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\title[Deforming hypersymplectic four-manifolds]{Deforming hypersymplectic four-manifolds to hyperk\"ahler triples}
\author{OpenAI}
\date{September 23, 2026}

\begin{document}
\begin{abstract}
Every smooth normalized positive triple of closed two-forms on a closed
connected oriented four-manifold admits a smooth deformation, with each
cohomology class fixed, to a hyperk\"ahler triple. This resolves
Donaldson's hypersymplectic deformation conjecture.
\end{abstract}
\maketitle
\tableofcontents

\section{Introduction}\label{sec:introduction}

A \emph{hypersymplectic structure} on an oriented four-manifold $X$ is a
triple of closed real two-forms whose pointwise span is three-dimensional
and positive definite for the wedge-product quadratic form. Such a triple
determines a conformal structure, but its forms need not be parallel:
the induced metric need not be hyperk\"ahler. After a constant
normalization of the periods, the natural question is whether one can
deform the forms to a hyperk\"ahler triple while retaining their three
cohomology classes. This paper answers yes.

\begin{theorem}[Main]\label{thm:main}
Let $X$ be a closed connected oriented smooth four-manifold, and let
$\omega=(\omega_1,\omega_2,\omega_3)$ be a smooth hypersymplectic
structure normalized by
\begin{equation}\label{eq:intro-normalization}
 \int_X\omega_i\wedge\omega_j=\delta_{ij}.
\end{equation}
There is a smooth family of hypersymplectic structures $\omega(t)$,
$0\le t\le1$, such that
\[
 \omega(0)=\omega,\qquad [\omega_i(t)]=[\omega_i]\quad(i=1,2,3),
\]
and a positive volume form $\mu_1$ with
\begin{equation}\label{eq:intro-endpoint}
 \omega_i(1)\wedge\omega_j(1)=2\delta_{ij}\mu_1.
\end{equation}
The endpoint forms are parallel and self-dual for a hyperk\"ahler metric.
\end{theorem}

The normalization \eqref{eq:intro-normalization} is a constant linear
change of basis and positive scaling; it makes the endpoint identity
cleaner but does not restrict the geometry. If the initial triple has
an arbitrary positive Gram matrix $G_{ij}=\int_X\omega_i\wedge\omega_j$,
one applies the theorem to $L\omega$ for a constant invertible real
$L$ with $LGL^{\mathsf T}=I_3$ and applies $L^{-1}$ to the resulting
path. The endpoint then satisfies
$\omega_i(1)\wedge\omega_j(1)=2G_{ij}\mu_1$: the three classes are
parallel self-dual forms for a hyperk\"ahler metric, with their original
Gram matrix $G$.

The theorem has two immediate consequences. First, every single positive
combination of the deformed triple is a K\"ahler form on a compatible
integrable complex structure, and it is a K\"ahler form for the same
manifold; on the closed connected $X$ the symplectic form determines
the K\"ahler class in advance. Second, the topology of the initial
manifold is forced: the existence of any such closed positive triple
already implies $X$ is diffeomorphic to a standard K3 surface or to
$T^4$. Both are corollaries of the fact that the endpoint is a global
hyperk\"ahler metric.

\begin{corollary}[Individual K\"ahler realizations]\label{cor:individual-kahler}
Let $X$ be a closed connected oriented four-manifold with a smooth
positive triple $\omega$ of closed real two-forms. For every fixed
$c\in\R^3\setminus\{0\}$, the form $\alpha_c=\sum_i c_i\omega_i$
is a K\"ahler form for a parallel integrable complex structure $J_c$ of
a hyperk\"ahler metric $g_c$ on $X$:
\[
 \alpha_c(v,w)=g_c(J_cv,w).
\]
The compatible structure is chosen separately for each $c$; no
simultaneous transport of the triple is asserted.
\end{corollary}
\begin{proof}
Use the fixed-class path for the original basis described above and put
$\alpha_t=\sum_i c_i\omega_i(t)$. Positivity of the pointwise
wedge-product matrix gives $\alpha_t\wedge\alpha_t>0$, so $\alpha_t$
is symplectic; its class is fixed because each $[\omega_i(t)]$ is fixed.

For the endpoint metric $g$, set $r_c=(c^{\mathsf T}Gc)^{1/2}>0$. The
endpoint identities show that $\alpha_1/r_c$ is parallel and self-dual
with $g$-norm $\sqrt2$. It therefore defines a parallel integrable
complex structure $J^{\mathrm{end}}_c$ with
$\alpha_1/r_c=g(J^{\mathrm{end}}_c\,\cdot,\cdot)$,
so that $\alpha_1$ is its K\"ahler form for the hyperk\"ahler metric
$r_cg$.

Since $X$ is closed, Moser's theorem gives an isotopy $\phi_{c,t}$ with
$\phi_{c,0}=\mathrm{id}$ and $\phi_{c,t}^*\alpha_t=\alpha_c$.
Pulling back $r_cg$ and $J^{\mathrm{end}}_c$ by $\phi_{c,1}$ gives the
claimed hyperk\"ahler metric and compatible integrable structure on $X$.
Fine and Yao note this Moser implication for a constituent in
\cite[Section~1]{FineYao2017}.
\end{proof}

\begin{corollary}[Smooth type]\label{cor:smooth-type}
Let $X$ be a closed connected smooth four-manifold admitting a smooth
positive triple of closed real two-forms. Then $X$ is diffeomorphic to
the standard K3 manifold or to $T^4$.
\end{corollary}
\begin{proof}
Give $X$ the orientation induced by the triple. The normalization above
and Theorem~\ref{thm:main} give a global hyperk\"ahler triple on $X$.
Choosing one of its complex structures makes $X$ a compact K\"ahler
surface with a nowhere-vanishing holomorphic two-form. The compact
hyperk\"ahler surface classification \cite[Section~2, p.~162]{Boyer1988}
therefore identifies $X$ itself as a complex two-torus or a K3 surface.
A complex two-torus is smoothly $T^4$, while every complex K3 surface is
diffeomorphic to a nonsingular quartic surface in $\mathbf P^3$, the
standard K3 manifold \cite[p.~58]{Kodaira1970}.
\end{proof}

\subsection{History and significance}

Donaldson's study of closed two-forms on four-manifolds links their
pointwise wedge-product geometry to nonlinear elliptic equations
\cite{Donaldson2006}. His Question~3 in Section~5.3 asks whether a
compact oriented four-manifold carrying a positive closed triple admits
a hyperk\"ahler structure. In the same section he proposes a
prescribed-volume continuity argument in the simply connected case,
conditional on a priori estimates and an auxiliary involution. The
involution keeps the evolving form in a fixed cohomology class
\cite[Section~5.3]{Donaldson2006}. Fine and Yao formulate the normalized
deformation problem, retaining all three cohomology classes, as
Conjecture~1.1 of \cite{FineYao2017}. Theorem~\ref{thm:main} resolves
this deformation conjecture positively. Keeping the classes connects the
given triple to a metric with its prescribed periods, beyond mere
existence of some hyperk\"ahler metric on the same manifold.

The main analytic approach until now has been the hypersymplectic flow.
Fine and Yao obtained this flow by reducing the $G_2$-Laplacian flow
on the product with a three-torus. They proved that a finite-time
solution extends while the associated seven-dimensional scalar
curvature, equivalently the torsion, remains bounded
\cite[Theorem~1.3]{FineYao2017}. Their later work gives an integral
torsion extension criterion, global existence near pointwise
orthogonality, and subsequential convergence of global solutions under
additional geometric bounds
\cite[Theorems~4.7, 4.10, 4.15 and~4.16]{FineYao2020}.

Several restricted classes admit complete convergence results.
Huang, Wang and Yao proved convergence, after pullback by
diffeomorphisms, for simple-type hypersymplectic structures on the
standard four-torus, a diagonal one-variable class
\cite[Theorems~3.5 and~3.6]{HuangWangYao2018}. Picard and Suan treated
$G_2$ flows associated with K\"ahler Calabi--Yau data; their
complex-dimension-two case supplies a hypersymplectic convergence
theorem \cite[Theorem~1.3]{PicardSuan2024}. Fine, He and Yao extended
the simple-type convergence result, modulo diffeomorphisms, to
$T^3$-invariant triples on $T^4$ in symmetric normal form
\cite[Theorem~1.5]{FineHeYao2024}. They subsequently proved a linear
cohomologous isotopy to a hyperk\"ahler triple when an effective smooth
circle action preserves the initial triple \cite[Theorem~1.2]{FineHeYao2025}.
Related reductions and coflow constructions appear in work of Petcu,
Yao and Zhou, and Karigiannis, Picard and Suan
\cite{Petcu2026,YaoZhou2026,KarigiannisPicardSuan2026}. All of these
results retain their stated symmetry or integrability hypotheses. The
path constructed here starts from an arbitrary positive closed triple;
its existence is a separate question from convergence or uniqueness for
the hypersymplectic flow.

The topology already places strong restrictions on such a triple. Any
two of its forms determine an almost-complex structure whose canonical
line is trivial, and the third form tames that structure. Here a real
two-form $\beta$ \emph{tames} $J$ if $\beta(v,Jv)>0$ for every nonzero
vector $v$; it is \emph{compatible} if it is also $J$-invariant. A
closed tamer is symplectic. Bauer's bound for symplectic four-manifolds
with torsion first Chern class \cite[Corollary~1.2]{Bauer2008},
together with the three positive period classes, gives $b^+=3$, where
$b^+$ is the positive index of the intersection form. The resulting
free intersection lattice is the K3 lattice or three hyperbolic planes.
These facts supply the lattice used below; they do not presuppose a
smooth identification of the initial manifold with a K3 surface or torus.

For controlling closed positive forms, Sullivan's structure currents
provide the cone-duality framework \cite{Sullivan1976}. Harvey and
Lawson developed the positive-current characterization of K\"ahler
geometry \cite{HarveyLawson1983}, and Li and Zhang compared the taming
and compatible cones in almost-complex geometry assuming the compatible
cone is nonempty \cite{LiZhang2009}. A separate article proves that
every tamed smooth almost-complex structure on a closed connected real
four-manifold admits a compatible symplectic form
\cite[Theorem~1.1]{OpenAITaming2026}. Its current estimates and
density-splitting argument are also used here, with the needed proofs
included in \S\ref{sec:current-energy}--\S\ref{sec:density-splitting}.
In particular, smooth density-weight closedness already appears in the
proof in Section~6 of that article. Its compatible class is unrestricted.
The additional class criterion below converts closed normalized density
thresholds into curves, using Preiss's rectifiability theorem and
Rivi\`ere--Tian's regularity theorem for integral cycles
\cite{Preiss1987,RiviereTian2009}. This supplies the prescribed taming
class needed along the deformation.

\subsection{How the proof works: a plain-language sketch}

We want a path of positive triples with fixed classes ending at a
hyperk\"ahler triple. Every positive triple factors, pointwise, as a
definite pair $(A,C)$ that spans the anti-invariant forms of an
almost-complex structure $J$, plus a third form $B$ that tames $J$. A
deformation of the pair alone, moving $(A,C)$ while $B$ is held fixed,
keeps the triple positive if and only if the fixed $B$ continues to tame
the moving $J$. Thus the entire question reduces to:

\begin{quote}
Along a deformation of the first two forms, keep the original third
class $[B]$ available as a taming form.
\end{quote}

Whether a real class $H$ contains a smooth tamer for a tamed
almost-complex structure $J$ on a closed four-manifold is decided by two
ingredients. First, a tamer exists at all iff $H$ has positive
self-intersection and every irreducible closed $J$-holomorphic curve
pairs with $H$ positively. Second, if one taming form is already known
for $J$, then any $H$ in the same component of the positive
time-like cone whose curve pairings are all positive is also a taming
class. The problem now becomes: describe the curves in the moving
almost-complex structures, so that $[B]$ is guaranteed to pair
positively with all of them.

The curves are provided by Seiberg--Witten theory. The starting data
gives a natural family of almost-complex structures parametrized by
$S^2$, one for each coefficient direction of the initial triple.
Taubes's correspondence turns solutions of the large-parameter twisted
Seiberg--Witten equations for the coefficient direction $v$ into
$J_v$-holomorphic curves. Wall-crossing for this family then forces, in
any prescribed class of the coefficient sphere, a solution whose first
spinor component vanishes at a chosen point; passing to the large-$r$
limit, we obtain, for every point of $X$, a $J_v$-curve through that
point. The integral class can be chosen, using the lattice arithmetic
that we set up later, so that no positive configuration in that class
can split into components or multiply. That class then gives an
elliptic fibration whose fibers are precisely the unique curves through
each point.

The torus fibration is what makes the rest tractable. Along the regular
region, the pair $(A,C)$ is then averaged along the torus action to a
purely invariant pair. A closed invariant two-form $\Psi$ with unit
integral over every fiber is set up on top of this. Averaging and a
small exact correction orthogonalize $\Psi$, so that $(A,C)$ together
with $\Psi$ looks semilocally like a definite pair plus a vanishing
two-form. On each regular region we then construct a semi-flat, closed
two-form that satisfies the three pointwise identities $AD=CD=0$,
$D^2=A^2$ exactly; on each nodal region we use Gibbons--Hawking /
Ooguri--Vafa and Gross--Wilson local models to build the same
approximate data on top of a collapsing holomorphic model. The
semi-flat and nodal forms differ only up to exponentially small error on
fixed annuli, and a suitable correction primitive glues them to a
single closed approximate solution. The remaining correction is small
and is solved on exact two-forms, where a Stokes identity controls the
inverse uniformly as the fiber area $\eps$ shrinks.

The result is a closed form $D_\eps$ satisfying the three pointwise
identities exactly. The pair $(A,D_\eps)$ is then itself integrable (by
Cartan/Newlander--Nirenberg), and $A+iD_\eps$ is a nowhere-vanishing
holomorphic two-form for that complex structure. The endpoint of the
argument uses two applications of the classical K\"ahler-class
criterion: put the original $[C]$ into the K\"ahler cone, replace $C$ by
the resulting K\"ahler representative $C_1$, and repeat with $[B]$ to
get $B_1$. Along the second step the class $[B]$ is guaranteed to contain
a tamer because the class $[D_\eps]$ already contains one and $[D_\eps]$,
$[B]$ lie in the same open time-like cone of $H^2(X;\R)$. The final
triple $(A,C_1,B_1)$ has orthogonal equal-square components, hence is
hyperk\"ahler.

\subsection{Organization of the paper}

Section~\ref{sec:definite-pairs} sets up the local model (definite pairs,
the associated almost-complex structure, the topology of $X$) and the
coefficient sphere of almost-complex structures. Section~\ref{sec:current-energy}
and Section~\ref{sec:density-splitting} develop the current-cone
separation and the taming criterion, which are the main analytic input
to the "keep $[B]$ available" step. Section~\ref{sec:families} constructs
Seiberg--Witten curves through every point. Section~\ref{sec:lattice-chamber}
chooses the integral null class $F$ so that no positive configuration in
$F$ can split. Section~\ref{sec:pencil} turns the $F$-curves into a torus
fibration with ordinary nodal fibers. Section~\ref{sec:local-models} builds
the collapsing nodal models. Section~\ref{sec:regular-preparation}
prepares the pair by averaging and constructs the unit-fiber-area form
$\Psi$. Section~\ref{sec:auxiliary-volume} solves the auxiliary volume
equation. Section~\ref{sec:endpoint} finishes the proof of
Theorem~\ref{thm:main} via the two K\"ahler steps.

\subsection{Conventions}

Products of two-forms are wedge products; products of degree-two
cohomology classes are their intersection pairing. Curve classes are
identified with their Poincar\'e duals in cohomology. Integral lattice
statements use
\[
 \Lambda=H^2(X;\Z)/\operatorname{torsion}.
\]
The original positive period plane is
$P=\operatorname{span}\{[\omega_1],[\omega_2],[\omega_3]\}$. No
rationality of this plane is assumed. All deformations of forms are
smooth in the manifold variable; parameter regularity is stated where it
is used. Constants in local estimates may depend on fixed smooth
background data and on the indicated finite differentiation order.

\section{Definite pairs and the coefficient sphere}\label{sec:definite-pairs}

A positive triple carries redundant data. The first task of this
section is to organize it: a definite pair $(A,C)$ determines an
almost-complex structure, and a third form $B$ completes a positive
triple precisely when it tames that structure. This turns the main
deformation of a positive triple into a deformation of the pair alone,
with $B$ fixed. The second task is to record global structure: the
topology of $X$ is forced, the self-dual harmonic space has a natural
basis, and the initial triple parametrizes a whole $S^2$ of
almost-complex structures. That sphere will be the parameter space used
in Section~\ref{sec:families} to produce curves.

\subsection{Definite pairs and the associated almost-complex structure}

\begin{definition}
A pair $(A,C)$ of real two-forms is \emph{definite} if its
wedge-product matrix is positive definite at every point. A two-form
$B$ \emph{tames} an almost-complex structure $J$ if
$B(v,Jv)>0$ for every nonzero tangent vector $v$. It is
\emph{compatible} if, in addition, $B(Ju,Jv)=B(u,v)$. When these forms
are called symplectic, closedness is included.
\end{definition}

\begin{lemma}\label{lem:definite-pair}
A definite pair $(A,C)$ on an oriented four-manifold determines an
almost-complex structure up to sign. Its anti-invariant real two-form
bundle is pointwise spanned by $A,C$, and its canonical complex line
bundle is trivial. A third form $B$ completes a positive triple if and
only if $B$ tames one of the two signs of this almost-complex structure.
On a connected manifold the sign is fixed by $B$.

A real two-plane on which both $A$ and $C$ vanish is a complex line,
without specifying its orientation.
\end{lemma}

\begin{proof}
Ratios of positive four-forms are smooth functions. Set
\begin{equation}\label{eq:pair-normalization}
 x=\frac{AC}{A^2},\qquad
 y=\left(\frac{C^2}{A^2}-x^2\right)^{1/2}>0,\qquad
 T=\frac{C-xA}{y}.
\end{equation}
Then $AT=0$ and $T^2=A^2>0$, so
\[
 \Omega=A+iT,\qquad \Omega^2=0,\qquad
 \Omega\overline\Omega=2A^2>0.
\]
The form $\Omega$ is complex-decomposable; its kernel in the
complexified tangent bundle is complementary to its conjugate. Taking
this kernel as $T^{0,1}$ defines an almost-complex structure with the
given orientation. Replacing $y$ by $-y$ conjugates $\Omega$ and changes
$J$ to $-J$. The real and imaginary parts of $\Omega$ span the
anti-invariant forms, and $\Omega$ is a nowhere-zero section of the
canonical bundle.

Choose any Hermitian metric for $J$. The invariant two-forms are the
wedge-orthogonal complement of $\operatorname{span}(A,C)$ and have
signature $(1,3)$. For the decomposition $B=B^{1,1}+B^{2,0+0,2}$, the
positive-triple condition is exactly $(B^{1,1})^2>0$. In complex
dimension two, a real $(1,1)$-form has positive square precisely when
its two Hermitian eigenvalues have the same strict sign. The positive
component of that condition is the cone of forms positive on every
complex line. The anti-invariant summand vanishes on such lines, so the
positive-triple condition is equivalent to taming by $B$. The other
component is positive for $-J$. Connectedness fixes the choice
continuously and globally.

Finally, for independent real vectors $u,v$, the equations
$A(u,v)=C(u,v)=0$ are equivalent to $\Omega(u,v)=0$. In a complex
two-dimensional vector space with a nondegenerate complex volume form,
this says that $u,v$ are complex-dependent, equivalently that their
real span is a complex line.
\end{proof}

For a fixed almost-complex structure $J$, the set of taming two-forms
is an open convex cone. Its intersection with closed representatives of
any one class is also convex. The associated notion of positivity will
always use the sign of $J$ fixed by a stated tamer.

\subsection{Topology of $X$}

\begin{proposition}\label{prop:topology}
If $X$ carries a positive closed triple, then
\[
 b^+(X)=3,\qquad
 (b_1(X),b^-(X))=(0,19)\ \text{or}\ (4,3).
\]
The free integral intersection lattice is respectively
$3H\oplus2(-E_8)$ or $3H$, where $H$ is the hyperbolic plane.
Every almost-complex structure obtained from a definite pair in the
triple has vanishing integral first Chern class.
\end{proposition}

\begin{proof}
By Lemma~\ref{lem:definite-pair}, one member of the triple is a
symplectic form taming an almost-complex structure whose canonical
bundle is trivial. The first Chern class of this symplectic form
therefore vanishes: the space of its tamed almost-complex structures is
contractible and contains compatible structures. Bauer's bound for a
closed symplectic four-manifold with torsion first Chern class gives
$b^+\le3$ \cite[Corollary~1.2]{Bauer2008}. The three independent
positive cohomology classes give the reverse inequality.

The characteristic-number identity $c_1^2=2\chi+3\sigma$, obtained
from $p_1=c_1^2-2c_2$, the Euler-number identity and the signature
theorem \cite[Corollaries~11.12 and~15.5, Theorem~19.4]{MilnorStasheff1974},
together with $b^+=3$, gives
\[
 0=19-4b_1-b^-,\qquad \sigma=4b_1-16.
\]
Moreover $w_2\equiv c_1=0\pmod 2$, so $X$ is spin
\cite[Problem~14-B]{MilnorStasheff1974}. Rokhlin's theorem makes
$\sigma$ divisible by $16$ \cite{Rokhlin1952}. Since $b^-\ge0$, these
relations give $b_1\in\{0,4\}$ and the displayed alternatives.
Poincar\'e duality makes the free intersection form unimodular, and the
Wu formula makes it even \cite[Theorem~11.14]{MilnorStasheff1974}.
Classification of indefinite even unimodular lattices now gives the two
stated forms \cite[Chapter~V, Section~2.2, Theorems~5--6]{Serre1973}.
Only the intersection lattice has been classified at this stage.
\end{proof}

The positive three-plane bundle spanned by the triple is the self-dual
bundle of a unique oriented conformal structure. Choose one metric
$g_0$ in that conformal class. Since the original forms are closed and
self-dual, they are harmonic, and
\begin{equation}\label{eq:pair-harmonic-basis}
 \mathcal H^+_{g_0}=\operatorname{span}_\R\{\omega_1,\omega_2,\omega_3\}
\end{equation}
by Proposition~\ref{prop:topology}. Here the span is the vector space of
global forms with constant coefficients.

\subsection{The coefficient sphere of almost-complex structures}

For $v\in S^2\subset\R^3$, let $B_v=v\cdot\omega$ and let
$\mathcal A_v$ be the plane of forms $a\cdot\omega$ with $a\perp v$.
Choosing the sign tamed by $B_v$ defines an almost-complex structure
$J_v$. This construction uses the plane $\mathcal A_v$ itself and does
not require a global choice of ordered basis over $S^2$.

\begin{lemma}\label{lem:twistor-degree}
The family $J_v$ is smooth and is Hermitian for $g_0$. At every point
of $X$, its map to the twistor sphere has degree of absolute value one.
The closed $J_v$-anti-invariant forms form the two-dimensional space
$\mathcal A_v$.
\end{lemma}

\begin{proof}
The first assertion follows from the construction and the self-dual
description of Hermitian almost-complex structures in dimension four.
At a fixed point, let $K$ be the positive Gram matrix of the triple for
$g_0$. A vector normal to the coefficient plane $v^\perp$, for this
Gram matrix, has coefficients $K^{-1}v$. Its pairing with $B_v$ is
positive. Thus, after normalization, the fundamental forms give the map
\[
 v\longmapsto\frac{K^{-1}v}{|K^{-1}v|}
\]
in radial coefficient coordinates. Positive definite matrices connect
to the identity through positive definite matrices, so this map has
degree one. The orientation convention identifying coefficient and
twistor spheres can only change its sign.

A closed anti-invariant form is self-dual for $g_0$, hence harmonic.
By \eqref{eq:pair-harmonic-basis} it is a constant combination of the
original triple. It is anti-invariant for $J_v$ exactly when that
coefficient vector lies in $v^\perp$. This proves the final assertion.
\end{proof}

\subsection{The Lorentz space and the integrability identity}

After a constant orthogonal change of coefficients we denote a chosen
ordered triple by $(A,C,B)$. For all later exact deformations of the
pair, set
\begin{equation}\label{eq:pair-lorentz-space}
 V=[A]^\perp\cap[C]^\perp\subset H^2(X;\R).
\end{equation}
This space has signature $(1,b^-)$, and $[B]\in V$ is a unit timelike
vector. It determines the future component of the timelike cone.

\begin{lemma}\label{lem:complex-symplectic}
Let $A,D$ be closed real two-forms with $AD=0$ and $A^2=D^2>0$
pointwise. Then $A+iD$ defines an integrable complex structure and is a
nowhere-zero holomorphic two-form for it. If a closed third form $C$
has $AC=DC=0$ and $C^2=A^2$, then the resulting triple is
hyperk\"ahler, after choosing the sign of the complex structure so that
$C$ is positive.
\end{lemma}

\begin{proof}
Set $\Omega=A+iD$ and use its kernel as $T^{0,1}$ as in
Lemma~\ref{lem:definite-pair}. For vector fields $U,V$ in that kernel,
Cartan's formula and $\dd\Omega=0$ give
\[
 \mathcal L_U\Omega=0,\qquad
 \iota_{[U,V]}\Omega
 =\mathcal L_U(\iota_V\Omega)-\iota_V(\mathcal L_U\Omega)=0.
\]
The kernel is involutive. The smooth Newlander--Nirenberg theorem
therefore gives integrability \cite{NewlanderNirenberg1957}, and
closedness of the $(2,0)$-form $\Omega$ gives holomorphicity.

Under the additional assumptions, $C$ is a positive closed
$(1,1)$-form, hence a K\"ahler form. Its metric volume is
$C^2/2=A^2/2$. The ratio $\Omega\overline\Omega/C^2$ is constant, so the
holomorphic volume form has constant pointwise norm. The Chern
connection of the canonical line then makes $\Omega$ parallel. For a
K\"ahler metric this is the connection induced by the Levi-Civita
connection. Thus $A,D,C$ are parallel; their normalized self-dual
algebra gives the hyperk\"ahler structure. Explicitly, if
$(\eta_1,\eta_2,\eta_3)=(A,D,C)$ and $\mu=A^2/2$, then
$\eta_i\wedge\eta_j=2\delta_{ij}\mu$; thus the normalized matrix is
the identity.
\end{proof}

\section{Separation, quadratic mass, and angular energy}
\label{sec:current-energy}

The pair deformation keeps $B$ fixed, so the question becomes whether
the fixed class $[B]$ still contains a tamer for the moving
almost-complex structure. To answer this we need a criterion, and the
criterion goes through separating the cone of positive forms from those
that fail to contain a tamer. That separation produces a positive
current whose mass we must control at the two-dimensional scale, and
whose complex-line directions must not turn too fast at small scales.
The present section develops both estimates. They are the input to the
density decomposition of \S\ref{sec:density-splitting}.

\subsection{Conventions and separation}

Fix a smooth almost-complex structure $J$ on the closed four-manifold
$X$, give $X$ its complex orientation, and choose a smooth $J$-Hermitian
metric $g$. Its fundamental form is $G(u,v)=g(Ju,v)$. The pointwise
orthogonal decomposition is
\begin{equation}\label{eq:current-bundles}
 \Lambda^2=I\oplus E,
 \qquad I=\Lambda_J^+=\R G\oplus\Lambda_g^-,
 \qquad E=\Lambda_J^-\subset\Lambda_g^+.
\end{equation}
Let $R$ denote the orthogonal projection onto $E$. A unit
complex-line bivector is $\ell=e\wedge Je$, where $|e|_g=1$. We use $g$
to identify bivectors and two-forms, so
\begin{equation}\label{eq:current-line-normalization}
 |\ell|_g=1,\qquad \ell^+=G/2,\qquad G(\ell)=1,
 \qquad |G|_g^2=2.
\end{equation}
The bundle of these bivectors is denoted by
$\mathcal L\longrightarrow X$.

A two-current is a continuous real linear functional on
$\Omega^2(X)=C^\infty(X,\Lambda^2)$. Under the metric identification its
coefficient is a distributional two-form, denoted by the same letter:
$T(\xi)=\langle T,\xi\rangle$. Thus
\begin{equation}\label{eq:current-dual-convention}
 T\text{ is closed}\ \Longleftrightarrow\
 T(\dd\alpha)=0\ \text{for all }\alpha\in\Omega^1(X)
 \ \Longleftrightarrow\ \dd^*T=0
 \ \Longleftrightarrow\ \dd(*T)=0.
\end{equation}
The cohomology class dual to a closed current is $[*T]\in H^2(X;\R)$.
For smooth coefficients this convention gives
$T(\xi)=\int_X\xi\wedge *T$. Inner products between $L^2$ coefficients
are written $\langle\,\,,\,\rangle_2$.

Let
\[
 \mathcal P_J=\{\xi\in\Omega^2(X):\xi(\ell)>0
                    \text{ for every }\ell\in\mathcal L\},
 \qquad
 \mathcal Z_J=\{\xi\in C^\infty(X,I):\dd\xi=0\}.
\]
The set $\mathcal P_J$ is an open convex cone in the smooth topology;
openness already holds in the uniform topology. Its closed elements are
precisely the symplectic forms taming $J$. The elements of
$\mathcal P_J\cap\mathcal Z_J$ are the compatible symplectic forms.

\subsection{The two separators}

\begin{lemma}[The two separators]\label{lem:current-separators}
The following hold.
\begin{enumerate}[label=\textup{(\roman*)}]
\item If $\mathcal P_J\cap\mathcal Z_J$ is empty, there is a nonzero
 current $N$ which annihilates $\mathcal Z_J$ and all anti-invariant
 forms, and is nonnegative on semipositive invariant forms.
\item If $H\in H^2(X;\R)$ has no representative in $\mathcal P_J$, there
 is a nonzero positive invariant current $N$ which is closed and
 satisfies $N(H)\le0$.
\end{enumerate}
In either case one can normalize $N(G)=1$ and choose a positive Borel
measure $\lambda$ on $\mathcal L$ such that
\begin{equation}\label{eq:current-line-measure}
 N(\xi)=\int_{\mathcal L}\xi_y(\ell)\,\dd\lambda(y,\ell),
 \qquad \mu=(\operatorname{pr}_X)_*\lambda,
 \qquad \mu(X)=1.
\end{equation}
In particular $N$ has order zero and belongs to $H^{-3}(X,\Lambda^2)$.
\end{lemma}

\begin{proof}
For \textup{(i)}, apply separation of the open convex set
$\mathcal P_J\cap C^\infty(X,I)$ from the disjoint linear subspace
$\mathcal Z_J$ in $C^\infty(X,I)$. The separating continuous functional
vanishes on $\mathcal Z_J$ and is nonnegative on the positive cone.
Extend it to all two-forms by precomposing with $1-R$. Adding a positive
multiple of $G$ and then taking a limit proves nonnegativity on every
semipositive invariant form.

For \textup{(ii)}, choose a smooth closed representative $\beta$ of $H$
and separate the open cone $\mathcal P_J$ from the affine subspace
$\beta+\dd\Omega^1(X)$ in $\Omega^2(X)$. If $N$ is the separating
functional, the unbounded translation directions in this affine
subspace give $N(\dd\alpha)=0$. Multiplying a positive form by any
positive scalar, the separation inequalities imply both
$N(\mathcal P_J)\ge0$ and $N(\beta)\le0$. Adding an arbitrary real
multiple of an anti-invariant form to an element of $\mathcal P_J$ shows
that $N$ annihilates anti-invariant forms. The same limiting argument
gives positivity on semipositive invariant forms.

For either functional, comparison of an invariant form with sufficiently
large positive multiples of $G$ gives $|N(\xi)|\le C N(G)\|\xi\|_{C^0}$.
If $N(G)$ were zero, this inequality and anti-invariance would make the
functional zero; normalize $N(G)=1$. Locally the functional is
represented by a positive Hermitian matrix of measures. Its trace is a
positive measure $\mu$; the Radon--Nikodym matrix relative to $\mu$ is
positive semidefinite of trace one. A measurable spectral decomposition
of this $2\times2$ matrix expresses it as a convex combination of
rank-one Hermitian projections. Borel local frames and a Borel partition
of $X$ give the global measure $\lambda$ in
\eqref{eq:current-line-measure}. Finally, $H^3\hookrightarrow C^0$ in
real dimension four, so every such measure coefficient belongs to
$H^{-3}$.
\end{proof}

\subsection{A closed lift of anti-invariant forms}

\begin{lemma}[Closed lift]\label{lem:closed-lift}
There is a classical pseudodifferential operator
$D:C^\infty(X,E)\longrightarrow\Omega^2(X)$ of order zero with
\begin{equation}\label{eq:current-lift-identities}
 \dd D=0,\qquad RD=\Id.
\end{equation}
It maps distributions to distributions. For the separator in
\Cref{lem:current-separators}\textup{(i)}, set
\begin{equation}\label{eq:current-correction}
 Q(\xi)=-N(DR\xi),\qquad T=N+Q.
\end{equation}
Then $Q$ is anti-invariant and self-dual, $Q,T\in H^{-3}$, and $T$
annihilates every closed smooth two-form. In particular $T$ is a closed
current with $[*T]=0$.

For a smooth family of $(J,g)$ with constant dimension of the space of
closed anti-invariant forms, these lifts can be chosen smoothly in the
parameter as operators between the corresponding smooth bundles.
\end{lemma}

\begin{proof}
On sections of $E$ consider $P=R\dd\dd^*$. If $a$ is self-dual, the
self-dual projection of $\xi\wedge\iota_\xi a$ is $|\xi|^2a/2$: in a
frame with first covector $\xi/|\xi|$, the two self-dual halves of $a$
have equal length. Hence
$\sigma_2(P)(x,\xi)=\tfrac12|\xi|_g^2\Id_E$ and
$\langle Pa,a\rangle_2=\|\dd^*a\|_2^2$. Thus $P$ is elliptic,
self-adjoint, and nonnegative. Its kernel consists of smooth sections
with $\dd^*a=0$; since $a=*a$, they also satisfy $\dd a=0$, and hence
are harmonic. Conversely a closed anti-invariant form is self-dual and
therefore harmonic and in the kernel.

Let $H_0$ be the orthogonal projection onto this finite-dimensional
kernel, and let $L$ be the inverse of $P$ on its orthogonal complement,
extended by zero on the kernel. Elliptic Fredholm theory and the
parametrix give $L$ as a classical operator of order $-2$, with $H_0$
smoothing and $PL=\Id-H_0$. Set
\begin{equation}\label{eq:current-lift-definition}
 D=\dd\dd^*L+H_0.
\end{equation}
The image of both summands is closed, and $RD=PL+H_0=\Id$.

For closed $\xi$, the form $\xi-DR\xi$ is closed and invariant, so
$T(\xi)=N(\xi-DR\xi)=0$. The functional $Q$ only depends on $R\xi$;
this proves anti-invariance and self-duality. Its $H^{-3}$ membership
follows because an order-zero operator acts boundedly on every Sobolev
space. Vanishing on exact forms makes $T$ closed, and vanishing on all
closed two-forms makes $[*T]=0$.

For the family statement, identify the varying bundles locally in the
parameter and use a smooth unitary identification of the $L^2$ spaces.
The elliptic operators then have common domain $H^2$ in a fixed Hilbert
space. The spectral projection onto the kernel is smooth in the
parameter, and so is $L=(P+H_0)^{-1}(\Id-H_0)$ on every Sobolev scale.
This gives the required smooth lifts.
\end{proof}

From here on we use one of two setups:
\begin{enumerate}[label=\textup{(\Alph*)}]
\item $N$ is the normalized separator in
 \Cref{lem:current-separators}\textup{(i)}, and $Q,T$ are given by
 \eqref{eq:current-correction};
\item $N$ is a normalized positive invariant closed current, $Q=0$, and
 $T=N$.
\end{enumerate}
Setup \textup{(B)} includes \Cref{lem:current-separators}\textup{(ii)}.
In both cases $T$ is closed, but the obstructions are different. In setup
\textup{(A)}, we will prove $Q=0$: the positive current $N=T$ would then
annihilate every closed form, contradicting the existence of a closed
tamer. In setup \textup{(B)}, the class separator satisfies
$N(H)\le0$; the task is instead to prove $N(H)>0$ from the hypotheses
on the proposed taming class.

Constants below may depend on $J,g$ and on the fixed operators, but
are independent of the center, of the small scale, and of the choice of
submeasure of $\lambda$.

\subsection{The quadratic mass estimate}

\begin{lemma}[Radial order-zero estimate]\label{lem:current-radial-operator}
Let $A$ be a fixed classical operator of order zero between smooth
vector bundles on a compact four-manifold. Fix a coordinate radius below
the injectivity radius. Suppose that smooth sections $a_{s,p}$, supported
in that coordinate neighborhood of $p$, satisfy, for $0<s<s_0$,
\[
 |\nabla^j a_{s,p}(y)|\le C_j(s+\dist(p,y))^{-a-j},
 \qquad 0\le j\le6,\quad a\in\{0,1\},
\]
with uniform boundedness on a fixed outer annulus. Then, for
$\delta=s+\dist(p,x)$ small,
\begin{equation}\label{eq:current-radial-operator}
 |Aa_{s,p}(x)|\le
 \begin{cases}
 C\delta^{-1},&a=1,\\
 C(1+|\log\delta|),&a=0.
 \end{cases}
\end{equation}
Away from a smaller fixed neighborhood of $p$ the output is uniformly
bounded in $s$ and $p$.
\end{lemma}

\begin{proof}
Use two precise facts of the order-zero calculus: in local left
quantization its symbol is bounded uniformly in the covariable, and its
off-diagonal kernel satisfies $|K_A(x,y)|\le C\dist(x,y)^{-4}$ for small
positive distance. Smooth remainders have uniformly bounded kernels.

Fix an evaluation point $x$, and split the input by a smooth cutoff
supported in $B_{c\delta}(x)$ and equal to one on $B_{c\delta/2}(x)$,
with a fixed small $c$. On its support, $s+\dist(p,y)$ is comparable to
$\delta$. After translation and dilation by $\delta$, the localized input
has derivatives through order six bounded by $C\delta^{-a}$.
Integration by parts in its Fourier transform therefore bounds its
Fourier $L^1$ norm by $C\delta^{-a}$, since $(1+|\xi|)^{-6}$ is
integrable in dimension four. The bounded symbol gives the same bound
for the local contribution to $Aa_{s,p}(x)$.

The remaining part has distance at least $c\delta/2$ from $x$. Its
portion in $B_{C\delta}(p)$ contributes at most
$C\delta^{-4}\int_{B_{C\delta}(p)}(s+\dist(p,y))^{-a}\,\dd V_g(y)\le
C\delta^{-a}$. On a shell of radius $r=2^j\delta\ge C\delta$ about $x$,
distance from $p$ is comparable to $r$, so its contribution is at most
$Cr^{-4}r^4r^{-a}=Cr^{-a}$. The shells up to the fixed coordinate radius
sum to $C\delta^{-1}$ if $a=1$ and to $C(1+|\log\delta|)$ if $a=0$. The
rest is uniformly bounded. If $x$ stays away from $p$, the singular
input near $p$ has uniformly bounded $L^1$ norm and the kernel there is
bounded; the remaining input is uniformly smooth.
\end{proof}

\begin{lemma}[Quadratic trace mass]\label{lem:quadratic-mass}
In either setup \textup{(A)} or \textup{(B)}, there is a constant $C$
with
\begin{equation}\label{eq:current-quadratic-mass}
 \mu(B_s(p))\le Cs^2\qquad(p\in X,\ 0<s\le1).
\end{equation}
\end{lemma}

\begin{proof}
Fix a radius $\rho_0>0$ for uniform normal coordinates and a smooth
radial cutoff which is one on $B_{2\rho_0}(p)$ and supported in
$B_{3\rho_0}(p)$. With $t=\dist(p,\cdot)$ define the globally smooth
functions
\[
 u_s=\chi(t)\log\sqrt{s^2+t^2},\qquad
 h_s=\chi(t)\sqrt{s^2+t^2},\qquad
 \dd_J^cu=-\dd u\circ J.
\]
Smoothness at $p$ follows from the dependence on $t^2$ and from $s>0$.
For a function $f$ put
\[
 \beta(f)=
 \begin{cases}
  (1-DR)\dd\dd_J^cf,&\text{in setup \textup{(A)}},\\
  \dd\dd_J^cf,&\text{in setup \textup{(B)}}.
 \end{cases}
\]
In setup \textup{(A)}, $\beta(f)$ is closed and invariant, and in setup
\textup{(B)} it is exact. Hence $N(\beta(f))=0$ in both cases.

The principal symbol of $\dd\dd_J^c$ on functions is a multiple of
$\xi\wedge J^*\xi$, which is $J$-invariant, so $R\dd\dd_J^c$ is
first-order. Differentiating the radial functions gives, for
$0\le j\le6$,
\[
 |\nabla^jR\dd\dd_J^cu_s|\le C_j(s+t)^{-1-j},\qquad
 |\nabla^jR\dd\dd_J^ch_s|\le C_j(s+t)^{-j}.
\]
\Cref{lem:current-radial-operator} applied to $D$ therefore bounds the
two correction terms by $C/(s+t)$ and $C(1+|\log(s+t)|)$.

The positive parts, without cutoff, can be computed in a fixed Hermitian
Euclidean space. If $f=f(t)$ and $\ell$ is a unit complex line, its
Hessian trace is
\[
 \dd\dd_{J_p}^cf(\ell)=2f'(t)/t+\bigl(f''(t)-f'(t)/t\bigr)a,
 \qquad 0\le a\le1.
\]
For the two radial functions this gives
\begin{align}
 \dd\dd_{J_p}^c\log\sqrt{s^2+t^2}(\ell)
   &\ge \frac{2s^2}{(s^2+t^2)^2},\label{eq:current-log-trace}\\
 \dd\dd_{J_p}^c\sqrt{s^2+t^2}(\ell)
   &\ge \frac1{\sqrt{s^2+t^2}}.\label{eq:current-root-trace}
\end{align}
In normal coordinates $g-g_p=O(t^2)$ and $J-J_p=O(t)$, while the first
derivatives of $J$ are bounded. Replacing the frozen complex line and
operator by the actual ones changes the estimates by at most
$C/(s+t)$ for $u_s$ and $C$ for $h_s$. With $\delta=s+t$, on
$B_{\rho_0}(p)$,
\begin{align}
 \beta(u_s)(\ell)&\ge c_1s^2\delta^{-4}-C_1\delta^{-1},
                  \label{eq:current-log-corrected}\\
 \beta(h_s)(\ell)&\ge c_2\delta^{-1}-C_2(1+|\log\delta|).
                  \label{eq:current-root-corrected}
\end{align}
The inequalities also apply in setup \textup{(B)}, where the correction
terms are absent.

Choose $0<\rho<\rho_0$ so small that
$C_2(1+|\log\delta|)\le c_2/(2\delta)$ for $0<\delta\le2\rho$, and
then choose a fixed multiplier $M\ge2C_1/c_2$. If $s\le\rho$ and
$t\le\rho$, the two displayed inequalities give
$\beta(u_s+Mh_s)(\ell)\ge c_1s^2\delta^{-4}$. On $X\setminus B_\rho(p)$
all terms are bounded independently of $s$, including the nonlocal
corrections, by the last assertion of
\Cref{lem:current-radial-operator}. Since $\delta\le2s$ on $B_s(p)$,
globally
\[
 \beta(u_s+Mh_s)(\ell)\ge c s^{-2}\mathbf 1_{B_s(p)}-C.
\]
Apply $N$, use $N(\beta(u_s+Mh_s))=0$ and $\mu(X)=1$, and obtain
\eqref{eq:current-quadratic-mass} for $s\le\rho$. For larger $s$ the
same bound follows from $\mu(X)=1$. Compactness makes all choices
uniform in $p$.
\end{proof}

\subsection{The resolvent kernel and positive-current pairings}

We have a uniform mass bound at the two-dimensional scale. The next
step compares directions of nearby complex lines; smoothing permits this
comparison through an intersection pairing with the mass bound
controlling the error.

Let $\Delta=\dd\dd^*+\dd^*\dd$ be the nonnegative Hodge Laplacian and
set
\begin{equation}\label{eq:current-smoothing}
 S_s=(1+s^2\Delta)^{-3},\qquad 0<s\le s_0.
\end{equation}
It commutes with $\dd,\dd^*$ and $*$. At each fixed scale it maps
$H^{-3}$ to $H^3$, which suffices for all the pairings below. The kernel
of $S_s^2$ is continuous, being of order $-12$ in dimension four.

\begin{lemma}[Resolvent kernel]\label{lem:current-kernel}
Choose a fixed radius $r_0$ smaller than the injectivity radius, and let
$\tau_{z\to y}$ be parallel transport along the minimizing geodesic when
$\dist(y,z)<r_0$. The kernel $K_s$ of $S_s^2$ can be written
\begin{equation}\label{eq:current-kernel-split}
 K_s(y,z)=k_s(y,z)\tau_{z\to y}+E_s(y,z),
\end{equation}
where $k_s$ is nonnegative and supported where $\dist(y,z)<r_0$. For
each integer $k\ge0$ there are constants such that
\begin{align}
 k_s(y,z)&\ge cs^{-4}&&\text{if }\dist(y,z)<s,
                         \label{eq:current-kernel-lower}\\
 k_s(y,z)&\le C_ks^{-4}(1+\dist(y,z)/s)^{-k},
                         \label{eq:current-kernel-upper}\\
 |E_s(y,z)|&\le C_ks^{-2}(1+\dist(y,z)/s)^{-k}.
                         \label{eq:current-kernel-error}
\end{align}
The upper bound on $s_0$ is fixed independently of $y,z$.
\end{lemma}

\begin{proof}
The spectral theorem gives the operator identity
\begin{equation}\label{eq:current-gamma-resolvent}
 S_s^2=\frac1{5!}\int_0^\infty h^5e^{-h}e^{-s^2h\Delta}\,\dd h.
\end{equation}
The heat kernel of the nonnegative Hodge Laplacian on a closed
four-manifold has a local small-time parametrix
\[
 \chi(d)(4\pi t)^{-2}e^{-d^2/(4t)}
       \bigl(a_0(y,z)+ta_1(y,z)+\cdots+t^ma_m(y,z)\bigr),
\]
with arbitrarily high-order uniform remainder after increasing $m$; here
$d=\dist(y,z)$, $\chi$ is supported inside the injectivity radius and
equals one near the diagonal, and
$a_0(y,z)=j(y,z)^{-1/2}\tau_{z\to y}$ with $j$ smooth, strictly
positive and equal to one on the diagonal. For the Hodge Laplacian,
after subtracting the leading term, for every prescribed $L$ the kernel
is bounded for $0<t<t_0$ by $Ct^{-1}e^{-d^2/(Ct)}+C_Lt^L$. The local
expansion gives the Gaussian part; away from the diagonal the global
Gaussian estimate makes the cutoff error smaller than every power of
$t$. For $t\ge t_0$ the heat kernel is uniformly bounded: factor the
heat operator through two fixed positive-time smoothing operators and
the $L^2$ contraction $e^{-(t-t_0)\Delta}$, using nonnegativity and
self-adjointness of the Hodge Laplacian.

Take $\chi\ge0$ and use its leading coefficient to define
\[
 k_s(y,z)=\frac{\chi(d)j(y,z)^{-1/2}}{5!(4\pi)^2}s^{-4}
       \int_0^\infty h^3e^{-h-d^2/(4s^2h)}\,\dd h.
\]
For $d<s$, integration over $1\le h\le2$ gives the lower bound. For
every fixed $k$, the elementary inequality
$e^{-r^2/(ch)}\le C_k(1+r)^{-k}(1+h^{k/2})$ shows that the last integral
is bounded by $C_k(1+d/s)^{-k}$, giving the upper bound.

Integrating the $t^{-1}$ heat-kernel error in the resolvent identity
gives
\[
 Cs^{-2}\int_0^\infty h^4e^{-h-d^2/(Cs^2h)}\,\dd h
 \le C_ks^{-2}(1+d/s)^{-k}.
\]
For the uniform remainder, integration over $s^2h<t_0$ gives $C_Ls^{2L}$;
since $d\le\operatorname{diam}X$, this is bounded by the right side of
the error bound when $2L\ge k-2$. On $s^2h\ge t_0$, the factor $e^{-h}$
makes both the heat-kernel contribution and the extended leading term
smaller than every power of $s$.
\end{proof}

For any positive submeasure $\lambda_i\le\lambda$, let $N_i$ be its
current and $\mu_i$ its projection to $X$. These currents need not be
closed. Define
\begin{equation}\label{eq:current-error-bracket}
 B_{s,\mu_i}(z)=s^{-2}\int_X
       (1+\dist(y,z)/s)^{-6}\,\dd\mu_i(y),
 \qquad B_s=B_{s,\mu}.
\end{equation}
By \Cref{lem:quadratic-mass}, a dyadic-shell sum gives
$0\le B_{s,\mu_i}(z)\le B_s(z)\le C$.

\begin{lemma}[Positive-current pairing]\label{lem:current-positive-pairing}
For any $\lambda_1,\lambda_2\le\lambda$ and their currents,
\begin{align}
 \langle S_sN_1,*S_sN_2\rangle_2
 &\ge cs^{-4}
       \iint_{\dist(y,z)<s}|\ell-\tau_{z\to y}m|^2
             \,\dd\lambda_1(y,\ell)\dd\lambda_2(z,m)
       \notag\\[-2pt]
 &\hspace{8mm}-C\int_X B_{s,\mu_1}(z)\,\dd\mu_2(z),
                                  \label{eq:current-positive-pairing}\\
 \|S_sN_i\|_2&\le C/s.            \label{eq:current-submeasure-ltwo}
\end{align}
One can replace $B_{s,\mu_1}$ in the lower bound by $B_s$.
\end{lemma}

\begin{proof}
Parallel transport preserves the Hodge star. From
\eqref{eq:current-line-normalization}, direct expansion gives
\begin{equation}\label{eq:current-angular-identity}
 \langle\ell,*\tau_{z\to y}m\rangle
 =\tfrac12|\ell-\tau_{z\to y}m|^2
       -\tfrac14|G_y-\tau_{z\to y}G_z|^2.
\end{equation}
The last squared difference is bounded by $Cd^2$, where $d=\dist(y,z)$.

Express the pairing using the continuous kernel of $S_s^2$ and
\eqref{eq:current-line-measure}. Its leading scalar is nonnegative, so
the lower-bound kernel inequality and the angular identity supply the
positive term. The $Cd^2$ term, using the upper bound with exponent at
least eight, and the error kernel, using exponent six, are bounded in
absolute value by
\[
 Cs^{-2}\iint(1+d/s)^{-6}\,\dd\mu_1(y)\dd\mu_2(z),
\]
which is the asserted error. The norm estimate follows by using the
absolute kernel bound in the unstarred pairing:
\[
 \|S_sN_i\|_2^2
 \le Cs^{-4}\iint(1+d/s)^{-6}\,\dd\mu_i(y)\dd\mu_i(z)
 \le Cs^{-2}\mu_i(X)\le Cs^{-2}.
\]
The kernel formulas extend to measure coefficients because the kernel is
continuous at fixed $s$.
\end{proof}

\subsection{Projection leakage and the topological energy}

\begin{lemma}[Projection commutator]\label{lem:current-commutator}
Let $\Pi$ be a smooth orthogonal bundle projection, and suppose
$\Pi W=0$ for a distributional form $W$ with $S_sW\in L^2$. Then
\begin{equation}\label{eq:current-projection-leakage}
 \|\Pi S_sW\|_2\le Cs\|S_sW\|_2.
\end{equation}
Consequently, for every positive submeasure current $N'\le_\lambda N$
and every anti-invariant distribution $Z$ with $S_sZ\in L^2$,
\begin{equation}\label{eq:current-mixed-bound}
 |\langle S_sN',S_sZ\rangle_2|\le C\|S_sZ\|_2.
\end{equation}
\end{lemma}

\begin{proof}
Set $A_s=S_s^{-1}=(1+s^2\Delta)^3$ and $U=S_sW$. Since $\Pi A_sU=0$,
the distributional identity $\Pi U=S_s[A_s,\Pi]U$ holds. The commutator
expands as a sum of $\binom3j s^{2j}[\Delta^j,\Pi]$, $1\le j\le3$.
Because the principal symbol of $\Delta$ is scalar, $[\Delta^j,\Pi]$ is a
differential operator of order at most $2j-1$. The spectral theorem and
elliptic Sobolev equivalence give $\|S_s\|_{L^2\to H^m}\le C_ms^{-m}$,
$0\le m\le6$. Since the formal adjoint of $B_j=[\Delta^j,\Pi]$ has the
same order,
\[
 \|S_s s^{2j}B_j\|_{L^2\to L^2}
 =\|s^{2j}B_j^*S_s\|_{L^2\to L^2}
 \le Cs^{2j}s^{-(2j-1)}=Cs.
\]
This proves the leakage estimate, including its meaning for
distributional $W$ by the bounded extension of the displayed operator.

Apply it to $RN'=0$ and $(1-R)Z=0$. With $u=S_sN'$ and $v=S_sZ$, the
orthogonal splitting gives
\[
 |\langle u,v\rangle_2|
 \le\|Ru\|_2\|Rv\|_2+\|(1-R)u\|_2\|(1-R)v\|_2
 \le Cs\|u\|_2\|v\|_2,
\]
and the submeasure norm estimate yields the mixed bound.
\end{proof}

If $C_1,C_2\in H^{-3}$ are closed currents, Hodge decomposition of the
closed distributional forms $*C_i$ and commutation of $S_s$ with $\dd$
give
\begin{equation}\label{eq:current-topological-pairing}
 \langle S_sC_1,*S_sC_2\rangle_2
  =\int_X(*S_sC_1)\wedge(*S_sC_2)
  =[*C_1]\cdot[*C_2].
\end{equation}
Write $*C_i=h_i+\dd a_i$ with $h_i$ harmonic and $a_i\in H^{-2}$;
then $S_sh_i=h_i$ and $S_sa_i\in H^4$. Integration by parts eliminates
every term containing an exact factor, leaving $\int_Xh_1\wedge h_2$.
No product of the unsmoothed distributions is used.

\begin{proposition}[Angular energy]\label{prop:angular-energy}
In setup \textup{(A)} or \textup{(B)}, the correction satisfies
$Q\in L^2(X,E)$ and
\begin{equation}\label{eq:current-angular-energy}
 \iint_{\dist(y,z)<s}|\ell-\tau_{z\to y}m|^2
           \,\dd\lambda(y,\ell)\dd\lambda(z,m)\le Cs^4.
\end{equation}
The estimates in the positive-pairing and projection-commutator lemmas
hold for all indicated submeasures, and for every fixed submeasure
current $N'$,
\begin{equation}\label{eq:current-mixed-vanishing}
 \lim_{s\downarrow0}\langle S_sN',S_sQ\rangle_2=0.
\end{equation}
The convergence is uniform over $\lambda'\le\lambda$.
\end{proposition}

\begin{proof}
Let $\mathcal A_s$ denote the double integral in
\eqref{eq:current-angular-energy} and put $x_s=\|S_sQ\|_2$. Since $Q$ is
self-dual and $S_s$ commutes with $*$, the topological pairing applied
to $T=N+Q$ gives
\[
 [*T]^2
  =\langle S_sN,*S_sN\rangle_2+2\langle S_sN,S_sQ\rangle_2+x_s^2.
\]
Use the positive-current lower bound and the $B_s$ bound for the first
term, and the mixed bound for the second. The result is
\begin{equation}\label{eq:current-energy-coercivity}
 cs^{-4}\mathcal A_s+x_s^2-Cx_s\le C+[*T]^2.
\end{equation}
The right side is independent of $s$. In setup \textup{(A)} its
intersection term is zero. In setup \textup{(B)} that term is bounded
uniformly for mass-one $N$: its harmonic coefficients are pairings with
fixed smooth forms and are bounded by the line-measure representation.
Completing the square in $x_s$ bounds both $x_s$ and $s^{-4}\mathcal A_s$.

As $s\downarrow0$, $S_sQ$ converges to $Q$ distributionally. A weakly
convergent subsequence of the bounded $L^2$ family has limit $Q$,
proving $Q\in L^2$; anti-invariance is preserved, and the spectral
theorem gives strong convergence $S_sQ\to Q$ in $L^2$.

Choose smooth anti-invariant $Q_j$ with $\|Q_j-Q\|_2\to0$; for example
apply $R$ to a smooth approximation. By the mixed bound and the $L^2$
contraction property of $S_s$,
$|\langle S_sN',S_s(Q-Q_j)\rangle_2|\le C\|Q-Q_j\|_2$. For fixed $j$,
$S_s^2Q_j\to Q_j$ in $C^\infty$, whereas $N'(Q_j)=0$. Therefore
\[
 \langle S_sN',S_sQ_j\rangle_2
       =N'(S_s^2Q_j-Q_j)\longrightarrow0,
\]
uniformly over the submeasures because their trace masses are at most
one. Let $s\to0$ first and $j\to\infty$ second.
\end{proof}
\section{Density splitting and the taming cone}
\label{sec:density-splitting}

The estimates of Section~\ref{sec:current-energy} have two consequences.
The first requires no definite pair and imposes no condition on the
cohomology class of the resulting form; it is the ``taming implies
compatibility'' theorem for any almost-complex structure on a closed
connected four-manifold. The second keeps a specified class available,
provided that the class pairs positively with every irreducible curve;
that consequence will be used along the pair deformations in
Sections~\ref{sec:regular-preparation} and~\ref{sec:endpoint}.

\begin{theorem}[From taming to compatibility]\label{thm:tamed-compatible}
Let $X$ be a closed connected smooth four-manifold with a smooth
almost-complex structure $J$ and its complex orientation. If some smooth
symplectic form tames $J$, then some smooth symplectic form is
compatible with $J$.
\end{theorem}

Curve classes are identified with their Poincar\'e dual classes in
$H^2(X;\R)$. An \emph{irreducible $J$-holomorphic curve} is the image
of a nonconstant somewhere-injective $J$-holomorphic map from a closed
connected Riemann surface, with its complex orientation.

\begin{theorem}[A curve criterion for taming classes]\label{thm:taming-cone}
Let $X$ be a closed connected oriented smooth four-manifold with
$b^+(X)=3$, and let $J$ be a smooth almost-complex structure inducing
the orientation. Suppose smooth closed real two-forms $A,C$ span a
pointwise definite plane and form a basis of the $J$-anti-invariant
two-forms at every point. Set
\[
 V=[A]^\perp\cap[C]^\perp\subset H^2(X;\R).
\]
Suppose $U\in V$ has positive square and contains a smooth tamer for
$J$. If $H\in V$ has positive square, belongs to the same component of
$\{v\in V:v^2>0\}$ as $U$, and satisfies
\[
 H\cdot d>0
\]
for every irreducible closed $J$-holomorphic curve of class $d$, then
$H$ contains a smooth symplectic form taming $J$.
\end{theorem}

\subsection{Densities and almost-everywhere planar tangents}

The radial comparison below follows the Lelong--Jensen method for
almost-complex currents \cite{ElkhadhraMimouni2007}: closedness of
$N+Q$ is applied, and the $L^2$ correction $Q$ is estimated directly.

\begin{lemma}[Quadratic density]\label{lem:density-existence}
At every $p\in X$, the limit
\begin{equation}\label{eq:density-definition}
 \vartheta(p)=\lim_{s\downarrow0}s^{-2}\mu(B_s(p))
\end{equation}
exists and is finite. The function $\vartheta$ is measurable and
bounded, and vanishes $dV_g$-almost everywhere.
\end{lemma}

\begin{proof}
Put $\nu=\mu+|Q|\,dV_g$. Then by Cauchy--Schwarz and the volume growth
of a four-dimensional ball,
\[
 \int_{B_s(p)}|Q|\,dV_g
 \le \vol_g(B_s(p))^{1/2}\|Q\|_{L^2(B_s(p))}\le Cs^2,
\]
so $\nu(B_s(p))\le Cs^2$, uniformly in $p$.

Fix $p$, identify a normal-coordinate ball with a Euclidean ball, and
freeze $J_p$ and $G_p$ as constant tensors $J_0,G_0$. Writing
$t=|\zeta|=\dist(p,\exp_p\zeta)$, define
$b_p(s)=s^{-2}T(\mathbf 1_{B_s(p)}G_0)$. These evaluations are
well-defined because $T$ has measure coefficients. Since $G(\ell)=1$,
$G-G_0=O(t)$ and $Q$ annihilates the invariant form $G$,
\begin{equation}\label{eq:density-frozen-ratio}
 b_p(s)=s^{-2}\mu(B_s(p))+O(s),
\end{equation}
because both error integrals are bounded by $Cs^{-2}s\,\nu(B_s(p))$.

In frozen coordinates
$b_p(s)=s^{-2}N(\mathbf 1_{B_p(s)}G_0)+o(1)$. Approximating the
Euclidean ball from inside and outside by continuous test functions and
dominated convergence give
\[
 b_p(s)\longrightarrow\lim_{s\downarrow0}s^{-2}\mu(B_s(p)).
\]
Indeed, the measure of a manifold ball is comparable to the measure of
the corresponding Euclidean ball, the ratio of the two measures is
bounded, and by uniqueness of the differentiation limit the two
expressions have the same limit.

Measurability follows from a fixed sequence of radii tending to zero;
boundedness from \eqref{eq:density-frozen-ratio}. For the last
assertion, localize the measures in a relatively compact coordinate
chart and apply the differentiation theorem for Radon measures
\cite[Theorem~4.19]{Kinnunen2026}. At volume-almost every center, the
measure-to-volume ratio of small Euclidean balls is bounded; a
geodesic ball of radius $s$ lies in a coordinate ball of comparable
radius with smooth volume $O(s^4)$, so
$\mu(B_s(p))=O_p(s^4)$ and $s^{-2}\mu(B_s(p))\to0$. A finite chart
cover finishes the proof.
\end{proof}

Write $N=M\mu$, where $M$ is the measurable positive Hermitian bivector
of trace one obtained by disintegrating $\lambda$: thus
$M(p)=\int\ell\,d\lambda_p(\ell)$ for a probability measure $\lambda_p$
on complex lines, $\mu$-almost everywhere. The following tangent
analysis identifies $M(p)$ as a point mass at a single line, and so
gives a strong form of $\lambda$-almost-everywhere planarity.

\begin{proposition}[Planar tangents]\label{prop:density-tangent-measure}
For $\mu$-almost every $p$, the measure $\lambda_p$ is concentrated on a
single complex line $\ell_p$. Consequently $N$ is a positive (1,1)
current of bidimension $(2,2)$ with a single complex tangent plane
$\lambda$-almost everywhere.
\end{proposition}

\begin{proof}
Fix $p$ with a finite density $\vartheta(p)$ and such that the measure
$\gamma_p$ obtained by the homothety $s^{-2}(z\mapsto p+sz)_*\mu$ has
a tangent measure $\gamma=M(p)\mathbf 1_{P_p}$ in a complex
two-dimensional vector space.
Closedness of $T=N+Q$ passes to the limit as $s\downarrow0$: smooth
test forms converge in $C^1$ after rescaling, and the $Q$ term is
controlled by the gradient estimate of \eqref{eq:density-q-gradient}.
Constant linear maps depend only on finitely many fixed test forms, so
the evaluation condition
\[
 \partial(M(p)\gamma)=0
\]
holds for $\gamma$ as a tensor-valued distribution on $\R^4$.

Identify $M(p)$ with a constant skew-symmetric matrix. The equation
$\partial(M(p)\gamma)=0$ says that all distributional derivatives of
$\gamma$ in the column space of that matrix vanish: equivalently,
$\gamma$ is translation invariant along $\im M(p)$. The nonzero positive
Hermitian $M(p)$ has real rank two or four. Rank four would make
$\gamma$ a nonzero multiple of Lebesgue measure, which contradicts the
ball-mass bound $s^{-2}\mu(B_s(p))\to\vartheta(p)<\infty$. Thus rank is
two. Trace-one normalization makes $M(p)=\ell_p$, the unit bivector of
a complex line $P_p$. The rank-one positive Hermitian $M(p)$ is an
extreme point of the trace-one positive matrices: if its quadratic form
vanishes on a vector, positivity forces almost every matrix in any
convex decomposition to vanish there too. Hence $\lambda_p$ is
concentrated on $\ell_p$.
\end{proof}

\subsection{Closed density weights}

\begin{proposition}[Closed density weights]\label{prop:density-weights}
For either separator construction, let $\vartheta$ be the density of
\Cref{lem:density-existence}. If $b:\R\to\R$ is smooth and bounded with
$b(0)=0$, then the current $b(\vartheta)N$ is closed. The currents
\begin{equation}\label{eq:density-weighted-restrictions}
 N^+=\mathbf 1_{\{\vartheta>0\}}N,\qquad
 I_a=\mathbf 1_{\{\vartheta>a\}}\frac{\pi}{\vartheta}N\quad(a>0)
\end{equation}
are also closed. In particular, if
$N^d=\mathbf 1_{\{\vartheta=0\}}N$, then $N^d+Q$ is closed.
\end{proposition}

\begin{proof}
First smooth the density. Fix nonnegative $h\in C_c^\infty((0,1))$
normalized by $\int_0^1h=1$, and set
\begin{equation}\label{eq:density-smoothed}
 f_s(y)=s^{-2}\int_Xh\bigl(\dist(y,z)^2/s^2\bigr)\,d\mu(z),
 \qquad M_s(z)=\mu(B_s(z)).
\end{equation}
$f_s$ is smooth (the support lies inside the injectivity radius and the
cutoff vanishes near its boundary), and the mass bound makes it
uniformly bounded. Integration against the radial distribution of
$\mu$ gives
\[
 f_s(y)=-\int_0^1h'(u)\frac{\mu(B_{s\sqrt u}(y))}{s^2}\,du
 \longrightarrow-\vartheta(y)\int_0^1u\,h'(u)\,du=\vartheta(y).
\]
There are no atoms by quadratic growth, so no endpoint term appears at
$u=0$. Thus $f_s\to\vartheta$ also volume-almost everywhere.

Fix a smooth one-form $\alpha$. To prove closedness of $b(\vartheta)N$,
apply closedness of $T=N+Q$ before passing to the limit:
\begin{equation}\label{eq:density-weight-product-rule}
 0=T\bigl(d(b(f_s)\alpha)\bigr)
  =T\bigl(b(f_s)d\alpha\bigr)+T\bigl(b'(f_s)\,df_s\wedge\alpha\bigr).
\end{equation}
Dominated convergence gives $b(f_s)N\to b(\vartheta)N$ and
$b(f_s)Q\to0$ as currents (the second uses $b(0)=0$, the a.e.
convergence of $f_s$, and domination by $C|Q|$). It remains to show the
derivative term vanishes; $b'$ is bounded on the common bounded range
of the $f_s$, and the $N$-part is bounded by a constant times
$\int |df_s|_\ell\,d\lambda$, the $Q$-part by a constant times
$\int|Q|\,|df_s|\,dV_g$. We need the two gradient estimates
\begin{equation}\label{eq:density-two-gradient-estimates}
 \int_X|Q|\,|df_s|\,dV_g=o(1),
 \qquad \int|df_s|_\ell\,d\lambda(y,\ell)=o(1).
\end{equation}

\noindent\emph{The derivative against $Q$.} Differentiating the radial
kernel gives $|df_s(z)|\le Cs^{-3}M_s(z)$. Fubini and the volume bound
give $\int_XM_s\,dV_g\le Cs^4$; and $M_s/s^2$ is uniformly bounded and
$\to0$ volume-a.e. Thus
\begin{equation}\label{eq:density-q-gradient}
 s^{-3}\int_X|Q|M_s\,dV_g
 \le\left(\int_X|Q|^2\frac{M_s}{s^2}\,dV_g\right)^{1/2}
    \left(s^{-4}\int_XM_s\,dV_g\right)^{1/2}=o(1),
\end{equation}
the first factor by dominated convergence and the second by
boundedness. This proves the first estimate of
\eqref{eq:density-two-gradient-estimates}.

\noindent\emph{The tangential derivative.} Closedness converts
tangential differentiation into normal derivatives; the radial kernel
then supplies a transverse displacement factor multiplied by the
difference of nearby line directions, and the angular-energy bound
controls the two together. Fix $(y,\ell)$; choose an orthonormal basis
of $T_yX$ with first two vectors spanning $\ell$, and normal
coordinates $\zeta_1,\ldots,\zeta_4$. Write
$k_s^y(z)=h(|\zeta|^2/s^2)$, $q_i=\partial_{\zeta_i}k_s^y$,
$\omega_{12}=d\zeta_1\wedge d\zeta_2$. First variation of squared
distance yields
\begin{equation}\label{eq:density-first-variation}
 df_s(y)(e_i)=-s^{-2}\int q_i(z)\,d\mu(z),\qquad i=1,2.
\end{equation}
Replace this scalar integral by $T(q_i\omega_{12}) = (N+Q)(q_i\omega_{12})$:
on $B_s(y)$, normal coordinates and parallel transport differ by
$O(s^2)$, so
\[
 \omega_{12}(m)=\langle\ell,\tau_{zy}m\rangle+O(s^2),\qquad
 |1-\omega_{12}(m)|\le C\bigl(s^2+|\ell-\tau_{zy}m|^2\bigr).
\]
Since $|q_i|\le C/s$, the integrated replacement error is at most
\begin{equation}\label{eq:density-replacement-error}
 s^{-2}\!\int\!\Bigl|\int q_i d\mu-T(q_i\omega_{12})\Bigr|d\lambda
 \le Cs^{-3}\Bigl(s^2\!\iint_{\dist<s}\!d\mu\,d\mu+\mathcal A_s\Bigr)
 +Cs^{-3}\!\int_X\!|Q|M_s\,dV_g=o(1),
\end{equation}
where the double mass is $O(s^2)$, $\mathcal A_s$ is the angular
integral of $O(s^4)$, and the last term tends to zero by
\eqref{eq:density-q-gradient}.

For $i=1$, closedness of $T$ on $d(k_s^y d\zeta_2)$ gives
$T(q_1\omega_{12})=-\sum_{a=3}^4T(q_a\,d\zeta_a\wedge d\zeta_2)$;
for $i=2$ the analogous identity holds for $d(k_s^y d\zeta_1)$. These
are legitimate smooth compactly supported tests because the cutoff
vanishes near the boundary of the coordinate ball. The mixed forms
satisfy, for $k=1,2$ and $a=3,4$,
\[
 |(d\zeta_a\wedge d\zeta_k)(m)|\le C\bigl(s^2+|\ell-\tau_{zy}m|\bigr),
\]
and the normal derivative of the radial kernel obeys
$|q_a(z)|\le Cs^{-1}|\operatorname{pr}_{\ell^\perp}\zeta/s|$.
Replacing $T$ by $N+Q$, the $Q$-contribution again tends to zero by
\eqref{eq:density-q-gradient}, the $s^2$ geometric error is $O(s)$, and
Cauchy--Schwarz bounds the rest by
\begin{equation}\label{eq:density-angular-transverse-product}
 Cs^{-3}\mathcal A_s^{1/2}\mathcal T_s^{1/2}=s^{-3}\cdot O(s^2)\cdot
 o(s)=o(1).
\end{equation}
Combining \eqref{eq:density-first-variation},
\eqref{eq:density-replacement-error}, and
\eqref{eq:density-angular-transverse-product} proves the second estimate
of \eqref{eq:density-two-gradient-estimates}.

\noindent\emph{The closed weighted limit.} The two derivative estimates
permit the limit in \eqref{eq:density-weight-product-rule}: the
derivative term $\to0$, the first term $\to b(\vartheta)N(d\alpha)$, so
this last vanishes for every smooth $\alpha$---$b(\vartheta)N$ is
closed, as claimed.

For the remaining weights, take a smooth nondecreasing
$\chi:\R\to[0,1]$ vanishing on $(-\infty,0]$ and equal to one on
$[1,\infty)$; the bounded functions $\chi(nt)\to\mathbf 1_{\{t>0\}}$
pointwise on $[0,\infty)$ and vanish at zero. Fix $a>0$ and set
\[
 b_{a,n}(t)=
 \begin{cases}
 \pi\,\chi(n(t-a))/t,&t>0,\\
 0,&t\le0.
 \end{cases}
\]
These are smooth, vanish near zero, are uniformly bounded by $\pi/a$,
and converge at every $t\ge0$ to $\mathbf 1_{\{t>a\}}\pi/t$. Dominated
convergence and closedness of each smooth-weight current prove
\eqref{eq:density-weighted-restrictions}. No derivative bound uniform
in $n$ is used: the smoothing limit was taken first for each $n$.
Finally $N^d+Q=T-N^+$ is closed.
\end{proof}

\subsection{Vanishing of the diffuse error and the compatible form}

On the zero-density set the negative kernel error vanishes identically
in the limit. The resulting intersection pairing first closes the
compatibility question and then provides the nonnegative diffuse square
needed for the class criterion.

\begin{lemma}[Vanishing diffuse kernel error]\label{lem:density-diffuse-error}
Set $B_s(z)=s^{-2}\int_X(1+\dist(y,z)/s)^{-6}\,d\mu(y)$. The functions
are uniformly bounded, and $B_s(z)\to0$ whenever $\vartheta(z)=0$. In
particular,
\begin{equation}\label{eq:density-diffuse-error}
 \int_XB_s(z)\,d\mu^d(z)\longrightarrow0,\qquad
 \mu^d=\mathbf 1_{\{\vartheta=0\}}\mu.
\end{equation}
\end{lemma}

\begin{proof}
For each dyadic shell, quadratic growth gives the summable bound
$s^{-2}(1+2^j)^{-6}\mu(B_{2^{j+1}s}(z))\le C\,2^{-4j}$. The same
estimate holds once the radius exceeds a fixed small radius, by
increasing $C$ and using finite total mass. This proves uniform
boundedness. If $\vartheta(z)=0$, each fixed shell contribution tends to
zero because $s^{-2}\mu(B_{2^{j+1}s}(z))\to0$; the summable bound then
gives pointwise convergence. A second application of dominated
convergence proves \eqref{eq:density-diffuse-error}.
\end{proof}

For closed currents $S,S'$, the smoothing operator
$S_s=(1+s^2\Delta_g)^{-3}$ preserves cohomology classes and commutes
with the Hodge star. Hodge decomposition therefore gives
\begin{equation}\label{eq:density-smoothed-intersection}
 \langle S_sS,*S_sS'\rangle_{L^2}=\mathfrak c(S)\cdot\mathfrak c(S'),
\end{equation}
independently of $s$; the Sobolev regularity after smoothing is
sufficient for this identity.

\begin{proof}[Proof of \Cref{thm:tamed-compatible}]
Suppose that no compatible symplectic form exists. The separator of
\Cref{lem:current-separators} gives a nonzero positive current $N$,
normalized to trace mass one, and anti-invariant $Q$ such that
$T=N+Q$ annihilates all smooth closed two-forms; in particular $T$ is
closed with $\mathfrak c(T)=0$. By \Cref{prop:density-weights}, $T^d=N^d+Q$
is closed. \eqref{eq:density-smoothed-intersection} gives
\begin{align}
 0&=\langle S_sT,*S_sT^d\rangle\notag\\
 &=\langle S_sN,*S_sN^d\rangle+\langle S_sN,S_sQ\rangle
  +\langle S_sQ,S_sN^d\rangle+\|S_sQ\|_2^2.
 \label{eq:density-compatibility-pairing}
\end{align}
The self-duality of $Q$ accounts for the signs. Both mixed terms tend to
zero by \Cref{prop:angular-energy}. The positive-current lower bound of
\Cref{lem:current-positive-pairing}, applied with second submeasure
$\lambda^d=\mathbf 1_{\{\vartheta=0\}}\lambda$, gives
$\langle S_sN,*S_sN^d\rangle\ge-C\int_XB_s\,d\mu^d=o(1)$. Since
$S_sQ\to Q$ in $L^2$, taking a limit inferior in
\eqref{eq:density-compatibility-pairing} forces $\|Q\|_2^2=0$. Thus
$T=N$ is a nonzero positive current annihilating all closed forms.

Let $\eta$ be a taming symplectic form. Compactness of the bundle of
unit complex lines and strict taming give $\eta_y(\ell)\ge c_\eta>0$, so
$N(\eta)\ge c_\eta\mu(X)>0$, contradicting the annihilation property.
This proves existence. The argument has used neither rectifiability nor
a theorem representing integral cycles by curves.
\end{proof}

\begin{lemma}[Nonnegative diffuse square]\label{lem:density-diffuse-square}
If $Q=0$, then $N^d$ is closed and $\mathfrak c(N^d)^2\ge0$.
\end{lemma}

\begin{proof}
Closedness is \Cref{prop:density-weights}. By
\eqref{eq:density-smoothed-intersection} and
\Cref{lem:current-positive-pairing},
\[
 \mathfrak c(N^d)^2=\langle S_sN^d,*S_sN^d\rangle
 \ge-C\int_XB_s\,d\mu^d,
\]
and the right side tends to zero by \Cref{lem:density-diffuse-error}.
\end{proof}

\subsection{Integral thresholds and the curve criterion}

To use the hypothesis on curve classes, we must turn the positive-density
part into integral cycles. The factor $\pi/\vartheta$ in $I_a$ is
chosen to give multiplicity one.

\begin{lemma}[Threshold cycles]\label{lem:density-integral-thresholds}
For every $a>0$, the current $I_a$ from
\Cref{eq:density-weighted-restrictions} is an integral two-cycle with
positive $J$-complex tangent orientation. If $J$ is tamed, it is a
finite sum of currents of integration over irreducible closed
$J$-holomorphic curves, with positive integer multiplicities. The
identity
\begin{equation}\label{eq:density-layer-cake}
 N^+=\frac1\pi\int_0^\infty I_a\,da
\end{equation}
holds as an absolutely convergent integral of currents.
\end{lemma}

\begin{proof}
The density $\vartheta$ is bounded and vanishes with $dV_g$-zero mean.
By \Cref{prop:density-tangent-measure}, $\lambda$-almost everywhere the
current has a single complex tangent line with weight $\vartheta$. The
normalization of $\lambda$ makes $(\operatorname{pr}_X)_*\lambda=\mu$;
hence the trace measure of $I_a$ equals
$\mathbf 1_{\{\vartheta>a\}}\pi\,dV_g$ and is rectifiable. Preiss's
rectifiability theorem \cite{Preiss1987} upgrades the a.e.~planar
tangents to rectifiability; closedness of $I_a$ and positivity then
express it as an integral sum of simple curves. Positivity of the
$J$-complex orientation is pointwise: the tangent plane is the chosen
complex line. Taming prevents a nonzero closed component with infinite
mass; factoring each component through its somewhere-injective image
gives a finite sum of irreducible closed curves with positive integer
multiplicities.

Finally, pointwise on $\{\vartheta>0\}$,
$\int_0^\infty\mathbf 1_{\{\vartheta>a\}}\pi/\vartheta\,da=\pi$; for a
smooth two-form $\alpha$, Fubini is justified by
\[
 \int_0^\infty|I_a(\alpha)|\,da\le\pi\|\alpha\|_{C^0}\mu^+(X).
\]
This proves \eqref{eq:density-layer-cake} with its stated normalization.
\end{proof}

\begin{proof}[Proof of \Cref{thm:taming-cone}]
Assume that $H$ has no taming representative. The class separator of
\Cref{lem:current-separators}\textup{(ii)} gives a nonzero closed
positive current $N$, with $Q=0$, satisfying
\begin{equation}\label{eq:density-cone-separator}
 H\cdot\mathfrak c(N)\le0.
\end{equation}
Its positive-density and diffuse parts are separately closed by
\Cref{prop:density-weights}. Anti-invariance of $A,C$ gives
$\mathfrak c(N^d)\cdot[A]=N^d(A)=0=\mathfrak c(N^d)\cdot[C]$; thus
$w=\mathfrak c(N^d)$ lies in $V$ and $w^2\ge0$ by
\Cref{lem:density-diffuse-square}.

If $N^d\ne0$, choose a taming form $\eta$ in $U$. Strict taming on the
compact bundle of unit complex lines gives $U\cdot w=N^d(\eta)>0$; in
particular $w\ne0$. Since $[A],[C]$ span a positive two-plane and
$b^+=3$, the intersection form on $V$ has signature $(1,b^-)$. The
conditions $w^2\ge0$ and $U\cdot w>0$ put $w$ in the nonzero closed time
cone determined by $U$. Every timelike $H$ in that component has
$H\cdot w>0$: writing $H=h_0u+h_-$, $w=w_0u+w_-$ with a unit timelike
$u$ and orthogonal negative parts, $h_0>|h_-|$, $w_0\ge|w_-|$, $w_0>0$,
so $H\cdot w\ge h_0w_0-|h_-|\,|w_-|>0$. If $N^d=0$, its pairing is
instead zero; this distinguishes the nonzero diffuse case needed for
strict positivity.

For every $a>0$, \Cref{lem:density-integral-thresholds} writes $I_a$ as
a positive integral sum of irreducible closed $J$-curves; the hypothesis
on $H$ gives $H\cdot\mathfrak c(I_a)\ge0$, with strict inequality
whenever $I_a\ne0$. If $N^+\ne0$, the layer identity has positive total
mass, so the set of $a$ for which $I_a\ne0$ has positive Lebesgue
measure; pairing \eqref{eq:density-layer-cake} with a smooth
representative of $H$ gives $H\cdot\mathfrak c(N^+)>0$. The pairing is
zero when $N^+=0$. At least one of $N^d,N^+$ is nonzero, so
$H\cdot\mathfrak c(N)=H\cdot\mathfrak c(N^d)+H\cdot\mathfrak c(N^+)>0$,
contradicting \eqref{eq:density-cone-separator}. A smooth closed taming
representative is symplectic by pointwise nondegeneracy.
\end{proof}

\section{The coefficient sphere and curves through a point}
\label{sec:families}

We now leave the gauge theory behind and turn to the geometry of the
original triple. The coefficient sphere $S^2$ of a normalized triple
parametrizes a whole sphere of almost-complex structures $J_v$. The
fixed spin-c structure of any one of them can be carried over the
sphere, and the Seiberg--Witten invariants of this family give, in the
large-parameter limit, holomorphic curves in any prescribed nonzero
integral class with zero self-intersection. The main output of this
section is that those curves pass through every prescribed point.

Curve classes are identified with their Poincar\'e duals; an integral
class written in $\Lambda=H^2(X;\Z)/\mathrm{torsion}$ may be lifted to
$H^2(X;\Z)$ when specifying a line bundle, and curve configurations
and their total classes are only ever used modulo torsion. By
\Cref{prop:topology},
\begin{equation}\label{eq:families-topology}
 b^+=3,\qquad
 (b_1,\sigma)=(0,-16)\ \text{or}\ (4,0),\qquad
 2\chi+3\sigma=0.
\end{equation}
Let $J_v$, $v\in S^2$, be the coefficient sphere of
\Cref{lem:twistor-degree}, with the sign chosen so that
$B_v=v\cdot\omega$ tames $J_v$. Write
$P=\operatorname{span}\{[\omega_1],[\omega_2],[\omega_3]\}$ for the
original positive period plane.

A \emph{positive curve configuration} is a finite sum of irreducible
closed pseudoholomorphic curves with positive integer multiplicities.

\begin{theorem}\label{thm:family-curves}
Let $F\in\Lambda$ be nonzero with $F^2=0$, and write
$F^+=\operatorname{pr}_PF$. For every $p\in X$, there exists a finite
positive $J_v$-holomorphic configuration of total class $F$ whose
support contains $p$. Its parameter is necessarily the coefficient
direction
\[
 v=\frac{F^+}{|F^+|},
\]
where $P$ is identified with $\R^3$ by the normalized ordered period
basis. Thus the almost-complex structure is the same for all
prescribed points.
\end{theorem}

The strategy is: (a) build a smooth $J_v$-compatible family $\eta_v$
with normalized degree one, (b) use the family Seiberg--Witten
invariants to force a zero of the first spinor component at the
prescribed point, and (c) extract a $J_v$-holomorphic curve from the
large-parameter limit. Steps (a) and (c) are geometric; step (b) is
the wall-crossing computation for the fixed spin-c structure.

\subsection{A smooth compatible family}

\begin{lemma}\label{lem:compatible-family}
There is a smooth family of closed $J_v$-compatible forms
$\eta_v$. Orthogonal projection to $P$ satisfies
\[
 \operatorname{pr}_P[\eta_v]=a(v)[B_v],\qquad a(v)>0,
\]
and consequently the normalized period projection of this family has
degree one in the coefficient coordinates.
\end{lemma}

\begin{proof}
Fix a metric $g_0$ in the conformal class determined by the original
triple. All $J_v$ are orthogonal for $g_0$, and the three closed
self-dual forms $\omega_i$ span its self-dual harmonic forms because
$b^+=3$. Denote by $R_v$ the projection to the $J_v$-anti-invariant
two-forms, and consider $L_v^{\mathrm{ell}}=R_v\dd\dd^*$ on the
anti-invariant forms, where the adjoint is taken with respect to
$g_0$. This is the nonnegative elliptic operator in
\Cref{lem:closed-lift}, with kernel the closed anti-invariant forms;
the kernel is smooth in $v$ and has rank two over $S^2$ by
\Cref{lem:twistor-degree}.

Compatibility is equivalent to taming by a closed $-J$-invariant form.
By \Cref{thm:tamed-compatible}, the compatible cone is nonempty for
$J_v$; compactness of $S^2$ and the smooth dependence of the operator
kernel on $v$ then give a smooth choice of compatible forms $\eta_v$.
Each $\eta_v$ is $J_v$-invariant and closed, so it is orthogonal to
the anti-invariant classes. Therefore its projection to $P$ is a
multiple of $[B_v]$. Moreover
$\int_X\eta_v\wedge B_v>0$: the invariant part of $B_v$ is positive,
and its anti-invariant part wedges trivially with $\eta_v$. Since
$[B_v]^2=1$, the multiple is positive. The remaining assertions follow
from the fact that $\operatorname{pr}_P$ acts as the identity on the
unit sphere of $P$.
\end{proof}

\subsection{The fixed spin-c structure and the equations}

Fix the canonical $\mathrm{Spin}^c$ structure of one $J_v$, with
isomorphism type $\mathfrak s_0$. The structures $J_v$ form a
connected family, so all their canonical structures have the same
fiberwise isomorphism type. For $E\in H^2(X;\Z)$, let
$\mathfrak s_E=\mathfrak s_0\otimes L_E$, where
$c_1(L_E)=E$; the family extends over the contractible space of
metrics with no twist from the parameter base. Its determinant class
and dimensions are
\begin{equation}\label{eq:families-dimensions}
 c_1(\det\mathfrak s_E)=2E,\qquad d(\mathfrak s_E)=E^2,\qquad
 \operatorname{ind}_{\C}\not D_E=\frac{4E^2-\sigma}{8}.
\end{equation}
Fix a homology orientation once and for all.

\subsection{Wall-crossing at the harmonic wall}

\begin{lemma}[Wall-crossing]\label{lem:families-wall-crossing}
Let $\mathfrak s$ be the fixed spin-c structure of the family $J_v$
over $S^2$, and let $q=d(\mathfrak s)/2$ be its expected complex
dimension. Then the family invariant in the chamber after crossing the
harmonic wall gives
\begin{equation}\label{eq:families-wall-formula}
 \operatorname{FSW}_{\mathfrak s}(S^2)
  =s_{b_1/2}(K-O),
\end{equation}
where $K,O$ are the ranks of the complex Dirac index, and
$\operatorname{FSW}$ denotes the family Seiberg--Witten invariant of
the universal spinor evaluation.
\end{lemma}

\begin{proof}
The critical wall-crossing theorem and its section-class version are
\cite[Theorem~4.10 and Proposition~4.11]{LiLiu2001}; we only check
the normalization and degree. A transverse homotopy between two
families meets the harmonic wall at isolated parameters in
$S^2\times[0,1]$, because its normal rank is $b^+=3$. Above such a
parameter the reducibles form the torus $\operatorname{Pic}^0(X)$.
Represent the complex Dirac index on this torus as $K-O$, with ranks
$k,o$, using a stabilization when necessary. Write
$x=c_1(\mathcal O(1))$ on the projective bundle of lines
$\pi:\mathbb P(K)\to\operatorname{Pic}^0(X)$. The obstruction bundle
contributes $e(\mathcal O(1)\otimes O)=\sum_ix^{o-i}c_i(O)$.
Projective pushforward therefore gives
\begin{align}
 \pi_*\Bigl(x^q\sum_ix^{o-i}c_i(O)\Bigr)
 &=\sum_is_{q+o-(k-1)-i}(K)c_i(O)\notag\\
 &=s_{b_1/2}(K-O),
 \label{eq:families-projective-pushforward}
\end{align}
because \eqref{eq:families-topology}\eqref{eq:families-dimensions}
imply $q+o-(k-1)=b_1/2$. The normal-crossing signs sum to the degree
difference, giving \eqref{eq:families-wall-formula}.

The based-gauge normalization makes the universal flat line trivial
at $p$; no Picard-base line is added to $x$. Writing a determinant
connection as a canonical connection plus twice a twisting connection
does not alter this conclusion: the remaining circle acts on the
spinor with weight one, as used in
\eqref{eq:families-projective-pushforward}.
\end{proof}

\subsection{The evaluation line over the coefficient sphere}

For $E^2=0$, write $\operatorname{FSW}_E(U_p)$ for the spherical
pairing of the based-gauge evaluation line $U_p$ in the large-parameter
chamber, and $\operatorname{SW}_X(E)$ for the ordinary dimension-zero
invariant of $\mathfrak s_E$. Denote by $\pi$ the projection of the
regular family moduli space to $S^2$.

\begin{lemma}[Evaluation identity]\label{lem:families-evaluation-line}
The evaluation line has first Chern class $U_p+e\pi^*h$, where
$h=c_1(\mathcal O(1))$ is the twistor parameter. Its spherical pairing
satisfies
\begin{equation}\label{eq:families-evaluation-identity}
 \operatorname{FSW}_E(U_p)=U_p\bigl([M_E^{\mathrm{reg}}]\bigr)
 =\operatorname{SW}_X(E)+e\operatorname{FSW}_{\mathfrak s_E}(S^2).
\end{equation}
For every prescribed point $p$, if $\operatorname{FSW}_E(U_p)\ne0$
then the exact family has an evaluation zero at $p$.
\end{lemma}

\begin{proof}
The evaluation line over the coefficient sphere is a line bundle
$\mathcal O(1)\otimes\mathcal E_E|_p$. The evaluation transforms with
the same gauge weight as the spinor, and its sign on a reducible link
can be checked on $\ell\to\mathcal E_E|_p$: it lies in
$\operatorname{Hom}(\ell,\mathcal E_E|_p)=\mathcal O(1)\otimes
\mathcal E_E|_p$. On the quotient, the evaluation section is a
section of the tensor product of the based-gauge evaluation line with
the pullback of $\mathcal E_E|_{\{p\}\times S^2}$; its first Chern
class is $U_p+e\pi^*h$. Pairing $\pi^*h$ with the family moduli cycle
gives $\operatorname{SW}_X(E)$; equivalently, restrict the family to a
regular base point. Formula \eqref{eq:families-evaluation-identity}
is now immediate.

For the last assertion, suppose the exact family had no evaluation
zero. Fixed-$r$ compactness (see \Cref{lem:families-taubes-compactness})
would give the same absence for all sufficiently small perturbations,
because otherwise a sequence of their zeros would limit to an exact
zero. Applying the preceding identity to a regular such perturbation
gives a contradiction.
\end{proof}

\subsection{Compactness and the family curves}

\begin{lemma}[Taubes compactness]\label{lem:families-taubes-compactness}
Fix a compact smooth parameter family of backgrounds and a compact
smooth family of allowed choices of taming pairs, and let $r\to\infty$.
After passing to a subsequence, the solutions of the perturbed
Seiberg--Witten equations converge to a positive $J_\infty$-holomorphic
configuration of the prescribed class; the support contains all limit
points of first-component zeros.
\end{lemma}

\begin{proof}
This is the standard compactness argument of Taubes
\cite[Proposition~5.2 and Section~6]{Taubes1996}, with the essential
uniform estimate \eqref{eq:families-taubes-off-zeros} below. The
first-component estimate, the off-support decay, the flux
identification, and the disk lower bound are carried out in the
original \cite[Equations~(7.1)--(7.5) and Lemma~7.1]{Taubes1996}; the
uniform estimates hold because the smooth background family is
compact and the choices of $r$ are made after the background and the
approximation tolerance are fixed.
\end{proof}

We state the two standard inputs we use explicitly. First, the first
spinor component $a_n$ of a solution has a uniform integral two-form
$T_n$ with bounded mass and periods on closed classes equal to those
of $E$:
\begin{equation}\label{eq:families-taubes-current}
 T_n(\zeta)=\frac{i}{2\pi}\int_XF_{a_n}\wedge\zeta,
 \qquad
 |T_n|\le C(r_n)
\end{equation}
for a constant $C(r_n)$ that stays bounded in $r_n$; the identification
follows from the topological identity for the determinant line. Second,
off the first-component zero set, the $L^1$-to-$C^1$ distance from the
canonical connection is
\begin{equation}\label{eq:families-taubes-off-zeros}
 |\mathcal A-\mathcal A_{\mathrm{can}}|
 +|F_{\mathcal A}-F_{\mathcal A_{\mathrm{can}}}|
 \le Cr^{-1}+Cr\exp\bigl(-\sqrt r\,
          \dist(x,\alpha^{-1}(0))/C\bigr)
\end{equation}
when the distance is at least $r^{-1/2}$
\cite[Proposition~6.1]{Taubes1999}.

Fix the disk bound. Write
$f_n=\frac{i}{2\pi}F_{a_n}
      =\frac{i}{4\pi}(F_{\mathcal A_n}-F_{\mathcal A_{\mathrm{can},n}})$.
The sharp estimate
\cite[Proposition~4.2, Equation~(4.9)]{Taubes1999}, specialized to
$|\eta_n|=\sqrt2$, is
\begin{equation}\label{eq:families-taubes-sharp}
 |F_{\mathcal A_n}^-|\le\frac{r_n}{2\sqrt2}(1-|\alpha_n|^2)+C.
\end{equation}
Its constants depend on the background family in a fixed smooth way.
If $\ell$ is a unit, positively oriented $J_n$-complex bivector, then
$\ell^+=\eta_n/2$, $|\ell^-|=1/\sqrt2$, and the curvature equation
together with the second displayed identity give the disk bound
\begin{equation}\label{eq:families-disk-lower-bound}
 u^*f_n\ge-C_0\,\dd A_u
\end{equation}
for every $J_n$-holomorphic disk $u$, with $C_0\ge1$ independent of $n$
and of the parameter.

\begin{proof}[Proof of \Cref{thm:family-curves}]
The projection $F^+$ is nonzero: otherwise $F$ would be a nonzero
square-zero vector in the negative-definite space $P^\perp$. Lift $F$
to an integral class and use the fixed spin-c structure
$\mathfrak s_F$.

The ordinary invariant vanishes. Choose the opposite coefficient
direction $w=-F^+/|F^+|$, so that $F[B_w]<0$. If
$\operatorname{SW}_X(F)\ne0$, ordinary invariance and fixed-parameter
monopole compactness would give solutions of the equations for
arbitrarily large $r$ at this fixed background. Taubes compactness
would give a nonempty positive $J_w$-holomorphic configuration of
class $F$, contradicting its negative $B_w$-area.

The spherical $U_p$ pairing is nonzero. For $b_1=0$ this follows
directly from \Cref{lem:families-wall-crossing} with $K-O=0$; for
$b_1=4$, the projective pushforward \eqref{eq:families-projective-pushforward}
reduces the family invariant to the same computation.
Since the ordinary invariant vanishes,
\Cref{lem:families-evaluation-line} forces a first-component zero over
$p$ in the exact family for each sufficiently large $r$. Extracting a
convergent parameter subsequence and applying
\Cref{lem:families-taubes-compactness} gives a configuration through
$p$ in some $J_v$, with total class $F$.

The two closed forms whose coefficients are perpendicular to $v$ are
$J_v$-anti-invariant and restrict to zero on every component of this
configuration. Thus $F^+$ is parallel to $[B_v]$. Its $B_v$-area is
positive, so the proportionality factor is positive, which proves the
asserted direction.
\end{proof}

\subsection{Root exclusion}

The unmarked wall calculation also excludes the square-minus-two
classes needed in the lattice argument.

\begin{proposition}\label{prop:root-exclusion}
If $b_1=0$, no class $E\in\Lambda$ with $E^2=-2$ is orthogonal to $P$.
\end{proposition}

\begin{proof}
Lift $E$ to an integral line-bundle class. Its ordinary expected
dimension is $-2$, and its spherical family has dimension zero. By
\Cref{lem:families-wall-crossing}, its large-parameter unmarked
family invariant is $\pm1$. For each sufficiently large $r$, take
small regular perturbations and then a fixed-$r$ limit to obtain an
actual solution of the equations at some parameter $v_r$. Pass to a
sequence with $v_r\to v$, and apply
\Cref{lem:families-taubes-compactness}. The limit is a nonempty
positive $J_v$-holomorphic configuration of class $E$. Every
nonconstant component has positive $B_v$-area, since $B_v$ tames
$J_v$. This contradicts $E[B_v]=0$, which follows from $E\perp P$.
\end{proof}

\section{An integral null class with a uniform chamber}
\label{sec:lattice-chamber}

Curve existence (\Cref{thm:family-curves}) is available for every
nonzero integral class with $F^2=0$. What is needed now is one such
class for which a positive curve configuration cannot split into
components of smaller degree. This choice uses only the original
positive period plane $P$, so the resulting restrictions on curve
classes survive every later pair deformation that preserves the two
periods. The argument has three parts: (a) a lattice-theoretic choice
of a primitive null class $F$ avoiding finitely many obstructions;
(b) adjunction and positivity of intersections, which together turn
the lattice properties into geometric properties of the curves; and
(c) an orthogonal adaptation of the coefficients making $[B]$ the
direction of $F^+$.

\subsection{The lattice and the chamber class}

Let
\[
 \Lambda=H^2(X;\Z)/\mathrm{torsion},\qquad
 \Lambda_{\Q}=\Lambda\otimes_{\Z}\Q,\qquad
 \Lambda_{\R}=\Lambda\otimes_{\Z}\R.
\]
A \emph{curve class} is the image in $\Lambda$ of the integral
Poincar\'e dual of a closed oriented curve, with its complex
orientation. All equalities of integral classes in this section are
equal to torsions; this does not affect intersections or integrals of
closed real forms. The original positive period plane is $P$, and
$d^+$ denotes the orthogonal projection of $d\in\Lambda_{\R}$ to $P$.

\begin{proposition}[Choice of chamber class]\label{prop:chamber-class}
There is a primitive class $F\in\Lambda$ with $F^2=0$ such that the
following implication holds for every $d\in\Lambda$ and every real
number $c>0$:
\begin{equation}\label{eq:lattice-chamber-implication}
 d^2\ge-2,\qquad d^+=cF^+,\qquad d\cdot F\le0
 \quad\Longrightarrow\quad
 c\in\Z_{>0},\qquad d\cdot F=0.
\end{equation}
Moreover $F^+\ne0$.
\end{proposition}

\begin{proof}
\noindent\emph{Step~1: a rational null subspace with many integral
directions.} Because $P^\perp$ is negative definite and $\dim P=3$,
the restriction of the intersection form to the $(3,b^-+1)$-dimensional
subspace $P^\perp\oplus\R F_0$ for a null $F_0\in\Lambda$ is indefinite,
so it contains an integral null vector. Take a rational null subspace
$L$ of $\Lambda$ of maximal dimension, namely dimension three: the
intersection form restricted to $L_{\R}$ has signature $(1,2)$ and
intersects $\Lambda$ in a lattice. Let $l\in L$ be a primitive vector
with $l\cdot l=0$ and such that $L_{\R}=l^\perp\cap L_{\R}\oplus\R l$.
Let
\[
 \mathcal F=\{F\in L\cap\Lambda:\ F^2=0,\ F\hbox{ primitive},\
          F\cdot l=1\}.
\]
Each $F\in\mathcal F$ is a nonzero null vector and belongs to $L$. Since
$l$ is null and $F\cdot l=1$, we have $F^+\notin P^\perp$: otherwise
$F\cdot l=F^+\cdot l+F^-\cdot l=F^+\cdot l=F^+\cdot l$ and $F^+\in P$
would force $F^+\cdot l=0$, a contradiction. In particular $F^+\ne0$.
The set $\mathcal F$ is infinite by completing any rational null vector
to a hyperbolic pair with $l$.

\noindent\emph{Step~2: projecting and pinning $d^+$.} Fix $F\in\mathcal F$
and suppose $d\in\Lambda$, $c>0$ satisfy the three left-hand conditions
of \eqref{eq:lattice-chamber-implication}. Then
\[
 r:=d-cF\in P^\perp,\qquad r^2=d^2-2c(d\cdot F)\ge-2,
\]
because $F\cdot F=0$, so $r$ lies in the compact set
\[
 K=\{r\in P^\perp:r^2\ge-2\}.
\]

\noindent\emph{Step~3: only finitely many cosets.} Let
$q:\Lambda_{\R}\to\Lambda_{\R}/L_{\R}$ be the quotient map. Because
$L_{\R}$ is rational and $L$ is saturated, a basis of $L$ extends to an
integral basis of $\Lambda$; the remaining basis vectors project to a
lattice basis in the real quotient. Hence $q(\Lambda)$ is discrete and
$\ker(q|_\Lambda)=L$. Since $q(d)=q(r)\in q(K)$, only finitely many
cosets $d+L$ can occur in the displayed expression for $r$. This finite
set is independent of $F$.

\noindent\emph{Step~4: avoiding irrational points on a torus.} Orthogonal
projection restricts to an isomorphism
\[
 \pi:L_{\R}\longrightarrow P:
\]
its kernel is null and lies in the negative-definite space $P^\perp$,
and both spaces have dimension three. Hence $\pi(L)$ is a lattice in
$P$. Each of the finitely many cosets just found determines one point of
\[
 \mathbb T=P/\pi(L),
\]
because adding an element of $L$ changes the projection by an element of
$\pi(L)$. For $F\in\mathcal F$ define the circle
\[
 \mathbb S_F=\{t\pi(F)+\pi(L):t\in\R\}\subset\mathbb T.
\]
Two such circles with different directions intersect only at torsion
points: if $t\pi(F)-s\pi(F')\in\pi(L)$ with $F,F'$ not collinear, choose
an integral linear functional on $L$ vanishing on $F$ and nonzero on
$F'$; applying it to $tF-sF'\in L$ forces $s\in\Q$, so the common point
is torsion. In particular a non-torsion point of $\mathbb T$ lies on at
most one circle $\mathbb S_F$.

The elements of $\mathcal F$ have distinct directions: two collinear
elements with $l$-pairing one are equal. Since $\mathcal F$ is infinite,
we can choose $F\in\mathcal F$ whose circle contains none of the
non-torsion points among the finitely many possible coset points.

\noindent\emph{Step~5: integrality and vanishing.} For this $F$, every
$d,c$ satisfying the hypotheses gives a torsion point
\[
 d^++\pi(L)=c\pi(F)+\pi(L).
\]
So there is an integer $k>0$ with $kcF\in L$. Pairing with $l$ shows
$kc\in\Z$, so $c\in\Q$. It follows that $r=d-cF\in P^\perp\cap\Lambda_\Q$,
and the lattice property of $P^\perp$ now gives
\begin{equation}\label{eq:lattice-integral-coefficient}
 c=l\cdot d\in\Z_{>0}.
\end{equation}
If $m=d\cdot F<0$, then $m$ is a negative integer and
\[
 r^2=d^2-2cm\ge-2+2c\ge0.
\]
Negative definiteness of $P^\perp$ forces $r=0$, which would imply
$m=0$. Contradiction; hence $d\cdot F=0$, and the implication is proved.
\end{proof}

\subsection{Curve signs}

Fix $F$ as in \Cref{prop:chamber-class}. Make a constant orthogonal
change of the coefficient basis of the original triple and denote the
result again by $(A,C,B)$, choosing
\begin{equation}\label{eq:lattice-adapted-periods}
 [B]=\frac{F^+}{\lVert F^+\rVert},\qquad
 [A]\cdot F=[C]\cdot F=0,\qquad
 \beta=[B]\cdot F=\lVert F^+\rVert>0.
\end{equation}
Here the norm is the positive intersection norm on $P$. The three
classes remain orthonormal, and $V=[A]^\perp\cap[C]^\perp$ has
signature $(1,b^-)$. All later statements are made for arbitrary
definite pairs with these two periods $[A],[C]$; the choice is fixed by
$F$ and the original period plane $P$, independent of any later pair.

\begin{corollary}[Curve signs]\label{cor:curve-signs}
Let $J$ be an almost-complex structure with definite pair $(A,C)$
satisfying the adapted periods \eqref{eq:lattice-adapted-periods} and
tamed by $B$. For every irreducible closed $J$-holomorphic curve with
class $d\in\Lambda$ (identified with its Poincar\'e dual), let
$m=F\cdot d$ and write $[B]\cdot d=\beta c$ with the constant
$\beta=[B]\cdot F>0$. Then:
\begin{enumerate}[label=\textup{(\roman*)}]
\item $m\ge0$.
\item $m=0$ if and only if $d=cF$ for some real $c$ (in which case
$c\in\Z$, and $c>0$ if $d\ne0$).
\end{enumerate}
\end{corollary}

\begin{proof}
By \Cref{lem:definite-pair}, the definite pair trivializes the
canonical bundle, so $c_1(TX,J)=0$ for either sign of $J$. The
standard adjunction and singularity theorem
\cite[Theorem~2.8]{Wendl2020} give
\begin{equation}\label{eq:lattice-adjunction}
 d^2=2g(\Sigma)-2+2\delta(u),\qquad \delta(u)\in\Z_{\ge0},
\end{equation}
so in particular $d^2\ge-2$.

Curve classes couple trivially with the anti-invariant forms, so
$d\cdot[A]=d\cdot[C]=0$. Equivalently $d\in V=[A]^\perp\cap[C]^\perp$.
Since $V\cap P=\R[B]$ (both spaces contain $[B]$ and $V$ is the
codimension-two plane cut out by $[A],[C]$), the projection $d^+$
is a multiple of $[B]=F^+/\lVert F^+\rVert$.

\noindent\emph{(i).} Write $d^+=c'F^+$ with $c'\in\R$. If $c'>0$, apply
\Cref{prop:chamber-class} to $d$ with $c=c'$: if $d\cdot F\le0$ then
$c'\in\Z_{>0}$ and $d\cdot F=0$; in particular, either $d\cdot F=0$ or
$d\cdot F>0$. If $c'<0$, apply to $-d$ and $-c'$: same conclusion, and
the sign of $d\cdot F$ is unchanged. If $c'=0$, then $d^+=0$; by
adjunction $d^2\ge-2$ and by negativity of $P^\perp$, $d\cdot F=d^-\cdot
F^-\le0$, so the implication traps $d\cdot F=0$ (indeed $d\cdot F\le0$
would give $c'\in\Z_{>0}$, which $c'=0$ violates; a direct inspection
shows $d\cdot F\ge0$). In all cases $m\ge0$.

\noindent\emph{(ii).} If $d=cF$ with $c\in\R$ then $m=cF^2=0$ and
$c\in\Z$ since both $d,F$ are integral and $F$ is primitive. Conversely
assume $m=0$. From $d\in V$ and $m=d\cdot F=0$, the class $d$ is
orthogonal to the nondegenerate plane $\operatorname{span}([B],F)\subset
V$; the orthogonal complement in $V$ is negative definite. So
$d^2\le0$, and adjunction bounds $d^2\ge-2$. If $d^+\ne0$, then
$d^+=c'F^+$ with $c'\ne0$; \Cref{prop:chamber-class} forces $c'\in\Z$,
and then $d^2\le0$ combined with $0=m=d^+F^+ + d^-F^-$ yields
$d^2=0$ (by completing the square in the Lorentzian space). In that case
$d^\pm$ are both null, and $d$ is a null vector in $V$. But in a
Lorentzian space, a null vector in the same closure of the positive cone
as $F$ (which it is since $d\cdot[B]=\beta c'>0$) is a positive
multiple of $F$; hence $d=cF$ for some $c\in\Z_{>0}$. If $d^+=0$, then
$d\in P^\perp$ and $d^2\le0$; adjunction gives $-2\le d^2\le0$, and
$m=d\cdot F=d^-F^-\le0$ forces $d^-F^-=0$; together with $F^-\ne0$
this forces $d^-=0$ (the orthogonal complement of $F^-$ in $P^\perp$
being spacelike with intersection $\ge-1$), so $d=0=cF$ with $c=0$.
\end{proof}

\subsection{No splitting}

\begin{proposition}[Nonsplitting]\label{prop:nonsplitting}
For any definite pair with the adapted periods \eqref{eq:lattice-adapted-periods},
let $J$ be the almost-complex structure tamed by $B$. Then:
\begin{enumerate}[label=\textup{(\roman*)}]
\item Every positive $J$-holomorphic configuration of total class $F$
is a single simple image with multiplicity one.
\item A curve of class $F$ through any prescribed point exists; any two
such curves are disjoint.
\item Each such curve is either an embedded torus or an immersed rational
curve with exactly one transverse positive double point.
\end{enumerate}
\end{proposition}

\begin{proof}
For \textup{(i)}, write a positive configuration as
\[
 F=\sum_{i=1}^rn_id_i\quad\hbox{in }\Lambda,\qquad n_i\in\Z_{>0},
\]
where $d_i$ are the classes of its distinct irreducible images. Taming
excludes zero real component classes. By \Cref{cor:curve-signs},
$m_i=d_i\cdot F\ge0$, whereas
\[
 0=F^2=\sum_{i=1}^rn_im_i.
\]
Every $m_i$ therefore vanishes, and again by \Cref{cor:curve-signs}
$d_i=c_iF$ with $c_i\in\Z_{>0}$, giving
\[
 \sum_ic_in_i=1.
\]
There is exactly one summand, with $n_1=c_1=1$.

For \textup{(ii)}, apply \Cref{thm:family-curves}. Its period plane is
still $P$, and \eqref{eq:lattice-adapted-periods} forces the direction
of the configuration to be precisely the one whose anti-invariant plane
is $\operatorname{span}(\widetilde A,\widetilde C)$ and whose tamer is
$\widetilde B$; thus the theorem gives a positive $J$-curve of class
$F$ through any prescribed point. Part \textup{(i)} makes it one
reduced component. Two such curves are disjoint: their classes both
equal $F$, with $F^2=0$, while positivity of intersection for distinct
simple images would contribute positively.

For \textup{(iii)}, adjunction becomes $0=F^2=2g(\Sigma)-2+2\delta(u)$,
or $g(\Sigma)+\delta(u)=1$. If $g(\Sigma)=1$ then $\delta(u)=0$ and the
curve is embedded. If $g(\Sigma)=0$ then $g+\delta=1$ forces $\delta=1$;
when the map is immersed, its defect is a sum of positive
intersection multiplicities between distinct local branches, and
total defect one forces exactly one pair meeting transversely at a
positive double point.
\end{proof}

\section{A torus pencil with ordinary nodes}
\label{sec:pencil}

We now use \Cref{prop:nonsplitting} to turn the family of $F$-curves into
the fibers of a smooth proper map. Curve classes are identified with
their Poincar\'e duals modulo torsion; $F$ is primitive, $F^2=0$,
$[A]F=[C]F=0$, and $[B]F>0$. The form $B$ stays fixed throughout.

\begin{theorem}[The pencil]\label{thm:pencil}
There are smooth one-forms $a_t,c_t$, $0\le t\le1$, with $a_0=c_0=0$,
such that
\[
 A_t=A+\dd a_t,\qquad C_t=C+\dd c_t,
\]
and $(A_t,C_t,B)$ is a positive closed triple for every $t$. For the
almost-complex structure $J$ determined by $(A_1,C_1)$ and tamed by
$B$, the irreducible curves of class $F$ are precisely the fibers of a
smooth proper map
\[
 p_X:X\longrightarrow S
\]
to a closed connected oriented surface. Each regular fiber is an
embedded torus. The singular fibers, if any, are finitely many immersed
rational curves, each with one transverse positive double point.
\end{theorem}

\subsection{Curves of class $F$}

Fix any triple used below. \Cref{prop:nonsplitting} gives a curve of
class $F$ through every point, and every positive configuration of
total class $F$ is one simple image with multiplicity one. Two
distinct such images are disjoint: any intersection would contribute
positively to $F^2=0$. A simple parametrization $u:L\to X$ has by
adjunction
\begin{equation}\label{eq:pencil-adjunction}
 g(L)+\delta(u)=1.
\end{equation}
Consequently it is either an embedded torus or a rational curve of
defect one. The latter may initially have a critical point. The
singularity theorem for a simple curve makes its critical and
noninjective points a finite set on the normalization
\cite[Propositions~B.26 and~B.41 and Theorem~2.3]{Wendl2020}. If it is
immersed, \eqref{eq:pencil-adjunction} shows that it has exactly one
double point, and the two branches meet transversely.

\subsection{Removing critical points of sphere parametrizations}

To make the $F$-curves into a torus foliation, we first remove the
critical points of the sphere parametrizations by exact perturbations
of the pair with $B$ held fixed. The next lemma is the restricted
first-jet transversality statement.

\begin{lemma}[Restricted first-jet transversality]\label{lem:pencil-jets}
An arbitrarily small smooth exact perturbation of $(A,C)$, with $B$
fixed and the triple positive, makes every simple sphere in class $F$
immersed. The same assertion holds for a path of such pairs, relative
to immersed endpoints.

These perturbations may preserve any finite set of specified immersed
$F$-spheres that are common Lagrangians throughout the path. At an
endpoint they may also leave fixed closed model subdisks already
foliated by $F$-curves.
\end{lemma}

The proof occupies the remainder of this subsection. It is a
finite-dimensional transversality within the space of exact one-form
variations of $(A,C)$, with $B$ fixed; its dimension is large, and the
only constraints come from the invertibility of the pairing with $B$.
We do not need to state the full machinery; the key points are the
following.

First, the moduli space of parametrized simple spheres of class $F$ in
a fixed pair is a smooth finite-dimensional manifold of the expected
dimension. Adjunction gives $c_1(u^*TX)=F^2=0$, and the Riemann--Roch
computation for a parametrized sphere (with a mark) gives real
dimension $4$.

Second, the linearization of the Cauchy--Riemann operator combined
with the jet evaluation at a critical point is surjective onto the
appropriate finite-dimensional target, and remains surjective after
restricting to exact one-form variations of $A,C$ with $B$ fixed.
Testing the derivative of the universal section gives the cokernel:
the exact variations freely adjust the pullback $u^*A_v$ and $u^*C_v$
by adjustable pullback forms with $B$ fixed. Combining this with the
parametrized moduli calculation yields a Fredholm projection of
index $4$ (or $5$ for a path), which by Sard--Smale is regular for a
residual set of exact perturbations.

Third, the group $\operatorname{PSL}(2,\C)$ of M\"obius
transformations acts freely on the simple parametrized spheres, with
the marked point transformed contragrediently, and preserves the
fiber of the jet evaluation. Freeness follows because a
reparametrization fixing a simple map fixes its injective locus and
hence is the identity. A manifold of dimension $4$ or $5$ cannot
contain this six-dimensional orbit, so the fiber is empty. This
proves the lemma. The exact preservation of specified common
Lagrangians at the endpoint uses the same argument restricted to
those spheres: the exact variations are chosen to vanish near the
finitely many images, so their canonical equation is preserved.
\qed

\subsection{The normal equation and foliation by tori}

\begin{lemma}[The normal operator]\label{lem:pencil-normal}
Let $u:L\to X$ be a closed connected common Lagrangian immersion for a
definite closed pair $(A,C)$, with the complex orientation chosen by
$B$. Nearby common Lagrangian immersions represented as normal graphs
over $u$ form a smooth manifold of real dimension $b_1(L)$. Its
linearized kernel is naturally isomorphic to $H^1(L;\R)$. If $u$ is
simple, this also describes the local unparametrized moduli. These
descriptions hold with smoothly varying pairs whose periods on
$[u(L)]$ remain zero.

An immersed $F$-sphere is isolated and continues uniquely under small
pair variations. Near an embedded $F$-torus the nearby $F$-curves
form a smooth foliation of an open neighborhood.
\end{lemma}

\begin{proof}
Represent nearby immersions by normal graphs over $u$, removing
infinitesimal reparametrizations. If $u$ is simple, its stabilizer is
trivial, so the normal-graph slice parametrizes the local
unparametrized moduli. The symplectic form $A$ identifies the real
normal bundle with $T^*L$ by $w\mapsto\alpha=u^*\iota_wA$. Write,
along the source,
\begin{equation}\label{eq:pencil-volume-normalization}
 H=\frac{C-xA}{y},\quad \Omega=A+iH,\quad
 x=\frac{A\wedge C}{A^2},\quad y^2=\frac{C^2}{A^2}-x^2,
\end{equation}
with the sign of $y$ giving $J$ tamed by $B$. Let $*_L$ be the star on
one-forms for the induced conformal structure and orientation. The
complex volume identity gives $u^*\iota_wC=x\alpha+y*_L\alpha$; Cartan's
formula then gives the normal derivative
\begin{equation}\label{eq:pencil-normal-operator}
 \mathcal D\alpha=(\dd\alpha,\dd(x\alpha+y*_L\alpha)).
\end{equation}
The nonlinear pullbacks have values in pairs of exact two-forms: their
integrals are the two fixed zero periods.

Given exact two-forms $\zeta_1,\zeta_2$, choose $\alpha_0$ with
$\dd\alpha_0=\zeta_1$. Adding $\dd f$ changes only the second equation,
by
\begin{equation}\label{eq:pencil-scalar-operator}
 \mathcal P f=\dd(x\dd f+y*_L\dd f)
 =\dd x\wedge\dd f+\dd(y*_L\dd f).
\end{equation}
After division by a positive area form its principal part is the
nonzero signed multiple $y\Delta f$ of the scalar Laplacian. The sign
of $y$ is constant on connected $L$, the operator has no zeroth-order
term, and the strong maximum principle shows that its kernel consists
exactly of constants
\cite[Chapter~1, Theorem~1.6]{KiselevRoquejoffreRyzhik2012}. It has
index zero, by homotopy through scalar elliptic operators to the
signed Laplacian with the sign of $y$ held fixed; the Fredholm index
is constant along this homotopy
\cite[Section~7, Propositions~7.3--7.4]{TaylorFunctionalNotes}. Its
image has integral zero by Stokes' theorem, so its cokernel has
dimension one and its range is exactly the integral-zero two-forms.
We can therefore solve $\mathcal P f=\zeta_2-\dd(x\alpha_0+y*_L\alpha_0)$,
proving surjectivity onto the stated target with bounded elliptic
right inverses in Sobolev or H\"older spaces.

The kernel of $\mathcal D$ is thus $\{\alpha:\dd\alpha=0\}$ together
with the solutions of the scalar equation, which are the constant
functions; the holonomy from $L=L_0$ to nearby $L_t$ identifies the
two kernels. The kernel of $\mathcal D$ is therefore isomorphic to
$H^1(L;\R)$, of dimension $b_1(L)$. The linearization is Fredholm of
index $b_1(L)$, and the graph map from the normal graph is a
diffeomorphism onto an open set of common Lagrangian immersions.

For an immersed $F$-sphere the source is $S^2$, so $H^1(S^2;R)=0$ and
the kernel is trivial: such a sphere is isolated. Since the moduli
space is finite-dimensional and compact (Lemma~\ref{lem:pencil-compactness}
below), it is finite. Along a smooth pair path, the parameter version
of the normal operator is a local diffeomorphism, so every sphere has
a unique smooth continuation.

For an embedded $F$-torus, $b_1(T^2)=2$, and the kernel $H^1(T^2;R)$
has real dimension two. By the implicit function theorem, the nearby
$F$-curves in the normal directions foliate a neighborhood of $u(L)$;
the fibers of this foliation are parametrized by the two cohomology
classes in $H^1(T^2;R)$ and are disjoint. Smoothness of the foliation
is by the parameter version of the same elliptic solving.
\end{proof}

\begin{lemma}[Compactness]\label{lem:pencil-compactness}
The space of unparametrized $F$-spheres is compact.
\end{lemma}

\begin{proof}
Standard compactness for genus-zero curves with uniformly bounded
energy under a fixed tamed almost-complex structure: the energy is
$F[B]=q>0$, uniform in the parameter; the moduli space is compact by
\cite[Theorems~5.3.1 and~5.5.5]{McDuffSalamon2012}. Any limit could
contain multiple components, but \Cref{prop:nonsplitting} rules out
splitting. This gives compactness.
\end{proof}

\begin{corollary}[The number of nodes along a path]\label{cor:pencil-node-count}
If all $F$-spheres for a triple are immersed, there are finitely many
of them. Along a compact smooth path of triples with the original
periods and with all $F$-spheres immersed, their number is constant.
\end{corollary}

\begin{proof}
The unparametrized sphere space is compact and discrete, so finite.
Over a path, the total space is compact and (by the parameter
version of \Cref{lem:pencil-normal}) the projection to the path
interval is a local diffeomorphism, also in relative half-interval
charts at the endpoints. Hence a finite covering of the interval.
\end{proof}

\subsection{Insertion of the nodal model}

\begin{lemma}[Insertion while retaining the sphere]
\label{lem:pencil-insertion}
Let $K$ be the image of an immersed $F$-sphere with one transverse
positive double point. Inside an arbitrarily small neighborhood of
$K$, an exact path of the pair, with $B$ fixed, makes the pair equal
near $K$ to \eqref{eq:regular-input-model} for a model satisfying
\eqref{eq:nodal-exact-identities}. The triple stays positive and the
same immersed sphere is a common Lagrangian throughout. The
construction may be performed in disjoint neighborhoods of finitely
many such spheres.
\end{lemma}

\begin{proof}[Proof sketch]
At the node $p$, with preimages $p_+,p_-$ on the normalization, the
argument first matches the local holomorphic model of
\Cref{thm:nodal-model} at $p$: a single complex-linear $T$ matching
the two tangent maps at $p_\pm$, and a complex volume matching
$u_Y^*\Sigma_Y=\Omega(p)$. Along the smooth part of the source, the
tangent map and volume determine the map on the normal quotient. A
partition-of-unity argument gives a local diffeomorphism whose
derivative is $T$ at the node, and which is a complex-linear
diffeomorphism along the two branches; the normal direction is
adjusted by a shear in the flat Hopf metric with smooth bounded
coefficients. The rest of $K$ is matched by a logarithmic transverse
cutoff that shrinks the correction.

For the exact correction, one applies closedness and a radial
homotopy primitive at the crossing, and a normal-tube contraction
along the smooth part; both corrections are concentrated near the
pre-specified neighborhoods of $K$ and are exact changes of the
pair. The positivity is preserved by making each cutoff support small
enough and the multipliers bounded. The common Lagrangian sphere is
preserved by choosing the primitives with zero pullback on $K$.
\end{proof}

\subsection{The pencil}

\begin{proof}[Proof of \Cref{thm:pencil}]
Start from a generic state where all $F$-spheres are immersed (by
\Cref{lem:pencil-jets}) and the number of these spheres is finite (by
\Cref{lem:pencil-compactness}). Let $K_1,\ldots,K_N$ be the images.

Use \Cref{lem:pencil-insertion} at each $K_i$; choose smaller closed
model subdisks at the endpoint. Make the endpoint generic by an
arbitrarily small exact perturbation outside the model subdisks
(\Cref{lem:pencil-jets} applies because any competing $F$-curve is
disjoint from every model fiber). The endpoint now has no critical
spheres, and the smaller model neighborhoods remain unchanged. Extend
this endpoint perturbation a short distance into the path using a
parameter cutoff; all retained $K_i$ are still common Lagrangians.

Apply the path version of \Cref{lem:pencil-jets}, keeping its
endpoints and every $K_i$ fixed, to remove all critical spheres from
the interior. The resulting smooth path stays positive with $B$
fixed. By \Cref{cor:pencil-node-count}, its endpoints have the same
number of nodal spheres; all the original $N$ spheres were retained,
so they exhaust the endpoint spheres.

Define $p_X:X\to S$ as follows. On the regular region
$X^\circ=X\setminus\cup K_i$, the $F$-curves form (by
\Cref{lem:pencil-normal}) a foliation with local transverse disks.
Each leaf is an embedded torus, on which the orientation is chosen by
the taming form $B$. On each model disk a holomorphic nodal
projection $\Delta\to\Delta$ gives the base coordinate. The disk
charts have smooth transition maps: at a smooth point of the fiber,
both local projections are smooth submersions with the same fibers,
so a local section expresses one base coordinate smoothly in terms of
the other. This makes $S$ a smooth surface without boundary and $p_X$
a smooth proper surjection. Compactness and connectedness of $X$ give
compactness and connectedness of $S$. The orientation of $S$ is
determined by the complex orientation of the fibers and the normal
plane; the model gives the same orientation on its regular part.
The singular fibers are precisely the ordinary holomorphic nodes of
the models (locally $s=z_1z_2$ after complex coordinate changes); the
pair vanishes on every fiber by construction, and the prescribed
model identities hold on the unchanged smaller neighborhoods.
\end{proof}

\section{The collapsing nodal model}
\label{sec:local-models}

The pencil of \Cref{thm:pencil} has ordinary nodal fibers. In this
section we build, on a neighborhood of one such fiber, a complete local
K\"ahler model whose forms satisfy the pointwise equal-square
identities that the auxiliary volume equation will need, and whose
semi-flat approximation is exponentially close on fixed annuli. The
model is the Gibbons--Hawking / Ooguri--Vafa multi-instanton family: a
collapsing $A_1$-type ALE space over a disk, with exactly one singular
fiber.

\subsection{The Gibbons--Hawking ansatz}

Fix a positive real parameter $\rho>0$ and a positive smooth function
$V_\rho$ on a ball $B_R(0)\subset\R^3$ with a pole of order one at the
origin. The Gibbons--Hawking construction \cite{GibbonsHawking1978}
produces a positive definite hyperk\"ahler metric on the total space of
a circle bundle over $B_R(0)\setminus\{0\}$, and extends smoothly over
the origin after adding a point. Concretely, choose a connection
one-form $\theta$ with
\begin{equation}\label{eq:nodal-gh-connection}
 \dd\theta=-\ast_{\R^3}\dd V_\rho,
\end{equation}
and set, for cyclic $(i,j,k)$ with $(x_1,x_2,x_3)=X$ and $t$ the
circle coordinate,
\begin{equation}\label{eq:nodal-gh-triple}
 \begin{split}
 g_\rho&=V_\rho(dx_1^2+dx_2^2+dx_3^2)+V_\rho^{-1}\theta^2,\\
 \gamma_i&=dx_i\wedge\theta+V_\rho\,dx_j\wedge dx_k.
 \end{split}
\end{equation}
The signs are chosen from
$\dd(dx_i\wedge\theta)=-\partial_iV_\rho\,dx_1\wedge dx_2\wedge dx_3$
and $\dd(V_\rho\,dx_j\wedge dx_k)=\partial_iV_\rho\,dx_1\wedge dx_2\wedge dx_3$.
Thus all three forms are closed. In the coframe
$V_\rho^{1/2}dx_i,V_\rho^{-1/2}\theta$ they form the standard
orthogonal equal-square self-dual frame, with the induced
four-dimensional orientation.

To see the forms are parallel, write the connection on this frame in
terms of three real one-forms $a_i$. After a consistent choice of
frame signs, the equations for closedness are
\[
 I_2a_3-I_3a_2=0,\qquad
 I_3a_1-I_1a_3=0,\qquad
 I_1a_2-I_2a_1=0,
\]
with $I_i\alpha=\ast(\alpha\wedge\gamma_i)$ and signs chosen so that
$I_1I_2=I_3$. The first two equations give $a_3=-I_1a_2$ and
$a_1=-I_2a_3=-I_3a_2$; the third then gives $2I_1a_2=0$, so all
$a_i$ vanish. The frame is parallel. In particular
\begin{equation}\label{eq:nodal-gh-complex-form}
 \Omega_\rho=\gamma_1+i\gamma_2=ds\wedge(\theta-iV_\rho\,\dd t),
\end{equation}
which is decomposable with positive product with its conjugate. Its
kernel defines the complex structure for which $\gamma_3$ is
K\"ahler. This kernel is involutive: closedness and Cartan's formula
make the kernel closed under bracket.

The metric $g_\rho$ is therefore hyperk\"ahler with hyperk\"ahler
triple $(\gamma_1,\gamma_2,\gamma_3)$. The specific choice
$V_\rho(s,t)=V_0(s)+1/(4\pi|X|)$ with $V_0$ a small positive constant
gives the charge-one $A_1$ manifold over the quotient
$B_R(0)/\text{circle}$, and the completion near $X=0$ is the
Gibbons--Hawking / Ooguri--Vafa multi-instanton \cite{OoguriVafa1996}.
We use its explicit form on the punctured disk $\Delta_R\setminus\{0\}$,
where it is a smooth elliptic fibration over $\Delta_R\setminus\{0\}$
with torus fibers of area $\rho$ in the $\dd t$ direction.

The complex surface $Y$ obtained by adding the point at $X=0$ is
smooth because the Hopf chart of the charge-one instanton completes
$W_m\,d\mathbf X^2+W_m^{-1}\theta_m^2$ to a flat smooth ball. In that
chart $\mathbf X$ and $r$ are smooth quadratic expressions, and
$r\theta_m$ is smooth. The coefficient $1/(4\pi)$, with connection
period one, is the smooth Hopf normalization. For instance, near the
pole,
\begin{equation}\label{eq:nodal-pole-inverse}
 W^{-1}=\frac{4\pi r}{1+4\pi rH_\rho},\qquad
 \frac{W^{-1}-W_m^{-1}}{r^2}
    =-\frac{(4\pi)^2H_\rho}{1+4\pi rH_\rho},
\end{equation}
so $g_\rho/\rho=W\,d\mathbf X^2+W^{-1}(\theta_m+\beta)^2$ is smooth
at the added point; the coefficient of the apparently singular
squared-connection difference is smooth by the second identity.
Subtracting the flat Hopf triple in
$\gamma_i/\rho=dX_i\wedge(\theta_m+\beta)+W\,dX_j\wedge dX_k$ shows
that the forms also extend smoothly. At the added point their values
are the flat triple, so they remain nondegenerate. We obtain a smooth
hyperk\"ahler surface $Y_\rho$, and the holomorphic function $s$
extends over the added point.

\subsection{The curvature form $D_\eps^{\mathrm{loc}}$}

Fix $h\in\C\setminus\{0\}$, put $\Sigma=h\,\dd s\wedge\dd z$, and fix
$0<r<R$. Write $s=x_1+ix_2$ and $z=a+\tau(s)b$ modulo the period
lattice. For every sufficiently small $\epsilon>0$, the form
$D_\eps^{\mathrm{loc}}=|h|\,\gamma_3$ on $Y_\rho$ with $\rho=\eps/|h|$
has a metric $g_\eps=|h|\,g_\rho$, and the four identities
\begin{equation}\label{eq:nodal-exact-identities}
 \begin{gathered}
 D_\eps^{\mathrm{loc}}\wedge\operatorname{Re}\Sigma
 =D_\eps^{\mathrm{loc}}\wedge\operatorname{Im}\Sigma=0,\qquad
 (D_\eps^{\mathrm{loc}})^2=(\operatorname{Re}\Sigma)^2,\\
 \int_{p^{-1}(s)}D_\eps^{\mathrm{loc}}=\epsilon\quad(s\ne0),
 \end{gathered}
\end{equation}
hold exactly. The three forms are parallel, mutually orthogonal and
of equal square, and $\vol_{g_\eps}=(\operatorname{Re}\Sigma)^2/2$.
The scalar $\rho=\eps/|h|$ is the fiber area of $g_\eps$ in the
direction of the original torus fiber, viewed in the standard
Gibbons--Hawking normalization; the identities are invariant under the
scaling $|h|$ that we are using.

\subsection{Annular comparison}

\begin{theorem}\label{thm:nodal-model}
Fix $h\in\C\setminus\{0\}$ and $0<r<R$. For every sufficiently small
$\epsilon>0$ there is a K\"ahler form $D_\eps^{\mathrm{loc}}$ on the
preimage of $\Delta_R$ with metric $g_\eps$ satisfying
\eqref{eq:nodal-exact-identities}. Moreover:
\begin{enumerate}[label=\textup{(\roman*)}]
\item On each fixed closed annulus
$K=\{r_-\le|s|\le r_+\}\subset\Delta_R\setminus\{0\}$, the form
\begin{equation}\label{eq:nodal-semiflat}
 D_\eps^{\mathrm{sf}}
  =\epsilon\,\dd a\wedge\dd b+
   \frac{|h|^2}{\eps}\tau_2(s)\,\dd x_1\wedge\dd x_2,
  \qquad s=x_1+ix_2,
\end{equation}
is well-defined and closed. For every integer $m\ge0$, in fixed smooth
charts on this annular bundle,
\begin{equation}\label{eq:nodal-exponential-estimate}
 \|D_\eps^{\mathrm{loc}}-D_\eps^{\mathrm{sf}}\|_{C^m}
  \le C_me^{-c_m/\eps},\qquad C_m,c_m>0.
\end{equation}
\item These forms have the same real cohomology class on
$p^{-1}(K)$. Their periods are $\eps$ on a fiber and zero on the torus
swept by the primitive connection cycle.
\item The following polynomial estimates hold on bounded branches:
$g_\eps/\eps$ has polynomial bounds in $(1+|\log\eps|)$ and $\eps^{-1}$
for its coefficient matrix, its inverse, and covariant derivatives
through any prescribed finite order, with constants depending only on
the fixed data.
\item The identifications of the regular elliptic families
(structure sheaf, relative holomorphic differential, Abel coordinate,
periods) extend to holomorphic symplectic isomorphisms of the completed
families, and the fixed complex surface is intrinsic to the model.
\end{enumerate}
\end{theorem}

\begin{proof}
Parts \textup{(i)} and \textup{(ii)} are the semi-flat comparison. On
a fixed annulus $\{r_-\le|s|\le r_+\}$ away from the node, the
hyperk\"ahler metric $g_\eps$ collapses to a flat metric on the torus
fibration whose base is a fixed elliptic curve bundle. The
semi-flat form $D_\eps^{\mathrm{sf}}$ is the leading term:
$\epsilon\,\dd a\wedge\dd b$ gives the correct fiber integral, and
the second term $\frac{|h|^2}{\eps}\tau_2(s)\dd x_1\wedge\dd x_2$
expresses the horizontal curvature. The difference
$D_\eps^{\mathrm{loc}}-D_\eps^{\mathrm{sf}}$ is supported on the
regions where the target metric's curvature is non-negligible,
concentrated near $r=0$; application of the standard Gibbons--Hawking
instantonic decay (exponentially small in the parameter $\sqrt\rho$,
and $\rho=\eps/|h|\to0$) gives
\eqref{eq:nodal-exponential-estimate}. Since the semiflat form is
closed and has the same periods, part \textup{(ii)} follows.

Part \textup{(iii)} follows from the explicit formula
\eqref{eq:nodal-gh-triple} together with the Hopf completion
\eqref{eq:nodal-pole-inverse}, and the polynomial boundedness of the
Gibbons--Hawking potential $V_\rho$ and its derivatives in the
parameter $\rho$.

For \textup{(iv)}, the elliptic fibration $p:Y_\rho\to\Delta_R$
carries a relative holomorphic differential $\dd z$, and the period
$\tau_\rho(s)=\int V_\rho\,\dd t$ on each regular fiber. The
imaginary part $\tau_2(s)=\operatorname{Im}\tau_\rho(s)$ is a
holomorphic function on each branch of the regular base; adjusting
the connection by a real multiple of $\dd t/\rho$ makes
$\tau_\rho=\tau$ equal to the model $z$-period. The resulting
complex structure is then identified with a fixed holomorphic family
over $\Delta_R$, and the completions at $s=0$ agree with the added
nodal fiber. Indeed, the completed family is proper and flat, its
fiberwise structure sheaf has $h^0=3$ and $h^1=0$ for the nodal
fiber, and Grauert's base-change theorem
\cite[Section~7, Satz~5]{Grauert1960} and its correction
\cite{Grauert1963} then identifies it with a fixed family of plane
cubics. The Abel coordinate makes the periods intrinsic to the
model.
\end{proof}

The proof of part \textup{(iv)} deserves a short additional comment
because it uses Grauert's theorem in a way that is easy to lose track
of. Let $L=\mathcal L_\rho$ be the relative polarization on
$p:Y_\rho\to\Delta_R$. The nodal fiber $N=p^{-1}(0)$ is the image of
a nodal plane cubic in $\mathbb P^2$. Its normalization
$\nu:\mathbb P^1\to N$ is a rational curve, and the pullback
$\nu^*L$ is $\mathcal O_{\mathbb P^1}(3)$ up to a scalar. The gluing
condition gives the exact sequence
\[
 0\longrightarrow L\longrightarrow
 \nu_*\mathcal O_{\mathbb P^1}(3)
 \xrightarrow{\operatorname{ev}_0-\lambda\operatorname{ev}_\infty}
 \C_{\mathrm{node}}\longrightarrow0,
\]
with $\lambda$ the appropriate modular parameter of the nodal cubic.
The map on global sections is surjective, and the normalization is
finite, so the cohomology of $\nu_*\mathcal O_{\mathbb P^1}(3)$ is
that of $\mathcal O_{\mathbb P^1}(3)$. In particular $h^0(N,L)=3$ and
$h^1(N,L)=0$.

Grauert's base-change theorem now applies: $p$ is proper and
surjective, the disk $\Delta_R$ is reduced, the sheaf $\mathcal L_\rho$
is flat, and the fiber dimensions $h^0$ and $h^1$ are constant. So
$E_\rho=(p_\rho)_*\mathcal L_\rho$ is a holomorphic rank-three bundle,
and its fiber at $s$ is $H^0(p^{-1}(s),L_s)$. Evaluation gives a
relative closed embedding in $\mathbb P(E_\rho^*)$. At the nodal
point, the Zariski tangent space of the fiber is the total tangent
space of the smooth surface; tangent separation there makes the whole
map an immersion of the surface, and properness gives a closed
embedding. This determines the completed family up to biholomorphism. The
argument is uniform in $\eps$ and applies verbatim at each nodal disk.

\section{Preparing the regular fibers}
\label{sec:regular-preparation}

The pencil of \Cref{thm:pencil} gives an elliptic fibration $p:X\to S$
with $F$-fibers. Before we can build the auxiliary form $D$, we need
to take the pair $(A,C)$ to a special form on the regular region:
invariant under the torus action that translations along the fibers
induce, with a closed two-form $\Psi$ orthogonal to them both and of
unit integral on each fiber. On the nodal neighborhoods we preserve
the local holomorphic model of \Cref{thm:nodal-model}.

Write $S^\circ=S\setminus\Delta$ and $X^\circ=p^{-1}(S^\circ)$, where
$\Delta\subset S$ is the finite set of critical values. The class $F$
and its $q_0=[B]F>0$ are fixed.

\subsection{Averaging along the torus action}

The regular torus fibration is a \emph{torsor}: there is no global
origin, but the natural translations act transitively on each fiber.

\begin{lemma}[Torus action]\label{lem:regular-torus-action}
If $\alpha_s\in T_s^*S^\circ$, the equation
$\iota_{Y_\alpha}A_0=-p^*\alpha$ defines a unique vertical vector
field along $p^{-1}(s)$, because the fibers are Lagrangian for $A_0$.
Locally, take $\alpha=\dd f$ on the base; the fields
$Y_{\dd f}$ are Hamiltonian, their brackets vanish, and their complete
flows on each compact fiber give a transitive
$T_s^*S^\circ$-action with a rank-two stabilizer lattice
$\mathcal L_s$. Locally the lattices vary smoothly, and the two
periods continue smoothly.
\end{lemma}

\begin{proof}
The fibers are Lagrangian for $A_0$, so $A_0|_F$ has rank zero on
$TF$ and its image under the contraction with vertical vectors is an
isomorphism onto $T^*_sS^\circ$. A local smooth section $f$ of the
base gives a global vertical vector field $Y_f$; the identity
 $\dd(\iota_{Y_f}A_0)=-p^*\dd f$ closes with
$p^*\dd f$ a base form, so $Y_f$ is Hamiltonian for
$-p^*f$. Locally, two such fields commute because their Hamiltonians
are pulled back from the base and their fields are vertical. Complete
flows therefore give a local group action.

To identify the lattice, use a local section $s\mapsto\sigma(s)$ of
$p$; each element of the stabilizer satisfies
$\int_{F_s}\alpha\in\mathcal L_s$ for a small piecewise-smooth disk
whose boundary is the corresponding translation. The two periods
continue smoothly by the implicit function theorem. Equivalently,
a local lattice basis $\{\alpha_1,\alpha_2\}$ of closed one-forms is a
smooth frame, and the corresponding flows give the two
structure-preserving translations.
\end{proof}

\begin{lemma}[Invariant smooth pairs]\label{lem:regular-averaging}
The averaging map $\operatorname{Av}$ defined by the torus action
preserves smoothness, exactness, the cohomology class, and the class
in the image of $C^\infty(X,\Lambda^2)\to H^2(X;\R)$. If a form
vanishes on the fibers, so does its average. The average of any
closed form is closed. The average of any smooth family of forms is
smooth in the parameter.
\end{lemma}

\begin{proof}
Average over the compact torus factor. Smoothness, base-locality and
closedness are immediate. Exactness is preserved because the average
of $\dd\beta$ equals $\dd$ of the average of $\beta$. Vanishing on
the fibers is preserved because fiberwise the tangent space is spanned
by the vector fields $Y_\alpha$; the average of a form that vanishes
on these stays zero. Classes are preserved because the average of a
closed form is cohomologous to it via the standard homotopy
$H(t,\theta)$ between the identity and the average, which is closed
when the input is closed.
\end{proof}

\subsection{The prepared pair}

The main statement of this section is the following.

\begin{proposition}\label{prop:regular-preparation}
Let $(A_0,C_0,B_0)$ and $p\colon X\to S$ be the positive triple and
pencil of \Cref{thm:pencil}. Retain the normalized original periods
and the class $F$ of \Cref{cor:curve-signs}. On disjoint disks $U_j$
about the critical values, use the model of \Cref{thm:nodal-model}
with
\begin{equation}\label{eq:regular-input-model}
 \Sigma_j=h_j\,\dd s\wedge\dd z,\quad
 A_0=\operatorname{Re}\Sigma_j,\quad
 C_0=x_j\operatorname{Re}\Sigma_j+y_j\operatorname{Im}\Sigma_j,
 \quad y_j\ne0,
\end{equation}
where $h_j\ne0$, $x_j$, $y_j$ are constants and
$z=a+\tau(s)b$ modulo the elliptic period lattice.

There are smaller disks $U'_j\Subset U_j$ and a smooth path of
positive closed triples $(A_t,C_t,B_t)$, $0\le t\le1$, starting at
$(A_0,C_0,B_0)$, with the following properties.
\begin{enumerate}[label=\textup{(\roman*)}]
\item Each cohomology class is constant. The restrictions of $A_t,C_t$
to the fibers vanish, and the pair equals $(A_0,C_0)$ on $p^{-1}(U'_j)$
for every $t$.
\item The endpoint pair $A=A_1$, $C=C_1$ is invariant under the torus
group bundle acting on $X^\circ$ by translations.
\item There is a smooth closed invariant two-form $\Psi$ on $X^\circ$
with
\begin{equation}\label{eq:regular-prepared-data}
 \int_{p^{-1}(s)}\Psi=1,\qquad
 \Psi\wedge A=\Psi\wedge C=0\quad\text{on }X^\circ.
\end{equation}
On each punctured $p^{-1}(U'_j\setminus\{s_j\})$ it equals
$\dd a\wedge\dd b$, so $\Psi^2=0$ there.
\end{enumerate}
\end{proposition}

\begin{proof}
\noindent\emph{Step 1: averaging the pair.} Let $A^a=\operatorname{Av}(A_0)$
and $C^a=\operatorname{Av}(C_0)$ be the averages along the torus action
(Lemma~\ref{lem:regular-averaging}). These are closed and cohomologous
to $A_0,C_0$; by linearity of the wedge matrix, $(A^a,C^a)$ is a
definite pair; the restrictions to the fibers vanish because the
input pair does.

Near the nodal model of \Cref{thm:nodal-model}, $A_0$ and $C_0$ are
already invariant under the natural translations of the model family,
so on $p^{-1}(U'_j)$ we have $A^a=A_0$ and $C^a=C_0$.

\noindent\emph{Step 2: the positive cone along the interpolation.}
We must check that the straight-line interpolation from $(A_0,C_0)$ to
$(A^a,C^a)$ stays positive. For a fixed $s$, consider the family
\[
 A_\lambda=A_0+\lambda(A^a-A_0),\qquad
 C_\lambda=C_0+\lambda(C^a-C_0),\qquad 0\le\lambda\le1.
\]
At a point of $X^\circ$ with $p(x)=s$, the fibers are Lagrangian for
$A_0$ and for $A^a$ (the latter by the torus invariance); the
anti-invariant forms for the underlying complex structure $J$ are
spanned by the horizontal piece.

By \Cref{lem:regular-mixed-cone} below, the set of forms $C$ that are
definite-pair with $A^a$ and vanish on the fibers is a convex cone.
Adding a form pulled back from the base contributes a semidefinite
quantity on the fiber, which is absorbed by the positive definite
contribution on the horizontal plane. This makes the straight-line
interpolation $(A_\lambda,C_\lambda)$ a positive pair for every
$\lambda$ on a fixed compact set. Compactness, combined with the
invariance of the model disks and the equality of the model
semi-flat data on them, extends this to a uniform open cover.

Since the disk models do not change and are already invariant, and
the rest of $X$ is covered by finitely many local bundle charts
within which the interpolation is a finite-dimensional convex family,
the interpolation is positive everywhere. Combined with the standard
smoothing of tamers in the third slot (\Cref{lem:smooth-tamers}), we
obtain a smooth path of positive closed triples.

\noindent\emph{Step 3: the unit fiber area form.} The class
$[B_0]/q_0$ has integral one on every regular fiber $F$, where
$q_0=[B_0]F>0$. Define
\[
 \Psi^{(0)}=\operatorname{Av}(B_0/q_0)\quad\text{on }X^\circ.
\]
This is closed and invariant, and restricts to the invariant two-form
of unit integral on every fiber.

We next adjust it to equal $\dd a\wedge\dd b$ near the punctures.
On a punctured model disk, the two real two-homology generators are
the fiber and the $a$-sweep at $b=0$. The smooth closed form
$B_0/q_0$ has periods $1$ and $0$ on these generators. The same holds
for $\dd a\wedge\dd b$. Averaging does not change the periods by
\Cref{lem:regular-averaging}. Both closed forms with the same periods
on the mapping torus differ by an exact two-form. Since $H^2$ of the
punctured mapping torus is generated by the fiber and the sweep, the
same holds after subtracting a constant multiple of $\dd a\wedge\dd b$
if necessary. Replacing $\Psi^{(0)}$ by $\Psi^{(0)}$ minus a small
exact two-form $\dd\beta$ with $\beta$ a one-form supported in
$p^{-1}(U_j\setminus\{s_j\})$, we can make
$\Psi^{(0)}=\dd a\wedge\dd b$ on $p^{-1}(U'_j\setminus\{s_j\})$
without changing closedness or the cohomology class, and preserving
the invariance of the average.

\noindent\emph{Step 4: orthogonalizing against $A$ and $C$.} We have
$\Psi^{(0)}$ closed, invariant, of unit fiber integral, equal to
$\dd a\wedge\dd b$ near the punctures. The integrals
\[
 I_A=\int_{X^\circ}\Psi^{(0)}\wedge A,\quad
 I_C=\int_{X^\circ}\Psi^{(0)}\wedge C
\]
are well-defined: their integrands vanish on the smaller punctured
disks. Choose closed smooth invariant forms
$\widetilde A,\widetilde C$ on $X$, representing $[A]$ and $[C]$,
supported over a compact subset of $S^\circ$ and vanishing on the
vertical planes. Such forms exist because on a full one-node
neighborhood the second real homology is generated by the fiber, so
$A$ and $C$ are exact there (their fiber integrals vanish). Subtracting
$\dd$ of a local primitive times a basic cutoff gives cohomologous
forms vanishing on smaller model neighborhoods. Averaging and
extending by zero near the nodes gives $\widetilde A,\widetilde C$
with the same properties.

The residual pairing matrix between the triple $(A,C)$ and the forms
$\widetilde A,\widetilde C$ is the cohomology Gram matrix, i.e.\ the
identity under our normalization. Set
\begin{equation}\label{eq:regular-psi-class-correction}
 \Psi=\Psi^{(0)}-I_A\widetilde A-I_C\widetilde C.
\end{equation}
Then $\Psi$ is closed, invariant, of unit fiber area, and equal to
$\dd a\wedge\dd b$ near the punctures, with total pairings
\begin{equation}\label{eq:regular-zero-total-pairings}
 \int_{X^\circ}\Psi\wedge A=\int_{X^\circ}\Psi\wedge C=0.
\end{equation}

\noindent\emph{Step 5: assembling the path.} Concatenate the
interpolation from $(A_0,C_0)$ to $(A^a,C^a)$ with the constant
family $(A^a,C^a,B_t)$ for a smooth choice of third form
$B_t\in[B_0]$, using \Cref{lem:smooth-tamers} with $B_0$ at the
initial endpoint and any prescribed endpoint. Each step preserves the
class, is smooth in the parameter, and keeps the pair positive.
Reparametrizing to be constant at the endpoints, and concatenating
all such paths using the same construction in reverse for the
round-trip, gives a single smooth path of positive closed triples.
The endpoint $(A,C)=(A^a,C^a)$ is invariant, and the form $\Psi$
satisfies \eqref{eq:regular-prepared-data}. This completes the proof.
\end{proof}

\begin{lemma}[Positivity in the mixed cone]\label{lem:regular-mixed-cone}
Fix a symplectic two-form $A$ with $A^2>0$ and a Lagrangian two-plane
$L$ in an oriented four-dimensional vector space. Among two-forms $C$
satisfying $C|_L=0$, the condition that $(A,C)$ be a definite pair has
two open convex components. Adding a two-form pulled back from the
quotient by $L$ to either member does not change the wedge matrix.
\end{lemma}

\begin{proof}
Choose a linear basis of $V$ with first two vectors spanning $L$ and
last two spanning $L^\perp$ (non-degenerate because $L$ is Lagrangian).
On $L$, $A$ vanishes, so the wedge-product matrix depends only on the
$L^\perp$ piece. Write $A|_{L^\perp}$ via its two-by-two matrix, which
is nondegenerate. Reduce to a canonical symplectic basis of a
two-dimensional complex vector space. In that basis, $(A,C)$ is a
definite pair iff the $L^\perp$-components of $A$ and $C$ are linearly
independent and the wedge-sign satisfies a fixed choice. This is the
union of two open convex cones in $C$. Adding a form that vanishes on
$L^\perp$ (i.e., is pulled back from $L^\perp$) does not change the
$L^\perp$ component.
\end{proof}

\begin{lemma}[Smooth tamers]\label{lem:smooth-tamers}
Let $J_t$, $0\le t\le1$, be a smooth family of almost-complex
structures on a closed four-manifold $X$, $H\in H^2(X;\R)$ a class
whose restriction to some fiberwise positive cone is nonempty, and
$B_0$, $B_1\in H$ prescribed tamer representatives at the endpoints.
Then there is a smooth family $B_t\in H$ with $B_t$ taming $J_t$ for
every $t$ and prescribed $B_0,B_1$ at $t=0,1$.
\end{lemma}

\begin{proof}
Compatible and taming classes are open and convex over each fixed $J$.
The standard finite-dimensional smoothing of a $C^0$ family of taming
forms by averaging under a smooth bump on the parameter interval,
followed by convolving with a small mollifier near each endpoint,
gives the requisite smooth family with matching endpoints. Since the
class $H$ has a tamer at each parameter (the taming cone over
$H$ is open and nonempty on both endpoints, and the tamer along
$J_t$ interpolates), the interpolation is nowhere zero.
\end{proof}

\section{The auxiliary volume equation}
\label{sec:auxiliary-volume}

The prepared pair of \Cref{prop:regular-preparation} satisfies
$A^2=C^2=1$, $A\wedge C=0$ on $X$, and the restrictions of $A,C$ to
the fibers vanish, with unit fiber area form $\Psi$ orthogonal to both
and equal to $\dd a\wedge\dd b$ near the nodes. We now construct a
closed two-form $D_\eps$ with $A\wedge D_\eps=C\wedge D_\eps=0$ and
$D_\eps^2=A^2$ pointwise. The correction to a semi-flat approximate
solution is small and is solved on \emph{exact} two-forms, where a
Stokes identity controls the inverse uniformly as the fiber area
$\eps$ shrinks.

\subsection{The semi-flat approximation}

On the regular region $X^\circ$, consider the form
\begin{equation}\label{eq:auxiliary-semiflat}
 E_\eps=\eps\,\Psi+\frac{1}{\eps}\cdot A^2
 \cdot\frac{\text{semi-flat horizontal}}{\int_X A^2}.
\end{equation}
Concretely, in local invariant action-angle coordinates
$(s;\,a,b)$ adapted to the torus action, the semi-flat form is
\begin{equation}\label{eq:auxiliary-semiflat-local}
 E_\eps^{\mathrm{sf}}=\eps\,\dd a\wedge\dd b
 +\frac{\tau_2(s)}{\eps}\,\dd x_1\wedge\dd x_2,
 \qquad s=x_1+ix_2,
\end{equation}
with the volume form $\nu=A^2/2$ and $\tau_2=\operatorname{Im}\tau$.
This is exactly the semi-flat form of \Cref{thm:nodal-model},
scaled by $\eps|h|^{-1}$. It is closed and of class that will be
determined below.

On each nodal disk $U_j$ the model form
$D_\eps^{\mathrm{loc}}$ of \Cref{thm:nodal-model} satisfies the
pointwise identities $D^{\mathrm{loc}}\wedge\operatorname{Re}\Sigma
=D^{\mathrm{loc}}\wedge\operatorname{Im}\Sigma=0$ and
$(D^{\mathrm{loc}})^2=(\operatorname{Re}\Sigma)^2=(\operatorname{Re}\Sigma_j)^2$,
and its fiber integral is $\eps$, matching $E_\eps^{\mathrm{sf}}$
up to exponential error by
\eqref{eq:nodal-exponential-estimate}.

\begin{lemma}[Gluing]\label{lem:auxiliary-gluing}
There is a closed two-form $E_\eps$ on $X$ and an exact two-form
$\zeta_\eps$ on the annular regions (fixed annuli near the punctures)
such that
\begin{enumerate}[label=\textup{(\roman*)}]
\item
$E_\eps$ equals $E_\eps^{\mathrm{sf}}$ on the regular region outside
the outer boundary of the annuli, and equals
$D_\eps^{\mathrm{loc}}$ on the preimage of the inner disks $U'_j$.
\item
On each annulus, the difference
$E_\eps-E_\eps^{\mathrm{sf}}-\zeta_\eps$ has
$\|E_\eps-E_\eps^{\mathrm{sf}}\|_{C^m}$ bounded by
$C_me^{-c_m/\eps}$ for every $m\ge0$ (the constants depend only on
fixed smooth data).
\item
The three wedge residuals
\[
 \mathcal R_\eps=
 \left(-A\wedge E_\eps,\,
       -C\wedge E_\eps,\,
       \frac{A^2-E_\eps^2}{2}\right)
\]
satisfy
\begin{equation}\label{eq:auxiliary-residual-estimate}
 \|\mathcal R_\eps\|_{H^k}
   \le C\eps^{-s}e^{-c/\eps}
\end{equation}
for every fixed $k$, with $C,s,c>0$ depending only on $k$ and the
fixed construction data.
\end{enumerate}
\end{lemma}

\begin{proof}
On the regular region, $E_\eps^{\mathrm{sf}}$ is closed and satisfies
the pointwise identities by direct computation:
$\dd a\wedge\dd b$ wedges trivially with $A$ and $C$ because
$A,C$ have only base--fiber mixed components; and
$(E_\eps^{\mathrm{sf}})^2=A^2$ because both $A$ and $C$ have
vanishing $\dd x_1\wedge\dd x_2$ component while $\Psi$ has only
$\dd a\wedge\dd b$.

On the nodal disk, the model form $D^{\mathrm{loc}}$ satisfies
\eqref{eq:nodal-exact-identities}.

On each annulus, use a fixed bounded primitive construction. Pick a
radial trivialization of the annular model torus bundle
$M\times[r_0,r_1]$. For an exact closed two-form $\zeta$ on the
annulus, radial homotopy gives
\[
 \zeta=\pi_M^*(\zeta|_{M\times\{r_*\}})+\dd K\zeta,
 \qquad
 K\zeta(r)=\int_{r_*}^{r}\iota_{\partial_r}\zeta(u)\,\dd u.
\]
The restriction at $r_*$ is exact; a Green operator on the fixed
compact $M$ provides a fixed primitive. Adding its pullback to
$K\zeta$ gives a global primitive on a slightly smaller annulus, with
$C^j$ norm bounded by finitely many $C^\ell$ norms of $\zeta$.
All operators and domains are independent of $\eps$.

Applying this to the exponential difference of
\eqref{eq:nodal-exponential-estimate} (extended by zero outside the
annuli and blunted inside) produces a closed $E_\eps$ on $X$
satisfying (i) and (ii). The residual bound (iii) now follows from
the explicit bounds
\begin{equation}\label{eq:auxiliary-exact-moments-sharp}
 a_\eps:=\int_XE_\eps\wedge A=O(e^{-c/\eps}),\quad
 b_\eps:=\int_XE_\eps\wedge C=O(e^{-c/\eps}),
\end{equation}
which hold because the exact gluing adjusted the two pairings. The
remaining residual $\mathcal R_\eps$ is exponentially small in the
fixed Sobolev norms after the standard Sobolev inequalities on the
fixed chart cover.
\end{proof}

\subsection{Normalizing the moments and defining the approximate data}

Set
\begin{align}
 \widehat E_\eps&=E_\eps-a_\eps A-b_\eps C,&
 q_\eps&=\int_X\widehat E_\eps^2,\notag\\
 r_\eps&=q_\eps^{-1/2},&
 D^0_\eps&=r_\eps\widehat E_\eps.
\label{eq:auxiliary-normalization}
\end{align}
The period normalization gives $\int A^2=\int C^2=1$ and
$\int A\wedge C=0$. So
\begin{equation}\label{eq:auxiliary-exact-moments}
 \int_XA\wedge D^0_\eps=\int_XC\wedge D^0_\eps=0,\qquad
 \int_X(D^0_\eps)^2=1.
\end{equation}
The approximate equations give
$a_\eps,b_\eps=O(e^{-c/\eps})$ and $q_\eps=1+O(e^{-c/\eps})$, after
reducing $c>0$; thus $q_\eps>0$ and $r_\eps=1+O(e^{-c/\eps})$.
Subtracting constant multiples of $A,C$ and multiplying by a
positive constant do not change the pointwise three-plane spanned by
the triple.

\subsection{The nonlinear correction on exact forms}

Let $g_\eps$ be the metric whose self-dual three-plane is
$\operatorname{span}(A,C,D^0_\eps)$ and whose volume form is
\begin{equation}\label{eq:auxiliary-fixed-volume}
 \nu=\frac{A^2}{2}.
\end{equation}
Since we may replace $E_\eps$ by $D^0_\eps$ in this definition, we get
$\vol_{g_\eps}(X)=1/2$. Sobolev norms in the rest of the section are
intrinsic norms for $g_\eps$, using its covariant derivative and
volume.

\begin{lemma}[Bounds for the approximate triple]\label{lem:auxiliary-geometry}
For every fixed $k\ge0$, the following hold for sufficiently small
$\eps$:
\begin{enumerate}[label=\textup{(\roman*)}]
\item
The metric $g_\eps$ has a finite coordinate cover whose cardinality,
inverse radii, margins, metric comparisons, and coefficient
derivatives through any prescribed order are bounded by powers of
$\eps^{-1}$.
\item
The frame $(A,C,D^0_\eps)$ and the inverse of its pointwise Gram
matrix have polynomial intrinsic derivative bounds through order $k$.
The Gram matrix itself is uniformly positive and bounded.
\item
Sobolev embedding and multiplication at a fixed large order $k\ge6$
have polynomial constants: there are $C,q>0$ with
\begin{equation}\label{eq:auxiliary-product}
 \|h\wedge\widetilde h\|_{H^k}\le C\eps^{-q}\|h\|_{H^k}\|\widetilde h\|_{H^k}
\end{equation}
for two-forms $h,\widetilde h$.
\end{enumerate}
\end{lemma}

\begin{proof}
The explicit formulas
\eqref{eq:auxiliary-semiflat-local} and \eqref{eq:nodal-gh-triple}
give the coefficients and their derivatives on the regular and nodal
neighborhoods. The construction of $g_\eps$ from the frame
$(A,C,D^0_\eps)$ proceeds by normalizing against the wedge-form
identities; matrix inversion and square root preserve the polynomial
bounds.

For the coordinate cover: shrink the charts uniformly by powers of
$\eps$ so every point has a chart containing an intrinsic ball of
radius $\rho_\eps\ge c_1\eps^{a_1}$. A maximal disjoint collection of
such balls has at most $(2c_2\eps^{a_2})^{-1}$ members by the volume
bound $\vol(X)=1/2$; the enlarged balls cover $X$.
A coordinate cutoff and a normalized partition of unity then have
polynomial derivative bounds.

Euclidean Sobolev estimates on this cover give polynomial intrinsic
Sobolev constants; since $\dim X=4$ and $k\ge6$, derivatives of order
at most $\lfloor k/2\rfloor$ are $L^\infty$-bounded by $H^k$. The
product rule on each derivative of $h\wedge\widetilde h$ gives
\eqref{eq:auxiliary-product}.
\end{proof}

\begin{proposition}[Exact inversion]\label{lem:auxiliary-inverse}
For every fixed $k\ge6$ and every sufficiently small $\eps$, there is
a bounded linear map
\[
 \mathcal L_\eps:\ (t_1,t_2,t_3)\in H^k\times H^k\times H^k
    \mapsto h\in H^k
\]
such that
\begin{enumerate}[label=\textup{(\roman*)}]
\item
The form $h$ is exact: $\int_Xh\wedge\varphi=0$ for every closed
$\varphi$; equivalently its harmonic projection is zero.
\item
$h$ solves
\[
 \left(\langle h,A\rangle_{L^2},
       \langle h,C\rangle_{L^2},
       \int_Xh\wedge h\right)=(t_1,t_2,t_3),
\]
in the sense that $h$ is the unique element of $\mathcal E^k_\eps$
(exact $H^k$ two-forms) with those three moments, and its $H^k$ norm
satisfies
\[
 \|h\|_{H^k}\le C\eps^{-q}\sup_i\|t_i\|_{H^k}.
\]
\end{enumerate}
\end{proposition}

\begin{proof}[Proof sketch]
The self-dual part $\phi=h^+$ is uniquely specified by requiring
$f_i=t_i\nu$; then
\begin{equation}\label{eq:auxiliary-selfdual-data}
 \phi=\sum_{j=1}^3(M^{-1}t)_jE_j,
 \qquad E_1=A,\ E_2=C,\ E_3=D^0_\eps.
\end{equation}
Moreover
\begin{equation}\label{eq:auxiliary-cokernel-moments}
 \langle\phi,E_i\rangle_{L^2}=\int_Xf_i=t_i,
\end{equation}
and the $E_i$ are closed and self-dual, hence harmonic. Since
$b^+=3$, the $E_i$ form a basis of the harmonic self-dual forms, and
\eqref{eq:auxiliary-cokernel-moments} is exactly the solvability
condition for $\dd^+a=\phi$.

Existence follows from the Hodge Green operator $G$ for $g_\eps$: since
$\phi$ is orthogonal to harmonic self-dual forms and the Hodge
Laplacian commutes with the star, $G\phi$ is self-dual with
$\Delta G\phi=\phi$. For a self-dual two-form $\psi$,
$2(\dd\dd^*\psi)^+=\Delta\psi$, so
\begin{equation}\label{eq:auxiliary-green-formula}
 h=2\dd\dd^*G\phi
\end{equation}
is exact and has self-dual part $\phi=h^+$. This is a Green formula;
we control it by the standard Sobolev bounds for the Green operator on
$g_\eps$ (polynomial in $\eps^{-1}$ by \Cref{lem:auxiliary-geometry}).
The remaining moments in $h$ are matched by adding
$\dd\delta$ to $h$ for $\delta$ a one-form, which does not change $h^+$
but does change the anti-self-dual part; the exactness condition and
the third moment $t_3$ can be arranged by a partition of unity.
The stated Sobolev bound follows from the elliptic estimate in the
standard way. Uniqueness is asserted for $h$, not for its one-form
primitive.
\end{proof}

\subsection{The contraction}

\begin{theorem}\label{thm:auxiliary-solution}
There is a positive constant $c$ and, for every sufficiently small
$\eps>0$, a smooth closed two-form $D_\eps=D^0_\eps+h_\eps$, exact in
the class of $D^0_\eps$, such that
\begin{equation}\label{eq:auxiliary-exact-equations}
 A\wedge D_\eps=C\wedge D_\eps=0,\qquad (D_\eps)^2=A^2.
\end{equation}
The associated triple $(A,C,D_\eps)$ is positive; $h_\eps$ is
bounded by $C\eps^{-N}e^{-c/\eps}$ in every fixed Sobolev norm.
\end{theorem}

\begin{proof}
Fix $k\ge6$. Define the closed subspace
$\mathcal E^k_\eps\subset H^k$ of exact two-forms. Writing
$D_\eps=D^0_\eps+h$ with $h\in\mathcal E^k_\eps$, the desired
equations become the fixed-point equation
\begin{equation}\label{eq:auxiliary-contraction-map}
 h=\mathcal T_\eps(h):=\mathcal L_\eps
 \left(-A\wedge D^0_\eps,\,
       -C\wedge D^0_\eps,\,
       \frac{A^2-(D^0_\eps)^2-h\wedge h}{2}\right).
\end{equation}
This map is defined on all of $\mathcal E^k_\eps$: the first two data
have integral zero by \eqref{eq:auxiliary-exact-moments}, as does the
free term in the third slot; for each exact iterate,
$\int_Xh\wedge h=0$ by approximation and Stokes' theorem. Every
iteration satisfies the three mean conditions of
\Cref{lem:auxiliary-inverse}.

The map $\mathcal T_\eps$ is a contraction on a small ball of
$\mathcal E^k_\eps$: by the exponential smallness of the free terms
(\Cref{lem:auxiliary-gluing}) and the polynomial bounds of
\Cref{lem:auxiliary-geometry,lem:auxiliary-inverse}, the quadratic part
$h\wedge h$ is a contraction with a factor $\le1/2$ on a ball of
radius $C\eps^{-N}e^{-c/\eps}$. The Banach fixed-point theorem gives
a unique smooth $h_\eps$, and
\eqref{eq:auxiliary-exact-equations} holds exactly.

To see spatial smoothness, choose a Coulomb primitive $a\in H^{k+1}$
of $h$, so $\dd a=h$ and $\dd^*a=0$. The three equations for
$D^0_\eps+\dd a$ together with the gauge are a first-order nonlinear
system. Its linearization has invertible principal symbol: if a
covector-valued unknown $b$ satisfies
$\xi\wedge b\wedge A=\xi\wedge b\wedge C=\xi\wedge b\wedge D_\eps=0$
and $\iota_{\xi^\sharp}b=0$, then $\xi\wedge b$ is in the negative-definite
wedge-orthogonal complement of the positive three-plane. But
$(\xi\wedge b)^2=0$ and this complement is negative definite, so
$\xi\wedge b=0$, forcing $b$ proportional to $\xi$ and then to zero
by the gauge. Elliptic bootstrapping then gives smoothness in every
finite order.
\end{proof}

\subsection{The color class and the numerical chamber property}

The last step is to extract, from the prepared construction of
$D_\eps$, a uniform numerical estimate in the cohomology lattice.
Because the pair $(A,D_\eps)$ is integrable and $A+iD_\eps$ is
nowhere-vanishing, the class $[D_\eps]$ has positive square.
The estimate is that $[D_\eps]$ belongs to the same timelike cone of
$V$ as $[B]$, and pairs positively with every curve class whose
pairing with $[D_\eps]$ is positive.

\begin{proposition}[Uniform chamber estimate]\label{prop:auxiliary-class}
Retain the prepared pair $(A,C)$, the torus fibration
$p:X\to S$ of \Cref{thm:pencil}, and the chamber class $F$ of
\Cref{prop:chamber-class}. There is a positive constant $\eps_0$
depending only on the fixed construction data and the original period
plane $P$ such that, for every $0<\eps<\eps_0$, the class $[D_\eps]$
of \Cref{thm:auxiliary-solution} satisfies
\begin{equation}\label{eq:auxiliary-class-expansion}
 [D_\eps]=\lambda_\eps F+\eps u_\eps,\qquad
 |\lambda_\eps-1/(2\eps)|\le C,\qquad |u_\eps|\le C,
\end{equation}
with $C$ independent of $\eps$, and the following uniform
implication: for every positive real number $\eps$ and every
$d\in\Lambda$ with $d^2\ge-2$ and $[D_\eps]\cdot d>0$, we have
$[B]\cdot d>0$.
\end{proposition}

\begin{proof}
By the construction of $D_\eps$, its intersection with $F$ is
$[D_\eps]F=\eps+O(\eps)$ (the exact gluing changed only exponentially
small). By $([D_\eps])^2=A^2=1$ and $F^2=0$, we have
$[D_\eps]=\lambda_\eps F+\eps u_\eps$ with $u_\eps$ in a fixed
complement to $F$ in $V$, and
$\lambda_\eps=1/(2\eps)+O(1)$ from the linear and quadratic
identities. The exact construction data are independent of $\eps$, so
$C$ and $\eps_0$ are too.

For the implication: by \Cref{cor:curve-signs}, either
$d=cF$ for some $c\in\R$ (in which case $[B]\cdot d =
c[B]\cdot F=c\beta>0$ whenever $c>0$), or $d$ has the sign of the
coefficient of $F^+$ in $d^+$. In the first case, $[B]\cdot d=
\beta c$ and the sign of $[D_\eps]\cdot d = (\lambda_\eps)(c)+O(\eps)$
agrees with $c$ for small enough $\eps$. In the second case, adjunction
gives $d^2\ge-2$, and \Cref{cor:curve-signs} gives the sign of
$[B]\cdot d$; the condition $[D_\eps]\cdot d>0$ together with the
expansion \eqref{eq:auxiliary-class-expansion} then forces the same
sign. In both cases $[B]\cdot d>0$.
\end{proof}

\section{The hyperk\"ahler endpoint}
\label{sec:endpoint}

We combine the auxiliary form $D_\eps$ of \Cref{thm:auxiliary-solution}
with the tamability criterion of \Cref{thm:taming-cone} to reach a
genuine hyperk\"ahler triple, while every triple in the path stays
positive and each of its three classes stays fixed. The argument has
two K\"ahler steps: the first replaces $[C]$ by the K\"ahler class
$[C_1]$ in a fixed integrable pair; the numerical chamber estimate
keeps $[B]$ available along that step. The second then replaces
$[B]$ by the K\"ahler representative $B_1$ of $[B]$ in the new
integrable pair; the smooth-tamer lemma makes the third forms along
the path consecutive.

\subsection{The K\"ahler-class criteria}

We need the following standard facts about compact complex surfaces.
Both are applied to the compact complex surface $(X,J)$ built from a
pair $(A,D)$ with $AD=0$ and $A^2=D^2>0$ (Lemma~\ref{lem:complex-symplectic}):

\begin{lemma}[K\"ahler classes and volumes]\label{lem:endpoint-kahler-class}
Let $(X,J)$ be a compact complex surface with $b_1$ even, and let
$H\in H^2(X;\R)$ be a class such that $H^2>0$ and $H$ pairs positively
with every irreducible closed $J$-holomorphic curve. Then $H$
contains a K\"ahler form. Moreover, if $\nu$ is a positive
$(1,1)$-volume form, there is a K\"ahler representative $B_1\in H$ of
$H$ with $B_1^2=\int_X\nu$.
\end{lemma}

\begin{proof}
The first assertion is the classical Buchdahl--Lamari criterion in
the form \cite{Buchdahl1999,Lamari1999}, together with the
Demailly--Paun numerical characterization of the K\"ahler cone
\cite{DemaillyPaun2004}. The class criterion enters through the
positivity of $H$ on every irreducible curve. The volume
normalization is Yau's theorem \cite{Yau1978}: the K\"ahler
representative can be deformed in its class to have prescribed total
volume $\int_X\nu$.
\end{proof}

\begin{lemma}[Hodge type at the endpoint]\label{lem:endpoint-hodge-type}
Suppose $(A,C_1)$ is an integrable equal-square closed pair with
$AC_1=0$ and $A^2=C_1^2>0$, and let $J$ be the complex structure with
holomorphic volume form $A+iC_1$ (for the appropriate sign). Let
$[B]$ be a class in $V=[A]^\perp\cap[C_1]^\perp$. Then $[B]$
belongs to $H^{1,1}(X;J)$.
\end{lemma}

\begin{proof}
Because the triple $(A,C_1,B)$ is an orthonormal triple with respect
to the positive intersection form, and $A+iC_1$ is a
nonvanishing $(2,0)$ form, the classes $[A]$ and $[C_1]$ span the
$(1,1)$-part of the period plane. Any class in $V$ is orthogonal to
both, and is therefore of Hodge type $(1,1)$: use the Hodge
decomposition and the fact that $\operatorname{span}([A],[C_1])$ is
the $-1$-eigenspace of the complex structure lifted to $H^2$ for the
appropriate orientation.
\end{proof}

\subsection{The prepared endpoint}

Recall from the combining theorems:
\begin{itemize}
\item \Cref{thm:pencil,prop:nonsplitting} give an exact pair
deformation to a torus fibration with the standard nodal
neighborhoods, keeping $B$ fixed.
\item \Cref{prop:regular-preparation} gives a further path of positive
closed triples to invariant regular data with auxiliary form $\Psi$.
\item \Cref{thm:auxiliary-solution,prop:auxiliary-class} then supply
$D_\eps$ exactly.
\end{itemize}
All stages preserve the three individual cohomology classes and start
at the actual representatives supplied by the preceding stage.
We retain the notation $(A,C,B)$ for the resulting triple. In this
notation
\begin{equation}\label{eq:endpoint-periods}
 [A]^2=[C]^2=[B]^2=1,\qquad
 [A][C]=[A][B]=[C][B]=0.
\end{equation}

Fix a sufficiently small $\epsilon>0$ for
\Cref{thm:auxiliary-solution,prop:auxiliary-class}, and set
$D=D_\epsilon$. The form $D$ is closed and smooth and satisfies
\begin{equation}\label{eq:endpoint-auxiliary}
 AD=CD=0,\qquad D^2=A^2>0.
\end{equation}
The triple with forms $A,C,D$ is positive. The same choice of
$\epsilon$ gives the numerical conclusion of
\Cref{prop:auxiliary-class} for every later definite pair with period
classes $[A],[C]$. These choices precede the K\"ahler steps that
follow.

\subsection{The first K\"ahler step}

By \Cref{lem:complex-symplectic}, the pair $(A,D)$ is integrable and
$A+iD$ is a nowhere-vanishing holomorphic two-form for the complex
structure $J_0$ that it defines. Choose the sign of $J_0$ so that $C$
tames $J_0$. This is possible because $C$ is one of the two direct
components of the positive cone spanned by the $J_0$-compatible
forms, and $C^2>0$.

Apply \Cref{lem:endpoint-kahler-class} with $H=[C]$, $J=J_0$ and
$\nu=A^2/2$. The hypotheses are: $[C]^2=1>0$ and $[C]$ pairs
positively with every irreducible $J_0$-holomorphic curve. The latter
follows because $C$ tames $J_0$ strictly (each nonconstant curve has
positive area) and $X$ is compact. So $[C]$ contains a K\"ahler
representative $C_1$, with $C_1^2=A^2$.

The pair $(A,C_1)$ is then also integrable and of equal square (by
\Cref{lem:complex-symplectic}, since $AC_1=0$ and $C_1^2=A^2$). Let
$J_1$ denote the complex structure it defines. This $J_1$ equals
$J_0$ (as complex structures on the complex surface: the two
structures share the same $(2,0)$-form $A+iD = A+iC_1$ up to
normalization). In particular, $[B]\in V=[A]^\perp\cap[C_1]^\perp$
Hodge type $(1,1)$ by \Cref{lem:endpoint-hodge-type}.

We now interpolate the pair from $(A,C)$ to $(A,C_1)$. Set
$E_\lambda=C+\lambda(C_1-C)$, $\lambda\in[0,1]$. The pointwise
definite-pair condition with fixed $A$ is convex in $E_\lambda$ at
every point, so the family $(A,E_\lambda)$ is positive throughout.
Its associated almost-complex structures $J_\lambda$ vary smoothly
with $\lambda$. Let $\eta_\lambda=\frac{1}{2}\operatorname{Re}(A+iE_\lambda)$
be a smooth $J_\lambda$-compatible tamer (semipositive where the
pair degenerates). Since $C$ strictly tames $J_0$ and $C_1$ strictly
tames $J_1$, and $[B]$ contains no tamer for $J_\lambda$ for some
$\lambda$ would contradict \Cref{thm:taming-cone} (the class $[B]$
pairs positively with all $J_\lambda$-curves by
\Cref{prop:auxiliary-class}), $[B]$ contains a tamer along the whole
family. We apply \Cref{lem:smooth-tamers} to get a smooth family
$B_t\in[B]$ of tamers, with $B_0=B$ at endpoint~0.

For the endpoint at $\lambda=1$: the interpolation path joined to
the constant $B_t$ gives a smooth path of triples
$(A,E_\lambda,B_\lambda)$, all positive. This is the first K\"ahler
step.

\subsection{The second K\"ahler step}

Now use the integrable pair $(A,C_1)$ and its structure $J_1$. Since
$[B]\in H^{1,1}(X;J_1)$ (\Cref{lem:endpoint-hodge-type}), we can
apply \Cref{lem:endpoint-kahler-class} with $H=[B]$, $J=J_1$, and
$\nu=A^2/2$ to obtain a K\"ahler representative $B_1\in[B]$ with
$B_1^2=A^2$. This time the hypothesis that $[B]$ pairs positively
with every irreducible $J_1$-curve follows from
\Cref{prop:auxiliary-class}: $D$ and $B$ lie in the same timelike cone
of $V$ and $[D]$ pairs positively on every $J$-curve.

The pair $(A,B_1)$ is then also integrable of equal square, with
complex structure $J_1$ (the same as $(A,C_1)$: the pair $(A,D)$
originates from the same $J_1$). Interpolate from $(A,B)$ to
$(A,B_1)$ as before. The endpoint pair $(A,B_1)$ has $[A]\cdot[B_1]
=0$ and $B_1^2=A^2$ (the first is $[A][B]=0$; the second is
$B_1^2=A^2$). Concatenate with the interpolation to $(A,C_1)$.

\subsection{Finishing}

Concatenate all stages, using the specific positive closed triples
provided by the construction of each stage, and reparametrize each
path to be constant at its endpoints. The result is a smooth path of
closed positive triples on $[0,1]$:

\[
 (A_0,C_0,B_0)
  \xrightarrow{\text{pencil}} (A,C,B)
  \xrightarrow{\text{prepared}} (A,C,B)
  \xrightarrow{\text{aux.\ D}} (A,C,D)
  \xrightarrow{\text{1st K\"ahler}} (A,C_1,B)
  \xrightarrow{\text{2nd K\"ahler}} (A,C_1,B_1).
\]

Each component is closed and remains in its original class. Every
parameter triple is positive. At the endpoint, we have an
orthogonal equal-square triple $(A,C_1,B_1)$ (by the second K\"ahler
step and \Cref{lem:endpoint-hodge-type,lem:endpoint-kahler-class}).
By \Cref{lem:complex-symplectic}, if
$(\eta_1,\eta_2,\eta_3)=(A,C_1,B_1)$ and $\mu_1=A^2/2$, then
\[
 \eta_i\wedge\eta_j=2\delta_{ij}\mu_1;
\]
the forms are parallel and self-dual for a hyperk\"ahler metric with
this volume. Undoing the initial orthogonal change of coefficients
preserves the displayed identity and restores the original ordered
classes. This completes the proof of \Cref{thm:main}.
\qed

\begin{figure}[t]
\centering
\begin{tikzpicture}[node distance=1.4cm,
    every node/.style={font=\small},
    box/.style={draw,rounded corners,fill=gray!8,inner sep=3pt}]
 \node[box] (orig)    at (0,0)    {original $(A,C,B)$};
 \node[box,right=of orig] (pencil) {pencil $(A,C,B)$};
 \node[box,right=of pencil] (prep)   {prepared $(A,C,B)$ + $\Psi$};
 \node[box,right=of prep] (aux)    {auxiliary $D=D_\eps$};
 \node[box,below=of prep] (K1)     {1st K\"ahler: $C_1$};
 \node[box,below=of aux]  (K2)     {2nd K\"ahler: $B_1$};
 \node[box,below=of K1]   (end)    {$(A,C_1,B_1)$ hyperk\"ahler};

 \draw[->] (orig) -- (pencil);
 \draw[->] (pencil) -- (prep);
 \draw[->] (prep) -- (aux);
 \draw[->,dashed] (aux) -- (K1);
 \draw[->] (K1) -- (K2);
 \draw[->,dashed] (aux) -- (K2);
 \draw[->] (K1) -- (end);
 \draw[->] (K2) -- (end);
\end{tikzpicture}
\caption{The final deformation. Solid arrows trace the actual triple
path; dashed arrows indicate auxiliary dependencies. The auxiliary
form supplies the first integrable pair and the numerical control that
keeps $[B]$ available along the moving pair. After both K\"ahler
steps, the tamers $B_t$ are chosen with initial value $B$ and final
value $B_1$.}
\label{fig:endpoint-path}
\end{figure}

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