# Complete scalar-flat Kähler 4-manifolds with a continuous symmetry

Brian Weber

(Date: September 1, 2023)

###### Abstract.

We study complete scalar-flat Kähler manifolds with a Killing field and a mild
asymptotic condition. We show that topological and geometric rigidities exist
that powerfully restrict the manifold’s behavior at infinity. We create a
rough classification for scalar-flat Kähler manifolds with a Killing field.

##  1\. Introduction

Kähler manifolds with zero scalar curvature and a continuous symmetry have
received a great deal of attention. Numerous metrics of this kind are known in
the compact [26], [3] and non-compact cases [15] [52] and a variety of ansätze
and construction methods have been built [29] [25] [2] [19]. Treatments are
available from perspectives such as twistor theory, Einstein-Weyl geometry,
toric Kähler geometry, and under asymptotic rigidity assumptions; see for
example [47], [14], [6], [20], [33], [53]. Even still, the variety of complete
metrics of this type has not been understood.

Without any asymptotic conditions a full classification of the complete case
is probably out of reach. In this paper we take the asymptotic condition to be
a lower bound on the norm of the Killing field. This condition appears
naturally, for example, in the study of ALE or ALF metrics. Additionally, in
creating singularity models for collapsing Kähler manifolds, an F-structure
[10] on the collapsing sequence appears. In the limit this leads, under the
right controls, to a Killing field on a complete manifold whose norm is
bounded strictly away from zero.

To explain our results, let $(M^{4},J,g_{4},V)$ be a Kähler manifold where $V$
is a symplectomorphic Killing field. If $M^{4}$ is simply connected we can
solve $dz=-i_{V}\omega$ up to a constant; if not we can always solve for $z$
locally, or pass to the universal cover where we can solve for $z$ globally.
The function $z$ is called a Killing potential or the momentum for $V$. Level-
sets $M^{3}_{z}=\\{z=const\\}$ are manifolds away from zeros of $V$, and the
intrinsic volume form of $M^{3}_{z}$ is
$dVol_{3}=dVol_{4}({\frac{\nabla{}z}{|\nabla{}z|}},\cdot,\cdot,\cdot)$. The
Killing field $V$ passes to each $M^{3}_{z}$, where we may take a Riemannian
quotient to obtain an orbifold $M^{2}_{z}$, called the Kähler reduction at
$z=const$. Each $M^{2}_{z}$ is a smooth Riemannian orbifold that inherits a
metric and complex structure from $M^{4}$; we denote its Gaussian curvature by
$K$ and its intrinsic volume form by $dVol_{2}$. By pullback, the function $K$
and $2$-form $dVol_{2}$ also exists on the parent manifold $M^{4}$. In the
scalar-flat case, away from zeros of $V$, we prove the relations

(1) $\displaystyle
K\,dVol_{4}=d\left[\frac{\triangle_{4}z}{|V|}dVol_{3}\,-\,Jd\log|V|^{2}\wedge*dVol_{2}\right],\quad\text{and}$
(2) $\displaystyle 2|\mathop{\rm
Ric}{}_{4}|^{2}\,dVol_{4}=d\left[\mathcal{L}_{\frac{\partial}{\partial{}z}}\left(\left(\frac{\triangle_{4}z}{|V|}\right)^{2}V_{\flat}\wedge{}dVol_{2}\right)-Jd\left(\frac{\triangle_{4}z}{|V|}\right)^{2}\wedge*dVol_{2}\right].$

Eq. (1) is a global form of the “Toda-lattice equation” (see Section 4) and
Eq. (2) is ultimately a Chern-Gauss-Bonnet identity. Using an integration
relation between $M^{4}$ and the reductions $M^{2}_{z}$ (Lemma 2.4), the
following evolution equations are more-or-less immediate consequences where
$M^{2}_{z}$ is compact and $|V|\neq 0$:

(3)
$\displaystyle\int_{M^{2}_{z}}K\,dVol_{2}=\frac{d}{dz}\int_{M^{2}_{z}}\triangle_{4}z\,dVol_{2},$
(4) $\displaystyle\int_{M^{2}_{z}}|\mathop{\rm
Ric}{}_{4}|^{2}\,dVol_{2}=\frac{1}{2}\frac{d^{2}}{dz^{2}}\int_{M^{2}_{z}}(\triangle_{4}z)^{2}\,dVol_{2}.$

Figure 1. Schematic depiction of the Kähler reduction. Each level-set
$M^{3}_{z}$ inherits a Killing field. When $z$ is nonsingular then $M^{3}_{z}$
is a manifold and its quotient $M^{2}_{z}$ is a Riemannian orbifold.

Before stating the theorems we address the issue of finding the potential
function $z$. A topological obstruction exists in $H^{1}_{dR}(M^{4})$, but the
function $\triangle_{4}z$ and field $\partial/\partial{}z$ exist regardless.
We approach this more systematically in Section 2, but for now we state that
$\triangle_{4}z=*(dV\wedge\omega)$ which exists globally, and we will see
$\partial/\partial{}z=-|V|^{-2}(i_{V}\omega)^{\sharp}$ which exists everywhere
except at zeros of $V$. The level-sets $M^{3}_{z}$ and quotients $M^{2}_{z}$
exist even if $z$ does not exist. This is because the distribution

(5) $\mathcal{D}\;=\;\\{v\in{}TM^{4}\;\big{|}\;\omega(v,V)=0\\}$

is always integrable, as a consequence of $d(i_{V}\omega)=0$. Supposing $V$ is
Hamiltonian and a potential $z$ exists, then
$\mathcal{D}=(\nabla{}z)^{\perp}$. But if $V$ is Hamiltonian or not, we may
still assume every leaf of $\mathcal{D}$ is an embedded 3-manifold, except
where a leaf intersects the singular locus $\\{V=0\\}$. One can, for example,
pass to the universal cover of $M^{4}$, where $V$ is indeed Hamiltonian. Lemma
2.6 gives an explicit criterion for Hamiltonicity. Equations (1) and (2) hold
everywhere except at zeros of $V$, even if $V$ is not Hamiltonian.

We address another a possible complication in the formation of the orbifolds
$M^{2}_{z}$. The field $V$ passes to each $M^{3}_{z}$, but if its orbits are
not closed then the Riemannian quotient $M^{2}_{z}$ is non-Hausdorff; for this
reason we only consider the case of Killing fields with closed orbits. This is
not too serious a restriction however. If the orbits are non-closed, then the
closure of its diffeomorphism action has rank at least 2, and we can find a
second Killing field. The case of two symmetries has been dealt with elsewhere
(see for example [15], [53] and references therein), so when orbits are non-
closed, one can refer to those works. If orbits of $V$ are closed but
unbounded, we can take a discrete quotient along its diffeomorphism flow, and
obtain closed orbits. We assume throughout this paper that the diffeomorphism
flow of $V$ is parameterized by the unit circle with parameter $t\in[0,2\pi)$.

######  Theorem 1.1.

Assume a compact set $K\subset{}M^{4}$ and number $\epsilon>0$ exist so
$|V|>\epsilon$ on $M^{4}\setminus{}K$. Then every integral leaf $M^{3}_{z}$ is
compact, or every integral leaf is non-compact.

If the asymptotic condition $|V|>\epsilon$ is replaced by only $|V|>0$, then
Theorem 1.1 is false. See §7.4 for an example.

The geometric Euler number of $M^{2}_{z}$, also called its orbifold Euler
number, is

(6) $\chi^{g}(M^{2}_{z})\;=\;\frac{1}{2\pi}\int_{M^{2}_{z}}K\,dVol_{2}$

(see [37]). If $\chi(M^{2})$ is its topological Euler number, we have the
well-known relation
$\chi(M^{2})=\chi^{g}(M^{2})+\sum_{i}(1-|\Gamma_{i}|^{-1})$ where the sum is
taken over all orbifold points $p_{i}$ and $|\Gamma_{i}|$ is the cardinality
of the orbifold group at $p_{i}$. In particular
$\chi(M^{2})\geq\chi^{g}(M^{2})$ with equality if and only if $M^{2}$ is a
manifold.

When $M^{3}_{z}$ is non-singular then $M^{3}_{z}\rightarrow{}M^{2}_{z}$ is an
$\mathbb{S}^{1}$ fibration and $M^{3}_{z}$ is a Seifert fibered space. When
compact, the fibration’s geometric Euler number is

(7)
$e^{g}(M^{3}_{z},M^{2}_{z})\;=\;\frac{1}{4\pi^{2}}\int_{M^{3}_{z}}|V|^{-4}V_{\flat}\wedge{}dV_{\flat}$

under the normalizing condition that orbits of $V$ are closed and
parameterized by $t\in[0,2\pi)$. If $M^{3}_{z}\rightarrow{}M^{2}_{z}$ is not
just a fibration but a fiber bundle with Euler number $e$, then
$e=e^{g}(M^{3}_{z},M^{2}_{z})$. A second interpretation of (7) is that its
integrand is the natural Chern-Simons 3-form on $M^{3}_{z}$ associated to the
fibration; see [36], and the proof of Proposition 3.7 below.

######  Theorem 1.2.

Assume the leaves $M^{3}_{z}$ are compact. Away from zeros of $V$ the
quantities $\frac{d}{dz}\int\,dVol_{2}$ and $e^{g}(M^{3}_{z},M^{2}_{z})$ are
constant and

(8)
$\frac{d}{dz}\int_{M^{2}_{z}}\,dVol_{2}\;=\;2\pi{}e^{g}(M^{3}_{z},M^{2}_{z}).$

Assume in addition that $M^{4}$ is scalar-flat. Away from zeros of $V$, the
quantities $\frac{d^{2}}{dz^{2}}\int|V|^{2}dVol_{2}$,
$\frac{d}{dz}\int\triangle_{4}z\,dVol_{2}$, and $\chi^{g}(M^{2}_{z})$ are
constant and

(9)
$\frac{d^{2}}{dz^{2}}\int_{M^{2}_{z}}|V|^{2}\,dVol_{2}\;=\;\frac{d}{dz}\int_{M^{2}_{z}}\triangle_{4}z\,dVol_{2}\;=\;4\pi\chi^{g}(M^{2}_{z}).$

When $M^{4}$ is scalar flat, we also have

(10)
$\frac{d^{2}}{dz^{2}}\int_{M^{2}_{z}}(\triangle_{4}z)^{2}\,dVol_{2}\;=\;2\int_{M^{2}_{z}}|\mathop{\rm
Ric}|^{2}\,dVol_{2}.$

Based on Theorems 1.1 and 1.2, we create the following rough classification
for scalar-flat Kähler manifolds $M^{4}$ with a symplectomorphic Killing field
$V$:

  * •

Type I: $M^{4}$ is compact.

  * •

Type II: $M^{4}$ is non-compact, and $|V|>\epsilon>0$ outside a compact set.

    * –

Type IIa: the distribution $\mathcal{D}$ has compact leaves

    * –

Type IIb: the distribution $\mathcal{D}$ has non-compact leaves

  * •

Type III: a divergent sequence of points $p_{i}$ exist with
$\lim_{i}|V(p_{i})|=0$.

    * –

Type IIIa: outside some compact set, $|V|>0$

    * –

Type IIIb: a divergent sequence of points $p_{i}$ exist so that
$|V(p_{i})|=0$.

In addition to Theorems 1.1 and 1.2, another motivation for dividing Type II
from Type III is that Type II metrics appear as “deepest bubble” blowup limits
of collapsing extremal Kähler manifolds. A full classification of Type II
metrics would have consequences for the study of Kähler 4-manifolds generally.
About compact scalar-flat manifolds a great deal has been written; see for
example [31] [26] [40] [4] [32] [11]. Hope for classifying Type III metrics
seems out of reach. This paper focuses on the Type II case.

######  Lemma 1.3 (cf. Lemma 2.8).

Assume $M^{4}$ is Type IIa. Then $M^{4}$ has at most two ends, the ends
characterized by $z$ being unbounded from above, or else from below.

Recall that a manifold end is an unbounded open set with compact boundary. In
the Type IIa case $|dz|=|V|$ is asymptotically bounded away from zero, level-
sets are compact, and $V$ is holomorphic so its zeros cannot accumulate; thus
the set of critical values $\mathcal{C}=\\{z_{0},\dots,z_{N}\\}$ of $z$ is
finite. For $i\in\\{1,\dots,N\\}$ define
$M^{4}_{i}=\\{a\in{}M^{4}\;|\;z_{i-1}<z(a)<z_{i}\\}$. Set
$M^{4-}=\\{a\in{}M^{4}\;|\;z(a)<z_{0}\\}$ and
$M^{4+}=\\{a\in{}M^{4}\;|\;z_{N}<z(a)\\}$. We have

(11)
$M^{4}\;=\;\overline{M^{4-}}\,\cup\,\overline{M^{4}_{1}}\,\cup\,\dots\,\cup\,\overline{M^{4}_{N}}\cup\,\overline{M^{4+}}$

and by Lemma 1.3, $M^{4+}$ and $M^{4-}$ are the two possible manifold ends.

######  Theorem 1.4 (cf. Corollary 4.4).

Assume $M^{4}$ is Type IIa and $z$ has a non-empty set of critical points
$\mathcal{C}$. Then on any $M^{4}_{i}$, on $M^{4+}$, or on $M^{4-}$, the
level-sets $M^{3}_{z}$, reductions $M^{2}_{z}$, and fibrations
$M^{3}_{z}\rightarrow{}M^{2}_{z}$ are isotopic under the diffeomorphism flow
of $\frac{\partial}{\partial{}z}$.

If $M^{4+}\neq\varnothing$, then $\chi^{g}(M^{2}_{z})\geq 0$ and
$e^{g}(M^{3}_{z},M^{2}_{z})\geq 0$ on $M^{4+}$. If $M^{4-}\neq\varnothing$,
then $\chi^{g}(M^{2}_{z})\geq 0$ and $e^{g}(M^{3}_{z},M^{2}_{z})\leq 0$ on
$M^{4-}$.

In the Type III case level-sets need not be isotopic; see §7.4. A
classification of the topologies of Type IIa manifold ends follows from Lemma
1.3 and Theorem 1.4, combined with the classification of Seifert fibered
spaces. Type IIa manifolds have one or two ends, and each end has the
asymptotic topology of a ray crossed with a Seifert fibered space with
$\chi^{g}(M^{2}_{z})\geq 0$ and either $e^{g}(M^{3}_{z},M^{2}_{z})\geq 0$ or
$e^{g}(M^{3}_{z},M^{2}_{z})\leq 0$. They fall into the following families.

If $\chi^{g}(M^{2}_{z})=0$ and $e^{g}(M^{3}_{z},M^{2}_{z})=0$, the Thurston
geometry of $M^{3}_{z}$ is Euclidean, $M^{3}_{z}$ is a compact quotient of
$\mathbb{R}^{3}$ by a crystallographic group. There are only three cases that
are orientable with orientable base: the 3-torus, a fibration over
$\mathbb{S}^{2}$ with four exceptional orbits each of multiplicity $2$, and a
fibration over $\mathbb{S}^{2}$ with three exceptional orbits each of
multiplicity $3$.

If $\chi^{g}(M^{2}_{z})=0$ and $e^{g}(M^{3}_{z},M^{2}_{z})\neq 0$, the
Thurston geometry of $M^{3}_{z}$ is Nil, so $M^{3}_{z}$ is a quotient of
$\mathbb{R}^{3}$ by a discrete subgroup of the Heisenberg group (see §4 of
[41]). This is the family of nilmanifolds, which are classified by
$e^{g}(M^{3}_{z},M^{2}_{z})$.

If $\chi^{g}(M^{2}_{z})>0$ and $e^{g}(M^{3}_{z},M^{2}_{z})=0$ then the
Thurston geometry of $M^{3}_{z}$ is $\mathbb{S}^{2}\times\mathbb{R}$. By
Theorem 1.5 we must have $\int_{M^{4}}|\mathop{\rm Ric}|^{2}=\infty$. Only two
oriented compact 3-manifolds of this type, which are
$\mathbb{S}^{2}\times\mathbb{S}^{1}$ and $\mathbb{R}P^{3}\\#\mathbb{R}P^{3}$
(which has $\mathbb{S}^{2}\times\mathbb{S}^{1}$ as a double cover). However
$\mathbb{R}P^{3}\\#\mathbb{R}P^{3}$ has base $\mathbb{R}P^{2}$ which is non-
orientable, so the only possibility for $M^{3}_{z}$ is
$\mathbb{S}^{2}\times\mathbb{S}^{1}$.

Lastly if $\chi^{g}(M^{2}_{z})>0$ and $e^{g}(M^{3}_{z},M^{2}_{z})\neq 0$ then
$M^{3}_{z}$ has spherical Thurston geometry, so $M^{3}_{z}$ is covered by a
3-sphere and $M^{2}_{z}$ is a 2-sphere with up to three orbifold points. There
are infinitely many topologies in this case, which fall into finitely many
families.

More detailed information on any of these cases is available; see for example
[39], [37], [24], or [41]. For “Thurston geometries” generally see [45].

We prove two theorems in the case Ricci curvature has a bound.

######  Theorem 1.5 (cf. Prop. 6.2).

Assume $M^{4}$ is Type IIa and $\int_{M^{4}}|\mathop{\rm Ric}|^{2}<\infty$. If
$e^{g}(M^{3}_{z},M^{2}_{z})=0$ on $M^{4+}$ (resp. $M^{4-}$) then
$\chi^{g}(M^{2}_{z})=0$ on $M^{4+}$ (resp. $M^{4-}$).

######  Proposition 1.6 (cf. Prop. 6.3).

Assume $M^{4}$ is Type IIa, non-flat, and Ricci flat. Then $M^{4}$ is one-
ended, $\triangle_{4}z$ is a non-zero constant, and
$2e^{g}(M^{3}_{z},M^{2}_{z})\cdot\triangle_{4}z=\chi^{g}(M^{2}_{z})$ for non-
critical $z$. If $\triangle_{4}z>0$ then $M^{4-}=\varnothing$. If
$\triangle_{4}z<0$ then $M^{4+}=\varnothing$.

We have two results in the case that $V$ is globally bounded from below.

######  Proposition 1.7 (cf. Prop. 5.2).

Assume $M^{4}$ is Type I and its Killing field is nowhere-zero. Then $M^{4}$
is flat.

######  Theorem 1.8 (cf. Theorem 6.6).

Assume $M^{4}$ is Type IIa and non-flat, $|V|$ has a global lower bound
$|V|>\epsilon>0$, and $|\mathop{\rm Ric}|$ has an upper bound.

Then $M^{4}$ is 2-ended, $z$ ranges from $-\infty$ to $\infty$, we have
$\chi^{g}(M^{2}_{z})\equiv 2$, $e^{g}(M^{3}_{z},M^{2}_{z})\equiv 0$ and
$L^{2}(|\mathop{\rm Ric}|)=\infty$. All level-sets are diffeomorphic to
$\mathbb{S}^{2}\times\mathbb{S}^{1}$, and $M^{4}$ is diffeomorphic to
$\mathbb{R}\times\mathbb{S}^{2}\times\mathbb{S}^{1}$.

The main example of a Type IIa metric with a global lower bound on $|V|$ is
the scalar-flat cross product $\mathbb{S}^{2}\times{}\mathbb{H}^{2}$ where the
hyperbolic factor has a hyperbolic Killing field; see §7.2. Theorem 1.8 says
any Type IIa manifold with $V$ bounded away from $0$ qualitatively resembles
this. Theorem 1.8 is false if only $|V|>0$.

The following conjecture, if true, classifies the case where $|V|>\epsilon>0$
and level-sets are non-compact. Compare with [51] which proved the Ricci-flat
equivalent, and Theorem 1.6 of [54] for the case with two symmetries instead
of one.

######  Conjecture 1.9.

Assume $M^{4}$ is Type IIb and $|V|$ has a global lower bound
$|V|>\epsilon>0$. Then $M^{4}$ is flat.

In closing, we remark that scalar-flat Kähler metrics with a continuous
symmetry are locally determined by solutions of the LeBrun equations [30]

(12) $u_{xx}+u_{yy}+(e^{u})_{zz}=0\quad\quad w_{xx}+w_{yy}+(we^{u})_{zz}=0.$

The nonlinearity of these equations and their fundamentally local nature make
global questions difficult to address, however. Another approach that has been
explored is the momentum or Hamiltonian constructions in Kähler geometry [25],
[2], [3]. New in our approach are Eqs. (1) and (2), the global nature of which
provide new controls that are especially useful in the non-complete case.

In Section 2 we establish basic properties of the coordinate system, and prove
Lemma 1.3 and Theorems 1.1 and 1.4. In Sections 3 and 4 we prove Eq. (1), and
prove Theorem 1.2 except (10). In Section 5 we use the Chern-Gauss-Bonnet
formalism to prove (2) and (10), and prove Proposition 1.7. In Section 6 we
prove Theorems 1.5, 1.6 and 1.8. Section 7 provides examples.

##  2\. Basic properties of $M^{4}$

Here we lay out the basic facts about Kähler 4-manifolds with a Killing field,
and prove Theorem 1.2. Section 2.1 establishes facts about our preferred
framing, the “mixed frame”
$\\{\nabla{}x,\nabla{}y,\nabla{}z,\frac{\partial}{\partial{}t}\\}$. Section
2.2 studies volume forms and proves the “Integration Lemma” referred to in the
Introduction. Section 2.3 establishes some global properties of $M^{4}$ and
includes the proof of Theorem 1.2. We first record some standard Kähler
identities we will use frequently. For any function $f$ and 1-form $\alpha$ we
have

(13)
$\begin{array}[]{ll}dVol_{4}=\frac{1}{2}\omega\wedge\omega,&d\alpha\wedge\omega=-\frac{1}{2}div_{4}(J\alpha^{\sharp})\,\omega\wedge\omega,\\\
(dJdf)^{+}=-\frac{1}{2}\left(\triangle_{4}{}f\right)\omega,&df\wedge{}Jdg\wedge\omega=-\frac{1}{2}\left<\nabla{}f,\,\nabla{}g\right>\omega\wedge\omega.\end{array}$

Because $V$ is symplectomorphic we have $\mathcal{L}_{V}\omega=0$, and because
$d\omega=0$ and $\mathcal{L}_{V}=di_{V}+i_{V}d$ we have $0=d(i_{V}\omega)=0$.
Therefore we can solve $dz=-i_{V}\omega$ locally, although an obstruction to a
global solution may exist in the deRham space $H^{1}_{dR}(M^{4})$. We
frequently use the following relations:

(14) $dz\;=\;-i_{V}\omega,\quad\quad
V_{\flat}\;=\;i_{\nabla{}z}\omega,\quad\quad
dz\;=\;JV_{\flat},\quad\quad\nabla{z}\;=\;-JV.$

where $V_{\flat}=\left<V,\,\cdot\right>$ and we use the convention
$J(\eta)=\eta\circ{}J$ when $\eta\in\bigwedge^{1}$. This causes a sign to
appear sometimes, due to the fact that $J(V_{\flat})=-(JV)_{\flat}$.

###  2.1. The mixed frame

We use the LeBrun coordinates $(x,y,z,t)$ of [30]. To construct these
coordinates let $M^{3}_{0}$ be a non-singular leaf of $\mathcal{D}$; on
$M^{3}_{0}$ we create three local coordinates $(x,y,t)$. To create the
function $t$, choose any transversal to $V$ within $M^{3}_{0}$, declare $t=0$
on this transversal, and push the value of $t$ along the action of $V$. We’ve
assumed the flow of $V$ is parameterized by the unit circle, so we may take
$t\in[0,2\pi)$. On the quotient $M^{2}_{0}$ of $M^{3}_{0}$, we may place local
isothermal coordinates $x$ and $y$. Pulling back along the quotient, we obtain
functions $x$, $y$ on $M^{3}_{0}$. Because the flow of $V$ is by isometries,
we retain the relations $\left<\nabla{}x,\nabla{}y\right>=0$ and
$|\nabla{}x|^{2}=|\nabla{}y|^{2}$.

On any neighnorhood of $M^{3}_{0}$ that is deformation retractable onto
$M^{3}_{0}$ we can solve $dz=-i_{V}\omega=JV_{\flat}$ for $z$, so we have a
locally defined function $z$ with $\nabla{}z=-JV$. Finally we push the
functions $x$, $y$, and $t$ along the diffeomorphism flow of $\nabla{}z$ to
obtain functions $(x,y,t)$ on a region of $M^{4}$. We remark that because $t$
parameterizes the flow of $V$ and
$\frac{\partial{}x}{\partial{}t}=\frac{\partial{}y}{\partial{}t}=\frac{\partial{}z}{\partial{}t}=0$,
we have the relation $\frac{\partial}{\partial{}t}=V$.

For computations we find the “mixed frame”
$\\{\nabla{}x,\nabla{}y,\nabla{}z=-JV,\frac{\partial}{\partial{}t}=V\\}$ to be
most convenient. Lemmas 2.1 and 2.2 establish its basic properties.

######  Lemma 2.1.

$\text{Hess}\,z$, $\nabla{}V_{\flat}$, and $\nabla{}V$ are $J$-invariant in
the sense that

(15)
$\nabla_{Jv}J\nabla{}z=-\nabla_{v}\nabla{}z,\;\;\nabla_{Jv}JV_{\flat}=-\nabla_{v}\nabla{}V_{\flat},\;\;\text{Hess}\,z(J\cdot,J\cdot)=\text{Hess}\,z(\cdot,\cdot)$

where $v$ is any vector field. We have the following relation:

(16)
$(\text{Hess}\,z)(J\cdot,\,\cdot)\;=\;\nabla{}V_{\flat}\;=\;\frac{1}{2}dV_{\flat}.$

The Lie derivatives of $g$, $\omega$, and $J$ along $\nabla{}z$ are

(17)
$\mathcal{L}_{\nabla{}z}g=2\text{Hess}\,z,\quad\mathcal{L}_{\nabla{}z}\omega=2\nabla{}V_{\flat}=dV_{\flat},\quad\text{and}\quad\mathcal{L}_{\nabla{}z}J\;=\;0$

###### Proof.

The first two expression in (15) are immediate from the third. To prove the
third, the fact that $J$ is covariant-constant and compatible with the metric
gives

(18) $\displaystyle\left<\nabla_{v}\nabla{}z,w\right>$
$\displaystyle=\left<\nabla_{v}V,Jw\right>=-\left<\nabla_{Jw}V,v\right>=\left<\nabla_{Jw}\nabla{}z,Jv\right>=\left<\nabla_{Jv}\nabla{}z,Jw\right>.$

The expression $(\text{Hess}\,z)(J\cdot,\,\cdot)=\nabla{}V_{\flat}$ is
immediate from $J$-invariance of $\text{Hess}\,z$. To see
$dV_{\flat}=2\nabla{}V_{\flat}$ we compute

(19) $\displaystyle dV_{\flat}(A,B)$
$\displaystyle=A\left<V,B\right>-A\left<V,B\right>-\left<V,[A,B]\right>$
$\displaystyle=\left<\nabla_{A}V,\,B\right>-\left<\nabla_{A}V,\,B\right>=2\left<\nabla_{A}V,\,B\right>$

where we used $[A,B]=\nabla_{A}B-\nabla_{B}A$. For the Lie derivative of the
metric, we use the torsion-free condition again to find

(20)
$\displaystyle\left(\mathcal{L}_{\nabla{}z}g\right)(A,B)=\nabla{}z\left<A,B\right>-\left<[\nabla{}z,A],B\right>-\left<A,[\nabla{}z,B]\right>=2\left<\nabla_{A}z,\,B\right>$

To take the Lie derivative of $\omega$, we use
$\mathcal{L}_{\nabla{}z}=di_{\nabla{}z}+i_{\nabla{}z}d$, Eq. (14), and the
fact that $d\omega=0$ to find

(21) $\mathcal{L}_{\nabla{}z}\omega=dV_{\flat}=2\nabla{}V_{\flat}.$

Finally for $\mathcal{L}_{\nabla{}z}J$ we use the Leibniz rule
$\mathcal{L}_{\nabla{}z}\omega=\mathcal{L}_{\nabla{}z}g\circ{}J+g\circ\mathcal{L}_{\nabla{}z}J$
to obtain

(22)
$2\nabla{}V_{\flat}=2(\text{Hess}\,z)(J\cdot,\,\cdot)+g\left(\left(\mathcal{L}_{\nabla{}z}J\right)\cdot,\,\cdot\right).$

Then the fact that $2\text{Hess}\,z(J\cdot,\,\cdot)=2\nabla{}V_{\flat}$ means
$\mathcal{L}_{\nabla{}z}J=0$. ∎

######  Lemma 2.2 (The coordinate system).

The fields $\nabla{}x$, $\nabla{}y$, $\nabla{}z$, and
$\frac{\partial}{\partial{}t}$ are mutually orthogonal. The fields $\nabla{}z$
and $\nabla{}t$ are orthogonal. We have $\frac{\partial}{\partial{}t}=V$ and
$\frac{\partial}{\partial{}z}=|\nabla{}z|^{-2}\nabla{}z=-|V|^{-2}JV=-|V|^{-2}(i_{V}\omega)^{\sharp}$.
The functions $x$ and $y$ on $M^{4}$ are pluriharmonic conjugates (meaning
$dx=Jdy$).

###### Proof.

By construction the functions $x$ and $y$ are isothermal coordinates on
$M^{2}_{0}$ so $dx=Jdy$ on $M^{2}_{0}$. Because $J$ and $g$ are $V$-invariant,
the complex structure and metric on $M^{2}_{0}$ are inherited from $M^{4}$.
From the coordinate construction we have
$\mathcal{L}_{\nabla{}z}x=\mathcal{L}_{\nabla{}z}y=0$ and from Lemma 2.1 we
have $\mathcal{L}_{\nabla{}z}J=0$. Thus $\mathcal{L}_{\nabla{}z}\left(dx-
Jdy\right)=0$ and so we retain $dx=Jdy$. In particular $dJdx=0$ and $dJdy=0$
so $x$ and $y$ are both pluriharmonic.

Because $x$, $y$, $z$ are constant along $V$ and $t$ is a unit-speed
parameterization of its flow, we have $\frac{\partial}{\partial{}t}=V$. The
fact that $\nabla{}z$ is orthogonal to $\nabla{}x$ is immediate from
$0=\mathcal{L}_{\nabla{}z}x=\left<\nabla{}z,\nabla{}x\right>$. We similarly
have $\mathcal{L}_{\nabla{}z}y=0=\mathcal{L}_{\nabla{}z}t$ so $\nabla{}z$ is
also orthogonal to $\nabla{}y$ and $\nabla{}t$. The fact that
$\frac{\partial}{\partial{}t}$ is orthogonal to $\nabla{}x$ is due to
$\left<\frac{\partial}{\partial{}t},\nabla{}x\right>=\frac{\partial{}x}{\partial{}t}=0$.
Similarly $\frac{\partial}{\partial{}t}$ is orthogonal to $\nabla{}y$ and
$\nabla{}z$.

In any coordinate system $\\{x^{1},\dots,x^{n}\\}$ the matrices
$g(\nabla{}x^{i},\nabla{}x^{j})$ and
$g\left(\frac{\partial}{\partial{}x^{i}},\frac{\partial}{\partial{}x^{i}}\right)$
are inverses. Because $\nabla{}z$ is orthogonal to $\nabla{}t$, $\nabla{}x$,
$\nabla{}y$, the matrix $g(\nabla{}x^{i},\nabla{}x^{j})$ splits into a
$1\times 1$ block and a $3\times{}3$ block. From the $1\times 1$ block we get
$\frac{\partial}{\partial{}z}=|\nabla{}z|^{-2}\nabla{}z$. ∎

###  2.2. Volume forms

We define the 3-form $dVol_{3}$ to be the intrinsic volume form of
$M^{3}_{z}\subset{}M^{4}$ and the 2-form $dVol_{2}$ to be the intrinsic volume
form of $M^{2}_{z}$. By pull-back along the action of $V$, $dVol_{2}$ can be
regarded as a 2-form on any of the $M^{3}_{z}$ or on $M^{4}$. The local
expressions are

(23) $\displaystyle dVol_{4}$
$\displaystyle\;=\;|V|^{-2}|\nabla{}x|^{-2}\,dz\wedge{}V_{\flat}\wedge{}dx\wedge{}dy$
$\displaystyle dVol_{3}$
$\displaystyle\;=\;i_{\frac{\nabla{}z}{|\nabla{}z|}}dVol_{4}\;=\;|V|^{-1}|\nabla{}x|^{-2}V_{\flat}\wedge{}dx\wedge{}dy$
$\displaystyle dVol_{2}$
$\displaystyle\;=\;i_{\frac{V}{|V|}}i_{\frac{\nabla{}z}{|\nabla{}z|}}dVol_{4}\;=\;|\nabla{}x|^{-2}dx\wedge{}dy$
$\displaystyle*dVol_{2}$
$\displaystyle\;=\;|\nabla{}z|^{-2}dz\wedge{}V_{\flat}$ $\displaystyle\omega$
$\displaystyle\;=\;|\nabla{}z|^{-2}dz\wedge{}V_{\flat}+|\nabla{}x|^{-2}dx\wedge{}dy\;=\;*dVol_{2}\,+\,dVol_{2}.$

######  Lemma 2.3.

We have the exterior derivatives

(24) $\displaystyle d\big{(}dVol_{2}\big{)}$
$\displaystyle=-\left(\frac{\partial}{\partial{}z}\log|\nabla{}x|^{2}\right)dz\wedge{}dVol_{2},$
$\displaystyle d\big{(}*dVol_{2}\big{)}$
$\displaystyle=\left(\frac{\partial}{\partial{}z}\log|\nabla{}x|^{2}\right)dz\wedge{}dVol_{2}.$

###### Proof.

From (23) we have $d(dVol_{2})=d(|\nabla{}x|^{-2})\wedge{}dx\wedge{}dy$ which
is the first equation in (24). The second equation follows from $d\omega=0$
and $\omega=*dVol_{2}+dVol_{2}$. ∎

######  Lemma 2.4 (Integration Lemma).

In addition to (23) we have

(25) $dVol_{4}=dz\wedge{}dt\wedge{}dVol_{2}\quad\text{and}\quad
dVol_{3}=|V|dt\wedge{}dVol_{2}.$

Consequently, if $f=f(x,y,z)$ is any $V$-invariant function, then

(26)
$\displaystyle\int_{M^{3}_{z}}f\,dVol_{3}=2\pi\int_{M^{2}_{z}}|V|\,f\,dVol_{2}.$

On the domain
$M^{4}_{z_{0},z_{1}}=\\{p\in{}M^{4}\,\big{|}\,z_{0}<z(p)<z_{1}\\}$ we have

(27)
$\int_{M^{4}_{z_{0},z_{1}}}f\,dVol_{4}\;=\;2\pi\int_{z_{0}}^{z_{1}}\left(\int_{M^{2}_{z}}f\,dVol_{2}\right)\,dz.$

###### Proof.

By Lemma 2.2 $\nabla{}z$ is perpendicular to $\nabla{}t$, $\nabla{}x$, and
$\nabla{}y$. Thus
$\frac{1}{|V|^{2}}V_{\flat}=dt-\frac{\left<\nabla{}x,\nabla{}t\right>}{|\nabla{}x|^{2}}dx-\frac{\left<\nabla{}y,\nabla{}t\right>}{|\nabla{}y|^{2}}dy$.
The equations (25) are immediate from this and (23). We have

(28) $\displaystyle\int_{M^{3}_{z}}f\,dVol_{3}$
$\displaystyle=\int_{M^{3}_{z}}f\,\frac{1}{|V|}V_{\flat}\wedge{}dVol_{2}=\int_{M^{3}_{z}}f\,|V|\,dt\wedge{}dVol_{2}.$

Because the parameterization is $t\in[0,2\pi)$ and $|V|$, $f$ are
$t$-invariant, this is

(29)
$\displaystyle\int_{M^{3}_{z}}f\,dVol_{3}=\int_{M^{2}_{z}}\left(\int_{0}^{2\pi}f|V|\,dt\right)\,dVol_{2}=2\pi\int_{M^{3}_{z}}f|V|\,dVol_{2}$

Finally, using $dVol_{4}=\frac{1}{|V|}dz\wedge{}dVol_{3}$, we have

(30)
$\small\int_{z_{0}<z<z_{1}}f\,dVol_{4}=\int_{z_{0}}^{z_{1}}\left(\int_{M^{3}_{z}}\frac{1}{|V|}f\,dVol_{3}\right)dz\\\
=2\pi\int_{z_{0}}^{z_{1}}\left(\int_{M^{2}_{z}}f\,dVol_{2}\right)dz.$

∎

As a final remark on the LeBrun coordinate system, from [30] the metric is

(31) $g\;=\;we^{u}(dx^{2}\,+\,dy^{2})\,+\,w\,dz^{2}\,+\,w^{-1}(dt+\alpha)^{2}$

where the 1-form $\alpha$ depends on the transversal and the LeBrun functions
$u$, $w$ are

(32) $u\;=\;\log|\nabla{}z|^{2}\,-\,\log|\nabla{x}|^{2},\quad
w\;=\;|\nabla{}z|^{-2}.$

The metric (31) is scalar-flat Kähler if and only if $u$ and $w$ solve the
system (12).

###  2.3. Global properties

Recall that the distribution
$\mathcal{D}=\left\\{v\in{}TM^{4}\;\big{|}\;\omega(V,v)=0\right\\}$ is
integrable and its leaves are labeled by $M^{3}_{z}$.

######  Lemma 2.5.

The diffeomorphism flow of $|V|^{-2}JV$ preserves the distribution
$\mathcal{D}$. Consequently the diffeomorphism flow of
$\frac{\partial}{\partial{}z}$ (not $\nabla{}z$) permutes the leaves
$M^{3}_{z}$.

###### Proof.

Let $v$ be any vector field whose vectors lie in $\mathcal{D}$; this means
$i_{v}i_{V}\omega=0$. In the LeBrun system
$\frac{\partial}{\partial{z}}=-|V|^{-2}JV$. The identity
$i_{A}\mathcal{L}_{B}-\mathcal{L}_{B}i_{A}=i_{[A,B]}$ gives

(33) $\displaystyle
i_{v}\mathcal{L}_{\frac{\partial}{\partial{}z}}i_{V}\omega$
$\displaystyle\;=\;i_{[v,\frac{\partial}{\partial{}z}]}i_{V}\omega\;=\;\left<JV,\,\nabla_{v}\frac{\partial}{\partial{}z}\right>-\left<JV,\,\nabla_{\frac{\partial}{\partial{}z}}v\right>$
$\displaystyle\;=\;-\left<\nabla{}z,\,\nabla_{v}\frac{\partial}{\partial{}z}\right>-\left<JV,\,\nabla_{\frac{\partial}{\partial{}z}}v\right>.$

Because $\left<\nabla{}z,\partial/\partial{}z\right>=1$ and
$\left<JV,v\right>=i_{v}i_{V}\omega=0$, this is

(34) $\displaystyle
i_{v}\mathcal{L}_{\frac{\partial}{\partial{}z}}i_{V}\omega$
$\displaystyle\;=\;\left<\nabla_{v}\nabla{}z,\,\frac{\partial}{\partial{}z}\right>+\left<\nabla_{\frac{\partial}{\partial{}z}}JV,\,v\right>$
$\displaystyle\;=\;\left<\nabla_{\frac{\partial}{\partial{}z}}\nabla{}z,\,v\right>-\left<\nabla_{\frac{\partial}{\partial{}z}}\nabla{}z,\,v\right>\;=\;0$

where we used the symmetry of $Hess(z)$. We conclude that the diffeomorphism
flow of $\frac{\partial}{\partial{}z}$ leaves the distribution
$\mathcal{D}=\\{v\,\big{|}\,i_{v}i_{V}\omega=0\\}$ invariant. ∎

######  Lemma 2.6.

If $M^{4}$ is non-compact and $\mathcal{D}$ has a compact leaf, $V$ is
Hamiltonian.

###### Proof.

If $V$ has a zero then it is Hamiltonian (see for example [34]), so we need
only consider the case $V$ has no zeros and $-|V|^{2}JV$ is defined
everywhere.

Let $M^{3}_{0}$ be a compact integral surface of $\mathcal{D}$, and let
$\psi_{z}:M^{4}\rightarrow{}M^{4}$ be the diffeomorphism flow of
$-|V|^{-2}JV$, with flow parameter $z$. Because $-|V|^{-2}JV$ permutes the
integral surfaces of $\mathcal{D}$, if there is a point $p\in{}M^{3}_{0}$ for
which $\psi_{z}(p)\in{}M^{3}_{0}$ then in fact
$\psi_{z}:M^{3}_{0}\rightarrow{}M^{3}_{0}$. Possibly a value $z_{0}\neq 0$
exists where $\psi_{z_{0}}$ maps the integral surface $M^{3}_{0}$ to itself.
In this case $M^{4}$ is a mapping torus for
$\psi_{z_{0}}:M^{3}_{0}\rightarrow{}M^{3}_{0}$. But then $M^{4}$ is compact,
contradicting the hypotheses. Therefore $\psi_{z}$ never maps $M^{3}_{0}$ to
itself unless $z=0$. Assign to each point $p\in{}M^{4}$ the unique value of
$z$ for which $p\in\psi_{z}(M^{3}_{0})$ defines the Killing potential
$z:M^{4}\rightarrow\mathbb{R}$. ∎

######  Lemma 2.7.

Assume $M^{4}$ is connected. Then every leaf of $\mathcal{D}$ is connected.

###### Proof.

We show that if a leaf $M_{z_{0}}^{3}$ is locally connected, then nearby leafs
are also locally connected. For an argument by contradiction, assume not. Then
there is a point $p_{0}\in{}M^{3}_{z_{0}}$ so that any sufficiently small ball
$B_{\delta}$ containing $p_{0}$ and sufficiently small $\epsilon>0$, then
$M^{3}_{z_{0}+\epsilon}\cap{}B_{\delta}$ is disconnected.

By the isotopy criterion of Morse theory this is only possible if $p_{0}$ is a
critical point of $z$. Choosing a sequence $\delta_{i}\searrow 0$ sufficiently
small and expanding the metric by $\delta_{i}{}^{-1}$ the ball
$B_{\delta_{i}}$ becomes the ball $B_{1}$. For small enough $\delta_{i}$ local
balls are diffeomorphic to Euclidean balls, so we can take a limit of the
metric and complex structures in the $C^{\infty}$ sense on compact subsets.
The metric converges to the flat metric on $\mathbb{R}^{4}$ and the complex
structure converges to the complex structure on $\mathbb{C}^{2}$. The field
$V$ converges to a symplectomorphic Killing field on $\mathbb{C}^{2}$ with a
zero still at $p_{0}$. The structure of symplectomorphic Killing fields on
flat $\mathbb{C}^{2}$ is completely straighforward: after choosing appropriate
local complex coordinates $z^{1}=x^{1}+\sqrt{-1}y^{1}$,
$z^{2}=x^{2}+\sqrt{-1}y^{2}$ where $p_{0}$ has coordinates $(0,0)$, we can
write

(35) $\displaystyle V$
$\displaystyle\;=\;\alpha\,Im\left(z^{1}\frac{\partial}{\partial{}z^{1}}\right)+\beta\,Im\left(z^{2}\frac{\partial}{\partial{}z^{2}}\right)$
$\displaystyle\;=\;\alpha\left(-y^{1}\frac{\partial}{\partial{}x^{1}}+x^{1}\frac{\partial}{\partial{}y^{1}}\right)+\beta\left(-y^{2}\frac{\partial}{\partial{}x^{2}}+x^{2}\frac{\partial}{\partial{}y^{2}}\right)$

for some $\alpha,\beta\in\mathbb{R}$, not both zero. Thus the level-sets of
$z$ have the form
$\alpha\big{(}(x_{1})^{2}+(y^{1})^{2}\big{)}+\beta\big{(}(x^{2})^{2}+(y^{2})^{2}\big{)}$.
When $\alpha$ and $\beta$ have the same sign the level-sets are topologically
3-spheres. When $\alpha$ and $\beta$ have opposite signs, the level-sets are
connected hyperboloids of revolution. When one of $\alpha$, $\beta$ is zero
each level-set is a product of a circle with a plane. In all cases, the level-
sets are connected.

The convergence of both the metric and the potential in this blowup limit is
in the $C^{\infty}$ sense. Thus, just prior to the limit for sufficiently
small $\delta_{i}$, the critical point of the Killing potential must still be
a saddle, so its level-sets near the critical point have the same structure
and in particular are connected. This contradicts the assumption that level-
sets are locally disconnected near $p$.

Thus because the leafs of $\mathcal{D}$ cannot locally separate into more than
one component, all leafs must all have the same number of components. Because
$M^{4}$ is connected, they must all have one component. ∎

_Proof of Theorem 1.1_. We assume $M^{4}$ is complete and $|V|>\epsilon$
outside a compact set $K$. Assume one leaf of $\mathcal{D}$ is compact. We
show all leafs of $\mathcal{D}$ are compact.

Lemma 2.6 states $V$ is Hamiltonian so $z$ exists globally. Let $\Omega$ be a
precompact domain that contains $K$ and the compact level-set $M^{3}_{z_{0}}$
which we assumed exists (it does not matter if it is singular). Let $\gamma$
be a trajectory of $z$ in the sense that

(36)
$\dot{\gamma}\;=\;\frac{\partial}{\partial{}z}\;=\;\frac{1}{|\nabla{}z|^{2}}\nabla{z},\quad\text{and}\quad
z(\gamma(s))\;=\;s.$

Thus $\frac{d}{ds}=\frac{\partial}{\partial{}z}$ along $\gamma$. Assume
$\gamma:(a,b)\rightarrow{}M^{4}$ is maximally extended in the sense that the
domain of $\gamma$ cannot be extended in such a way that (36) remains true.

First we show $\gamma$ must intersect $\Omega$. Because $\gamma$ can always be
uniquely extended except when $|\nabla{}z|=0$, we have three possibilities
regarding $b$. The first is $b=+\infty$. The second is $b<\infty$ and some
$\delta>0$ exists so $\gamma:[b-\delta,\,b)\rightarrow{}M^{4}$ is contained in
a compact subset of $M^{4}$. The third is $b<\infty$ and
$\gamma:[b-\delta,\,b)\rightarrow{}M^{4}$ is contained in no compact subset of
$M^{4}$. In the second and third cases, we show $\gamma$ intersects $\Omega$.

Assume the second possibility, that $b$ is finite and $\gamma|_{[b-\delta,b)}$
remains bounded. We show that $\gamma$ intersects $\Omega$. If
$|(\nabla{}z)_{\gamma(s)}|>\epsilon$ for all $s\in[b-\delta,b)$ then $\gamma$
remains smooth with uniformly controlled tangent vector and therefore the path
can be extended beyond $b$. Thus $|\nabla{}z|$ must approach $0$ near $b$, so
$\gamma$ must intersect $K$ and therefore $\Omega$.

Assume the third possibility, that $b$ is finite but
$\gamma|_{[b-\epsilon,b)}$ is unbounded; we again show $\gamma$ intersects
$\Omega$. Assume not; then $|\nabla{}z|\geq\epsilon$ along $\gamma$. Fix any
$s_{0}\in(a,b)$. A sequence $\\{s_{i}\\}$ exists with $s_{i}\nearrow{}b$ and
$\text{dist}(\gamma(s_{0}),\gamma(s_{i}))\rightarrow\infty$. Then

(37)
$\displaystyle|s_{i}-s_{0}|=|z(\gamma(s_{i}))-z(\gamma(s_{0}))|=\left|\int_{s_{0}}^{s_{i}}\frac{dz}{ds}\,ds\right|=\left|\int_{s_{0}}^{s_{i}}\left<\frac{d}{ds},\,\nabla{}z\right>\,ds\right|$
$\displaystyle\;\;\;=\int_{s_{0}}^{s_{i}}\left|\frac{d}{ds}\right||\nabla{}z|\,ds\geq\epsilon\int_{s_{0}}^{s_{i}}\left|\frac{d}{ds}\right|\,ds=\epsilon\;len(\gamma;s_{0},s_{i})\geq\epsilon\,\text{dist}(\gamma(s_{0}),\gamma(s_{i}))$

where we used
$\left<\frac{d}{ds},\,\nabla{}z\right>=\left|\frac{d}{ds}\right||\nabla{}z|$,
which is due to $\frac{d}{ds}=|\nabla{}z|^{-2}\nabla{}z$ by Lemma 2.2. Because
$\text{dist}(\gamma(s_{0}),\gamma(s_{1}))$ is unbounded, this contradicts the
fact that $|s_{i}-s_{0}|\leq|b-s_{0}|$. This forces our assumption that
$|\nabla{}z|>\epsilon$ along $\gamma$ to be false. We conclude that $\gamma$
intersects $K$ and therefore $\Omega$.

We have shown that either $\gamma$ intersects $\Omega$ or else $b=\infty$. The
same considerations hold for the value $a$. Thus either $\gamma$ intersects
$\Omega$ or else $\gamma:(-\infty,\infty)\rightarrow{}M^{4}$. In the latter
case, because $(z\circ\gamma)(s)=s$ as $s$ ranges from $-\infty$ to $\infty$,
$z\circ\gamma$ takes every value including the value $z_{0}$. Thus $\gamma$
intersects $M^{3}_{z_{0}}$, so intersects $\Omega$. We conclude that every
trajectory intersects $\Omega$.

Now we prove that every level-set $M^{3}_{z}$ is compact. For a contradiction,
assume some $z^{\prime}$ exists so $M^{3}_{z^{\prime}}$ is not compact.
Levelsets are closed (by continuity of $z$) so compactness is equivalent to
boundedness. Thus $M^{3}_{z^{\prime}}$ is unbounded. Because
$M^{3}_{z^{\prime}}$ is unbounded and $\overline{\Omega}$ is compact, a
sequence of points $\\{p_{i}\\}\subset{}M^{3}_{z^{\prime}}$ exists with
$\text{dist}(p_{i},\overline{\Omega})\rightarrow\infty$. Let $\gamma_{i}$ be
the trajectory of $z$ passing through $p_{i}$; in particular
$\gamma_{i}(z^{\prime})=p_{i}\in{}M^{3}_{z^{\prime}}$. Because $\gamma_{i}$
intersects $\Omega$, we can find some $s_{i}$ so $\gamma_{i}$ remains outside
$\overline{\Omega}$ on $(s_{i},z^{\prime}]$ and
$\gamma_{i}(s_{i})\in\overline{\Omega}$. Because $s_{i}=z(\gamma_{i}(s_{i}))$
and $z$ is uniformly bounded on $\overline{\Omega}$, the numbers $s_{i}$ are
uniformly bounded. By (37) we have

(38)
$|z^{\prime}-s_{i}|\;\geq\;\epsilon\,\text{dist}(\gamma(s_{i}),\,p_{i})\;\geq\;\epsilon\,\text{dist}(p_{i},\,\overline{\Omega}).$

The left-hand side is bounded and the right-hand side is unbounded. This
contradiction shows $M^{3}_{z^{\prime}}$ is compact. ∎

Before proving our final lemma, Lemma 2.8, we recall some basic definitions.
An open subset $\Omega$ of a complete Riemannian manifold is called a manifold
end if it is unbounded but has a connected compact boundary. If
$K\subset{}M^{4}$ is compact then the number of ends of $M^{n}$ relative to
$K$ is the number of unbounded components of $M^{n}\setminus{}K$. We say that
$M^{n}$ has $k$ many ends if the number of ends of $M^{n}$ relative to any
compact subset is $\leq{}k$, and there exists some compact subset $K$ so that
$M^{n}$ has $k$ many ends relative to $K$.

When all level-sets $M^{3}_{z}$ are compact, recall the decomposition

(39)
$M^{4}\;=\;\overline{M^{4-}}\,\cup\,\overline{M^{4}_{1}}\,\cup\,\dots\,\cup\,\overline{M^{4}_{N}}\,\cup\,\overline{M^{4+}}$

where possibly $M^{4+}$ or $M^{4-}$ are empty.

######  Lemma 2.8 (cf. Lemma 1.3).

Assume $|V|>\epsilon>0$ outside some compact set, and at least one level-set
is compact. Then each levelset $M^{3}_{z}$ and fibration
$M^{3}_{z}\rightarrow{}M^{2}_{z}$ in $M_{i}^{4}$ (or in $M^{4+}$ or in
$M^{4-}$) is isotopic along the $\frac{\partial}{\partial{}z}$ diffeomorphism
flow. If $M^{4+}\neq\varnothing$ then $z(M^{4+})=[z_{N},\infty)$, and if
$M^{4-}\neq\varnothing$ then $z(M^{4-})=(-\infty,z_{0}]$ on $M^{4-}$.

###### Proof.

By Lemma 2.6 $z$ is Hamiltonian, by Lemma 2.7 all level-sets are connected,
and by Theorem 1.1 all level-sets are compact. Because levelsets of
$\mathcal{D}$ are compact and connected, $M^{4}$ can have at most $2$ ends.

Away from critical points of $z$ the diffeomorhism flow of
$\frac{\partial}{\partial{}z}$ permutes the levelsets, but because
$\frac{\partial}{\partial{}z}$ commutes with $V$, it also permutes the orbits
of $V$. Because $M^{3}_{z}\rightarrow{}M^{2}_{z}$ is a quotient with leaves
being the orbits of $V$, the flow of $\frac{\partial}{\partial{}z}$ permutes
the fibration structure $M^{3}_{z}\rightarrow{}M^{2}_{z}$ as well.

Assuming $M^{4+}\neq\varnothing$, then because $M^{4+}$ is a manifold end
trajectories of $z$ in the sense of (36) are unbounded. Because
$|V|>\epsilon>0$ outside of some compact set, by (37) along any such
trajectory the value of $z$ is unbounded. Therefore $z$ has no upper bound on
$M^{4+}$ and so $z(M^{4+})=[z_{N},\infty)$. The same argument holds for
$M^{4-}$ after replacing $z$ by $-z$. ∎

##  3\. The Killing 2-form

We gather required information about $dV_{\flat}$, the “Killing 2-form.” In
§3.2 we find a global expression for the LeBrun equation
$w_{xx}+w_{yy}+(we^{u})_{zz}=0$ of (12), and use this with information from
§3.1 to prove the first part of Theorem 1.2. We frequently we use the
decomposition $\bigwedge{}^{2}=\bigwedge^{+}\oplus\bigwedge^{-}$ where the
projectors are $\pi^{\pm}:\bigwedge^{2}\rightarrow\bigwedge^{\pm}$,
$\pi^{\pm}=\frac{1}{2}(Id\pm*)$. This leads to identities such as
$\eta=-*\eta+2\eta^{+}$ and $\eta^{-}=-*\eta+\eta^{+}$ whenever $\eta$ is a
2-form.

###  3.1. The 2-form $dV_{\flat}$

We gather some necessary information about $dV_{\flat}$.

######  Lemma 3.1.

We have

(40) $\displaystyle
dV_{\flat}\;=\;-\left(\frac{\partial}{\partial{}z}\log|\nabla{}z|^{2}\right)dz\wedge{}V_{\flat}-dz\wedge{}Jd\log|\nabla{}z|^{2}$
$\displaystyle\hskip 30.71497pt-
V_{\flat}\wedge{}d\log|\nabla{}z|^{2}-|\nabla{}z|^{2}\left(\frac{\partial}{\partial{}z}\log|\nabla{}x|^{2}\right)dVol_{2}$

###### Proof.

We use $dV_{\flat}=2\left<\nabla_{\cdot}V,\,\cdot\right>$ from Lemma 2.1.
Letting $v$ be any vector field,

(41) $\displaystyle
dV_{\flat}\left(\frac{\partial}{\partial{}z},\,v\right)=2\left<\nabla_{\frac{\partial}{\partial{}z}}V,\,v\right>=2|\nabla{}z|^{-2}\left<\nabla_{\nabla{}z}J\nabla{}z,\,v\right>$
$\displaystyle\quad\quad=-2|\nabla{}z|^{-2}\left<\nabla_{\nabla{}z}\nabla{}z,\,Jv\right>=-2|\nabla{}z|^{-2}\left<\nabla_{Jv}\nabla{}z,\,\nabla{}z\right>=-\left<Jv,\nabla\log|\nabla{z}|^{2}\right>.$

Therefore $i_{\frac{\partial}{\partial{z}}}dV_{\flat}=-Jd\log|\nabla{}z|^{2}$.

To compute $i_{V}dV_{\flat}$, we use the fact that
$\mathcal{L}_{V}V_{\flat}=0$ which is $i_{V}dV_{\flat}+d(i_{V}V_{\flat})=0$.
Thus $i_{V}dV_{\flat}=-di_{V}V_{\flat}=-d|V|^{2}=-|V|^{2}d\log|V|^{2}$. From
these expressions for $i_{\frac{\partial}{\partial{}z}}dV_{\flat}$ and
$i_{V}dV_{\flat}$, the first three terms of (40) follow.

Lastly we must check the action of $dV_{\flat}$ on pairs of vectors that are
perpendicular to both $\partial/\partial{}z$ and $V_{\flat}$, meaning we check
the action of $dV_{\flat}$ on $\nabla{}x$ and $\nabla{}y$. Using
$\left<V,\,\nabla{}x\right>=\left<V,\,\nabla{}y\right>=0$,
$\nabla{}y=J\nabla{}x$, and $V=J\nabla{}z$ we have

(42) $\displaystyle dV_{\flat}(\nabla{}x,\,\nabla{}y)$
$\displaystyle\;=\;2\left<\nabla_{\nabla{}x}V,\,\nabla{}y\right>\;=\;-2\left<V,\,\nabla_{\nabla{}x}J\nabla{}x\right>\;=\;-2\left<\nabla{}z,\,\nabla_{\nabla{}x}\nabla{}x\right>$
$\displaystyle\;=\;-2\left<\nabla{}x,\,\nabla_{\nabla{}z}\nabla{}x\right>\;=\;-\nabla{}z|\nabla{}x|^{2}$

Because $(dx\wedge{}dy)(\nabla{}x,\,\nabla{}y)=|\nabla{}x|^{4}$ we therefore
have

(43) $\displaystyle dV_{\flat}\Big{|}_{span\\{dx\wedge{}dy\\}}$
$\displaystyle=-|\nabla{}z|^{2}|\nabla{}x|^{-4}\left(\frac{\partial}{\partial{}z}|\nabla{}x|^{2}\right)\,dx\wedge{}dy$
$\displaystyle=-|\nabla{}z|^{2}\left(\frac{\partial}{\partial{}z}\log|\nabla{}x|^{2}\right)dVol_{2}.$

∎

######  Lemma 3.2.

We have the expressions

(44) $\displaystyle V_{\flat}\wedge{}dV_{\flat}$
$\displaystyle\;=\;dz\wedge{}V_{\flat}\wedge{}Jd\log|\nabla{}z|^{2}\,-\,|\nabla{}z|^{2}\left(\frac{\partial}{\partial{}z}\log|\nabla{}x|^{2}\right)\,V_{\flat}\wedge{}dVol_{2}$
$\displaystyle\;=\;*\bigg{(}-d|V|^{2}\,+\,(\triangle_{4}z)\,dz\bigg{)}.$

###### Proof.

The first equality in (44) is immediate from (40). To compute the second
equality, we use the identity $dV_{\flat}=-*dV_{\flat}+2(dV_{\flat})^{+}$.
Because $(dV_{\flat})^{+}=-(dJdz)^{+}=\frac{1}{2}(\triangle_{4}z)\omega$, from
(13), and because $**:\bigwedge^{3}\rightarrow\bigwedge^{3}$ is multiplication
by $(-1)$,

(45) $\displaystyle
V_{\flat}\wedge{}dV_{\flat}=-V_{\flat}\wedge*dV_{\flat}+2V_{\flat}\wedge(dV_{\flat})^{+}$
$\displaystyle\hskip
14.45377pt=**(V_{\flat}\wedge*dV_{\flat})-(\triangle_{4}z)**(V_{\flat}\wedge\omega)=*\left(i_{V}dV_{\flat}-(\triangle_{4}z)\,i_{V}\omega\right).$

Then $\mathcal{L}_{V}=di_{V}+i_{V}d$ and $\mathcal{L}_{V}V_{\flat}=0$ gives
$i_{V}dV_{\flat}=-d|V|^{2}$, and (14) gives $i_{V}\omega=-dz$. Entering these
into (45) provides the conclusion. ∎

######  Lemma 3.3.

We have the following expressions for $\triangle_{4}z$:

(46)
$\triangle_{4}z\;=\;|\nabla{}z|^{2}\frac{\partial}{\partial{}z}\left(\log|\nabla{}z|^{2}-\log|\nabla{}x|^{2}\right)\;=\;\left<\nabla{}z,\,\nabla{}u\right>\;=\;wu_{z}.$

###### Proof.

We have $V_{\flat}=-Jdz$ so that
$dV_{\flat}\wedge\omega=(\triangle_{4}z)dVol_{4}$. Using $dV_{\flat}$ from
Lemma 3.1 and $\omega$ from (23), we have

(47) $\displaystyle(\triangle_{4}z)\,dVol_{4}=dV_{\flat}\wedge\omega$
$\displaystyle=|\nabla{}z|^{2}\left(\frac{\partial}{\partial{}z}\log|\nabla{}z|^{2}-\frac{\partial}{\partial{}z}\log|\nabla{}x|^{2}\right)dVol_{4}$
$\displaystyle=\left<\nabla{}z,\,\nabla\log|\nabla{}z|^{2}-\nabla\log|\nabla{}x|^{2}\right>\,dVol_{4}.$

Using (32), this is
$\triangle_{4}z=\left<\nabla{}z,\nabla{}u\right>=|\nabla{}z|^{2}\frac{\partial{}u}{\partial{}z}=wu_{z}$.
∎

######  Lemma 3.4.

We have the two Lie derivatives

(48)
$\displaystyle\mathcal{L}_{\frac{\partial}{\partial{}z}}V_{\flat}\;=\;-Jd\log|\nabla{}z|^{2}\quad\text{and}\quad\mathcal{L}_{\frac{\partial}{\partial{}z}}dV_{\flat}\;=\;-dJd\log|\nabla{}z|^{2}.$

###### Proof.

Using
$\mathcal{L}_{\frac{\partial}{\partial{}z}}=di_{\frac{\partial}{\partial{}z}}+i_{\frac{\partial}{\partial{}z}}d$
and $i_{\frac{\partial}{\partial{}z}}V_{\flat}=0$, we have

(49)
$\mathcal{L}_{\frac{\partial}{\partial{}z}}V_{\flat}\;=\;i_{\frac{\partial}{\partial{}z}}dV_{\flat}\;=\;-|V|^{-2}Jd|\nabla{}z|^{2}\;=\;-Jd\log|\nabla{}z|^{2}.$

The second equation follows from the commutator identity
$[\mathcal{L}_{\frac{\partial}{\partial{}z}},d]=0$. ∎

######  Lemma 3.5.

We have Lie derivatives

(50) $\displaystyle\mathcal{L}_{\frac{\partial}{\partial{}z}}dVol_{4}$
$\displaystyle\;=\;\left(-\frac{\partial}{\partial{}z}\log|\nabla{}x|^{2}\right)\,dVol_{4}$
$\displaystyle\mathcal{L}_{\frac{\partial}{\partial{}z}}dVol_{3}$
$\displaystyle\;=\;\left(\frac{1}{2}\frac{\partial}{\partial{}z}\log|\nabla{}z|^{2}\,-\,\frac{\partial}{\partial{}z}\log|\nabla{}x|^{2}\right)\,dVol_{3}$

Especially important are Lie derivatives involving $dVol_{2}$. These are

(51)
$\displaystyle\mathcal{L}_{\frac{\partial}{\partial{}z}}dVol_{2}\;=\;\left(-\frac{\partial}{\partial{}z}\log|\nabla{}x|^{2}\right)\,dVol_{2}$
$\displaystyle\mathcal{L}_{\frac{\partial}{\partial{}z}}\big{(}|V|^{2}\,dVol_{2}\big{)}\;=\;(\triangle_{4}z)\,dVol_{2}$
$\displaystyle\mathcal{L}_{\frac{\partial}{\partial{}z}}\left(V_{\flat}\wedge{}dVol_{2}\right)\;=\;\left(\frac{1}{|V|}\triangle_{4}z\right)\,dVol_{3}$
$\displaystyle\mathcal{L}_{\frac{\partial}{\partial{}z}}*dVol_{2}\;=\;-\left(\frac{\partial}{\partial{}z}\log|V|^{2}\right)*dVol_{2}+dz\wedge{}Jd|V|^{-2}.$

###### Proof.

Because
$\mathcal{L}_{\frac{\partial}{\partial{}z}}=di_{\frac{\partial}{\partial{}z}}$
on any 4-form, and because $i_{\nabla{}z}\omega=V_{\flat}$, we have

(52) $\displaystyle\mathcal{L}_{\frac{\partial}{\partial{z}}}dVol_{4}$
$\displaystyle=di_{\frac{\partial}{\partial{z}}}dVol_{4}=d\left(|\nabla{}z|^{-2}(i_{\nabla{}z}\omega)\wedge\omega\right)=d\left(|\nabla{}z|^{-2}V_{\flat}\wedge\omega\right)$
$\displaystyle=d\left(|\nabla{}z|^{-2}V_{\flat}\right)\wedge\omega=d(|\nabla{}z|^{-2})\wedge{}V_{\flat}\wedge\omega+|\nabla{}z|^{-2}dV_{\flat}\wedge\omega$
$\displaystyle=-d(|\nabla{}z|^{-2})\wedge{}Jdz\wedge\omega-|\nabla{}z|^{-2}dJdz\wedge\omega.$

Then using (13) and Lemma 3.3 this is

(53) $\displaystyle\mathcal{L}_{\frac{\partial}{\partial{z}}}dVol_{4}$
$\displaystyle\;=\;-\left(\frac{\partial}{\partial{}z}\log|\nabla{}z|^{2}\right)\,dVol_{4}+|\nabla{}z|^{-2}(\triangle_{4}z)dVol_{4}$
$\displaystyle\;=\;-\left(\frac{\partial}{\partial{}z}\log|\nabla{}x|^{2}\right)dVol_{4}$

The second equation in (50) follows from the identity
$[\mathcal{L}_{A},i_{B}]=i_{[A,B]}$ along with
$dVol_{3}=i_{|\nabla{}z|\frac{\partial}{\partial{}z}}dVol_{4}=i_{\frac{\nabla{}z}{|\nabla{}z|}}dVol_{4}$.

In (51), the first equation follows from $[\mathcal{L}_{A},i_{B}]=i_{[A,B]}$
and $dVol_{2}=i_{|V|^{-1}V}dVol_{3}$. The second equation follows from the
Leibniz rule and Lemma 3.3. The third equation uses Lemma 3.4 to obtain

(54)
$\displaystyle\mathcal{L}_{\frac{\partial}{\partial{}z}}(V_{\flat}\wedge{}dVol_{2})$
$\displaystyle\;=\;\left(\mathcal{L}_{\frac{\partial}{\partial{}z}}V_{\flat}\right)\wedge{}dVol_{2}+V_{\flat}\wedge\left(\mathcal{L}_{\frac{\partial}{\partial{}z}}dVol_{2}\right)$
$\displaystyle\;=\;-Jd\log|V|^{2}\wedge{}dVol_{2}-\left(\frac{\partial}{\partial{}z}\log|\nabla{}x|^{2}\right)V_{\flat}\wedge{}dVol_{2}$
$\displaystyle\;=\;|\nabla{}z|^{-2}\triangle_{4}z\,V_{\flat}\wedge{}dVol_{2}$

Then $dVol_{3}=|\nabla{}z|^{-1}V_{\flat}\wedge{}dVol_{2}$ gives the desired
equation. For the final equation,

(55)
$\displaystyle\mathcal{L}_{\frac{\partial}{\partial{}z}}\big{(}*dVol_{2}\big{)}$
$\displaystyle=\left(di_{\frac{\partial}{\partial{}z}}+i_{\frac{\partial}{\partial{}z}}d\right)*dVol_{2}$
$\displaystyle=d\left(|V|^{-2}V_{\flat}\right)+\left(\frac{\partial}{\partial{}z}\log|\nabla{}x|^{2}\right)dVol_{2}$
$\displaystyle=d|V|^{-2}\wedge{}V_{\flat}+|V|^{-2}dV_{\flat}+\left(\frac{\partial}{\partial{}z}\log|\nabla{}x|^{2}\right)dVol_{2}.$

Using Lemma 3.1, this is

(56)
$\displaystyle\mathcal{L}_{\frac{\partial}{\partial{}z}}\big{(}*dVol_{2}\big{)}$
$\displaystyle=-\left(\frac{\partial}{\partial{}z}\log|V|^{2}\right)\,*dVol_{2}+dz\wedge{}Jd|V|^{-2}.$

∎

###  3.2. Volume of $M^{2}_{z}$

The LeBrun equations (12) are local, as they rely on a choice of isothermal
coordinates $x$, $y$ on $M^{2}_{z}$. However our first lemma shows the second
of the LeBrun equations has a more global expression.

######  Lemma 3.6.

The LeBrun equation $w_{xx}+w_{yy}+(we^{u})_{zz}=0$ is the same as

(57)
$\mathcal{L}_{\frac{\partial}{\partial{}z}}\mathcal{L}_{\frac{\partial}{\partial{}z}}dVol_{2}\,+\,\left(\triangle_{2}|V|^{-2}\right)dVol_{2}\quad\text{when
$V\neq 0$.}$

Therefore when $M^{2}_{z}$ is compact,
$\frac{d^{2}}{dz^{2}}\int_{M^{2}_{z}}dVol_{2}=0$ at regular values of $z$.

###### Proof.

Because $x,y$ are isothermal coordinates on $M^{2}_{z}$,
$\triangle_{2}=|\nabla{}x|^{-2}\left(\frac{\partial^{2}}{\partial{}x^{2}}+\frac{\partial^{2}}{\partial{}y^{2}}\right)$
and $dVol_{2}=|\nabla{}x|^{-2}dx\wedge{}dy$. From (32), $w=|\nabla{}z|^{-2}$
and $we^{u}=|\nabla{}x|^{-2}$. Thus

(58) $\displaystyle 0$ $\displaystyle\;=\;w_{xx}+w_{yy}+(we^{u})_{zz}$
$\displaystyle\;=\;\left(\frac{\partial^{2}}{\partial{}x^{2}}+\frac{\partial^{2}}{\partial{}y^{2}}\right)|\nabla{}z|^{-2}+\frac{\partial^{2}}{\partial{}z^{2}}|\nabla{}x|^{-2}$
$\displaystyle\;=\;|\nabla{}x|^{-2}\triangle_{2}|\nabla{}z|^{-2}\,+\,\frac{\partial^{2}}{\partial{}z^{2}}|\nabla{}x|^{-2}$

Because
$\mathcal{L}_{\frac{\partial}{\partial{}z}}dx=d\frac{\partial{}x}{\partial{}z}=0$
and
$\mathcal{L}_{\frac{\partial}{\partial{}z}}dy=d\frac{\partial{}y}{\partial{}z}=0$
we have $\mathcal{L}_{\frac{\partial}{\partial{}z}}(dx\wedge{}dy)=0$, so

(59)
$\displaystyle\mathcal{L}_{\frac{\partial}{\partial{}z}}\mathcal{L}_{\frac{\partial}{\partial{}z}}dVol_{2}$
$\displaystyle=\left(\frac{\partial^{2}}{\partial{}z^{2}}|\nabla{}x|^{-2}\right)dx\wedge{}dy=-\left(\triangle_{2}|V|^{-2}\right)\;dVol_{2}.$

When $M^{2}_{z}$ is compact, integrating both sides of (59) gives

(60)
$\displaystyle\int_{M^{2}_{z}}\mathcal{L}_{\frac{\partial}{\partial{}z}}\mathcal{L}_{\frac{\partial}{\partial{}z}}dVol_{2}\;=\;-\int_{M^{2}_{z}}\triangle_{2}|V|^{-2}\,dVol_{2}\;=\;0.$

(Because $M^{2}_{z}$ is smooth in the orbifold sense and because $|V|^{-2}$ is
smooth, there is no issue in using integration-by-parts despite orbifold
points; see for example [12] [46] [50] [17].) By Lemma 2.5 the diffeomorphism
flow of $\frac{\partial}{\partial{}z}$ fixes $\mathcal{D}$ and therefore
permutes the leaves $M^{3}_{z}$ and their reductions $M^{2}_{z}$. We conclude
that

(61)
$\displaystyle\frac{d^{2}}{dz^{2}}\int_{M^{2}_{z}}dVol_{2}\;=\;\int_{M^{2}_{z}}\mathcal{L}_{\frac{\partial}{\partial{}z}}\mathcal{L}_{\frac{\partial}{\partial{}z}}dVol_{2}\;=\;0.$

∎

######  Proposition 3.7 (cf. Theorem 1.2).

When $z$ is non-singular and $M^{2}_{z}$ is compact,

(62)
$\frac{d}{dz}\int_{M^{2}_{z}}\,dVol_{2}\;=\;\frac{1}{2\pi}\int_{M^{3}_{z}}\frac{1}{|V|^{4}}V_{\flat}\wedge{}dV_{\flat}\;=\;2\pi{}e^{g}(M^{3}_{z},M^{2}_{z})$

where $e^{g}(M^{3}_{z},M^{2}_{z})$ is the Euler number of the fibration
$M^{3}_{z}\rightarrow{}M^{2}_{z}$.

###### Proof.

The Lie group $\mathbb{S}^{1}$ acts on $M^{3}_{z}$, generated by the field
$\frac{\partial}{\partial{}t}=V$. Therefore

(63)
$\theta\;=\;\frac{1}{|V|^{2}}V_{\flat}\;\otimes\;\frac{\partial}{\partial{}t}\;\in\;\Gamma\left(\bigwedge{}^{1}\otimes\mathfrak{g}\right)$

is a connection 1-form on the fibration $M^{3}_{z}\rightarrow{}M^{2}_{z}$.
Because the Lie group is abelian its curvature 2-form is $F=d\theta$. Thus

(64)
$F\;=\;d\left(\frac{1}{|V|^{2}}V_{\flat}\right)\Big{|}_{M^{3}_{z}}\;\otimes\;\frac{\partial}{\partial{}t}\;\in\;\Gamma\left(\bigwedge{}^{2}\otimes\mathfrak{g}\right).$

Certainly then $Tr(F)=d(|V|^{-2}V_{\flat})$. Because the Lie algebra is
Abelian, the Chern-Simons form is simply

(65)
$Tr(\theta\wedge{}d\theta)\;=\;\frac{1}{|V|^{2}}V_{\flat}\wedge{}d\left(\frac{1}{|V|^{2}}V_{\flat}\right)\;=\;\frac{1}{|V|^{2}}V_{\flat}\wedge{}Tr(F).$

By definition the geometric Euler number is
$e^{g}(M^{3}_{z},M^{2}_{z})=\frac{1}{2\pi}\int_{M^{2}_{z}}Tr(F)$. Thus

(66)
$\chi(M^{3}_{z},M^{2}_{z})=\frac{1}{2\pi}\int_{M^{2}_{z}}Tr(F)=\frac{1}{4\pi^{2}}\int_{M^{3}_{z}}\frac{V}{|V|^{2}}\wedge{}Tr(F)=\frac{1}{4\pi^{2}}\int_{M^{3}_{z}}Tr(\theta\wedge{}d\theta)$

(the second equality is from the Integration Lemma, Lemma 2.4). When
$M^{3}_{z}$ is compact it is known that this is a rational number, and when
$M^{3}_{z}\rightarrow{}M^{2}_{z}$ is a compact fiber bundle it is an integer
(the usual Euler number of the bundle)—see, for example, Lemma 1 of [36].

From Lemma 3.2 and Lemma 3.5 we compute

(67)
$Tr(\theta\wedge{}d\theta)\;=\;\frac{1}{|V|^{-2}}\left(\frac{\partial}{\partial{}z}\log|\nabla{}x|^{2}\right)V_{\flat}\wedge{}dVol_{2}\;=\;\frac{1}{|V|^{2}}V_{\flat}\wedge{}\mathcal{L}_{\frac{\partial}{\partial{}z}}dVol_{2}.$

so using (67) and Lemma 2.4 we obtain

(68) $\displaystyle\chi(M^{3}_{z},M^{2}_{z})$
$\displaystyle=\frac{1}{4\pi^{2}}\int_{M^{3}_{z}}\frac{1}{|V|^{2}}V_{\flat}\wedge\mathcal{L}_{\frac{\partial}{\partial{}z}}dVol_{2}=\frac{1}{2\pi}\int_{M^{2}_{z}}\mathcal{L}_{\frac{\partial}{\partial{}z}}dVol_{2}$
$\displaystyle=\frac{1}{2\pi}\frac{d}{dz}\int_{M^{2}_{z}}dVol_{2}.$

By Lemma 3.6, $\frac{d}{dz}\int_{M^{2}_{z}}dVol_{2}$ is constant when the
level-sets $M^{3}_{z}$ are compact and $z$ is non-singular. ∎

##  4\. The Euler number of the base

We prove the evolution equation
$\frac{d^{2}}{dz^{2}}\int_{M^{2}_{z}}|V|^{2}dVol_{2}=4\pi\chi^{g}(M^{2}_{z})$
when $M^{4}$ is scalar-flat and the integral leaves $M^{3}_{z}$ are compact.
In §4.2 we perform an example.

###  4.1. Interpretation of the Toda-lattice equation

The equation $u_{xx}+u_{yy}+(e^{u})_{zz}=0$—the “Toda-lattice equation” or
“$SU(\infty)$ Toda-lattice equation”—has long been studied in physics, Kähler
geometry, and conformal geometry. Because it relies on the formation of $x$,
$y$ coordinates, this equation is fundamentally local. We prove the Toda-
lattice equation has the more global expression

(69) $\displaystyle 2K\,dVol_{4}$
$\displaystyle\;=\;d\left[-(*dVol_{2})\wedge{}Jd\log|\nabla{}z|^{2}+\frac{1}{|V|}\triangle_{4}z\,dVol_{3}\right]$

where $K$ is the Gaussian curvature of $M^{2}_{z}$, considered as a function
on $M^{4}$ by pulling back along the reduction.

######  Proposition 4.1.

In the scalar-flat case the equation $u_{xx}+u_{yy}+(e^{u})_{zz}=0$ is

(70) $\displaystyle 2K\,dVol_{2}$
$\displaystyle\;=\;\Big{(}\triangle_{2}\log|V|^{2}\Big{)}\;dVol_{2}+\mathcal{L}_{\frac{\partial}{\partial{}z}}\mathcal{L}_{\frac{\partial}{\partial{}z}}\Big{(}|V|^{2}\,dVol_{2}\Big{)}$
$\displaystyle\;=\;\Big{(}\triangle_{2}\log|V|^{2}\Big{)}\;dVol_{2}+\mathcal{L}_{\frac{\partial}{\partial{}z}}\Big{(}\triangle_{4}z\;dVol_{2}\Big{)}.$

###### Proof.

Recall $u=\log\frac{|\nabla{}z|^{2}}{|\nabla{}x|^{2}}$, $w=|\nabla{}z|^{-2}$
from (32). Thus $u_{xx}+u_{yy}+(e^{u})_{zz}=0$ is

(71)
$0\;=\;\left(\frac{\partial^{2}}{\partial{}x^{2}}+\frac{\partial^{2}}{\partial{}y^{2}}\right)\log|\nabla{}z|-\left(\frac{\partial^{2}}{\partial{}x^{2}}+\frac{\partial^{2}}{\partial{}y^{2}}\right)\log|\nabla{}x|+\frac{1}{2}\left(\frac{|\nabla{}z|^{2}}{|\nabla{}x|^{2}}\right)_{zz}.$

The classic Gaussian curvature expression is
$K=-h^{-2}\left(\partial_{xx}+\partial_{yy}\right)(\log{}h)$ where
$h=|\partial/\partial{}x|$ and $(x,y)$ is an isothermal coordinate system.
Using $|\nabla{}x|=|\partial/\partial{}x|^{-1}$,
$dVol_{2}=|\nabla{}x|^{-2}dx\wedge{}dy$, and the fact that
$\mathcal{L}_{\frac{\partial}{\partial{}z}}(dx\wedge{}dy)=0$, then Eq. (71)
becomes

(72) $\displaystyle 2K\;dVol_{2}$
$\displaystyle\;=\;2\left(|\nabla{}x|^{2}\left(\frac{\partial^{2}}{\partial{}x^{2}}+\frac{\partial^{2}}{\partial{}y^{2}}\right)\log|\nabla{}x|\right)\left(|\nabla{}x|^{-2}dx\wedge{}dy\right)$
$\displaystyle\;=\;\left(2\left(\frac{\partial^{2}}{\partial{}x^{2}}+\frac{\partial^{2}}{\partial{}y^{2}}\right)\log|\nabla{}z|+\frac{\partial^{2}}{\partial{}z^{2}}\frac{|\nabla{}z|^{2}}{|\nabla{}x|^{2}}\right)\;dx\wedge{}dy$
$\displaystyle\;=\;\left(\left(\frac{\partial^{2}}{\partial{}x^{2}}+\frac{\partial^{2}}{\partial{}y^{2}}\right)\log|\nabla{}z|^{2}\right)\;dx\wedge{}dy+\mathcal{L}_{\frac{\partial}{\partial{}z}}\mathcal{L}_{\frac{\partial}{\partial{}z}}\left(\frac{|\nabla{}z|^{2}}{|\nabla{}x|^{2}}\;dx\wedge{}dy\right).$

Then using $(\partial^{2}_{xx}+\partial^{2}_{yy})=h^{2}\triangle_{2}$ and
$dx\wedge{}dy=h^{-2}dVol_{2}$, we arrive at

(73) $\displaystyle 2K\,dVol_{2}$
$\displaystyle\;=\;\Big{(}\triangle_{2}\log|V|^{2}\Big{)}\;dVol_{2}+\mathcal{L}_{\frac{\partial}{\partial{}z}}\mathcal{L}_{\frac{\partial}{\partial{}z}}\Big{(}|V|^{2}\,dVol_{2}\Big{)}.$

For the second equation in (70), use
$\mathcal{L}_{\frac{\partial}{\partial{}z}}(|V|^{2}\,dVol_{2})=(\triangle_{4}z)dVol_{2}$
from (51). ∎

######  Proposition 4.2.

In the scalar-flat case the equation $u_{xx}+u_{yy}+(e^{u})_{zz}=0$ is

(74) $\displaystyle 2K\,dVol_{4}$
$\displaystyle\;=\;\Big{(}\triangle_{2}\log|V|^{2}\Big{)}\;dVol_{4}+\mathcal{L}_{\frac{\partial}{\partial{}z}}\Big{(}\triangle_{4}z\,dVol_{4}\Big{)},\quad\text{or}$
$\displaystyle 2K\,dVol_{4}$
$\displaystyle\;=\;d\left[-Jd\log|V|^{2}\wedge*dVol_{2}+\frac{1}{|V|}\triangle_{4}z\,dVol_{3}\right].$

###### Proof.

Taking the exterior product with $*dVol_{2}$ on both sides of (70), then using
the Leibniz rule, we have

(75) $\displaystyle 2K\,dVol_{4}$
$\displaystyle\;=\;\Big{(}\triangle_{2}\log|V|^{2}\Big{)}\;dVol_{4}+\mathcal{L}_{\frac{\partial}{\partial{}z}}\Big{(}\triangle_{4}z\,dVol_{2}\Big{)}\wedge*dVol_{2}$
$\displaystyle\;=\;\Big{(}\triangle_{2}\log|V|^{2}\Big{)}\;dVol_{4}+\mathcal{L}_{\frac{\partial}{\partial{}z}}\Big{(}\triangle_{4}z\,dVol_{2}\wedge*dVol_{2}\Big{)}$
$\displaystyle\hskip
101.17755pt-(\triangle_{4}z)\,dVol_{2}\wedge\mathcal{L}_{\frac{\partial}{\partial{}z}}\Big{(}*dVol_{2}\Big{)}.$

We show the third term on the right vanishes. Using the fourth equation from
(51) of Lemma 3.5 we have

(76) $\displaystyle
dVol_{2}\wedge\mathcal{L}_{\frac{\partial}{\partial{}z}}\Big{(}*dVol_{2}\Big{)}$
$\displaystyle=-\left(\frac{\partial}{\partial{}z}\log|V|^{2}\right)dVol_{4}+dz\wedge{}Jd|V|^{-2}\wedge{}dVol_{2}$
$\displaystyle=-\left(\frac{\partial}{\partial{}z}\log|V|^{2}\right)dVol_{4}+dz\wedge{}Jd|V|^{-2}\wedge\omega$

where we used $\omega=dVol_{2}+*dVol_{2}$ with $dz\wedge*dVol_{2}=0$. Then
using (13) to simplify $dz\wedge{}Jd|V|^{-2}\wedge\omega$ we obtain

(77) $\displaystyle
dVol_{2}\wedge\mathcal{L}_{\frac{\partial}{\partial{}z}}\Big{(}*dVol_{2}\Big{)}=-\left(\frac{\partial}{\partial{}z}\log|V|^{2}\right)dVol_{4}-\left<\nabla{}z,\,\nabla|V|^{-2}\right>\,dVol_{4}$
$\displaystyle\hskip
72.26999pt=-\left(\frac{\partial}{\partial{}z}\log|V|^{2}\right)dVol_{4}-\left(\frac{\partial}{\partial{}z}\log|V|^{-2}\right)\,dVol_{4}\;=\;0.$

We conclude that (75) is

(78) $\displaystyle 2K\,dVol_{4}$
$\displaystyle\;=\;\Big{(}\triangle_{2}\log|V|^{2}\Big{)}\;dVol_{4}+\mathcal{L}_{\frac{\partial}{\partial{}z}}\Big{(}\triangle_{4}z\,dVol_{4}\Big{)}$

which is the first equation of (74). For the second equation, note that
$\mathcal{L}_{\frac{\partial}{\partial{}z}}=di_{\frac{\partial}{\partial{}z}}$
on 4-forms, and that
$i_{\frac{\partial}{\partial{}z}}(dVol_{4})=|V|^{-1}dVol_{3}$ by (23). Thus

(79)
$\displaystyle\mathcal{L}_{\frac{\partial}{\partial{}z}}\Big{(}\triangle_{4}z\,dVol_{4}\Big{)}\;=\;d\left(\frac{1}{|V|}\triangle_{4}z\,dVol_{3}\right).$

For the other term on the right side of (78) we compute

(80)
$\displaystyle\left(\triangle_{2}\log|V|^{2}\right)\;dVol_{4}\;=\;\left(\triangle_{2}\log|V|^{2}\right)\;dVol_{2}\wedge*dVol_{2}$
$\displaystyle\quad\quad\;=\;d\left(i_{\nabla{}\log|V|^{2}}dVol_{2}\right)\wedge*dVol_{2}$
$\displaystyle\quad\quad\;=\;d\left(i_{\nabla{}\log|V|^{2}}dVol_{2}\wedge*dVol_{2}\right)+(i_{\nabla{}\log|V|^{2}}dVol_{2})\wedge{}d(*dVol_{2}).$

We show the second term on the right vanishes. From Lemma 24 we have
$d(*dVol)=\left(\frac{\partial}{\partial{}z}\log|\nabla{}x|^{2}\right)dz\wedge{}dVol_{2}$.
Then

(81) $\displaystyle(i_{\nabla{}\log|V|^{2}}dVol_{2})\wedge{}d(*dVol_{2})$
$\displaystyle\quad=\left(\frac{\partial}{\partial{}z}\log|\nabla{}x|^{2}\right)(i_{\nabla{}\log|V|^{2}}dVol_{2})\wedge{}dVol_{2}\wedge{}dz$
$\displaystyle\quad=\frac{1}{2}\left(\frac{\partial}{\partial{}z}\log|\nabla{}x|^{2}\right)i_{\nabla{}\log|V|^{2}}(dVol_{2}\wedge{}dVol_{2})\wedge{}dz\;=\;0$

where we used $i_{v}(\eta\wedge\eta)=2(i_{v}\eta)\wedge\eta$ whenever $\eta$
is a 2-form, and that $dVol_{2}\wedge{}dVol_{2}=0$ due to the fact that
$dVol_{2}$ is decomposable (it is a multiple of $dx\wedge{}dy$). Therefore

(82)
$\displaystyle\triangle_{2}\log|V|^{2}\,dVol_{4}\;=\;d\Big{(}(i_{\nabla{}\log|V|^{2}}dVol_{2})\wedge*dVol_{2}\Big{)}.$

Because we have the two expressions

(83) $\displaystyle i_{\nabla{}\log|V|^{2}}dVol_{2}$
$\displaystyle=-\left(\frac{\partial}{\partial{}y}\log|V|^{2}\right)dx+\left(\frac{\partial}{\partial{}x}\log|V|^{2}\right)dy$
$\displaystyle Jd\log|V|^{2}$
$\displaystyle=\left(\frac{\partial}{\partial{}y}\log|V|^{2}\right)dx-\left(\frac{\partial}{\partial{}x}\log|V|^{2}\right)dy-\left(\frac{\partial}{\partial{}z}\log|V|^{2}\right)V_{\flat},$

then using $*dVol_{2}=|V|^{-2}dz\wedge{}V_{\flat}$ we obtain

(84)
$\displaystyle\left(i_{\nabla{}\log|V|^{2}}dVol_{2}\right)\wedge*dVol_{2}\;=\;-Jd\log|V|^{2}\wedge*dVol_{2}.$

Inserting these back into (78), we obtain as claimed

(85) $\displaystyle 2K\,dVol_{4}$
$\displaystyle\;=\;d\left(-Jd\log|V|^{2}\wedge*dVol_{2}+\frac{1}{|V|}\triangle_{4}z\,dVol_{3}\right).$

∎

######  Proposition 4.3 (cf. Theorem 1.2).

Assume $M^{4}$ is scalar-flat and the level-sets $M^{3}_{z}$ are compact. Then
at non-critical values of $z$ the value $\chi^{g}(M^{2}_{z})$ is constant and

(86)
$\frac{d^{2}}{dz^{2}}\int_{M^{2}_{z}}|V|^{2}\,dVol_{2}\;=\;\frac{d}{dz}\int_{M^{2}_{z}}\triangle_{4}z\,dVol_{2}\;=\;4\pi\chi^{g}(M^{2}_{z}).$

###### Proof.

We use the convenient notation
$M^{4}_{z_{0},z_{1}}=\\{p\in{}M^{4}\,\big{|}\,z_{0}<z(p)<z_{1}\\}$. If $z$ is
non-critical at some value, by smoothness and the fact that levelsets are
compact, $z$ is non-critical on an interval $z_{0}<z<z_{1}$ around that value.
Integrating both sides of (74) gives

(87) $\displaystyle\int_{M^{4}_{z_{0},z_{1}}}2K\,dVol_{4}$
$\displaystyle\hskip
36.135pt=\int_{M^{4}_{z_{0},z_{1}}}d\left[\left(\frac{1}{|\nabla{}z|}\triangle_{4}{}z\right)dVol_{3}+dz\wedge{}V_{\flat}\wedge{}Jd|\nabla{}z|^{-2}\right].$

For the left-hand side of (87), Lemma 2.4 gives

(88)
$\displaystyle\int_{M^{4}_{z_{0},z_{1}}}2K\,dVol_{4}=4\pi\int_{z_{0}}^{z_{1}}\left(\int_{M^{2}_{z}}K\,dVol_{2}\right)\,dz=8\pi^{2}\int_{z_{0}}^{z_{1}}\chi^{g}(M^{2}_{z})\,dz.$

due to the Gauss-Bonnet theorem on orbifolds which states
$\chi^{g}(M^{2}_{z})=\frac{1}{2\pi}\int_{M^{2}_{z}}K\,dVol_{2}$ (as discussed
in the introduction). For the right-hand side of (87), by Stokes theorem and
the Integration Lemma 2.4 we have

(89)
$\displaystyle\int_{M^{4}_{z_{0},z_{1}}}d\left[\left(\frac{1}{|\nabla{}z|}\triangle_{4}{}z\right)dVol_{3}+dz\wedge{}V_{\flat}\wedge{}Jd|\nabla{}z|^{-2}\right]$
$\displaystyle\quad\quad=\int_{M^{3}_{z_{1}}}\left(\frac{1}{|\nabla{}z|}\triangle_{4}{}z\right)dVol_{3}-\int_{M^{3}_{z_{0}}}\left(\frac{1}{|\nabla{}z|}\triangle_{4}{}z\right)dVol_{3}$
$\displaystyle\quad\quad=2\pi\int_{M^{2}_{z_{1}}}\triangle_{4}{}z\,dVol_{2}-2\pi\int_{M^{2}_{z_{0}}}\triangle_{4}{}z\,dVol_{2}$

The terms of the form
$\int_{M^{3}_{z}}dz\wedge{}V_{\flat}\wedge{}J|\nabla{}z|^{-2}$ disappear
because $M^{3}_{z}$ is an integral leaf of $(\nabla{}z)^{\perp}$, so $dz$
vanishes on $M^{3}_{z}$. Combining (88) and (89), multiplying by
$\frac{1}{z_{1}-z_{0}}$, and taking a limit, the Fundamental Theorem of
Calculus gives

(90)
$\frac{d}{dz}\int_{M^{2}_{z}}\triangle_{4}{}z\;dVol_{2}\;=\;4\pi\chi^{g}(M^{2}_{z}).$

Lemma 3.5 gives
$(\triangle_{4}z)dVol_{2}=\mathcal{L}_{\frac{\partial}{\partial{}z}}(|V|^{2}dVol_{2})$.
Therefore

(91)
$\frac{d^{2}}{dz^{2}}\int_{M^{2}_{z}}|V|^{2}\,dVol_{2}=\frac{d}{dz}\int_{M^{2}_{z}}(\triangle_{4}z)\,dVol_{2}=2\int{}K\,dVol_{2}=4\pi\chi^{g}(M^{2}).$

Lastly we show the constancy of $\chi^{g}(M^{2}_{z})$. The left-hand side of
(90) is

(92)
$\small\frac{d}{dz}\int_{M^{2}_{z}}(\triangle_{4}z)dVol_{2}=\frac{1}{2\pi}\frac{d}{dz}\int_{M^{3}_{z}}\frac{\triangle_{4}z}{|V|}\,dVol_{3}=\frac{1}{2\pi}\int_{M^{3}_{z}}\mathcal{L}_{\frac{\partial}{\partial{}z}}\left(\frac{1}{|V|}\triangle_{4}z\,dVol_{3}\right).$

Away from $|V|=0$, $M^{3}_{z}$ is a manifold and
$\mathcal{L}_{\frac{\partial}{\partial{}z}}\left(\frac{1}{|V|}\triangle_{4}z\,dVol_{3}\right)$
is a smooth 3-form. Thus the quantities on both sides of (90) are smoothly
varying with $z$. But $\chi^{g}(M^{2}_{z})$ is rational (see, for example,
[41]), so therefore is a constant. ∎

######  Corollary 4.4 (cf. Theorem 1.4).

Assume $M^{4}$ has compact level-sets and $|V|>\epsilon>0$ for sufficiently
large $z$ (or large $-z$). Then for large $z$ (or $-z$) the numbers
$\chi^{g}(M^{2}_{z})$ and $e^{g}(M^{3}_{z},M^{3}_{z})$ are constant. Further,
we have $\chi^{g}(M^{2}_{z})\geq 0$ for large $z$ or $-z$. We have
$e^{g}(M^{3}_{z},M^{2}_{z})\geq 0$ for large $z$ and
$e^{g}(M^{3}_{z},M^{2}_{z})\leq 0$ for large $-z$.

###### Proof.

By Lemma 2.8 we have that the $M^{3}_{z}$ and fibrations
$M^{3}_{z}\rightarrow{}M^{2}_{z}$ are isotopic for large $z$; this implies the
constancy of $\chi^{g}(M^{2}_{z})$ and $e^{g}(M^{3}_{z},M^{2}_{z})$. To see
that $\chi^{g}(M^{2}_{z})\geq 0$ for large $z$, note that the range of $z$ is
infinite, so
$\frac{d^{2}}{dz^{2}}\int_{M^{2}_{z}}|V|^{2}dVol_{2}=4\pi\chi^{g}(M^{2}_{z})$
forces the constant $\chi^{g}(M^{2}_{z})$ to be non-negative, or else
$\int_{M^{2}_{z}}|V|^{2}dVol_{2}$ would become negative.

To see that $e^{g}(M^{3}_{z},M^{2}_{z})\geq 0$ on $M^{4+}$, note that
$\frac{d}{dz}\int_{M^{2}_{z}}dVol_{2}=2\pi{}e^{g}(M^{3}_{z},M^{2}_{z})$ for
all large $z$ forces $e^{g}(M^{3}_{z},M^{2}_{z})\geq 0$ or else
$Vol(M^{2}_{z})$ would become negative. ∎

Remark. We have no need to prove it here, but when the scalar curvature $s$ of
$M^{4}$ is non-zero, then (69) becomes

(93) $\displaystyle\left(2K-s\right)\,dVol_{4}$
$\displaystyle\;=\;d\left[-Jd\log|V|^{2}\wedge*dVol_{2}+\frac{1}{|V|}\triangle_{4}z\,dVol_{3}\right].$

This resembles a Chern-Gauss-Bonnet transgression. Despite this resemblance,
note that $(2K-s)dVol_{4}$ is not scale-invariant and this expression has no
obvious interpretation as a product of symmetric invariant polynomials.

###  4.2. The Euclidean example

We illustrate the results of the previous two sections with the example that
$M^{4}$ is flat $\mathbb{C}^{2}$. The choices of Killing field with compact
level-sets are given by the potential

(94) $z\;=\;-\frac{1}{2}\alpha|z^{1}|^{2}-\frac{1}{2}\beta|z^{2}|^{2}$

where $\alpha$, $\beta$ are both positive—if $\alpha$, $\beta$ have different
signs then the $M^{3}_{z}$ are non-compact. For orbits to close up after a
diffeomorphism flow of $2\pi$, $\alpha$ and $\beta$ must be coprime integers.
The easiest example is $\alpha=\beta=1$, where the Killing field is

(95)
$V=y^{1}\frac{\partial}{\partial{}x^{1}}-x^{1}\frac{\partial}{\partial{}y^{1}}+y^{2}\frac{\partial}{\partial{}x^{2}}-x^{2}\frac{\partial}{\partial{}y^{2}}$

and the orbits are Hopf circles. In this case $M^{3}_{z}$ is the sphere
$\mathbb{S}^{3}(\sqrt{-2z})\subset\mathbb{C}^{2}$ of radius $\sqrt{-2z}$ and
$M^{2}_{z}$ is the 2-sphere of radius $\sqrt{-z/2}$ and area $-2\pi{}z$. We
have $|V|^{2}=2z$ and $\triangle_{4}{}z=-4$. We compute

(96)
$\displaystyle\int_{M^{2}_{z}}|V|^{2}dVol_{2}\;=\;4\pi(z)^{2},\quad\frac{d^{2}}{dz^{2}}\int_{M^{2}_{z}}|V|^{2}dVol_{2}\;=\;8\pi\;=\;4\pi\chi^{g}(M^{2}_{z})$
$\displaystyle\int_{M^{2}_{z}}\triangle_{4}z\,dVol_{2}\;=\;8\pi{}z,\quad\frac{d}{dz}\int_{M^{2}_{z}}\triangle_{4}z\,dVol_{2}\;=\;8\pi\;=\;4\pi\chi^{g}(M^{2}_{z})$
$\displaystyle\int_{M^{2}_{z}}dVol_{2}\;=\;-2\pi{}z,\quad\frac{d}{dz}\int_{M^{2}_{z}}dVol_{2}\;=\;-2\pi\;=\;2\pi{}e^{g}(M^{3}_{z},M^{2}_{z}),$

recalling that the Hopf fibration has
$\chi^{g}(\mathbb{S}^{3},\mathbb{S}^{2})=-1$. Thus we explicitly recover the
conclusions of Propositions 4.3 and 3.7.

Referring to the construction of Proposition 3.7, we have $V_{\flat}=rJdr$
where $r=(|z^{1}|^{2}+|z^{2}|^{2})^{1/2}$ and $dV_{\flat}=-2\omega$. Then the
integrand of (62) is the quantity
$|V|^{-4}V_{\flat}\wedge{}dV_{\flat}=-2r^{-3}dVol_{3}$ (the “$-$” sign comes
from orientation). Using $r=\sqrt{-2z}$, we have

(97) $\small
e^{g}(M^{3}_{z},M^{2}_{z})=\frac{1}{4\pi^{2}}\int_{M^{3}_{z}}|V|^{-4}V_{\flat}\wedge{}dV_{\flat}=-\frac{1}{4\pi^{2}}\int_{\mathbb{S}^{3}(\sqrt{-2z})}\frac{2}{\big{(}\sqrt{-2z}\big{)}^{3}}\,dVol_{3}=-1.$

This explicitly recovers Eq. (62).

##  5\. The Chern-Gauss-Bonnet Formalism

Here we compute the Chern-Gaus-Bonnet integrals based on the existence the
symplectomorphic Killing field $V$. We show that one of these integrals
reduces to

(98)
$\frac{d^{2}}{dz^{2}}\int_{M^{2}_{z}}(\triangle_{4}z)^{2}\,dVol_{2}=2\int_{M^{2}_{z}}|\mathop{\rm
Ric}|^{2}\,dVol_{2}$

when level-sets are compact and scalar curvature is zero. The computation
requires precise manipulation of the details of the theory; for example sign
choices are crucial and not always consistent in the literature. We take the
time to start from the beginning and be careful. We conclude in §5.4 by giving
explicit demonstrations of our results on the scalar flat LeBrun instantons
[29].

First we record a well-known Bochner identity.

######  Lemma 5.1.

$d(\triangle_{4}z)+2\mathop{\rm Ric}(\nabla{}z)=0$. In particular
$\frac{\partial}{\partial{}z}(\triangle_{4}z)=-2\mathop{\rm
Ric}\left(\frac{\nabla{}z}{|\nabla{}z|},\,\frac{\nabla{}z}{|\nabla{}z|}\right)$.

###### Proof.

First we compute $\triangle{}V_{\flat}$. The Killing condition
$V_{i,j}=-V_{j,i}$ forces $V_{j,j}=0$ so the usual commutation identity gives

(99) $\displaystyle V_{i,jj}$
$\displaystyle\;=\;-V_{j,ij}\;=\;-V_{j,ji}-\mathop{\rm
Rm}{}_{ijjs}V_{s}\;=\;-\mathop{\rm Ric}{}_{is}V_{s}.$

Because $J$ is covariant constant and $\mathop{\rm Ric}$ is $J$-invariant,
this is

(100) $\triangle_{4}{}V+\mathop{\rm
Ric}(V)\;=\;0\quad\quad\text{or}\quad\quad\triangle_{4}{}dz\;+\;\mathop{\rm
Ric}(\nabla{}z)\;=\;0.$

Then we compute

(101) $z_{,ijj}\;=\;z_{,jij}\;=\;z_{,jji}+\mathop{\rm
Rm}{}_{ijjs}z_{,s}\;=\;z_{,jji}+\mathop{\rm Ric}{}_{is}z_{,s}$

which is $\triangle_{4}dz=d\triangle_{4}z+\mathop{\rm Ric}(\nabla{}z)$. Thus
$d\triangle_{4}z+2\mathop{\rm Ric}(\nabla{}z)=0$. ∎

###  5.1. The transgression construction

We review the basic theory and set notation; this section’s objective is to
record Eq. 112. References are [23] [27] [13]. Despite our Kähler setting it
will be easiest to assume the structure group is $SO(4)$, due to the ease of
representing the Hodge star on the Lie algebra $\mathfrak{o}(4)$.

Recall the bracket $[A,B]$ when $A\in\bigwedge{}^{k}\otimes\mathfrak{o}(4)$
and $B\in\bigwedge{}^{l}\otimes\mathfrak{o}(4)$. This is defined by extending
the rule
$[\eta\otimes\mathfrak{a},\mu\otimes\mathfrak{b}]=\eta\wedge\mu\otimes[\mathfrak{a},\mathfrak{b}]$
by linearity. We have the commutation identity $[A,B]=(-1)^{kl+1}[B,A]$.

Let $D$ be a connection
$D:\bigwedge^{k}\otimes\mathfrak{o}(4)\rightarrow\bigwedge^{k+1}\otimes\mathfrak{o}(4)$;
then $D$ obeys the Leibniz rule
$D(\eta\wedge{}A)=d\eta\wedge{}A+(-1)^{k}\eta\wedge{}DA$ when
$\mu\in\bigwedge{}^{k}$ and $A\in\bigwedge{}^{k}\otimes\mathfrak{g}$. $D$ can
be expressed locally as a $\mathfrak{g}$-valued $1$-form $\theta$, where we
have $DA=dA+[\theta,A]$. The curvature of $D$ is
$F=D\circ{}D=d\theta+\frac{1}{2}[\theta,\theta]\in\bigwedge{}^{2}\otimes\mathfrak{g}$.

We use two different, but related, Hodge-star operators: one that operates on
the vector space $\bigwedge^{2}$ and the other that operates on the Lie
algebra $\mathfrak{o}(4)$. They are abstractly the same but it is important
for practical reasons to distinguish them, so we use
$*:\bigwedge^{2}\rightarrow\bigwedge^{2}$ and write $*\eta$ when $\eta$ is a
2-form, and $\star:\mathfrak{o}(4)\rightarrow\mathfrak{o}(4)$ and write
$\mathfrak{a}\star$ when $\mathfrak{a}\in\mathfrak{o}(4)$. If
$A=\eta\otimes\mathfrak{a}$ is a $\mathfrak{o}(4)$-valued 2-form, we write

(102) $*A=(*\eta)\otimes\mathfrak{a},\quad\quad
A\star=\eta\otimes(\mathfrak{a}\star),\quad\quad*A\star=(*\eta)\otimes(\mathfrak{a}\star).$

In a basis, the operators $*$ and $\star$ have expressions in terms of the
Levi-Civita antisymmetric symbol $\epsilon$. If $\eta=\eta_{ij}$ is a 2-form
then $(*\eta)_{ij}=\frac{1}{2}\epsilon^{kl}{}_{ij}\eta_{kl}$ and if
$\mathfrak{a}\in\mathfrak{o}(4)$ is an antisymmetric matrix
$\mathfrak{a}=\mathfrak{a}^{i}_{j}$ then
$(\mathfrak{a}\star)^{i}_{j}=-\frac{1}{2}\mathfrak{a}^{k}_{l}\epsilon^{il}{}_{jk}$.
The operators $*$, $\star$ create projectors
$\pi^{\pm}:\bigwedge^{2}\rightarrow\bigwedge^{\pm}$,
$\pi^{\pm}:\mathfrak{o}(4)\rightarrow\mathfrak{o}(3)$ where
$\pi^{\pm}=\frac{1}{2}(1\pm*)$, $\pi^{\pm}=\frac{1}{2}(1\pm\star)$,
respectively.

It is well-known that the vector space of quadratic symmetric invariant
polynomials on $\mathfrak{o}(4)$ is spanned by just two polynomials, which in
polar form are
$\mathcal{P}_{1}(\mathfrak{a},\mathfrak{b})=-\frac{1}{2}Tr(\mathfrak{a}\mathfrak{b})$
(which is half the polarization of the matrix norm) and
$\mathcal{P}_{2}(\mathfrak{a},\mathfrak{b})=-\frac{1}{2}Tr(\mathfrak{a}\mathfrak{b}\star)$
(which is twice the polarization of the Pfaffian). If $\mathcal{P}$ is a
quadratic polynomial on $\mathfrak{o}(4)$, it extends to the algebra
$\bigoplus_{k}\bigwedge{}^{k}\otimes\mathfrak{o}(4)$ by setting

(103)
$\mathcal{P}(\eta\otimes\mathfrak{a},\,\mu\otimes\mathfrak{b})\;=\;\eta\wedge\mu\cdot\mathcal{P}(\mathfrak{a},\mathfrak{b})$

and extending by bilinearlity. For $A\in\bigwedge^{k}\otimes\mathfrak{o}(4)$,
$A\in\bigwedge^{l}\otimes\mathfrak{o}(4)$ we have the commutator and Leibniz
identities

(104) $\displaystyle\mathcal{P}(A,\,B)\;=\;(-1)^{kl}\mathcal{P}(B,\,A),$
$\displaystyle
d\mathcal{P}(A,\,B)\;=\;\mathcal{P}(DA,\,B)+(-1)^{k}\mathcal{P}(A,\,DB).$

Below, we will make use of the identity

(105)
$\mathcal{P}(\eta\otimes\mathfrak{a},\;\eta\wedge\mu\otimes\mathfrak{b})\;=\;0$

whenever $\eta\in\bigwedge^{1}$ and $\mu$ is any other form.

Now assume $\bar{\theta}\in\bigwedge^{1}\otimes\mathfrak{g}$ has the
particular form $\bar{\theta}=\eta\otimes{}\mathfrak{a}$ where $\eta$ is a
1-form and $\mathfrak{a}$ is a section $\mathfrak{a}\in\mathfrak{g}$. Then
$D_{1}=d+[\theta-\bar{\theta},\,\cdot]$ is another connection. Interpolating
between the original connection $D=d+\theta$ with curvature $F$ and the new
connection $D_{1}=d+\theta-\bar{\theta}$ with curvature $F_{1}$, for each $t$
we have a connection $D_{t}=d+\theta-t\bar{\theta}$ with curvature

(106) $\displaystyle F_{t}$
$\displaystyle=d(\theta-t\bar{\theta})+\frac{1}{2}[\theta-t\bar{\theta},\,\theta-t\bar{\theta}]=d\theta-
td\bar{\theta}+\frac{1}{2}[\theta,\,\theta]-[\bar{\theta},\,\theta]+t^{2}[\bar{\theta},\,\bar{\theta}].$

Because we assumed $\bar{\theta}=\eta\otimes{}A$ we have
$[\bar{\theta},\bar{\theta}]=\eta\wedge\eta\otimes[A,A]=0$. Therefore

(107) $\displaystyle F_{t}$
$\displaystyle\;=\;d\theta+\frac{1}{2}[\theta,\,\theta]-t\left(d\bar{\theta}+[\theta,\bar{\theta}]\right)\;=\;F\,-\,tD\bar{\theta}.$

Then when $\mathcal{P}$ a symmetric invariant polynomial we have

(108)
$\mathcal{P}(F_{1},F_{1})\,-\,\mathcal{P}(F,F)\;=\;\int_{0}^{1}\frac{d}{dt}\mathcal{P}(F_{t},F_{t})\,dt\;=\;2\int_{0}^{1}\mathcal{P}\left(\frac{d}{dt}F_{t},F_{t}\right)\,dt.$

By (106) we have $F_{t}=F-tD\bar{\theta}$ so
$\frac{d}{dt}F_{t}=-D\bar{\theta}$. Thus

(109) $\displaystyle\mathcal{P}(F_{1},F_{1})\,-\,\mathcal{P}(F,F)$
$\displaystyle\;=\;-2\int_{0}^{1}\mathcal{P}\left(D\bar{\theta},F-tD\bar{\theta}\right)\,dt$
$\displaystyle\;=\;-2\mathcal{P}\left(D\bar{\theta},F\right)+\mathcal{P}\left(D\bar{\theta},D\bar{\theta}\right).$

To simplify this, we use the Leibniz rule (104) and the fact that $DF=0$ to
obtain

(110)
$\mathcal{P}\left(D\bar{\theta},F\right)=d\left(\mathcal{P}\left(\bar{\theta},F\right)\right)$

Then by the Leibniz rule we have

(111) $\displaystyle\mathcal{P}\left(D\bar{\theta},D\bar{\theta}\right)$
$\displaystyle\;=\;d\left(\mathcal{P}\left(\bar{\theta},D\bar{\theta}\right)\right)+\mathcal{P}\left(\bar{\theta},D\circ{}D\bar{\theta}\right)$
$\displaystyle\;=\;d\left(\mathcal{P}\left(\bar{\theta},D\bar{\theta}\right)\right)+\mathcal{P}\left(\bar{\theta},F\wedge\bar{\theta}\right)$
$\displaystyle\;=\;d\left(\mathcal{P}\left(\bar{\theta},D\bar{\theta}\right)\right)+\mathcal{P}\left(\eta\otimes{A},F\wedge\eta\otimes{A}\right)\;=\;d\left(\mathcal{P}\left(\bar{\theta},D\bar{\theta}\right)\right)$

where we also used (105). Equation (109) is therefore

(112)
$\mathcal{P}(F,F)\;=\;\mathcal{P}(F_{1},F_{1})+d\Big{(}2\mathcal{P}(\bar{\theta},\,F)-\mathcal{P}(\bar{\theta},D\bar{\theta})\Big{)}.$

The curvature tensor $F\in\bigwedge^{2}\otimes\mathfrak{o}(4)$ is usually
expressed as a matrix of 2-forms $F=(F^{i}_{j})$. For the Levi-Civita
connection, $F$ relates to the Riemann tensor $\mathop{\rm Rm}$ by

(113) $F^{i}_{j}\;=\;-\frac{1}{2}\mathop{\rm
Rm}{}_{stj}{}^{i}\,\eta^{s}\wedge\eta^{t}$

where we use the sign convention $sec(e_{i},e_{j})=\mathop{\rm Rm}_{ijji}$.
The decomposition $\bigwedge^{2}=\bigwedge^{+}\oplus\bigwedge^{-}$ breaks
$\mathop{\rm Rm}\in\bigwedge^{2}\otimes\bigwedge^{2}$ into the four components
$\mathop{\rm Rm}^{\pm\pm}\in\bigwedge^{\pm}\otimes\bigwedge^{\pm}$. For
example $\mathop{\rm Rm}^{+-}=\frac{1}{4}(1+*)F(1-\star)$. It is well-known
[44], [5] that

(114) $\displaystyle\mathop{\rm
Rm}{}^{++}=W^{+}+\frac{s}{12}Id_{\bigwedge^{+}},\quad\mathop{\rm
Rm}{}^{--}=W^{-}+\frac{s}{12}Id_{\bigwedge^{-}},$ $\displaystyle\hskip
57.81621pt\mathop{\rm Rm}{}^{+-}+\mathop{\rm Rm}{}^{-+}=\frac{1}{2}\mathop{\rm
R\text{\i}\makebox[0.0pt]{\raisebox{5.0pt}{\tiny$\circ$\;\,}}c}\mathop{\makebox{$\wedge$}\bigcirc}{}g.$

It is typical [18] to use $F^{+}$ to indicate $\frac{1}{2}(F+F\star)$—as
opposed to $\frac{1}{2}(F+*F)$—so for example $F^{+}=\mathop{\rm
Rm}{}^{++}+\mathop{\rm Rm}^{-+}$. For Kähler metrics, Proposition 2 of [16]
states
$W^{+}=-\frac{s}{12}\left(Id_{\bigwedge^{+}}+\frac{3}{2}\omega\otimes\omega\right)$.
(To resolve a sign issue with Derdzhinski’s Eq. (36) of [16], because our
convention on the Riemann tensor is the opposite to the one used there, a
tensor $T_{ijkl}:\bigwedge^{+}\rightarrow\bigwedge^{+}$ operates by
$T(\zeta)_{ij}=\frac{1}{2}T_{ijst}\zeta^{ts}$—notice the index reversal—so
that
$Id_{\bigwedge^{+}}=-\frac{1}{2}\omega\otimes\omega-\frac{1}{2}\eta\otimes\eta-\frac{1}{2}\mu\otimes\mu$
where $\eta,\mu\in\bigwedge^{+}$ are orthogonal and of length $\sqrt{2}$.) We
have the well known facts for Kähler metrics

(115) $\mathop{\rm
Rm}{}^{++}\;=\;-\frac{s}{8}\omega\otimes\omega,\quad\quad|\mathop{\rm
Rm}{}^{++}|^{2}\;=\;\frac{1}{4}s^{2}.$

###  5.2. Construction of the Chern-Gauss-Bonnet integrands

This section’s objective is to record Eq. (127). Recall the decomposition of
$dJdz\in\bigwedge^{2}$ into its $\bigwedge^{\pm}$ components gives

(116) $dJdz\;=\;-\frac{1}{2}(\triangle_{4}{}z)\,\omega\,+\,(dJdz)^{-}.$

Because $Jdz=-V_{\flat}$ we have
$(dV_{\flat})^{+}=\frac{1}{2}(\triangle_{4}z)\omega$. Let
$\\{e_{1},e_{2},e_{3},e_{4}\\}$ be any orthonormal frame and
$\\{\eta^{1},\eta^{2},\eta^{3},\eta^{4}\\}$ its coframe. Because
$\nabla{}V_{\flat}$ is antisymmetric, then $\nabla{}V$ is antisymmetric in the
sense that $V^{i}{}_{,j}=\eta^{i}(\nabla_{e_{j}}V)$ is an anti-symmetric
matrix, so it is a section with values in $\mathfrak{o}(4)$. By (16) we have
$(dJdz)_{ij}=-2V^{s}{}_{,j}g_{si}$. Using
$\nabla{}V^{+}=\frac{1}{2}(\nabla{}V+\nabla{}V\star)$ and (116) we have

(117)
$(\nabla{}V)^{+}\;=\;-\frac{1}{4}(\triangle{}z)\,J\quad\text{or}\quad\left[(\nabla{}V)^{+}\right]^{i}_{j}\;=\;-\frac{1}{4}(\triangle{}z)\,J_{j}{}^{i}$

where $J=J_{s}{}^{t}\eta^{s}\otimes{}e_{t}$ is the complex structure. (The
“$-$” sign in (117) comes from $\omega_{sj}g^{si}=-J_{j}{}^{i}$.)

Let $\\{e_{1},e_{2},e_{3},e_{4}\\}$ be an orthonormal frame adapted to $V$ in
the sense that $[V,e_{i}]=0$ and let
$\\{\eta^{1},\eta^{2},\eta^{3},\eta^{4}\\}$ be the associated coframe. Let
$\nabla=d+\theta$ be Levi-Civita differentiation, so $\theta$ is an
$\mathfrak{o}(4)$-valued 1-form. Letting $w$ be any vector, $i_{w}\theta$ is a
matrix with components $i_{w}\theta^{i}_{j}=\eta^{i}(\nabla_{w}e_{j})$.
Recalling that $\nabla{}V$ is antisymmetric, we create a new
$\mathfrak{o}(4)$-valued 1-form $\bar{\theta}$ by

(118) $\bar{\theta}\;=\;\frac{1}{|V|^{2}}V_{\flat}\otimes\nabla{}V.$

Then $\bar{\theta}\in\bigwedge{}^{1}\otimes\mathfrak{o}(4)$ is a tensor away
from $V=0$. Therefore $\theta-\bar{\theta}$ is a connection 1-form. It has
connection $D_{1}=d+[\theta-\bar{\theta},\,\cdot]$ and curvature $F_{1}$. We
show that $V$ is a null vector of $F_{1}$. First we verify that the connection
$\theta-\bar{\theta}$ has a null vector:

(119) $\displaystyle i_{V}(\theta-\bar{\theta})^{i}_{j}$
$\displaystyle\;=\;i_{V}\theta^{i}_{j}-i_{V}\left(\eta^{1}\otimes\frac{1}{|V|}\eta^{i}(\nabla_{e_{j}}V)\right)$
$\displaystyle\;=\;\eta^{i}(\nabla_{V}e_{j})-\eta^{i}(\nabla_{e_{j}}V)\;=\;\eta^{i}([V,\,e_{j}])\;=\;0.$

Next we verify that $\theta$ and $\bar{\theta}$ are $V$-invariant in the sense
that $\mathcal{L}_{V}\theta=0$. Very simply,
$\mathcal{L}_{V}\theta^{i}_{j}=\mathcal{L}_{V}(\eta^{j}(\nabla{}e_{j}))$; then
adaptedness of the frame gives $\mathcal{L}_{V}(\eta^{i})=0$ and
$\mathcal{L}_{V}(e_{j})=0$, and we have $\mathcal{L}_{V}(\nabla)=0$ because
$V$ is Killing. Thus the Leibniz rule gives $\mathcal{L}_{V}\theta=0$.
Essentially the same argument gives $\mathcal{L}_{V}\bar{\theta}=0$.

Now we establish that $F_{1}$ has a null vector. Because
$i_{V}(\theta-\bar{\theta})=0$ we have
$i_{V}[\theta-\bar{\theta},\theta-\bar{\theta}]=0$. Because
$i_{V}(\theta-\bar{\theta})=0$ and $\mathcal{L}_{V}(\theta-\bar{\theta})=0$,
then the fact that $\mathcal{L}_{V}=di_{V}+i_{V}d$ on $\mathfrak{o}(4)$-valued
1-forms gives $i_{V}d(\theta-\bar{\theta})=0$. Therefore

(120) $\displaystyle i_{V}F_{1}$
$\displaystyle=i_{V}d(\theta-\bar{\theta})\,-\,\frac{1}{2}i_{V}[\theta-\bar{\theta},\,\theta-\bar{\theta}]=0.$

Because $F_{1}$ has a null vector, the 4-form $\mathcal{P}(F_{1},F_{1})$ has a
null vector so certainly $\mathcal{P}(F_{1},F_{1})=0$. Therefore (112) gives

(121) $\displaystyle\mathcal{P}(F,F)$
$\displaystyle=d\,\bigg{(}2\mathcal{P}(\bar{\theta},\,F)-\mathcal{P}(\bar{\theta},\,D\bar{\theta})\bigg{)}.$

We apply this to the usual Chern-Gauss-Bonnet formulas for the signature
$\tau(M^{4})$ and Euler invariant $\chi(M^{4})$. The 4-forms
$\mathcal{P}_{\tau}=\frac{1}{12\pi^{2}}\mathcal{P}_{1}(F,F)$ and
$\mathcal{P}_{\chi}=\frac{1}{8\pi^{2}}\mathcal{P}_{2}(F,F)$ have the
expressions111Some presentations have a factor of 1 rather than $\frac{1}{4}$
on the Weyl components. This depends on whether the Weyl tensors are normed as
operators $\bigwedge^{\pm}\rightarrow\bigwedge^{\pm}$ as in [16], or as
tensors as $|W|^{2}=W_{ijkl}W^{ijkl}$. The latter is 4 times the former. We
use the latter, hence the factor of $\frac{1}{4}$.

(122)
$\displaystyle\mathcal{P}_{\tau}\;=\;\frac{1}{12\pi^{2}}\left(\frac{1}{4}|W^{+}|^{2}-\frac{1}{4}|W^{-}|^{2}\right)\,dVol,$
$\displaystyle\mathcal{P}_{\chi}\;=\;\frac{1}{8\pi^{2}}\left(\frac{1}{24}s^{2}\,-\,\frac{1}{2}|\mathop{\rm
R\text{\i}\makebox[0.0pt]{\raisebox{5.0pt}{\tiny$\circ$\;\,}}c}|^{2}\,+\,\frac{1}{4}|W^{+}|^{2}\,+\,\frac{1}{4}|W^{-}|^{2}\right)\,dVol.$

By (121) we have transgressions

(123)
$\displaystyle\mathcal{P}_{\tau}\;=\;d\mathcal{TP}_{\tau}\quad\text{where}\quad\mathcal{TP}_{\tau}\;=\;-\frac{1}{24\pi^{2}}\left(2\bar{\theta}^{i}_{j}\wedge{}F^{j}_{i}-\bar{\theta}^{i}_{j}\wedge{}D\bar{\theta}^{j}_{i}\right),\quad\text{and}$
$\displaystyle\mathcal{P}_{\chi}\;=\;d\mathcal{TP}_{\chi}\quad\text{where}\quad\mathcal{TP}_{\chi}\;=\;-\frac{1}{16\pi^{2}}\left(2\bar{\theta}^{i}_{j}\wedge(F\star)^{j}_{i}-\bar{\theta}^{i}_{j}\wedge(D\bar{\theta}\star)^{j}_{i}\right)$

which hold whenever $V\neq 0$. Using $|\mathop{\rm
Rm}|^{2}=\frac{1}{6}s^{2}+2|\mathop{\rm
R\text{\i}\makebox[0.0pt]{\raisebox{5.0pt}{\tiny$\circ$\;\,}}c}|^{2}+|W^{+}|^{2}+|W^{-}|^{2}$
and $|\mathop{\rm
Rm}{}^{++}|=\big{|}\frac{1}{12}s\,Id_{\bigwedge^{+}}+W^{+}\big{|}^{2}=\frac{1}{12}s^{2}+|W^{+}|^{2}$,
we obtain the useful combination

(124)
$\displaystyle-8\pi^{2}(3\mathcal{P}_{\tau}+2\mathcal{P}_{\chi})\;=\;\big{(}|\mathop{\rm
R\text{\i}\makebox[0.0pt]{\raisebox{5.0pt}{\tiny$\circ$\;\,}}c}|^{2}\,-\,|\mathop{\rm
Rm}{}^{++}|^{2}\big{)}\,dVol.$

Other combinations of $\mathcal{P}_{\chi}$ and $\mathcal{P}_{\tau}$ are
possible, but this combination is particularly useful because of the
simplicity of its transgression:

(125)
$\displaystyle-8\pi^{2}(3\mathcal{TP}_{\tau}+2\mathcal{TP}_{\chi}),=2\bar{\theta}^{i}_{j}\wedge\left(2(F^{+})^{j}_{i}-(D\bar{\theta}^{+})^{j}_{i}\right)$

where the $D\bar{\theta}^{+}$ term is easy to work with due to (117). We
abbreviate

(126) $\displaystyle\mathcal{P}_{\mathop{\rm Ric}}$
$\displaystyle\;\triangleq\;-8\pi^{2}(3\mathcal{P}_{\tau}+2\mathcal{P}_{\chi}),\quad\text{which
is transgressed by}$ $\displaystyle\mathcal{TP}_{\mathop{\rm Ric}}$
$\displaystyle\;\triangleq\;2\bar{\theta}^{i}_{j}\wedge\left(2(F^{+})^{j}_{i}-(D\bar{\theta}^{+})^{j}_{i}\right)$

so $\mathcal{P}_{\mathop{\rm Ric}}=d\mathcal{TP}_{\mathop{\rm Ric}}$ when
$V\neq 0$. Lastly we consider the Kähler case. In that case $|\mathop{\rm
Rm}{}^{++}|^{2}=\frac{1}{4}s^{2}$ and $|\mathop{\rm
R\text{\i}\makebox[0.0pt]{\raisebox{5.0pt}{\tiny$\circ$\;\,}}c}|^{2}=|\mathop{\rm
Ric}|^{2}-\frac{1}{4}s^{2}$ and we have the key expression

(127) $\displaystyle\mathcal{P}_{\mathop{\rm Ric}}$
$\displaystyle=\left(|\mathop{\rm
Ric}|^{2}\,-\,\frac{1}{2}s^{2}\right)\,dVol_{4}=d\left[2\bar{\theta}^{i}_{j}\wedge\left(2(F^{+})^{j}_{i}-(D\bar{\theta}^{+})^{j}_{i}\right)\right]=d\mathcal{TP}_{\mathop{\rm
Ric}}.$

######  Proposition 5.2 (cf. Theorem 1.7).

Let $M^{4}$ be a compact, scalar-flat Kähler manifold with a Killing field
$V$. If $V$ is nowhere zero, then $M^{4}$ is flat.

###### Proof.

The scalar-flat Kähler condition implies $s$, $W^{+}$, and $\mathop{\rm
Rm}{}^{++}$ are all zero. The fact that $V\neq 0$ implies the transgression on
the right-hand side (127) is smooth. Integration by parts gives

(128) $\int_{M^{4}}|\mathop{\rm
Ric}|^{2}\,dVol_{4}\;=\;\int_{M^{4}}d\left[2\bar{\theta}^{i}_{j}\wedge\left(2(F^{+})^{j}_{i}-(D\bar{\theta}^{+})^{j}_{i}\right)\right]\;=\;0$

so that $\mathop{\rm Ric}\equiv 0$. Similarly,
$\mathcal{P}_{\tau}=-\frac{1}{96\pi^{2}}|W^{-}|^{2}$ is smoothly transgressed
by (123), so integrating gives $W^{-}\equiv 0$. We conclude that $\mathop{\rm
Rm}\equiv 0$. ∎

###  5.3. Simplifying the Transgression

We simplify $\mathcal{TP}_{\mathop{\rm Ric}}$ in three steps, at the end
arriving at a very usable expression. The aim is to record Eq. (154).

Given a Kähler metric, recall the Ricci tensor is $J$-invariant and
$\rho=\mathop{\rm Ric}(J\cdot,\cdot)$ is a $(1,1)$-form called the Ricci form.
We use $\rho_{0}=\mathop{\rm
R\text{\i}\makebox[0.0pt]{\raisebox{5.0pt}{\tiny$\circ$\;\,}}c}(J\cdot,\cdot)$
for the primitive Ricci form, and note that
$\rho_{0}=\rho-\frac{1}{4}s\omega$. For convenience we set
$\eta=\frac{1}{|V|}V_{\flat}$.

######  Lemma 5.3 (Step 1).

If the metric is Kähler and $V$ is Hamiltonian then

(129) $\displaystyle\mathcal{TP}_{\mathop{\rm Ric}}$
$\displaystyle\;=\;2\left(\frac{1}{|V|}\triangle_{4}z\right)\,\eta\wedge\rho+\frac{1}{2}\left(\frac{1}{|V|}\triangle_{4}z\right)^{2}\eta\wedge{}d\eta.$

If $V$ is not Hamiltonian, (129) remains true after replacing $\triangle_{4}z$
by $*\left[(dV_{\flat})^{+}\wedge\omega\right]$.

###### Proof.

In the Hamiltonian case, (117) gives
$(dV_{\flat})^{+}=\frac{1}{2}(\triangle_{4}z)\omega$, which gives
$\triangle_{4}z=*\left[(dV_{\flat})^{+}\wedge\omega\right]$. In the non-
Hamiltonian case, we still have that $(dV_{\flat})^{+}$ is proportional to
$\omega$ and that
$(dV_{\flat})^{+}=\frac{1}{2}*\left(dV_{\flat}\wedge\omega\right)\cdot\omega$.
In everything that follows we begin our computations with $dV_{\flat}$ or
$\nabla{}V$, and we can simply take $\triangle_{4}z$ to be shorthand for the
quantity $*\left(dV_{\flat}\wedge\omega\right)$.

We first evaluate $Tr(\bar{\theta}\wedge{}F^{+})$. Using (117) we have

(130) $\displaystyle Tr(\bar{\theta}\wedge{}F^{+})$
$\displaystyle\;=\;\frac{1}{|V|}\eta\wedge(\nabla{}V)^{s}_{t}(F^{+})_{s}^{t}$
$\displaystyle\;=\;\frac{1}{|V|}\eta\wedge(\nabla{}V^{+})^{s}_{t}(F^{+})_{s}^{t}\;=\;-\frac{1}{4}(\triangle_{4}z)\frac{1}{|V|}\eta\wedge(F^{+})^{t}_{s}J_{t}{}^{s}.$

We have $F^{+}=-\frac{1}{2}(\mathop{\rm Rm}^{++}+\mathop{\rm Rm}^{-+})$.
Because $(\mathop{\rm Rm}^{+-})^{j}_{i}(\nabla{}V^{+})^{i}_{j}=0$ we can write

(131) $\displaystyle(F^{+})^{t}_{s}J_{t}{}^{s}$
$\displaystyle\;=\;-\frac{1}{2}\left(\mathop{\rm Rm}{}^{++}+\mathop{\rm
Rm}{}^{-+}\right)_{ijs}{}^{t}J_{t}{}^{s}\,\cdot\,\eta^{i}\wedge\eta^{j}$
$\displaystyle\;=\;-\frac{1}{2}\left(\mathop{\rm Rm}{}^{++}+\mathop{\rm
Rm}{}^{-+}+\mathop{\rm
Rm}{}^{+-}\right)_{ijst}\omega^{ts}\,\cdot\,\eta^{i}\wedge\eta^{j}$

By (115) we have $\mathop{\rm Rm}^{++}=-\frac{s}{8}\omega\otimes\omega$ and
using $\omega_{st}\omega^{st}=4$ we have

(132) $-\frac{1}{2}(\mathop{\rm
Rm}{}^{++})_{ijst}\omega^{ts}=-\frac{s}{4}\omega_{ij}$

By (114) we have $(\mathop{\rm Rm}{}^{-+}+\mathop{\rm
Rm}{}^{+-})=\frac{1}{2}\mathop{\rm
R\text{\i}\makebox[0.0pt]{\raisebox{5.0pt}{\tiny$\circ$\;\,}}c}\,\mathop{\makebox{$\wedge$}\bigcirc}{}g$,
so

(133) $\displaystyle-\frac{1}{2}(\mathop{\rm Rm}{}^{+-}+\mathop{\rm
Rm}{}^{-+})_{ijst}\omega^{ts}\cdot\eta^{i}\wedge\eta^{j}$ $\displaystyle\hskip
28.90755pt=-\frac{1}{4}\left(\mathop{\rm
R\text{\i}\makebox[0.0pt]{\raisebox{5.0pt}{\tiny$\circ$\;\,}}c}{}_{it}g_{js}+\mathop{\rm
R\text{\i}\makebox[0.0pt]{\raisebox{5.0pt}{\tiny$\circ$\;\,}}c}{}_{js}g_{it}-\mathop{\rm
R\text{\i}\makebox[0.0pt]{\raisebox{5.0pt}{\tiny$\circ$\;\,}}c}{}_{is}g_{jt}-\mathop{\rm
R\text{\i}\makebox[0.0pt]{\raisebox{5.0pt}{\tiny$\circ$\;\,}}c}{}_{jt}g_{is}\right)\omega^{ts}\cdot\eta^{i}\wedge\eta^{j}$
$\displaystyle\hskip
28.90755pt=-(\rho_{0})_{ij}\cdot\eta^{i}\wedge\eta^{j}\;=\;-2\rho_{0}.$

Combining these, we have
$(F^{+})_{s}^{t}J_{t}{}^{s}=-2\rho_{0}-\frac{s}{2}\omega=-2\rho$. Therefore

(134)
$Tr(\bar{\theta}\wedge{}F^{+})\;=\;\frac{1}{2}\left(\frac{1}{|V|}\triangle_{4}z\right)\,\eta\wedge\rho.$

Next we compute $Tr(\bar{\theta}\wedge{}D\bar{\theta}^{+})$. Using the Leibniz
rule,

(135) $\displaystyle D\bar{\theta}^{+}$
$\displaystyle\;=\;D\left(\eta\otimes\frac{1}{|V|}DV^{+}\right)\;=\;d\eta\otimes\frac{1}{|V|}\nabla{}V^{+}\,-\,\eta\wedge{}D\left(\frac{1}{|V|}\nabla{}V^{+}\right).$

For the last term, we use
$|V|^{-1}\nabla{}V^{+}=-\frac{1}{4}|V|^{-1}(\triangle_{4}z)J$ and the fact
that $J$ is covariant-constant to obtain

(136) $\displaystyle D\bar{\theta}^{+}$
$\displaystyle\;=\;d\eta\otimes\frac{1}{|V|}\nabla{}V^{+}\,+\,\frac{1}{4}\eta\wedge{}d\left(\frac{1}{|V|}\triangle_{4}z\right)\otimes{}J.$

Therefore

(137) $\displaystyle Tr(\bar{\theta}\wedge{}D\bar{\theta}^{+})$
$\displaystyle\;=\;\eta\wedge{}d\eta\,\cdot\,|V|^{-2}Tr\left((\nabla{}V)(\nabla{}V^{+})\right)$
$\displaystyle\quad\hskip
28.90755pt+\eta\wedge\eta\wedge{}d\left(\frac{1}{|V|}\triangle_{4}z\right)\,\cdot\,Tr\left(\frac{1}{|V|}(\nabla{}V)(J)\right)$
$\displaystyle\;=\;\eta\wedge{}d\eta\,\cdot\,|V|^{-2}Tr\left((\nabla{}V)(\nabla{}V^{+})\right)$

because $\eta\wedge\eta=0$. Then using (117) and
$Tr(JJ)=J_{i}{}^{j}J_{j}{}^{i}=-|J|^{2}=-4$, we have

(138)
$Tr\left((\nabla{}V)(\nabla{}V^{+})\right)\;=\;Tr\left((\nabla{}V^{+})(\nabla{}V^{+})\right)\;=\;-|\nabla{}V^{+}|^{2}\;=\;-\frac{1}{4}\left(\triangle_{4}z\right)^{2}$

and we conclude that

(139)
$Tr(\bar{\theta}\wedge{}D\bar{\theta})\;=\;-\frac{1}{4}\left(\frac{1}{|V|}\triangle_{4}z\right)\,\eta\wedge{}d\eta.$

Combining the two pieces, we obtain as promised

(140) $\displaystyle TP_{\mathop{\rm Ric}}$
$\displaystyle\;=\;4\bar{\theta}^{i}_{j}\wedge(F^{+})^{j}_{i}-2\bar{\theta}^{i}_{j}\wedge(D\bar{\theta}^{+})^{j}_{i}$
$\displaystyle\;=\;2\left(\frac{1}{|V|}\triangle_{4}z\right)\,\eta\wedge\rho+\frac{1}{2}\left(\frac{1}{|V|}\triangle_{4}z\right)^{2}\eta\wedge{}d\eta.$

∎

######  Lemma 5.4 (Step 2).

If the metric is Kähler and $V$ is Hamiltonian we have

(141) $\displaystyle\mathcal{TP}_{\mathop{\rm Ric}}$
$\displaystyle\;=\;\frac{1}{2}*d\left(\frac{1}{|V|}\triangle_{4}z\right)^{2}+\left(\frac{1}{|V|}\triangle_{4}z\right)\left[s+\frac{1}{2}\left(\frac{1}{|V|}\triangle_{4}z\right)^{2}\right]\,dVol_{3}.$

If $V$ is not Hamiltonian, (141) remains true after replacing $\triangle_{4}z$
with $*\left[(dV_{\flat})^{+}\wedge\omega\right]$.

###### Proof.

We evaluate $\eta\wedge\rho$ using the Bochner identity
$d(\triangle_{4}z)+2\mathop{\rm Ric}(\nabla{}z,\cdot)=0$ of Lemma 5.1. This
with $*\rho=\rho^{+}-\rho^{-}=-\rho+2\rho^{+}$ and
$\rho^{+}=\frac{s}{4}\omega$ gives

(142) $\rho\;=\;-*\rho\,+\,\frac{s}{2}\omega.$

We have the usual identity $*(\mu\wedge*\gamma)=\gamma(\mu^{\sharp},\cdot)$
when $\mu$ is a $1$-form and $\gamma$ is a 2-form. Using
$\eta=|V|^{-1}V_{\flat}$ and the fact that
$**:\bigwedge{}^{3}\rightarrow\bigwedge^{3}$ is multiplication by $-1$ gives

(143) $\displaystyle\eta\wedge\rho$
$\displaystyle\;=\;-\eta\wedge*\rho\,+\,\frac{s}{2}\,\eta\wedge\omega\;=\;\frac{1}{|V|}*\left(i_{V}\rho\,-\,\frac{s}{2}\,i_{V}\omega\right).$

Then $\rho(\cdot,\cdot)=\mathop{\rm Ric}(J\cdot,\cdot)$ along with (14) and
Lemma 5.1 gives

(144) $\displaystyle\eta\wedge\rho$
$\displaystyle\;=\;\frac{1}{|V|}*\left(-\mathop{\rm
Ric}(\nabla{}z)\,+\,\frac{s}{2}\,dz\right)\;=\;\frac{1}{2}\frac{1}{|V|}*\Big{(}d(\triangle_{4}{}z)\,+\,s\,dz\Big{)}.$

To compute $\eta\wedge{}d\eta$, note
$\eta\wedge{}d\eta=|V|^{-2}V_{\flat}\wedge{}dV_{\flat}$. Then Eq. (44) of
Lemma 3.2 is

(145)
$\eta\wedge{}d\eta\;=\;-\frac{1}{|V|^{2}}*d|V|^{2}+\frac{1}{|V|^{2}}(\triangle_{4}z)*dz$

Placing these expressions for $\eta\wedge\rho$ and $\eta\wedge{}d\eta$ into
(140), we have

(146) $\displaystyle 2\mathcal{TP}_{\mathop{\rm Ric}}$
$\displaystyle\;=\;4\left(\frac{1}{|V|}\triangle_{4}z\right)\left(\frac{1}{2}\frac{1}{|V|}*\Big{(}d(\triangle_{4}{}z)\,+\,s\,dz\Big{)}\right)$
$\displaystyle\quad\quad+\left(\frac{1}{|V|}\triangle_{4}z\right)^{2}\left(-\frac{1}{|V|^{2}}*d|V|^{2}+\frac{1}{|V|^{2}}(\triangle_{4}z)*dz\right).$

Collecting terms,

(147) $\displaystyle 2\mathcal{TP}_{\mathop{\rm Ric}}$
$\displaystyle\;=\;2\left(\frac{1}{|V|}\triangle_{4}z\right)*\left[\frac{1}{|V|}d(\triangle_{4}z)-(\triangle_{4}{}z)\frac{1}{|V|^{2}}d|V|\right]$
$\displaystyle\quad\quad+\left(\frac{1}{|V|}\triangle_{4}z\right)\left[2s+\left(\frac{1}{|V|}\triangle_{4}z\right)^{2}\right]\frac{1}{|V|}\,*dz$
$\displaystyle\;=\;*d\left(\frac{1}{|V|}\triangle_{4}z\right)^{2}+\left(\frac{1}{|V|}\triangle_{4}z\right)\left[2s+\left(\frac{1}{|V|}\triangle_{4}z\right)^{2}\right]\,dVol_{3}.$

∎

######  Lemma 5.5 (Step 3).

If the metric is Kähler and $V$ is Hamiltonian we have

(148) $\displaystyle\mathcal{TP}_{\mathop{\rm Ric}}$
$\displaystyle\;=\;\frac{1}{2}\mathcal{L}_{\frac{\partial}{\partial{}z}}\left((\triangle_{4}z)^{2}\frac{1}{|V|}\,dVol_{3}\right)\,-\,\frac{1}{2}(*dVol_{2})\wedge{}Jd\left(\frac{1}{|V|}\triangle_{4}z\right)^{2}$
$\displaystyle\hskip
57.81621pt+s\left(\frac{1}{|V|}\triangle_{4}z\right)\,dVol_{3}.$

If $V$ is not Hamiltonian, replace $\frac{\partial}{\partial{}z}$ with
$-|V|^{-2}JV_{\flat}$ and $\triangle_{4}z$ with $*(dV^{+}\wedge\omega)$.

###### Proof.

We begin with

(149)
$\displaystyle*d\left(\frac{\triangle_{4}z}{|V|}\right)^{2}=\frac{\partial}{\partial{}z}\left(\frac{\triangle_{4}z}{|V|}\right)^{2}*dz+\frac{\partial}{\partial{}x}\left(\frac{\triangle_{4}z}{|V|}\right)^{2}*dx+\frac{\partial}{\partial{}y}\left(\frac{\triangle_{4}z}{|V|}\right)^{2}*dy.$

The last two terms on the right simplify: whenever $f=f(z,x,y)$ then

(150) $f_{x}*dx-f_{y}*dy=-(*dVol_{2})\wedge{}Jdf$

The first term also simplifies using $*dz=V_{\flat}\wedge{}dVol_{2}$. Then
(149) becomes

(151)
$*d\left(\frac{\triangle_{4}z}{|V|}\right)^{2}=\frac{\partial}{\partial{}z}\left(\frac{\triangle_{4}z}{|V|}\right)^{2}\,V_{\flat}\wedge{}dVol_{2}-(*dVol_{2})\wedge{}Jd\left(\frac{\triangle_{4}z}{|V|}\right)^{2}.$

Use the Leibniz rule on the first term and Lemma 3.5, we obtain

(152)
$\displaystyle\frac{\partial}{\partial{}z}\left(\frac{\triangle_{4}z}{|V|}\right)^{2}\;V_{\flat}\wedge{}dVol_{2}$
$\displaystyle\quad\quad\;=\;\mathcal{L}_{\frac{\partial}{\partial{}z}}\left(\left(\frac{1}{|V|}\triangle_{4}z\right)^{2}V_{\flat}\wedge{}dVol_{2}\right)-\left(\frac{1}{|V|}\triangle_{4}z\right)^{2}\mathcal{L}_{\frac{\partial}{\partial{}z}}\left(V_{\flat}\wedge{}dVol_{2}\right)$
$\displaystyle\quad\quad\;=\;\mathcal{L}_{\frac{\partial}{\partial{}z}}\left(\frac{1}{|V|}(\triangle_{4}z)^{2}\,dVol_{3}\right)-\left(\frac{1}{|V|}\triangle_{4}z\right)^{3}\,dVol_{3}.$

Combining, we obtain

(153)
$\displaystyle*d\left(\frac{1}{|V|}\triangle_{4}z\right)^{2}+\left(\frac{1}{|V|}\triangle_{4}z\right)^{3}dVol_{3}$
$\displaystyle\hskip
14.45377pt=\mathcal{L}_{\frac{\partial}{\partial{}z}}\left(\frac{1}{|V|}(\triangle_{4}z)^{2}\,dVol_{3}\right)-*dVol_{2}\wedge{}Jd\left(\frac{1}{|V|}\triangle_{4}z\right)^{2}.$

Substituting this into (141) gives the result. ∎

######  Proposition 5.6 (cf. Theorem 1.2).

Assume $M^{4}$ is scalar-flat and assume the leaf $M^{3}_{z}$ is compact and
non-singular. Then

(154)
$\frac{d^{2}}{dz^{2}}\int_{M^{2}_{z}}(\triangle_{4}{}z)^{2}\,dVol_{2}\;=\;2\int_{M^{2}_{z}}|\mathop{\rm
Ric}|^{2}\,dVol_{2}$

###### Proof.

With $s=0$, inserting (148) into (127) gives

(155) $\small 2|\mathop{\rm
Ric}|^{2}\,dVol_{4}=d\left[\mathcal{L}_{\frac{\partial}{\partial{}z}}\left((\triangle_{4}z)^{2}\frac{1}{|V|}\,dVol_{3}\right)-\frac{dz}{|V|}\wedge\frac{V_{\flat}}{|V|}\wedge{}Jd\left(\frac{1}{|V|}\triangle_{4}z\right)^{2}\right].$

Choose $z_{0}<z_{1}$, and denote
$M^{4}_{z_{0},z_{1}}=\\{p\in{}M^{4}\big{|}z_{0}<z(p)<z_{1}\\}$. Integrating,

(156) $\displaystyle\int_{M^{4}_{z_{0},z_{1}}}|\mathop{\rm
Ric}|^{2}\,dVol_{4}\;=\;\frac{1}{2}\int_{M^{3}_{z_{1}}}\mathcal{L}_{\frac{\partial}{\partial{}z}}\left((\triangle_{4}z)^{2}\frac{1}{|V|}\,dVol_{3}\right)$
$\displaystyle\hskip
122.85876pt-\frac{1}{2}\int_{M^{3}_{z_{0}}}\mathcal{L}_{\frac{\partial}{\partial{}z}}\left((\triangle_{4}z)^{2}\frac{1}{|V|}\,dVol_{3}\right)$
$\displaystyle\hskip
43.36243pt\;=\;\frac{1}{2}\frac{\partial}{\partial{}z}\Big{|}_{z=z_{1}}\int_{M^{3}_{z}}(\triangle_{4}z)^{2}\frac{1}{|V|}\,dVol_{3}-\frac{1}{2}\frac{\partial}{\partial{}z}\Big{|}_{z=z_{0}}\int_{M^{3}_{z}}(\triangle_{4}z)^{2}\frac{1}{|V|}\,dVol_{3}$

where the term
$|V|^{-2}dz\wedge{}V_{\flat}\wedge{}Jd(|V|^{-1}\triangle_{4}z)^{2}$ vanished
because $dz\equiv 0$ restricted to $M^{3}_{z}=\\{z=const\\}$. The Integration
Lemma, Lemma 2.4, gives

(157)
$\int_{M^{3}_{z}}(\triangle_{4}z)^{2}\frac{1}{|V|}\,dVol_{3}\;=\;2\pi\int_{M^{2}_{z}}(\triangle_{4}z)^{2}\,dVol_{2}$

Then we consider the left-hand side. From Lemma 2.4

(158) $\int_{M^{4}_{z_{0},z_{1}}}|\mathop{\rm
Ric}|^{2}\,dVol_{4}\;=\;2\pi\int_{z_{0}}^{z_{1}}\left(\int_{M^{2}_{z}}|\mathop{\rm
Ric}|^{2}\,dVol_{2}\right)dz$

From this, we multiply both sides of (156) by $\frac{1}{z_{1}-z_{0}}$ and take
the limit $z_{1}-z_{0}\rightarrow 0$. The Fundamental Theorem of Calculus then
gives

(159) $\displaystyle 2\int_{M^{2}_{z}}|\mathop{\rm Ric}|^{2}\,dVol_{2}$
$\displaystyle\;=\;\frac{d^{2}}{dz^{2}}\int_{M^{2}_{z}}(\triangle_{4}z)^{2}\,dVol_{2}$

∎

###  5.4. The LeBrun instanton examples.

We produce nontrivial examples demonstrating Lemma 5.4 and Proposition 5.6. To
be able to easily compute Ricci curvatures, we use the framework of [35]. In
that framework, $C=C(w)$ and $F=F(w)$ and the metric is

(160) $\displaystyle
g\;=\;C\left(\frac{1}{2F}dw^{2}\,+\,F(\eta^{1})^{2}+(\eta^{2})^{2}+(\eta^{3})^{2}\right),$
$\displaystyle J(dw)\;=\;-2F\eta^{1},\quad J(\eta^{2})\;=\;-\eta^{3},$
$\displaystyle\omega\;=\;\frac{1}{2}Cdw\wedge\eta^{1}+C\eta^{2}\wedge\eta^{3}$

where $\eta^{1}$, $\eta^{2}$, $\eta^{3}$ form the left-invariant coframe on
$\mathbb{S}^{2}$ with $d\eta^{i}=-\epsilon^{i}{}_{jk}\eta^{j}\wedge\eta^{k}$.
This framework is very flexible: when $C=C_{0}e^{-w}$, the metric is Kähler
for any $F$.

From [29], the LeBrun instanton metric of mass $m$ on the surface $O(-k)$ is

(161) $\displaystyle
g=\frac{1}{f(r)}(dr)^{2}+r^{2}\left(f(r)(\eta^{1})^{2}+(\eta^{2})^{2}+(\eta^{3})^{2}\right),\text{
where}$ $\displaystyle
f(r)=\left(1-\frac{m^{2}}{r^{2}}\right)\left(1+\frac{(k-1)m^{2}}{r^{2}}\right)$

Changing coordinates by $r=e^{-w/2}$ to express this in the form (160), we
find $C=e^{-w}$ and $F=1+m^{2}(k-2)e^{w}-m^{4}(k-1)e^{2w}$ with coordinate
range $w\in(-\infty,-\log\,m^{2}]$. For example $k=1$ is the Burns metric and
$k=2$ is the Eguchi-Hanson metric. For any $w<-\log(m^{2})$ the level-set
$M^{3}_{w}$ is the lens space $L(k,1)$ and the symplectic reduction
$M^{2}_{w}$ is a smooth 2-sphere. For $w=const$ we have
$Vol_{3}(M^{3}_{w})=\frac{2}{k}\pi^{2}e^{-w}$ and
$Vol_{2}(M^{2}_{w})=\pi{}e^{-w}.$ From (13) of [35]

(162) $|\mathop{\rm Ric}|^{2}\;=\;16m^{4}(k-2)^{2}e^{4w}.$

Letting $\psi,\theta,\varphi$ be the Euler coordinates on $\mathbb{S}^{3}$,
the Killing field is $V=\frac{2}{k}\frac{\partial}{\partial\psi}$ (see §6 of
[53]) and using the fact that $i_{V}\eta^{1}=\frac{1}{k}$ we obtain the
momentum function $z=-\frac{1}{2k}(e^{-w}-m^{2})$ with range
$z\in(-\infty,0]$. We compute for general $F=F(w)$

(163) $\displaystyle
dJdz=-\frac{1}{k}\frac{\partial}{\partial{}w}(e^{-w}F)dw\wedge\eta^{1}+\frac{2}{k}{}e^{-w}F\eta^{1}\wedge\eta^{2},$
$\displaystyle
dJdz\wedge\omega=-\frac{1}{k}e^{w}\left(\frac{\partial}{\partial{}w}(e^{-w}F)-e^{-w}F\right)\omega\wedge\omega$

so therefore $\triangle_{4}z=-\frac{4}{k}-\frac{2(k-2)}{k}m^{2}e^{w}$.
Changing variables from $w$ to $z$, we find

(164) $\displaystyle\triangle_{4}z\;=\;-2\frac{-4z+m^{2}}{-2kz+m^{2}},$
$\displaystyle Vol_{2}(M^{2}_{z})\;=\;\pi(-2kz+m^{2}),$
$\displaystyle\int_{M^{2}_{z}}\triangle_{4}z\,dVol_{2}\;=\;2\pi(4z-m^{2}),$
$\displaystyle\int_{M^{2}_{z}}(\triangle_{4}z)^{2}\,dVol_{2}\;=\;4\pi\frac{(-4z+m^{2})^{2}}{-2kz+m^{2}},$
$\displaystyle\int_{M^{2}_{z}}|\mathop{\rm
Ric}|^{2}\,dVol_{2}\;=\;16\pi{}m^{4}(k-2)^{2}\frac{1}{(-2kz+m^{2})^{3}}.$

We have $M^{2}_{z}\approx\mathbb{S}^{2}$ for each $z$ so
$\chi^{g}(M^{2}_{z})=2$. Each $M^{3}_{z}$ is the lense space $L(k,1)$ and
$M^{3}_{z}\rightarrow{}M^{2}_{z}$ has the usual fiber bundle structure (the
standard projection along a $k$-to-1 quotient of the Hopf fibration), so the
Euler number of this $\mathbb{S}^{1}$ fiber bundle is
$e^{g}(M^{3}_{z},M^{2}_{z})=-k$. Direct computation from (164) gives

(165)
$\displaystyle\frac{d}{dz}Vol_{2}(M^{2}_{z})\;=\;-2k\pi\;=\;2\pi{}e^{g}(M^{3}_{z},M^{2}_{z})$
$\displaystyle\frac{d}{dz}\int_{M^{2}_{z}}\triangle_{4}z\,dVol_{2}\;=\;8\pi{}\;=\;4\pi\chi^{g}(M^{2}_{z})$

which recovers (86) of §4 and (62) from §3. Further, we have

(166)
$\displaystyle\frac{d}{dz}\int_{M^{2}_{z}}(\triangle_{4}z)^{2}\,dVol_{2}\;=\;8\pi\frac{(-4z+m^{2})(4kz-(k-4)m^{2})}{(-2kz+m^{2})^{2}}$
$\displaystyle\frac{d^{2}}{dz^{2}}\int_{M^{2}_{z}}(\triangle_{4}z)^{2}\,dVol_{2}\;=\;32\pi{}m^{4}(k-2)^{2}\frac{1}{(-2kz+m^{2})^{3}}\;=\;2\int_{M^{2}_{z}}|\mathop{\rm
Ric}|^{2}\,dVol_{2}$

which recovers the Chern-Gauss-Bonnet relation Eq. (154) of Proposition 5.6.

##  6\. Type II metrics

We prove Theorem 1.5, Proposition 1.6, and Theorem 1.8.

###  6.1. The basic inequalities

Recall that Type IIa metrics have compact level-sets and have a lower bound
$|V|>\epsilon>0$ outside a compact set. Theorems 1.5 and 1.6 both follow from
the basic relationship between $\int_{M^{2}_{z}}dVol_{2}$,
$\int_{M^{2}_{z}}(\triangle_{4}z)dVol_{2}$, and
$\int_{M^{2}_{z}}(\triangle_{4}z)^{2}dVol_{2}$ from Hölder’s inequality that

(167)
$\left(\int_{M^{2}_{z}}\triangle_{4}z\,dVol_{2}\right)^{2}\leq\int_{M^{2}_{z}}dVol_{2}\;\int_{M^{2}_{z}}(\triangle_{4}z)^{2}dVol_{2}.$

The quantities in (167) are controlled by other factors:
$\int_{M^{2}_{z}}dVol_{2}$ is controlled by $e^{g}(M^{3}_{z},M^{2}_{z})$ by
Proposition 3.7, $\int_{M^{2}_{z}}\triangle_{4}z\,dVol_{2}$ is controlled by
$\chi^{g}(M^{2}_{z})$ by Proposition 4.3, and
$\int_{M^{2}_{z}}(\triangle_{4}z)^{2}dVol_{2}$ is controlled by
$\int_{M^{4}}|\mathop{\rm Ric}|^{2}dVol_{4}$ by Proposition 5.6. It will be
convenient to abbreviate $\chi^{g}=\chi^{g}(M^{2}_{z})$ and
$e^{g}=e^{g}(M^{3}_{z},M^{2}_{z})$.

######  Lemma 6.1 (The Basic Inequalities).

Assume $M^{4}$ is Type IIa, $\mathcal{C}=\\{z_{0},\dots,z_{N}\\}$ are the
critical values of $z$, and either $M^{4+}$ or $M^{4-}$ is non-empty.

If $M^{4+}$ and/or $M^{4-}$ is non-empty, we have

(168) $\displaystyle\chi^{g}\geq 0,\;e^{g}\geq 0\quad\text{on $M^{4+}$},$
$\displaystyle\chi^{g}\geq 0,\;e^{g}\leq 0\quad\text{on $M^{4-}$}.$

Assuming the set of critical values $\mathcal{C}$ is non-empty, set

(169) $\displaystyle
C_{0}=\lim_{z\searrow{}z_{N}}\int_{M^{2}_{z}}|V|^{2}\,dVol_{2},\quad
C_{0}^{\prime}=\lim_{z\nearrow{}z_{0}}\int_{M^{2}_{z}}|V|^{2}\,dVol_{2},$
$\displaystyle
C_{1}=\lim_{z\searrow{}z_{N}}\int_{M^{2}_{z}}\triangle_{4}z\,dVol_{2},\quad
C_{1}^{\prime}=\lim_{z\nearrow{}z_{0}}\int_{M^{2}_{z}}\triangle_{4}z\,dVol_{2},$
$\displaystyle C_{2}=\lim_{z\searrow{}z_{N}}\int_{M^{2}_{z}}dVol_{2},\hskip
30.71497ptC_{2}^{\prime}=\lim_{z\nearrow{}z_{0}}\int_{M^{2}_{z}}dVol_{2}.$

Then we have

(170) $\displaystyle\int_{M^{2}_{z}}|V|^{2}dVol_{2}$
$\displaystyle\;\leq\;\left(2\pi\chi^{g}\cdot{}(z-z_{N})^{2}+C_{1}\cdot(z-z_{N})+C_{0}\right)\quad\text{on
$M^{4+}$}$ $\displaystyle\int_{M^{2}_{z}}|V|^{2}\,dVol_{2}$
$\displaystyle\;\leq\;\left(2\pi\chi^{g}\cdot{}(z-z_{0})^{2}+C_{1}^{\prime}\cdot(z-z_{0})+C_{0}^{\prime}\right)\quad\text{on
$M^{4-}$}$

and

(171)
$\displaystyle|4\pi\chi^{g}\cdot(z-z_{N})+C_{1}|\leq\sqrt{\big{(}2\pi{}e^{g}\cdot(z-z_{N})+C_{2}\big{)}\int_{M^{2}_{z}}(\triangle_{4}z)^{2}\,dVol_{2}}\quad\text{on
$M^{4+}$}$
$\displaystyle|4\pi\chi^{g}\cdot(z-z_{0})+C_{1}^{\prime}|\leq\sqrt{\big{(}2\pi{}e^{g}\cdot(z-z_{0})+C_{2}^{\prime}\big{)}\int_{M^{2}_{z}}(\triangle_{4}z)^{2}\,dVol_{2}}\quad\text{on
$M^{4-}$}.$

If $|z|$ is sufficiently large then $|V|>\epsilon$; for these $z$ we have

(172) $\displaystyle 2\pi{}e^{g}\cdot(z-z_{N})+C_{1}$
$\displaystyle\;\leq\;\epsilon^{-2}\int_{M^{2}_{z}}|V|^{2}\,dVol_{2}\quad\text{on
$M^{4+}$}$ $\displaystyle 2\pi{}e^{g}\cdot(z-z_{0})+C_{1}^{\prime}$
$\displaystyle\;\leq\;\epsilon^{-2}\int_{M^{2}_{z}}|V|^{2}\,dVol_{2}\quad\text{on
$M^{4-}$}.$

###### Proof.

Proposition 3.7 gives

(173) $\displaystyle
Vol(M^{2}_{z})\;=\;2\pi{}e^{g}\cdot(z-z_{N})+C_{0}{}\quad\text{on $M^{4+}$},$
$\displaystyle
Vol(M^{2}_{z})\;=\;2\pi{}e^{g}\cdot(z-z_{0})+C_{0}^{\prime}{}\quad\text{on
$M^{4-}$}.$

Because $Vol(M^{2}_{z})>0$, this forces $e^{g}\geq 0$ on $M^{4+}$ and
$e_{g}\leq 0$ on $M^{4-}$. Proposition 4.3 gives on $M^{4+}$

(174)
$\displaystyle\int_{M^{2}_{z}}|V|^{2}=2\pi\chi^{g}\cdot(z-z_{N})^{2}+C_{1}\cdot(z-z_{N})+C_{2}\quad\text{and}$
$\displaystyle\hskip
28.90755pt\;\int_{M^{2}_{z}}\triangle_{4}z\,dVol_{2}=4\pi\chi^{g}\cdot(z-z_{N})+C_{1},$

and on $M^{4-}$

(175)
$\displaystyle\int_{M^{2}_{z}}|V|^{2}=2\pi\chi^{g}\cdot(z-z_{0})^{2}+C_{1}^{\prime}\cdot(z-z_{0})+C_{2}^{\prime}\quad\text{and}$
$\displaystyle\hskip
28.90755pt\int_{M^{2}_{z}}\triangle_{4}z\,dVol_{2}=4\pi\chi^{g}\cdot(z-z_{0})+C_{1}^{\prime}.$

Because $\int_{M^{2}_{z}}|V|^{2}dVol_{2}\geq 0$, this forces $\chi^{g}\geq 0$
on both $M^{4+}$ and $M^{4-}$. This proves (168) and (170). Also (171) follows
from Hölder’s inequality (167).

Because the metric is Type II, the set $\\{|V|<\epsilon\\}$ has compact
closure, and by Lemma 2.8 the sets
$M^{4}_{z_{0},z_{1}}=\\{a\in{}M^{4}\;|\;z_{0}<z(a)<z_{1}\\}$ are compact, but
exhaust $M^{4}$ as $z_{1}\rightarrow{}\infty$ and $z_{0}\rightarrow-\infty$.
Therefore sufficiently small $z_{0}$ and sufficiently large $z_{1}$ exist so
that $|V|>\epsilon$ on $M^{4}\setminus{}M^{4}_{z_{0},z_{1}}$. Therefore when
$|z|>\text{max}\\{|z_{0}|,|z_{1}|\\}$

(176)
$Vol(M^{2}_{z})=\int_{M^{2}_{z}}|V|^{-2}|V|^{2}\,dVol_{2}\leq\epsilon^{-2}\int_{M^{2}_{z}}|V|^{2}\,dVol_{2}$

which verifies (172). ∎

###  6.2. Proofs of Theorem 1.5 and Proposition 1.6

######  Proposition 6.2 (cf. Theorem 1.5).

Assume $M^{4}$ is Type IIa and $\int_{M^{4}}|\mathop{\rm Ric}|^{2}<\infty$. If
$e^{g}(M^{3}_{z},M^{2}_{z})=0$ on $M^{4+}$ (resp. $M^{4-}$), then
$\chi^{g}(M^{2}_{z})=0$ on $M^{4+}$ (resp. $M^{4-}$).

###### Proof.

Assume $M^{4+}$ is non-empty and $e^{g}=0$ (the argument for $M^{4-}$ is
entirely similar). Eq. (171) gives

(177)
$\left(4\pi\chi^{g}\cdot(z-z_{N})+C_{1}\right)^{2}\;\leq\;C_{2}\,\int_{M^{2}_{z}}(\triangle_{4}z)^{2}\,dVol_{2}.$

Set

(178) $\displaystyle
C_{3}=\lim_{z\rightarrow{}z_{N}{}^{+}}\frac{d}{dz}\int_{M^{3}_{z}}(\triangle_{4}z)^{2}\,dVol_{2},\quad\text{and}$
$\displaystyle
C_{4}=\lim_{z\rightarrow{}z_{N}{}^{+}}\int_{M^{3}_{z}}(\triangle_{4}z)^{2}\,dVol_{2}.$

Because $z$ has no critical points on $M^{4+}$, Proposition 5.6 along with the
Integration Lemma gives

(179)
$\frac{d}{dz}\int_{M^{3}_{z}}(\triangle_{4}z)^{2}\,dVol_{2}\;=\;C_{3}\,+\,\int_{M^{4}_{z_{N},z}}|\mathop{\rm
Ric}|^{2}\,dVol_{4}.$

Because $\frac{d}{dz}\int_{M^{3}_{z}}(\triangle_{4}z)^{2}\,dVol_{2}$ is
monotonically increasing, sending $z\rightarrow\infty$ on the right side of
(179), that
$\frac{d}{dz}\int_{M^{3}_{z}}(\triangle_{4}z)^{2}dVol_{2}\leq{}C_{3}+\int_{M^{4+}}|\mathop{\rm
Ric}|^{2}dVol_{4}$. Thus

(180)
$\int_{M^{2}_{z}}(\triangle_{4}z)^{2}\,dVol_{2}\;\leq\;\left(C_{3}+\int_{M^{4+}}|\mathop{\rm
Ric}|^{2}\right)z\,+\,C_{4}$

for large $z$. The fact that the left side of (177) is quadratic when
$\chi^{g}\neq 0$ but the right side has linear bounds forces $\chi^{g}=0.$ ∎

######  Proposition 6.3 (cf. Prop 1.6).

Assume $M^{4}$ is Type IIa and Ricci-flat. Then $M^{4}$ is one-ended so either
$M^{4-}=\varnothing$ of $M^{4+}=\varnothing$, and $\triangle_{4}z$ is constant
on $M^{4}$. For non-critical $z$ we have
$2e^{g}(M^{3}_{z},M^{2}_{z})\cdot\triangle_{4}z=\chi^{g}(M^{2}_{z})$. Further,
if $M^{4}$ is non-flat then $\triangle_{4}z\neq 0$. If $\triangle_{4}z>0$ then
$M^{4-}=\varnothing$ and if $\triangle_{4}z<0$ then $M^{4+}=\varnothing$.

###### Proof.

Because $\mathop{\rm Ric}=0$, Lemma 5.1 shows $d(\triangle_{4}z)=0$ so
$\triangle_{4}z$ is constant. Then at any non-critical value of $z$, by
Proposition 4.3 we have
$(\triangle_{4}z)\frac{d}{dz}Vol(M^{2}_{z})=2\pi\chi^{g}(M^{2}_{z})$ and by
Proposition 3.7 we have
$\frac{d}{dz}Vol(M^{2}_{z})=4\pi{}e^{g}(M^{3}_{z},M^{2}_{z})$. Therefore
$2e^{g}(M^{3}_{z},M^{2}_{z})\triangle_{4}{}z=\chi^{g}(M^{2}_{z})$ as claimed.

To see that $M^{4}$ must be one-ended, by Proposition 4.3 we have

(181) $\displaystyle\frac{d}{dz}\int_{M^{2}_{z}}|V|^{2}dVol_{2}$
$\displaystyle\;=\;\int_{M^{2}_{z}}\triangle_{4}z\,dVol_{2}$
$\displaystyle\;=\;4\pi\chi^{g}\cdot(z-z_{N})\,+\,\int_{M^{2}_{z_{N}}}\triangle_{4}z\,dVol_{2}\quad\text{on
$M^{4+}$}$ $\displaystyle\frac{d}{dz}\int_{M^{2}_{z}}|V|^{2}dVol_{2}$
$\displaystyle\;=\;\int_{M^{2}_{z}}\triangle_{4}z\,dVol_{2}$
$\displaystyle\;=\;4\pi\chi^{g}\cdot(z-z_{0})\,+\,\int_{M^{2}_{z_{0}}}\triangle_{4}z\,dVol_{2}\quad\text{on
$M^{4-}$.}$

If $\triangle_{4}z<0$ then the first equation forces $\chi^{g}=0$ on $M^{4+}$,
but then $\frac{d}{dz}\int|V|^{2}dVol_{2}$ is a negative constant, an
impossibility due to the fact that $z$ has no upper bound by Lemma 2.8 and
$\int|V|^{2}dVol_{2}>0$; therefore $M^{4+}=\varnothing$. Likewise if
$\triangle_{4}z>0$ then the second equation forces $\chi^{g}=0$ on $M^{4-}$,
but then $\frac{d}{dz}\int|V|^{2}dVol_{2}$ is a positive constant, an
impossibility because $z$ has no lower bound; therefore $M^{4-}=\varnothing$.

Lastly we consider the possibility that $\triangle_{4}z=0$. If $M^{4}$ has two
ends, then a line exists in $M^{4}$ so $M^{4}$ is flat by the Cheeger-Gromoll
splitting theorem [9]. Therefore $M^{4}$ has only one end. By Lemma 1.3, if
$M^{4-}=\varnothing$ then $z$ obtains a minimum, and if $M^{4+}=\varnothing$
then $z$ obtains a maximum. But because $\triangle_{4}z=0$, this contradicts
the maximum principle. We conclude that a non-flat, Ricci-flat Type IIa metric
with $\triangle_{4}z=0$ cannot exist. ∎

###  6.3. A global bound $|V|>\epsilon$

We study the case that $M^{4}$ has compact level-sets and a global bound
$|V|>\epsilon$. By Lemma 2.8, every level-set $M^{3}_{z}$ is isotopic so in
particular are diffeomorphic, and the manifold $M^{4}$ a product
$M^{3}\times\mathbb{R}$ where $z:M^{4}\rightarrow\mathbb{R}$ is projection
onto the second factor. Any one-variable function
$\varphi:\mathbb{R}\rightarrow\mathbb{R}$ can be regarded as a function on
$M^{4}$ via $\varphi\circ{}z:M^{4}\rightarrow\mathbb{R}$. If the level-sets
are compact and $M^{4}$ is Type IIa, then $\varphi\circ{}z$ has compact
support if and only if $\varphi$ has compact support. Before proving our main
results we make two $L^{2}$-style integral estimates.

######  Lemma 6.4.

Assume $M^{4}$ has compact level-sets, and $\varphi=\varphi(z)$ is a smooth
1-variable function with compact support on $M^{4}$. Then

(182)
$\displaystyle\int\varphi(\triangle_{4}z)^{2}\,dVol_{4}=-\int\varphi^{\prime}|\nabla{}z|^{2}(\triangle_{4}z)\,dVol_{4}+2\int\varphi\mathop{\rm
Ric}(\nabla{}z,\nabla{}z)\,dVol_{4}.$

###### Proof.

When $\varphi$ has compact support, integration by parts gives

(183) $\displaystyle\int\varphi^{2}(\triangle_{4}z)^{2}\,dVol_{4}$
$\displaystyle=-2\int\left<\nabla\varphi,\nabla{}z\right>\varphi\triangle_{4}z\,dVol_{4}-\int\varphi^{2}\left<\nabla{}z,\,\nabla(\triangle_{4}z)\right>\,dVol_{4}.$

To simplify this , we use $\nabla(\triangle_{4}z)=-2\mathop{\rm
Ric}(\nabla{}z,\cdot)$ from Lemma 5.1, and we use
$\nabla\varphi=\varphi^{\prime}\nabla{}z$ to obtain
$\left<\nabla\varphi,\nabla{}z\right>=\left<\varphi^{\prime}\nabla{}z,\nabla{}z\right>=\varphi^{\prime}|\nabla{}z|^{2}$.
Substituting into (183) gives(182). ∎

######  Lemma 6.5.

Assume $M^{4}$ has compact levelsets, a global lower bound $|V|>\epsilon>0$,
and a global upper bound $|\mathop{\rm Ric}|\leq\Lambda^{2}$. If $\varphi$ is
a 1-variable function with compact support in $[z^{\prime},z^{\prime\prime}]$
so that $M^{3}_{z^{\prime}}$ is non-empty. Then

(184) $\displaystyle\int\varphi(\triangle_{4}z)^{2}\,dVol_{4}$
$\displaystyle\leq\int_{z^{\prime}}^{z^{\prime\prime}}|\varphi^{\prime}|\left(2\Lambda^{2}|z-z^{\prime}|+C^{\prime}\right)\left(2\pi\chi^{g}(z-z_{0})^{2}+C_{0}\right)\,dz$
$\displaystyle+2\Lambda^{2}\int_{z^{\prime}}^{z^{\prime\prime}}|\varphi|\left(2\pi\chi^{g}(z-z_{1})^{2}+C_{1}\right)\,dz$

where the constants are defined as follows: $\chi^{g}$ is the constant
$\chi^{g}(M^{2}_{z})$, $z_{0}$ may be chosen arbitrarily,
$C_{0}=\sup_{M^{3}_{z_{0}}}|\triangle_{4}z|$, $z_{1}$ is any value where
$\int_{M^{2}_{z_{1}}}|\nabla{}z|^{2}dVol_{2}=\inf_{z}\int_{M^{2}_{z}}|\nabla{}z|^{2}dVol_{2}$
reaches a minimum, and $C_{1}=\int_{M^{2}_{z_{1}}}|\nabla{}z|^{2}dVol_{2}$.

###### Proof.

Pick any $z_{0}$. By Lemma 5.1
$\frac{\partial\triangle_{4}z}{\partial{}z}=-2\mathop{\rm
Ric}\left(\frac{\nabla{}z}{|\nabla{}z|},\frac{\nabla{}z}{|\nabla{}z|}\right)$,
so

(185)
$\sup_{M^{3}_{z}}|\triangle_{4}z|\;\leq\;2\Lambda^{2}|z-z_{0}|+\sup_{M^{3}_{z_{0}}}|\triangle_{4}z|.$

Abbreviating $C_{0}=\sup_{M^{3}_{z_{0}}}|\triangle_{4}z|$, Lemma 6.4 gives

(186) $\displaystyle\int\varphi(\triangle_{4}z)^{2}\,dVol_{4}$
$\displaystyle\leq\int|\varphi^{\prime}||\nabla{}z|^{2}\left(2\Lambda^{2}|z-z_{0}|+C_{0}\right)\,dVol_{4}$
$\displaystyle\quad\quad+2\Lambda^{2}\int|\varphi||\nabla{}z|^{2}\,dVol_{4}$

Then the Integration Lemma (Lemma 2.4) gives

(187) $\displaystyle\int\varphi(\triangle_{4}z)^{2}\,dVol_{4}$
$\displaystyle\leq\int_{z^{\prime}}^{z^{\prime\prime}}|\varphi^{\prime}|\left(2\Lambda^{2}|z-z^{\prime}|+C^{\prime}\right)\left(\int_{M^{2}_{z}}|\nabla{}z|^{2}\,dVol_{2}\right)dz$
$\displaystyle+2\Lambda^{2}\int_{z^{\prime}}^{z^{\prime\prime}}|\varphi|\left(\int_{M^{2}_{z}}|\nabla{}z|^{2}\,dVol_{2}\right)dz$

Proposition 4.3 gives
$\frac{d^{2}}{dz}\int_{M^{2}_{z}}|\nabla{}z|^{2}dVol_{2}=4\pi\chi^{g}(M^{2}_{z})$.
Because the $M^{2}_{z}$ are all isotopic we can use the constant $\chi^{g}$
for $\chi^{g}=\chi^{g}(M^{2}_{z})$.

Because $z\mapsto\int_{M^{2}_{z}}|\nabla{}z|^{2}dVol_{2}$ is an upward opening
parabola or constant, there exists some number $z_{1}$ so that
$\int_{M^{2}_{z}}|\nabla{}z|^{2}dVol_{2}$ reaches a minimum of $C_{0}>0$ at
$z=z_{1}$; therefore
$\int_{M^{2}_{z}}|\nabla{}z|^{2}dVol_{2}=2\pi\chi^{g}(z-z_{1})^{2}+C_{1}$.
Therefore (187) gives

(188) $\displaystyle\int\varphi(\triangle_{4}z)^{2}\,dVol_{4}$
$\displaystyle\leq\int_{z^{\prime}}^{z^{\prime\prime}}|\varphi^{\prime}|\left(2\Lambda^{2}|z-z_{0}|+C_{0}\right)\left(2\pi\chi^{g}(z-z_{1})^{2}+C_{1}\right)\,dz$
$\displaystyle+2\Lambda^{2}\int_{z^{\prime}}^{z^{\prime\prime}}|\varphi|\left(2\pi\chi^{g}(z-z_{1})^{2}+C_{1}\right)\,dz.$

∎

######  Theorem 6.6 (cf. Theorem 1.8).

Assume $M^{4}$ is non-flat, but scalar-flat and has a pointwise bound
$|\mathop{\rm Ric}|\leq\Lambda^{2}$. Assume $M^{4}$ has compact level-sets,
and a global bound $|V|>\epsilon>0$.

Then $M^{4}$ is 2-ended, $L^{2}(|\mathop{\rm Ric}|)=\infty$, the level-sets
$M^{3}_{z}$ and fibrations $M^{3}_{z}\rightarrow{}M^{2}_{z}$ are all mutually
isotopic, $e^{g}(M^{3}_{z},M^{2}_{z})=0$, and $\chi^{g}(M^{2}_{z})>0$.
Topologically, each $M^{2}_{z}$ is $\mathbb{S}^{2}$, each $M^{3}_{z}$ is
$\mathbb{S}^{2}\times{}\mathbb{S}^{1}$, the fibration
$M^{3}_{z}\rightarrow{}M^{2}_{z}$ is the projection
$\mathbb{S}^{2}\times\mathbb{S}^{1}\rightarrow\mathbb{S}^{2}$, and $M^{4}$ is
$\mathbb{S}^{2}\times\mathbb{S}^{1}\times\mathbb{R}$.

###### Proof.

By Lemma 2.8 $M^{4}$ is 2-ended, $z$ ranges from $-\infty$ to $\infty$, and
the values $e^{g}=e^{g}(M^{3}_{z},M^{2}_{z})$ and
$\chi^{g}=\chi^{g}(M^{2}_{z})$ are constant. To see that $e^{g}=0$, by
Proposition 3.7 we have

(189) $Vol(M^{2}_{z})\;=\;2\pi{}e^{g}(M^{3}_{z},M^{2}_{z})\,z\;+\;C_{3}.$

Because $Vol(M^{2}_{z})>0$ for all $z$, necessarily $e^{g}=0$. Thus
$Vol(M^{2}_{z})$ is constant.

We use a proof by contradiction to prove strict inequality $\chi^{g}>0$.
Assume to the contrary that $\chi^{g}=0$. Then by Proposition 4.3 constants
$C_{0}$ and $C_{1}$ exist so

(190) $\displaystyle\int_{M^{2}_{z}}\triangle_{4}z\,dVol_{2}$
$\displaystyle\;=\;C_{2},\quad\text{and}\quad\int|V|^{2}\,dVol_{2}\;=\;C_{2}\cdot{}z\,+\,C_{1}.$

Because $\int_{M^{2}_{z}}|V|^{2}dVol_{2}>0$ and $z$ ranges in
$(-\infty,\infty)$, necessarily $C_{2}=0$ so

(191)
$\displaystyle\int_{M^{2}_{z}}\triangle_{4}z\,dVol_{2}\;=\;0,\quad\int_{M^{2}_{z}}|V|^{2}\,dVol_{2}\;=\;C_{1}$

Assume $\eta:M^{4}\rightarrow\mathbb{R}$ has compact support. By (127) we have
$\int\eta|\mathop{\rm
Ric}|^{2}dVol_{4}=-\int{}d\eta\wedge\mathcal{TP}{}_{\mathop{\rm Ric}}$. Using
the expression for $\mathcal{TP}_{\mathop{\rm Ric}}$ in Lemma 5.5, we have

(192) $2\int\eta|\mathop{\rm
Ric}|^{2}\;=\;-\int{}d\eta\wedge\mathcal{L}_{\frac{\partial}{\partial{}z}}\left((\triangle_{4}z)^{2}\frac{1}{|V|}\,dVol_{3}\right).$

The Leibniz rule gives

(193) $2\int\eta|\mathop{\rm
Ric}|^{2}\;=\;\int{}(\triangle_{4}z)^{2}\frac{1}{|V|}\;\mathcal{L}_{\frac{\partial}{\partial{}z}}\big{(}d\eta\big{)}\wedge{}dVol_{3}$

where we used the fact that
$\mathcal{L}_{\frac{\partial}{\partial{}z}}=di_{\frac{\partial}{\partial{}z}}$
on 4-forms is a total derivative, which vanish under the integral. Now assume
$\eta=\eta(z)$ is a function of one variable that has compact support. Then

(194)
$\mathcal{L}_{\frac{\partial}{\partial{}z}}\big{(}d\eta\big{)}\;=\;\frac{\partial^{2}\eta}{\partial{}z^{2}}dz$

so abbreviating $\frac{\partial^{2}\eta}{\partial{}z^{2}}=\eta^{\prime\prime}$
we have

(195) $\displaystyle 2\int\eta|\mathop{\rm Ric}|^{2}$
$\displaystyle\;=\;\int{}(\triangle_{4}z)^{2}\frac{1}{|V|}\;\eta^{\prime\prime}dz\wedge{}dVol_{3}\;=\;\int\eta^{\prime\prime}(\triangle_{4}z)^{2}\,dVol_{4}.$

Using Lemma 6.5, with $\varphi=\eta^{\prime\prime}$ and the assumption that
$\chi^{g}=0$, we obtain

(196) $\displaystyle 2\int\eta|\mathop{\rm Ric}|^{2}$
$\displaystyle\;\leq\;C_{1}\int_{z^{\prime}}^{z^{\prime\prime}}|\eta^{\prime\prime\prime}|\left(2\Lambda^{2}|z-z_{0}|+C_{0}\right)+2\Lambda^{2}C_{1}\int_{z^{\prime}}^{z^{\prime\prime}}|\eta^{\prime\prime}|dz$

where $\eta$ has support in $[z^{\prime},z^{\prime\prime}]$, $z_{0}$ may be
chosen arbitrarily, and $C_{0}=\sup_{M^{2}_{z_{0}}}|\triangle_{4}z|$. We
select $z_{0}=0$ and choosing $\delta>0$ we select

(197) $\eta=\begin{cases}0,&z\notin[-\delta^{-1},\,\delta^{-1}]\\\
(1-(\delta\,z)^{2})^{4},&z\in[-\delta^{-1},\,\delta^{-1}].\end{cases}$

Then $\eta\in{}C^{3}_{c}$ so we may use the estimate (196). The integration in
(196) may be carried out explicitly. We obtain

(198) $2\int\eta|\mathop{\rm
Ric}|^{2}\;\leq\;\frac{4608(3\sqrt{7}+4\sqrt{21})\Lambda^{2}C_{1}}{2401}\Lambda^{2}C_{1}\delta+\frac{1808}{49}C_{0}C_{1}\delta^{2}.$

Noting that $\lim_{\delta\rightarrow 0^{+}}\eta(z)\equiv 1$, Eq. (198) gives
in the limit

(199) $\int|\mathop{\rm Ric}|^{2}\;=\;0$

and we have proven that if $\chi^{g}(M^{2}_{z})=0$, then $M^{4}$ is Ricci-
flat. But $M^{4}$ is 2-ended so it has a line, so by the Splitting Theorem [9]
the manifold is flat, contradicting the assumption that $M^{4}$ is non-flat.
We conclude that $\chi^{g}>0$.

We have proven that $e^{g}=0$ and $\chi^{g}>0$. We must prove
$\int|\mathop{\rm Ric}|^{2}=\infty$. By Propositions 3.7 and 4.3 we have
constants $C_{3}>0$ and $C_{1}$ so

(200) $\displaystyle C_{3}\;=\;\int_{M^{2}_{z}}\,dVol_{2}\quad\text{and}\quad
4\pi\chi^{g}\cdot{}z+C_{1}\;=\;\int_{M^{2}_{z}}\triangle_{4}z\,dVol_{2}.$

By Hölder’s inequality therefore

(201)
$(4\pi\chi^{g}\cdot{}z\,+\,C_{1})^{2}\;\leq\;C_{3}\int_{M^{2}_{z}}(\triangle_{4}z)^{2}\,dVol_{2}$

which means the convex function
$\int_{M^{2}_{z}}(\triangle_{4}z)^{2}\,dVol_{2}$ has quadratic growth (or
greater). In particular
$\frac{d}{dz}\int_{M^{2}_{z}}(\triangle_{4}z)^{2}\,dVol_{2}$ has no upper
bound. But

(202) $\displaystyle\frac{d}{dz}\int_{M^{2}_{z}}(\triangle_{4}z)^{2}dVol_{2}$
$\displaystyle\;=\;\left.\frac{d}{dz}\right|_{z=0}\int_{M^{2}_{z}}(\triangle_{4}z)^{2}dVol_{2}\,+\,\int_{0}^{z}\left(\int_{M^{2}_{z}}|\mathop{\rm
Ric}|^{2}dVol_{2}\right)\,dz$
$\displaystyle\;=\;\left.\frac{d}{dz}\right|_{z=0}\int_{M^{2}_{z}}(\triangle_{4}z)^{2}\,dVol_{2}\,+\,\int_{M^{4}_{0,z}}|\mathop{\rm
Ric}|^{2}\,dVol_{4}$

where we used the Integration Lemma (Lemma 2.4) in the second line. Because
$\frac{d}{dz}\int_{M^{2}_{z}}(\triangle_{4}z)^{2}\,dVol_{2}$ has no upper
bound, so $\int_{M^{4}_{0,z}}|\mathop{\rm Ric}|^{2}\,dVol_{4}$ has no upper
bound. We conclude that $\int_{M^{4}}|\mathop{\rm Ric}|^{2}dVol_{4}=\infty$.

Now the Seifert fibration $M^{3}_{z}\rightarrow{}M^{2}_{z}$ has positive
orbifold Euler number for its base and zero fibration Euler number. A glance
at the classification of Seifert fibered spaces—see [48] or Theorem 5.3 of
[41]—states that $M^{3}_{z}$ has Thurston covering geometry
$\mathbb{S}^{2}\times\mathbb{R}$. Because $M^{3}_{z}$ is compact, orientable,
has orientable fibers, and an oriented base, it has the topology of
$\mathbb{S}^{2}\times\mathbb{S}^{1}$ and the fibration
$M^{3}_{z}\rightarrow{}M^{2}_{z}$ is the projection onto the $\mathbb{S}^{2}$
fiber. By Lemma 2.8 and the fact that $z$ is unbounded from above and below
means $M^{4}\approx\mathbb{S}^{2}\times\mathbb{S}^{1}\times\mathbb{R}$ where
the projection onto the last factor $M^{4}\rightarrow\mathbb{R}$ is the
function $z$. ∎

##  7\. Examples

We present some illustrating examples of scalar-flat metrics with a Killing
field. We refer the reader also to Sections 4.2 and 5.4 in the text, which
presented flat examples and LeBrun instanton examples in detail.

###  7.1. Type I examples

The easiest example is $\mathbb{S}^{2}\times\Sigma^{2}$ where $\mathbb{S}^{2}$
has its unit round metric and $\Sigma^{2}$ is a surface of negative genus.
More examples come from blowing (inserting $\mathbb{P}^{1}$ divisors of
negative self-intersection) up at zeros of the Killing field, see for example
[42] and references therein. Compact scalar-flat metrics have been studied
extensively and there are too many references to mention; see for example
[26], [49], [43], [25], [32].

###  7.2. The flat and hyperbolic examples

The simplest example is $\mathbb{C}^{2}$ with two rotational fields generated
by the diffeomorphism flow
$\psi_{t}.(z_{1},z_{1})=(e^{2\pi\alpha}z_{1},e^{2\pi\beta}z_{2})$. If $\alpha$
and $\beta$ have the same sign this is a Type IIa example. If $\alpha$ or
$\beta$ is zero, or if they have opposite signs, the levelsets are non-compact
and this is a Type IIb example.

A wider range of phenomena is captured in the
$\mathbb{S}^{2}\times\mathbb{H}^{2}$ examples. With $\theta_{1}\in[0,2\pi)$
the flow parameter on $\mathbb{S}^{2}$ and $\theta_{2}\in[0,2\pi)$ the flow
parameter on $\mathbb{H}^{2}$, the metric is

(203)
$g=g_{\mathbb{S}^{2}}+g_{\mathbb{H}^{2}}=\Big{(}(dr^{1})^{2}+(\sin(r)d\theta_{1})^{2}\Big{)}+\Big{(}(dr^{2})^{2}+(f(r)d\theta_{2})^{2}\Big{)}$

where $f(r^{2})$ satisfied $f^{\prime\prime}(r^{2})=f(r^{2})$; the three cases
are $f(r^{2})=\sinh(r^{2})$, $r^{2}\in[0,\infty)$ where the Killing field is
elliptic, $f(r^{2})=e^{r^{2}}$, $r^{2}\in(-\infty,\infty)$ where the Killing
field is parabolic, and $f(r^{2})=\cosh(r^{2})$, $r^{2}\in(-\infty,\infty$
where the Killing field is hyperbolic. On the $\mathbb{S}^{1}$ factor
$r^{1}\in[-\pi,\pi]$. We have $\theta_{1},\theta_{2}\in[0,2\pi)$. We find
potential functions in the usual way by solving