total plus generators preserve each joint type. Similarly,
$\sum_su_s(J_s^{(i)}+J_s^{(j)})^2=4U_D$, so $U_D$ vanishes on minus
spin zero and is $Q_1(U)$ on minus spin one. Its latter spectrum is the
spectrum of $U$. The displayed combined formula for $\mathcal P$ therefore
preserves the joint types. On $(3,1)$,
\[
\mathcal P\ge4-2x-2y>0.
\]
On $(2,0)$ the denominator in $Q_2$ is $24$, so under the trace
identification $\widetilde Q=(2/5)Q_2(T)=(2/5)B_F(T)$ and $U_D=0$.
This proves the stated trace action.
\end{proof}
\subsection{The weighted square and its first-derivative form}
The undetermined Hessian component $Y$ is now isolated, and its
curvature-weighted quadratic form is positive. We complete its weighted
square and integrate by parts to replace the Hessian by first derivatives.
The prescribed trace component must be subtracted in full.
Set
\[
B=\frac{\langle B_F(T)Z,Z\rangle}{6},\qquad
C_N=\frac{\langle B_F(N)Z,Z\rangle}{6},\qquad
E_B=\frac{|Q_{5/2}(T)Z|^2}{6},
\qquad N=(T^2)_0.
\]
For a scalar weight \(m\), write \(L_m=dm\otimes Z\), with no
symmetrization, and let \((L_m)_Y\) denote its \((3,1)\) projection.
\begin{proposition}[Weighted Hessian inequality]
\label{prop:weighted-hessian}
For every global Lipschitz scalar weight \mbox{$m\ge m_0>0$},
\begin{equation}
0\le\mathbf E\left[
mG_v(p,k,l)
-\frac56\sum_i m_i\,
\frac{\operatorname{tr}\bigl(Z_i\sum_jZ_j^2\bigr)}6
+\frac{\langle\mathcal P(L_m)_Y,(L_m)_Y\rangle}{24m}
\right],
\tag{10}\label{eq:10}
\end{equation}
where \(i,j\) in the middle term are derivative indices and
\[
G_v=\left(-42-3b^2-\frac{10}{3}v^2\right)e
+75B+\frac{25}{3}C_N-\frac{100}{3}E_B
+5v^2|p|^2.
\]
\end{proposition}
\begin{proof}
By Lemma~\ref{lem:hessian-operator},
\[
\mathbf E\left[
\frac{m}{6}\left\langle
\mathcal P\left(Y+\frac{(L_m)_Y}{2m}\right),
Y+\frac{(L_m)_Y}{2m}\right\rangle\right]\ge0.
\]
The last term in its expansion is the last term of \eqref{eq:10}.
Since \(\mathcal P\) preserves the type splitting and
\(X=Y+D_0\), the other terms are
\[
\frac16\mathbf E\bigl[
m\langle\mathcal PX,X\rangle+\langle\mathcal PX,L_m\rangle
-m\langle\mathcal PD_0,D_0\rangle-\langle\mathcal PD_0,L_m\rangle
\bigr].
\]
We compute these terms for smooth \(m\) first. Compression may be omitted
inside any of the displayed inner products, because both arguments
belong to \(\mathcal H\).
\paragraph{The identity term.}
Integration by parts and \eqref{eq:1} give
\[
\frac16\mathbf E\left[m|\nabla Z|^2
+\sum_i m_i\langle\nabla_iZ,Z\rangle\right]
=-\frac16\mathbf E\bigl[m\langle\Delta Z,Z\rangle\bigr]
=\mathbf E\bigl[m(-13e+20B)\bigr].
\]
This will be multiplied by the coefficient \(4\) in \(\mathcal P\).
\paragraph{The derivative-slot curvature terms.}
Write \(A_{ikjl}=(T_D)_{ikjl}\). The two terms to integrate are
\[
\mathbf E\left[
mA_{ikjl}\langle\nabla_iZ_j,\nabla_kZ_l\rangle
+A_{ikjl}\langle\nabla_iZ_j,m_kZ_l\rangle\right].
\]
Integrate the derivative \(k\) on the second paired tensor.
The derivative of \(m\) cancels the second term exactly.
Moreover \(\nabla_kA_{ikjl}=0\), by divergence freedom and skewness in
\(i,k\). The result is
\[
-\mathbf E\bigl[mA_{ikjl}\langle\nabla_k\nabla_iZ_j,Z_l\rangle\bigr]
=\frac12\mathbf E\bigl[
mA_{ikjl}\langle[\nabla_i,\nabla_k]Z_j,Z_l\rangle\bigr].
\]
This calculation uses neither symmetry of \(X\) nor parallelness of
\(T_D\), and the factor \(1/2\) comes only from skewing the two
covariant derivatives.
The curvature in the derivative slot has the contraction
\[
\frac12(T_D)_{ikjl}R_{ikjh}=3v^2\delta_{lh}.
\]
To see its normalization, substitute the quaternion expansions: the
contraction over \(i,k\) removes all mixed plus/minus terms and gives
\(\frac12\sum_s t_s(1+t_s)\delta_{lh}=3v^2\delta_{lh}\).
The matrix-slot curvature gives the quadratic form of
\[
-\frac12\sum_s t_s(1+t_s)I_s\Sigma_s.
\]
Since \(t_s^2=N_{ss}+2v^2\), its compression is
\[
-\frac12\bigl(8Q_{5/2}(T)+8Q_{5/2}(N)+2v^2C\bigr)
=4v^2-4Q_{5/2}(T)-4Q_{5/2}(N).
\]
After division by \(6\), the full \(T_D\) contribution is therefore
\[
\mathbf E\bigl[m(7v^2e-4B-4C_N)\bigr].
\]
For \(U_D\), the same integration gives \(3\mathbf E[mb^2e]\).
Its matrix-slot curvature contraction is zero, because the minus forms
are orthogonal to the plus forms defining the curvature action on
\(F\).
\paragraph{The nonparallel matrix-slot coefficient.}
Put \(A=B_F(T)\) temporarily. One integration by parts gives
\[
\mathbf E\left[m\langle A\nabla_iZ,\nabla_iZ\rangle
+m_i\langle A\nabla_iZ,Z\rangle\right]
=-\mathbf E\bigl[m\langle A\Delta Z,Z\rangle\bigr]
-\mathbf E\bigl[m\langle(\nabla_iA)\nabla_iZ,Z\rangle\bigr].
\]
The operator \(\nabla_iA\) is symmetric, so the last term is
\[
\frac12\mathbf E\left[
\left\langle\left(m\Delta A+\sum_i m_i\nabla_iA\right)Z,Z\right\rangle
\right].
\]
Here \(\nabla_iA=B_F(Z_i)\) and
\(\Delta A=6B_F(T)-6B_F(N)\), since \(B_F\) is a linear parallel
construction. Also, pairing \(B_F(T)Q_{5/2}(T)Z\) with \(Z\) gives
\(|Q_{5/2}(T)Z|^2\), because the latter vector belongs to
\(G\). After division by \(6\), the contribution is
\[
\mathbf E\left[
m(-10B-3C_N+20E_B)
+\frac1{12}\sum_i m_i\langle B_F(Z_i)Z,Z\rangle\right].
\]
Since
\(\langle B_F(Z_i)Z,Z\rangle
=\operatorname{tr}(Z_i\sum_jZ_j^2)\),
multiplication by \(-5/3\) produces exactly the middle term in
\eqref{eq:10}. In particular, the variation of \(B_F(T)\) has been
retained explicitly; \(\mathcal P\) is not assumed parallel.
\paragraph{The prescribed trace component.}
By Lemmas~\ref{lem:hessian-components} and \ref{lem:hessian-operator},
\[
\mathcal PD_0=\frac5{12}\operatorname{Tr}^*(4-B_F(T))\Delta T,
\qquad \operatorname{Tr}L_m=\sum_i m_iZ_i.
\]
The characteristic polynomial of a trace-free \(3\times3\) matrix gives
\(B_F(T)T=N\) and \(B_F(T)N=v^2T\). Thus
\[
B_0=(4-B_F(T))(T-N)=(4+v^2)T-5N.
\]
Using \(\Delta T=6(T-N)\), the two trace subtractions, before division
by \(6\), are
\[
-\frac52\mathbf E\left[
m\langle B_0,\Delta T\rangle+\sum_i m_i\langle B_0,Z_i\rangle\right]
=\frac52\mathbf E\bigl[m\langle\nabla B_0,Z\rangle\bigr].
\]
Since \(\nabla N=2B_F(T)Z\) and
\(\langle T,Z_i\rangle=6vp_i\), their normalized contribution is
\[
\mathbf E\left[
m\left(\left(10+\frac52v^2\right)e-25B+5v^2|p|^2\right)\right].
\]
Multiplying the preceding contributions by their coefficients in
\(\mathcal P\) and collecting gives \(G_v\) as stated.
Finally, a global Lipschitz weight $m\ge m_0>0$ can be approximated
uniformly and in $H^1$ by smooth positive weights with a common positive
lower bound. The first derivatives converge in $L^2$, so their quadratic
products in the square cost converge in $L^1$; the uniformly bounded
reciprocals of the weights pass to the same limit. All other terms use
only the weight and its first derivative. Hence the identity and inequality
extend to every weight in the stated class.
\end{proof}
\begin{lemma}[The quadratic-form matrices]
\label{lem:hessian-matrices}
At a point where \(v>0\), use the three vector parameters \(p,k,l\) in
\eqref{eq:2}. Then \(G_v\) is the quadratic form, with scalar entries
multiplying vector dot products, whose matrix is
\begin{equation}
\begin{pmatrix}
-70-5b^2+100\sigma v-25v^2
&-100\sqrt3\rho v&-35\rho v^2/\sqrt3\\
-100\sqrt3\rho v
&-210-15b^2-300\sigma v-190v^2&-75v-35\sigma v^2\\
-35\rho v^2/\sqrt3
&-75v-35\sigma v^2&-42-3b^2-10v^2
\end{pmatrix}.
\tag{11}\label{eq:11}
\end{equation}
\end{lemma}
\begin{proof}
Expanding the diagonal and off-diagonal matrix entries in \eqref{eq:2}
gives
\[
\frac{B}{v}\ \longleftrightarrow\
\begin{pmatrix}
4\sigma/3&-4\rho/\sqrt3&0\\
-4\rho/\sqrt3&-4\sigma&-1\\
0&-1&0
\end{pmatrix},
\qquad
\frac{C_N}{v^2}\ \longleftrightarrow\
\begin{pmatrix}
4/3&0&-\rho/\sqrt3\\
0&-4&-\sigma\\
-\rho/\sqrt3&-\sigma&0
\end{pmatrix}.
\]
Here one uses
\(\sum s_i\widetilde s_i=0\),
\(\sum\widetilde s_i^3=-6\rho\), and
\(N/v^2=\operatorname{diag}(\sigma s_i+\rho\widetilde s_i)\).
The energy metric in these coordinates is
\(H=\operatorname{diag}(5/3,5,1)\).
If \(\mathsf B\) denotes the matrix of \(B\), the matrix of \(E_B\) is
\(\mathsf B H^{-1}\mathsf B\): \(B_F(T)\) compressed to
\(G\) is the self-adjoint operator whose quadratic form is
\(B\). Using \(\sigma^2+\rho^2=1\), this gives
\[
\frac{E_B}{v^2}\ \longleftrightarrow\
\begin{pmatrix}
16/15&0&4\rho/(5\sqrt3)\\
0&21/5&4\sigma/5\\
4\rho/(5\sqrt3)&4\sigma/5&1/5
\end{pmatrix}.
\]
Substitution into the definition of \(G_v\) gives \eqref{eq:11} by
matrix addition. The invariant expression for \(G_v\) is defined
everywhere; on the zero set of \(T\), all its derivative parameters
vanish almost everywhere, as established for the norm identity.
\end{proof}
\subsection{The two costs of a variable weight}
For the remaining argument choose the fixed positive weights
\begin{equation}
M=1-\frac38(v-b),\qquad m=M^2,
\qquad
\overline M=1-\frac38(b-v),\qquad \overline m=\overline M^2.
\tag{12}\label{eq:12}
\end{equation}
They are one rational choice for the later coupling argument. Since
$v+b<2$, both $M$ and $\overline M$ lie between $1/4$ and $7/4$.
Almost everywhere,
\[
dm=-\frac34M(p-q),\qquad d\overline m=-\frac34\overline M(q-p).
\]
Proposition~\ref{prop:weighted-hessian} applies to these global Lipschitz
weights. Their derivatives produce the two costs estimated next.
The Hessian has been eliminated from the integrated inequality. Two
terms caused by the weight remain: a cubic contraction multiplied by
$dm$, and a projected quadratic form in $dm\otimes Z$. We now bound
both by first-derivative energies, while keeping the sign needed for
the later scalar relaxation.
\begin{lemma}[The cubic contraction]
\label{lem:cubic-contraction}
For every real tangent vector \(a\), with
\(Z_a=\sum_i a_iZ_i\),
\begin{equation}
\left|\frac{\operatorname{tr}\bigl(Z_a\sum_jZ_j^2\bigr)}6\right|
\le \frac45\sqrt{\frac35}\,|a|e^{3/2}.
\tag{13}\label{eq:13}
\end{equation}
\end{lemma}
\begin{proof}
By homogeneity, assume \(|a|=1\); the case \(Z_a=0\) is immediate.
Put \(\nu=|Z_a|/\sqrt6\), and diagonalize the normalized matrix
\(s=Z_a/\nu\). Apply the algebraic parametrization \eqref{eq:2} with
this \(s\), rather than with \(T/v\), and denote its vector parameters
by \(\widehat p,\widehat k,\widehat l\).
The diagonal entries of \(Z_a=\nu s\) imply
\(\widehat p_a=\nu\) and \(\widehat k_a=0\).
The part corresponding to the single component
\(\widehat p_a a\) has energy
\(e_1=5\nu^2/3\). It is orthogonal in the energy metric to the
remaining parameters. In the displayed matrix for \(B/v\), it has no
cross term with those parameters, because \(\widehat k_a=0\) and the
\(\widehat p,\widehat l\) entry is zero. Its contribution to
\(\frac16\sum_j\operatorname{tr}(sZ_j^2)\) has absolute value at most
\(4e_1/5\).
For any symmetric trace-free \(3\times3\) matrix \(S\), its
characteristic polynomial gives
\[
|(S^2)_0|^2=\frac{|S|^4}{6}.
\]
Since \(|s|=\sqrt6\), Cauchy--Schwarz yields
\(|\operatorname{tr}(sS^2)|\le |S|^2\).
Applied to the remaining tensor, this bounds its contribution by
\(e-e_1\). Therefore
\[
\left|\frac{\operatorname{tr}(Z_a\sum_jZ_j^2)}6\right|
\le\nu\left(e-\frac{e_1}{5}\right)
=\sqrt{\frac35}\,e^{3/2}
\sqrt{\frac{e_1}{e}}\left(1-\frac{e_1}{5e}\right).
\]
For \(0\le r\le1\), the function \(\sqrt r(1-r/5)\) is increasing
and has maximum \(4/5\). This proves \eqref{eq:13}.
\end{proof}
\begin{lemma}[Real-vector projection identities]
\label{lem:real-vector-projections}
Let \(P_+\) be the projection in \(\mathcal H\) onto total plus spin
\(3\), retaining both possible minus spins. For any real tangent vector
\(a\) and any \(Z\in G\),
\[
|P_+(a\otimes Z)|^2=\frac7{12}|a|^2|Z|^2,
\]
\[
\left\langle\widetilde QP_+(a\otimes Z),P_+(a\otimes Z)\right\rangle
=\frac12|a|^2\langle Q_{5/2}(T)Z,Z\rangle.
\]
\end{lemma}
\begin{proof}
Complexify at the point, writing the outer tangent representation as
\(A\otimes B\), with \(A,B\) the plus and minus spin halves, and
\(G_{\mathbb C}=H_{5/2}\otimes B'\).
For a real vector \(a\), its reduced density on \(A\) is
\(\frac12|a|^2\mathrm{Id}_A\). Indeed, the three traceless Hermitian
operators on \(A\) are scalar multiples of \(-iI_s\), and each has
zero expectation because \(a^{\mathsf T}I_sa=0\).
Its trace is \(|a|^2\), which fixes the scalar part.
The complete vector is \(a\otimes Z\), a product between the outer
factor and the entire inner factor. Consequently, for every operator
\(O\) acting only on \(A\otimes H_{5/2}\),
\[
\langle a\otimes Z,O(a\otimes Z)\rangle
=\frac{|a|^2}{2}
\langle Z,((\operatorname{Tr}_A O)\otimes\mathrm{Id}_{B'})Z\rangle.
\]
This partial trace retains every correlation of the inner plus factor
with \(B'\); neither \(a\) nor \(Z\) is assumed to be a product within
its own plus and minus factors.
The operator \(\operatorname{Tr}_A P_+\) is scalar on the irreducible
space \(H_{5/2}\). Its trace is the dimension \(7\) of spin \(3\),
whereas \(H_{5/2}\) has dimension \(6\). Thus it equals
\(7\mathrm{Id}/6\), proving the first identity.
The family
\(\operatorname{Tr}_A(P_+\widetilde Q(T)P_+)\) is equivariant and
linear in the spin-\(2\) variable \(T\). Since spin \(2\) occurs once
in \(\operatorname{End}(H_{5/2})\), it is a scalar multiple of
\(Q_{5/2}(T)\). To determine the scalar, take
\(T=\operatorname{diag}(-1,-1,2)\) and the inner highest weight
\(5/2\). The outer weight \(1/2\) gives total weight \(3\) and
\(Q_3(T)\)-eigenvalue \(-1\). The outer weight \(-1/2\) projects to
total weight \(2\), where that eigenvalue is
\((12-3\cdot2^2)/15=0\). The normalized outer trace is therefore
\(-1/2\). Since \(Q_{5/2}(T)\) has eigenvalue \(-1\) on its highest
weight, the second identity has coefficient \(1/2\), as asserted.
\end{proof}
\begin{proposition}[The projection cost]
\label{prop:hessian-projection-cost}
For the weight \eqref{eq:12},
\begin{equation}
\frac{\langle\mathcal P(L_m)_Y,(L_m)_Y\rangle}{24m}
\le\frac9{64}|p-q|^2
\left[\frac7{12}(4+y)e-\frac54B\right].
\tag{14}\label{eq:14}
\end{equation}
The bracket on the right is nonnegative. Both assertions remain valid
under the closed inequalities \(v+b\le2\) and \(x+y\le2\).
\end{proposition}
\begin{proof}
Let \(P_-\) be the projection onto total minus spin \(1\), so the
highest Hessian projection is \(P_+P_-\). On that type,
\(-U_D\le y\), and hence \(\mathcal P\le A\), where
\[
A=4+y-\frac52\widetilde Q.
\]
This operator acts only on the plus factors and commutes with \(P_-\).
On the entire high-plus subspace,
\[
A\ge4+y-2x\ge0;
\]
the last inequality follows from \(x+y\le2\).
It follows in operator order that \(P_-AP_-\le A\) there.
Thus, for every real \(a\),
\[
\langle\mathcal P(a\otimes Z)_Y,(a\otimes Z)_Y\rangle
\le \langle AP_+(a\otimes Z),P_+(a\otimes Z)\rangle
=6|a|^2\left[\frac7{12}(4+y)e-\frac54B\right],
\]
where the last equality is Lemma~\ref{lem:real-vector-projections}.
This operator argument is valid for arbitrary correlations after the
plus projection and does not discard a mixed minus-factor term.
Almost everywhere,
\(dm=-\frac34M(p-q)\).
Substituting \(a=dm\), dividing by \(24m\), and using \(m=M^2\)
gives \eqref{eq:14}.
Finally, for any unit real \(a\), the same equality and the lower bound
for \(A\) give
\[
\frac7{12}(4+y)e-\frac54B
\ge \frac7{12}(4+y-2x)e\ge0.
\]
In particular, subsequent upper estimates may replace \(|p-q|^2\) by
\((|p|+|q|)^2\) in \eqref{eq:14} without reversing the inequality.
\end{proof}
==== END Positively-curved-Einstein-four-manifolds//sections/05-hessian.tex ====
==== BEGIN Positively-curved-Einstein-four-manifolds//sections/06-coupling.tex ====
\section{Coupling the two Weyl blocks}\label{sec:coupling}
Assume that neither Weyl block vanishes identically and that their second
moments agree:
\[
\mathbf E v^2=\mathbf E b^2.
\]
We retain the notation
\[
\delta=2-v-b,\qquad K=4+v^2+b^2,\qquad
M=1-\frac38(v-b),\qquad m=M^2,
\]
and place a bar over an expression to exchange the two blocks. In
particular, $\bar M=1-\frac38(b-v)$ and $\bar m=\bar M^2$.
The weights in \eqref{eq:12} have opposite slopes along $v+b=2$.
Their derivatives introduce quartic terms in the first derivatives of
curvature. We combine the two Hessian inequalities with one identity that
controls those costs, the balanced variance defect, and tests of the weak
norm equations. This produces a single integrand with nonnegative mean.
The task is then to bound its quartic, quadratic, and potential parts by
three expressions that complete one square.
\subsection{The combined integral inequality}
Start with the sum of the two integrands in \eqref{eq:10}. The three
additions below are written explicitly in \eqref{eq:16}--\eqref{eq:18}.
Here $h\ge0$ is a scalar coefficient and $\kappa$ is a scalar multiplier,
whose rational values are specified below. Their roles and integral signs
are as follows; the degrees refer to first
derivatives of curvature in the resulting integrand.
\begin{center}
\small
\begin{tabular}{@{}p{0.24\textwidth}p{0.42\textwidth}p{0.25\textwidth}@{}}
\toprule
Added expression & Integral and derivative degrees & Purpose\\
\midrule
$\Delta\delta$ identity &
$\dfrac h3\mathbf E[\delta(|Y|^2+|\bar Y|^2)]\ge0$; degrees $4,2,0$ &
Controls the quartic weight costs\\
Balanced variance defect & Nonnegative integral; degrees $2,0$ &
Controls $v-b$ using moment balance\\
$\kappa$ norm tests & Zero integral; degrees $2,0$ &
Adjusts the quadratic and potential terms\\
\bottomrule
\end{tabular}
\end{center}
We use the following verified rational choice for the constants and
functions in this combination and its final comparison. It is one
sufficient choice; no optimality or uniqueness is asserted.
Set
\[
D=158,\qquad h=\frac{17}{10},\qquad
c=\frac25-\frac3{10}vb,
\qquad a=v-1,\quad s=b-1,\quad
n=\frac\delta2,\quad t=\frac{v-b}{2}.
\]
Thus $c\ge 1/10$ on $v,b\ge0$, $v+b\le2$. Define
\begin{equation}
\begin{aligned}
\kappa(v,b)={}&-\frac D6(a-s)+z_1a^2
+\left(60-\frac{2D}{9}+2z_1\right)as+z_2s^2\\
&+\sum_{i=0}^3z_{3+i}a^is^{3-i}
+\sum_{i=0}^4z_{7+i}a^is^{4-i},\\
4(z_1,\ldots,z_{11})={}&
(-43,-45,-63,21,183,208,61,46,101,79,-41),\\
4\Theta(v,b)={}&n^2(1-n)(105+5n)
+t^2(4+360n-586n^2)-4t^4,\\
R_v={}&\frac{4(1+v)}{1+6v+v^2},\qquad
R_b=\frac{4(1+b)}{1+6b+b^2}.
\end{aligned}
\tag{15}\label{eq:15}
\end{equation}
Here and below an exchanged expression is obtained by exchanging all
block variables, and $\bar\kappa=\kappa(b,v)$.
The multiplier $\kappa$ changes the integrand through a zero-integral
norm test. The functions $\Theta$, $c$, $R_v$, and $R_b$ enter only the
comparison for the final square; they are not additional integrands.
Neither $\kappa$ nor $\Theta$ is required to have a sign.
The first addition is
\begin{equation}
\begin{aligned}
h\bigg[&(e+\bar e)
\left(\lambda-\frac{e-|p|^2}{v}
-\frac{\bar e-|q|^2}{b}\right)
+40\delta(B+\bar B)\\
&\hspace{24mm}
-30\delta\sum_{\pm}v^2(1-2\sigma v+v^2)\bigg],
\end{aligned}
\tag{16}\label{eq:16}
\end{equation}
where
\[
\lambda=6(\sigma v^2+\bar\sigma b^2-v-b)-26\delta.
\]
The notation $\sum_\pm$ includes the displayed term and its exchange.
To verify its sign, use
\[
\Delta e=\frac{2}{6}|\nabla Z|^2+26e-40B
\]
and \eqref{eq:3} in
\[
\mathbf E\bigl[(e+\bar e)\Delta\delta\bigr]
=\mathbf E\bigl[\delta\Delta(e+\bar e)\bigr].
\]
The decomposition $\nabla Z=Y+D_0$ gives
$|D_0|^2=(5/12)|\Delta T|^2$. Moreover,
\[
\frac{2|D_0|^2}{6}
=30v^2(1-2\sigma v+v^2).
\]
Consequently the integral of \eqref{eq:16} equals
$\frac h3\mathbf E[\delta(|Y|^2+|\bar Y|^2)]$, which is nonnegative.
The second addition is
\begin{equation}
D\left[\frac{|p-q|^2}{K}
-\left(1-\frac{\delta^2}{4}\right)(v-b)^2\right].
\tag{17}\label{eq:17}
\end{equation}
Indeed, the equality of the second moments implies
\[
\mathbf E(v-b)=\frac12\mathbf E[\delta(v-b)],
\]
and hence
\[
(\mathbf E(v-b))^2\le\frac14\mathbf E[\delta^2(v-b)^2].
\]
Applying \eqref{eq:4} to $v-b$ proves that \eqref{eq:17} has
nonnegative integral. Finally add the zero-integral expression
\begin{equation}
\sum_\pm\left[
\kappa(v,b)\left\{6v(1-\sigma v)+\frac{e-|p|^2}{v}\right\}
+\nabla\kappa(v,b)\cdot p\right].
\tag{18}\label{eq:18}
\end{equation}
This is the weak identity \eqref{eq:3} tested against $\kappa(v,b)$.
Let $\mathcal I$ denote the resulting total integrand. Thus
$\mathbf E\mathcal I\ge0$. All these integrals are legitimate without
a positive lower bound for either norm: $e/v$ and $\bar e/b$ are
integrable by the regularization proving \eqref{eq:3}, while the energies
and all scalar coefficients are bounded. Bounded Lipschitz test functions
are admissible by smooth approximation, uniformly and in $H^1$.
For the fixed weights in \eqref{eq:12}, $dm$ is linear in $p,q$.
Thus the cubic contraction multiplied by $dm$ and the quadratic form
in $L_m=dm\otimes Z$ have degree four in first derivatives of curvature.
Grouping the exact total integrand by this degree gives
$\mathcal I=\mathcal I_4+\mathcal I_2+\mathcal I_0$, where
\[
\begin{aligned}
\mathcal I_4={}&
\sum_\pm\left[
-\frac56\sum_i m_i\,
\frac{\operatorname{tr}\bigl(Z_i\sum_j Z_j^2\bigr)}6
+\frac{\langle\mathcal P(L_m)_Y,(L_m)_Y\rangle}{24m}
\right]\\
&-h(e+\bar e)
\left(\frac{e-|p|^2}{v}+\frac{\bar e-|q|^2}{b}\right),\\[1mm]
\mathcal I_2={}&
\sum_\pm\left[
mG_v+h\lambda e+40h\delta B
+\kappa(v,b)\frac{e-|p|^2}{v}
+\nabla\kappa(v,b)\cdot p
\right]
+\frac DK|p-q|^2,\\[1mm]
\mathcal I_0={}&
-D\left(1-\frac{\delta^2}{4}\right)(v-b)^2\\
&+\sum_\pm\left[
6\kappa(v,b)v(1-\sigma v)
-30h\delta v^2(1-2\sigma v+v^2)
\right].
\end{aligned}
\]
Here $\lambda$ is the original symmetric expression in \eqref{eq:16}.
The exchange in $\sum_\pm$ includes the operator, the $Y$-projection,
the weight, the multiplier and its arguments, and all derivative labels.
On a block's zero set, only its corresponding derivative summands vanish
almost everywhere; the exchanged contributions remain. The quotients
use the zero value assigned in \eqref{eq:3}, and $G_v$ uses its invariant
expression there. Products such as $\sigma v^2$ use their continuous
zero value; no continuity is asserted for the singular coefficients.
\begin{samepage}
\begin{proposition}[Three bounds for the combined integrand]
\label{prop:coupling-estimates}
For this decomposition, define
\[
r=R_v|p|^2+R_b|q|^2,\qquad
j=(|p|^2+|q|^2)
\left(\frac{|p|^2}{v}+\frac{|q|^2}{b}\right).
\]
Then, almost everywhere,
\[
\mathcal I_4\le-cj,\qquad
\mathcal I_2\le2\Theta r,\qquad
\mathcal I_0\le-\frac{\Theta^2}{c}.
\]
The corresponding derivative summands are assigned value zero on a
zero set, as in \eqref{eq:3}. The total integrand and the bounds are
integrable, and $\mathbf E\mathcal I\ge0$.
\end{proposition}
\end{samepage}
Section~\ref{sec:classification} combines these estimates with $r^2\le j$ to obtain
\[
\mathcal I\le-\frac{(\Theta-cr)^2}{c}-c(j-r^2).
\]
The strictness needed for vanishing is proved there.
We prove the three estimates in turn. The computations first take place
on $v,b>0$, $v+b\le2$, $x+y\le2$. The polynomial certificates used in
the proof are supplied in Section~\ref{sec:certificates}.
\subsection{The quartic terms}
Put
\[
e_0=\frac53|p|^2,\qquad e_d=e-e_0\ge0,\qquad z=|p|+|q|.
\]
Since $dm=-\frac34M(p-q)$, \eqref{eq:13} bounds the cubic contraction
multiplied by $dm$ in \eqref{eq:10} by
$\frac12\sqrt{3/5}\,Mz e^{3/2}$. The nonnegative bracket in
\eqref{eq:14} is bounded above by $g e_0+H_d e_d$, where
\[
g=\frac7{12}(4+y)-\sigma v+\frac{\rho^2v^2}{4},\qquad
H_d=\frac7{12}(4+y)+\frac54v+4.
\]
In fact, the pure defect contribution to $-5B/4$ is at most
$(5/4)v e_d$, and its mixed contribution is at most
$2v|\rho|\sqrt{e_0e_d}$. The latter is at most
$(\rho^2v^2/4)e_0+4e_d$. Nonnegativity of the original bracket permits
the replacement $|p-q|^2\le z^2$. We obtain
\begin{equation}
\begin{aligned}
\mathcal I_4\le{}&
\sum_\pm\left[
\frac12\sqrt{\frac35}\,zMe^{3/2}
+\frac9{64}z^2(g e_0+H_de_d)\right]\\
&-h(e+\bar e)
\left[\frac{(2/5)e_0+e_d}{v}
+\frac{(2/5)\bar e_0+\bar e_d}{b}\right].
\end{aligned}
\tag{19}\label{eq:19}
\end{equation}
We first show that this upper bound decreases when either defect
increases, with $p,q$ fixed. Denote the right-hand side by $\mathcal Q$.
For $e>0$, direct differentiation gives
\[
\begin{aligned}
\partial_{e_d}\mathcal Q={}&
\frac34\sqrt{\frac35}\,zM\sqrt e
+\frac9{64}z^2H_d\\
&-h\left[
\frac{(2/5)e_0+e_d}{v}
+\frac{(2/5)\bar e_0+\bar e_d}{b}
+\frac{e+\bar e}{v}\right].
\end{aligned}
\]
Set $u=\sqrt{\bar e/e}$. The bounds
\[
\begin{gathered}
z\le\sqrt{\frac35}(\sqrt e+\sqrt{\bar e}),\qquad
M\le\frac74-\frac34v,\qquad H_d\le\frac{15}{2}+\frac23v,\\
\frac25e_0+e_d\ge\frac25e,\qquad
\frac25\bar e_0+\bar e_d\ge\frac25\bar e,
\qquad \frac vb\ge\frac{v}{2-v}
\end{gathered}
\]
therefore imply
\[
\frac ve\partial_{e_d}\mathcal Q
\le A_0(1+u)^2+L_0(1+u)
-h\left[\frac75+
\left(1+\frac{2v}{5(2-v)}\right)u^2\right],
\]
where
\[
A_0=\frac{27}{320}v\left(\frac{15}{2}+\frac23v\right),
\qquad L_0=\frac9{20}v\left(\frac74-\frac34v\right).
\]
All substitutions in the negative bracket use lower bounds; thus they
give the required upper bound after multiplication by $-h$.
Write
\[
C_0=\frac75h-A_0-L_0,\qquad
D_0^{\mathrm{sc}}=h-A_0+\frac{2hv}{5(2-v)}.
\]
Lemma~\ref{lem:scalar-certificates} gives $C_0>0$ and
\[
4C_0\left((h-A_0)(2-v)+\frac25hv\right)
-(2-v)(2A_0+L_0)^2>0.
\]
For $0<v<2$ this says
$4C_0D_0^{\mathrm{sc}}>(2A_0+L_0)^2$; hence the homogenized
quadratic
\[
-C_0+(2A_0+L_0)u-D_0^{\mathrm{sc}}u^2
\]
is negative for every real $u$. No ordering of the two energies is
needed. Before taking an endpoint limit, write the derivative bound as
\[
\partial_{e_d}\mathcal Q\le\frac1v
\left[-C_0e+(2A_0+L_0)\sqrt{e\bar e}
-D_0^{\mathrm{sc}}\bar e\right].
\]
It extends continuously to $e=0$, where necessarily $e_0=e_d=0$.
At $e=\bar e=0$ its value is zero; otherwise the derivative remains
negative. The exchanged argument applies to $\bar e_d$. Decreasing
the defects one at a time consequently gives
\[
\mathcal Q(e_0+e_d,\bar e_0+\bar e_d)
\le\mathcal Q(e_0,\bar e_0+\bar e_d)
\le\mathcal Q(e_0,\bar e_0).
\]
We may therefore put both defects equal to zero. Write
\[
V=x-v,\qquad W=y-b,\qquad
0\le V\le v,\quad0\le W\le b,\quad V+W\le\delta.
\]
We claim
\[
g\le g_v:=\frac7{12}(4+b+W)-v+2V,
\qquad
g_b=\frac7{12}(4+v+V)-b+2W.
\]
For $r_0=x/v\in[1,2]$, use
\[
\sigma=\frac{3r_0-r_0^3}{2},\quad
1-\sigma=\frac{(r_0-1)^2(r_0+2)}2,\quad
1+\sigma=\frac{(2-r_0)(r_0+1)^2}{2}.
\]
Since $r_0v\le2$,
\[
\frac{(r_0-1)(r_0+2)}2
\left(1+\frac{v(1+\sigma)}4\right)\le2.
\]
After substituting $v\le2/r_0$, the gap in this inequality is
\[
\begin{aligned}
&2-\frac{(r_0-1)(r_0+2)}2
\left(1+\frac{1+\sigma}{2r_0}\right)\\
&\quad=\frac{2-r_0}{8r_0}
\left[4r_0(r_0+3)-(r_0-1)(r_0+2)(r_0+1)^2\right]\ge0.
\end{aligned}
\]
The bracket is concave on $[1,2]$ and has endpoint values $16$ and $4$.
This also covers $r_0=2$ without division by $2-r_0$. Multiplying the preceding bound
by $V=v(r_0-1)$ proves the claim for $g$, using
$\rho^2=1-\sigma^2$.
Set $P=\sqrt{e_0}$ and $Q=\sqrt{\bar e_0}$. We will prove
\begin{equation}
\begin{aligned}
&\left(\frac{67}{50}+\frac{27}{100}vb\right)
(P^2+Q^2)(bP^2+vQ^2)\\
&\qquad\ge
\frac{27}{128}vb(P+Q)^2(g_vP^2+g_bQ^2)
+\frac34vb(P+Q)(MP^3+\bar M Q^3).
\end{aligned}
\tag{20}\label{eq:20}
\end{equation}
At zero defects, $z=\sqrt{3/5}(P+Q)$. Thus the positive terms of
\eqref{eq:19} are at most $2/5$ times the right-hand side of
\eqref{eq:20}, divided by $vb$. Since
\[
\frac{10}{9}\left(h-\frac{67}{50}-\frac{27}{100}vb\right)=c,
\]
\eqref{eq:20} gives the first estimate in
Proposition~\ref{prop:coupling-estimates}; the factor $9/25$ that remains
is exactly the conversion from $(e_0,\bar e_0)$ to $(|p|^2,|q|^2)$.
To prove \eqref{eq:20}, exchange the blocks, if necessary, so that
$P>0$ and $0\le u=Q/P\le1$; the case $P=Q=0$ is immediate.
In $g_v+g_bu^2$ the coefficients of $V,W$ are nonnegative, and their
difference is $(17/12)(1-u^2)\ge0$. Replacing $(V,W)$ by
$(\delta,0)$ therefore enlarges this expression. This is a relaxation,
not an assertion that the new pair is geometrically attained. The
enlarged values are
\[
g_v=\frac{19}{3}-3v-\frac{17}{12}b,
\qquad g_b=\frac72-\frac{19}{12}b.
\]
Put $\omega=27/128$, $\mu=9/32$, and define
\[
\begin{aligned}
A&=\frac{67}{50}(1+u^2),\\
G&=\frac{27}{100}(1+u^2)u^2
+3\omega(1+u)^2+\mu(1+u)(1-u^3),\\
H&=\frac{27}{100}(1+u^2)
+\frac\omega{12}(17+19u^2)(1+u)^2
-\mu(1+u)(1-u^3),\\
L&=\omega(1+u)^2\left(\frac{19}{3}+\frac72u^2\right)
+\frac34(1+u)(1+u^3).
\end{aligned}
\]
Dividing the difference between the left and enlarged right sides of
\eqref{eq:20} by $vbP^4$ gives exactly
\[
\frac Av+\frac{Au^2}{b}+Gv+Hb-L.
\]
Here $A,L>0$, and Lemma~\ref{lem:scalar-certificates} gives $G,H>0$.
For
\[
Z_*=L^2-4A(G+u^2H)
\]
the same lemma gives $Z_*\le0$ on $[0,1/16]$ and
$Z_*^2\le64A^2u^2GH$ on $[1/16,1]$. In the latter interval,
$Z_*\le|Z_*|\le8Au\sqrt{GH}$. In both intervals it follows that
\[
L\le2\sqrt A(\sqrt G+u\sqrt H).
\]
Taking square roots is legitimate because both sides are nonnegative.
The arithmetic--geometric mean inequalities applied to $A/v+Gv$ and
$Au^2/b+Hb$ finish the proof of \eqref{eq:20}.
\subsection{The quadratic terms}
The quartic bound is established. We next bound the quadratic part by its \emph{axial value}: the algebraic
comparison with $\sigma=\bar\sigma=1$, $\rho=\bar\rho=0$, and
$k=l=\bar k=\bar l=0$, holding $v,b,p,q$ fixed. The corresponding spectral
shapes are $(-v,-v,2v)$ and $(-b,-b,2b)$. In the plus contribution, replacing
$\bar\sigma$ by $1$ raises the expression by
$6h(1-\bar\sigma)b^2e\ge0$. After this replacement, the shape-dependent
quadratic expression is
\[
mG_v+h\lambda_v e+40h\delta B
+\frac{\kappa}{v}(e-|p|^2)
+(\partial_v\kappa)|p|^2+(\partial_b\kappa)p\cdot q,
\]
where $\lambda_v=6(\sigma v^2+b^2-v-b)-26\delta$. The last two terms and
the separate variance term are unchanged by the shape comparison.
Define
\[
P_0=300m+30hv+160h\delta,\qquad
P_k=300m+160h\delta,\qquad P_l=35mv.
\]
On the closed spectral triangle, $m\ge1/16$ and $h,\delta,v\ge0$;
therefore $P_0=P_k+30hv>0$. This supplies the positive denominator in
the Schur comparison below.
By \eqref{eq:11} and \eqref{eq:16}, the loss in the $pp$ entry when
$\sigma<1$ is $(1-\sigma)vP_0/3$, and the mixed entries with $(k,l)$
are $-v\rho(P_k,P_l)/\sqrt3$. The negative of the pure $(k,l)$ matrix
is exactly
\[
\mathsf D_\sigma=
\begin{pmatrix}K_1&K_{12}\\K_{12}&K_2\end{pmatrix}
-\frac\kappa v\operatorname{diag}(5,1),
\]
where
\[
\begin{aligned}
K_2&=m(42+3b^2+10v^2)-h\lambda_v,\\
K_1&=5K_2+m(300v\sigma+140v^2)+160h\delta v\sigma,\\
K_{12}&=m(75v+35v^2\sigma)+40h\delta v.
\end{aligned}
\]
Thus the required completion of squares follows from
\begin{equation}
P_0\left[
v\begin{pmatrix}K_1&K_{12}\\K_{12}&K_2\end{pmatrix}
-\kappa\operatorname{diag}(5,1)\right]
-(1+\sigma)v^2
\begin{pmatrix}P_k\\P_l\end{pmatrix}
\begin{pmatrix}P_k&P_l\end{pmatrix}\succeq0.
\tag{21}\label{eq:21}
\end{equation}
Let $\mathsf A_\sigma$ be this matrix and $w=(P_k,P_l)^T$.
The identity $P_0=P_k+30hv$ gives
\[
-\partial_\sigma\mathsf A_\sigma
=v^2\left[
\begin{pmatrix}30hv\\-P_l\end{pmatrix}
\begin{pmatrix}30hv&-P_l\end{pmatrix}
+\operatorname{diag}(0,6hvP_0)\right]\succeq0.
\]
Lemma~\ref{lem:matrix-certificates} proves \eqref{eq:21} at
$\sigma=1$, and this matrix monotonicity proves it for every
$-1\le\sigma\le1$. When $\sigma<1$, put
$\ell=(1-\sigma)vP_0/3>0$ and $d=-v\rho w/\sqrt3$.
The Schur complement is
\[
\mathsf D_\sigma-\frac{dd^T}{\ell}
=\mathsf D_\sigma-\frac{(1+\sigma)v}{P_0}ww^T
=\frac{\mathsf A_\sigma}{P_0v}\succeq0.
\]
At $\sigma=1$, one has $\rho=0$ and
\[
\mathsf D_1=\frac{\mathsf A_1+2v^2ww^T}{P_0v}\succeq0,
\]
so no division by $1-\sigma$ is needed. No sign of $\kappa$ is assumed.
Applying the exchanged argument proves the announced axial reduction.
At the axial comparison, \eqref{eq:2} and the matrix for $B/v$ give
$e_{\rm ax}=5|p|^2/3$ and $B_{\rm ax}=4v|p|^2/3$. Put
$\lambda_{\rm ax}=6(v^2+b^2-v-b)-26\delta$. Since $\delta=-a-s$,
\[
\begin{aligned}
\bigl(h\lambda e+40h\delta B\bigr)_{\rm ax}
&=\frac53h\bigl(\lambda_{\rm ax}+32v\delta\bigr)|p|^2\\
&=\frac53h\bigl(6(a^2+s^2)+32a\delta\bigr)|p|^2.
\end{aligned}
\]
Together with the axial $pp$ entry of \eqref{eq:11}, this gives
$X_v|p|^2$ for the axial part of $mG_v+h\lambda e+40h\delta B$, where
\[
X_v=m(-70+100v-25v^2-5b^2)
+\frac53h\bigl(6(a^2+s^2)+32a\delta\bigr),
\qquad X_b=X_v(b,v).
\]
Collecting the remaining quadratic terms gives the desired estimate
precisely when
\begin{equation}
\begin{gathered}
\begin{pmatrix}\mathcal Q_v&J\\J&\mathcal Q_b\end{pmatrix}
\preceq2\Theta\operatorname{diag}(R_v,R_b),\\
\mathcal Q_v=X_v+\frac DK+\partial_v\kappa+\frac{2\kappa}{3v},
\qquad
\mathcal Q_b=X_b+\frac DK+\partial_b\bar\kappa
+\frac{2\bar\kappa}{3b},\\
J=\frac{\partial_b\kappa+\partial_v\bar\kappa}{2}-\frac DK.
\end{gathered}
\tag{22}\label{eq:22}
\end{equation}
The polynomial matrix certificate in Lemma~\ref{lem:matrix-certificates}
will prove this inequality.
The leading terms of $\kappa$ can now be read in their intended role.
Along the formal axial boundary $v=1+t$, $b=1-t$, direct substitution
in the displayed formulas gives
\[
\mathcal Q_v,\mathcal Q_b,J=O(t^2)\quad(t\to0),
\qquad \Theta=t^2-t^4.
\]
Thus the constant and tangential linear terms cancel at the meeting
point of the two axial blocks. This explains the local shape of the
rational choice; the certificate proves its inequality on the full
stated domain.
\subsection{The potential and the zero sets}
The quadratic estimate has been reduced to its explicit matrix certificate.
It remains to bound the derivative-free potential and to extend the
three inequalities across the zero sets. The potential is symmetric under exchanging the two blocks. For its
pointwise estimate, suppose $v\ge b$ and define
\[
\begin{aligned}
E_0={}&D\left(1-\frac{\delta^2}{4}\right)(v-b)^2
+30h\delta(v^2a^2+b^2s^2)
+6(av\kappa+sb\bar\kappa),\\
F={}&cE_0-\Theta^2,\qquad
L_v=10h\delta v-\kappa,\qquad
L_b=10h\delta b-\bar\kappa.
\end{aligned}
\]
Direct collection from \eqref{eq:16}--\eqref{eq:18} gives
\[
\mathcal I_0=-E_0-6v^2(1-\sigma)L_v
-6b^2(1-\bar\sigma)L_b.
\]
Since $y-b\le\delta$ and $r_0=y/b\le2$,
\[
b^2(1-\bar\sigma)
=\frac12b^2(r_0-1)^2(r_0+2)\le2\delta^2.
\]
It therefore suffices to establish
\begin{equation}
L_v\ge0,\qquad F\ge0,\qquad
F+12c\delta^2L_b\ge0\qquad(v\ge b).
\tag{23}\label{eq:23}
\end{equation}
Indeed, if $L_b\ge0$ then $\mathcal I_0\le-E_0\le-\Theta^2/c$.
If $L_b<0$, the displayed spectral bound instead gives
\[
\mathcal I_0\le-E_0-12\delta^2L_b\le-\Theta^2/c.
\]
Lemmas~\ref{lem:matrix-certificates} and
\ref{lem:potential-certificate} prove all three inequalities in
\eqref{eq:23}.
The use of an ordering in \eqref{eq:22} and \eqref{eq:23} is purely
algebraic. All weights and their derivatives have already been formed
from the fixed global labels of the two blocks. Their summed targets
are symmetric. At a point with $v<b$, apply the ordered inequality to
$(b,v)$ and exchange the two vector entries. At $v=b$, either labeling
gives the same conclusion. No maximum, minimum, or selected label is
differentiated.
Finally, on the actual zero set $\{v=0\}$, one has $x=0$ and $y<2$ by
strict sectional positivity, and the components $Z,p,k,l$ vanish almost everywhere.
To pass the algebraic bounds to this set, first set all plus derivative
parameters $p,k,l$ to zero in the positive-variable inequalities.
Then $e,B,C_N$ and every numerator divided by $v$ are identically zero
before taking any limit. Keep $b>0$ and all minus spectral and derivative
data fixed. Choose axial plus spectra $(-v,-v,2v)$, for which $x=v$,
with $0<v<\min\{2-y,2-b\}$ tending to zero. This sequence remains in
both required spectral inequalities. The remaining quadratic inequality
is the minus diagonal, whose denominator $bK(1+6b+b^2)$ stays positive;
the quartic bound becomes $-c|q|^4/b$. All cross terms with $p$ vanish.
This argument does not assert continuity of the singular plus
coefficients. The other zero set is treated by
exchange, and if both norms vanish all derivative components vanish.
The potential inequalities are polynomial and extend continuously.
The quartic quotients are bounded in absolute value by a constant times
$e/v+\bar e/b$, and are therefore integrable. This completes the proof
of Proposition~\ref{prop:coupling-estimates}, subject only to the explicit
polynomial verifications in the next section.
==== END Positively-curved-Einstein-four-manifolds//sections/06-coupling.tex ====
==== BEGIN Positively-curved-Einstein-four-manifolds//sections/07-certificates.tex ====
\section{Exact polynomial certificates}\label{sec:certificates}
This section proves the finite polynomial inequalities used in
Section~\ref{sec:coupling}. The coefficients are rational throughout.
We give the conversion formulas, domains, and coefficient bounds so
that each sign follows from a finite polynomial expansion.
The method uses the nonnegative Bernstein basis on intervals and
simplices; see Boudaoud--Caruso--Roy~\cite{BoudaoudCarusoRoy2008} and
Roy~\cite[Sections~1 and~4]{Roy2007}. We prove the conversion formulas
and their matrix-valued consequence here. Positivity is verified for
these specific polynomials, with their boundary zeros retained.
The accompanying standard-library Python checker recomputes the saved
exact coefficient data and checks the specified algebraic identities.
Its coverage record distinguishes those executed reconstructions from
the collection identities and geometric reductions proved in the text.
The source package accompanies this PDF under the stable article identifier
\begin{center}
\small\ttfamily Positively-curved-Einstein-four-manifolds-September-23-2026
\end{center}
The arithmetic package is at \path{verification/}; from
the paper directory, run
\begin{verbatim}
python3 -B verification/verify.py --check
\end{verbatim}
The adjacent \path{COVERAGE.md} records the precise scope of that check.
A detached copy of this PDF does not contain the checker.
\subsection{Bernstein coefficients on intervals and triangles}
\begin{lemma}[Bernstein conversion]\label{lem:bernstein-conversion}
Let $p$ be a polynomial of degree at most $N$ on an interval $[r,s]$,
and write
\[
p(r+(s-r)z)=\sum_{j=0}^N a_jz^j.
\]
Its degree-$N$ Bernstein expansion is
\[
p(r+(s-r)z)=\sum_{i=0}^N h_i\binom Ni z^i(1-z)^{N-i},
\qquad
h_i=\sum_{j=0}^i a_j\frac{\binom ij}{\binom Nj}.
\]
In particular, $h_i\ge\eta$ for all $i$ implies $p\ge\eta$ on
$[r,s]$.
For a polynomial on a triangle with ordered vertices
$p_\ell=(r_\ell,s_\ell)$, set
\[
V=r_3+(r_1-r_3)z+(r_2-r_3)w,\qquad
W=s_3+(s_1-s_3)z+(s_2-s_3)w.
\]
If $P(z,w)=p(V,W)=\sum_{d,e}P_{de}z^dw^e$ has degree at most $N$,
then its normalized degree-$N$ triangular coefficients are
\[
[p]_{ij}=s_{ij}(P):=
\sum_{d=0}^i\sum_{e=0}^j
\frac{\binom id\binom je}
{\binom N{d+e}\binom{d+e}d}\,P_{de},
\qquad i,j\ge0,\quad i+j\le N.
\]
On $z,w\ge0$, $z+w\le1$, the polynomial is the sum of these coefficients
times the nonnegative basis functions
\[
\frac{N!}{i!j!(N-i-j)!}
z^iw^j(1-z-w)^{N-i-j}.
\]
These basis functions sum to one. The same expansion applies entrywise
to a symmetric polynomial matrix: positive semidefiniteness of every
coefficient matrix implies positive semidefiniteness throughout the
triangle.
\end{lemma}
\begin{proof}
For an interval, expand
$z^j=z^j(z+(1-z))^{N-j}$ and compare coefficients in the Bernstein
basis. On a triangle, expand
\[
z^dw^e=z^dw^e(z+w+(1-z-w))^{N-d-e}.
\]
This gives the displayed factors after dividing each coefficient by the
corresponding multinomial coefficient. Equivalently, substitute
\[
(v,b)=\frac{u_1p_1+u_2p_2+u_3p_3}{u_1+u_2+u_3}
\]
and multiply by $(u_1+u_2+u_3)^N$. The normalized coefficient of
$u_1^iu_2^ju_3^{N-i-j}$ is $[p]_{ij}$. The positivity statements
follow from the nonnegative partition of unity, also for matrix
quadratic forms.
\end{proof}
\subsection{The quartic scalar certificates}
\begin{lemma}[Scalar certificates]\label{lem:scalar-certificates}
For the polynomials in Section~\ref{sec:coupling}, the following bounds
hold:
\[
\begin{gathered}
C_0>0,\qquad
4C_0\left((h-A_0)(2-v)+\frac25hv\right)
-(2-v)(2A_0+L_0)^2>0\quad(0\le v\le2),\\
G,H>0\quad(0\le u\le1),\qquad
Z_*\le0\quad(0\le u\le1/16),\\
Z_*^2\le64A^2u^2GH\quad(1/16\le u\le1).
\end{gathered}
\]
\end{lemma}
\begin{proof}
Apply the interval formula of Lemma~\ref{lem:bernstein-conversion} in
the degrees and on the intervals indicated below. Every coefficient
has at least the displayed positive lower bound. In the first two rows
the bound applies on each of the two intervals separately.
\begin{center}
\small
\begin{tabular}{@{}lccc@{}}
\toprule
Polynomial & Degree & Interval & Lower bound\\
\midrule
$C_0$ & $2$ & $[0,1],\ [1,2]$ & $1/2$\\
$4C_0((h-A_0)(2-v)+\frac25hv)
-(2-v)(2A_0+L_0)^2$
& $5$ & $[0,1],\ [1,2]$ & $1/2$\\
$G$ & $4$ & $[0,1]$ & $9/10$\\
$H$ & $4$ & $[0,1]$ & $1/4$\\
$-Z_*$ & $8$ & $[0,1/16]$ & $13/100$\\
$64A^2u^2GH-Z_*^2$ & $16$ & $[1/16,1]$ & $12/100$\\
\bottomrule
\end{tabular}
\end{center}
For the first two rows the coefficients are in fact strictly greater
than $1/2$. For example, their respective minima on the two intervals are
\[
\left(\frac{3971}{3200},\frac{1063}{1600}\right),
\qquad
\left(\frac{101027}{20000},\frac{283729}{400000}\right).
\]
All entries in this table are obtained by substituting the displayed
formulas for $A_0,L_0,C_0$ and for $A,G,H,L,Z_*$ into the coefficient sum in
Lemma~\ref{lem:bernstein-conversion}. Thus each bound is an inequality
between rational coefficients. The lemma proves the required signs.
\end{proof}
\subsection{The two matrix certificates}
For the quadratic inequality \eqref{eq:22}, introduce
\[
\begin{aligned}
r_v&=1+6v+v^2,\qquad r_b=1+6b+b^2,
\qquad \alpha=\frac54,\\
H_v&=8vK\Theta(1+v)
-r_v\left\{vD+K\left[v(X_v+\partial_v\kappa)
+\frac{2\kappa}{3}\right]\right\},\\
H_b&=H_v(b,v),\\
S_*&=4(\alpha+\alpha^{-1})vb
+\frac{\alpha b(1-v)^2+\alpha^{-1}v(1-b)^2}{2}.
\end{aligned}
\]
The diagonal differences in \eqref{eq:22} are
$H_v/(vKr_v)$ and $H_b/(bKr_b)$. Since
$r_v=8v+(1-v)^2$ and $r_b=8b+(1-b)^2$, weighted arithmetic--geometric
mean gives
\[
\sqrt{vb r_vr_b}
\le\frac{\alpha b r_v+\alpha^{-1}v r_b}{2}=S_*.
\]
It is therefore sufficient to prove
\[
\begin{pmatrix}H_v&S_*KJ\\S_*KJ&H_b\end{pmatrix}\succeq0.
\]
Indeed its diagonal entries are nonnegative and its determinant gives
$H_vH_b\ge S_*^2K^2J^2\ge vb r_vr_bK^2J^2$. Dividing by the positive
denominators proves that the matrix of differences in \eqref{eq:22}
is positive semidefinite. This argument uses $J^2$ and imposes no sign
condition on $J$.
For a symmetric $2\times2$ matrix, record its entries in the order
$(11,22,12)$ as a triple $(r,h',s')$. In this notation the two triples
to be certified are
\[
\begin{aligned}
\mathcal A={}&P_0\bigl(v(K_1,K_2,K_{12})-\kappa(5,1,0)\bigr)
-2v^2(P_k^2,P_l^2,P_kP_l),\qquad \sigma=1,\\
\mathcal B={}&(H_v,H_b,S_*KJ).
\end{aligned}
\]
The first triple is precisely the matrix in \eqref{eq:21} at
$\sigma=1$.
\begin{lemma}[Matrix and boundary certificates]
\label{lem:matrix-certificates}
The symmetric matrix represented by $\mathcal A$ is positive semidefinite on
\[
\mathcal D=\{(v,b):v,b\ge0,\ v+b\le2\}.
\]
The symmetric matrix represented by $\mathcal B$ is positive semidefinite on
\[
\mathcal D_+=\{(v,b)\in\mathcal D:v\ge b\}.
\]
On $\mathcal D_+$ one also has
\[
L_v\ge0,\qquad F+12c\delta^2L_b\ge0.
\]
\end{lemma}
\begin{proof}
Use the vertices
\[
\begin{gathered}
O=(0,0),\quad S=(1,0),\quad C=(1,1),\quad
U=(2,0),\quad W_0=(0,2),\\
A=(1/2,1/2),\quad B=(3/2,1/2),\quad E=(1,1/2).
\end{gathered}
\]
A three-letter word specifies the vertices of an ordered triangle.
The triangles $OSC$, $OCW_0$, and $SCU$ cover $\mathcal D$.
The triangles $OAS$, $ASE$, $ACE$, $SBE$, $CBE$, and $SUB$ cover
$\mathcal D_+$; the latter is also the single triangle $OCU$, or the
union $OSC\cup SCU$. All edges are included. The two subdivisions are
shown below.
\input{figures/spectral-domains.tex}
For a matrix row, the last two columns below bound, respectively,
\[
\min([r]_{ij},[h']_{ij}),\qquad
[r]_{ij}[h']_{ij}-[s']_{ij}^{2}.
\]
Thus the last column is the determinant of each coefficient matrix.
For a scalar row, the first bound applies directly to its coefficient.
The bounds apply to every nonzero coefficient; a zero matrix exception
means that all three of its entries vanish.
\begin{center}
\small
\begin{tabular}{@{}lcccrr@{}}
\toprule
Object & Triangle & $N$ & Zeros &
\shortstack{Diagonal or\\scalar bound} & Determinant bound\\
\midrule
$\mathcal A$ & $OSC$ & $8$ & $0$ & $16000$ & $53000000$\\
& $OCW_0$ & $8$ & $0$ & $890$ & $3900000$\\
& $SCU$ & $8$ & $0$ & $2400$ & $3500000$\\
\midrule
$\mathcal B$ & $OAS$ & $9$ & $0$ & $43$ & $1050$\\
& $ASE$ & $9$ & $0$ & $43$ & $490$\\
& $ACE$ & $10$ & $1$ & $2$ & $21$\\
& $SBE$ & $9$ & $0$ & $43$ & $1300$\\
& $CBE$ & $10$ & $2$ & $13/100$ & $4$\\
& $SUB$ & $9$ & $0$ & $5$ & $290$\\
\midrule
$L_v$ & $OCU$ & $6$ & $1$ & $9/10$ & ---\\
$F+12c\delta^2L_b$ & $OSC$ & $9$ & $6$ & $19/100$ & ---\\
& $SCU$ & $9$ & $7$ & $9/1000$ & ---\\
\bottomrule
\end{tabular}
\end{center}
Here are explicit expansion instructions for all entries of the table.
They also fix the coefficient ordering. Write
$a=V-1$, $s=W-1$ in \eqref{eq:15} and form $\kappa(V,W)$ and
its two partial derivatives. A partial derivative is obtained by
reducing the appropriate exponent of each monomial and multiplying by
that exponent. Differentiate before substituting the affine expressions
$V,W$. Put
\[
\kappa_* =\kappa(V,W),\qquad \bar\kappa_* =\kappa(W,V).
\]
For clarity, the derivative of the exchanged polynomial with respect to
its first variable is $(\partial_2\kappa)(W,V)$, and its derivative
with respect to its second variable is $(\partial_1\kappa)(W,V)$.
Ordinary coefficient arrays are multiplied by convolution:
\[
(PQ)_{de}=\sum_{f=0}^d\sum_{g=0}^e P_{fg}Q_{d-f,e-g}.
\]
Use
\[
\begin{aligned}
m_*&=\left(1-\frac38(V-W)\right)^2,
&d_0&=2-V-W,\\
P_{k,*}&=300m_*+160h d_0,
&P_{l,*}&=35m_*V,\\
Y_*&=m_*(42+3W^2+10V^2)
-h\bigl[6(V^2+W^2-V-W)-26d_0\bigr].
\end{aligned}
\]
For $\mathcal A$, apply $s_{ij}$ from
Lemma~\ref{lem:bernstein-conversion} to the three entries of
\[
\begin{aligned}
&(P_{k,*}+30hV)
\bigl\{V(A_1,Y_*,A_{12})-\kappa_*(5,1,0)\bigr\}\\
&\hspace{22mm}-2V^2(P_{k,*}^2,P_{l,*}^2,P_{k,*}P_{l,*}),
\end{aligned}
\]
where
\[
A_1=5Y_*+m_*(300V+140V^2)+160h d_0V,
\qquad
A_{12}=m_*(75V+35V^2)+40h d_0V.
\]
These are the original entries of \eqref{eq:21} at $\sigma=1$, expressed
only in the affine polynomials $V,W$.
For $\mathcal B$, put
\[
\begin{aligned}
K_*&=4+V^2+W^2,\qquad \Theta_* =\Theta(V,W),\\
X_*&=m_*(-70+100V-25V^2-5W^2)
+\frac53h\bigl(6(a^2+s^2)+32a d_0\bigr),\\
H_*&=8VK_*\Theta_*(1+V)\\
&\quad-(1+6V+V^2)
\left\{VD+K_*\left[V\bigl(X_*+(\partial_1\kappa)(V,W)\bigr)
+\frac{2\kappa_*}{3}\right]\right\}.
\end{aligned}
\]
Apply $s_{ij}$ to $H_*$, to its exchange in $V,W$, and to
\[
S_*(V,W)\left\{
\frac{K_*}{2}\bigl((\partial_2\kappa)(V,W)
+(\partial_2\kappa)(W,V)\bigr)-D\right\}.
\]
For example, on $CBE$ the triples at $(i,j)=(8,0),(8,1),(8,2)$ are
\[
\left(\frac{11041}{810},\frac{4063}{810},\frac{60557}{16200}\right),
\quad
\left(\frac{4328}{405},\frac{839}{405},\frac{3977}{2700}\right),
\quad
\left(\frac{191}{81},\frac{191}{81},-\frac{1189}{1350}\right).
\]
For the two scalar rows, apply $s_{ij}$ to
$10h d_0V-\kappa_*$ and, respectively, to
\[
\begin{aligned}
c(V,W)\bigg\{&D\left(1-\frac{d_0^2}{4}\right)(V-W)^2
+30h d_0(V^2a^2+W^2s^2)\\
&+6(Va\kappa_*+Ws\bar\kappa_*)
+12d_0^2(10h d_0W-\bar\kappa_*)\bigg\}-\Theta_*^2.
\end{aligned}
\]
Together with the coefficient-sum formula, these prescriptions produce
the displayed rational bounds by addition and multiplication of finite
arrays.
The zero exceptions are as follows, with the third homogeneous index
always $N-i-j$. For $\mathcal B$, they are the vertex $C$ in $ACE$
and $CBE$, namely $(0,10,0)$ and $(10,0,0)$, and also $(9,1,0)$ in
$CBE$. For $L_v$ the only zero is $(0,6,0)$, the vertex $C$ in $OCU$.
For $F+12c\delta^2L_b$ on $OSC$, the zeros are exactly $i+j\le2$.
On $SCU$ they are exactly $9-j\le2$, together with $(i,j)=(0,6)$.
There are no other exceptions.
Every nonzero matrix coefficient has positive diagonal entries and
positive determinant, and each exceptional matrix is zero. Thus all
coefficient matrices are positive semidefinite. The scalar coefficients
are all nonnegative. Lemma~\ref{lem:bernstein-conversion} and the stated
domain coverage prove the assertions.
\end{proof}
\subsection{The remaining potential polynomial}
\begin{lemma}[The polynomial $F$]\label{lem:potential-certificate}
The polynomial $F=cE_0-\Theta^2$ is nonnegative on $\mathcal D$.
\end{lemma}
\begin{proof}
Use $v=1-n+t$, $b=1-n-t$. The full domain becomes
\[
0\le n\le1,\qquad |t|\le1-n.
\]
Write $F_*(n,t)=F(1-n+t,1-n-t)$ for the pullback to these coordinates.
It is even in $t$, and direct expansion gives
$F_*=\sum_{j=0}^4 f_j(n)t^{2j}$, where
\[
\begin{aligned}
240f_0={}&n^3(1-n)\bigl(53511n^4-151593n^3
+222537n^2-83607n+10256\bigr),\\
60f_1={}&n\bigl(19389n^5-418926n^4+523719n^3
-289776n^2+67782n+10564\bigr),\\
60f_2={}&-1260129n^4+1370598n^3-451164n^2+23442n+74,\\
10f_3={}&-12200n^2+2601n+75,\qquad
f_4=-\frac{23}{5}.
\end{aligned}
\]
For $n<1$, put $z=t^2/(1-n)^2\in[0,1]$. The degree-four Bernstein
coefficients in $z$ are
\[
h_i(n)=\sum_{j=0}^i
\frac{\binom ij}{\binom4j}f_j(n)(1-n)^{2j},
\qquad 0\le i\le4.
\]
Define
\[
d_0(n)=n^3(1-n),\qquad d_1(n)=n(1-n),\qquad
d_2(n)=d_3(n)=d_4(n)=1-n.
\]
Each quotient $q_i=h_i/d_i$ is a polynomial. Applying the interval
conversion formula to it on each of the four intervals
\[
[0,1/4],\quad[1/4,1/2],\quad[1/2,3/4],\quad[3/4,1]
\]
gives the following lower bounds for every coefficient:
\begin{center}
\begin{tabular}{@{}cccc@{}}
\toprule
$i$ & Divisor $d_i$ & Degree of $q_i$ & Lower bound\\
\midrule
$0$ & $n^3(1-n)$ & $4$ & $3$\\
$1$ & $n(1-n)$ & $6$ & $31$\\
$2$ & $1-n$ & $7$ & $1/5$\\
$3$ & $1-n$ & $7$ & $12/5$\\
$4$ & $1-n$ & $7$ & $9/5$\\
\bottomrule
\end{tabular}
\end{center}
The complete prescription for each entry is to insert the displayed
$f_j$ into $h_i$, cancel the indicated polynomial factor, and apply
the coefficient sum in Lemma~\ref{lem:bernstein-conversion} in the
listed degree. In particular, the identities $h_i=d_iq_i$ hold as
polynomial identities, including at the endpoints.
Since every $d_i$ is nonnegative on $[0,1]$, every $h_i$ is
nonnegative. Therefore
\[
F_*=\sum_{i=0}^4h_i(n)\binom4i z^i(1-z)^{4-i}\ge0
\qquad(n<1).
\]
At $n=0$, the coordinate $z=t^2$ still covers the entire permitted
interval $|t|\le1$; the vanishing of $h_0,h_1$ causes no difficulty.
At $n=1$, the domain contains only $t=0$, and directly
$F_*(1,0)=F(0,0)=f_0(1)=0$. Thus no numerical division by a vanishing factor
is required, and all of $\mathcal D$ is covered.
\end{proof}
Lemmas~\ref{lem:scalar-certificates}, \ref{lem:matrix-certificates},
and \ref{lem:potential-certificate} establish every polynomial
inequality used in Section~\ref{sec:coupling}. Hence the three estimates
of Proposition~\ref{prop:coupling-estimates} hold on the stated
positive-variable spectral domain. The compatible limit in
Section~\ref{sec:coupling} extends them to the zero sets of the standing
strictly positive geometry.
==== END Positively-curved-Einstein-four-manifolds//sections/07-certificates.tex ====
==== BEGIN Positively-curved-Einstein-four-manifolds//sections/08-classification.tex ====
\section{Half-conformal flatness and the metric classification}
\label{sec:classification}
The coupled estimates have been proved for the standing strictly
positive geometry, including its zero sets. Their strictness now rules
out two nonzero Weyl blocks. We then isolate the remaining harmonic
alternative, exclude it by a conditional lower volume estimate, and
identify the parallel curvature tensor. The orientation-cover argument
comes only after the oriented conclusion has been established.
\begin{proposition}[Half-conformal flatness]\label{prop:one-block-vanishes}
At least one of $T,U$ vanishes identically; equivalently, the metric is
half-conformally flat.
\end{proposition}
\begin{proof}
Suppose otherwise. Proposition~\ref{prop:balanced-moments} gives $\mathbf E v^2=\mathbf E b^2$, so Proposition~\ref{prop:coupling-estimates} applies. Its integrable total expression $\mathcal I$ has nonnegative integral and satisfies
\[
\mathcal I\le-cj+2\Theta r-\Theta^2/c
=-\frac{(\Theta-cr)^2}{c}-c(j-r^2),\qquad c\ge\frac1{10},
\]
where
\[
r=R_v|p|^2+R_b|q|^2,\qquad
j=(|p|^2+|q|^2)(|p|^2/v+|q|^2/b).
\]
The zero-set conventions established with \eqref{eq:3} apply to the quotients. For $v>0$,
\[
1+6v+v^2-4\sqrt v(1+v)=(\sqrt v-1)^4,
\]
so $R_v\le1/\sqrt v$, with equality only at $v=1$. Cauchy--Schwarz therefore gives $r^2\le j$.
This inequality is strict almost everywhere on the set where $p$ or $q$ is nonzero. Indeed a Lipschitz function has zero gradient almost everywhere on each of its level sets: at a differentiability and density point of a level set, a nonzero differential would contradict density. Thus, outside a null set, an active gradient $p$ has $v>0$ and $v\ne1$, and similarly for $q$. At such a point
\[
r< |p|^2/\sqrt v+|q|^2/\sqrt b\le\sqrt j.
\]
Terms with vanishing gradients are read as zero. It follows that $\mathcal I\le0$ almost everywhere, and that $\mathcal I<0$ almost everywhere where either gradient is nonzero. Since $\mathbf E\mathcal I\ge0$, both gradients vanish almost everywhere. Connectedness and continuity make $v,b$ constant. Both constants are positive, so the separate gaps \eqref{eq:5} imply $v+b\ge2$, contradicting $v+b\le x+y<2$.
\end{proof}
Once one Weyl block vanishes, $(M,g)$ belongs to the classically
classified compact half-conformally flat Einstein class with positive
scalar curvature~\cite{Hitchin1981,FriedrichKurke1982,Obata1962}.
We retain the lower volume and parallel curvature argument below as an
independent proof of this final step under our standing hypothesis of
strictly positive sectional curvature.
\subsection{The remaining harmonic alternatives}
After reversing orientation if necessary, one-block vanishing leaves
$U=0$. If $T$ also vanishes, the sectional curvature is constantly one.
For a nonzero surviving block, the characteristic identities reduce the
possibilities to two integer values.
\begin{lemma}[The one-block harmonic alternatives]
\label{lem:one-block-alternatives}
Suppose $U=0$ and $T\not\equiv0$. Then
\[
\frac{V_M}{\pi^2}=\frac{8-2n_+}{3},\qquad
s_0=\mathbf E v^2=\frac{3n_+}{4-n_+},\qquad n_+=1\text{ or }2.
\]
In the second case, $s_0=3$ and $V_M=4\pi^2/3$.
\end{lemma}
\begin{proof}
Proposition~\ref{prop:characteristic-identities} gives $n_-=0$ and
$\chi=2+n_+$, $\tau=n_+$. Solving its two identities yields the displayed
volume and moment formulas. Positivity of volume gives $n_+<4$; the
nonzero-block gap gives $s_0\ge1$, while $v<2$ gives $s_0<4$.
The finite nonnegative integer $n_+$ is therefore $1$ or $2$.
Substitution gives the stated values in the second case.
\end{proof}
\subsection{Excluding the exceptional harmonic count}
The value $n_+=2$ would force exactly $s_0=3$ and $V_M=4\pi^2/3$.
We now prove a strict lower bound under that same global moment
hypothesis. The statement does not require one-block vanishing; that
conclusion is used only to produce the exceptional value to which the
bound will be applied.
\begin{proposition}[Conditional lower volume bound]
\label{prop:conditional-lower-volume}
If $s_0=\mathbf E(v^2+b^2)=3$, then $V_M>4\pi^2/3$.
\end{proposition}
Recall from Section~\ref{sec:volume} that $\mathcal E$ is expectation over
the unit tangent bundle, and that $\mathcal R$ is the normal Jacobi
curvature matrix along its geodesics. With
$\mathcal J''+\mathcal R\mathcal J=0$, $\mathcal J(0)=0$ and
$\mathcal J'(0)=\mathrm{Id}$, put $D(t)=\det\mathcal J(t)$.
We use the topology statement in Lemma~\ref{lem:volume-topology}, the
geodesic averages in Lemma~\ref{lem:geodesic-averages}, and the discrete
determinant in Lemma~\ref{lem:discrete-jacobi}.
\begin{lemma}[Lower polar bound]\label{lem:lower-polar}
One has
\[
V_M\ge2\pi^2\int_0^{\pi/\sqrt3}\mathcal E D(t)\,dt.
\]
\end{lemma}
\begin{proof}
The open ball of radius $\pi/\sqrt3$ lies in the injectivity domain at
every center by Lemma~\ref{lem:volume-topology}. Its volume is the
corresponding integral of $D$ and is at most $V_M$. Averaging over centers
gives the bound. Increasing open radii justify the endpoint; it does not
change the integral.
\end{proof}
\begin{lemma}[Lower radial comparison]\label{lem:lower-radial}
If $s_0=3$ and $0<t<\pi/\sqrt3$, then, with $\theta=\sqrt3t$,
\[
\mathcal E D(t)\ge
t^3\left(\frac{\sin\theta}{\theta}\right)^{5/6}
e^{-\theta^2/36}.
\]
Moreover,
\[
-\log\left(\frac{\sin\theta}{\theta}\right)-\frac{\theta^2}{6}
\ge\frac12\left(\frac\theta\pi\right)^4
+\frac13\left(\frac\theta\pi\right)^6.
\]
\end{lemma}
\begin{proof}
For each integer $N\ge2$, put $a=t/N$. Let $L_0$ be the scalar
Dirichlet tridiagonal matrix on the interior nodes $1,\ldots,N-1$,
with diagonal entries $2$ and adjacent off-diagonal entries $-1$.
Put $G=a^2L_0^{-1}$. Its entries are nonnegative; explicitly,
\[
G_{ij}=a^2\frac{\min(i,j)(N-\max(i,j))}{N},\qquad
\operatorname{tr}G=\frac{a^2(N^2-1)}6\longrightarrow\frac{t^2}{6}.
\]
Write $\mathcal R_N$ for the block diagonal matrix with blocks
$\mathcal R(ia)$ and define
$H_N=(G^{1/2}\otimes\mathrm{Id})\mathcal R_N
(G^{1/2}\otimes\mathrm{Id})$.
For fixed $t<\pi/\sqrt3$ and large $N$,
\[
0\prec H_N\preceq q_t\mathrm{Id},\qquad q_t<1,
\qquad
\|H_N\|\le\frac{3a^2}{2-2\cos(\pi/N)},\qquad
\operatorname{tr}H_N=3\operatorname{tr}G.
\]
In particular,
\[
-\log\det(\mathrm{Id}-H_N)
=\sum_{k\ge1}\frac{\operatorname{tr}H_N^k}{k}
\le\frac{3\operatorname{tr}G}{1-q_t}.
\]
This bound is uniform over all geodesics and large $N$ at the fixed
time. It justifies taking expectations in the series and gives a
uniform positive lower bound for the determinant ratios, independent
of their increasing matrix dimensions.
By \eqref{eq:8}, $\mathcal E|\mathcal R|^2=3+3s_0/2=15/2$.
For any ordered product of $k\ge2$ curvature matrices,
\[
|\operatorname{tr}(\mathcal R_1\cdots\mathcal R_k)|
\le3^{k-2}|\mathcal R_1|\,|\mathcal R_k|.
\]
Cauchy--Schwarz and invariance of geodesic-flow measure bound its
expectation by $(5/6)3^k$. Neither commutation nor independence of the
matrices is needed. Expanding the block trace of
$((G\otimes\mathrm{Id})\mathcal R_N)^k$, its scalar coefficients are
products of nonnegative entries of $G$. Their sum is
$\operatorname{tr}(G^k)$, so
\[
\mathcal E\operatorname{tr}H_N^k
\le\frac56\,3^k\operatorname{tr}(G^k)\quad(k\ge2).
\]
The term $k=1$ is exactly $3\operatorname{tr}G$. Therefore
\[
\mathcal E\log\det(\mathrm{Id}-H_N)
\ge\frac56\log\det(\mathrm{Id}-3G)-\frac12\operatorname{tr}G.
\]
Jensen's inequality supplies a lower bound for the expected determinant
by the exponential of this expression.
The flat block determinant is $N^3$, and the normalization is exactly
\[
\det(\mathrm{Id}-H_N)=\frac{\det\mathcal L_N}{N^3}.
\]
Since $a=t/N$, Lemma~\ref{lem:discrete-jacobi} gives
$\det(\mathrm{Id}-H_N)\to D(t)/t^3$, uniformly over geodesics at this
fixed $t$: on the compact unit tangent bundle the curvature coefficients
and their first derivatives are uniformly bounded. The scalar version gives
\[
\det(\mathrm{Id}-3G)\longrightarrow\frac{\sin\theta}{\theta}.
\]
Passing to these limits proves the first assertion. Equivalently, the
uniform logarithmic bound permits passing averaged logarithms directly;
applying Jensen before the limit already suffices.
Finally, in the scalar positive logarithmic series keep only the
$k=2,3$ terms and, in each, the largest eigenvalue of $G$, which tends
to $t^2/\pi^2$. Subtracting $3\operatorname{tr}G$ and passing to the
limit yields the second assertion.
\end{proof}
For the exact evaluation, use the new lower-comparison variable
$u=\sqrt3t/\pi$. Lemma~\ref{lem:sine-minorant} gives $a_0=157/50<\pi$
and
\[
0\le S(\min(u,1-u))\le\sin(\pi u)\qquad(0\le u\le1).
\]
The two strict rational integral bounds below include the prefactor;
the rows use their polynomial representatives up to the shared endpoint.
\begin{center}
\begin{tabular}{@{}llll@{}}
\toprule
Integrand and prefactor & Interval & Degree & Lower bound\\
\midrule
$\dfrac{a_0^3}{27}u^2(u^4/2+u^6/3)S(\min(u,1-u))$
&$[0,1/2]$&15&$698/10^6$\\
same &$[1/2,1]$&15&$40554/10^6$\\
\bottomrule
\end{tabular}
\end{center}
They follow by the rational polynomial integration formula used in
Section~\ref{sec:volume}. Adding the base term
$2(a_0^2-4)/9=4883/3750$ gives a total greater than
$1343/1000>4/3$.
\begin{proof}[Proof of Proposition~\ref{prop:conditional-lower-volume}]
Write the first bound in
Lemma~\ref{lem:lower-radial} relative to $t^3\sin\theta/\theta$.
Its exponential factor is
\[
\exp\left[\frac16\left(
-\log\frac{\sin\theta}{\theta}-\frac{\theta^2}{6}\right)\right]
\ge1+\frac16\left[\frac12\left(\frac\theta\pi\right)^4
+\frac13\left(\frac\theta\pi\right)^6\right].
\]
Integrate Lemma~\ref{lem:lower-polar}, first up to any
$T<\pi/\sqrt3$ and then let $T$ increase to the endpoint. All integrands
are nonnegative, so monotone convergence suffices; no uniform endpoint
bound for a logarithm is required. The changes of variable
$\theta=\sqrt3t$, $u=\theta/\pi$ give
\[
\frac{V_M}{\pi^2}
\ge\frac29(\pi^2-4)
+\frac{\pi^3}{27}\int_0^1
u^2(u^4/2+u^6/3)\sin(\pi u)\,du.
\]
The last two rational integral bounds and their constant term make
this strictly greater than $1343/1000>4/3$, proving the assertion.
\end{proof}
Lemma~\ref{lem:one-block-alternatives} makes the $n_+=2$ case equal to
$V_M=4\pi^2/3$ under $s_0=3$, contrary to
Proposition~\ref{prop:conditional-lower-volume}. Hence only $n_+=1$
remains when exactly one block is nonzero.
\subsection{Parallel curvature and global identification}
\begin{proposition}[The remaining parallel tensors]\label{prop:parallel-models}
Up to reversing orientation, either $T=U=0$, or $U=0$ and $T$ is parallel
with eigenvalues $2,-1,-1$.
\end{proposition}
\begin{proof}
In the nonzero one-block case, the preceding exclusion gives $n_+=1$,
so Lemma~\ref{lem:one-block-alternatives} gives $\mathbf E v^2=1$.
The equality case of Proposition~\ref{prop:weyl-gap} yields
$\nabla T=0$ and $v=\sigma=1$, hence eigenvalues $2,-1,-1$.
The exchanged orientation is identical. If both blocks vanish, the
curvature-operator decomposition gives constant sectional curvature one.
\end{proof}
The model calculation is the Hopf-submersion curvature computation,
following the framework of O'Neill~\cite{ONeill1966}. We include it to
fix the metric scaling precisely.
\begin{lemma}[Curvature of the standard models]\label{lem:model-curvature}
The two tensors in Proposition~\ref{prop:parallel-models} are the curvature tensors of the unit round sphere and of twice the Fubini--Study metric induced by the Hopf quotient of the unit sphere $S^5\subset\mathbb C^3$.
\end{lemma}
\begin{proof}
The first assertion follows immediately from the curvature-operator decomposition. In the second case choose at one point the orthogonal complex structure $I$ corresponding to the eigenline of $T$ with eigenvalue $2$. On an orthonormal pair $X,Y$, the sectional curvature is
\[
K(X,Y)=\tfrac12\bigl(1+3\langle IX,Y\rangle^2\bigr).
\]
Sectional curvatures determine an algebraic curvature tensor by polarization. Only this choice at a single tangent space is needed; no global complex structure is assumed.
For comparison, let $g_H$ be the quotient metric on $S^5/S^1=\mathbb{CP}^2$ obtained by isometric horizontal lift. At $z\in S^5$, the unit vertical field is $iz$. For horizontal lifts of base vector fields, the horizontal part of the sphere connection is the lifted base connection by the metric formula, while
\[
(\nabla_XY)^{\mathrm{vert}}=-\langle iX,Y\rangle iz.
\]
This follows by differentiating $\langle Y,iz\rangle=0$. Substitution in the curvature formula yields
\[
K_{g_H}(X,Y)=1+3\langle iX,Y\rangle^2.
\]
To track the factor three, the vertical bracket is
$[X,Y]^{\mathrm{vert}}=-2\langle iX,Y\rangle iz$ and invariant horizontal lifts satisfy $\nabla_{iz}Y=iY$. The differentiated vertical term and the vertical bracket therefore contribute, respectively, one and two copies of the same squared inner product. Equivalently, the extra horizontal term in the sphere curvature expression is $3\langle iX,Y\rangle iY$, whose pairing with $X$ is $-3\langle iX,Y\rangle^2$ relative to the base expression. Scaling $g_H$ by two gives the required sectional curvatures.
The quotient metric has parallel curvature: a unitary transformation that fixes a complex line and is minus the identity on its orthogonal complement induces an isometry fixing that point with tangent differential $-\operatorname{Id}$. It acts by $-1$ on the covariant five-tensor $\nabla R$, so $\nabla R=0$. Transitivity gives this conclusion at every point. The round sphere likewise has parallel curvature.
\end{proof}
Parallel curvature turns matching tensors at one point into matching
Jacobi equations along paths. This is the continuation principle behind
curvature-based isometry results such as Ambrose's~\cite{Ambrose1956}.
We give the continuation and its inverse explicitly in the present
compact, simply connected setting.
\begin{lemma}[Global continuation of the model isometry]\label{lem:global-isometry}
The local identifications furnished by Lemma~\ref{lem:model-curvature} extend to global isometries.
\end{lemma}
\begin{proof}
Lemma~\ref{lem:volume-topology} shows that $M$ is simply connected. Both model spaces are simply connected as well. For the Hopf quotient this can be seen directly: local sections lift any loop to a path in $S^5$. Join its endpoints in the circle fibre to close the path, contract the resulting loop in $S^5$, and project the contraction. The added fibre path projects to a constant, so the original loop contracts.
Choose a tangent-space isometry matching the full curvature tensor at one point. On sufficiently small normal balls the exponential maps give an isometry: in parallel frames the Jacobi equations have identical coefficients because both curvature tensors are parallel. At a new point, the differential again matches curvature and defines the same construction on a new normal ball. Compactness supplies a common positive radius for these local constructions on the two manifolds.
An isometry germ is determined by its value and differential through exponential maps. Hence it continues uniquely along each path. Subdivide a homotopy square into small pieces lying in normal balls; uniqueness on overlaps shows that continuation is invariant under homotopy. Simple connectivity therefore gives a global local isometry. Continue the inverse germ in the same way on the model. The two compositions agree with the identity near the original points and then everywhere by continuation. They are inverse global isometries.
\end{proof}
\begin{corollary}[The oriented classification]
\label{cor:oriented-classification}
Every connected oriented smooth closed Einstein four-manifold with
strictly positive sectional curvature is, after positive scaling,
isometric to the round $S^4$ or to $\mathbb{CP}^2$ with its
Fubini--Study metric.
\end{corollary}
\begin{proof}
Strict sectional positivity gives a positive Einstein constant, so the
metric can be normalized to $\operatorname{Ric}=3g$.
Proposition~\ref{prop:parallel-models} and
Lemmas~\ref{lem:model-curvature}--\ref{lem:global-isometry} identify
that normalized metric with the unit round sphere or twice the Hopf
quotient metric. Undoing the positive normalization proves the stated
oriented conclusion.
\end{proof}
\subsection{The orientation cover and the third model}
\begin{proof}[Completion of Theorem~\ref{thm:classification}]
Strict sectional positivity makes the Ricci tensor positive definite,
so the Einstein constant is positive.
If $M$ is orientable, choose an orientation and apply
Corollary~\ref{cor:oriented-classification}.
Otherwise let $p:\widetilde M\to M$ be the orientation double cover.
It is connected, closed and oriented; for the standard construction and
connectedness see~\cite[Proposition~3.25]{Hatcher2002}.
The pulled-back metric $\widetilde g=p^*g$ has the same Einstein and
strict sectional-positivity properties. The nontrivial deck map
$\tau$ switches the two local orientations over each point, so it is
a free orientation-reversing involution, and $p\circ\tau=p$ makes it
an isometry of $\widetilde g$. After positive scaling,
Corollary~\ref{cor:oriented-classification} identifies this cover with one of its
two standard models. Conjugating $\tau$ by the model isometry preserves
its freeness and its orientation sign.
The $\mathbb{CP}^2$ case is impossible. Its integral cohomology ring is
$\mathbb Z[h]/(h^3)$ with $|h|=2$, so $h^2$ generates
$H^4(\mathbb{CP}^2;\mathbb Z)$
\cite[Theorem~3.19]{Hatcher2002}. Any self-diffeomorphism sends $h$ to
$\pm h$ and therefore fixes $h^2$. It has degree $+1$, contradicting
the required orientation reversal.
On the unit round $S^4\subset\mathbb R^5$, distance determines ambient
inner products by $\langle x,y\rangle=\cos d(x,y)$. An isometry therefore
preserves these inner products: its values on the five coordinate unit
vectors determine an $A\in O(5)$, and comparison with their images gives
$\tau(x)=Ax$ for every $x\in S^4$. Since $\tau^2=\operatorname{id}$,
we have $A^2=I$. An orthogonal involution has an orthogonal decomposition
into its $+1$ and $-1$ eigenspaces. A nonzero $+1$ eigenvector would give
a fixed point on the sphere, so freeness forces $A=-I$. The scaled base
is consequently the round quotient
$S^4/\{x\sim -x\}=\mathbb{RP}^4$.
The antipodal map has degree $(-1)^5=-1$ on $S^4$, so it reverses
orientation~\cite[Section~2.2, property~(f)]{Hatcher2002}.
Its quotient is a closed connected smooth manifold
and inherits the Einstein and positive-sectional-curvature properties
from the round sphere through the local isometry. Thus this third model
does occur, and it is the only nonorientable case.
\end{proof}
==== END Positively-curved-Einstein-four-manifolds//sections/08-classification.tex ====
==== BEGIN Positively-curved-Einstein-four-manifolds//sections/appendix-a-analytic.tex ====
\section{Analytic foundations and characteristic formulas}
\label{app:analytic}
This appendix proves the analytic inputs stated in
Section~\ref{sec:analytic}. Throughout, $M$ is the standing closed,
connected, oriented four-manifold with $\operatorname{Ric}=3g$ and
strictly positive sectional curvature. Only that normalized geometry
and the conventions of Section~\ref{sec:preliminaries} are used; none
of the volume, Hessian, coupling, or classification conclusions enters.
\subsection{Elliptic facts, density, and Poisson solvability}
\label{app:analytic-preliminaries}
We use the following basic analytic facts on a closed smooth
four-manifold. The inclusion $H^1\subset L^4$ is continuous, and
$H^1\subset L^s$ is compact for $2\le s<4$. Smooth scalar and bundle
operators with Laplacian leading symbol satisfy the local
$W^{2,p}$ estimates for $1<p<\infty$, together with their iterated
Sobolev consequences. All operators below have these hypotheses,
and all nonlinear terms are put on the right side before applying
the estimates. For scalar operators we use the strong interior estimates,
Dirichlet solvability, and improvement of strong-solution integrability
in Gilbarg--Trudinger~\cite[Theorem~9.11, pp.~235--236;
Theorem~9.15 and Lemma~9.16, pp.~241--242]{GilbargTrudinger}.
The initial weak-to-strong passage for the nonlinear minimizer is
justified in the next subsection, rather than inferred from an estimate
for strong solutions.
For a weak bundle equation with $L^2$ data, write the operator in a smooth
local frame: its principal part is scalar, while its first-order coupling
applied to an $H^1$ section belongs to $L^2$. A cutoff reduces each
component to a scalar equation for $\Delta_g-C$ on a smooth ball,
with $C>0$. Strong zero-boundary solvability and coercive energy
uniqueness identify that component with its $W^{2,2}$ solution.
Once this initial regularity is established, componentwise interior
estimates and interpolation absorb the smooth first-order couplings;
differentiation gives the higher regularity used for Poisson solutions
and eigenfunctions.
Smooth functions are dense in $H^1(M)$: use local mollification in a
finite coordinate cover and a smooth partition of unity. The same
construction gives uniform and $H^1$ approximation for a Lipschitz
function, as used in the weak norm testing rule.
For completeness, every smooth mean-zero function $h$ has a unique smooth
mean-zero solution of $\Delta\phi=h$. On the mean-zero subspace of $H^1$,
the Dirichlet energy is coercive. Indeed, otherwise a sequence with
unit $L^2$ norm and gradients tending to zero has, by compactness in
$L^2$, a mean-zero limit with unit norm and zero weak gradient. On the
connected manifold that limit is constant, a contradiction. The
variational equation
\[
\mathbf E\langle d\phi,d\psi\rangle=-\mathbf E(h\psi)
\]
therefore has a unique mean-zero weak solution by the Hilbert-space
representation theorem. The local regularity above makes it smooth.
The same energy identity proves uniqueness. This is the Poisson fact
used in Proposition~\ref{prop:weighted-variance}.
\subsection{The sharp Sobolev inequality}
\label{app:sharp-sobolev}
We give the normalized four-dimensional Einstein case of Ilias's
inequality~\cite[Theorem~3]{Ilias}. Its nonlinear Hessian proof belongs
to the Bochner approach to elliptic rigidity; compare
Bidaut-V\'eron--V\'eron~\cite{BidautVeronVeron}.
\begin{proof}[Proof of Proposition~\ref{prop:sharp-sobolev}]
Fix $2<s<4$ and minimize
\[
Q_s(w)=
\frac{\mathbf E(w^2+\tfrac12|\nabla w|^2)}
{(\mathbf E|w|^s)^{2/s}}
\]
under $\mathbf E|w|^s=1$. A minimizing sequence is bounded in
$H^1$; weak compactness, compact inclusion into $L^s$, and lower
semicontinuity give a minimizer. Taking its absolute value does
not increase the energy, so choose $w\ge0$.
The ordinary Sobolev embedding makes the minimum $m_s$ positive,
and the constant test function gives $m_s\le1$.
The Euler equation is
\[
-\Delta w+2w=2m_s w^{s-1}.
\]
After setting $u=m_s^{1/(s-2)}w$, it becomes
\[
-\Delta u+2u=2u^{s-1}.
\]
We record the regularity needed to use a negative power of $u$.
We first pass from the weak minimizer to a strong solution. Choose a
smooth coordinate ball $B$ and a cutoff $\eta\in C_c^\infty(B)$.
For $z=\eta u$ the weak equation gives
\[
(\Delta_g-2)z
=-2\eta u^{s-1}+2\langle d\eta,du\rangle+(\Delta_g\eta)u
=:F\in L^{r_0}(B),
\qquad r_0=\min\{2,4/(s-1)\}>4/3.
\]
The strong Dirichlet existence theorem
\cite[Theorem~9.15]{GilbargTrudinger} gives a solution
$w_D\in W^{2,r_0}(B)\cap W^{1,r_0}_0(B)$ with this right side.
The domain and coefficients are smooth, the operator is uniformly
elliptic, and its zeroth-order coefficient is $-2$.
In dimension four, $r_0>4/3$ implies
$W^{2,r_0}\subset W^{1,q}$ for some $q>2$; compatibility of Sobolev
traces therefore gives $w_D\in H^1_0(B)$.
The difference $z-w_D\in H^1_0(B)$ solves
$(\Delta_g-2)(z-w_D)=0$ weakly. Testing by $z-w_D$ shows that it
vanishes. Thus $u$ is locally $W^{2,r_0}$, and the strong interior
estimates and integrability improvement now apply
\cite[Theorem~9.11 and Lemma~9.16]{GilbargTrudinger}.
Initially $u\in L^4$, while $u^{s-2}\in L^r$ with
$r=4/(s-2)>2$. If $u\in L^q$, the equation has right side in
$L^p$ with $1/p=1/q+1/r$. The $W^{2,p}$ estimate and Sobolev
embedding improve $1/q$ by a fixed positive amount smaller than
$1/2-1/r$. Initially $p=4/(s-1)>1$. Iteration reaches $p>2$,
giving boundedness; at a borderline exponent $p=2$, use any
finite larger $q$ before the next iteration.
After boundedness, the equation gives $u\in W^{2,p}$ for every finite
$p$. Since $s-1>1$, the derivative of $t^{s-1}$ is bounded on the
bounded nonnegative range of $u$. Differentiating the equation once,
and applying the same local regularity argument to each first
coordinate derivative, gives $u\in W^{3,p}$ for every finite $p$.
In particular, $u\in C^{2,\alpha}$ for some $\alpha>0$.
For $C\ge2$ we have
\[
(-\Delta_g+C)u=(C-2)u+2u^{s-1}\ge0.
\]
The classical strong maximum principle gives $u>0$, since
$u\ge0$ and $u\not\equiv0$. A direct barrier proof uses a positive
exponential radial function on an annulus tangent to the zero set:
the weak maximum principle forces a nonzero inward derivative at the
tangency point, contradicting nonnegativity and differentiability.
Compactness now gives a positive minimum. The nonlinearity is smooth
on this positive range, so further differentiation and elliptic
regularity give smoothness of $u$.
Put $F=u^{-1}$. Its equation is
\[
F\Delta F-2|\nabla F|^2+2F^2=2F^{4-s}.
\]
Since
$\operatorname{div}(\operatorname{Hess}_0F)
=\frac34\,d(\Delta F+4F)$, differentiation gives
\[
\begin{aligned}
\operatorname{div}(F^{-3}\operatorname{Hess}_0F)
&=\frac34F^{-4}
\bigl[F\,d\Delta F+(\Delta F)dF
-4\operatorname{Hess}F(\nabla F,\,\cdot\,)+4F\,dF\bigr]\\
&=\frac32(4-s)F^{-1-s}\,dF .
\end{aligned}
\]
Pairing with $dF$ and integrating yields
\[
-\mathbf E F^{-3}|\operatorname{Hess}_0F|^2
=\frac32(4-s)\mathbf E F^{-1-s}|dF|^2.
\]
Both sides have opposite signs, so $dF=0$. The positive constant
solution of the equation for $u$ is $u=1$. The normalization of
$w$ then gives $w=1$ and $m_s=1$.
Thus the claimed inequality holds with $4$ replaced by every
$s\in(2,4)$ and outer exponent $2/s$. For fixed $f\in H^1$,
the ordinary inclusion $H^1\subset L^4$ and dominated convergence
give $\|f\|_{L^s(d\nu)}\to\|f\|_{L^4(d\nu)}$ as $s\uparrow4$.
This proves the stated critical inequality.
\end{proof}
\subsection{Harmonic forms and the characteristic identities}
\label{app:characteristic-identities}
We derive the Euler and signature formulas using the complex of forms and
an explicit signed heat computation. This is the heat-supertrace method
of McKean--Singer~\cite{McKeanSinger1967}; general formulations and local
coefficient calculations appear in Gilkey~\cite[Theorems~1.7.6 and
4.8.16]{Gilkey1984}.
We first record the spectral heat construction for a smooth symmetric
Laplace-type operator $L=-\Delta+E$ on a Euclidean bundle over $M$. Adding a sufficiently
large constant gives a coercive quadratic form on $H^1$.
Its weak inverse is compact and self-adjoint on $L^2$ by
compact embedding, and elliptic regularity makes its
eigenfunctions smooth. The spectrum is discrete with finite
multiplicities and tends to infinity. Spectral heat evolution
is smoothing for every positive time: polynomial powers of
the shifted eigenvalue times $e^{-t\lambda}$ are bounded,
and elliptic estimates for powers of $L$ identify the
corresponding Sobolev bounds.
For completeness, these estimates also justify smooth kernel
and trace expansions. For a sufficiently large integer $k$,
evaluation of any prescribed finite derivative after
$(L+C)^{-k}$ is a bounded functional on $L^2$, uniformly in
its base point. Applied to an orthonormal eigenbasis, the
Riesz representation theorem gives a uniform bound for the
sum of the squared derivative values divided by
$(\lambda+C)^{2k}$. Cauchy--Schwarz and the exponential factor
then give uniform convergence of the spectral kernel series
and all its derivatives for $t$ in compact subintervals of
$(0,\infty)$. Integrating the diagonal gives the usual heat
trace. Uniqueness with prescribed initial point mass follows
by testing the kernel against the eigenfunctions.
\begin{lemma}[The form-complex Laplacians]
\label{lem:form-complex-laplacians}
Consider the elliptic complex
\[
\Lambda^0\xrightarrow{\ d\ }\Lambda^1
\xrightarrow{\ \sqrt2\,d^+\ }\Lambda^+,
\qquad d^+=P_+d.
\]
Its three Laplacians are the Hodge Laplacians on the indicated
bundles. Written as $-\Delta+E$, their potentials are
\[
E_0=0,\qquad E_1=3\operatorname{Id},\qquad
E_+=4\operatorname{Id}-2T.
\]
Its alternating heat trace is $1+n_+$.
\end{lemma}
\begin{proof}
The middle complex Laplacian is
\[
dd^*+2(d^+)^*d^+
=dd^*+d^*d+d^*\!\star d.
\]
The last term is zero by $d^2=0$ and the formula for the
codifferential. On self-dual two-forms, Hodge star exchanges
$dd^*$ and $d^*d$, so $2P_+dd^*$ is their sum. This identifies
the last Laplacian; the scalar one is immediate. Their elliptic
principal symbols are consequently those of the metric Laplacian.
Commuting covariant derivatives in the one-form Hodge Laplacian
gives the Ricci endomorphism, equal to $3\operatorname{Id}$.
For the plus potential, write a self-dual form as
$\eta=\sum_s\eta_s I_s/\sqrt2$. Its Hodge energy is
$2\mathbf E|d^*\eta|^2$, which equals
\[
\mathbf E\left|\sum_s I_s\nabla\eta_s\right|^2
=\mathbf E\left(
|\nabla\eta|^2+\frac12\langle C\nabla\eta,\nabla\eta\rangle
\right),
\]
where $C=\sum I_s\otimes\Sigma_s$ now uses the spin-one
generators on the plus coefficients. The equality follows by
expanding the quaternion products. Integrating the $C$ term
and taking its skew derivative commutator gives the potential
\[
-\frac12\sum_s(1+t_s)\Sigma_s^2.
\]
In the spin-one representation,
$\Sigma_s^2=4(e_se_s^{\mathsf T}-\operatorname{Id})$.
The potential is therefore $4\operatorname{Id}-2T$,
as asserted.
The scalar kernel consists of the constants. A harmonic one-form
satisfies
\[
0=\mathbf E\bigl(|\nabla\eta|^2+3|\eta|^2\bigr),
\]
so the middle kernel vanishes. The last kernel has dimension
$n_+$. On each positive eigenspace of the complex Laplacian,
the complex differential commutes with the Laplacian, and its
adjoint divided by the eigenvalue is a contracting homotopy.
That finite-dimensional complex has zero alternating dimension.
Summing the heat factors over the positive spectrum leaves only
the kernel contribution $1+n_+$.
\end{proof}
\begin{lemma}[Local signed heat coefficient]
\label{lem:signed-heat-coefficient}
For the three operators in
Lemma~\ref{lem:form-complex-laplacians}, let $\operatorname{str}$
denote the trace on $\Lambda^0$ minus that on $\Lambda^1$ plus
that on $\Lambda^+$. Write $\Omega$ for their connection
curvatures. The constant local coefficient of their alternating
heat trace, including the factor $(4\pi)^{-2}$, is
\[
\frac1{16\pi^2}
\left(\frac12\operatorname{str}E^2
+\frac1{12}\operatorname{str}\sum_{k,i}\Omega_{ki}^2\right).
\]
\end{lemma}
\begin{proof}
Fix the source point of the kernel. In normal coordinates
$z$ around it, use radially parallel orthonormal frames on
each of the three bundles and set
$\mu=\sqrt{\det g}$, $h=\mu^{-1/2}$. The ansatz
\[
(4\pi t)^{-2}e^{-|z|^2/(4t)}
\sum_{j\ge0}t^jU_j(z)
\]
gives $U_0=h\operatorname{Id}$ and the transport equations
\[
\left(z\cdot\partial+j+\frac12z\cdot\partial\log\mu\right)U_j
=-LU_{j-1}.
\]
Here the radial connection coefficient is zero.
The scalar part follows also from
$\Delta|z|^2=8+2z\cdot\partial\log\mu$.
Solving the radial equation gives
\[
U_j(z)=-h(z)\int_0^1 s^{j-1}
\bigl(h^{-1}LU_{j-1}\bigr)(sz)\,ds.
\]
These transport solutions compute the actual local asymptotics.
Indeed, truncate at any large $N$ and apply a cutoff inside a
normal ball. The resulting kernel has initial delta mass,
and the transport equations leave a residual uniformly
$O(t^{N-2})$; derivatives of the cutoff give exponentially
small terms. The heat evolution of a metric connection
with bounded potential has a bounded sup norm on bounded
time intervals, by the maximum principle for the squared
norm. Duhamel integration therefore makes the difference
from the true kernel $O(t^{N-1})$. Compactness permits a
uniform normal-ball radius and uniform constants in the
source point. Taking $N$ large justifies the diagonal
coefficient calculation below.
In these frames write the connection as $\partial_i+A_i$.
Every $A_i$ is skew-symmetric and radial gauge gives
\[
A_i(z)=\frac12z^k\Omega_{ki}(0)+O(|z|^2).
\]
For example, the exact radial-gauge identity
$A_i(z)=\int_0^1s z^k\Omega_{ki}(sz)\,ds$ follows by contracting
the curvature equation with $z^k$ and integrating along
the radius.
The signed ranks and potentials satisfy, at every point,
\[
\operatorname{str}\operatorname{Id}=1-4+3=0,
\qquad
\operatorname{str}E=-12+\operatorname{tr}(4\operatorname{Id}-2T)=0.
\]
In coordinate form,
\[
L=-g^{ij}\bigl[
(\partial_i+A_i)(\partial_j+A_j)
-\Gamma^k_{ij}(\partial_k+A_k)\bigr]+E.
\]
Apply this to $U_0=h\operatorname{Id}$. All purely scalar terms
have zero supertrace. Every term linear in $A_i$ or a
coordinate derivative of $A_i$ has zero ordinary trace in
each bundle, by skewness. The potential has zero supertrace
pointwise. Thus the exact surviving expression is
\[
\operatorname{str}(LU_0)(z)
=-h(z)g^{ij}(z)\operatorname{str}(A_i(z)A_j(z)).
\]
In particular, derivatives of the varying potential have
already disappeared before the radial integral is taken.
The transport formula now gives
\[
\begin{aligned}
\operatorname{str}U_1(z)
&=h(z)\int_0^1 g^{ij}(sz)
\operatorname{str}(A_i(sz)A_j(sz))\,ds\\
&=\frac1{12}z^kz^\ell
\operatorname{str}\sum_i\Omega_{ki}(0)\Omega_{\ell i}(0)
+O(|z|^3).
\end{aligned}
\]
The coefficient is
$\frac14\int_0^1s^2\,ds=\frac1{12}$.
Scalar and metric corrections do not change this quadratic term.
At the center, $A_i(0)=0$, $\Gamma(0)=0$, and
$\sum_i\partial_iA_i(0)=\frac12\sum_i\Omega_{ii}(0)=0$
as an endomorphism. Consequently
\[
U_1(0)=c\operatorname{Id}-E(0),\qquad
(LU_1)(0)=-\sum_i\partial_i^2U_1(0)+E(0)U_1(0),
\]
where $c$ is the same scalar for all three bundles.
Thus there are no additional connection terms at the center.
Taking the Laplacian of the signed quadratic expression gives
\[
\operatorname{str}\sum_i\partial_i^2U_1(0)
=\frac16\operatorname{str}\sum_{ki}\Omega_{ki}^2,
\qquad
\operatorname{str}(EU_1(0))=-\operatorname{str}E^2.
\]
Since the transport equation at the center gives
$U_2(0)=-\frac12(LU_1)(0)$, we obtain
\[
\operatorname{str}U_2(0)
=\frac12\operatorname{str}E^2
+\frac1{12}\operatorname{str}\sum_{ki}\Omega_{ki}^2.
\]
The lower coefficients have zero signed diagonal trace.
Multiplication by the heat prefactor $(4\pi t)^{-2}$
now proves the assertion.
\end{proof}
\begin{proof}[Proof of Proposition~\ref{prop:characteristic-identities}]
By Hodge theory, harmonic forms represent de Rham cohomology. The
scalar kernel is one-dimensional, and the one-form kernel vanishes by
Lemma~\ref{lem:form-complex-laplacians}. Hodge star gives the corresponding
dimensions in degrees four and three. On harmonic two-forms, the
intersection form is positive definite on the self-dual part and negative
definite on the anti-self-dual part. Thus its signature is $n_+-n_-$,
and the alternating sum of the Betti numbers is $2+n_++n_-$. This
identifies the integers in Proposition~\ref{prop:characteristic-identities}
with the topological signature and Euler characteristic without using
simple connectivity.
The potential-square supertrace is
\[
\operatorname{str}E^2
=-\operatorname{tr}(9\operatorname{Id}_{\Lambda^1})
+\operatorname{tr}\bigl((4\operatorname{Id}_{\Lambda^+}-2T)^2\bigr)
=12+24v^2.
\]
On one-forms, the connection curvature has plus and minus
generators $I_s,J_s$. Orthogonality of the two-form bases,
together with $I_s^2=J_s^2=-\operatorname{Id}$, gives
\[
\operatorname{tr}_{\Lambda^1}\sum_{ki}\Omega_{ki}^2
=-4\sum_s\bigl((1+t_s)^2+(1+u_s)^2\bigr)
=-24(1+v^2+b^2).
\]
On plus two-forms only the plus generators act, and each
speed-two spin-one generator has square trace $-8$.
Therefore
\[
\operatorname{tr}_{\Lambda^+}\sum_{ki}\Omega_{ki}^2
=-8\sum_s(1+t_s)^2=-24(1+2v^2).
\]
The scalar connection is flat. Hence
$\operatorname{str}\sum\Omega_{ki}^2=24(b^2-v^2)$, and
Lemma~\ref{lem:signed-heat-coefficient} gives the signed
constant density
\[
\frac1{16\pi^2}(6+10v^2+2b^2).
\]
Integrate this against $d\operatorname{vol}_g$ and compare
with the constant alternating heat trace from
Lemma~\ref{lem:form-complex-laplacians}:
\[
16\pi^2(1+n_+)
=\operatorname{Vol}(M)\,
\mathbf E(6+10v^2+2b^2).
\]
Reverse orientation to get the analogous formula with
$n_-,b,v$. Adding and subtracting the two formulas gives
exactly \eqref{eq:6}.
On anti-self-dual forms the potential is
$4\operatorname{Id}-2U$. If $U=0$, the energy of a harmonic
anti-self-dual form is the integral of
$|\nabla\eta|^2+4|\eta|^2$, so its kernel is zero.
The assertion for $T=0$ follows in the same way.
\end{proof}
==== END Positively-curved-Einstein-four-manifolds//sections/appendix-a-analytic.tex ====
==== BEGIN Positively-curved-Einstein-four-manifolds//sections/appendix-b-topology.tex ====
\section{Minimizing geodesics and topology}
\label{app:topology}
We prove Lemma~\ref{lem:volume-topology} in the standing setting:
$M$ is a closed, connected, oriented smooth four-manifold with
$\operatorname{Ric}=3g$ and strictly positive sectional curvature.
Along a unit-speed geodesic, let $\mathcal R$ be the normal Jacobi
curvature matrix in a parallel orthonormal frame. The hypotheses give
\[
0<\mathcal R<3\operatorname{Id},\qquad
\operatorname{tr}\mathcal R=3.
\]
The trace supplies the diameter estimate, the upper bound supplies the
conjugate-radius estimate, and strict positivity supplies the shortening
variations. Orientation is used in the normal-holonomy argument. These
are the classical methods of Myers, Synge, and Klingenberg
\cite{Myers1941,Synge1936,Klingenberg1959}; the proof below retains the
endpoint variations and proves simple connectivity independently of
all volume and classification conclusions.
\begin{proof}[Proof of Lemma~\ref{lem:volume-topology}]
For a unit-speed geodesic of length $L$, the index form on normal fields
with zero endpoint values is
\[
I(Y,Y)=\int_0^L\bigl(|Y'|^2-\langle\mathcal RY,Y\rangle\bigr)\,dt.
\]
It is nonnegative on a minimizing segment, by the second variation of
energy. Summing it on three parallel normal unit fields multiplied by
$\sin(\pi t/L)$ gives
\[
\frac{3L}{2}\left(\frac{\pi^2}{L^2}-1\right),
\]
so a minimizing segment has length at most $\pi$.
It cannot contain an interior conjugate point: a nonzero Jacobi field
vanishing at the initial point and that interior point, extended by
zero, has index zero but is not in the nullspace of the index form.
The latter assertion follows by integration by parts against a field
whose value pairs nontrivially with the jump in its derivative.
A nonnegative quadratic form, however, pairs a zero-energy vector to
zero against every vector, a contradiction.
The ordinary Dirichlet Poincar\'e inequality and $\mathcal R<3\mathrm{Id}$
show that a nonzero field with zero endpoints has positive index when
$L\le\pi/\sqrt3$. Thus no conjugate point occurs at such a distance.
We show that no cut point occurs at a smaller distance either.
Let $c$ be the infimum of all cut radii; compactness and normal balls give
$c>0$. Suppose $c<\pi/\sqrt3$ and choose cut segments with lengths tending
to $c$, all below the conjugate-radius bound. Each endpoint has two
distinct minimizing segments from its starting point. To see this,
extend a cut segment a little and take limits of minimizers to the
extended endpoints. If the limiting minimizer were the original
segment alone, local invertibility of the exponential map would force
the extended segment to remain minimizing, contrary to its cut time.
After passing to a subsequence, compactness gives two length-$c$
minimizing segments with common endpoints $p,q$. Their initial vectors
cannot coalesce: the endpoint map
\[
(p,w)\longmapsto(p,\exp_p w)
\]
is a local diffeomorphism at every vector in question, so two converging
branches in the same inverse-function neighborhood would coincide.
This map also supplies two distinct smooth local geodesic branches for
nearby pairs of endpoints.
Their terminal unit velocities must be opposite. If these velocities
are $u_1,u_2$ with $u_2\ne-u_1$, move $q$ initially in direction
$-(u_1+u_2)$. The first length variation of both branches is
$-(1+\langle u_1,u_2\rangle)<0$. Both branches would then have length
less than $c$. Every geodesic of length less than the global cut infimum
is minimizing and lies in the injectivity domain, so two distinct such
branches are impossible. Moving $p$ instead proves that their initial
velocities are opposite as well. The first segment followed by the
reverse of the second is therefore a smooth closed geodesic.
Parallel transport around this closed geodesic fixes the tangent and
preserves orientation. Its action on the three-dimensional normal
space belongs to $\mathrm{SO}(3)$, and hence fixes a nonzero vector.
Let $V$ be the resulting periodic parallel normal field. Move both
endpoints along geodesics with initial velocities $V(p),V(q)$, so their
accelerations at the variation origin vanish. The local geodesic
branches vary smoothly. Their variation fields $J$ are normal: their
tangential components are affine and have zero boundary values.
For either branch, $W=V-J$ has zero boundary values and
$I(J,W)=0$, while $I(W,W)\ge0$ because the unvaried segment is minimizing.
Thus
\[
I(J,J)\le I(V,V)=-\int\langle\mathcal RV,V\rangle<0.
\]
The first length variations vanish, and these negative index values
are the second length variations, since there are no endpoint
acceleration or tangential terms. Both branch lengths therefore become
less than $c$. They remain distinct in their disjoint inverse-function
neighborhoods, and lengths below $c$ force them to be minimizing. This
is the same contradiction as above, proving the injectivity bound.
Finally, a nontrivial free homotopy class would have a shortest
representative. Indeed, constant-speed minimizing sequences have a
uniform Lipschitz bound; compactness gives a uniformly convergent
subsequence, and sufficiently close loops are freely homotopic by the
short normal geodesics between corresponding points. A uniform normal
ball radius also gives a positive lower bound on lengths of nontrivial
loops. Local geodesic replacement shows that the minimizing loop is
a nonconstant smooth closed geodesic. Its normal holonomy again fixes
a nonzero parallel field, whose periodic variation has strictly
negative second length variation. The variation preserves its free
homotopy class, a contradiction. Hence the fundamental group is
trivial.
\end{proof}
==== END Positively-curved-Einstein-four-manifolds//sections/appendix-b-topology.tex ====