# Pseudoconvex submanifolds in Kähler 4-manifolds

Brian Weber  Brian Weber

(April 26, 2021; August 1, 2022)

# Pseudoconvex submanifolds in Kähler 4-manifolds

Brian Weber  Brian Weber

(April 26, 2021; August 1, 2022)

###### Abstract

On Kähler 4-manifolds, not necessarily compact or of finite topological type,
we obtain relationships between the fundamental group of compact embedded
Levi-flat or pseudoconvex submanifold and the fundamental group of the ambient
manifold $M^{4}$. When a Levi-flat submanifold $V^{3}$ has finite fundamental
group then $\pi_{1}(M^{4})=\iota_{*}\pi_{1}(V^{3})$; when a non-separating
pseudoconvex submanifold $V^{3}$ has finite fundamental group, then
$\pi_{1}(M^{4})=\iota_{*}\pi_{1}(V^{3})\rtimes\mathbb{Z}$. As applications, if
a Kähler manifold (compact or not) has an embedded holomorphic
$\mathbb{P}^{1}$ of positive self-intersection, it must intersect all other
holomorphic $P^{1}$ of non-negative self-intersection, the fundamental group
of $M^{4}$ is trivial, and no ALE or ALF ends exist. If a Levi-flat
submanifold and an embedded holomorphic $\mathbb{P}^{1}$ of positive self-
intersection both exist, they intersect. The total number of ALE plus ALF ends
is zero or one regardless of what other kinds of ends exist. We provide
examples, such as a 2-ended scalar-flat Kähler metric conformal to the Taub-
NUT.

##  1 Introduction

We show that pseudoconvex or Levi-flat submanifolds of finite fundamental
group within a Kähler 4-manifold (not necessarily closed or of finite
topological type) force strong restrictions on the fundamental group on the
ambient manifold. As applications, we show the existence of any such a
submanifold forces restrictions on the geometry of its ends, on whether any
other pseudoconvex submanifolds can exist, on whether embedded holomorphic
$\mathbb{P}^{1}$ submanifolds of positive self-intersection can exist, and
restrictions on the fundamental group of any Kähler 4-manifold that contains
such a $\mathbb{P}^{1}$.

We consider Kähler manifolds-with-boundary $(M^{4},J,g)$ that satisfy the
following pseudconvexity condition ($P$):

  * ($P$)

$M^{4}$ is a topologically closed manifold or manifold-with-boundary, not
necesarily compact. If $M^{4}$ has boundary, every boundary component is of
class $C^{2}$, pseudoconvex with respect to the outward pointing normal, and
compact. There may be finite or infinitely many boundary components.

Sometimes we will say “pseudoconvex Kähler manifold” to mean a manifold that
satisfies (P). Notably $M^{4}$ need not be compact or even have finite
topological type. Embedded pseudoconvex submanifolds
$\iota:V^{3}\hookrightarrow{}M^{4}$, except where otherwise stated, will be
assumed to have the following finiteness condition ($F$):

  * ($F$)

The submanifold $\iota:V^{3}\hookrightarrow{}M^{4}$ is embedded, compact, of
class at least $C^{2}$, and is Levi-flat or pseudoconvex with respect to some
choice of normal. Either $V^{3}$ intersects no point of the boundary, or it is
a boundary component. If it is a boundary component, we choose the outward
pointing normal. Finally $\pi_{1}(V^{3})$ is a finite subgroup.

Theorem 1.3 shows that when $M^{4}$ satisfies ($P$) and $V^{3}\subset{}M^{4}$
is a Levi-flat submanifold satisfying ($F$) then in fact
$\iota_{*}\pi_{1}(V^{3})=\pi_{1}(M^{4})$. Further, any such submanifold
$V^{3}$ is separating, and if there are two Levi-flat submanifolds
$V_{0}^{3}$, $V_{1}^{3}$ of class $C^{2}$, and either one of them has finite
fundamental group, they intersect. If $V^{3}\subset\partial{}M^{4}$ is a
boundary component it is the only boundary component. If $V^{3}\subset{}M^{4}$
is in the interior and has finite fundamental group, then $M^{4}$ has no
boundary.

Theorem 1.4 weakens the Levi-flat condition on the submanifold $V^{3}$ to the
requirement that $V^{3}$ be pseudoconvex rather than Levi-flat. If $M^{4}$ is
compact then $V^{3}$ is separating, and the component on which $V^{3}$ forms a
pseudoconvex boundary has finite $\pi_{1}(M^{4})$. If $V^{3}$ is not
separating then necessarily the manifold is non-compact, and its fundamental
group is $\pi_{1}(M^{4})=\iota_{*}(\pi_{1}(V^{3}))\rtimes\mathbb{Z}$.

Using the techniques developed to prove Theorems 1.3 and 1.4, we prove a
related result, Proposition 1.5, dealing with the case that two pseudoconvex
submanifolds $V_{0}^{3}$ and $V_{1}^{3}$ exist within $M^{4}$. We cannot
retain the very strong conclusion that they intersect—consider the case of
small balls around distinct points in $\mathbb{C}^{2}$. But we prove that no
component of $M^{4}\setminus\\{V_{0}^{3}\cup{}V_{1}^{3}\\}$ cannot have
pseudoconvex copies of both $V_{0}^{3}$ and $V_{1}^{3}$ on its boundary
(whether or not it might also have pseudo-concave copies of $V_{0}^{3}$ or
$V_{1}^{3}$ on its boundary). This proposition has some strong consequences.
For example a pseudoconvex Kähler manifold (compact or not) can have at most
one embedded holomorphic $\mathbb{P}^{1}$ of positive self-intersection. No
ALE or ALF Kähler 4-manifold can have a holomorphic, embedded $\mathbb{P}^{1}$
of positive self-intersection.

After this proposition and the two theorems, we lay out a few immediate
consequence in the form of four corollaries. A boundaryless Kähler manifold,
compact or not, can have at most one embedded holomorphic $\mathbb{P}^{1}$ of
positive self-intersection. Any Kähler 4-manifold, compact or not, with an
embedded holomorphic $\mathbb{P}^{1}$ of positive self-intersection has
trivial fundamental group. A classic result dating back to [8] is that if
every end of a Kähler manifold is ALE or ALF, then it has just one end. We
expand this classic result to show that such a manifold has only one end of
ALE or ALF type, regardless of whatever other kinds of ends it might have,
even if it has infinitely many ends.

The paper’s final section has examples demonstrating the sharpness of some of
these results. This includes examples displaying several singular and non-
singular Levi-flat submanifolds inside $\mathbb{P}^{2}$ and
$\mathbb{P}^{2}\\#\overline{\mathbb{P}}{}^{2}$—all of which, in the smooth
case, intersect all curves of self-intersection $+1$. Another example is an
explicit construction of a complete, 2-ended scalar-flat Kähler metric, one of
whose ends is ALE. This is a Kähler metric conformally equivalent to the
classic Taub-NUT metric. Ordinarily in dimension 4 a Kähler metric cannot be
conformal to another Kähler metric, but this is possible in the Taub-NUT case
because of an unusually large number of metric-compatible complex structures
available, including two that are not part of its hyperKähler structure. See
Section 5.3.

###  1.1 Definitions and basic concepts

The operator
$\sqrt{-1}\partial\bar{\partial}:\bigwedge^{0}\rightarrow\bigwedge^{2}$ is
real, and can be expressed
$\sqrt{-1}\partial\bar{\partial}{}f=-\frac{1}{2}d(J(df))$; throughout we use
the convention $J\eta=\eta\circ{}J$ for 1-forms $\eta$ and we often use $dJdf$
for $d(J(df))$. A function is called pluriharmonic when
$\partial\bar{\partial}{}f=0$ or equivalently $dJdf=0$.

Let $V^{3}\subset{}M^{4}$ be a surface of class $C^{2}$, and let $\hat{n}$ be
a unit 1-form defined in a neighborhood of $V^{3}$ that is perpendicular to
$V^{3}$ in the sense that $\hat{n}(X)=0$ for every $X\in{}T_{p}V^{3}$. Then
the 2-form
$L=-\frac{1}{2}dJ\hat{n}\big{|}_{\hat{n}^{\perp},J\hat{n}^{\perp}}\in\bigwedge^{2}V^{3}$
is called the Levi form of $V^{3}$. Ostensibly $L$ depends a specific choice
of $\hat{n}$ in a neighborhood of $L$, but it is easily shown to be
independent of this choice, except that the sign of $L$ reverses if $\hat{n}$
is replaced by $-\hat{n}$.

The Levi form of a $C^{2}$ surface with respect to some normal is said to be
non-negative, positive, or zero depending on whether $L(X,JX)\geq 0$, $>0$, or
$=0$, respectively, for all sections $X\in{}TV^{3}$ where also
$JX\in{}TV^{3}$. A surface $V^{3}$ with a choice of normal is defined to be
pseudoconvex, strictly pseudoconvex, or Levi-flat if $L\geq 0$, $L>0$, or
$L\equiv 0$. On a Kähler manifold, by dualizing we can evaluate $L$ on 1-forms
rather than vectors. Non-negativity means $L(J\eta,\eta)\geq 0$ when
$\eta,J\eta\perp\\{\hat{n},J\hat{n}\\}$—the reason for the switch in the place
of $J$ from $L(X,JX)$ to $L(J\eta,\eta)$ is because of our convention for $J$:
$J\eta$ means $+\eta\circ{}J$ instead of $-\eta\circ{}J$. Given 1-forms
$\eta,\gamma$ that are perpendicular to $\hat{n}$ and $J\hat{n}$ on $V^{3}$,
$L(\eta,\gamma)$ is

$\displaystyle
L(\eta,\gamma)\;=\;-\frac{1}{2}dJ\hat{n}\big{(}\eta,\gamma\big{)}\;=\;-\frac{1}{2}*\big{(}\eta\wedge\gamma\wedge*dJ\hat{n}\big{)}$
(1)

and we sometimes write $L(\eta\wedge\gamma)$ in place of $L(\eta,\gamma)$.

Given one or more boundary components, there are always bounded harmonic
functions $f:M^{4}\rightarrow[0,1]$ that equal $0$ or $1$ on each boundary
component. Our main technical result is that when $f$ is such a function then

$\displaystyle
2\int_{M^{4}}|\partial\bar{\partial}{}f|^{2}\;=\;-\int_{\partial{}M^{4}}L\big{(}*(Jdf\wedge{}df)\big{)}.$
(2)

See (11). Clearly in the pseudoconvex case where $L\geq 0$, we obtain some
very coercive estimates—in particular harmonicity $\triangle{}f=0$ forces
pluriharmonicity $\partial\bar{\partial}{}f=0$, and if $f$ is non-constant
then $L\geq 0$ forces $L\equiv 0$. Ordinarily harmonic functions are very easy
to find and pluriharmonic functions are very hard to find. The boundary
condition $L\geq 0$ makes the concepts much more closely tied together. The
following lemmas are almost immediate consequences of (2). Unlike our theorems
these two lemmas do not require $M^{4}$ to obey condition ($P$), which
requires all boundary components to be compact. Our lemmas only require some
boundary components to be compact (but otherwise must obey ($P$)).

###### Lemma 1.1

Assume $(M^{4},g,J)$ satisfies ($P$) with the exception that its boundary
components need not be compact, but that only a single boundary component
$V^{3}\subseteq\partial{}M^{4}$ is compact. Let $f:M^{4}\rightarrow[0,1]$ be
an harmonic function that equals $1$ on $V^{3}$ and $0$ on all other boundary
components.

Then $f$ is pluriharmonic. If $f$ is non-constant (eg. if $\partial{}M^{4}$
has more than one component, or $M^{4}$ is non-parabolic) then every component
of $\partial{}M^{4}$ is Levi-flat.

###### Lemma 1.2

Assume $(M^{4},g,J)$ satisfies ($P$), and $V^{3}$ is a boundary component
satisfying ($F$).

Then $V^{3}$ is the only boundary component of $M^{4}$,
$\iota_{*}(\pi_{1}(V^{3}))=\pi_{1}(M^{4})$, and the Riemannian manifold
$(M^{4},g)$ is either parabolic or compact (in particular it has no ALE or ALF
ends).

Lemmas 1.1 and 1.2 easily apply to the case of Kähler manifolds that have an
ALE or ALF end. Such manifold ends are always non-parabolic, because the
Green’s functions decay along such ends asymptotically like $dist^{-2}$ or
$dist^{-1}$, respectively. Such ends also meet the criteria from [12] or [7],
for example. The paper [18] contains results similar to these lemmas, arrived
at differently.

###  1.2 Results

This paper focuses on how Levi-flat or pseudoconvex submanifolds within (or on
the boundary of) a Kähler 4-manifold forces topological constraints on the
ambient manifold. This is analogous, in a certain sense, to the large body of
literature on the subject of global curvature constraints forcing topological
restrictions on manifolds. The global condition in our setting is not global
curvature controls, but rather the existence of a Kähler structure. We use
$\iota$ for the natural inclusion $\iota:V^{3}\rightarrow{}M^{4}$ of a
submanifold into its ambient manifold.

######  Theorem 1.3 (Levi-flat submanifolds)

Assume $(M^{4},J,g)$ satisfies ($P$) and $V^{3}$ is a Levi-flat submanifold
satisfying ($F$); in particular, $\pi_{1}(V^{3})$ is finite.

If $V^{3}$ is a boundary component then it is the only boundary component and
$\iota_{*}\pi_{1}(V^{3})=\pi_{1}(M^{4})$.

If $V^{3}$ is not a boundary component but an interior submanifold, then
$M^{4}\setminus{}V^{3}$ has two components, $\partial{M}^{4}=\varnothing$, and
$\iota_{*}\pi_{1}(V^{3})=\pi_{1}(M^{4})$.

Finally, if $V_{0}^{3}$, $V_{1}^{3}$ are any two compact Levi-flat
submanifolds that do not intersect $\partial{}M^{4}$ and both obey ($F$)
except that only one but not both of $\pi_{1}(V_{0}^{3})$,
$\pi_{1}(V_{1}^{3}$) need be finite, then $V_{0}^{3}$ and $V_{1}^{3}$
intersect.

Our second result weakens the requirement that the submanifold be Levi-flat to
the condition that it be pseudoconvex (with respect to either of its normals).

######  Theorem 1.4 (Pseudoconvex submanifolds)

Assume $(M^{4},J,g)$ satisfies ($P$), and $V^{3}$ is a pseudoconvex
submanifold satisfying ($F$); in particular $\pi_{1}(V^{3})$ is finite.

If $V^{3}$ is a boundary component, then it is the only boundary component and
$\iota_{*}\pi_{1}(V^{3})=\pi_{1}(M^{4})$.

If $V^{3}$ is not a boundary component but an interior submanifold, then
exactly one of the following holds:

  * i)

$V^{3}$ is a separating surface; then letting
$M_{0}^{4}\subset{}M^{4}\setminus{}V^{3}$ be the component with the
pseudoconvex copy of $V^{3}$ on its boundary, we have
$\partial{}M_{0}^{4}=V^{3}$ and $\iota_{*}\pi_{1}(V^{3})=\pi_{1}(M^{4}_{0})$.

  * ii)

$V^{3}$ is a non-separating surface; then
$\pi_{1}(M^{4})=\iota_{*}\pi_{1}(V^{3})\rtimes\mathbb{Z}$,
$\partial{}M^{4}=\varnothing$, and the first betti number is unity:
$b^{1}(M^{4})=1$. In addition to $\partial{M}^{4}$ being boundaryless (and
therefore geodesically complete), it is also non-compact.

In particular, if a compact Kähler manifold has a pseudoconvex submanifold
$V^{3}$ with finite fundamental group, $V^{3}$ is always separating. Another
immediate consequence is that any complete Kähler 4-manifold with one end that
is ALE or ALF, it has finite fundamental group.

Theorem 1.3 says, among other things, that non-intersecting Levi-flat
submanifolds cannot exist in $M^{4}$ if either of their fundamental groups is
finite. If we weakening the assumption to pseudoconvexity then non-
intersecting submanifolds certainly can exist, for instance concentric spheres
in $\mathbb{C}^{2}$. Nevertheless we can still draw some conclusions, which we
summarize in the following proposition.

######  Proposition 1.5 (Two pseudoconvex submanifolds)

Let $(M^{4},g,J)$ be a manifold (possibly non-compact) that satisfies ($P$).
Assume $V_{0}^{3}$, $V_{1}^{3}$ are non-intersecting compact submanifolds that
are pseudoconvex with respect to some orientation, and assume one of the
fundamental groups, either $\pi_{1}(V^{3}_{0})$ or $\pi_{1}(V^{3}_{1})$, is
finite.

Then any component of $M^{4}\setminus\\{V_{0}^{3}\cup{}V_{1}^{3}\\}$ has, on
its boundary, at most _one_ pseudoconvex copy of $V_{0}^{3}$ or $V_{1}^{3}$
but never both.

There is no apriori requirement that $V_{0}^{3}$ or $V_{1}^{3}$ be separating,
no restrictions on the number of components
$M^{4}\setminus\\{V_{0}^{3}\cup{}V_{1}^{3}\\}$ might have (it might have one,
two, or three), and no restriction on the boundedness of the components. It is
also possible such a component may have pseudo-concave boundary components,
which, by hypothesis ($P$), could only come from other copies of $V_{0}^{3}$
or $V_{1}^{3}$.

The proposition says nothing about, for instance, pairs of pseudoconvex
submanifolds such as be concentric 3-spheres in $\mathbb{C}^{2}$ or perhaps
small non-intersecting 3-spheres around distinct points in a Kähler manifold.
But as mild sounding as Proposition 1.5 may seem, it can have some strong
consequences. This theorem forbids any two ends of type ALE or ALF, regardless
of what other kinds of ends the manifold might have—the reason is that ALE or
ALF manifold ends have convex separating surfaces in the differential
geometric sense (meaning they have positive second fundamantal forms) which
are therefore also pseudoconvex. (For a precise definition of “ALE” and “ALF,”
see for example [3].) This theorem also gives extremely strong restrictions on
Kähler 4-manifolds with an embedded, holomorphic $\mathbb{P}^{1}$ submanifold
of positive self-intersection, as any such $\mathbb{P}^{1}$ has a neighborhood
which is pseudoconvex with respect to the inward pointing normal. Thus by
Proposition 1.5 no two such $\mathbb{P}^{1}$ submanifolds can exist, unless
they intersect. If the Kähler manifold has an ALE or ALF end, no such
$\mathbb{P}^{1}$ can exist at all. This and other phenomena are explored in
the Lemmas below.

###  1.3 Consequences

Our theorems produce certain topological restrictions on Kähler 4-manifolds
from the existence of Levi-flat or pseudoconvex submanifolds. Here we explore
some immediate consequences of our theorems in four corollaries. Much of the
value in these corollaries lies in the fact that they apply in the compact as
well as the non-compact case. In the compact case the content of these lemmas
is largely known. For example, if a closed surface has a $\mathbb{P}^{1}$ of
positive self-intersection it is already known to be a rational surface, and
because these are classified, our Corollary 1.7 for example was already known
in the compact case.

###### Corollary 1.6

Assume $(M^{4},J,g)$ is a complete Kähler manifold (not necessarily compact)
that has both a compact embedded Levi-flat surface $V^{3}$ and an embedded
holomorphic submanifold $N=\mathbb{P}^{1}$ of positive self-intersection. Then
they intersect: $V^{3}\cap{}N\neq\varnothing$.

###### Corollary 1.7

Let $(M^{4},J,g)$ be a pseudoconvex Kähler manifold (not necessarily compact)
that satisfies ($P$).

If $(M^{4},J,g)$ has an embedded holomorphic $\mathbb{P}^{1}$ of positive
self-intersection, then $\pi_{1}(M^{4})=\\{e\\}$ and
$\partial{}M^{4}=\varnothing$.

If $(M^{4},J,g)$ has an embedded holomorphic $\mathbb{P}^{1}$ of zero self-
intersection, then every compact component of $\partial{}M^{4}$ has infinite
$\pi_{1}(V^{3})$.

###### Corollary 1.8

Assume $(M^{4},J,g)$ is a complete Kähler manifold (not necessarily compact).
If $N=\mathbb{P}^{1}$ and $N^{\prime}=\mathbb{P}^{1}$ are embedded holomorphic
submanifolds, one of which has non-negative self-intersection and one of which
has positive self-intersection, then they intersect:
$N\cap{}N^{\prime}\neq\varnothing$.

###### Corollary 1.9

Let $(M^{4},g,J)$ be a complete Kähler manifold that has $k$ many ALE ends and
$l$ many ALF ends (apriori $k$ or $l$ might be infinite). Then $k+l$ is zero
or one. If $k+l=1$ then $\pi_{1}(M^{4})$ is finite. If $(M^{4},g,J)$ has a
holomorphic embedded $\mathbb{P}^{1}$ of self-intersection $\geq 0$, then
$k+l=0$.

An older theorem, originally a consequence of Kohn-Rossi’s work [8], states
that a Kähler manifold of finite topological type, all of whose ends are ALE
or ALF, can have just one end. More recently various other proofs of this
theorem have emerged, using techniques very far from Kohn-Rossi’s original
methods. In one work [18] potential theory and the open mapping theorem was
used. In other works [6], [9] a compactification of ALE manifolds in the
complex category was performed, which allowed the use of theorems designed for
the compact case. In still another [12] potential theory was used, along with
a curvature assumption used to control asymptotics of the Green’s function.

##  2 Interaction between harmonic functions and pseudoconvex boundary
components

We shall frequently refer to harmonic functions created by the method of
exhaustion; this is a common construction, and we run through it for the
purposes of adaptation to our manifolds-with-boundary. If
$V_{1}^{3},\dots,V_{K}^{3}$ is a finite collection of compact boundary
components we can create a bounded harmonic function $f$ with $f\equiv 1$ on
each $V_{i}^{3}$, $f\equiv 0$ on all other components of $\partial{}M^{4}$. To
build this function, let $\\{\Omega_{i}\\}_{i=1}^{\infty}$ be pre-compact
domains that each contain $\bigcup_{i=1}^{K}V_{i}^{3}$ and that exhaust
$M^{4}$. For each $i$ create harmonic functions
$f_{i}:\Omega_{i}\rightarrow[0,1]$ with boundary conditions $f_{i}=1$ on
$\bigcup_{i=1}^{K}V_{i}^{3}$ and $f_{i}=0$ on
$\partial\Omega_{i}\setminus\bigcup_{i=1}^{K}V_{i}^{3}$. Set
$f_{\infty}=\lim_{i}f_{i}$. Then $f_{\infty}$ is certainly a bounded harmonic
function $f_{\infty}:M^{4}\rightarrow[0,1]$ that is $1$ on
$\bigcup_{i=1}^{K}V_{i}^{3}$ and zero on all other boundary components (if
there are any others). Such a function is unique, in the sense that if
$f^{\prime}_{\infty}$ is an harmonic function created in the same way except
using a different exhaustion
$\Omega_{1}^{\prime}\subset\Omega_{2}^{\prime}\subset\dots$, then
$f^{\prime}_{\infty}=f_{\infty}$.

Possibly an harmonic function created this way is constant. This leads to the
distinction between parabolic and non-parabolic manifold ends. A non-compact
Riemannian manifold with non-empty but compact boundary—a manifold end—is
called parabolic if the harmonic function created by the method of exhaustion
is constant, and non-parabolic if it is non-constant. This terminology, now
standard, originated in [12]. If $M^{n}$ is a Riemannian manifold with non-
empty but compact boundary and $f_{\infty}$ is the harmonic function created
by this method of exhaustion, then the quantity
$\int_{M^{n}}|df_{\infty}|^{2}$ is always finite, and is called the harmonic
capacity of $M^{n}$. This follows from a simple argument using the classical
Hopf lemma: the pre-compact domains $\Omega_{i}$ all have finite capacity, and
an application of the Hopf lemma shows that the value of
$\int_{\Omega_{i}}|df_{i}|^{2}$ decreases as $i$ increases. For more details,
see for example [12].

###  2.1 Integration by Parts

We use the convention $\triangle{}f=+Tr\,\text{Hess}f$. On a Kähler manifold
with Kähler form $\omega$, the projection of $dJdf\in\bigwedge^{2}$ onto
$\bigwedge^{+}$ is $-\frac{1}{2}(\triangle{f})\omega$. From this we obtain

$\displaystyle*dJdf=-(\triangle{}f)\omega-dJdf.$ (3)

Assume $M^{4}$ is a manifold with compact boundary, $f$ is a $C^{2}$ function,
and $\varphi$ is a cutoff function—that is, a $C^{\infty}$ function with
compact support on $\overline{M}{}^{4}$. We do not require $\varphi$ to vanish
on any particular boundary component. Using (3), that
$2\sqrt{-1}\partial\bar{\partial}{}f=-dJdf$, along with integration by parts
we find

$\displaystyle\begin{aligned}
&4\int\varphi^{2}|\partial\bar{\partial}{}f|^{2}dVol\;=\;\int\varphi^{2}dJdf\wedge*dJdf\\\
&\quad\;=\;\frac{1}{2}\int\varphi^{2}(\triangle{}f)^{2}\omega\wedge\omega-\int\varphi^{2}dJdf\wedge{}dJdf\\\
&\quad\;=\;\int\varphi^{2}(\triangle{}f)^{2}\,dVol+2\int\varphi\,d\varphi\wedge{}Jdf\wedge{}dJdf-\int_{\partial{}M^{4}}\varphi^{2}\,Jdf\wedge{}dJdf.\end{aligned}$
(4)

######  Lemma 2.1 (Integration by Parts)

Let $(M^{4},g)$ be a Kähler manifold with $C^{2}$ boundary. Let
$V_{1}^{3},\dots,V_{K}^{3}\subset\partial{}M^{4}$ be finitely many components
of $\partial{}M^{4}$, each of which is compact (although $\partial{}M^{4}$
might have other components, possibly infinitely many, that might not be
compact). Let $f:M^{4}\rightarrow[0,1]$ be the harmonic function obtained from
the exhaustion method, where $f\equiv 1$ on each $V_{1}^{3},\dots,V_{K}^{3}$
and $f\equiv 0$ on every other boundary component. Let $\varphi$ be any
$C_{c}^{0,1}$ function with compact support on $\overline{M}{}^{4}$. Then

$2\int\varphi^{2}|\partial\bar{\partial}{}f|^{2}\;=\;\int\varphi\,d\varphi\wedge{}Jdf\wedge{}dJdf-\int_{\partial{}M^{4}}\varphi\,L\big{(}*(Jdf\wedge{}df)\big{)}\,dA$
(5)

where $L$ is the Levi form of $\partial{}M^{4}$.

Proof. When $f:M^{4}\rightarrow[0,1]$ is harmonic and $\varphi$ has compact
support, (4) is

$\displaystyle\begin{aligned}
&4\int\varphi^{2}|\partial\bar{\partial}{}f|^{2}\;=\;2\int\varphi\,d\varphi\wedge{}Jdf\wedge{}dJdf-\int_{\partial{}M^{4}}\varphi\,Jdf\wedge{}dJdf.\end{aligned}$
(6)

We relate the boundary term of (6) to the Levi form. First, we rewrite

$\displaystyle-\int_{\partial{}M}\varphi\,Jdf\wedge{}dJdf\;=\;-\int_{\partial{}M}\varphi*\big{(}\hat{n}\wedge{}Jdf\wedge{}dJdf\big{)}\,dA$
(7)

where $dA$ is the area form on $\partial{}M^{4}$, $\hat{n}$ is the outward
unit normal, and $*$ is the Hodge-star on $M^{4}$. Because
$f:M^{4}\rightarrow[0,1]$ is harmonic and either zero or one on each boundary
component, certainly $|df|>0$ and $f$ is a defining function for each
component. Therefore the Levi form is

$\displaystyle
L\;=\;\mp\frac{1}{2}dJ\left(|df|^{-1}df\right)\Big{|}_{df^{\perp},Jdf^{\perp}}\;=\;\mp\frac{1}{2}|df|^{-1}\,dJdf\Big{|}_{df^{\perp},Jdf^{\perp}},$
(8)

where $\mp$ is $-$ when $df$ is outward pointing (this is the case when $f=1$
on a boundary component, which is the case on all the $V_{i}^{3}$ components),
and is $+$ when $df$ is inward pointing (this is the case when $f=0$ on a
boundary component, which is the case on all components of $\partial{}M^{4}$
except the $V_{i}^{3}$).

The outward pointing 1-form is $\hat{n}=\pm|df|^{-1}df$. Using the fact that
$*\omega_{1}\wedge*\omega_{2}=\omega_{1}\wedge\omega_{2}$ whenever
$\omega_{1},\omega_{2}\in\bigwedge{}^{2}$, the integrand of (7) is

$\displaystyle\begin{aligned}
\hat{n}\wedge{}Jdf\wedge{}dJdf&\;=\;\pm\,df\wedge{}Jdf\wedge\left(|df|^{-1}dJdf\right)\\\
&\;=\;\pm*(df\wedge{}Jdf)\wedge*\left(|df|^{-1}dJdf\right)\\\
&\;=\;-2*\,L\big{(}*(df\wedge{}Jdf)\big{)}\\\
&\;=\;+2*\,L\big{(}*(Jdf\wedge{}df)\big{)}.\end{aligned}$ (9)

(The “$\pm$” from $\hat{n}=\pm|df|^{-1}df$ and the “$\mp$” of (8) combine to
make “$-$”.) Therefore (6) is precisely

$\displaystyle
2\int\varphi^{2}|\partial\bar{\partial}{}f|^{2}\;=\;\int\varphi\,d\varphi\wedge{}Jdf\wedge{}dJdf-\int_{\partial{}M^{4}}\varphi\,L\big{(}*(Jdf\wedge{}df)\big{)}.$
(10)

$\Box$

######  Lemma 2.2 (Improved integration by parts)

Let $(M^{4},g)$ be a Kähler manifold along with an harmonic function
$f:M^{4}\rightarrow[0,1]$ that satisfy the hypotheses of Lemma 2.1. Then if
$\int_{\partial{}M^{4}}L\big{(}*(Jdf\wedge{}df)\big{)}\,dA$ is integrable,

$2\int_{M^{4}}|\partial\bar{\partial}{}f|^{2}\;=\;-\int_{\partial{}M^{4}}L\big{(}*(Jdf\wedge{}df)\big{)}\,dA.$
(11)

Finally, if $\partial{}M^{4}$ is pseudoconvex then
$\int_{\partial{}M^{4}}L\big{(}*(Jdf\wedge{}df)\big{)}\,dA$ is integrable.

Proof. We prove this using (5) and the fact that the capacity $\int|df|^{2}$
is finite. The method is by choosing good cutoff functions. We are brief
because this kind of argument is very standard. For each $i$ let $\varphi_{i}$
be a cutoff function with $\varphi_{i}=1$ on a very large compact set
$\Omega_{i}$, $\varphi=0$ on an even larger compact set $\Omega_{i+1}$, and
the exhaustion $\Omega_{1}\subset\Omega_{2}\subset\dots$ is chosen so also
$|d\varphi_{i}|<i^{-1}$. It might be objected that our manifolds have
boundary, so the exponential map cannot provide the distance functions
necessary to build such cutoff functions $\varphi_{i}$. However a manifold
with $C^{2}$ boundary still has a distance function to any point, given by
approximation by minimizing paths. At interior points of the manifold such a
distance function retains the usual properties, in particular being Lipschitz
and having unit norm almost everywhere. This is enough to construct cutoff
functions in the usual way.

We estimate the first integral on the right of (5), using Hölder’s inequality,
by

$\displaystyle\begin{aligned}
&\left|\int_{M^{4}}\varphi_{i}{}d\varphi_{i}\wedge{}Jdf\wedge{}dJdf\right|\;\leq\;\max|d\varphi_{i}|\int_{M^{4}}\varphi_{i}|df||dJdf|\\\
&\hskip
72.26999pt\;\leq\;\frac{1}{2}\max|d\varphi_{i}|\left(\int_{M^{4}}|df|^{2}\,+\,\int_{M^{4}}\varphi_{i}^{2}|dJdf|^{2}\right).\end{aligned}$
(12)

Because $|d\varphi_{i}|<i^{-1}$, we have upper and lower estimates

$\displaystyle\begin{aligned}
&2\left(1-i^{-1}\right)\int\varphi^{2}_{i}|\partial\bar{\partial}{}f|^{2}\;\leq\;\frac{1}{2}i^{-1}\int|df|^{2}\,-\,\int_{\partial{}M^{4}}\varphi_{i}{}L\Big{(}*(Jdf\wedge{}df)\Big{)}\,dA\\\
&2\left(1+i^{-1}\right)\int\varphi^{2}_{i}|\partial\bar{\partial}{}f|^{2}\;\geq\;-\frac{1}{2}i^{-1}\int|df|^{2}\,-\,\int_{\partial{}M^{4}}\varphi_{i}{}L\Big{(}*(Jdf\wedge{}df)\Big{)}\,dA.\end{aligned}$
(13)

But the capacity term $\int|df|^{2}$ is finite. Therefore, as long as
$\int_{\partial{}M^{4}}L\big{(}*(Jdf\wedge{}df)\big{)}\,dA$ is integrable, we
take $i\rightarrow\infty$ and obtain (11).

To see this is integrable when $\partial{}M^{4}$ is pseudoconvex, recall from
the introduction that this means $L(J\eta,\eta)\geq 0$ when $\eta,J\eta$ are
perpendicular to $\hat{n}$ and $J\hat{n}$. Thus, because
$*(Jdf\wedge{}df)=J\eta\wedge\eta$ for some 1-form $\eta$, then
$L\big{(}*(Jdf\wedge{}df)\big{)}\geq 0$. Because the integrand is pointwise
non-negative,
$\int_{\partial{}M^{4}}\varphi_{i}{}L\big{(}*(Jdf\wedge{}df)\big{)}\,dA$ is
integrable (although it might equal $+\infty$). This concludes the proof.
$\Box$

Compare the argument of Lemma 2.2 to the argument of Lemma 3.1 of [11].

Proof of Lemma 1.1. Our assumption is that $\partial{}M^{4}$ is entirely
pseudoconvex, and $V^{3}\subseteq\partial{}M^{4}$ is a compact boundary
component. Create an harmonic function $f$ by the method of exhaustion so $f$
is unity on $V^{3}$ and zero on any other components of $\partial{}M^{4}$.
Possibly $f$ is constant, in which case $f\equiv 1$ and $df\equiv 0$.
Certainly in this case $\partial{M}^{4}$ has a single boundary component, and
is either compact or parabolic.

Possibly $f$ is non-constant. Of so, then by pseudoconvexity,
$L(*(Jdf\wedge{}df))\geq 0$ in the pointwise sense on $\partial{}M^{4}$.
Therefore (11) forces both $|\partial\bar{\partial}{}f|\equiv 0$ (so $f$ is
pluriharmonic), and $L(*(df\wedge{}Jdf))\equiv 0$. By the Hopf lemma $df$ is
never zero on $\partial{}M^{4}$ because every point on $\partial{}M^{4}$ is
either a global maximum or a global minimum of $f$. Thus
$L(*(df\wedge{}Jdf))\equiv 0$ forces $L\equiv 0$ on $\partial{}M{}^{4}$, which
is the same as $\partial{}M{}^{4}$ being Levi-flat. This concludes the proof
of Lemma 1.1.

######  Lemma 2.3 (Analytic maps into the strip)

Let $(M^{4},J)$ be a compact complex manifold with precisely two boundary
components $V_{0}^{3}$ and $V_{1}^{3}$. If $C_{0}\neq{}C_{1}$ there is no
analytic function $z:M^{4}\rightarrow\mathbb{C}$ with
$Re(z)\big{|}_{V_{0}^{3}}\equiv{}C_{0}$ and
$Re(z)\big{|}_{V_{1}^{3}}\equiv{}C_{1}$.

Remark. By “strip” we mean a locus
$\\{z\in\mathbb{C}\,|\,C_{0}\leq{}Re(z)\leq{}C_{1}\\}$.

Proof. This simple lemma follows from the open mapping theorem. With
$z:M^{4}\rightarrow\mathbb{C}$ being such an analytic function, by continuity
the image $z(M^{4})$ is connected and compact. By the open mapping theorem,
$\partial(z(M^{4}))\subseteq{}z(\partial{}M^{4})$.

But by hypothesis, the image of the boundary $z(\partial{}M^{4})$ is contained
within the the lines $\\{Re(z)=C_{0}\\}$ and $\\{Re(z)=C_{1}\\}$, and
intersects both of these lines. Because $z(M^{4})$ is connected and intersects
both lines, it intersects interior points of the strip. Because $z(M^{4})$ is
closed it is either the entire strip or else it has points of closure in the
interior of the strip. Because $z(M^{4})$ is compact it cannot be the entire
strip, so $z(M^{4})$ must have points of closure within the strip. This
contradicts the fact that the image’s boundary
$\partial(z(M^{4}))\subseteq{}z(\partial{}M^{4})$ lies completely within the
strip’s boundary lines, and concludes the proof. $\Box$

Proof of Lemma 1.2. The hypotheses are that $(M^{4},J,g)$ is Kähler with
pseudoconvex boundary, and has a boundary component
$V^{3}\subseteq\partial{}M^{4}$ that is compact and has finite
$\pi_{1}(V^{3})$.

We first show that $\iota_{*}(\pi_{1}(V^{3}))=\pi_{1}(M^{4})$. Let
$\widetilde{M}{}^{4}$ be the universal cover of $M^{4}$ with covering map
$p:\widetilde{M}{}^{4}\rightarrow{}M^{4}$. Let
$\widetilde{V}^{3}=p^{-1}(V^{3})$ be the pre-image of $V^{3}$. Each component
of $\widetilde{V}^{3}$ covers $V^{3}$ with deck group
$\iota_{*}(\pi_{1}(V^{3}_{i}))$, and because $\iota_{*}(\pi_{1}(V^{3}))$ is
finite each component is itself compact. The number of components is the
cardinality of the coset space $\pi_{1}(M^{4})/\iota_{*}(\pi_{1}(V^{3}))$.

For a proof by contradiction assume the inclusion
$\iota_{*}(\pi_{1}(V^{3}))\subseteq\pi_{1}(M^{4})$ is strict, so the pre-image
$\widetilde{V}{}^{3}=p^{-1}(V^{3})$ has at least two components which we label
$\widetilde{V}{}_{0}^{3},\widetilde{V}{}_{1}^{3}\subset{}\widetilde{V}{}^{3}$.
Let $f:\widetilde{M}{}^{4}\rightarrow[0,1]$ be the harmonic function created
by the method of exhaustion, with $f\equiv 1$ on $\widetilde{V}{}_{1}^{3}$ and
$f\equiv 0$ on all other components of $\partial\widetilde{M}{}^{4}$ including
$\widetilde{V}{}_{0}^{3}$. By hypothesis the boundary integral on the right
side of (11) is non-negative (indeed, even the integrand is non-negative in
the pointwise sense), so the equality

$2\int_{\widetilde{M}{}^{4}}|\partial\bar{\partial}{}f|^{2}\;=\;-\int_{\partial\widetilde{M}{}^{4}}L\big{(}*(Jdf\wedge{}df)\big{)}\,dA$
(14)

now forces both sides to equal zero. Because $f$ is not constant (it equals
$1$ on $\widetilde{V}_{1}^{3}$ and $0$ on $\widetilde{V}_{0}^{3}$), we have
$Jdf\wedge{}df$ nowhere zero on the boundary. This forces $L\equiv 0$ on
$\partial{}M^{4}$, and also forces $\partial\bar{\partial}{}f=0$ on $M^{4}$
which means $f$ is pluriharmonic. Because $dJdf=0$ (which is the same as
$\partial\bar{\partial}{}f=0$) and because $\widetilde{M}{}^{4}$ is simply
connected, the equation $dJdf=0$ means $Jdf=-dg$ for some real-valued function
$g$. The function $z=f+\sqrt{-1}g$ is therefore an analytic function.

Consider the submanifold

$\displaystyle
M_{\epsilon}^{4}\;=\;\left\\{p\in\widetilde{M}{}^{4}\;\big{|}\;f(p)\in[1-\epsilon,1]\right\\}.$
(15)

Because $f=1$ and $df\neq 0$ on $\widetilde{V}_{1}^{3}$ and
$\widetilde{V}_{1}^{3}$ this is a global maximum for $f$, we can choose
$\epsilon>0$ so small that $M_{\epsilon}^{4}$ is a collar neighborhood:
$M_{\epsilon}^{4}\approx\widetilde{V}_{1}^{3}\times[1-\epsilon,1]$. But then
$M_{\epsilon}^{4}$ is a manifold with two compact boundary components, and the
analytic function $z:M_{\epsilon}^{4}\rightarrow\mathbb{C}$ has
$Re(z)=f=1-\epsilon$ on one boundary and $Re(z)=f=1$ on the other boundary. By
Lemma 2.3 applied to $\widetilde{M}{}^{4}_{\epsilon}$ along with the analytic
function $z$, this is impossible. We conclude that
$f:\widetilde{M}{}^{4}\rightarrow[0,1]$ must be a constant function. Thus two
components $\widetilde{V}_{0}^{3}$ and $\widetilde{V}_{1}^{3}$ of
$p^{-1}(V^{3})$ cannot exist. Therefore
$\iota_{*}(\pi_{1}(V^{4}))=\pi_{1}(M^{4})$.

Next we show that actually $\partial{}M^{4}=V^{3}$. Let
$p:\widetilde{M}{}^{4}\rightarrow{}M^{4}$ again be the universal cover. We
have shown $\iota_{*}(\pi_{1}(V^{3}))=\pi_{1}(M^{4})$, so $p$ is a finite-
sheeted cover. By assumption the components of $\partial{}M^{4}$ are compact,
so all components of $\partial\widetilde{M}{}^{4}$ remain compact. Repeating
the argument above, we conclude that $\partial\widetilde{M}{}^{4}$ has just
one component, $p^{-1}(V^{3})$. Therefore $V^{3}$ is the only component of
$\partial{}M^{4}$.

Lastly we must show that $(M^{4},J,g)$ is parabolic or compact. The boundary
$\partial{}M^{4}$ has just one component, which is compact and pseudoconvex.
If $M^{4}$ is non-parabolic, then the harmonic function
$f:M^{4}\rightarrow[0,1]$ created by the method of exhaustion is non-constant
and a defining function on $\partial{}M^{4}$. Since the boundary
$\partial{}M^{4}$ is compact, we can repeat the previous argument—since $f$ is
non-constant and reaches a maximum at $\partial{}M^{4}$, we can still use a
collar neighborhood of $\partial{}M^{4}$ and an analytic function $z$ with
$Re(z)=f$, and draw a contradiction using Lemma 2.3. Therefore $M^{4}$ cannot
be non-parabolic, so it is parabolic or compact. This concludes the proof of
Lemma 1.2.

##  3 Levi-flat and Pseudoconvex Submanifolds

Up to this point, our results have dealt with boundary pseudoconvexity (see
[18] for similar results using different methods). We now move from boundary
pseudoconvexity to pseudoconvex submanifolds. To study submanifolds, we cut
$M^{4}$ along these submanifolds, which produces additional boundary
components, and using these new boundary components it is often possible to
apply our lemmas.

Proof of Theorem 1.3. The hypotheses are that $(M^{4},J,g)$ is a Kähler
manifold with pseudoconvex (or empty) boundary, has a compact embedded Levi-
flat submanifold $V^{3}$ with finite $\pi_{1}(V^{3})$, and either $V^{3}$ is a
boundary component or else $V^{3}\cap\partial{}M^{4}=\varnothing$.

If $V^{3}$ is a boundary component, then the fact that
$\iota_{*}(\pi_{1}(V^{3}))=\pi_{1}(M^{4})$ follows from Lemma 1.2. The rest of
the proof deals with the case $V^{3}\cap\partial{}M^{4}=\varnothing$.

Proof that $V^{3}$ is separating. For a contradiction, assume $V^{3}$ is non-
separating. This means $M^{4}{}^{\prime}=M^{4}\setminus{}V^{3}$ consists of a
just one component. Because $V^{3}$ and $M^{4}$ are orientable (so in
particular the submanifold $V^{3}$ has two distinct sides), the connected
manifold $M^{4}{}^{\prime}$ has two additional boundary components, each a
copy of $V^{3}$, which we call $V^{3}_{0}$ and $V^{3}_{1}$. Because the
submanifold $V^{3}$ was Levi-flat, the new boundary components $V_{0}^{3}$ and
$V_{1}^{3}$ are also Levi-flat. Therefore the boundary of $M^{4}{}^{\prime}$
continues to be pseudoconvex, so $M^{4}{}^{\prime}$ continues to obey
condition ($P$). Both $V^{3}_{0}$ and $V^{3}_{1}$ have finite fundamental
groups (as both are diffeomorphic to $V^{3}$). This contradicts Lemma 1.2,
which says $\partial{}M^{4}{}^{\prime}$ can have at most one such component.
This contradiction forces the submanifold $V^{3}$ to be separating.

Proof that $\partial{}M^{4}=\varnothing$. Because $V^{3}$ is separating,
$M^{4}\setminus{}V^{3}$ consists of two components which we call
$M_{0}^{4}{}^{\prime}$ and $M_{1}^{4}{}^{\prime}$. In addition to inheriting
any boundary components of the original manifold $M^{4}$, the new manifolds-
with-boundary $M_{0}^{4}{}^{\prime}$, $M_{1}^{4}{}^{\prime}$ each has as an
additional boundary component which is a copy of $V^{3}$, which we call
$V_{0}^{3}\subseteq\partial{}M^{4}_{0}{}^{\prime}$ and
$V_{1}^{3}\subseteq\partial{}M^{4}_{1}{}^{\prime}$. The components
$M_{i}^{4}{}^{\prime}$ continue to satisfy ($P$), because the new boundary
components are Levi-flat. But by Lemma 1.2, $V_{i}^{3}$ is the only boundary
component of $M_{i}^{4}{}^{\prime}$, $i=1,2$. Because each
$M_{i}^{4}{}^{\prime}$ has no boundary components besides $V_{i}^{3}$, the
original manifold $M^{4}$ can have no boundary components at all.

Case that $V^{3}_{0}$, $V^{3}_{1}$ are non-intersecting. Specifically, the
assumption is that $V^{3}_{0}$, $V^{3}_{1}$ are non-intersecting and Levi-
flat, neither intersects $\partial{}M^{4}$, and at least one of them has
finite fundamental group.

Without loss of generality we may assume $V^{3}_{0}$ has finite fundamental
group. From above, we know $\partial{}M^{4}=\varnothing$ and we know
$V^{3}_{0}$ is separating, although we do not know if $V^{3}_{1}$ is
separating or not. Then $M^{4}\setminus(V_{0}^{3}\cup{}V_{1}^{3})$ has either
has three components (if $V_{1}^{3}$ is separating) or two components (if
$V_{1}^{3}$ is not separating). In either case, there is one component
$M^{4}{}^{\prime}$ of $M^{4}\setminus(V_{0}^{3}\cup{}V_{1}^{3})$ that has on
its boundary both a copy of $V_{0}^{3}$ and also a copy of $V_{1}^{3}$ (if
$V_{1}^{3}$ is not separating, then $M^{4}{}^{\prime}$ actually has two copies
of $V_{1}^{3}$ on its boundary).

Because both $V_{0}^{3}$ and $V_{1}^{3}$ are Levi-flat, the component
$M^{4}{}^{\prime}$ continues to have pseudoconvex boundary. But because the
copy of $V_{0}^{3}$ on its boundary has finite fundamental group, Lemma 1.2
states this is the only boundary component, contradicting the existence of
$V_{1}^{3}$.

Proof that $\iota_{*}(\pi_{1}(V^{3}))=\pi_{1}(M^{4})$. The assumption is that
$V^{3}$ is a Levi-flat submanifold that does not intersect $\partial{}M^{4}$,
and that $\pi_{1}(V^{3})$ is finite.

For a proof by contradiction suppose the inclusion
$\iota_{*}(\pi_{1}(V^{3}))\subseteq\pi_{1}(M^{4})$ is strict. Because this
inclusion is strict, after passing to the universal cover
$p:\widetilde{M}{}^{4}\rightarrow{}M^{4}$ the preimage
$\widetilde{V}{}^{3}=p^{-1}(V^{3})$ has more than one component. Let
$\widetilde{V}{}^{3}_{0},\widetilde{V}{}^{3}_{1}\subset{}p^{-1}(V^{3})$ be two
distinct components. Each component of $p^{-1}(V^{3})$ is itself a cover of
$V^{3}$, and so the $V_{i}^{3}$ both have finite fundamental group. But the
result above states that no two Levi-flat submanifolds can exist in $M^{4}$
unless they intersect; this contradicts the fact that
$\widetilde{V}{}^{3}_{0}$ and $\widetilde{V}{}^{3}_{1}$ are distinct
components of $p^{-1}(V^{3})$. We conclude that
$\iota_{*}\pi_{1}(V^{3})=\pi_{1}(M^{4})$, and conclude the proof of Theorem
1.3.

Proof of Proposition 1.5. By hypothesis we have two non-intersecting
pseudoconvex submanifolds $V_{0}^{3}$ and $V_{1}^{3}$. We must show that if
$M^{4}{}^{\prime}$ is any component of
$M^{4}\setminus(V_{0}^{3}\cup{}V_{1}^{3})$, then $\partial{}M^{4}{}^{\prime}$
contains has at most one pseudoconvex copy of $V_{0}^{3}$ or $V_{1}^{3}$.

Assuming not. Then a component $M^{4}{}^{\prime}$ of
$M^{4}\setminus(V_{0}^{3}\cup{}V_{1}^{3})$ has copies of both $V_{0}^{3}$ and
$V_{1}^{2}$ on its boundary so that one copy of $V_{0}^{3}$ and one copy of
$V_{1}^{3}$ is pseudoconvex with respect to the outward normal. There are
exactly three possibilities:

  * i)

$\partial{}M^{4}{}^{\prime}$ has one pseudoconvex copy each of $V_{0}^{3}$ and
$V_{1}^{3}$, and no pseudo-concave components.

  * ii)

$\partial{}M^{4}{}^{\prime}$ has one pseudoconvex copy each of $V_{0}^{3}$ and
$V_{1}^{3}$ and one pseudo-concave copy of either $V_{0}^{3}$ or $V_{1}^{3}$
but not both.

  * iii)

$\partial{}M^{4}{}^{\prime}$ has one pseudoconvex and one pseudo-concave copy
each of $V_{0}^{3}$, $V_{1}^{3}$.

See Figure 1 for a depiction. Of course, $\partial{}M^{4}{}^{\prime}$ may
contain additional pseudoconvex components, inherited from the original
manifold $M^{4}$.

(a) Case (i): $\partial{}M^{4}{}^{\prime}$ has one pseudoconvex copy each of
$V_{0}^{3}$ and $V_{1}^{3}$, and no pseudo-concave copies.

(b) Case (ii): $\partial{}M^{4}{}^{\prime}$ has one pseudoconvex copy of both
$V_{0}^{3}$ and $V_{1}^{3}$, and one pseudo-concave copy of $V_{0}^{3}$.

(c) Case (iii): $\partial{}M^{4}{}^{\prime}$ has one pseudoconvex and one
pseudo-concave copy each of $V_{0}^{3}$ and $V_{1}^{3}$.

Figure 1: Schematic depiction of possibilities (i), (ii), (iii) for
$M^{4}\setminus(V_{0}^{3}\cup{}V_{1}^{3})$.

Proof in case (i). In this case since $M^{4}{}^{\prime}$ has an entirely
pseudoconvex boundary, with compact boundary components. The fact that either
$\pi_{1}(V_{0}^{3})$ or $\pi_{1}(V_{1}^{3})$ is finite contradicts Lemma 1.2.

Construction of the pseudoconvex manifold $M_{\infty}^{4}$. In cases (ii) and
(iii) the manifold $M^{4}{}^{\prime}$ does not have pseudoconvex boundary,
because $\partial{}M^{4}$ contains pseudo-concave copies of $V_{0}^{3}$ and/or
$V_{1}^{3}$.

To remedy this, we create a sequence of manifolds-with-boundary
$M_{0}^{4}\subset{}M_{1}^{4}\subset{}M_{2}^{4}\subset\dots$ that exhaust the
entirety of a new pseudoconvex manifold $M_{\infty}^{4}$. Set
$M_{0}^{4}=M^{4}{}^{\prime}$. To create $M_{1}^{4}$, start with one copy of
$M_{0}^{4}$, and glue another copy of $M_{0}^{4}$ onto the matching boundary
or boundaries created by the pseudo-concave copy of $V_{0}^{3}$ (in case (ii))
or the matching boundaries created by both $V_{0}^{3}$ and $V_{1}^{3}$ (in
case (iii)). The manifold $M_{1}^{4}=M_{0}^{4}\cup{}M_{0}^{4}$ still has
pseudo-concave boundary components: in case (ii) it has a pseudo-concave copy
of $V_{0}^{3}$ and in case (iii) a pseudo-concave copy of both $V_{0}^{3}$ and
$V_{1}^{3}$ (see the second image of Figure 2). The gluing occurs by matching
boundary components precisely as they had been matched in the original
manifold and therefore the new manifold remains analytic, and continues to
have, for example, an integrable complex structure.

|  (a) In cases (ii) and (iii) the boundary $\partial{}M_{0}^{4}$ has both
pseudo-concave and pseudoconvex copies of $V_{0}^{3}$ and/or $V_{1}^{3}$.
---|---
|  (b) The manifold-with-boundary $M_{1}^{4}$ is created by gluing the pseudo-
concave boundary components of $M_{0}^{4}$ to the matching pseudoconvex
boundary components on another copy of $M_{0}^{4}$. Certainly
$M_{1}^{4}=M_{0}^{4}\cup{}M_{0}^{4}$ continues to have both pseudo-concave and
pseudo-convex boundary components.
|  (c) Having formed $M^{4}_{i-1}$, we glue the pseudo-concave components of
$\partial{}M_{i-1}^{4}$ to the matching pseudoconvex components of yet another
$M_{0}^{4}$ to create $M_{i}^{4}=M_{0}^{4}\cup\dots\cup{}M_{0}^{4}$ (union of
$i$ many copies of $M_{0}^{4}$). Continuing indefinitely, we limit to
$M_{\infty}^{4}=M_{0}^{4}\cup{}M_{0}^{4}\cup\dots$ whose boundary is now
entirely pseudoconvex.
Figure 2:  Schematic depiction of the pseudoconvex manifold $M_{\infty}^{4}$.

For an induction process, starting with $M^{4}_{i-1}$ we attach a copy of
$M_{0}^{4}$ to $M^{4}_{i-1}$ along the pseudo-concave copy of $V_{0}^{3}$ (or
of both $V_{0}^{3}$ and $V_{1}^{3}$) to the matching pseudoconvex copy of
$V_{0}^{3}$ (or of $V_{0}^{3}$ and $V_{1}^{3}$) of $M_{0}^{4}$. Now we have
Kähler manifolds-with-boundary $M_{i}^{4}=M_{0}^{4}\cup\dots\cup{}M_{0}^{4}$
(union of $i$ many copies of $M_{0}^{4}$), each $M_{i}^{4}$ created from
$M_{i-1}^{4}$ by attaching one copy of $M_{0}^{4}$.

Picking any basepoint $p_{0}$, we now take a pointed limit in the Gromov-
Hausdorff sense, and arrive at a Kähler manifold-with-boundary
$M_{\infty}^{4}=M_{0}^{4}\cup{}M_{0}^{4}\cup\dots$. This manifold has
pseudoconvex boundary (the third image in Figure 2), but no pseudo-concave
boundary components. The new manifold $\partial{}M_{\infty}^{4}$ has exactly
one pseudoconvex copy each of $V_{0}^{3}$ and $V_{1}^{3}$.

Proof that cases (ii) and (iii) cannot occur. By gluing to together copies of
$M_{0}^{4}$, we have constructed the pseudoconvex manifold-with-boundary
$M_{\infty}^{4}$ which has one pseudoconvex copy each of $V_{0}^{3}$ and
$V_{1}^{3}$ on its boundary, and no pseudo-concave boundary components. Both
boundary components $V_{0}^{3}$, $V_{1}^{3}$ are compact, and one of the
groups $\pi_{1}(V_{0}^{3})$, $\pi_{1}(V_{1}^{3})$ is finite. This contradicts
Lemma 1.2 and concludes the proof of Proposition 1.5.

Proof of Theorem 1.4. The hypotheses are that $(M^{4},J,g)$ has pseudoconvex
boundary or no boundary, and $V^{3}$ is a boundary component or else a compact
embedded submanifold that is pseudoconvex with respect to one of its two
normals. We also assume the group $\pi_{1}(V^{3})$ is finite.

If $V^{3}$ is a boundary component, then Lemma 1.2 says it is the only
boundary component and $\iota_{*}(\pi_{1}(V^{3}))=\pi_{1}(M^{4})$. If $V^{3}$
is separating, then $M^{4}\setminus{}V^{3}$ has a component
$M_{0}^{4}{}^{\prime}$ on which $V^{3}$ is pseudoconvex. In this case, Lemma
1.2 provides the conclusion that $\partial{}M_{0}^{4}=V^{3}$ and
$\iota_{*}\pi_{1}(V^{3})=\pi_{1}(M_{0}^{4})$. The rest of the proof deals with
the case that $V^{3}$ is non-separating.

Assuming $V^{3}$ is non-separating, we make a standard argument to show that
$\pi_{1}(M^{4})$ has a factor of $\mathbb{Z}$. After removing $V^{3}$ we are
left with a connected manifold $M_{0}^{4}=M^{4}\setminus{}V^{3}$, and the
former submanifold $V^{3}$ becomes two distinct boundary components, one copy
being pseudoconvex and the other being pseudo-concave. For each
$i\in\mathbb{Z}$ let $M_{i}^{4}$ be a copy of $M_{0}^{4}$. Then gluing the
pseudoconvex copy of $V^{3}$ in $M_{i}^{4}$ to the pseudo-concave copy of
$V^{3}$ in $M_{i+1}^{4}$ and so on, we reach in the limit (a pointed Gromov-
Hausdorff limit) the manifold

$\displaystyle
M_{\infty}^{4}\;=\;\dots\,\cup\,M_{-2}^{4}\,\cup\,M_{-1}^{4}\,\cup\,M_{0}^{4}\,\cup\,M_{1}^{4}\,\cup\,M_{2}^{4}\,\cup\,\dots$
(16)

Each $M_{i}^{4}$ is a copy of $M_{0}^{4}$, and each gluing surface maps
unambiguously to $V^{3}\subset{}M^{4}$. Therefore we retain a map
$M_{\infty}^{4}\rightarrow{}M^{4}$. Each point within any $M_{i}^{4}$—in
particular any point on any of the gluing loci—has a small neighborhood which
is evenly covered. Therefore this is a covering map. The map
$d:M_{\infty}^{4}\rightarrow{}M_{\infty}^{4}$ generates the deck group of this
covering.

We have constructed a cover of $M^{4}$ with transitive deck group
$\left<d\right>\approx\mathbb{Z}$. Covering space theory now says
$\mathbb{Z}\approx\pi_{1}(M^{4})/\pi_{1}(M^{4}_{\infty})$, which is the same
as the existence of a short exact sequence
$0\rightarrow\pi_{1}(M^{4}_{\infty})\rightarrow\pi_{1}(M^{4})\rightarrow\mathbb{Z}\rightarrow
0$. But $\mathbb{Z}$ is projective so this sequence splits. We conclude that
$\pi_{1}(M^{4})\approx\pi_{1}(M^{4}_{\infty})\rtimes\mathbb{Z}$.

Next we verify that neither $M^{4}$ nor $M_{\infty}^{4}$ has any boundary
components. Consider the pseudoconvex manifold
$\dots\cup{}M_{-2}^{4}\cup{}M_{-1}^{4}\cup{}M_{0}^{4}\subset{}M_{\infty}^{4}$.
This has a boundary component that is a pseudoconvex copy of $V^{3}$. Because
$V^{3}$ has $\pi_{1}(V^{3})$ finite, Lemma 1.2 says this copy of $V^{3}$ is
its only boundary component. If $M^{4}$ were itself to have a boundary
component, this component would also be a component of the boundary of
$M_{0}^{4}$. Therefore $M_{\infty}^{4}$ would have additional boundary
components. We have just shown this is impossible, so we conclude
$\partial{}M^{4}=\varnothing$.

Next we show $\pi_{1}(M^{4}_{\infty})=\iota_{*}(\pi_{1}(V^{3}))$. Consider the
inclusion
$\iota_{*}\pi_{1}(V^{3})\subseteq\pi_{1}(M^{4})\approx\pi_{1}(M_{\infty}^{4})\rtimes\mathbb{Z}$.
Because $\pi_{1}(V^{3})$ is a finite group, its inclusion must be an inclusion
into the subgroup $\pi_{1}(M_{\infty}^{4})\times\\{0\\}$—to see this, note
that it is impossible for any element $a\in\pi_{1}(V^{3})$ to map to any
element of the form $(b,n)\in\mathbb{Z}\rtimes\pi_{1}(M_{\infty}^{4})$ where
$n\neq 0$, for any such element would be infinite cyclic. This produces an
inclusion $\iota_{*}\pi_{1}(V^{3})\subseteq\pi_{1}(M_{\infty}^{4})$ (simply by
leaving off the “0” factor).

For a proof by contradiction, assume the inclusion is strict:
$\iota_{*}(\pi_{1}(V^{3}))\subset\pi_{1}(M_{\infty}^{4})$. Letting
$p:\widetilde{M_{\infty}^{4}}\rightarrow{}M_{\infty}^{4}$ be the universal
cover, the pre-image $p^{-1}(V^{3})$ has more than one component. Referring to
the construction of $M^{4}_{\infty}$, for any $k$ consider the submanifold

$M_{k}^{4}{}^{\prime}\;\;\;\;=\;\;\;\;\dots\cup{}M_{k-2}^{4}\cup{}M_{k-1}^{4}\cup{}M_{k}^{4}\;\;\;\;\subset\;\;\;\;M_{\infty}^{4}.$
(17)

The submanifolds
$\dots\subset{}M_{k-1}^{4}{}^{\prime}\subset{}M_{k}^{4}{}^{\prime}\subset\dots$
constitute an exhaustion of $M_{\infty}^{4}$. Also note that
$M_{k}^{4}{}^{\prime}$ and $M_{l}^{4}{}^{\prime}$ are diffeomorphic, by the
deck action $d^{l-k}$.

Because $M_{\infty}^{4}$ has no boundary, as we saw above, the submanifold
$M_{k}^{4}{}^{\prime}$ has precisely one boundary component, which is a
pseudoconvex copy of $V^{3}$. The pre-image
$\widetilde{M_{k}^{4}{}^{\prime}}=p^{-1}(M_{k}^{4}{}^{\prime})$ has more than
one boundary component (because the pre-image of its boundary $p^{-1}(V^{3})$
has more than one component). However, conceivably
$\widetilde{M_{k}^{4}{}^{\prime}}$ also has more than one component, so we
cannot immediately apply Lemma 1.2.

So let $\gamma\subset{}M_{\infty}^{4}$ be a path that represents an element in
$\pi_{1}(M^{4}_{\infty})$ not in $\iota_{*}(\pi_{1}(V^{3}))$. There exists
some $k$ for which the loop $\gamma$ is contained within
$M^{4}_{k}{}^{\prime}$. We may move the basepoint to the boundary copy of
$V_{0}^{3}$, so without loss of generality we may assume $\gamma$ lies
entirely within $M_{k}^{4}{}^{\prime}$ and its two ends both terminate on the
same point of the boundary copy of $V_{0}^{3}$.

Now lift both $M_{k}^{4}{}^{\prime}$ and the path $\gamma$ along $p$. Then the
lifted path $\tilde{\gamma}$ is entirely inside
$\widetilde{M_{k}^{4}{}^{\prime}}$, and because $\gamma$ represented an
element of $\pi_{1}(M_{\infty}^{4})$ not in $\iota_{*}(\pi_{1}(V^{3}))$, the
path $\gamma$ no longer a loop but a path with distinct ends. Because the
basepoint in $M^{4}$ was originally on the boundary component $V_{0}^{3}$, the
two ends of the lifted path now terminate on (the pseudo-concave sides of) two
different boundary components. Therefore at least one path-connected component
of $\widetilde{M^{4}_{k}{}^{\prime}}$ has two or more compact pseudoconvex
copies of $V_{0}^{3}$ on its boundary. These two boundary components are
compact and have finite $\pi_{1}$, contradicting Lemma 1.2. Therefore
$\iota_{*}(\pi_{1}(V^{3}))=\pi_{1}(M^{4}_{\infty})$, as claimed. We conclude
that $\pi_{1}(M^{4})\approx\iota_{*}(\pi_{1}(V^{3}))\rtimes\mathbb{Z}$.

Finally we prove that if $V^{3}$ is non-separating, then $M^{4}$ is both
complete and non-compact. We have already seen that the Kähler manifold
$M^{4}$ has no boundary, so it is geodesically complete. Because
$\pi_{1}(M^{4})=\iota_{*}(\pi_{1}(V^{3}))\rtimes\mathbb{Z}$ and
$\iota_{*}(\pi_{1}(V^{3}))$ is finite, certainly
$H_{1}(M^{4};\mathbb{R})=\mathbb{R}$, so the first betti number of $M^{4}$ is
one. Thus it is impossible that $M^{4}$ is compact, because the first betti
number is even on all compact Kähler manifolds. This concludes the proof of
Theorem 1.4.

##  4 Boundaries, holomorphic submanifolds, and manifold ends

In this section we prove the four corollaries, but first we require some
information on neighborhoods of embedded holomorphic $\mathbb{P}^{1}$
submanifolds of non-negative self-intersection. From the differentiable
perspective such neighborhoods are completely canonical: an embedded
$\mathbb{P}^{1}$ has a neighborhood diffeomorphic to its normal bundle. The
normal bundle is a complex line bundle over $\mathbb{P}^{1}$, and
differentiably these are fully characterized by the self-intersection number
of the zero-section.

But to apply our theorems we require more: neighborhood boundaries must have a
sign on their Levi forms. In the case of zero self-intersection, an embedded
$\mathbb{P}^{1}$ has a neighborhood that is a holomorphic product of
$\mathbb{P}^{1}$ with a disk in $\mathbb{C}$ [16]. Certainly the boundary of
such a neighborhood is Levi-flat and has the topology of
$\mathbb{S}^{2}\times\mathbb{S}^{1}$. In the case of positive self-
intersection the situation is known to be much more complicated: neighborhoods
of $\mathbb{P}^{1}$ of positive self-intersection fall within an infinite-
dimensional moduli space of complex structures [14]. However all we require is
pseudo-concavity, and pseudo-concave neighborhoods are indeed known to exist.
The following result, an important thoerem in its own right, we record for our
purposes as a lemma.

######  Lemma 4.1 (cf. [1] [13] [17] [16])

Let $(M^{4},J)$ be a complex manifold and assume $\mathbb{P}^{1}$ is an
embedded holomorphic $\mathbb{P}^{1}$ of positive self-intersection. Then a
neighborhood $\Omega$ of this $\mathbb{P}^{1}$ exists, whose boundary has
negative Levi-form with respect to the outward pointing normal. Further,
$\Omega$ deformation-retracts onto $\mathbb{P}^{1}$, $\partial\Omega$ is
diffeomorphic to a lens space, and we can make this neighborhood as small as
desired: given any open set $\Omega^{\prime}$ containing $\mathbb{P}^{1}$, we
can assume $\overline{\Omega}\subset\Omega^{\prime}$.

Likewise, if such a $\mathbb{P}^{1}$ has zero self-intersection, there exist a
small neighborhood $\Omega$ of $\mathbb{P}^{1}$ with Levi-flat boundary and
boundary topology $\mathbb{S}^{2}\times\mathbb{S}^{1}$.

Proof of Corollary 1.6. The assumption is that $(M^{4},J,g)$ has a compact
Levi-flat surface $V^{3}$, as well as a $\mathbb{P}^{1}$ of positive self
intersection.

If $\mathbb{P}^{1}$ and $V^{3}$ do not intersect, then using the metric we can
build a small tubular neighborhood around $\mathbb{P}^{1}$ that does not
intersect $N^{3}$. From Lemma 4.1, inside this tubular neighborhood is a
domain $\Omega$ containing $\mathbb{P}^{1}$ that is pseudo-concave and
compact, and $\partial\Omega$ has finite fundamental group. Thus
$M^{4}\setminus\Omega$ has a pseudoconvex boundary component. The boundary of
$M^{4}\setminus\Omega$ is compact, pseudoconvex, and has finite fundamental
group, so Theorem 1.4 says $\pi_{1}(M^{4})$ is finite.

Therefore $(M^{4}\setminus\Omega)\setminus{}V^{3}$ is disconnected (by an easy
Mayer-Vietoris argument). Let $M^{4}{}^{\prime}$ be the component of
$(M^{4}\setminus\Omega)\setminus{}V^{3}$ whose boundary
$\partial{}M^{4}{}^{\prime}$ is the union of the compact pseudoconvex
submanifold $\partial\Omega$, which has finite fundamental group, and a copy
of $V^{3}$, which is compact and Levi-flat. This contradicts Thoerem 1.4 and
establishes the Corollary.

Proof of Corollary 1.7. This corollary studies two possibilities: that a
holomorphic embedded $\mathbb{P}^{1}$ of positive self-intersection exists,
and that a holomorphic embedded $\mathbb{P}^{1}$ of zero self-intersection
exists.

First assume $(M^{4},J,g)$ has a $\mathbb{P}^{1}$ of positive self-
intersection. By Lemma 4.1 we can place a sufficiently small neighborhood
$\Omega$ around $\mathbb{P}^{1}$ so that $\Omega$ does not intersect
$\partial{}M^{4}$ and so $M^{4}\setminus\Omega$ is still pseudoconvex. The
boundary component of $M^{4}\setminus\Omega$ corresponding to $\partial\Omega$
is a sphere or lens space, so has finite fundamental group. Because there is
one compact pseudoconvex boundary component with finite fundamental group, by
Theorem 1.4 this is the only component of $\partial(M^{4}\setminus\Omega)$.
Therefore $\partial{}M^{4}=\varnothing$.

Still assuming $\mathbb{P}^{1}$ has positive self-intersection, assume
$\pi_{1}(M^{4})$ is not trivial. Then the universal cover
$\widetilde{M}{}^{4}$ of $M^{4}$ is non-trivial, so there exists two (or more)
embedded holomorphic $\mathbb{P}^{1}$ submanifolds of positive self-
intersection. By Lemma 4.1, around these we can find non-intersecting pseudo-
concave neighborhoods $\Omega_{1}$ and $\Omega_{2}$ so that
$\partial\Omega_{1}$, $\partial\Omega_{2}$ both have finite fundamental
groups. Then $\widetilde{M}{}^{4}\setminus(\Omega_{1}\cup\Omega_{2})$ has
pseudoconvex boundary and two of its compact components have finite
fundamental group. This violates Theorem 1.4.

To verify the second assertion, assume some $\mathbb{P}^{1}$ has zero self-
intersection. For a contradiction assume there is a compact boundary component
$V^{3}\subseteq\partial{}M^{4}$ with finite $\pi_{1}(V^{3})$. By Lemma 4.1,
$\mathbb{P}^{1}$ has a small neighborhood $\Omega$ that does not intersect
$V^{3}$, and is Levi-flat and compact. However because $\pi_{1}(V^{3})$ is
finite, the manifold $M^{4}\setminus\Omega$ violates Theorem 1.4. This
establishes the corollary.

Proof of Corollary 1.8. The assumption is that $(M^{4},J,g)$ is Kähler
manifold, without boundary (not necessarily compact), that $N=\mathbb{P}^{1}$
has non-negative self-intersection, and that $N^{\prime}=\mathbb{P}^{1}$ has
positive self-intersection..

For a proof by contradiction assume $N\cap{}N^{\prime}=\varnothing$. By Lemma
4.1, $N$ and $N^{\prime}$ have small, non-intersecting neighborhoods $\Omega$
and $\Omega^{\prime}$; the boundary of $\Omega$ is Levi-flat and compact and
the boundary of $\Omega^{\prime}$ is compact, pseudo-concave, and has finite
fundmantal group. Therefore $M^{4}\setminus(\Omega\cup\Omega^{\prime})$ has
two compact pseudoconvex boundary components, one of which has finite
fundamental group. This violates Theorem 1.4, and establishes the corollary.

Proof of Corollary 1.9. The assumption is that $(M^{4},J,g)$ has $k$ many ALE
ends and $l$ many ALF ends; apriori $k$ or $l$ might be infinite.

Any ALE or ALF end has a geometrically convex separating surface (by
“geometrically convex” we mean non-negative second fundamental form), and
because these ends are geometrically convex they are pseudoconvex (this is
because, in the Kähler setting, the Levi form is the $J$-averaging of the
second fundamental form, restricted to the distribution perpendiculat to the
normal $\hat{n}$ and $J\hat{n}$). Label these separating submanifolds
$V_{1}^{3},V_{2}^{3},\dots$; there might be finitely or infinitely many of
these. Each $V_{i}^{3}$ separates the manifold into two pieces: a manifold end
(with a single boundary component, which is pseudo-concave), and a manifold
with a new pseudoconvex boundary component. The boundary components
$V_{i}^{3}$ are each quotients of spheres, so have finite fundmanental group.
If there are two such ends, then a component of
$M^{4}\setminus(V_{1}^{3}\cup{}V_{2}^{3})$ has two pseudoconvex compact
boundary components with finite fundamental group, in violation of Theorem
1.4. Therefore at most one such end exists, so $k+l$ is zero or one.

Next we suppose $(M^{4},J,g)$ has an embedded holomorphic $\mathbb{P}^{1}$
submanifold with non-negative self-intersection. By Lemma 4.1 there is a
neighborhood $\Omega$ so that $M^{4}\setminus\Omega$ has a compact,
pseudoconvex boundary. But then if $M^{4}$ has an ALE or ALF end, we can
separate that end, and have a compact pseudoconvex boundary remaining, that is
topologically either $\mathbb{S}^{3}$ or a Lens space so has finite fundmantal
group. There are now two compact pseudoconvex boundary components and one has
finite fundamental group, contradicting Theorem 1.4.

Lastly if $k+l=1$, $\pi_{1}(M^{4})$ is finite by Lemma 1.2. This establishes
the corollary.

##  5 Examples

###  5.1 A Kähler surface with non-intersecting Levi-flat submanifolds

A Kähler $(M^{4},J,g)$ cannot have Levi-flat submanifolds $V_{0}^{3}$,
$V_{1}^{3}$ with $V_{0}^{3}\cap{}V_{0}^{3}=\varnothing$ when either
submanifold has finite fundamental group (by Theorem 1.3). We give an example
showing that if both submanifold have infinite fundmantal group, then two such
submanifolds can exist.

Let $M^{4}$ be the complex surface $\mathbb{P}^{1}\times\mathbb{P}^{1}$.
Giving this the usual product metric, this is a Kähler manifold. Pick two
points $p,q\in\mathbb{P}^{1}$, and consider the complex submanifolds
$\\{p\\}\times\mathbb{P}^{1}$ and $\\{q\\}\times\mathbb{P}^{1}$. Expand these
into small neighborhoods

$\displaystyle\Omega_{\epsilon}\;=\;D_{\epsilon}(p)\times\mathbb{P}^{1},\quad\Omega_{\epsilon}^{\prime}\;=\;D_{\epsilon}(q)\times\mathbb{P}^{1}$
(18)

where the disks are defined as follows: Let
$f:\mathbb{P}^{1}\setminus\\{p,q\\}\rightarrow\mathbb{C}$ be the harmonic
function with a simple pole of $-\infty$ at $p$ and a simple pole of $+\infty$
at $q$. Then define $D_{\epsilon}(p)=\\{f<\log(\epsilon)\\}$ and
$D_{\epsilon}(q)=\\{f>-\log(\epsilon)\\}$.

Since $f$, interpreted as an extended real-valued function
$f:\mathbb{P}^{1}\times\mathbb{P}^{1}\rightarrow\mathbb{R}\cup\\{-\infty,\infty\\}$
is harmonic on one factor and constant on the other, it is pluriharmonic.
Consequently the boundaries $\partial\Omega_{\epsilon}=\\{f=\log\epsilon\\}$
and $\partial\Omega_{\epsilon}^{\prime}=\\{f=-\log\epsilon\\}$ are both Levi-
flat. For all sufficiently small $\epsilon$ these neighborhoods are non-
intersecting. They are both topologically equivalent to
$\mathbb{S}^{2}\times\mathbb{S}^{1}$, so have infinite fundamental group.

###  5.2 Levi-flat surfaces in $\mathbb{P}^{2}\\#\overline{\mathbb{P}}{}^{2}$

In addition to the example in $\mathbb{P}^{1}\times\mathbb{P}^{1}$, we give
examples of Levi-flats within
$M^{4}=\mathbb{P}^{2}\\#\overline{\mathbb{P}}{}^{2}$. This is a rational
surface, and has holomorphic $\mathbb{P}^{1}$ submanifolds of self-
intersection $+1$.

In $\mathbb{C}^{2}=\\{(z_{1},z_{2})\\}$ an example of a Levi-flat hypersurface
is $\\{Im(z_{1})=\alpha\\}$ where $\alpha$ is any fixed real number. Embedding
this into $\mathbb{P}^{2}$ by $(z_{1},z_{2})\mapsto[1;z_{1};z_{2}]$ and then
taking the closure produces a closed Levi-flat hypersurface $V^{3}_{\alpha}$,
except that $V_{\alpha}^{3}$ has a singular point which lies on the “sphere at
infinity” $\mathbb{P}^{1}=[0;z_{1};z_{2}]$. We examine how $V_{\alpha}^{3}$
interacts with the curves of positive self-intersection. One easily checks
that the $\mathbb{P}^{1}$ at infinity entirely lies within $V^{3}_{\alpha}$,
which also contains the singular point. The other rational curves of positive
self-intersection are (closures in $\mathbb{P}^{2}$ of) the 1-dimensional
subspaces within $[1;z_{1};z_{2}]$. Each of these intersects $V_{\alpha}^{3}$
either in a linear subspace and lies completely within $V_{\alpha}^{3}$ (these
are the various $\mathbb{P}^{1}$ given by $z_{1}=const$ where
$Im(const)=\alpha$), or in a circle (these are the various $\mathbb{P}^{1}$
given by the linear subspaces of $\mathbb{C}^{2}$ not of the form
$z_{1}=const$), or only at its singular point (these are the various
$\mathbb{P}^{1}$ given by any subspace of the form $z_{1}=const$ where
$Im(const)\neq\alpha$). The space $V_{\alpha}^{3}$ is foliated by the complex
lines $z_{1}=const$, $Im(const)=\alpha$, and each of these intersects the
singular point.

Next we look at the singular point more closely. As usual there are three
natural charts for $\mathbb{P}^{2}$. We defined $V_{\alpha}^{3}$ in the chart
$(z_{1},z_{2})\mapsto[1;z_{1};z_{2}]$ with $V_{\alpha}^{3}$ being
$\\{[1,\alpha^{\prime}+\sqrt{-1}\alpha;z_{2}]\\}$,
$\alpha^{\prime}\in\mathbb{R}$, $z_{2}\in\mathbb{C}$. In the chart
$[z_{0};1;z_{2}]$ the surface $V_{\alpha}^{3}$ appears as
$Im(1/z_{0})=\alpha$, which remains non-singular and in this chart is a copy
of $\mathbb{S}^{1}\times\mathbb{R}^{2}$ (except for the case $\alpha=0$).
Finally in the chart $[z_{0};z_{1};1]$ the hypersurface $V_{\alpha}^{3}$
appears as $Im(z_{1}/z_{0})=\alpha$, which is a complex cone at $(0,0)$. Since
$z_{1}/z_{0}$ is invariant under simultaneous multiplication of both
variables, we see a foliation of this cone by complex lines. The lines
themselves are parameterized by $\mathbb{S}^{1}$: this is because with
$Im(z_{1}/z_{0})$ fixed, the complex line on which $(z_{0},z_{1})$ lies is
determined by
$Re(z_{1}/z_{0})\in\mathbb{R}\cup\\{\infty\\}\approx\mathbb{S}^{1}$.

From the singular surface $V_{\alpha}^{3}$ we create a smooth Levi-flat
hypersurface which is diffeomorphic to $\mathbb{S}^{2}\times\mathbb{S}^{1}$.
Blow up $\mathbb{P}^{2}$ at the singular point of $V_{\alpha}^{3}$. In
particular, this desingularizes $V_{\alpha}^{3}$, as its cone point was
removed, and replacing it with a copy of $\mathbb{S}^{1}$ (due to the fact
that the complex lines of $V_{\alpha}^{3}$ passing through the cone point were
paramterized by $\mathbb{S}^{1}$). One can check on charts that the new
hypersurface, which is now within
$\mathbb{P}^{2}\\#\overline{\mathbb{P}}^{2}$, is smooth. Let
$V_{\alpha}^{3}{}^{*}$ be this new, non-singular surface.

As we have seen, the intersection of the desingularized surface
$V_{\alpha}^{3}{}^{*}$ with the exceptional divisor is a circle. On
$V^{3}_{\alpha}{}^{*}$ itself, the desingularization can be seen by removing a
neighborhood of the singular point—the boundary of this neighborhood is a
torus—and then gluing in a solid torus whose central circle is the
intersection locus with the exceptional divisor. The boundary torus is glued
by a torus automorphism—the automorphism exchanges a diagonal circle with a
contractible meridian and vice-versa—and so the desingularized surface has
infinite fundamental group. This surface continues to intersect every rational
curve of positive self-intersection. It is now foliated by curves of zero
self-intersection.

The original singular hypersurfaces $V_{\alpha}^{3}\subset\mathbb{P}^{2}$,
which are simply connected, are themselves of interest. These all intersect
every $\mathbb{P}^{1}$ of positive self-intersection, so each $V_{\alpha}^{3}$
obeys Corollary 1.6, even though they are not $C^{2}$. The singular
hypersurfaces $V_{\alpha}^{3}$ all intersect one another (along the
$\mathbb{P}^{1}$ at infinity), so the conclusion of Theorem 1.3 also holds for
the surfaces $V_{\alpha}^{3}$.

In the future, one would like to extend some version of Theorem 1.3 to the
case of potentially singular Levi-flat surfaces, as singular Levi-flats seem
to be plentiful. A number of technical hurdles stand in the way, such as
questions of internal topology, possible nodes or self-intersections, how the
Levi-flat condition interacts with or persists across the singularity, and
points of contact between submanifolds possibly being at singular points.

See [15] for additional examples of Levi-flat submanifolds in ruled surfaces.

###  5.3 A scalar-flat 2-ended Kähler instanton as a Taub-NUT conformal
transform

Our final example shows that a Kähler 4-manifold, even a scalar-flat Kähler
4-manifold, can have two ends even if one end is ALE.

The classic Euclidean Taub-NUT metric of Hawking [5] is a startling example in
a number of ways, one of these ways being that although Kähler itself, it is
conformal to two very different Kähler metrics. Normally it is impossible that
a Kähler metric in dimension 4 be conformal to another Kähler metric, but here
it is possible due to the unusually large number of complex structures the
Taub-NUT metric is compatible with. The classic metric is

$g\;=\;\frac{1}{4}\frac{r+m}{r-m}dr^{2}+4m^{2}\frac{r-m}{r+m}(\sigma^{1})^{2}+(r^{2}-m^{2})\big{(}(\sigma^{2})^{2}+(\sigma^{3})^{2}\big{)}$
(19)

on $r\in[m,\infty)$ where $m$ is a constant, and $\sigma^{1}$, $\sigma^{2}$,
$\sigma^{3}$ are the usual left-invariant 1-forms on $\mathbb{S}^{3}$
normalized in the usual way so
$d\sigma^{i}=-\epsilon^{i}{}_{jk}\sigma^{j}\wedge\sigma^{k}$. It is well known
that this metric is hyperKähler. It is half-conformally flat, but in the usual
orientation $\\{dr,\sigma^{1},\sigma^{2},\sigma^{3}\\}$ the “wrong” half:
$s=0$ and $Ric=0$ but

$W^{+}\;=\;\frac{8m}{(r+m)^{3}}\left(\frac{3}{2}\omega\otimes\omega-
Id_{\bigwedge^{-}}\right),\quad\quad W^{-}\;=\;0$ (20)

(in those days physicists were seeking “self-dual” metrics). The hyperKähler
structure is carried by three sections
$\omega_{1},\omega_{2},\omega_{3}\in\bigwedge^{-}$ which dualize to compatible
complex structures $I_{1}$, $I_{2}$, $I_{3}$ that obey the quaternionic
relations (and give the orientation opposite to
$\\{dr,\sigma^{1},\sigma^{2},\sigma^{3}\\}$).

However the metric (20) is compatible with two additional complex structures,
not related to $I_{1}$, $I_{2}$, or $I_{3}$, which we label $J^{+}$ and
$J^{-}$. Letting $e_{1},e_{2},e_{3}$ be the left-invariant fields on
$\mathbb{S}^{3}$ dual to the forms $\sigma^{i}$, and given any function
$f=f(r)$, let $J_{f}$ be the almost complex structure

$J_{f}\;=\;-f\frac{\partial}{\partial{}r}\otimes\sigma^{1}+\frac{1}{f}e_{1}\otimes{}dr-
e_{2}\otimes\sigma^{3}+e_{3}\otimes\sigma^{2}.$ (21)

In fact this is always integrable. To see this, note that

$\bigwedge{}^{0,1}=span_{\mathbb{C}}\left\\{\frac{1}{f}dr-\sqrt{-1}\sigma^{1},\;\sigma^{2}-\sqrt{-1}\sigma^{3}\right\\},$
(22)

and then on the basis we compute

$\displaystyle
d\left(\frac{1}{f}dr-\sqrt{-1}\sigma^{1}\right)=2\sqrt{-1}\sigma^{2}\wedge\sigma^{3}=-2\sigma^{2}\wedge\left(\sigma^{2}-\sqrt{-1}\sigma^{3}\right)$
(23) $\displaystyle
d\left(\sigma^{2}-\sqrt{-1}\sigma^{3}\right)\;=\;2\sigma^{1}\wedge\sigma^{3}+2\sqrt{-1}\sigma^{1}\wedge\sigma^{2}\;=\;2\sqrt{-1}\sigma^{1}\wedge\left(\sigma^{2}-\sqrt{-1}\sigma^{3}\right).$

Therefore
$d\left(\bigwedge^{0,1}\right)\subset\bigwedge^{1}\wedge\bigwedge^{0,1}$ so
$J_{f}$ is integrable. Then two complex structures compatible with $g$ are

$\displaystyle
J^{+}\;=\;-4m\frac{r-m}{r+m}\frac{\partial}{\partial{}r}\otimes\sigma^{1}+\frac{1}{4m}\frac{r+m}{r-m}e_{1}\otimes{}dr-
e_{2}\otimes\sigma^{3}+e_{3}\otimes\sigma^{2}$ (24) $\displaystyle
J^{-}\;=\;\;\;\,4m\frac{r-m}{r+m}\frac{\partial}{\partial{}r}\otimes\sigma^{1}-\frac{1}{4m}\frac{r+m}{r-m}e_{1}\otimes{}dr-
e_{2}\otimes\sigma^{3}+e_{3}\otimes\sigma^{2}.$

Their corresponding $(1,1)$ forms are

$\displaystyle
g\left(J^{+}\cdot,\,\cdot\right)\;=\;\;\;\,m\,dr\wedge\sigma^{1}+(r^{2}-m^{2})\sigma^{2}\wedge\sigma^{3}\;\in\;\bigwedge{}^{+},$
(25) $\displaystyle
g\left(J^{-}\cdot,\,\cdot\right)\;=\;-m\,dr\wedge\sigma^{1}+(r^{2}-m^{2})\sigma^{2}\wedge\sigma^{3}\;\in\;\bigwedge{}^{-}.$

Using the facts that $d\sigma^{1}=-2\sigma^{2}\wedge\sigma^{3}$ and
$d(\sigma^{2}\wedge\sigma^{3})=0$ we easily compute

$d\left(g\left(J^{\pm}\cdot,\,\cdot\right)\right)\;=\;d\big{(}\log(r\mp{}m)^{2}\big{)}\wedge{}g\left(J^{\pm}\cdot,\,\cdot\right),$
(26)

so neither form is Kähler, but both are conformally Kähler. Making the
conformal change with factors $(r\mp{}m)^{2}$ gives the two new metrics

$g^{\pm}\;=\;(r\mp{}m)^{-2}g.$ (27)

The corresponding forms $\omega^{\pm}=g^{\pm}(J^{\pm}\cdot,\cdot)$ are now
both closed, so we have two new Kähler metrics $g^{+}$ and $g^{-}$. Both are
conformally related to $g$ so they retain half-conformal flatness. In
particular they are Bach-flat, so by Proposition 4 part (v) of [4], they are
both extremal Kähler metrics. The pair of Kähler structures
$\\{(g^{+},J^{+},\omega^{+}),(g^{-},J^{-},\omega^{-})\\}$ on $M^{4}$ is an
ambiKähler structure; see [2].

The metric $g^{+}$ has Kähler form $\omega^{+}\in\bigwedge{}^{+}$ and by
Derdzinski’s Theorem, Proposition 2 of [4], we have that
$W^{+}=\frac{s}{12}\left(\frac{3}{2}\omega^{+}\otimes\omega^{+}-Id_{\bigwedge^{+}}\right)$.
Therefore (20) and the conformal invariance of $W^{+}$ shows that
$s_{g^{+}}=\frac{96m}{r+m}$ and we see this metric is strictly extremal
(meaning it is extremal but its scalar curvature is not constant). Clearly
$g^{+}$ is one-ended, as the conformal factor $(r+m)^{-2}$ is non-singular.
One easily checks $g^{+}$ is complete. However, unlike the Taub-NUT, it is not
ALF. Asymptotically it is cusp-like:
$g^{+}\approx\frac{1}{4}r^{-2}(dr^{2}+(4m)^{2}(\sigma^{1})^{2})+(\sigma^{2})^{2}+(\sigma^{3})^{2}$,
which as $r\rightarrow\infty$ approaches the metric product of a pseudo-sphere
times a sphere (locally, not globally).

The $g^{-}$ metric is more interesting to us, as it is two-ended. It has
Kähler form $\omega^{-}\in\bigwedge^{-}$. Since $W^{-}=0$, Derdzinski’s
theorem states $g^{-}$ is scalar-flat. Alternatively, one could use the
conformal change formulas to compute $s=0$. In addition to $s=0$ and
$W^{-}=0$, the conformal change formulas show

$\displaystyle
Ric_{g^{-}}=4\left(\frac{r-m}{r+m}\right)^{2}\left(-(\eta^{0})^{2}-(\eta^{1})^{2}+(\eta^{2})^{2}+(\eta^{3})^{2}\right)$
(28) $\displaystyle
W^{+}_{g^{-}}=8\frac{(r-m)^{2}}{(r+m)^{3}}\left(\frac{3}{2}|\omega|_{g^{-}}^{-2}\omega\otimes\omega-
Id_{\bigwedge^{+}}\right)$

where $\eta^{0}=|dr|^{-1}_{g^{-}}dr$ and
$\eta^{i}=|\sigma^{i}|^{-1}_{g^{-}}\sigma^{i}$ are unit 1-forms. In
particular, all curvature components are uniformly bounded. Notice that the
Ricci curvature does not decay to 0 along the cusp-like end (as
$r\rightarrow\infty$), but does decay to zero along the AE end (as
$r\searrow{}m$).

However $g^{-}$ is two-ended, as the conformal factor $(r-m)^{-2}$ is singular
at the “nut” at $r=m$. In dimension 4, it is well-known that an inverse-
quadratic conformal factor removes a point and replaces it with an
asymptotically Euclidean end; see Theorem 6.5 of [10] for example. In addition
to this new AE end that replaced the nut, there remains the end at $r=\infty$.
This end, in the $g^{-}$ metric, is asymptotically identical to the $g^{+}$
end: cusp-like, in the sense that it is (locally) very close to the metric
product of the sphere with the psuedosphere.

Thus the metric $g^{-}$, conformally related to the Taub-NUT metric, is a
scalar-flat Kähler 2-ended metric, with one asymptotically Euclidean end and
one cusp-like end.

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