# Growth of the Higgs Field for Kapustin-Witten solutions
on ALE and ALF gravitational instantons

theoremdefinition theorem propositiondefinition proposition
corollarydefinition corollary lemmadefinition lemma paragraph section

Growth of the Higgs Field for Kapustin-Witten solutions
on ALE and ALF gravitational instantons

Michael<EMAIL_ADDRESS>
Institute for Mathematics, Heidelberg University, Im Neuenheimer Feld 205,
Heidelberg, Germany.

### Abstract.

The $\theta$-Kapustin-Witten equations are a family of equations for a
connection $A$ on a principal $G$-bundle $E\to W^{4}$ and a one-form $\phi$,
called the Higgs field, with values in the adjoint bundle
$\operatorname{ad}E$. They give rise to second-order partial differential
equations that can be studied more generally on Riemannian manifolds $W^{n}$
of dimension $n$. For $G=SU(2)$, we report a dichotomy that is satisfied by
solutions of the second-order equations on Ricci-flat ALX spaces with
sectional curvature bounded from below. This dichotomy was originally
established by Taubes for $W^{n}=\mathbb{R}^{n}$; the alternatives are: either
the asymptotic growth of the averaged norm of the Higgs field over geodesic
spheres is larger than a positive power of the radius, or the commutator
$[\phi\wedge\phi]$ vanishes everywhere. As a consequence, we are able to
confirm a conjecture by Nagy and Oliveira, namely, that finite energy
solutions of the $\theta$-Kapustin-Witten equations on ALE and ALF
gravitational instantons with $\theta\neq 0$ are such that
$[\phi\wedge\phi]=0$, $\nabla^{A}\phi=0$, and $A$ is flat.

##  1 Introduction

Let $G=SU(2)$ and consider a principal $G$-bundle $E$ over a complete
Riemannian manifold $(W^{n},g)$ of dimenson $n$. Throughout, we assume that
$W^{n}$ is an ALX manifold. Suffice it to say for now that we take this to
mean $W^{n}$ is a non-compact manifold with fibered ends such that the
$k$-dimensional fibers have bounded volume. Consequently, the volume of
geodesic balls asymptotically grows like $r^{n-k}$.

Denote by
$(A,\phi)\in\mathcal{A}(E)\times\Omega^{1}(W^{n},\operatorname{ad}E)$ a pair
consisting of a connection on $E$ and an $\operatorname{ad}E$-valued one-form.
We write $\star$ for the Hodge star operator and equip
$\Omega^{k}(W^{n},\operatorname{ad}E)$ with the density-valued inner product
${\langle a,b\rangle=\operatorname{Tr}a\wedge\star b}$. Upon integration this
provides the usual $L^{2}$-product $\langle
a,b\rangle_{L^{2}(W)}=\int_{W^{n}}\langle a,b\rangle$ on
$\Omega^{k}(W^{n},\operatorname{ad}E)$. Throughout, we assume that $A$ and
$\phi$ have enough derivatives and are locally square-integrable.

In this article we report on a property of the pair $(A,\phi)$ whenever it
satisfies the following second order differential equation.

$\displaystyle{\nabla^{A}}^{\dagger}\nabla^{A}\phi+\frac{1}{2}\star[\star[\phi\wedge\phi]\wedge\phi]+\operatorname{Ric}\phi=0\
.$ (1)

Here ${\nabla^{A}}^{\dagger}$ is the formal adjoint of $\nabla^{A}$ with
respect to the $L^{2}$-product and the Ricci curvature is viewed as an
endomorphism of $\Omega^{1}(W,\operatorname{ad}E)$.

The differential equation (1) is of particular relevance in the context of the
Kapustin-Witten equations. To see this, consider for the moment the case of a
four-manifold $W^{4}$ and define the Laplace-type differential operator on
$\Omega^{1}(W,\operatorname{ad}E)$:

$\displaystyle\tilde{\Delta}_{A}(\phi)=-d_{A}d_{A}^{\star}\phi+\star
2d_{A}(d_{A}\phi)^{-}\ ,$ (2)

where $d_{A}^{\star}=\star d_{A}\star$ is the usual codifferential and
$(\cdot)^{\pm}$ denotes the (anti-)self-dual part of a given two-form. Compare
this operator with the $\theta$-Kapustin-Witten equations for $(A,\phi)$,
which are given by

$\displaystyle\big{(}\cos\tfrac{\theta}{2}\
(F_{A}-\tfrac{1}{2}[\phi\wedge\phi])-\sin\tfrac{\theta}{2}\ d_{A}\phi\
\big{)}^{+}$ $\displaystyle=0$ (3) $\displaystyle\big{(}\sin\tfrac{\theta}{2}\
(F_{A}-\tfrac{1}{2}[\phi\wedge\phi])+\cos\tfrac{\theta}{2}\ d_{A}\phi\
\big{)}^{-}$ $\displaystyle=0$ (4) $\displaystyle d_{A}^{\star}\phi\ $
$\displaystyle=0\ .$ (5)

Clearly, if $(A,\phi)$ is a solution of the $\theta=0$ version of the
Kapustin-Witten equations, then $\tilde{\Delta}_{A}\phi=0$. Moreover, using a
Bochner-Weitzenböck identity that relates $\tilde{\Delta}_{A}$ and the Bochner
Laplacian ${\nabla^{A}}^{\dagger}\nabla^{A}$, as well as the remaining part of
the $0$-Kapustin-Witten equations $F_{A}^{+}=[\phi\wedge\phi]^{+}$, one finds
that harmonicity of $\phi$ with respect to $\tilde{\Delta}_{A}$ is equivalent
to equation (1). In fact a very similar argument shows that the same is true
if $(A,\phi)$ is a solution of the $\theta$-Kapustin-Witten equations [Tau13,
Tau17, NO21].

The Kapustin-Witten equations arise from a family of topologically twisted,
four-dimensional, supersymmetric gauge theories that exhibit surprisingly deep
connections to several complementary areas of mathematics. They were first
studied by Kapustin and Witten in the context of the geometric Langlands
program [KW07]. Witten later realized that solutions of the Kapustin-Witten
equations are also related to topological invariants of knots, specifically to
Khovanov homology and its generalizations [Wit11, GW12]. Since then the moduli
space of solutions to the Kapustin-Witten equations has been subject to
extensive study [Tau13, Tau17, Tau18, He15, He19, HM19, Tan19, Tau19, Tau21].

######  Remark.

Note that we use a slightly different normalization than is otherwise common
in the literature: our $\theta$ coincides with $2\theta_{GU}$ in [GU12].
Though this is mostly a matter of taste, when viewed as dimensional reductions
of the five-dimensional Haydys-Witten equations, the normalization used here
has a geometric interpretation as incidence angle between Haydys’ preferred
vector field and the hyperplane orthogonal to the reduction.

Let us now return to general $n$-manifolds. In what follows, we are guided by
the intuition that if $\phi$ satisfies (1), then it is harmonic with respect
to some well-behaved Laplace-type operator. In particular, one should expect
that it satisfies an appropriate analogue of the mean-value principle. Hence,
fix some point $p\in W^{4}$ and denote by $B_{r}$ the closed geodesic ball of
radius $r$ centered at $p$. Consider the non-negative function $\kappa$
defined on $[0,\infty)$ by

$\displaystyle\kappa^{2}(r)=\frac{1}{r^{n-k-1}}\int_{\partial
B_{r}}\left\lVert\phi\right\rVert^{2}\ .$ (6)

As a consequence of the asymptotic volume growth on $W^{n}$, $\kappa(r)$ is
related to the average value of $\left\lVert\phi\right\rVert$ on geodesic
spheres $\partial B_{r}$ with large radius $r$. The mean-value principle for
Laplace-type differential operators then suggests that $\phi$ should satisfy
an inequality of the form $\left\lVert\phi(p)\right\rVert\leq\kappa(r)$ for
$r>0$. Although contributions from non-trivial curvature in the interior of
$B_{r}$ in general preclude this naive mean-value inequality, the controlled
asymptotics of ALX spaces retains enough control to deduce analogous bounds
for points that are far away from $p$.

A classical consequence of the mean-value principle is a relation between the
asymptotic behaviour of $\kappa$ at large radius and the values of $\phi$ in
the interior of the ball. For example, if the naive mean-value inequality was
satisfied at every point $p\in W^{n}$ and $\kappa(r)\to 0$ as $r\to\infty$,
then $\phi$ would be identically zero everywhere.

For $W^{n}=\mathbb{R}^{n}$, a result by Taubes generalizes this kind of
statement to a dichotomy between the growth of $\kappa(r)$ at infinity and the
vanishing of $[\phi\wedge\phi]$ on all of $W^{n}$ [Tau17]. Here we prove that
this dichotomy holds more generally if $W^{n}$ is an ALX gravitational
instanton (this is section 8 below):

######  Theorem A.

Let $W^{n}$ be a complete, Ricci flat ALX manifold of dimension $n\geq 2$ with
asymptotic fibers of dimension $k\leq n-1$ and sectional curvature bounded
from below. Consider $(A,\phi)$ as above and assume the pair satisfies the
second-order differential equation (1). Then either

  1. (i)

there is an $a>0$ such that $\liminf_{r\to\infty}\frac{\kappa(r)}{r^{a}}>0$,
or

  2. (ii)

$[\phi\wedge\phi]=0$.

If the fields $(A,\phi)$ are solutions of the $\theta$-Kapustin-Witten
equations and have square-integrable field strength we can say slightly more
(cf. section 9).

######  Theorem B.

Let $W^{4}$ be a complete, Ricci flat ALX manifold of dimension $4$ with
asymptotic fibers of dimension $k\leq 3$ and sectional curvature bounded from
below. Assume $(A,\phi)$ are solutions of the $\theta$-Kapustin-Witten
equations and satisfy $\int_{W^{4}}\left\lVert F_{A}\right\rVert^{2}<\infty$,
then either

  1. (i)

there is an $a>0$ such that $\liminf_{r\to\infty}\frac{\kappa(r)}{r^{a}}>0$,
or

  2. (ii)

$[\phi\wedge\phi]=0$, $\nabla^{A}\phi=0$, and $A$ is self-dual if $\theta=0$,
flat if $\theta\in(0,\pi)$, and anti-self-dual if $\theta=\pi$.

As an immediate consequence of Theorem B we are able to confirm a conjecture
of Nagy and Oliveira. For this we introduce the Kapustin-Witten energy

$\displaystyle E_{\text{KW}}=\int_{W^{4}}\left(\left\lVert
F_{A}\right\rVert^{2}+\left\lVert\nabla^{A}\phi\right\rVert^{2}+\left\lVert[\phi\wedge\phi]\right\rVert^{2}\right)\
.$ (7)

######  Corollary (Nagy-Oliveira Conjecture [NO21]).

Let $(A,\phi)$ be a finite energy solution of the $\theta$-Kapustin-Witten
equations with $\theta\neq 0\pmod{\pi}$ on an ALE or ALF gravitational
instanton and let $G=SU(2)$. Then $A$ is flat, $\phi$ is
$\nabla^{A}$-parallel, and $[\phi\wedge\phi]=0$.

###### Proof.

Under the given assumptions, the main result of Nagy and Oliveira [NO21, Main
Theorem 1] states that $\phi$ has bounded norm and thus, in particular,
bounded average over spheres. It follows that
$\liminf_{r\to\infty}\frac{\kappa(r)}{r^{a}}\to 0$ for any $a>0$, while the
finite energy condition subsumes square-integrability of $F_{A}$. Therefore,
Theorem B implies that $[\phi\wedge\phi]=0$, $\nabla^{A}\phi=0$, and that $A$
is flat. ∎

######  Remark.

The preceding argument is due to Nagy and Oliveira, who established the
corollary for $W^{4}=\mathbb{R}^{4}$ and $S^{1}\times\mathbb{R}^{3}$ [NO21,
Corollary 1.3]. Nagy and Oliveira relied on a version of Theorem B that
applies to $W^{4}=\mathbb{R}^{4}$ and was provided by Taubes alongside the
original dichotomy [Tau17]. Their conjecture stemmed from the expectation that
Taubes’ results can be extended to ALX spaces in general.

The main insight of this article is that Taubes’ proof strategy for the
special case $W^{n}=\mathbb{R}^{n}$ carries over to general ALX spaces. This
is a consequence of the well-behaved asymptotic volume growth, where problems
that arise from non-zero curvature in the interior can be excised. Hence, the
proof of Theorem A closely follows the one provided by Taubes in [Tau17].

We proceed as follows: In section 2 we collect the relevant definitions and
recall several classical results that will be used throughout. Then, in
section 3, we investigate the derivative of $\kappa$ and introduce the
relevant analogue of Almgren’s frequency function, as well as a function that
captures contributions from the mean curvature of the geodesic sphere. The key
finding of that section is that $\kappa$ is asymptotically almost non-
decreasing, which is a prerequisite for most of the heavy lifting in
subsequent sections. In section 4, we present a somewhat unusual version of
unique continuation that is satisfied by $\kappa$. The main insight is the
content of section 5, where we explain that slow asymptotic growth of $\kappa$
results in bounds for the frequency function. All these results are refined
with respect to the components of the one-form $\phi=\sum_{i}\phi_{i}dx^{i}$
by introducing in section 6 what we call the correlation tensor. Using a
second line of arguments, we also determine a priori bounds of the type
$\left\lVert\phi(x)\right\rVert\leq\kappa(r)$ in section 7, which are the
anticipated analogues of the mean value inequalities mentioned already above.
Finally, all these ingredients are combined into a proof of Theorem A in
section 8, while the proof of Theorem B occupies section 9.

### Acknowledgements

I thank Fabian Hahner for helpful comments on a draft of this article. This
work is funded by the Deutsche Forschungsgemeinschaft (DFG, German Research
Foundation) under Germany’s Excellence Strategy EXC 2181/1 - 390900948 (the
Heidelberg STRUCTURES Excellence Cluster).

##  2 Background

For the purposes of this article, an ALX space is a non-compact complete
Riemannian manifold that asymptotically looks like a fibration of closed
manifolds, where the fibers have bounded volume. This is made precise in the
following definition.

######  Definition 2.1 (ALXk Manifold).

Let $(W^{n},g)$ be a complete Riemannian manifold of dimension $n$ and fix
$p\in W^{n}$. Let $\pi_{Y}:Y^{n-1}\to B^{n-k-1}$ be a fibration over an
$n-k-1$-dimensional closed Riemannian base $(B,g_{B})$ with $k$-dimensional
closed Riemannian fibers $(X,g_{X})$. Equip $(0,\infty)_{r}\times Y$ with the
model metric ${g_{\infty}=dr^{2}+g_{X}+r^{2}g_{B}}$. We say $W^{n}$ is an ALXk
manifold if its end is modeled on $(0,\infty)\times Y$, that is, if there
exists $R>0$ such that there is a diffeomorphism $\varphi:W^{n}\setminus
B_{R}(p)\to(R,\infty)\times Y^{n-1}$ that satisfies for $j=0,1$, and $2$

$\displaystyle\lim_{r\to\infty}r^{j}\left\lVert(\nabla^{\text{LC}})^{j}\left(g-\varphi^{\ast}g_{\infty}\right)\right\rVert_{L^{\infty}(\partial
B_{r})}=0\ .$ (8)

######  Proposition .

If $(W^{n},g)$ is an ALXk space then

$\displaystyle\operatorname{vol}B_{r}(p)\sim
r^{n-k}\operatorname{vol}X\quad(r\to\infty)\ .$ (9)

######  Definition 2.2.

We call an ALX space $W^{n}$ a _gravitational instanton_ if it is Ricci flat
and its sectional curvature is bounded from below.

######  Remarks.

  * •

ALX spaces are usually considered in the context of four-manifolds and the “X”
is a place-holder for the following cases: ALE or Asymptotically Locally
Euclidean ($k=0$), ALF or Asymptotically Locally Flat ($k=1$), ALG ($k=2$),
ALH ($k=3$), where the last two are named by induction.

  * •

We do not demand that ALX gravitational instantons are hyper-Kähler, while we
add the possibly non-standard condition of bounded sectional curvature.

######  Examples.

  * •

The prototypical example of an ALE manifold is Euclidean space
$\mathbb{R}^{n}$. In this case there is the obvious diffeomorphism
$\mathbb{R}^{n}\setminus\\{0\\}\simeq(0,\infty)\times S^{n-1}$ via spherical
coordinates, the metric is
$\varphi_{\ast}g=dr^{2}+r^{2}g_{S^{n-1}}=g_{\infty}$, and the volume of balls
grows with the radius as $r^{n}$.

  * •

A prototypical example of an ALF space is $S^{1}\times\mathbb{R}^{n-1}$ with
the product metric $g=dt^{2}+g_{\mathbb{R}^{n-1}}$. Again, spherical
coordinates on the $\mathbb{R}^{n-1}$ factor provide a diffeomorphism to
$(0,\infty)\times S^{1}\times S^{n-2}$ with metric
$\varphi_{\ast}g=dr^{2}+dt^{2}+r^{2}g_{S^{n-2}_{r}}$, such that $S^{1}$ has
constant size, while $S^{n-2}_{r}$ is the sphere of radius $r$ centered at
$0\in\mathbb{R}^{n-1}$. Once the volume of $S^{1}$ is “filled”, the volume of
a geodesic ball approaches a growth of order $r^{n-1}$.

  * •

Famous examples of less trivial four-dimensional ALF gravitational instantons
are (multi-centered) Taub-NUT spaces. These are $S^{1}$-fibrations over
$\mathbb{R}^{3}$, where the $S^{1}$-fiber has asymptotically finite volume.

######  Theorem  (Global Laplacian Comparison Theorem [Cal58]).

If $\operatorname{Ric}\geq(n-1)K$ and $r(x)=d(p,x)$ denotes the geodesic
distance function based at a point $p$, then

$\displaystyle\Delta r\leq(n-1)\Delta_{K}r\ ,$ (10)

where on the right hand side $\Delta_{K}$ is the Laplacian on the unique
complete, $n$-dimensional, simply connected space of constant sectional
curvature $K$.

######  Remark.

Since $r$ is not necessarily differentiable the global Laplacian comparison
must be understood in a weak sense, e.g. in the weak sense of barriers as in
the work of Calabi [Cal58]. However, for our purposes it is sufficient to
consider the smooth locus of $r$, where the inequality holds as stated.

######  Proposition  (Mean Curvature Comparison on ALX spaces).

Let $W^{n}$ be an ALXk space and fix a point $p\in W^{n}$. The Laplacian of
the distance function $r(x)=d(p,x)$, or equivalently the mean curvature of the
geodesic sphere of radius $r$ based at $x$, has the following asymptotic
behaviour.

$\displaystyle\Delta r\sim\left\\{\quad\begin{matrix}\frac{n-1}{r}&(r\to
0)\\\\[10.00002pt] \frac{n-k-1}{r}&(r\to\infty)\end{matrix}\right.$ (11)

Furthermore, if $\operatorname{Ric}\geq 0$, then it is bounded from above by

$\displaystyle\Delta r\leq\frac{n-1}{r}\ .$ (12)

###### Proof.

For a start, note that $r(x)$ is smooth on
$M\setminus\\{p,\mathrm{Cut}(p)\\}$, where $\mathrm{Cut}(p)$ is the cut locus
of $p$. It is a standard result that the cut locus on a complete Riemannian
manifold has measure zero, so $r$ is differentiable almost everywhere. The
Gauss lemma tells us that $\nabla^{\mathrm{LC}}r=\partial_{r}$ is the radial
vector field of unit norm and is normal to geodesic spheres. As an aside, note
that $\Delta r=\operatorname{tr}(\nabla^{\mathrm{LC}})^{2}r$ is the trace of
the second fundamental form of the geodesic sphere and as such is identical to
its mean curvature.

The asymptotic behaviour for $r\to 0$ follows e.g. by a direct calculation in
Riemann normal coordinates. In particular, use
$g_{ij}=\delta_{ij}+\mathcal{O}(r^{2})$ and $\Gamma^{i}_{jk}=\mathcal{O}(r)$
and then observe that at leading order the result is identical to the
Euclidean case, while higher order corrections are $\mathcal{O}(r)$:

$\displaystyle\Delta r=\frac{n-1}{r}+\mathcal{O}(r)\ .$ (13)

When $r\to\infty$ the ALXk condition (8) implies that $\Delta
r\sim\Delta_{\infty}r$, where $\Delta_{\infty}$ denotes the Laplacian
associated to $g_{\infty}$ on $(0,\infty)\times Y^{n-1}$. Under the
diffeomorphism to $(0,\infty)\times Y^{n-1}$ the distance function is
identified with the coordinate on the first factor. Since the model metric is
block diagonal and only depends on $r$ via the $r^{2}$ factor in front of
$g_{B}$, we can calculate $\Delta_{g_{\infty}}r$ explicitly. Let
$(e_{i})_{i=1,\ldots,n}$ be an orthonormal frame of $(0,\infty)\times Y$ such
that $e_{1}=\partial_{r}$, $e_{2},\ldots,e_{k+1}$ are tangent to the fibers,
and $e_{k+2},\ldots,e_{n}$ are tangent to the base. Write $\nabla$ for the
Levi-Civita connection associated to $g_{\infty}$. By a direct calculation
$\nabla_{e_{i}}\partial_{r}=\frac{1}{r}e_{i}$ for $i=k+2,\ldots,n-1$ and zero
otherwise. Hence,

$\displaystyle\Delta_{g_{\infty}}r=\operatorname{tr}\nabla^{2}r=g^{ij}g(\nabla_{e_{i}}\partial_{r\
},e_{j})=\frac{\operatorname{tr}g_{B}}{r}=\frac{n-k-1}{r}\ .$ (14)

Finally, the upper bound in the case that $\operatorname{Ric}$ is non-negative
follows directly from the Laplacian Comparison Theorem (section 2). ∎

######  Theorem  (Bishop-Gromov’s Volume Comparison).

Let $(M,g)$ be a complete Riemannian manifold and assume
$\operatorname{Ric}\geq(n-1)K$. Denote by $\operatorname{vol}B_{r}(p)$ the
volume of the geodesic ball of radius $r$ based at $p\in M$. Similarly write
$\operatorname{vol}_{K}B_{r}(p_{K})$ for the volume of a geodesic ball with
the same radius inside the unique complete, $n$-dimensional, simply connected
space of constant sectional curvature $K$ at an arbitrary point $p_{K}$. Then
the function defined by

$\displaystyle
r\mapsto\frac{\operatorname{vol}B_{r}(p)}{\operatorname{vol}_{K}B_{r}(p_{K})}$
(15)

is non-decreasing and approaches $1$ as $r\to 0$. In particular
$\operatorname{vol}B_{r}(p)\leq\operatorname{vol}_{K}B_{r}(p_{K})$ and
$\operatorname{vol}\partial B_{r}(p)\leq\operatorname{vol}_{K}\partial
B_{r}(p_{K})$.

######  Lemma .

For any point $x$ in the interior of $B_{r}(p)$ there is a smooth, positive
Green’s function $G_{x}$ for the Dirichlet-Laplace problem on $B_{r}(p)$ with
singularity at $x$, i.e. $\Delta G_{x}(y)=\delta_{x}(y)$ and $G_{x}(\partial
B_{r}(p))=0$. If $W^{n}$ is a Ricci non-negative ALXk space with effective
dimension $n-k>2$, then for any $\epsilon>0$ there is a distance $D$ such that
whenever $d(x,y)>D$ the Green’s function is bounded by

$\displaystyle G_{x}(y)\leq\frac{(1+\epsilon)\
c}{\operatorname{vol}X\>d(x,y)^{n-k-2}}\ ,$ (16)

where the constant $c$ depends only on $n$.

###### Proof.

The existence of a positive Green’s function on compact, connected manifolds
with boundary is standard. The bound follows immediately from Theorem 5.2 in
Li-Yau’s seminal work [LY86]. Their theorem states

$\displaystyle G_{x}(y)\leq
c\int_{r^{2}}^{\infty}\frac{1}{\operatorname{vol}B_{\sqrt{t}}(x)}dt\ ,$ (17)

where $r=d(x,y)$ denotes the Riemannian distance between $x$ and $y$ and the
constant $c$ depends only on $n$.

Let $\epsilon>0$. By section 2 there is a distance $R\geq 0$, such that
whenever $r\geq R$ we find

$\displaystyle G_{x}(y)\leq
c\int_{r^{2}}^{\infty}\frac{1+\epsilon}{\operatorname{vol}X\
t^{(n-k)/2}}dt=\frac{(1+\epsilon)\ c}{\operatorname{vol}X\>r^{n-k-2}}\ .$ (18)

∎

##  3 The Frequency Function

On our way to show that $\kappa$ must have some minimal asymptotic growth, the
first step is to realize that its decay rate becomes arbitrarily small at
large radii. To see this we investigate the derivative of $\kappa$, which is
given in the upcoming proposition. The function $N(r)$ that arises in that
context is an analogue of the frequency function as introduced by Almgren
[Alm79] and we will refer to it by that name. The function $D(r)$ captures the
average deviation of the mean curvature of the geodesic sphere from its limit
at infinity.

######  Proposition .

Assume the pair $(A,\phi)$ satisfies (1). Whenever $\kappa$ is non-zero its
derivative is

$\displaystyle\frac{d\kappa}{dr}$
$\displaystyle=\frac{\left(N+D\right)\kappa}{r}\ ,$ (19)

where $N$ and $D$ are given by

$\displaystyle N(r)$
$\displaystyle=\frac{1}{r^{n-k-2}\kappa^{2}}\int_{B_{r}}\left(\left\lVert\nabla^{A}\phi\right\rVert^{2}+\left\lVert[\phi\wedge\phi]\right\rVert^{2}+\langle\operatorname{Ric}\phi,\phi\rangle\right)$
(20) $\displaystyle D(r)$
$\displaystyle=\frac{1}{2r^{n-k-2}\kappa^{2}}\int_{\partial B_{r}}\left(\Delta
r-\frac{n-k-1}{r}\right)\left\lVert\phi\right\rVert^{2}\ .$ (21)

Moreover, if $\operatorname{Ric}\geq 0$, then $N$ is non-negative, $D$ is
bounded from above by $k/2$, $\lim_{r\to 0}D=k/2$ and $\lim_{r\to\infty}D=0$.
As a consequence, if $\kappa$ is not identically zero near $r=0$, then it is
increasing on small enough neighbourhoods of $0$. Similarly, if $\kappa$ is
not asymptotically zero as $r\to\infty$, then it is asymptotically almost non-
decreasing in the sense that for any $\epsilon>0$ there is some (large) radius
$R$, such that $\frac{d\kappa}{dr}\geq-\frac{\epsilon\kappa}{r}$ for all
$r\geq R$.

For notational convenience we will say that $\kappa$ is _$\epsilon$ -almost
non-decreasing_ whenever its derivative is bounded below by
$\frac{d\kappa}{dr}\geq-\frac{\epsilon\kappa}{r}$.

###### Proof.

Denote by $X$ the radial unit vector field on $B_{r}$ and observe that

$\displaystyle\kappa^{2}(r)=\frac{1}{r^{n-k-1}}\int_{\partial
B_{r}}\left\lVert\phi\right\rVert^{2}=\frac{1}{r^{n-k-1}}\int_{B_{r}}\mathcal{L}_{X}\
\left\lVert\phi\right\rVert^{2}\ .$ (22)

By the product and Leibniz’ integral rule the derivative is then given by

$\displaystyle\frac{d}{dr}\kappa^{2}(r)$
$\displaystyle=-\frac{n-k-1}{r}\kappa^{2}+\frac{1}{r^{n-k-1}}\int_{B_{r}}\mathcal{L}_{X}\circ\mathcal{L}_{X}\
\left\lVert\phi\right\rVert^{2}\ .$ (23)

We can write the integral on the right hand side equivalently as an integral
over the trace of the (asymmetric) second Lie derivative
${\mathcal{L}^{2}_{Y,Z}:=\mathcal{L}_{Y}\circ\mathcal{L}_{Z}}$. To see this
denote by $(r,\theta_{i})$ polar normal coordinates on $B_{r}$ and note that
in these coordinates the metric is block-diagonal, i.e.
$g=dr^{2}+g_{S^{n-1}}$. Since for any top-form $\omega$ the pullback of
$\imath_{\partial_{\theta_{i}}}\omega$ to the boundary of the geodesic ball is
zero, one finds

$\displaystyle\int_{B_{r}}\mathcal{L}^{2}_{X,X}\left\lVert\phi\right\rVert^{2}=\int_{B_{r}}(\mathcal{L}_{X,X}^{2}+g_{S^{n-1}}^{ij}\mathcal{L}^{2}_{\partial_{\theta_{i}},\partial_{\theta_{j}}})\left\lVert\phi\right\rVert^{2}=\int_{B_{r}}\operatorname{tr}_{TM}\mathcal{L}^{2}\left\lVert\phi\right\rVert^{2}\
.$ (24)

Next, for any vector field $Y$ and top-form $\omega$ we may express the action
of the Lie derivative in terms of the Levi-Civita connection as
${\mathcal{L}_{Y}\omega=\nabla_{Y}\omega+\operatorname{div}Y\ \omega}$. Using
this we may write the second Lie derivative as

$\displaystyle\mathcal{L}_{Y,Z}^{2}\ \left\lVert\phi\right\rVert^{2}$
$\displaystyle=(\nabla_{Y}+\operatorname{div}Y)\
\nabla_{Z}\left\lVert\phi\right\rVert^{2}+\mathcal{L}_{Y}(\operatorname{div}Z\
\left\lVert\phi\right\rVert^{2})\ .$ (25)

Furthermore, we use ad-invariance and metric compatibility to write
$\nabla_{Y}\langle\phi,\phi\rangle=2\langle\phi,\nabla_{Y}^{A}\phi\rangle$,
and use that the formal adjoint is given by
${\nabla_{Y}^{A}+\operatorname{div}Y=-(\nabla_{Y}^{A})^{\dagger}}$. This leads
to

$\displaystyle\int_{B_{r}}\operatorname{tr}_{TM}\mathcal{L}^{2}\left\lVert\phi\right\rVert^{2}$
$\displaystyle=2\int_{B_{r}}\left(\left\lVert\nabla^{A}\phi\right\rVert^{2}-\langle\phi,{\nabla^{A}}^{\dagger}\nabla^{A}\phi\rangle\right)+\int_{B_{r}}\operatorname{tr}_{TM}\left(\mathcal{L}_{\cdot}\operatorname{div}(\cdot)\
\left\lVert\phi\right\rVert^{2}\right)$ (26)
$\displaystyle=2\int_{B_{r}}\left(\left\lVert\nabla^{A}\phi\right\rVert^{2}+\left\lVert[\phi\wedge\phi]\right\rVert^{2}+\langle\phi,\operatorname{Ric}\phi\rangle\right)+\int_{\partial
B_{r}}\Delta r\ \left\lVert\phi\right\rVert^{2}\ ,$ (27)

where we used the second order differential equation (1) in the first term and
that the only non-zero contribution in the second term contains the mean
curvature of the geodesic sphere since $\operatorname{div}X=\Delta r$.

All in all, as long as $\kappa\neq 0$, the derivative is given by

$\displaystyle\frac{d\kappa}{dr}$
$\displaystyle=\frac{1}{2\kappa}\frac{d}{dr}\kappa^{2}=\frac{1}{\kappa
r^{n-k-1}}\int_{B_{r}}\left(\left\lVert\nabla^{A}\phi\right\rVert^{2}+\left\lVert[\phi\wedge\phi]\right\rVert^{2}+\langle\phi,\operatorname{Ric}\phi\rangle\right)+\frac{1}{2\kappa
r^{n-k-1}}\int_{\partial B_{r}}\left(\Delta
r-\tfrac{n-k-1}{r}\right)\left\lVert\phi\right\rVert^{2}\ ,$ (28)

which upon identifying the terms on the right hand side with $N$ and $D$ is
the desired result.

Now assume $\operatorname{Ric}\geq 0$. On the one hand, $N$ is then clearly
non-negative. On the other hand, the results for $\Delta r$ from section 2
immediately provide both the global upper bound and the limits of $D$.

Combining these facts with the formula for $\frac{d\kappa}{dr}$ leads to the
conclusion that $\kappa$ is (almost) non-decreasing at both ends: Since $D$ is
continuous and $\lim_{r\to 0}D=k/2$, $D$ must be positive on some small
interval $[0,s)$. Thus, if $\kappa$ is non-zero somewhere in that interval
then it is increasing. The asymptotic bound for $r\to\infty$ works out
similarly. In that case there is an interval $[R,\infty)$ for any $\delta>0$
on which

$\displaystyle D\geq-\frac{\delta}{1+\delta}\frac{n-k-1}{2}\ .$ (29)

After a suitable choice of $\delta$ this provides the desired bound
$\frac{d\kappa}{dr}\geq-\frac{\epsilon\kappa}{r}$ for any $\epsilon>0$, which
concludes the proof. ∎

In the preceding proposition we already encountered lower bounds for
$\frac{d\kappa}{dr}$ near $r=0$ and $r\to\infty$. But when we keep track of
$N$ it becomes clear that $\frac{d\kappa}{dr}$ satisfies stronger bounds than
recorded so far. This is the content of the following two corollaries. The
first records a global growth limitation, while the second determines
asymptotic lower and upper bounds, both in dependence of the frequency
function $N$.

######  Corollary .

Assume $\kappa\neq 0$ on $[r_{0},r_{1}]$, then

$\displaystyle\kappa(r_{1})\ \leq\ \kappa(r_{0})\
\exp\int_{r_{0}}^{r_{1}}\frac{N(t)+k/2}{t}dt\ .$ (30)

###### Proof.

Recall from section 3 that $D\leq k/2$, so the derivative of $\kappa$ is
bounded by ${\frac{d\kappa}{dr}\leq\frac{(N+k/2)\kappa}{r}}$. By Grönwall’s
inequality $\kappa$ then can’t become larger than a solution of the underlying
differential equation, which is the stated bound. ∎

######  Corollary .

Let $\epsilon>0$ and $R$ be such that $\left|D\right|\leq\epsilon$ on
$[R,\infty)$. If $\kappa$ has no zeroes in $[r_{0},r_{1}]\subset[R,\infty)$
then it is bounded at $r_{1}$ from both sides as follows

$\displaystyle\kappa(r_{0})\
\exp\int_{r_{0}}^{r_{1}}\frac{N(t)-\epsilon}{t}dt\ \leq\ \kappa(r_{1})\ \leq\
\kappa(r_{0})\ \exp\int_{r_{0}}^{r_{1}}\frac{N(t)+\epsilon}{t}dt\ .$ (31)

Consequently, bounds for the frequency function on $[r_{0},r_{1}]$ have the
following effect:

  * •

if $a\leq N$ then
$\kappa(r_{1})\geq\big{(}\frac{r_{1}}{r_{0}}\big{)}^{a-\epsilon}\kappa(r_{0})$

  * •

if $N\leq b$ then
$\kappa(r_{1})\leq\big{(}\frac{r_{1}}{r_{0}}\big{)}^{b+\epsilon}\kappa(r_{0})$

###### Proof.

Since $\left|D\right|\leq\epsilon$ and $\kappa$ is non-zero on
$[r_{0},r_{1}]$, its derivative is bounded in both directions as follows

$\displaystyle\frac{(N-\epsilon)\
\kappa}{r}\leq\frac{d\kappa}{dr}\leq\frac{(N+\epsilon)\ \kappa}{r}\ .$ (32)

Grönwall’s inequality states that $\kappa$ is then bounded in either direction
by the corresponding solutions of the underlying differential equations, which
is exactly (31).

Clearly, if the frequency function is bounded below by some $a\geq 0$ the
first inequality in (31) reduces to

$\displaystyle\kappa(r_{1})\geq\kappa(r_{0})\
\exp\int_{r_{0}}^{r_{1}}\frac{a-\epsilon}{s}\;ds=\left(\frac{r_{1}}{r_{0}}\right)^{a-\epsilon}\kappa(r_{0})\
.$ (33)

The same argument, but based on the second inequality, yields the
corresponding upper bound if $N$ is bounded from above. ∎

We will later also need the derivative of $N$, which is given directly by the
product and Leibniz’ integral rule.

$\displaystyle\frac{d}{dr}N$ $\displaystyle=\frac{1}{r^{n-k-2}\kappa^{2}}\
\int_{\partial
B_{r}}\bigg{(}\lVert\nabla^{A}\phi\rVert^{2}+\left\lVert[\phi\wedge\phi]\right\rVert^{2}+\langle\operatorname{Ric}\phi,\phi\rangle\bigg{)}-\big{(}n-k-2+2(N+D)\big{)}\frac{N}{r}$
(34)

As an immediate consequence, we see that if $N$ is ever small, then it can’t
have been very much larger at nearby smaller radii $s<r$. This observation is
recorded more precisely in the following proposition.

######  Proposition .

Assume $\operatorname{Ric}\geq 0$. If $N\leq 1$ on some interval
$[r_{0},r_{1}]$, then
$N(r_{0})\leq\Bigl{(}\frac{r_{1}}{r_{0}}\Bigr{)}^{n}N(r_{1})$. Moreover,
whenever $N(r)<1$ at some $r\in(0,\infty)$ then $N\leq 1$ on the interval
$[N(r)^{1/n}\>r,r]$.

###### Proof.

Since $\operatorname{Ric}$ is non-negative the same is true for the first term
in (34). Moreover, in that case $D\leq\frac{k}{2}$ by section 3. Assume now
that $N\leq 1$ on all of $[r_{0},r_{1}]$. Then $N$ satisfies the following
differential inequality for any $r\in[r_{0},r_{1}]$

$\displaystyle\frac{dN}{dr}\geq-\frac{n\>N}{r}\ .$ (35)

Grönwall’s inequality states that then for any pair $s\leq r$ in
$[r_{0},r_{1}]$ the following inequality holds

$\displaystyle N(r)\geq\left(\frac{s}{r}\right)^{n}N(s)\ ,$ (36)

which proves the first part of the statement.

Now assume $N(r)<1$ for some $r\in(0,\infty)$. By continuity $r$ must be
contained in some interval $[r_{0},r_{1}]$ on which $N\leq 1$, so we can use
the preceding inequality in the form $N(s)\leq(r/s)^{n}N(r)$. The right hand
side is less than or equal to $1$ as long as $s\geq N(r)^{1/n}\>r$. ∎

##  4 Asymptotically Unique Continuation

Any function that is both non-negative and non-decreasing (trivially) has the
following property: if it is non-zero at a particular point $r_{0}$, it will
remain non-zero for any subsequent point $r>r_{0}$. Although $\kappa$ is not
non-decreasing, there is some (large) radius $R$ beyond which it behaves in
that same way. This is a consequence of the fact that the decay rate of
$\kappa$ becomes arbitrarily small at infinity.

To make this precise, recall that $\kappa$ is continuous, non-negative, and
$\epsilon$-almost non-decreasing at large radius. With respect to the last
property fix some $\epsilon>0$ with associated radius $R\geq 0$. Assume there
is an $r_{1}\in[R,\infty)$ at which $\kappa(r_{1})\neq 0$. This is the case,
for example, if $\kappa$ is not asymptotically equivalent to the zero
function. Since $\kappa$ is $\epsilon$-almost non-decreasing, section 3
provides a strictly positive lower bound for any larger radius $r\geq r_{1}$

$\displaystyle\kappa(r)\geq\left(\frac{r_{1}}{r}\right)^{\epsilon}\kappa(r_{1})\
,$ (37)

which prevents $\kappa$ from vanishing at any larger radius, at least as along
as $r_{1}\neq 0$. Note that, if $D$ is bounded from below, then there is a
choice of $\epsilon$ for which $R=0$ such that the conclusion holds for any
$r_{1}\in(0,\infty)$. In any case, the set on which $\kappa$ is non-zero must
include an interval $(r_{0},\infty)$ with some (possibly infinite) $r_{0}\geq
0$.

Let us now investigate the behaviour near $r_{0}$, where the growth rate of
$\kappa$ is controlled by the frequency function $N$ and the mean curvature
deviation $D$. Observe that if $r_{0}\neq 0$ and $N$ is bounded from above on
$(r_{0},r_{1}]$, then the assumptions $\kappa(r_{0})=0$ and $\kappa(r_{1})\neq
0$ lead to a contradiction. To see this use section 3 with the upper bound
$N\leq b$, which yields

$\displaystyle\kappa(r)\geq\left(\frac{r}{r_{1}}\right)^{b+k/2}\kappa(r_{1})\quad\text{for
all }r\in(r_{0},r_{1}]\ .$ (38)

The right hand side is strictly positive and thus prevents $\kappa$ from
vanishing at $r_{0}$.

This is an instance of Aronszajn’s unique continuation theorem, which states
that a non-trivial function that satisfies a second order, elliptic
differential inequality cannot exhibit zeroes of infinite order [Aro57]. A
particular consequence of this is that a non-negative and non-decreasing such
function on $(0,\infty)$ is then either identically zero or strictly positive
across the entire domain.

It was noted by Taubes that $\kappa$ – in fact its defining integral
$\int_{\partial B_{r}}\left\lVert\phi\right\rVert^{2}$ – satisfies Aronszajn’s
unique continuation theorem [Tau17a, Sec. 3]. We obtain the following,
slightly weaker version of the preceding statement:

######  Lemma .

Assume $\operatorname{Ric}\geq 0$ and that sectional curvature is bounded from
below. There is a radius $R\geq 0$, such that if $\kappa$ is non-zero at any
point in $[R,\infty)$, then it is strictly positive on all of $(0,\infty)$.
Moreover, $R=0$ if the mean curvature deviation $D$ is bounded from below.

Here it is possible that $\kappa$ is compactly supported, but only as long as
its support is restricted to a region where the mean curvature of the geodesic
spheres is (still) large. In view of the discussion above, the Lemma also
follows directly from a proof that $N$ is a priori bounded on any interval
$(r_{0},r_{1}]$ in which $\kappa$ does not have zeroes.

##  5 Slow Growth and Bounded Frequency

We have previously seen that an upper bound for $N$ leads to bounded growth of
$\kappa$. The goal of this section is to show that the converse is true when
$r\to\infty$. More precisely, we show that whenever $\kappa$ grows slower than
$\mathcal{O}(r^{a})$ between two large radii, then $N$ must have been bounded
from above on an interval leading up to the violation. Note that such
violations must occur for arbitrarily large radii when $\kappa$ is not
asymptotically bounded below by $r^{a}$, which is the situation of the second
alternative in Theorem A. Accordingly, the upcoming lemma and its refinement
in section 6 play a crucial role in the proof of the main theorem.

######  Lemma .

Assume $\kappa$ is not asymptotically zero. Fix an $\epsilon>0$ and denote by
$R$ the radius beyond which $\left|D\right|\leq\epsilon$. If there is a pair
of radii $r_{0}\leq r_{1}$ in $[R,\infty)$ for which
$\kappa(r_{1})\leq\left(\frac{r_{1}}{r_{0}}\right)^{a-\epsilon}\kappa(r_{0})$,
then there exists a radius $t\in[r_{0},r_{1}]$ such that

  1. (i)

$N(t)\leq a$.

Moreover, if $a<1$ the following holds on the interval $[\tilde{R},t]$, where
${\tilde{R}=\max\bigl{(}a^{\frac{1}{2n}}\>t,\;R\bigr{)}}$.

  1. (ii)

$N<\sqrt{a}$,

  2. (iii)

$\kappa\geq a^{\frac{\sqrt{a}+\epsilon}{2n}}\kappa(t)$.

###### Proof.

To see (i) assume to the contrary that $N>a$ on all of $[r_{0},r_{1}]$. Then
the first bullet in section 3 states that

$\displaystyle\kappa(r_{1})>\left(\frac{r_{1}}{r_{0}}\right)^{a-\epsilon}\kappa(r_{0})\
,$ (39)

which violates the assumption that $\kappa(r_{1})$ satisfies exactly the
opposite inequality. Hence, there is a $t\in[r_{0},r_{1}]$ at which $N(t)\leq
a$.

Assume now that $a<1$ and note that then the same is true for $N(t)$. In that
case we conclude via the second part of section 3 that $N\leq 1$ on
$[N(r)^{1/n}\>t,t]$. Since $N(t)\leq a<\sqrt{a}$, this interval contains as a
subinterval $[a^{1/2n}\>t,t]$. Then the first part of section 3 for
${s\in[a^{1/2n}\>t,t]}$ yields

$\displaystyle
N(s)\leq\left(\frac{r}{s}\right)^{n-k}N(t)\leq\frac{1}{\sqrt{a}}N(t)\leq\sqrt{a}\
,$ (40)

which proves that the same is true on the (possibly smaller) interval
$[\tilde{R},t]$ with ${\tilde{R}:=\max\bigl{(}a^{1/2n}\>t,R\bigr{)}}$.

Since $N\leq\sqrt{a}$ on $[\tilde{R},t]$ the second bullet of section 3
provides the bound

$\displaystyle\kappa(t)\leq\left(\frac{r}{t}\right)^{\sqrt{a}+\epsilon}\kappa(r)\leq
a^{-\frac{\sqrt{a}+\epsilon}{2n}}\kappa(r)\ ,$ (41)

where in the last step we used $a^{1/2n}\>t\leq\tilde{R}\leq r$ for any
$r\in[\tilde{R},t]$. ∎

##  6 The Correlation Tensor

There is an $\Omega_{p}^{1}\otimes\Omega_{p}^{1}$-valued function $T$ that
refines $\kappa^{2}$ to the effect that it resolves the behaviour of the
components of $\phi$. Note that, being a one-form, $\phi$ can be evaluated in
particular on the covariantly constant unit vector field on $B_{r}(p)$ that is
defined by parallel transport of a unit vector $v\in T_{p}W$ along radial
geodesics. The output is a smooth function $\phi_{v}$ on
$B_{r}(p)\setminus\operatorname{Cut}(p)$ that captures the evolution of the
$v$-component of $\phi$ along the geodesics emanating from $p$. This allows
the definition of what we will call the correlation tensor
$T:W^{4}\times(0,\infty)\to\Omega_{p}^{1}\otimes\Omega_{p}^{1}$. Its value for
$v,w\in T_{p}W$ is defined by

$\displaystyle T(p,r)(v,w)$ $\displaystyle=\frac{1}{r^{n-k-1}}\int_{\partial
B_{r}(p)}\langle\phi_{v},\phi_{w}\rangle\ .$ (42)

Note, in particular, that $\operatorname{tr}_{T_{p}W}T(p,r)=\kappa^{2}(p,r)$,
while the induced quadratic form $\kappa_{v}^{2}:=T(v,v)$ returns a version of
the function $\kappa^{2}$ that is based on the component $\phi_{v}$. Just like
$\kappa_{v}(p,r)$ has an interpretation as the mean value of $\phi_{v}$ on the
geodesic sphere, the value of $T(v,w)$ measures the correlation between the
components $\phi_{v}$ and $\phi_{w}$ on the geodesic sphere; hence its name.

An important observation is that $\kappa_{v}$ satisfies essentially the same
properties as $\kappa$. As a short motivation of that fact observe that if
$\tilde{\Delta}_{A}\phi=0$ and $v$ denotes the covariantly constant vector
field described above, then also
${\imath_{v}\tilde{\Delta}_{A}\phi=\tilde{\Delta}_{A}\phi_{v}=0}$. So it is
reasonable to expect that $\kappa_{v}$ behaves like a harmonic function in
exactly the same way that $\kappa$ does.

To make this more precise, if $\phi$ satisfies our main assumption (1) then
$\phi_{v}$ satisfies the following analogous second order equation

$\displaystyle{\nabla^{A}}^{\dagger}\nabla^{A}\phi_{v}+\frac{1}{2}\star[\star[\phi\wedge\phi_{v}]\wedge\phi]+\operatorname{Ric}(\phi)(v)=0\
.$ (43)

As a consequence the derivative of $\kappa_{v}$ comes with its own version of
the frequency function and mean curvature deviation, denoted $N_{v}$ and
$D_{v}$.

$\displaystyle\frac{d\kappa_{v}}{dr}=\frac{N_{v}+D_{v}}{r}\kappa_{v}$ (44)

The functions $N_{v}$ and $D_{V}$ are given by essentially the same
expressions as before, but with $\phi$ replaced by $\phi_{v}$ as follows.

$\displaystyle N_{v}$
$\displaystyle=\frac{1}{r^{n-k-2}\kappa_{v}^{2}}\int_{B_{r}}\left(\lVert\nabla^{A}\phi_{v}\rVert^{2}+\left\lVert[\phi\wedge\phi_{v}]\right\rVert^{2}+\left\langle\operatorname{Ric}(\phi)(v),\phi_{v}\right\rangle\right)$
(45) $\displaystyle D_{v}$
$\displaystyle=\frac{1}{r^{n-k-1}\kappa_{v}^{2}}\int_{\partial
B_{r}}\left(\Delta
r-\frac{n-k-1}{r}\right)\left\lVert\phi_{v}\right\rVert^{2}$ (46)

######  Proposition .

Let $v\in T_{p}W$. All previous results hold verbatim when we replace $\kappa$
and $N$ by $\kappa_{v}$ and $N_{v}$, respectively.

There are in fact analogous results for the correlation tensor $T$. To see
this define the (Frobenius) norm of $T$ with respect to the inner product on
$\Omega_{p}^{1}\otimes\Omega_{p}^{1}$ induced by the metric, i.e.

$\displaystyle\left\lVert
T\right\rVert^{2}=g^{\mu\rho}g^{\nu\sigma}T_{\mu\nu}T_{\rho\sigma}\ .$ (47)

In this expression the metric is evaluated at the point $p$. The norm of $T$
satisfies $\frac{1}{c}\kappa^{2}\leq\left\lVert T\right\rVert\leq\kappa^{2}$
for some constant $c$. The tensor $T$ is differentiable with respect to $r$
and there is then a (possibly larger) $c$ such that the following inequality
holds.

$\displaystyle\left\lVert\frac{dT}{dr}\right\rVert\leq c\
\frac{N+D}{r}\left\lVert T\right\rVert$ (48)

Below we will also use the notation $T^{\prime}:=\frac{dT}{dr}$.

From now on view $T$ as linear map from $T_{p}W\to T_{p}W$. If $T$ has a zero
eigenvalue at some radius $r$ and $v$ denotes the associated eigenvector, then
$\kappa_{v}^{2}(r)=0$. As a consequence of the unique continuation property in
section 4, whenever $r$ is too large $\kappa_{v}$ must then be identically
zero on an interval of the form $[R,\infty)$. This in turn implies that the
component $\phi_{v}$ vanishes on $W^{n}\setminus B_{R}$. We will deal with
such compactly supported components of $\phi$ later. In the definition of $T$
we restrict ourselves to the subspace of $T_{p}W$ that is orthogonal to the
zero eigenspace of $T$ at infinity. The correlation tensor then has strictly
positive eigenvalues on all of $(0,\infty)$, which will be assumed henceforth.

Denote by $\lambda:(0,\infty)\to\mathbb{R}$ the function that assigns to a
radius $r$ the smallest eigenvalue of $T(r)$. By the preceding paragraph we
may assume that this function is non-zero on all of $(0,\infty)$. Clearly
$\lambda\leq\kappa^{2}$ as the latter is the trace of $T$, so it should be
noted that under this assumption $\kappa$ is necessarily everywhere non-zero.

The following proposition states that $\lambda$ is nearly differentiable.

######  Proposition .

The finite differences of the smallest-eigenvalue function $\lambda$ at any
$r\in(0,\infty)$ satisfy the following bounds: Denote by $v$ a unit length
eigenvector of $T(r)$ with eigenvalue $\lambda(r)$ and let $\Delta>0$.

$\displaystyle\lambda(r+\Delta)-\lambda(r)$ $\displaystyle\leq\langle
v,T^{\prime}(r)\;v\rangle\ \Delta+\mathcal{O}(\Delta^{2})$ (49)
$\displaystyle\lambda(r)-\lambda(r-\Delta)$ $\displaystyle\geq\langle
v,T^{\prime}(r)\;v\rangle\ \Delta+\mathcal{O}(\Delta^{2})$ (50)

Moreover, $\lambda$ is locally Lipschitz continuous on $(0,\infty)$.

###### Proof.

For any $r\in(0,\infty)$ denote by $v_{r}$ an eigenvector of $T(r)$ with
eigenvalue $\lambda(r)$. For $\Delta\geq 0$ we then have

$\displaystyle\lambda(r+\Delta)-\lambda(r)=\langle
v_{r+\Delta},T(r+\Delta)v_{r+\Delta}\rangle-\langle
v_{r},T(r)v_{r}\rangle\leq\langle
v_{r},\big{(}T(r+\Delta)-T(r)\big{)}v_{r}\rangle\ .$ (51)

To get the upper bound on the right hand side note that the smallest
eigenvalue at $r+\Delta$ is $\lambda(r+\Delta)$, so the evaluation of
$T(r+\Delta)$ on $v_{r}$ can only ever result in the same or larger values.
The upper bound for the forwards finite difference is then a consequence of
Taylor’s theorem. Up to a minus sign the lower bound for the backwards finite
difference in the second inequality follows in exactly the same way.

The Lipschitz property of $\lambda$ follows by paying closer attention to the
remainder in Taylor’s theorem. Consider a pair $s<r\in(0,\infty)$ and Taylor’s
theorem at zeroth order

$\displaystyle T(r)=T(s)+R_{0}(r)\ .$ (52)

A standard estimate for the remainder states $\left\lVert
R_{0}(r)\right\rVert\leq\sup_{(r,s)}\left\lVert T^{\prime}\right\rVert\
(r-s)$, as long as the derivative is bounded on the given interval. As
mentioned before, the derivative of $T$ at any radius $\tilde{r}$ is bounded
by a multiple of $\frac{N+D}{\tilde{r}}\left\lVert T(\tilde{r})\right\rVert$.
Recall from section 3 that $D\leq\frac{k}{2}$. Also, as was discussed in the
context of the unique continuation property, $N$ is bounded on any compact
interval on which $\kappa$ is non-zero. It follows that $\lambda$ is Lipschitz
on any compact $[s,t]\subset(0,\infty)$, since

$\displaystyle\left|\lambda(r)-\lambda(s)\right|\leq\left|\langle
v_{s},\big{(}T(r)-T(s)\big{)}v_{s}\rangle\right|\leq c\left|r-s\right|\ .$
(53)

Equivalently (since $(0,\infty)$ is locally compact): $T$ is locally
Lipschitz. ∎

We can use the preceding proposition to show that, similar to $\kappa$, the
function $\lambda$ is asymptotically almost non-decreasing. The idea here is
that $\lambda$ consists of piecewise smooth segments, each coinciding with a
function $\kappa_{v}^{2}$ for some $v$. Any $\kappa_{v}$ is eventually
$\epsilon$-almost non-decreasing, so an analogues statement must be true for
$\lambda$. Observe that asymptotic bounds for the mean curvature deviation
$D_{v}$ only depend on the asymptotics of the mean curvature $\Delta r$, which
is independent of $v$. Thus, if $D$ is bounded from below by $-\epsilon$, then
the same is true for $D_{v}$.

Hence, let $R$ be such that $\left|D\right|\leq\epsilon$ for any larger radius
and consider points $r>s\geq R$. Define an equidistant partition of the
interval $[s,r]=\bigcup_{k=0}^{M-1}[r_{k},r_{k+1}]$ where
$r_{k}=s+\frac{k}{M}(r-s)$. Then, denoting by $v_{k}$ an eigenvector with
eigenvalue $\lambda(r_{k})$, the two estimates in section 6 imply

$\displaystyle\lambda(r)=\lambda(s)+\sum_{k=1}^{M}\langle
v_{k},T^{\prime}(r_{k})\;v_{k}\rangle\frac{r-s}{M}+\mathcal{O}\left(\frac{1}{M^{2}}\right)\
.$ (54)

This can be rewritten by using $\langle
v,T^{\prime}v\rangle=\frac{d}{dr}\kappa_{v}^{2}$ and the fact that each
$v_{k}$ is an eigenvector with the smallest eigenvalue at $r_{k}$, i.e.
$\kappa_{v_{k}}^{2}(r_{k})=\lambda(r_{k})$.

$\displaystyle\lambda(r)=\lambda(s)+2\sum_{k=1}^{M}\lambda(r_{k})\frac{\left(N_{v_{k}}(r_{k})\
+D_{v_{k}}(r_{k})\right)}{r_{k}}\
\frac{r-s}{M}+\mathcal{O}\left(\frac{1}{M^{2}}\right)$ (55)

Notably, since $\left|D\right|\leq\epsilon$ we find that $\lambda$ satisfies
the following inquality on $[R,\infty)$

$\displaystyle\frac{\lambda(r)-\lambda(s)}{r-s}$
$\displaystyle\geq-\frac{2\epsilon}{M}\sum_{k=1}^{M}\frac{\lambda(r_{k})}{r_{k}}+\mathcal{O}\left(\frac{1}{M^{2}}\right)\
.$ (56)

This is the finite difference analogue of the differential inequality
satisfied by $\kappa$ when it is $\epsilon$-almost non-decreasing. To make
this more apparent, observe that the right hand side of the last inequality
contains the arithmetic mean of the function $\lambda/r$ for the given
partition of $[s,r]$. If one takes $M\to\infty$ this expression approaches the
mean value of $\lambda/r$ on the given interval. Seeing that $\lambda$ is
continuous, it is then a consequence of the (integral) mean value theorem that
there is a point $t\in[s,r]$ at which the right hand side is given by
$\lambda(t)/t$, such that

$\displaystyle\frac{\lambda(r)-\lambda(s)}{r-s}\geq-2\epsilon\frac{\lambda(t)}{t}\
.$ (57)

If $\lambda$ is differentiable at $r$ the limit $s\to r$ exactly recovers the
differential inequality, as promised. We will correspondingly say that
$\lambda$ is $\epsilon$-almost non-decreasing whenever it satisfies the finite
difference inequality from above.

The fact that $\lambda$ is eventually $\epsilon$-almost non-decreasing allows
us to extend section 5 in such a way that it also provides bounds for $N_{v}$
and $\kappa_{v}$, where $v$ is a unit eigenvector of $T(t)$ associated to the
smallest eigenvalue $\lambda(t)$ at some distinguished $t\in[r_{0},r_{1}]$.

######  Lemma .

Fix $\epsilon>0$ and denote by $R$ the radius beyond which
$\left|D\right|\leq\epsilon$. Let $r_{0},r_{1}\in[R,\infty)$ be a pair of
radii such that $r_{1}$ is larger than any of $4r_{0}$,
$\left(\kappa^{2}(r_{0})/\lambda(r_{0})\right)^{\frac{1}{2a}}r_{0}$, and
$r_{0}^{1/({1-100\sqrt{a}})}$. If
${\kappa(r_{1})\leq\left(\frac{r_{1}}{r_{0}}\right)^{a-\epsilon}\kappa(r_{0})}$
and $a$ is sufficiently small (e.g. $a^{1/4n}<0.1$), then there exists a
radius $t\in[r_{1}^{1-100\sqrt{a}},r_{1}]$ of the following significance:

Let $v$ be an eigenvector of $T(t)$ associated to the smallest eigenvalue
$\lambda(t)$ and write ${\tilde{R}=\max\left(a^{\frac{1}{8n}}t,R\right)}$. On
all of $[\tilde{R},t]$

  1. (i)

$N\leq a^{1/4}$ and $\kappa\geq a^{\frac{a^{1/4}+\epsilon}{4n}}\kappa(t)$

  2. (ii)

$N_{v}\leq a^{1/4}$ and $\kappa_{v}\geq
a^{\frac{a^{1/4}+\epsilon}{4n}}\kappa_{v}(t)$

###### Proof.

Let $\epsilon>0$, denote by $R$ the radius beyond which
$\left|D\right|\leq\epsilon$, and assume $\sqrt{a}<1$. The proof proceeds by
establishing the existence of regions in $[r_{0},r_{1}]$ where both $N$ and
$N_{v}$ are less or equal to $\sqrt{a}$ at the same time. The stated bounds
then follow from section 5 (with $a$ replaced by $\sqrt{a}$ everywhere).

First, regarding the condition $N\leq\sqrt{a}$, observe that the set
${I=\left\\{\ r\in[\log r_{0},\log r_{1}]\mid N(\exp r)>\sqrt{a}\ \right\\}}$
makes up at most $\sqrt{a}$ of the length of the surrounding interval. To see
this, write $I=\coprod(a_{k},b_{k})$ and then go from $\kappa(r_{0})$ to
$\kappa(r_{1})$ by iteratively using section 3, with bounds $N>\sqrt{a}$ on
each $(a_{k},b_{k})$ and $N\geq 0$ on the intervals $[b_{k},a_{k+1}]$ in
between. This leads to

$\displaystyle\kappa(r_{1})\geq\left(\frac{r_{1}}{r_{0}}\right)^{-\epsilon}\prod_{k}\left(\frac{b_{k}}{a_{k}}\right)^{\sqrt{a}}\kappa(r_{0})\
.$ (58)

This inequality is only compatible with the assumption that
${\kappa(r_{1})\leq(r_{1}/r_{0})^{a-\epsilon}\kappa(r_{0})}$ if

$\displaystyle\sum_{k}(\log b_{k}-\log a_{k})\leq\sqrt{a}\;(\log r_{1}-\log
r_{0})\ .$ (59)

Equivalently, if $\left|I\right|\leq\sqrt{a}\left|[\log r_{0},\log
r_{1}]\right|$.

An analogous statement for points that satisfy the condition
$N_{v}\leq\sqrt{a}$ makes use of a slightly longer argument. As before, we set
out to investigate the measure of the set on which $N_{v}>\sqrt{a}$. Denote by
$L$ the largest integer such that $2^{L}r_{0}<r_{1}$ and consider the sequence
$\\{2^{\ell}r_{0}\\}_{\ell=0,1,\ldots,L-1}$. For each pair of neighbours in
this sequence we can apply equation (55) in the form

$\displaystyle\lambda(2s)\geq\lambda(s)+\frac{1}{M}\sum_{k=1}^{M}\lambda(s_{k})\left(N_{v_{k}}(r_{k})-\epsilon\right)+\mathcal{O}\left(\frac{1}{M^{2}}\right)\
,$ (60)

where $s_{k}=(1+k/M)\>s$ and $v_{k}$ denotes an eigenvector of $T(s_{k})$ with
eigenvalue $\lambda(s_{k})$. Next, note that section 3 with the lower bound
$N\geq 0$ provides

$\displaystyle\lambda(s_{k})=\kappa^{2}_{v_{k}}(s_{k})\geq\left(\frac{s_{k}}{s}\right)^{-2\epsilon}\kappa^{2}_{v_{k}}(s)\geq\frac{1}{4}\lambda(s)\
.$ (61)

Here the last inequality holds by virtue of $s_{k}/s<2$ and because
$\lambda(s)$ is the smallest eigenvalue of $T$ at $s$, such that
$\kappa^{2}_{v_{k}}(s)=T(s)(v_{k},v_{k})$ can’t be smaller. Also introduce for
the moment the notation $\langle N_{v}\rangle_{\ell}$ for the average of
$\\{N_{v_{k}}(s_{k})\\}_{k=1,\ldots,M}$ in the interval
$[2^{\ell}r_{0},2^{\ell+1}r_{0}]$. The inequality in (60) becomes

$\displaystyle\frac{\lambda(2^{\ell+1}r_{0})}{\lambda(2^{\ell}r_{0})}\geq
1+\frac{1}{4}\bigg{(}\langle
N_{v}\rangle_{\ell}-\epsilon\bigg{)}+\mathcal{O}\left(\frac{1}{M^{2}}\right)\
.$ (62)

In the last of these pairings we can use $r_{1}$ as endpoint of the interval
instead of $2^{L}r_{0}$ without changing the inequality, since
$r_{1}-2^{L-1}r_{0}>2^{L-1}r_{0}$ and $r_{1}\leq 2^{L+1}r_{0}$. Hence, the
product of all these ratios yields

$\displaystyle\frac{\lambda(r_{1})}{\lambda(r_{0})}\geq\prod_{\ell=0}^{L-1}\left(1+\frac{\langle
N_{v}\rangle_{\ell}-\epsilon}{4}+\mathcal{O}\left(\frac{1}{M^{2}}\right)\right)\
.$ (63)

Let $f$ denote the fraction of $\\{{0,1,\ldots,L-1\\}}$ for which $\langle
N_{v}\rangle_{\ell}+\mathcal{O}(1/M^{2})>\sqrt{a}$. Then the product
simplifies to

$\displaystyle\frac{\lambda(r_{1})}{\lambda(r_{0})}\geq\left(1+\frac{\sqrt{a}-\epsilon}{4}\right)^{fL}\left(1-\frac{\epsilon}{4}+\mathcal{O}\left(\frac{1}{M^{2}}\right)\right)^{1-fL}\
.$ (64)

On the left use that
$\lambda(r_{1})\leq\kappa^{2}(r_{1})\leq\left(\frac{r_{1}}{r_{0}}\right)^{2(a-\epsilon)}\kappa^{2}(r_{0})$
to replace $\lambda(r_{1})$. Upon taking logarithms on both sides the
resulting inequality reads

$\displaystyle
2(a-\epsilon)\log\left(\frac{r_{1}}{r_{0}}\right)+\log\left(\frac{\kappa^{2}(r_{0})}{\lambda(r_{0})}\right)\geq
fL\log\left(1+\frac{\sqrt{a}-\epsilon}{4}\right)+(1-f)L\log\left(1-\frac{\epsilon}{4}+\mathcal{O}\left(\frac{1}{M^{2}}\right)\right)\
.$ (65)

This can be simplified in several ways:  (1) Because $2^{L+1}r_{0}>r_{1}$ and
as long as we make sure that $r_{1}\geq 4r_{0}$ we can assume
$L>\frac{1}{2}\log r_{1}/r_{0}$.  (2) Since $\sqrt{a}-\epsilon<1$ the term
$\log\left(1+(\sqrt{a}-\epsilon)/4\right)$ is no smaller than
$(\sqrt{a}-\epsilon)/8$.  (3) By taking $M$ large, we can make sure that
$\log\left(1-\epsilon/4+\mathcal{O}(1/M^{2})\right)$ is larger than
$-\epsilon$.  Plugging everything in and solving for $f$ leads to the
following upper bound

$\displaystyle
f<32\sqrt{a}\left(1+\frac{\log\left(\kappa^{2}(r_{0})/\lambda(r_{0})\right)}{2a\log\left(r_{1}/r_{0}\right)}\right)\
.$ (66)

In particular, if
$r_{1}>\left(\kappa^{2}(r_{0})/\lambda(r_{0})\right)^{\frac{1}{2a}}r_{0}$ we
find that $f<64\sqrt{a}$.

Consider the subset $J\subset[\log r_{0},\log r_{1}]$ that consists of those
intervals ${[\log 2^{\ell}r_{0},\log 2^{\ell+1}r_{0}]}$ on which $\langle
N_{v}\rangle_{\ell}$ is larger than $\sqrt{a}$. The inequality for $f$ now
tells us that the length of $J$ can be at most $64\sqrt{a}\;\left|[\log
r_{1},\log r_{0}]\right|$. This means that the union of $J$ with the interval
$I$ from the first part has bounded measure

$\displaystyle\left|I\cup J\right|<65\sqrt{a}\;\left|[\log r_{0},\log
r_{1}]\right|\ .$ (67)

We conclude that whenever $\sqrt{a}$ is sufficiently small (for example
$\sqrt{a}<0.01$), then there exists a point $\log(t_{1})$ that is neither in
$I$ nor in $J$. In fact, if $r_{0}<r_{1}^{1-100\sqrt{a}}$, we can moreover
find a $t_{1}$ that is larger than $r_{1}^{1-100\sqrt{a}}$, since the measure
of $[r_{1}^{1-100\sqrt{a}},r_{1}]$ is still larger than $I$ and $J$ combined.

Choose and fix a $t_{1}$ with these properties. On the one hand this means
that $N(t_{1})\leq\sqrt{a}$ and section 5 implies that on the interval
$[\tilde{R}_{1},t_{1}]$ we have $N<a^{1/4}$ and ${\kappa\geq
a^{\frac{a^{1/4}+\epsilon}{4n}}\kappa(r)}$. Here, $\tilde{R}_{1}$ is the
larger of $a^{1/4n}t_{1}$ and $R$. On the other hand it means that $t_{1}$ is
from an interval $[2^{\ell}r_{0},2^{\ell+1}r_{0}]$ on which the average value
$\langle N\rangle_{\ell}$ is less than $\sqrt{a}$. Consequently, the partition
of that interval contains a radius $t_{2}$ at which $N_{v}(t_{2})<\sqrt{a}$.
We again deduce that $N_{v}<a^{1/4}$ and ${\kappa_{v}\geq
a^{\frac{a^{1/4}+\epsilon}{4n}}\kappa_{v}(r)}$ on $[\tilde{R}_{2},t_{2}]$,
where $\tilde{R}_{2}$ is the larger of $a^{1/4n}t_{2}$ and $R$.

Finally, $[\tilde{R}_{1},t_{1}]$ and $[\tilde{R}_{2},t_{2}]$ intersect
whenever $a$ is small enough. To see this explicitly, assume w.l.o.g that
$t_{1}>t_{2}$ and note that $t_{1}/t_{2}\leq 2$ since both are contained in an
interval of the form $[2^{\ell}r_{0},2^{\ell+1}r_{0}]$. It follows that
${\tilde{R}_{1}/t_{2}<2a^{1/4n}}$, which is less than one if for example
$a^{1/4n}<0.1$. The upper bound of $0.1$ is convenient, because we then also
have $\tilde{R}_{1}<a^{1/8n}t_{2}$, such that choosing $t=t_{2}$ and
${\tilde{R}=\max(a^{1/8n}t,R)}$ concludes the proof. ∎

##  7 A Priori Bounds

The final ingredient for the proof of Theorem A are pointwise a priori bounds
for $\left\lVert\phi\right\rVert$ in terms of $\kappa$, as well as estimates
for the volume of the subsets of $B_{r}$ on which
$\left\lVert\phi\right\rVert$ is small compared to that upper bound. These are
the promised analogues of the mean value inequality.

The proof relies mainly on the standard approach to the mean-value inequality
and the fact that according to section 3, $\kappa$ can’t decrease too rapidly
at infinity. However, due to the asymptotic nature of the latter statement and
the related lack of control in the interior, these bounds only hold for points
that lie in some large spherical geodesic shell surrounding $p$.

Below we use the notation $S(r_{1},r_{2})$ for the closed spherical geodesic
shell based at $p$ with inner radius $r_{1}$ and outer radius $r_{2}$, i.e.
all points that satisfy $r_{1}\leq d(p,x)\leq r_{2}$. Moreover, we write
$f_{\Omega_{i}}:=\frac{\operatorname{vol}\Omega_{i}}{\operatorname{vol}\Omega}$
for the fraction that a subset $\Omega_{i}$ occupies within its corresponding
surrounding set $\Omega$.

The following result holds verbatim if we replace $\phi$, $\kappa$ and $N$ by
the versions $\phi_{v}$, $\kappa_{v}$ and $N_{v}$ associated to a unit vector
$v\in T_{p}W$.

######  Lemma .

Let $W^{n}$ be an ALXk space of dimension $n$, $p\in W^{n}$,
$\operatorname{Ric}\geq 0$, and assume $(A,\phi)$ is a solution of (1). For
any $\epsilon\in(0,1/2)$ there exists a radius $R\geq 0$ and constants
$c_{i}>1$ that only depend on $n$ and $k$, such that for sufficiently large
$r\geq R$ the following holds.

  1. (i)

$\left\lVert\phi(x)\right\rVert\leq
c_{0}\;\frac{\kappa(r)}{{\sqrt{\operatorname{vol}X}}}$ for any $x\in
S\,\bigl{(}R+\frac{r-R}{8},\>r-\frac{r-R}{8}\bigr{)}$,

  2. (ii)

Let $t\in[R+\frac{6}{8}(r-R),R+\frac{7}{8}(r-R)]$ and assume that $N\leq 1$ on
$[t,r]$. Then the subset $\Omega_{1}\subseteq\partial B_{t}$ on which
${\left\lVert\phi(x)\right\rVert\leq\frac{1}{2}\frac{\kappa(r)}{\sqrt{\operatorname{vol}X}}}$
has relative volume $f_{\Omega_{1}}\leq c_{1}\epsilon+c_{2}\sqrt{N}\ .$

###### Proof.

Throughout the proof we fix some $\epsilon\in(0,1/2)$ and choose $R$ to be
some large enough radius, such that on the one hand $\kappa$ is
$\epsilon$-almost non-decreasing as provided by section 3 and on the other
hand $\operatorname{vol}B_{r}\leq(1+\epsilon)\operatorname{vol}X\>r^{n-k}$ for
any $r\geq R$ as in section 2. Assume for now simply that $r>R$ is some fixed
outer radius. We will collect conditions on $r$ as we go and will find that
they can always be met by choosing some larger $r$ to start with.

ad (i)
In this part of the proof we use a specific bump function $\beta$ on $W^{4}$
with compact support inside the geodesic shell $S(R,r)$, constructed as
follows. Denote by $\beta_{\mathbb{R}}$ a non-increasing function on
$\mathbb{R}$ that is equal to $1$ on $\left(-\infty,\frac{14}{16}\right)$ and
equal to $0$ on $\left(\frac{15}{16},\infty\right)$. Use this to define a
corresponding function on $W^{4}$ by the rule

$\displaystyle\beta(x)=\beta_{\mathbb{R}}\left(\,\left|\,\frac{d(p,x)-\frac{r+R}{2}}{\frac{r-R}{2}}\,\right|\,\right)\
.$ (68)

The function $\beta$ vanishes on the inner ball ${B_{R+\frac{r-R}{32}}(p)}$,
is equal to $1$ on the geodesic shell with inner radius ${R+\frac{r-R}{16}}$
and outer radius $r-\frac{r-R}{16}$, and is zero again outside of
${B_{r-\frac{r-R}{32}}(p)}$. Note in particular that $\beta$ and its
derivatives have compact support in the interior of $S(R,r)$. Furthermore, the
gradient and Laplacian of $\beta$ are bounded as follows

$\displaystyle\left|d\beta\right|$ $\displaystyle\leq\frac{32}{(r-R)}$
$\displaystyle\left|\Delta\beta\right|$
$\displaystyle\leq\frac{32}{(r-R)^{2}}\ .$ (69)

Denote by $G_{x}$ the positive Dirichlet Green’s function of the Laplacian on
$B_{r}(p)$ based at $x\in B_{r}(p)$. Recall from section 2 that, due to the
volume growth of $W^{n}$, for large enough distances the Green’s function and
its derivative are bounded from above as follows

$\displaystyle G_{x}(y)$
$\displaystyle\leq\frac{(1+\epsilon)c}{\operatorname{vol}X\>d(x,y)^{n-k-2}}$
$\displaystyle\left|dG_{x}(y)\right|$
$\displaystyle\leq\frac{(1+\epsilon)c}{\operatorname{vol}X\>d(x,y)^{n-k-1}}\
,$ (70)

where $c$ depends only on the dimension $n$.

With that in mind we now note that contracting equation (1) with
$\langle\cdot,\phi\rangle$ leads to

$\displaystyle\frac{1}{2}\Delta_{B}\left\lVert\phi\right\rVert^{2}+\lVert\nabla^{A}\phi\rVert^{2}+\left\lVert[\phi\wedge\phi]\right\rVert^{2}+\langle\operatorname{Ric}\phi,\phi\rangle=0\
,$ (71)

where $\Delta_{B}=\nabla^{\dagger}\nabla$ is the Bochner Laplacian associated
to the Levi-Civita connection. This equation implies
$\Delta_{B}\left\lVert\phi\right\rVert^{2}\leq 0$, so the function
$\left\lVert\phi\right\rVert^{2}$ is subharmonic111Note that the Bochner and
connection Laplacian differ by a sign:
$\Delta_{B}=-\operatorname{tr}\nabla^{2}$. and accordingly satisfies a version
of the mean-value inequality. To see this directly, multiply (71) by $\beta
G_{x}$ and integrate over $B_{r}(p)$

$\displaystyle\int_{B_{r}(p)}\ \beta
G_{x}\left(\Delta_{B}\left\lVert\phi\right\rVert^{2}+\lVert\nabla^{A}\phi\rVert^{2}+\left\lVert[\phi\wedge\phi]\right\rVert^{2}+\langle\operatorname{Ric}\phi,\phi\rangle\right)=0\
.$ (72)

Upon integration by parts in the first term, using that $\beta G_{x}=0$ on
$\partial B_{r}(p)$, and assuming that $x$ is an element of the geodesic shell
on which $\beta=1$, we find

$\displaystyle\left\lVert\phi(x)\right\rVert^{2}+\int_{B_{r}(p)}\
\left(g^{-1}(d\beta,dG_{x})+(\Delta_{B}\beta)G_{x}\right)\left\lVert\phi\right\rVert^{2}+\int_{B_{r}(p)}\
\beta
G_{x}\left(\lVert\nabla^{A}\phi\rVert^{2}+\left\lVert[\phi\wedge\phi]\right\rVert^{2}+\langle\operatorname{Ric}\phi,\phi\rangle\right)=0\
.$ (73)

Since $\operatorname{Ric}\geq 0$, the last term on the left hand side is non-
negative, so the equation provides the following estimate.

$\displaystyle\left\lVert\phi(x)\right\rVert^{2}\leq\left|\int_{B_{r}(p)}\
\left(g^{-1}(dG_{x},d\beta)+G_{x}\Delta_{B}\beta\right)\left\lVert\phi\right\rVert^{2}\right|$
(74)

Assume ${x\in S(R+\frac{r-R}{8},r-\frac{r-R}{8})}$, such that the distance
from $x$ to the support of $d\beta$ and $\Delta\beta$ is greater or equal to
$\frac{r-R}{16}$. Furthermore, make $r$ large enough that the previously
mentioned bounds on $G_{x}(y)$ hold for all points with distance
$d(x,y)\geq\frac{r-R}{16}$. Then, using the Cauchy-Schwarz inequality on the
first term and the estimates for $G$, $\left|dG\right|$, $\left|d\beta\right|$
and $\left|\Delta_{B}\,\beta\right|$, we arrive at

$\displaystyle\left\lVert\phi(x)\right\rVert^{2}\leq\frac{32(1+\epsilon)c}{\operatorname{vol}X\>(r-R)^{n-k}}\int_{S(R,r)}\left\lVert\phi\right\rVert^{2}\
.$ (75)

As final step use that in the given domain of integration $\kappa$ is
$\epsilon$-almost non-decreasing. Hence, section 3 with $N\geq 0$ provides the
estimate $\kappa(t)\leq\left(\frac{r}{t}\right)^{\epsilon}\kappa(r)$ for all
$t\leq r$. The integral in the last equation is thus bounded from above by

$\displaystyle\int_{S(R,r)}\left\lVert\phi\right\rVert^{2}=\int_{R}^{r}dt\
t^{n-k-1}\kappa^{2}(t)\leq\frac{r^{n-k}}{n-k-2\epsilon}\ \kappa^{2}(r)\ .$
(76)

Plugging this in, assuming $r>2R$, and using $\epsilon<1/2$ leads to

$\displaystyle\left\lVert\phi(x)\right\rVert^{2}\leq\frac{2^{n-k+6}c}{(n-k-1)\,\operatorname{vol}X}\kappa^{2}(r)\leq
c_{0}^{2}\frac{\kappa^{2}(r)}{\operatorname{vol}X}\ .$ (77)

We note that $c_{0}$ only depends on the dimension of $W^{n}$ and the
dimension of the fibers at infinity.

ad (ii)
Fix some radius $t\in[R+\frac{6}{8}(r-R),R+\frac{7}{8}(r-R)]$ and consider the
associated geodesic sphere $\partial B_{t}$. We are interested in the volume
of the subset of $\partial B_{t}$ where $\left\lVert\phi\right\rVert$ is small
compared to the bound in (i).

Hence, write $\Omega_{1}\subseteq\partial B_{t}$ for the points where
${\left\lVert\phi(x)\right\rVert\leq\frac{\kappa(r)}{2\sqrt{\operatorname{vol}X}}}$.
Also introduce $\Omega_{2}\subseteq\partial B_{t}$ to denote the set of points
at which
$\left\lVert\phi(x)\right\rVert\leq(1+\sqrt{N(r)})\frac{\kappa(r)}{\sqrt{\operatorname{vol}X}}$.
Since $N\geq 0$, $\Omega_{2}$ contains $\Omega_{1}$.

Split up the contributions to $\kappa^{2}(t)$ that arise from integration over
$\Omega_{1}$, $\Omega_{2}\setminus\Omega_{1}$, and their complement
$\Omega_{3}=\partial B_{t}\setminus\Omega_{2}$.

$\displaystyle\kappa^{2}(t)=\frac{1}{t^{n-k-1}}\left(\int_{\Omega_{1}}\left\lVert\phi\right\rVert^{2}+\int_{\Omega_{2}\setminus\Omega_{1}}\left\lVert\phi\right\rVert^{2}+\int_{\Omega_{3}}\left\lVert\phi\right\rVert^{2}\right)$
(78)

On $\Omega_{1}$ the integrand is bounded by
$\frac{\kappa^{2}(t)}{4\operatorname{vol}X}\,$, on
$\Omega_{2}\setminus\Omega_{1}$ we use
$(1+\sqrt{N})^{2}\frac{\kappa^{2}(r)}{\operatorname{vol}X}$, and the integral
over $\Omega_{3}$ can’t be larger than $\kappa^{2}(t)$ in any case.

With regard to the last of these bounds we now make use of the fact that
$\kappa$ is almost non-decreasing, which allows us to compare $\kappa^{2}(t)$
to $\kappa^{2}(r)$. To that end consider the derivative of $\kappa^{2}$.

$\displaystyle{\left.\kern-1.2pt\frac{d\kappa^{2}}{dr}\right|_{\tilde{r}}}=2\frac{N+D}{\tilde{r}}\kappa^{2}(\tilde{r})\geq-2\epsilon\frac{\tilde{r}^{1+2\epsilon}}{r^{2+2\epsilon}}\kappa^{2}(r)$
(79)

For the estimate on the right hand side we have used on the one hand that
$N\geq 0$ and $D\geq-\epsilon$, and on the other hand that $N\leq 1$ such that
section 3 implies that
$\kappa^{2}(\tilde{r})\geq(\tilde{r}/r)^{2+2\epsilon}\kappa^{2}(r)$.
Integration from $t$ to $r$ leads to

$\displaystyle\kappa^{2}(t)\leq\frac{\epsilon}{1+\epsilon}\left(2-\left(\frac{t}{r}\right)^{2+2\epsilon}\right)\kappa^{2}(r)\
.$ (80)

Since $t\leq r$, we may as well use the somewhat simpler statement
$\kappa^{2}(t)\leq 2\epsilon\kappa^{2}(r)$.

Plugging in the corresponding bounds on each of the $\Omega_{i}$, writing
$\operatorname{vol}\Omega_{i}=f_{\Omega_{i}}\operatorname{vol}\partial B_{t}$,
$f_{\Omega_{2}}\leq 1$, and using $\operatorname{vol}\partial
B_{t}\leq(1+\epsilon)\operatorname{vol}Xt^{n-k-1}$ thus leads to

$\displaystyle\kappa^{2}(t)$
$\displaystyle\leq\left(\frac{1}{4}f_{\Omega_{1}}+(1-f_{\Omega_{1}})\
(1+\sqrt{N(r)}\ )^{2}+2\epsilon\right)(1+\epsilon)\kappa^{2}(r)$ (81)
$\displaystyle\leq\left(1+2\epsilon-\frac{3}{4}f_{\Omega_{1}}\right)(1+\sqrt{N(r)}\
)^{2}(1+\epsilon)\kappa^{2}(r)\ ,$ (82)

which can be rearranged to

$\displaystyle f_{\Omega_{1}}$
$\displaystyle\leq\frac{4}{3}\left(1+2\epsilon-\frac{\kappa^{2}(t)}{\bigl{(}1+\sqrt{N(r)}\,\bigr{)}^{2}\,(1+\epsilon)\kappa^{2}(r)}\right)\
.$ (83)

To make this expression more useful we’ll now also need a lower bound for
$\kappa^{2}(t)$ in terms of $\kappa^{2}(r)$. This can again be achieved by
considering the derivative of $\kappa^{2}$. In this case we observe that the
following function is non-decreasing in $\tilde{r}$.

$\displaystyle\tilde{r}\mapsto\tilde{r}^{n-k-2}\kappa^{2}(\tilde{r})N(\tilde{r})=\int_{B_{\tilde{r}}(p)}\lVert\nabla^{A}\phi\rVert^{2}+\left\lVert[\phi\wedge\phi]\right\rVert^{2}+\langle\operatorname{Ric}(\phi),\phi\rangle$
(84)

Moreover, since $\kappa$ is $\epsilon$-almost non-decreasing for
$\tilde{r}>R$, we have $\left|D\right|\leq N+\epsilon$. This yields the
following upper bound for the derivative of $\kappa^{2}$:

$\displaystyle{\left.\kern-1.2pt\frac{d\kappa^{2}}{dr}\right|_{\tilde{r}}}=2\frac{N+D}{\tilde{r}}\kappa^{2}(\tilde{r})\leq
4\>\frac{r^{n-k-2}}{\tilde{r}^{n-k-1}}N(r)\kappa^{2}(r)+2\epsilon\>\frac{r^{2\epsilon}}{\tilde{r}^{1+2\epsilon}}\kappa^{2}(r)\
.$ (85)

Integration222For notational simplicity we assume here that $k\neq n-2$.
However, the result holds similarly for the case $k=n-2$, where the only
difference is that upon integration the formulas contain logarithms.  from $t$
to $r$ now leads to

$\displaystyle\kappa^{2}(r)-\kappa^{2}(t)$
$\displaystyle\leq\frac{4}{n-k-2}\left(\frac{r}{t}\right)^{n-k-2}N(r)\kappa^{2}(r)+\left(\left(\frac{r}{t}\right)^{2\epsilon}-1\right)\kappa^{2}(r)\
.$ (86)

Recall that $r/t<4/3$ and observe that thus ${(r/t)^{2\epsilon}-1<\epsilon}$
for any choice of $\epsilon\in(0,1/2)$. It follows that

$\displaystyle\kappa^{2}(t)\geq\Bigl{(}1-\epsilon-
cN(r)\Bigr{)}\>\kappa^{2}(r)\ ,$ (87)

where the constant $c$ only depends on $n$ and $k$. Plugging this lower bound
for $\kappa^{2}(t)$ into (83) and using that $\epsilon,N<1$, we conclude that

$\displaystyle f_{\Omega_{1}}$ $\displaystyle\leq
c_{2}\epsilon+c_{3}\sqrt{N(r)}\ ,$ (88)

where $c_{2}$ and $c_{3}$ only depend on $n$ and $k$. ∎

##  8 Proof of Taubes’ Dichotomy on ALX spaces

We are now in a position to prove Theorem A. As advertised before, the
arguments are in complete analogy to Taubes’ original proof [Tau17].

######  Theorem .

Let $W^{n}$ be an ALXk gravitational instanton of dimension $n\geq 2$ with
asymptotic fibers of dimension $k\leq n-1$ and fix a point $p\in W^{n}$.
Assume $(A,\phi)$ satisfies the second-order differential equation (1). Then

  1. (i)

either there is an $a>0$ such that
$\liminf_{r\to\infty}\frac{\kappa(r)}{r^{a}}>0$,

  2. (ii)

or $[\phi\wedge\phi]=0$.

###### Proof.

It is sufficient to consider the case where $\kappa$ is not asymptotically
zero, since otherwise $\kappa$ – and thus $\phi$ – has compact support due to
the (asymptotically) unique continuation property of section 4 and in that
case $[\phi\wedge\phi]$ must vanish. To see this, assume to the contrary that
$[\phi\wedge\phi]$ is non-zero on some neighbourhood of a point $p$. Since
$\phi$ is compactly supported, there is some radius $R$ such that $\phi$
vanishes on $W^{n}\setminus B_{R}(p)$ and $\kappa$ vanishes on $[R,\infty)$.
As a consequence, there is some positive constant $c$ such that for any $r>R$
we have

$\displaystyle\int_{B_{r}(p)}\nu\geq\int_{B_{r}(p)}\left\lVert[\phi\wedge\phi]\right\rVert^{2}=c>0\
.$ (89)

Here,
$\nu=\lVert\nabla^{A}\phi\rVert^{2}+\left\lVert[\phi\wedge\phi]\right\rVert^{2}+\langle\operatorname{Ric}(\phi),\phi\rangle$
denotes the integrand that defines $N$. The derivative of $\kappa^{2}$ at any
$r>R$ is given by

$\displaystyle\frac{d\kappa^{2}}{dr}=\frac{1}{r^{n-k-1}}\left(\int_{B_{r}(p)}\nu+\int_{\partial
B_{r}(p)}\left(\Delta
r-\frac{n-k-1}{r}\right)\left\lVert\phi\right\rVert^{2}\right)\ .$ (90)

The second integral vanishes, since $\phi=0$ at points with $d(x,p)>R$.
However, the first term is bounded from below by $\frac{c}{r^{n-k-1}}>0$,
which is in contradiction to the assumption that $\kappa$ is constant (namely
$0$) on all of $[R,\infty)$.

By the same reasoning one also finds that $[\phi,\phi_{v}]=0$ whenever
$\kappa_{v}$ has compact support. As a consequence it is sufficient to
consider the components of $\phi$ that are in the complement of the zero
eigenspace of $T$ at infinity (see the related discussion in section 6). Note
in particular that when only a single component $\phi_{v}$ is non-zero at
infinity, then $[\phi\wedge\phi]=0$ everywhere. Hence, assume from now on that
$T$ acts on a vector space of dimension at least 2 and that $\lambda>0$ on all
of $(0,\infty)$.

To prove the dichotomy stated in the theorem, assume that $\kappa$ is not
asymptotically bounded below by any positive power of $r$. This means that for
any small $a>0$ (say, for example, small enough that ${a^{1/8n}<0.1}$) we can
do the following: Set $\epsilon=a/2$ and let $R>1$ be a radius beyond which
$\left|D\right|\leq\epsilon$ and
$\operatorname{vol}B_{r}(p)\leq(1+\epsilon)\operatorname{vol}Xr^{n-k}$. Then
we can find an arbitrarily large $r_{1}\in[R,\infty)$ such that
${\kappa(r_{1})\leq\left(\frac{r_{1}}{R}\right)^{a-\epsilon}\kappa(R)}$. In
particular we may choose some $r_{1}$ that is larger than each of the four
numbers $4R$, $\left(\kappa^{2}(R)/\lambda(R)\right)^{1/2a}R$,
$\kappa(R)^{-1/a}$, and $R^{1/({1-100\sqrt{a}})}$. We are then in the
situation in which we can rely on section 6. The arguments in the upcoming
five parts show that this leads to a contradiction if $N\neq 0$ and $a$ is too
small.

### Part 1

Recall that section 6 provides the existence of a distinguished radius
$t\in[r_{1}^{1-100\sqrt{a}},r_{1}]\subseteq[R,r_{1}]$. In this part we collect
our previous results and slightly expand on our knowledge about the
eigenvalues of $T$ at and below $t$.

Write $u$ and $v\/$ for unit eigenvectors associated to the _largest_ and
_smallest_ eigenvalue of $T(t)$, respectively. Recall that these eigenvalues
coincide at $t$ with the values of $\kappa_{u}^{2}(t)$ and
$\kappa_{v}^{2}(t)$, so they satisfy
${\kappa_{v}^{2}(t)\leq\kappa_{u}^{2}(t)}$. If
$\kappa_{v}^{2}(t)\neq\kappa_{u}^{2}(t)$ the eigenvectors $v$ and $u$ are
guaranteed to be orthogonal. Otherwise $T(t)$ is a multiple of the identity
matrix and we choose an arbitrary pair of orthonormal vectors.

Denote by $\tilde{R}$ the larger of $a^{\frac{1}{8n}}t$ and $R$. section 6
establishes that on the interval $I:=[\tilde{R},t]$ the frequency functions
$N$ and $N_{v}$ are bounded from above by $a^{1/4}$ and provides associated
lower bounds for $\kappa$ and $\kappa_{v}$. As we show now, analogous
estimates hold for $N_{u}$ and $\kappa_{u}$.

First observe that the largest eigenvalue satisfies
$\kappa^{2}_{u}(t)\geq\frac{1}{n}\kappa^{2}(t)$ since $\kappa^{2}$ is the
trace of $T$. From this and the definition of $N$ and its $N_{u}$ version (cf.
section 3 and (45), respectively) it follows that ${N_{u}(t)\leq nN(t)}$. As a
consequence $N_{u}\leq na^{1/4}$ and, as long as we make sure that
$a^{1/4}<1/n$, a small variation of the second part of section 3 finds that
$N_{u}\leq a^{1/8}$ on all of $I$. Since $N_{u}$ is bounded from above, we can
now deduce bounds on $\kappa_{u}$ as usual via section 3.

$\displaystyle\kappa_{u}\geq a^{\frac{a^{1/8}+\epsilon}{8n}}\kappa_{u}(t)\ .$
(91)

In the current situation with $\epsilon=a/2$ this bound and its analogue for
$\kappa_{v}$ can be simplified considerably. For that observe that
$a^{\epsilon}=a^{a/2}$ is certainly larger than $1/2$, and similarly
$a^{\frac{a^{1/4}}{4n}}>(1/2)^{\frac{4}{4n}}>1/2$ and
$a^{\frac{a^{1/8}}{8n}}>1/2$. Applying these observations to the bounds for
$\kappa_{u}$ and $\kappa_{v}$ results in the main conclusions of this part.

The following estimates hold on all of $I=[\tilde{R},t]$:

  * •

$N_{v}\leq a^{1/4}$ and $\kappa_{v}\geq\frac{1}{4}\kappa_{v}(t)$

  * •

$N_{u}\leq a^{1/8}$ and $\kappa_{u}\geq\frac{1}{4}\kappa_{u}(t)$

### Part 2

The goal of this part is to show that there exist points in
${J=\bigl{[}\tilde{R}+\frac{6}{8}(t-\tilde{R}),\tilde{R}+\frac{7}{8}(t-\tilde{R})\bigr{]}}$
for which the integrals that appear in $N_{u}$ and $N_{v}$ are both small.
This interval is of significance, because it corresponds to radii that are
contained in the geodesic shell that appears in item (iii) of section 7.

Denote by
$\nu_{u}:=\left(\lVert\nabla^{A}\phi_{u}\rVert^{2}+\left\lVert[\phi\wedge\phi_{u}]\right\rVert^{2}\right)$
the integrand in $N_{u}$ and analogously for $\nu_{v\/}$. Consider the
following sets of radii in $J$.

$\displaystyle J_{u}$ $\displaystyle:=\left\\{\ s\in
J\mathrel{}\middle|\mathrel{}\int_{\partial
B_{s}}\nu_{u}\leq\frac{1}{2\left|J\right|}\;t^{n-k-2}\;\kappa_{u}^{2}(t)\;a^{1/16}\
\right\\}$ (92) $\displaystyle J_{v}$ $\displaystyle:=\left\\{\ s\in
J\mathrel{}\middle|\mathrel{}\int_{\partial
B_{s}}\nu_{v}\leq\frac{1}{2\left|J\right|}\;t^{n-k-2}\;\kappa_{v}^{2}(t)\;a^{1/16}\
\right\\}$ (93)

Note for later that $\left|J\right|=\left|I\right|/8$, where
$I=[\tilde{R},t]$, so the fact that $a^{1/8n}t\leq\tilde{R}$ and using
$a^{1/8n}<0.5$ yields $\left|J\right|\geq\frac{t}{16}$.

The measure of $J_{u}$ is bounded from below, since

$\displaystyle t^{n-k-2}\kappa^{2}_{u}(t)N_{u}(t)\geq\int_{J}ds\int_{\partial
B_{s}}\nu_{u}\geq\left(\left|J\right|-\left|J_{u}\right|\right)\;\frac{1}{2{\left|J\right|}}\;t^{n-k-2}\;\kappa_{u}^{2}(t)\;a^{1/16}\
,$ (94)

and similarly for $J_{v}$. Since both $N_{u}(t),N_{v}(t)\leq a^{1/8}$, we find

$\displaystyle\left|J_{u}\right|,\left|J_{v}\right|>\left(1-2a^{1/16}\right)\left|J\right|\
.$ (95)

Since $2a^{1/16}<0.5$ (recall that $a^{1/8n}<0.1$ and $n\geq 2$ in any case),
it follows that $J_{u}$ and $J_{v}$ must intersect. Hence, choose and fix from
now on a point $r\in J$ at which both (92) and (93) are satisfied.

### Part 3

As a next step we establish an $L^{2}$-bound for the function
$\operatorname{Tr}\phi_{u}\phi_{v}$ on $\partial B_{r}$, where $r$ is the
fixed radius we found at the end of Part 2. For this we view $\partial B_{r}$
with the induced metric as a compact Riemannian manifold, where we can rely on
a Hölder-Sobolev inequality.

First note that the derivative of $\operatorname{Tr}\phi_{u}\phi_{v}$ is
bounded by

$\displaystyle\left\lVert
d\operatorname{Tr}\phi_{u}\phi_{v}\right\rVert\leq\lVert\nabla^{A}\phi_{u}\rVert\;\lVert\phi_{v}\rVert+\lVert\phi_{u}\rVert\;\lVert\nabla^{A}\phi_{v}\rVert\
.$ (96)

Because $r$ lies inside the geodesic shell of section 7 we know that
$\left\lVert\phi_{u}\right\rVert\leq c_{0}\kappa_{u}(t)$ and
$\left\lVert\phi_{v}\right\rVert\leq c_{0}\kappa_{v}(t)$. In light of the fact
that both (92) and (93) are satisfied at $r$ (and using the Cauchy-Schwarz
inequality), we find that

$\displaystyle\int_{\partial B_{r}}\left\lVert
d\operatorname{Tr}\phi_{u}\phi_{v}\right\rVert^{2}\leq
16c_{0}^{2}t^{n-k-3}\kappa_{u}^{2}(t)\kappa_{v}^{2}(t)a^{1/16}\ .$ (97)

On the $n-1$ dimensional compact manifold $\partial B_{r}$ the Gagliardo-
Nirenberg-Sobolev inequality holds, such that

$\displaystyle\left\lVert\operatorname{Tr}\phi_{u}\phi_{v}\right\rVert_{L^{q}(\partial
B_{r})}\leq C\,\left\lVert
d\operatorname{Tr}\phi_{u}\phi_{v}\right\rVert_{L^{2}(\partial B_{r})}\ ,$
(98)

where $\frac{1}{q}=\frac{1}{2}-\frac{1}{n-1}$ and the constant $C$ only
depends on $n$. On the other hand, since $\partial B_{r}$ has finite measure
we can also use Hölder’s inequality with
$\frac{1}{p}=\frac{1}{q}+\frac{1}{n-1}(=\frac{1}{2})$, such that we obtain

$\displaystyle\left\lVert\operatorname{Tr}\phi_{u}\phi_{v}\right\rVert_{L^{2}(\partial
B_{r})}\leq\operatorname{meas}(\partial
B_{r})^{\frac{1}{n-1}}\left\lVert\operatorname{Tr}\phi_{u}\phi_{v}\right\rVert_{L^{q}(\partial
B_{r})}\leq C_{HS}\,r\left\lVert
d\operatorname{Tr}\phi_{u}\phi_{v}\right\rVert_{L^{2}(\partial B_{r})}\ ,$
(99)

where we have used Bishop-Gromov’s volume comparison section 2 to estimate the
area of the geodesic sphere and absorbed any radius-independent contributions
into the Hölder-Sobolev constant $C_{HS}$, which then still depends only on
$n$.

More explicitly, we conclude that we have the following $L^{2}$-bound for
$\operatorname{Tr}\phi_{u}\phi_{v}$ over $\partial B_{r}$:

$\displaystyle\int_{\partial
B_{r}}\left|\operatorname{Tr}\phi_{u}\phi_{v}\right|^{2}$ $\displaystyle\leq
C_{HS}\,c_{0}^{2}t^{n-k-1}\kappa_{u}^{2}(t)\kappa_{v}^{2}(t)a^{1/16}\ .$ (100)

### Part 4

Our next goal is to similarly determine an $L^{2}$-bound for the function
$\left|\phi_{u}\right|\cdot\left|\phi_{v}\right|$ on $\partial B_{r}$, where
$\left|\phi_{u}\right|$ denotes pointwise application of the norm induced by
the Killing form. It is a property of $\mathfrak{su}(2)$ that

$\displaystyle\left|[\phi_{u},\phi_{v}]\right|^{2}=4\left|\phi_{u}\right|^{2}\left|\phi_{v}\right|^{2}-4\operatorname{Tr}\left(\phi_{u}\phi_{v}\right)^{2}\
.$ (101)

Moreover, since $u$ and $v$ are orthonormal
$\left\lVert[\phi\wedge\phi_{v}]\right\rVert^{2}\geq\left|[\phi_{u},\phi_{v}]\right|^{2}\mu_{W^{n}}$,
so we can bound the following integral with the help of (93).

$\displaystyle\int_{\partial
B_{r}}\left|\phi_{u}\right|^{2}\left|\phi_{v}\right|^{2}-\int_{\partial
B_{r}}\left|\operatorname{Tr}(\phi_{u}\phi_{v})\right|^{2}=\frac{1}{4}\int_{\partial
B_{r}}\left|[\phi_{u},\phi_{v}]\right|^{2}\leq\frac{1}{4}\int_{\partial
B_{r}}\nu_{v}\leq 2\>t^{n-k-3}\>\kappa_{v}^{2}(t)\>a^{1/16}$ (102)

Together with the main result of Part 3 in equation (100) this leads to

$\displaystyle\int_{\partial
B_{r}}\left|\phi_{u}\right|^{2}\left|\phi_{v}\right|^{2}\leq\left(C_{HS}+\frac{2}{c_{0}^{2}t^{2}\kappa_{u}^{2}(t)}\right)c_{0}^{2}t^{n-k-1}\;\kappa_{u}^{2}(t)\;\kappa_{v}^{2}(t)\;a^{1/16}\
.$ (103)

Observe that things have been set up in such a way that
$t^{2}\kappa_{u}^{2}(t)>1$: First, $\kappa_{u}$ is almost non-decreasing, so
$\kappa_{u}^{2}(t)\geq(t/R)^{2\epsilon}\kappa_{u}^{2}(R)$. Second, we know
that $\kappa^{2}_{u}(R)\geq\lambda(R)$ since the latter is the smallest
eigenvalue of $T(R)$. Third, we have previously chosen $r_{1}$ large enough
that $\lambda(R)>(R/r_{1})^{2a}\kappa^{2}(R)$. Fourth, $t$ is larger than
$r_{1}^{1-\sqrt{100}a}$ while $R>1$. Plugging everything together yields

$\displaystyle t^{2}\kappa_{u}^{2}(t)\geq
r_{1}^{2(1-100\sqrt{a}-4a)}\kappa^{2}(r_{1})>r_{1}^{2a}\kappa^{2}(r_{1})\ ,$
(104)

where the last step follows via $a^{1/4n}<0.1$ and $n\geq 2$. Finally, recall
that we have also explicitly chosen $r_{1}$ to be large enough that the
rightmost expression is larger than $1$.

The upshot of this part is that there exists a constant $C>1$ that only
depends on $n$ and $k$, such that

$\displaystyle\int_{\partial
B_{r}}\left|\phi_{u}\right|^{2}\left|\phi_{v}\right|^{2}\leq
Ct^{n-k-1}\;\kappa_{u}^{2}(t)\;\kappa_{v}^{2}(t)\;a^{1/16}\ .$ (105)

From now on we allow the value of $C$ to increase from one equation to the
next.

### Part 5

In this final part, we combine the results of the previous four parts with
item (ii) of section 7. Thus, write $\Omega_{1}$ for the subset of $\partial
B_{r}$ where
${\left\lVert\phi_{u}\right\rVert\leq\frac{\kappa_{u}(t)}{2\operatorname{vol}X}}$.

The inequality in (105) remains true if we restrict the domain of integration
to $\partial B_{r}\setminus{\Omega_{1}}$, such that

$\displaystyle\frac{\kappa_{u}^{2}(t)}{4\operatorname{vol}X}\int_{\partial
B_{r}\setminus{\Omega_{1}}}\left|\phi_{v}\right|^{2}\leq\int_{\partial
B_{r}\setminus{\Omega_{1}}}\left|\phi_{u}\right|^{2}\left|\phi_{v}\right|^{2}\leq
C^{2}t^{n-k-1}\;\kappa_{u}^{2}(t)\;\kappa_{v}^{2}(t)\;a^{1/16}\ .$ (106)

More concisely, this leads to the inequality

$\displaystyle\int_{\partial
B_{r}\setminus{\Omega_{1}}}\left|\phi_{v}\right|^{2}\leq
C\operatorname{vol}Xt^{n-k-1}\kappa_{v}^{2}(t)a^{1/16}\ .$ (107)

Meanwhile, item $(i)$ of section 7 provides the upper bound
$\left|\phi_{v}\right|\leq c_{0}^{2}\kappa_{v}^{2}(t)$. Writing
$\operatorname{vol}\Omega_{1}=f_{\Omega_{1}}\operatorname{vol}B_{r}$ this
leads to

$\displaystyle\int_{{\Omega_{1}}}\left|\phi_{v}\right|^{2}\leq
f_{\Omega_{1}}(1+\epsilon)\operatorname{vol}X\>r^{n-k-1}c_{0}^{2}\kappa_{v}^{2}(t)\
.$ (108)

Item $(ii)$ of section 7 states that $f_{\Omega_{1}}\leq
c_{1}\epsilon+c_{2}\sqrt{N_{u}(t)}$. Recall from Part 1 that $N_{u}(t)\leq
a^{1/8}$ while $\epsilon=a/2$, such that $f_{\Omega_{1}}$ is bounded by some
multiple of $a^{1/16}$.

It follows that the sum of the two integrals satisfies

$\displaystyle{\int_{\partial B_{r}}\left|\phi_{v}\right|^{2}\leq
Ct^{n-k-1}\kappa_{v}^{2}(t)a^{1/16}}\ .$ (109)

This is equivalent to the statement that ${\kappa_{v}^{2}(r)\leq
Ca^{1/16}\kappa_{v}^{2}(t)}$. Finally, combining this with the bound
$\kappa_{v}^{2}(r)\geq\frac{1}{4}\kappa_{v}^{2}(t)$ from Part 1 culminates in
the inequality

$\displaystyle\kappa_{v}^{2}(t)\leq 4Ca^{1/16}\kappa_{v}^{2}(t)\ ,$ (110)

which is absurd, as we are free to choose $a$ arbitrarily small and in
particular such that $a^{1/16}<\frac{1}{4C}$.

∎

##  9 Proof of Taubes’ Dichotomy for Kapustin-Witten Solutions

In this section we prove Theorem B, which enhances section 8 for solutions of
the Kapustin-Witten equations on four-manifolds. We again closely follow
Taubes’ arguments, who proved an analogous statement in case the four-manifold
is Euclidean space $\mathbb{R}^{4}$.

######  Theorem .

Let $W^{4}$ be an ALX gravitational instanton of dimension $4$, with
asymptotic fibers of dimension $k\leq 3$, and such that sectional curvature is
bounded from below. Assume $(A,\phi)$ are solutions of the $\theta$-Kapustin-
Witten equations and if $\theta\not\equiv 0,\pi$ also assume that
$\int_{W^{4}}\left\lVert F_{A}\right\rVert^{2}<\infty$, then

  1. (i)

either there is an $a>0$ such that
$\liminf_{r\to\infty}\frac{\kappa(r)}{r^{a}}>0$,

  2. (ii)

or $[\phi\wedge\phi]=0$, $\nabla^{A}\phi=0$, and $A$ is self-dual if
$\theta=0$, flat if $\theta\in(0,\pi)$, and anti-self-dual if $\theta=\pi$.

###### Proof.

Since solutions of the Kapustin-Witten equations also satisfy equation (1),
the dichotomy of section 8 holds. It remains to show that in the case where
$[\phi\wedge\phi]$ is identically zero, also $\nabla^{A}\phi$ vanishes and $A$
is either (anti-)self-dual or flat as stated. Hence, assume from now on that
$[\phi\wedge\phi]=0$. At points where $\phi$ is non-zero the Higgs field can
then be written as $\phi=\omega\otimes\sigma$, where $\omega\in\Omega^{1}(M)$
and $\sigma\in\Gamma(M,\operatorname{ad}E)$, normalized such that
$\left\lVert\sigma\right\rVert=1$.

Consider first the case where $\theta=0$ (this also covers the case
$\theta=\pi$ by a reversal of orientation). The Kapustin-Witten equations then
state on the one hand $F_{A}^{+}=0$, so $A$ is anti-self-dual, and on the
other hand $(d_{A}\phi)^{-}=0$ and $d_{A}\star\phi=0$. The latter two
equations can only be satisfied if the constituents of $\phi$ satisfy
$\nabla^{A}\sigma=0$ and $(d\omega)^{-}=0=d\star\omega$. In particular
$\sigma$ is guaranteed to be covariantly constant at points where $\phi\neq
0$.

The zero locus of $\phi$ coincides with the zero locus of $\omega$. Since the
one-form $\omega$ satisfies the first order differential equations from above,
it is an example of what Taubes calls a $\mathbb{Z}/2$ harmonic spinor in
[Tau14]. In that article he investigates the zero locus of such $\mathbb{Z}/2$
harmonic spinors in general and Theorem 1.3 of [Tau14] states that the zero
locus has Hausdorff dimension 2. The relevance for us is that the complement
of the zeroes of $\omega$ in any given ball in $W^{4}$ is path connected. This
means that $\sigma$ can be defined at points where $\omega=0$ by parallel
transport along paths where $\omega$ is non-zero. Since $A$ is smooth and
$\sigma$ is $\nabla^{A}$-parallel, parallel transport along two different
paths results in the same value.

Since $\sigma$ is defined everywhere, the same is true for
$\omega=\frac{1}{2}\operatorname{Tr}(\phi\sigma)$. The elliptic differential
equations $(d\omega)^{-}=0=d\star\omega$ imply that
$\left\lVert\omega\right\rVert^{2}$ is subharmonic. Thus, by a classical
result of Yau [Yau76, Theorem 3 & Appendix (ii)], either
$\left\lVert\omega\right\rVert^{2}$ is constant or
$\lim_{r\to\infty}r^{-1}\int_{B_{r}}\left\lVert\omega\right\rVert^{2}>0$. The
latter is precluded by our assumptions and we conclude that
$\nabla^{A}\phi=0$.

In the case that $\theta\not\equiv 0\pmod{\pi}$, neither of the terms
$(d_{A}\phi)^{\pm}$ vanishes automatically. However, if we additionally assume
that $F_{A}$ is square-integrable, we can employ Uhlenbeck’s compactness
theorem for (anti-)self-dual connections to deduce that $A$ must be flat. To
see this, first note that we can find a coefficient $c(\theta)$ such that the
connection $A+c(\theta)\phi$ satisfies the $\pi/2$ version of the Kapustin-
Witten equations, so it is sufficient to consider the case $\theta=\pi/2$.

With that in mind and since $[\phi\wedge\phi]=0$, the Kapustin-Witten
equations state that $F_{A}=\star d_{A}\phi$. As a consequence the connection
$\hat{A}:=A+\phi$ is self-dual. Since $\int_{W^{4}}\lVert
F_{\hat{A}}\rVert^{2}<\infty$, this connection $\hat{A}$ is the pullback of a
smooth, regular connection on the one-point compactification of $W^{4}$, by
Uhlenbeck’s compactness theorem. It follows that at large radius the field
strength falls off faster than the volume of geodesic balls grows, i.e.
$\lVert F_{\hat{A}}\rVert\leq\frac{c}{r^{4-k}}$. Keeping this in mind, note
that $\nabla^{\hat{A}}\phi=\nabla^{A}\phi$, such that
$F_{\hat{A}}=2(d_{\hat{A}}\phi)^{+}$ and

$\displaystyle\int_{B_{r}}\lVert
F_{\hat{A}}\rVert^{2}=\int_{B_{r}}\operatorname{Tr}F_{\hat{A}}\wedge
d_{\hat{A}}\phi=\int_{\partial B_{r}}\operatorname{Tr}F_{\hat{A}}\wedge\phi\
,$ (111)

where we have used Stokes’ theorem and the Bianchi identity in the last
equality. With the bound on $\lVert F_{\hat{A}}\rVert$ the integral on the
right is bounded by a multiple of $\frac{\kappa}{r}$, which approaches $0$ for
$r\to\infty$, so $\hat{A}$ is flat. From here we are back in the situation
where $(d_{A}\phi)^{+}=0=d_{A}\star\phi$ and the same argument as before leads
to $\nabla^{A}\phi=0$.

∎

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