# Model knot solutions for the Twisted Bogomolny Equations

Panagiotis Dimakis  450 Serra Mall, Bldg 380, Stanford, CA 94305
<EMAIL_ADDRESS>
###### Abstract.

In this paper we prove existence for model knot solutions to the dimensionally
reduced twisted Kapustin-Witten equations on $\mathbb{R}^{3}_{+}$ for any
twisting parameter $t\in(0,\infty)$. We start with the explicit solutions for
$t=1$ derived in [9] and perform a continuity argument in $t$. This
corroborates a prediction of Gaiotto and Witten [[]p. 961]GW.

##  1\. Introduction

The Kapustin-Witten (KW) equations were introduced in [4] in a monumental
attempt to use quantum field theory in order to understand certain aspects of
the geometric Langlands correpsondence. These equations are defined on a four
dimensional manifold $M^{4}$ equipped with a $G$-bundle $E$. If $A$ is a
connection on $E$ and $\phi$ is an ad$(E)$-valued $1$-form, then the KW
equations take the form

$\begin{split}F_{A}-\phi\wedge\phi+\star d_{A}\phi&=0\\\
d_{A}\star\phi&=0.\end{split}$ (1.1)

In their paper, Kapustin and Witten derived these equations as a
specialization of a $1$-parameter family of equations to a physically
meaningful parameter value. The general family, which can be thought of as a
phase-shifted form of complexified self-duality equations, can be obtained in
the following way: If we define the complex connection $\mathcal{A}=A+i\phi$
and compute its curvature $F_{\mathcal{A}}$, this family of equations can be
written as

$e^{i\theta}F_{\mathcal{A}}=\bar{\star e^{i\theta}F_{\mathcal{A}}},$ (1.2)

where $\theta\in[0,\pi/2]$. The KW equations as written above, correspond to
$\theta=\pi/4$.

In subsequent papers, Witten in [9] and Gaiotto and Witten in [1] presented
strong evidence that one should be able to derive knot invariants through the
study of appropriate solutions of the Kapustin-Witten equations. More
specifically, they developed the following conjecture: the solution spaces of
the KW equations when $M^{4}=W^{3}\times\mathbb{R}^{+}$ , where
$W\times\\{0\\}$ contains a knot $K$, and where we impose a certain set of
singular boundary conditions along $(W\times\\{0\\})\setminus K$ and
separately along $K$, should contain enough information to capture the Jones
polynomial when $W=S^{3}$ and to define a generalization of the Jones
polynomial in general.

The simplest test case for this conjecture is when $W=\mathbb{R}^{3}$ and the
knot $K$ is a straight line. Witten in [9] was able to find explicit solutions
to equations (1.1) with gauge group $G=SU(2)$. Witten’s key observation was
that by dimensionally reducing the KW equations in the direction of the line
representing the knot, one ends up with a system of equations, called the
Extended Bogomolny equations (EBE), which have a Hermitian Yang-Mills
structure. As it will be explained below, this structure reduces the problem
to solving just one equation, called the moment map equation. By requiring
that the solution be rotationally and scale invariant and satisfy the
aforementioned asymptotic conditions along
$(\mathbb{R}^{2}\times\\{0\\})\setminus\\{0\\}$ and seperately along
$\\{0\\}$, Witten reduced the problem to a non-linear ordinary differential
equation, which he was able to solve explicitly. These solutions which are
called the model knot solutions, are indexed by a non-negative integer $k$,
which for physical reasons goes by the name magnetic charge of the knot. They
are the building blocks of the theory since any solution should look like one
of these model solutions near a point of an arbitrary knot.

The conjecture in the case of the straight unknot was settled in [5] through
the construction of a non linear Weitzenbock formula sensitive to the
asymptotic conditions and through the study of the mapping properties of the
linearization of equations (1.1). Despite the validity of the above conjecture
for the ”unknot”, there has been evidence, already explained in [[,
e.g.,]section 2.3]GW, the Kapustin-Witten equations as given in (1.1) together
with the asymptotic conditions used in [6] and [5] do not provide sufficient
information for the derivation of the Jones polynomial in the general case.

In [1] the authors suggested two possible alternatives to overcome this issue.
The first alternative, which goes by the name of complex symmetry breaking,
was to require a different asymptotic behavior for the solutions studied. We
will pursue ideas related to complex symmetry breaking in a companion paper to
the present one. The second alternative, which is what motivated the writing
of this paper, was to deform the equations (1.1) by changing the $\theta$
parameter in the family (1.2) and hope that for appropriately chosen angle
$\theta$ and without changing the asymptotic conditions, the solutions would
provide enough information to derive the Jones polynomial. In particular, they
conjectured the existence of local model knot solutions of arbitrary charge
$k$ with the same asymptotic behavior as the explicit solutions of Witten. The
main theorem of this paper proves the existence of such model knot solutions:

######  Theorem 1.1.

Given any natural number $k$ and angle $\theta\in(0,\pi/2)$, there exists a
model knot solution of charge $k$ to the $\theta$-deformed KW equations (1.2).

The idea of proof of this theorem is very similar in spirit to the deformation
argument of He -Mazzeo [3]. By dimensionally reducing the $\theta$-deformed KW
equations we obtain a system of equations, which we will call the Twisted
Extended Bogomolny Equations (TEBE), that continues to possess a Hermitian
Yang-Mills structure. As we will explain in greater detail below, this means
that there exist first order differential operators $\mathcal{D}_{i}$
,$i~{}=~{}1,2,3$ such that the TEBE equations assume the form

$\begin{split}&[\mathcal{D}_{i},\mathcal{D}_{j}]=0,~{}\text{i,j}=1,2,3,\\\
\Lambda(\cos^{2}\beta&[\mathcal{D}_{1},\mathcal{D}_{1}^{\dagger}]-\sin^{2}\beta[\mathcal{D}_{2},\mathcal{D}_{2}^{\dagger}])+[\mathcal{D}_{3},\mathcal{D}_{3}^{\dagger}]=0.\end{split}$

The parameter $\beta$ in these equations is related to the parameters $\theta$
and $t$ in the following way:

$\tan\theta=t=\tan(\pi/4+3\beta/2).$

Since the equations of the form $[\mathcal{D}_{i},\mathcal{D}_{j}]=0$ are
invariant under complex gauge transformations, we can choose particualrly
simple $\mathcal{D}_{i}$ that satisfy the first three equations and then find
a particular gauge so that the final equation is satisfied as well. In this
case, we reduce the problem to the study of the final equation

$\Lambda(\cos^{2}\beta[\mathcal{D}_{1},\mathcal{D}_{1}^{\dagger}]-\sin^{2}\beta[\mathcal{D}_{2},\mathcal{D}_{2}^{\dagger}])+[\mathcal{D}_{3},\mathcal{D}_{3}^{\dagger}]=0,$
(1.3)

where the unknown becomes a gauge transformation, or equivalently as we will
show in the next section, a hermitian metric. At this point, our approach
diverges from that of [3]. The main reason is that in the present context it
is not obvious how to construct a good approximate metric in order to perform
the corresponding continuity argument. We instead start by writing the
solution metric in the form (2.10) and consider the ansatz (2.12). The
motivation behind this ansatz is that a model solution of charge $k$ should
possess certain symmetries and also converge to the explicit model solution of
charge $k$ when $\beta\to 0$.

The proof of the main theorem is comprized of two parts. The first part is a
careful study of the linearization of (1.3) around an exact solution for each
$\beta$. We show that the linearized operator is an isomorphism between
appropriately chosen Banach spaces, and apply the inverse function theorem to
prove that if we already have a solution for $\beta$ then we have a solution
in an open neighborhood of $\beta$.

The second part contains the a priori estimates necessary for closedness. The
key observation is that once we assume that the solution has the particular
form (2.12) mentioned above together with specific asymptotics, we can find an
integrated identity that the solution has to satisfy. Using this identity, we
obtain crude global bounds which we progressively refine, first in a
neighborhood of zero and then everywhere else.

As the careful reader will notice, the a-priori estimates fail precisely at
$\theta=0,\pi/2$. Reading through the proof, it might seem at first that this
is a mathematical artifact in the sense that maybe the ansatz (2.12) is too
restrictive in general. However, the values $0$ and $\pi/2$ are special in the
following regard. If we re-write the family (1.2) as

$\begin{split}(F_{A}-\phi\wedge\phi-t\star d_{A}\phi)^{+}&=0\\\
(F_{A}-\phi\wedge\phi+t^{-1}\star d_{A}\phi)^{-}&=0\\\
d_{A}\star\phi&=0\end{split}$ (1.4)

where $t=\tan\theta$, the specialization of the equations at $t=\theta=0$ is
given by

$\begin{split}(F_{A}-\phi\wedge\phi)^{+}&=0\\\ (d_{A}\phi)^{-}&=0\\\
d_{A}\star\phi&=0.\end{split}$ (1.5)

In this special value, the equations decouple. These latter equations should
be considered as a four dimensional analog of the Hitchin equations. Currently
we do not know whether there should exist smooth model knot solutions for
$\theta=0$ or not. We hope to address the (non-)existence of the model knot
solutions as $\theta\to 0$ elsewhere. The same issue arises when
$\theta\to\pi/2$.

###  1.1. Acknowledgements

Before starting, I would like to thank Rafe Mazzeo for his patient guidance
and insights on this paper across countless meetings and especially for his
constant encouragement throughout the course of conceiving and writing this
paper.

##  2\. The twisted Bogomolny equations

###  2.1. Deriving the equations

Let us begin by introducing the Twisted Kapustin Witten (TKW) equations for a
general parameter $t\in(0,\infty)$. These equations are defined just like the
usual KW equations on a four dimensional manifold $M^{4}$ equipped with an
$SU(2)$-bundle $E$. If $\hat{A}$ is a connection on $E$ and $\Phi$ is an
ad$(E)$-valued $1$-form, then the TKW equations take the form

$\begin{split}F_{\hat{A}}-\Phi\wedge\Phi+\frac{t-t^{-1}}{2}d_{\hat{A}}\Phi+\frac{t+t^{-1}}{2}\star
d_{\hat{A}}\Phi&=0\\\ d_{\hat{A}}\star\Phi&=0.\end{split}$ (2.1)

We are primarily interested in the case where the four dimensional manifold
$M^{4}$ decomposes into a product as
$M^{4}\cong\mathbb{R}_{x_{1}}\times\Sigma_{z}\times\mathbb{R}_{y}^{+}$, with
local coordinates $(x_{1},z,y)$, where $\Sigma$ is a Riemann surface. We wish
to dimensionally reduce the equations in the $x_{1}$ direction, which means
that we want to consider $x_{1}$-invariant solutions. Expressing
$\hat{A}=A+A_{1}\,dx_{1}$, $\Phi=\phi+\phi_{1}\,dx_{1}$ and fixing the
orientation to be $\,dx_{1}\wedge\,dz\wedge\,dy$, the dimensionally reduced
Twisted Kapustin-Witten equations take the form

$\begin{split}F_{A}-\phi\wedge\phi+\frac{t-t^{-1}}{2}d_{A}\phi+\frac{t+t^{-1}}{2}(\star
d_{A}\phi_{1}+\star[\phi,A_{1}])&=0\\\
d_{A}A_{1}-[\phi,\phi_{1}]+\frac{t-t^{-1}}{2}(d_{A}\phi_{1}+[\phi,A_{1}])-\frac{t+t^{-1}}{2}\star
d_{A}\phi&=0\\\ d_{A}^{\star}\phi-[\phi_{1},A_{1}]&=0,\end{split}$ (2.2)

where $\star$ is the Hodge star on $\Sigma\times\mathbb{R}_{y}^{+}$. In [1]
Gaiotto and Witten observed that if we write $t=\tan(\pi/4-3\beta/2)$ and
under the condition $A_{1}-\tan\beta\phi_{1}=0$, the above dimensionally
reduced equations have a Hermitian Yang-Mills structure. In particular, if we
write $d_{A}=D_{\bar{z}}\,d{\bar{z}}+D_{z}\,dz+D_{y}\,dy$,
$\phi=\phi_{\bar{z}}\,d\bar{z}+\phi_{z}\,dz$ and define the operators

$\begin{split}&\mathcal{D}_{1}=(D_{\bar{z}}-\phi_{\bar{z}}\tan\beta)\,d\bar{z},~{}\mathcal{D}_{2}=(D_{z}+\phi_{z}\cot\beta)\,dz,~{}\mathcal{D}_{3}=D_{y}-i\frac{\phi_{1}}{\cos\beta}\\\
&\mathcal{D}_{1}^{\dagger}=(D_{z}-\phi_{z}\tan\beta)\,dz,~{}\mathcal{D}_{2}^{\dagger}=(D_{\bar{z}}+\phi_{\bar{z}}\cot\beta)\,d\bar{z},~{}\mathcal{D}_{3}^{\dagger}=D_{y}+i\frac{\phi_{1}}{\cos\beta},\end{split}$
(2.3)

the equations (2.2) take the following form

$\begin{split}&[\mathcal{D}_{i},\mathcal{D}_{j}]=0,~{}\text{i,j}=1,2,3,\\\
2i\Lambda(\cos^{2}\beta&[\mathcal{D}_{1},\mathcal{D}_{1}^{\dagger}]-\sin^{2}\beta[\mathcal{D}_{2},\mathcal{D}_{2}^{\dagger}])+[\mathcal{D}_{3},\mathcal{D}_{3}^{\dagger}]=0,\end{split}$
(2.4)

where $\Lambda:\Omega^{1,1}\rightarrow\Omega^{0}$ is the inner product with
the Kähler form $\frac{i}{2}\,dz\wedge\,d\bar{z}$. From this point onward, we
will call the latter equations the Twisted Extended Bogomolny Equations
(TEBE). A complete derivation of (2.4) from (2.2) can be found in the appendix
of [2].

######  Remark 2.1.

As we will demonstrate in the next subsection, the operators in (2.3) are in
unitary gauge.

###  2.2. Holomorphic versus unitary gauge

In this subsection we explain how to move between different gauge choices and
derive formulas for the connection and the field coefficients. We will focus
on the case where the Riemann surface $\Sigma$ is the complex plane since this
is the relevant case for the present paper. See [2] for a treatment of the
general case.

Fixing the bundle $E$ over $\mathbb{C}\times\mathbb{R}^{+}$ with hermitian
metric $H$, let $\mathcal{D}_{i}$ be first order differential operators acting
on sections $s\in C^{\infty}(\mathbb{C}\times\mathbb{R}^{+};E)$ such that

$\displaystyle\mathcal{D}_{1}fs=(\partial_{\bar{z}}f)s+f\mathcal{D}_{1}s,~{}\mathcal{D}_{2}fs=(\partial_{z}f)s+f\mathcal{D}_{2}s,~{}\mathcal{D}_{3}fs=(\partial_{y}f)s+f\mathcal{D}_{3}s$
$\displaystyle[\mathcal{D}_{i},\mathcal{D}_{j}]=0,$

where $f$ is a smooth function on the base. We define the adjoints of these
operators with respect to the metric $H$ in the following way:

$\displaystyle\partial_{\bar{z}}H(s,s^{\prime})$
$\displaystyle=H(\mathcal{D}_{1}^{\dagger}s,s^{\prime})+H(s,\mathcal{D}_{1}s^{\prime})$
$\displaystyle\partial_{z}H(s,s^{\prime})$
$\displaystyle=H(\mathcal{D}_{2}^{\dagger}s,s^{\prime})+H(s,\mathcal{D}_{2}s^{\prime})$
$\displaystyle\partial_{y}H(s,s^{\prime})$
$\displaystyle=H(\mathcal{D}_{3}^{\dagger}s,s^{\prime})+H(s,\mathcal{D}_{3}s^{\prime}).$

Here, the notation $H(s,s^{\prime}):=\bar{s}^{T}Hs^{\prime}$, so the inner
product is complex linear in the second argument. This definition differs from
the usual one but we stick to the current definition because it is consistent
with most of the work done on the KW equations. We denote the set of
differential operators $\\{\mathcal{D}_{1},\mathcal{D}_{2},\mathcal{D}_{3}\\}$
by $\Theta$. Next, we need to define the following operators:

$\begin{split}\mathcal{D}_{z}&=\sin^{2}\beta\mathcal{D}_{2}+\cos^{2}\beta\mathcal{D}_{1}^{\dagger},~{}\mathcal{D}_{\bar{z}}=\sin^{2}\beta\mathcal{D}_{2}^{\dagger}+\cos^{2}\beta\mathcal{D}_{1},\\\
\phi_{z}&=\sin\beta\cos\beta(\mathcal{D}_{2}-\mathcal{D}_{1}^{\dagger}),~{}\phi_{\bar{z}}=\sin\beta\cos\beta(\mathcal{D}_{2}^{\dagger}-\mathcal{D}_{1}),\\\
\mathcal{D}_{y}&=\frac{\mathcal{D}_{3}+\mathcal{D}_{3}^{\dagger}}{2},~{}\phi_{1}=\frac{i\cos\beta}{2}(\mathcal{D}_{3}-\mathcal{D}_{3}^{\dagger}).\end{split}$
(2.5)

From these operators we obtain a connection
$\nabla_{A}:=\mathcal{D}_{z}+\mathcal{D}_{\bar{z}}+\mathcal{D}_{y}$ and a
triple $(A,\phi,\phi_{1})$, called the Chern connection corresponding to the
data $(E,\Theta,H)$.

We say that the operators $\Theta$ are in unitary gauge if the triple
$(A,\phi,\phi_{1})$ consists of unitary matrices and that they are in
holomorphic gauge if $\mathcal{D}_{1}=\partial_{\bar{z}}\,d\bar{z}$ and
$\mathcal{D}_{3}=\partial_{y}$. It immediately follows from this definition
that the operators in (2.3) are in unitary gauge.

Working in holomorphic gauge, the commutation relations
$[\mathcal{D}_{i},\mathcal{D}_{j}]=0$ imply that
$\mathcal{D}_{2}=\partial_{z}+B_{z}$, where $B_{z}$ is a trace zero matrix
whose entries are holomorphic functions in $z$. As we discuss in the next
subsection, in order to generate model solutions of different charges,
different choices of $B_{z}$ will be needed. At this point it is instructive
to calculate the adjoints of $\Theta$ in holomorphic gauge and explicitly
write down the last equation in (2.4). We have that

$\begin{split}\partial_{\bar{z}}H(s,s^{\prime})&=H(\partial_{z}s,s^{\prime})+(\partial_{\bar{z}}H)(s,s^{\prime})+H(s,\partial_{\bar{z}}s^{\prime})\\\
&=H((\partial_{z}+H^{-1}\partial_{z}H,s^{\prime})+H(s,\partial_{\bar{z}}s^{\prime})\end{split}$

and therefore
$\mathcal{D}_{1}^{\dagger}=\partial_{z}+H^{-1}\partial_{z}H=H^{-1}\circ\partial_{z}\circ
H$. Similarly, we obtain that
$\mathcal{D}_{2}^{\dagger}=H^{-1}\circ(\partial_{\bar{z}}-B_{z}^{\dagger})\circ
H$ and $\mathcal{D}_{3}^{\dagger}=H^{-1}\circ\partial_{y}\circ H$. Plugging
these formulas into the last equation in (2.4) we obtain the following
equation in the metric $H$:

$4\partial_{\bar{z}}H^{-1}\partial_{z}H+4(\partial_{z}+B_{z})H^{-1}(\partial_{\bar{z}}-B_{z}^{\dagger})H+\partial_{y}H^{-1}\partial_{y}H=0.$
(2.6)

######  Proposition 2.2.

Assume that the data $(E,\Theta,H)$ satisfy (2.4). Then the data
$(E,\Theta^{g},H^{g})$ do as well, where
$\Theta^{g}:=(g^{-1}\circ\mathcal{D}_{1}\circ
g,g^{-1}\circ\mathcal{D}_{2}\circ g,g^{-1}\circ\mathcal{D}_{3}\circ g)$, and
$H^{g}:=g^{\dagger}Hg$.

###### Proof.

The operators $\Theta^{g}$ clearly satisfy the first three equations in (2.4).
In order to verify that the forth equation holds, we need to compute the
adjoints of the modified operators with respect to the modified metric
$H^{g}$. We have that

$\displaystyle\partial_{\bar{z}}H^{g}(s,s^{\prime})$
$\displaystyle=H(gs,gs^{\prime})$
$\displaystyle=H(\mathcal{D}_{1}^{\dagger}gs,gs^{\prime})+H(gs,\mathcal{D}_{1}gs^{\prime})$
$\displaystyle=H^{g}(g^{-1}\circ\mathcal{D}_{1}^{\dagger}\circ
gs,s^{\prime})+H^{g}(s,g^{-1}\circ\mathcal{D}_{1}\circ gs^{\prime}),$

and therefore,
$\mathcal{D}_{1}^{\dagger_{H^{g}}}=g^{-1}\circ\mathcal{D}_{1}^{\dagger}\circ
g$. Similarly,
$\mathcal{D}_{2}^{\dagger_{H^{g}}}=g^{-1}\circ\mathcal{D}_{2}^{\dagger}\circ
g$ and
$\mathcal{D}_{3}^{\dagger_{H^{g}}}=g^{-1}\circ\mathcal{D}_{3}^{\dagger}\circ
g$. Plugging these formulas into the last equation we obtain equation (2.6)
conjugated with $g$ and therefore the last equation is also satisfied.

∎

######  Remark 2.3.

As we mentioned in the previous subsection, the TEBE equations have a
Hermitian Yang-Mills structure. Let us briefly explain this now that we have
developed the required vocabulary. Denote by $\mathcal{G}_{\mathbb{C}}$ the
group of complex gauge transformations of $E$. As we saw, the differential
operators $\mathcal{D}_{i}$ transform under such gauge transformations in the
following way:

$\mathcal{D}_{i}^{g}=g^{-1}\circ\mathcal{D}_{i}\circ
g,~{}~{}\mathcal{D}_{i}^{\dagger_{H^{g}}}=g^{-1}\circ\mathcal{D}_{i}^{\dagger}\circ
g$

It is clear that the first three equations in (2.4) are invariant under the
action of the group $\mathcal{G}_{\mathbb{C}}$, while the last equation is
only invariant under the subgroup of unitary gauge transformations $g$ with
respect to the metric $H$, or more explicitly those which satisfy
$g^{\dagger}Hg=H$. Because of this, the last equation in (2.4) may be
considered as the moment map for the first three equations. From here on we
use this name to refer to this equation.

Finally, we show how to pass from holomorphic to unitary gauge. From the
definition it follows that in unitary gauge, if
$\mathcal{D}_{1}=\partial_{\bar{z}}+B$ then
$\mathcal{D}_{1}^{\dagger}=\partial_{z}-B^{\dagger}$. Writing
$\mathcal{D}_{1}=g_{1}^{-1}\circ\partial_{\bar{z}}\circ g_{1}$,
$B=g_{1}^{-1}(\partial_{\bar{z}}g_{1})$. Using the formulas above for the
adjoint, $D_{1}^{\dagger}=g_{1}^{-1}H^{-1}\circ\partial_{z}\circ Hg_{1}$ and
therefore, it must hold that
$g_{1}^{-1}(\partial_{\bar{z}}g_{1})=g_{1}^{\dagger}H(\partial_{z}H^{-1}(g_{1}^{-1})^{\dagger})$.
Writing $H=g^{\dagger}g$, we see that the only way this is possible is if
$g_{1}=g^{-1}$. Therefore, we obtain the following formulas for the operators
$\Theta$ is unitary gauge:

$\begin{split}\mathcal{D}_{1}&=g\circ\partial_{\bar{z}}\circ
g^{-1}=\partial_{\bar{z}}-(\partial_{\bar{z}}g)g^{-1}\\\
\mathcal{D}_{1}^{\dagger}&=(g^{-1})^{\dagger}\circ\partial_{z}\circ
g^{\dagger}=\partial_{z}+(g^{-1})^{\dagger}\partial_{z}g^{\dagger}\\\
\mathcal{D}_{2}&=g\circ(\partial_{z}+B_{z})\circ
g^{-1}=\partial_{z}-(\partial_{z}g)g^{-1}+gB_{z}g^{-1}\\\
\mathcal{D}_{2}^{\dagger}&=(g^{-1})^{\dagger}\circ(\partial_{\bar{z}}-B_{z}^{\dagger})\circ
g^{\dagger}=\partial_{\bar{z}}+(g^{-1})^{\dagger}\partial_{\bar{z}}g^{\dagger}-(g^{-1})^{\dagger}B_{z}^{\dagger}g^{\dagger}\\\
\mathcal{D}_{3}&=g\circ\partial_{y}\circ
g^{-1}=\partial_{y}-(\partial_{y}g)g^{-1}\\\
\mathcal{D}_{3}^{\dagger}&=(g^{-1})^{\dagger}\circ\partial_{y}\circ
g^{\dagger}=\partial_{y}+(g^{-1})^{\dagger}\partial_{y}g^{\dagger}.\end{split}$
(2.7)

Plugging in the expressions (2.7) in the formulas (2.5), we obtain the
connection and fields in unitary gauge

$\begin{split}A_{z}&=\sin^{2}\beta(-(\partial_{z}g)g^{-1}+g^{-1}B_{z}g)+\cos^{2}\beta(g^{-1})^{\dagger}\partial_{z}g^{\dagger}\\\
\phi_{z}&=\cos\beta\sin\beta(-(\partial_{z}g)g^{-1}+g^{-1}B_{z}g-(g^{-1})^{\dagger}\partial_{z}g^{\dagger}\\\
A_{y}&=\frac{1}{2}(-(\partial_{y}g)g^{-1}+(g^{-1})^{\dagger}\partial_{y}g^{\dagger}\\\
\phi_{1}&=-\frac{i}{2}\cos\beta((\partial_{y}g)g^{-1}+(g^{-1})^{\dagger}\partial_{y}g^{\dagger}),\end{split}$
(2.8)

with $A_{\bar{z}}=-A_{z}^{\dagger}$ and $\phi_{\bar{z}}=-\phi_{z}^{\dagger}$.

###  2.3. The moment map equation

In this subsection we focus on the moment map equation. In different parts of
the paper we will study the moment map in different gauges. In order to prove
openness it is more convenient to work in unitary gauge while the a-priori
estimates for closedness will be done in holomorphic gauge. Here, we restrict
to the latter and after producing an ansatz for the metric $H$, we write down
the moment map equations for this ansatz.

In order to explicitly write down the moment map equation, we need to choose
$B_{z}$. Since we want to perturb off of solutions to the extended Bogomolny
equations, it is natural to require that in the limit $\beta\to 0$, the moment
map equation we obtain is the same as the one appearing in [[]Section 2]MH2.
Undoubtedly, the simplest possible way of achieving this is to choose
$B_{z}=\csc\beta\begin{pmatrix}0&z^{k}\\\ 0&0\end{pmatrix}$. With this
particular choice, (2.6) becomes

$\begin{split}4\cos^{2}\beta~{}\partial_{\bar{z}}H^{-1}\partial_{z}H&+4\sin^{2}\beta~{}\left(\partial_{z}+\csc\beta\begin{pmatrix}0&z^{k}\\\
0&0\end{pmatrix}\right)H^{-1}\left(\partial_{\bar{z}}-\csc\beta\begin{pmatrix}0&0\\\
\bar{z}^{k}&0\end{pmatrix}\right)H\\\
&+\partial_{y}H^{-1}\partial_{y}H=0.\end{split}$ (2.9)

Following the notation of [[]Appendix C]GW, we express $H$ as

$H=\begin{pmatrix}Y^{-1}&-Y^{-1}\Sigma\\\
-Y^{-1}\bar{\Sigma}&Y+Y^{-1}|\Sigma|^{2}\end{pmatrix},$ (2.10)

where $Y$ is a positive and $\Sigma$ a complex-valued function. Then

$H^{-1}=\begin{pmatrix}Y+Y^{-1}|\Sigma|^{2}&Y^{-1}\Sigma\\\
Y^{-1}\bar{\Sigma}&Y^{-1}\end{pmatrix}.$

Using these in (2.6) yields a system of partial differential equations:

$\begin{split}\omega^{i}(\partial_{i}(Y^{-1}\bar{\partial}_{i}Y)+Y^{-2}\bar{\partial_{i}\Sigma}\partial_{i}\Sigma)+4|z|^{2k}Y^{-2}-2\sin\beta\,Y^{-2}(\bar{z}^{k}\partial_{z}\Sigma+z^{k}\partial_{\bar{z}}\bar{\Sigma})&=0\\\
\omega^{i}\partial_{i}(Y^{-2}\bar{\partial}_{i}\bar{\Sigma})-4\sin\beta\,\bar{z}^{k}\partial_{z}Y^{-2}&=0,\end{split}$
(2.11)

where $\omega^{1}=4\cos^{2}\beta$, $\omega^{2}=4\sin^{2}\beta$ and
$\omega^{3}=1$.

As presented in (2.11), this system is not easy to analyze. Fortunately it can
be simplified by rewriting $Y$ and $\Sigma$ in a suitable form. For the charge
$k$ model solution when $\beta=0$, $\Sigma\equiv 0$ and
$Y=yr^{k}e^{u(\sigma)}$, where $r=|z|$, $\sigma:=y/r$ and $u$ is real-valued
and satisfies

$u(\sigma)\to 1~{}\text{as}~{}\sigma\to 0,\quad u(\sigma)\sim
k\log\sigma~{}\text{as}~{}\sigma\to\infty.$

More generally then, we shall write $Y$ in this form, for some function
$u=u(\sigma)$ with the same asymptotics. To determine a good ansatz for the
form of $\Sigma$, first observe that if we insert this expression for $Y$ into
the second equation of (2.11) and expand in the lowest power in $y$, we must
have

$\Sigma\sim 2\sin\beta\,kz^{k+1}\sigma^{2}\ \ \mbox{as}\ \ \ y\to 0.$

Note also that (2.11) is invariant under dilations $(y,z)\to(\lambda y,\lambda
z)$, $\lambda>0$, as well as under rotations $z\to e^{i\theta}z$. Finally, it
is also invariant under reflections $z\to\bar{z}$, $\Sigma\to\bar{\Sigma}$
(the latter is needed only when $\zeta^{2}\neq 1$). The Nahm pole boundary
condition with a charge $k$ knot is invariant this entire set of symmetries.
This leads to the ansatz

$Y:=r^{k+1}e^{u(\sigma)},\quad\Sigma:=\sin\beta\,z^{k+1}v(\sigma),$ (2.12)

with

$\begin{split}&u(\sigma)\sim\log\sigma~{}\text{as}~{}\sigma\to 0\\\
&u(\sigma)\sim(k+1)\log\sigma~{}\text{as}~{}\sigma\to\infty.\\\ &v(\sigma)\sim
2k\sigma^{2}~{}\textrm{as}~{}\sigma\to 0.\end{split}$ (2.13)

Since $\Sigma$ should be bounded along the axis where $z=0$, we also expect
that

$|r^{k+1}v(\sigma)|\leq C,\ \mbox{i.e.,}\ \ |v(\sigma)|\leq C\sigma^{k+1}\
\mbox{as}\ \sigma\to\infty.$

Note that the supposed form of $Y$ is slightly different that the one above;
the present form is simpler for computations. Inserting (2.12) into (2.11)
produces the system

$\begin{split}(\sigma^{2}+1)u^{\prime\prime}+&\sigma u^{\prime}\\\
&+e^{-2u}\left[(\textrm{sin}^{2}\beta(2(k+1)v-\sigma
v^{\prime})-2)^{2}+\textrm{sin}^{2}\beta(\cos^{2}\beta\sigma^{2}+1)v^{\prime
2}\right]=0\\\ (\sigma^{2}+1)v^{\prime\prime}+&\sigma v^{\prime}\\\
&+4(k+1)(1-\sin^{2}\beta(k+1)v)-\left[4\sigma(1-\sin^{2}\beta(k+1)v)+2(\sigma^{2}+1)v^{\prime}\right]u^{\prime}=0.\end{split}$
(2.14)

We shall write these equations in a somewhat more tractable form in Section 4,
where we analyze them more carefully.

##  3\. Linear Analysis

###  3.1. The linearized equation

As the title of this subsection suggests, we start here by linearizing the
moment map equation

$\Omega_{H}:=2i\Lambda(\cos^{2}\beta\,[\mathcal{D}_{1},\mathcal{D}_{1}^{\dagger}]-\sin^{2}\beta\,[\mathcal{D}_{2},\mathcal{D}_{2}^{\dagger}])+[\mathcal{D}_{3},\mathcal{D}_{3}^{\dagger}]=0.$
(3.1)

To do this, we fix a background hermitian metric $H_{0}$ on the bundle $E$. To
linearize (3.2) around $H_{0}$, we consider variations $H=H_{0}e^{s}$ with
$s\in\Gamma(\mathbb{R}^{2}\times\mathbb{R}^{+},i\mathfrak{su}(E,H_{0}))$.
Substituting this $H$ into the equation yields

######  Proposition 3.1.

$\Omega_{H}=\Omega_{H_{0}}+\gamma(-s)\mathcal{L}_{H_{0}}s+Q(s),$ (3.2)

where

$\mathcal{L}_{H_{0}}s:=2i\Lambda(\cos^{2}\beta\,\mathcal{D}_{1}\mathcal{D}_{1}^{\dagger_{H_{0}}}-\sin^{2}\beta\,\mathcal{D}_{2}\mathcal{D}_{2}^{\dagger_{H_{0}}})s+\mathcal{D}_{3}\mathcal{D}_{3}^{\dagger_{H_{0}}}s,$
(3.3)

$\gamma(s):=\frac{e^{\text{ad}_{s}}-1}{\text{ad}_{s}}$ and $Q(s)$ is quadratic
in $s$.

###### Proof.

Keeping the $\mathcal{D}_{i}$ fixed,
$\mathcal{D}_{i}^{\dagger_{H}}=\mathcal{D}_{i}^{\dagger_{H_{0}}}+e^{-s}(\mathcal{D}_{i}^{\dagger_{H_{0}}}e^{s})$.
Indeed,

$\displaystyle\partial_{i}H(s^{\prime},s^{\prime\prime})$
$\displaystyle=H_{0}(\mathcal{D}_{i}^{\dagger_{H_{0}}}\circ
e^{s}s^{\prime},s^{\prime\prime})+H_{0}(e^{s}s^{\prime},\mathcal{D}_{i}s^{\prime\prime})$
$\displaystyle=H(e^{-s}\circ\mathcal{D}_{i}^{\dagger_{H_{0}}}\circ
e^{s}s^{\prime},s^{\prime\prime})+H(s^{\prime},\mathcal{D}_{i}s^{\prime\prime}).$

Using the identity
$e^{-s}(\mathcal{D}_{i}^{\dagger_{H_{0}}}e^{s})=\gamma(-s)\mathcal{D}_{i}^{\dagger_{H_{0}}}s$
we indeed get

$\displaystyle\Omega_{H}=$
$\displaystyle\Omega_{H_{0}}+\gamma(-s)\mathcal{L}_{H_{0}}s+$ $\displaystyle
2i\Lambda(\cos^{2}\beta\,(\mathcal{D}_{1}\gamma(-s))\mathcal{D}_{1}^{\dagger_{H_{0}}}-\sin^{2}\beta\,(\mathcal{D}_{2}\gamma(-s))\mathcal{D}_{2}^{\dagger_{H_{0}}})s+(\mathcal{D}_{3}\gamma(-s))\mathcal{D}_{3}^{\dagger_{H_{0}}}s.$

∎

In order to state our next result, we will need to consider adjoints of
operators with respect to the usual inner product on forms taking values in
the bundle $E$. By this, we mean that if $A$ is an operator and $a$ and $b$
are bundle valued forms, then

$\int\langle Aa,b\rangle_{H}=\int\langle a,A^{\star}b\rangle_{H}.$

The following lemma first appeared in [2]. We include a simplified proof here
for completeness. The sign diffference between the original result and the one
appearing below appears because contrary to [2] we stick with the
unconventional definition of Hermitian adjoints throughout the paper.

######  Lemma 3.2.

$\mathcal{L}_{H_{0}}:=-\cos^{2}\beta~{}\nabla_{1}^{\star}\nabla_{1}-\sin^{2}\beta~{}\nabla_{2}^{\star}\nabla_{2}+\mathcal{D}_{y}^{2}+\frac{\phi_{1}^{2}}{\cos^{2}\beta}-\frac{1}{2}[\Omega_{H},\star]$
(3.4)

where $\nabla_{i}:=\mathcal{D}_{i}+\mathcal{D}_{i}^{\dagger_{H_{0}}}$,
$\phi_{1}^{2}:=[\phi_{1},[\phi_{1},\star]]$ and $\mathcal{D}_{y}$ , $\phi_{1}$
were defined in (2.5).

###### Proof.

We start by recalling the Kahler identities [8]

$\begin{split}i[\Lambda,\mathcal{D}_{1}]&=-(\mathcal{D}_{1}^{\dagger_{H}})^{\star},~{}i[\Lambda,\mathcal{D}_{1}^{\dagger_{H}}]=\mathcal{D}_{1}^{\star}\\\
i[\Lambda,\mathcal{D}_{2}]&=(\mathcal{D}_{2}^{\dagger_{H}})^{\star},~{}i[\Lambda,\mathcal{D}_{2}^{\dagger_{H}}]=-\mathcal{D}_{2}^{\star}\\\
(\mathcal{D}_{3}^{\dagger_{H}})^{\star}&=\mathcal{D}_{3}.\end{split}$ (3.5)

The signs are reversed from the usual Kahler identities as written down in [2,
8] because of the unconventional definition of the Hermitian adjoints we use
in this paper. Using the Kähler identities, we obtain

$\begin{split}\nabla_{i}^{\star}\nabla_{i}&=\mathcal{D}_{i}^{\star}\mathcal{D}_{i}+(\mathcal{D}_{i}^{\dagger})^{\star}\mathcal{D}_{i}^{\dagger}\\\
&=-i\Lambda\mathcal{D}_{i}\mathcal{D}_{i}^{\dagger}+i\Lambda\mathcal{D}_{i}^{\dagger}\mathcal{D}_{i}\\\
&=-2i\Lambda\mathcal{D}_{i}\mathcal{D}_{i}^{\dagger}+i\Lambda[\mathcal{D}_{i},\mathcal{D}_{i}^{\dagger}].\end{split}$

Then, by (2.5),

$\begin{split}4\left(\mathcal{D}_{y}^{2}+\frac{\phi_{1}^{2}}{\cos^{2}\beta}\right)&=(\mathcal{D}_{3}+\mathcal{D}_{3}^{\dagger})^{2}-(\mathcal{D}_{3}-\mathcal{D}_{3}^{\dagger})^{2}\\\
&=2(\mathcal{D}_{3}\mathcal{D}_{3}^{\dagger}+\mathcal{D}_{3}^{\dagger}\mathcal{D}_{3})\\\
&=4\mathcal{D}_{3}\mathcal{D}_{3}^{\dagger}-2[\mathcal{D}_{3},\mathcal{D}_{3}^{\dagger}].\end{split}$

Putting these identities together, we obtain the required formula. ∎

######  Remark 3.3.

Notice that in the derivation of the formula for the
$\nabla_{i}^{\star}\nabla_{i}$ term, it would seem that the second line is
equal to $-i\Lambda[\mathcal{D}_{i},\mathcal{D}_{i}^{\dagger}]$. However, in
the Kahler identity
$i[\Lambda,\mathcal{D}_{1}^{\dagger_{H}}]=\mathcal{D}_{1}^{\star}$,
$D_{1}^{\dagger_{H}}$ denotes the hermitian adjoint acting on zero forms,
which differs from the same operator acting on $(0,1)$-forms by a sign.

###  3.2. Calculations in spherical coordinates

The aim of this subsection is to derive explicit formulas for the connection
and field coefficients that appear in the linearized equation when written in
spherical coordinates. The importance of these calculations will become
apparent in the next subsection. From this point onward, we will be working in
unitary gauge.

Writing the metric (2.10) as

$H=g^{\dagger}g=\begin{pmatrix}Y^{-1/2}&0\\\
-Y^{-1/2}\bar{\Sigma}&Y^{1/2}\end{pmatrix}\begin{pmatrix}Y^{-1/2}&-Y^{-1/2}\Sigma\\\
0&Y^{1/2}\end{pmatrix},$

and using the identities (2.7) and (2.8), we obtain the following explicit
formulas for the operators and the connection and field coefficients in
unitary gauge respectively:

$\begin{split}\mathcal{D}_{1}=g\circ\partial_{\bar{z}}\circ
g^{-1}&=\partial_{\bar{z}}+\frac{1}{2}\begin{pmatrix}Y^{-1}\partial_{\bar{z}}Y&2Y^{-1}\partial_{\bar{z}}\Sigma\\\
0&-Y^{-1}\partial_{\bar{z}}Y\end{pmatrix}\\\
\mathcal{D}_{1}^{\dagger}=(g^{-1})^{\dagger}\circ\partial_{z}\circ
g^{\dagger}&=\partial_{z}-\frac{1}{2}\begin{pmatrix}Y^{-1}\partial_{z}Y&0\\\
2Y^{-1}\partial_{z}\bar{\Sigma}&-Y^{-1}\partial_{z}Y\end{pmatrix}\\\
\mathcal{D}_{2}=g\circ\left(\partial_{z}+\csc\beta\begin{pmatrix}0&z^{k}\\\
0&0\end{pmatrix}\right)\circ
g^{-1}&=\partial_{z}+\frac{1}{2}\begin{pmatrix}Y^{-1}\partial_{z}Y&2Y^{-1}\partial_{z}\Sigma+2\csc\beta
Y^{-1}z^{k}\\\ 0&-Y^{-1}\partial_{z}Y\end{pmatrix}\\\
\mathcal{D}_{2}^{\dagger}=(g^{-1})^{\dagger}\circ\left(\partial_{\bar{z}}-\csc\beta\begin{pmatrix}0&0\\\
\bar{z}^{k}&0\end{pmatrix}\right)\circ
g^{\dagger}&=\partial_{\bar{z}}-\frac{1}{2}\begin{pmatrix}Y^{-1}\partial_{\bar{z}}Y&0\\\
2Y^{-1}\partial_{\bar{z}}\bar{\Sigma}+2\csc\beta
Y^{-1}\bar{z}^{k}&-Y^{-1}\partial_{\bar{z}}Y\end{pmatrix}\\\
\mathcal{D}_{3}=g\circ\partial_{y}\circ
g^{-1}&=\partial_{y}+\frac{1}{2}\begin{pmatrix}Y^{-1}\partial_{y}Y&2Y^{-1}\partial_{y}\Sigma\\\
0&-Y^{-1}\partial_{y}Y\end{pmatrix}\\\
\mathcal{D}_{3}^{\dagger}=(g^{-1})^{\dagger}\circ\partial_{y}\circ
g^{\dagger}&=\partial_{y}-\frac{1}{2}\begin{pmatrix}Y^{-1}\partial_{y}Y&0\\\
2Y^{-1}\partial_{y}\bar{\Sigma}&-Y^{-1}\partial_{y}Y\end{pmatrix}\end{split}$
(3.6)

and

$\begin{split}A_{\bar{z}}&=\frac{\cos^{2}\beta}{2}\begin{pmatrix}Y^{-1}\partial_{\bar{z}}Y&2Y^{-1}\partial_{\bar{z}}\Sigma\\\
0&-Y^{-1}\partial_{\bar{z}}Y\end{pmatrix}-\frac{\sin^{2}\beta}{2}\begin{pmatrix}Y^{-1}\partial_{\bar{z}}Y&0\\\
2Y^{-1}\partial_{\bar{z}}\bar{\Sigma}-2\csc\beta
Y^{-1}\bar{z}^{k}&-Y^{-1}\partial_{\bar{z}}Y\end{pmatrix}\\\
\phi_{\bar{z}}&=\sin\beta\cos\beta\begin{pmatrix}-Y^{-1}\partial_{\bar{z}}Y&-Y^{-1}\partial_{\bar{z}}\Sigma\\\
-Y^{-1}\partial_{\bar{z}}\bar{\Sigma}+\csc\beta
Y^{-1}\bar{z}^{k}&Y^{-1}\partial_{\bar{z}}Y\end{pmatrix}\\\
A_{z}&=\frac{\sin^{2}\beta}{2}\begin{pmatrix}Y^{-1}\partial_{z}Y&2Y^{-1}\partial_{z}\Sigma-2\csc\beta
Y^{-1}z^{k}\\\
0&-Y^{-1}\partial_{z}Y\end{pmatrix}-\frac{\cos^{2}\beta}{2}\begin{pmatrix}Y^{-1}\partial_{z}Y&0\\\
2Y^{-1}\partial_{z}\bar{\Sigma}&-Y^{-1}\partial_{z}Y\end{pmatrix}\\\
\phi_{z}&=\sin\beta\cos\beta\begin{pmatrix}Y^{-1}\partial_{z}Y&Y^{-1}\partial_{z}\Sigma-\csc\beta
Y^{-1}z^{k}\\\
Y^{-1}\partial_{z}\bar{\Sigma}&-Y^{-1}\partial_{z}Y\end{pmatrix}\\\
A_{y}&=\frac{1}{2}\begin{pmatrix}0&Y^{-1}\partial_{y}\Sigma\\\
-Y^{-1}\partial_{y}\bar{\Sigma}&0\end{pmatrix}\\\
\phi_{1}&=\frac{i\cos\beta}{2}\begin{pmatrix}Y^{-1}\partial_{y}Y&Y^{-1}\partial_{y}\Sigma\\\
Y^{-1}\partial_{y}\bar{\Sigma}&-Y^{-1}\partial_{y}Y\end{pmatrix}.\end{split}$
(3.7)

The first step is to express the linearized operator in Euclidean coordinates.
This allows us to read off the non-connection part of the linearized operator
with relative ease. To perform this calculation, it is convenient to use the
Euclidean coordinates $x_{2},x_{3}$ where $z:=x_{2}+ix_{3}$. We compute each
term in (3.4) explicitly.

$\displaystyle\mathcal{D}_{1}+\mathcal{D}_{1}^{\dagger}$
$\displaystyle=(\partial_{\bar{z}}+A_{\bar{z}}-\phi_{\bar{z}}\tan\beta)\,d\bar{z}+(\partial_{z}+A_{z}-\phi_{z}\tan\beta)\,dz$
$\displaystyle=(\partial_{2}+A_{2}-\phi_{2}\tan\beta)\,dx_{2}+(\partial_{3}+A_{3}-\phi_{3}\tan\beta)\,dx_{3}$
$\displaystyle=(D_{2}-\phi_{2}\tan\beta)\,dx_{2}+(D_{3}-\phi_{3}\tan\beta)\,dx_{3},$

where
$\phi_{2}\,dx_{2}+\phi_{3}\,dx_{3}=\phi_{z}\,dz+\phi_{\bar{z}}\,d\bar{z}$. The
adjoint operator $\nabla_{1}^{\star}$ is equal to

$\displaystyle\nabla_{1}^{\star}$
$\displaystyle=-(\partial_{2}+A_{2}-\phi_{2}\tan\beta)\,\langle
dx^{2},\star\rangle-(\partial_{3}+A_{3}-\phi_{3}\tan\beta)\,\langle
dx^{3},\star\rangle$
$\displaystyle=-(D_{2}-\phi_{2}\tan\beta)(\,dx^{2})^{\star}-(D_{3}-\phi_{3}\tan\beta)(\,dx^{3})^{\star}$

since in unitary gauge the metric is just the usual inner product. Therefore,

$\nabla_{1}^{\star}\nabla_{1}=-(D_{2}-\tan\beta\,\phi_{2})(D_{2}-\tan\beta\,\phi_{2})-(D_{3}-\tan\beta\,\phi_{3})(D_{3}-\tan\beta\,\phi_{3}).$

Similarly one obtains the equality

$\nabla_{2}^{\star}\nabla_{2}=-(D_{2}+\cot\beta\,\phi_{2})(D_{2}+\cot\beta\,\phi_{2})-(D_{3}+\cot\beta\,\phi_{3})(D_{3}+\cot\beta\,\phi_{3}).$

Plugging these formulae into the linearized operator and expanding, we obtain

$\mathcal{L}_{H}=D_{2}^{2}+D_{3}^{2}+D_{y}^{2}+\frac{[\phi_{1},[\phi_{1},\star]]}{\cos^{2}\beta}+[\phi_{2},[\phi_{2},\star]]+[\phi_{3},[\phi_{3},\star]].$

Now define

$\frac{N}{\rho^{2}}:=\frac{[\phi_{1},[\phi_{1},\star]]}{\cos^{2}\beta}+[\phi_{2},[\phi_{2},\star]]+[\phi_{3},[\phi_{3},\star]].$

Using (2.12), the operator $N$ is independent of $\rho$. In fact, we show
later that $N$ has a particularly special form.

We are now in a position to express $\mathcal{L}_{H}$ in spherical coordinates
$(\rho,\psi,\theta)$, where $y=\rho\cos\psi$ and $z=x_{2}+ix_{3}=\sin\psi\rho
e^{i\theta}$. This takes the form

$\mathcal{L}_{H}=D_{\rho}^{2}+\frac{2}{\rho}D_{\rho}+\frac{1}{\rho^{2}}(D_{\psi}^{2}+\cot\psi
D_{\psi}+\frac{1}{\sin^{2}\psi}D_{\theta}^{2})+N.$ (3.8)

To compute the connection coefficients $A_{\rho}$, $A_{\psi}$ and
$A_{\theta}$, insert (3.6) into (3.4) to get the explicit formula

$\begin{split}\mathcal{L}_{H}=\cos^{2}\beta&\left(\partial_{x_{1}}+\frac{1}{2}\begin{pmatrix}i\sin\theta\,Y^{-1}\partial_{r}Y&Y^{-1}\partial_{\bar{z}}\Sigma\\\
-Y^{-1}\partial_{z}\bar{\Sigma}&-i\sin\theta\,Y^{-1}\partial_{r}Y\end{pmatrix}\right)^{2}\\\
\cos^{2}\beta&\left(\partial_{x_{2}}-\frac{i}{2}\begin{pmatrix}\cos\theta\,Y^{-1}\partial_{r}Y&Y^{-1}\partial_{\bar{z}}\Sigma\\\
Y^{-1}\partial_{z}\bar{\Sigma}&-\cos\theta\,Y^{-1}\partial_{r}Y\end{pmatrix}\right)^{2}\\\
\sin^{2}\beta&\left(\partial_{x_{1}}+\frac{1}{2}\begin{pmatrix}-i\sin\theta\,Y^{-1}\partial_{r}Y&Y^{-1}\partial_{z}\Sigma+\csc\beta
Y^{-1}z^{k}\\\ -Y^{-1}\partial_{\bar{z}}\bar{\Sigma}-\csc\beta
Y^{-1}\bar{z}^{k}&i\sin\theta\,Y^{-1}\partial_{r}Y\end{pmatrix}\right)^{2}\\\
\sin^{2}\beta&\left(\partial_{x_{2}}+\frac{i}{2}\begin{pmatrix}\cos\theta\,Y^{-1}\partial_{r}Y&Y^{-1}\partial_{z}\Sigma+\csc\beta
Y^{-1}z^{k}\\\ Y^{-1}\partial_{\bar{z}}\bar{\Sigma}+\csc\beta
Y^{-1}\bar{z}^{k}&-\cos\theta\,Y^{-1}\partial_{r}Y\end{pmatrix}\right)^{2}\\\
&\left(\partial_{y}+\frac{1}{2}\begin{pmatrix}0&Y^{-1}\partial_{y}\Sigma\\\
-Y^{-1}\partial_{y}\bar{\Sigma}&0\end{pmatrix}\right)^{2}-\frac{1}{4}\left(\begin{pmatrix}Y^{-1}\partial_{y}Y&Y^{-1}\partial_{y}\Sigma\\\
Y^{-1}\partial_{y}\bar{\Sigma}&-Y^{-1}\partial_{y}Y\end{pmatrix}\right)^{2}.\end{split}$
(3.9)

Next, use the identities

$\begin{split}\partial_{x_{2}}&=\sin\psi\cos\theta\,\partial_{\rho}-\frac{\sin\theta\,}{\rho\sin\psi}\partial_{\theta}+\frac{\cos\psi\cos\theta}{\rho}\partial_{\psi}\\\
\partial_{x_{3}}&=\sin\psi\sin\theta\,\partial_{\rho}+\frac{\cos\theta}{\rho\sin\psi}\partial_{\theta}+\frac{\cos\psi\sin\theta}{\rho}\partial_{\psi}\\\
\partial_{y}&=\cos\psi\partial_{\rho}-\frac{\sin\psi}{\rho}\partial_{\psi}\end{split}$
(3.10)

to find the coefficients of the partial derivatives
$\partial_{\rho},~{}\partial_{\psi},~{}\partial_{\theta}$. After some
calculation, we obtain that the coefficient of $\partial_{\rho}$ equals

$\begin{pmatrix}0&A\\\ -\bar{A}&0\end{pmatrix}+\frac{2}{\rho},$

where

$A=\frac{1}{2}\left[\sin\psi\left(\cos^{2}\beta
e^{-i\theta}Y^{-1}\partial_{\bar{z}}\Sigma+\sin^{2}\beta
e^{i\theta}(Y^{-1}\partial_{z}\Sigma+\csc\beta Y^{-1}z^{k})\right)+\cos\psi
Y^{-1}\partial_{y}\Sigma\right].$ (3.11)

Similarly, the coefficient of $\partial_{\psi}$ is

$\frac{1}{\rho}\begin{pmatrix}0&B\\\
-\bar{B}&0\end{pmatrix}+\frac{\cot\psi}{\rho^{2}},$

where

$B=\frac{1}{2}\left[\cos\psi\left(\cos^{2}\beta
e^{-i\theta}Y^{-1}\partial_{\bar{z}}\Sigma+\sin^{2}\beta
e^{i\theta}(Y^{-1}\partial_{z}\Sigma+\csc\beta Y^{-1}z^{k})\right)-\sin\psi
Y^{-1}\partial_{y}\Sigma\right],$ (3.12)

and the coefficient of $\partial_{\theta}$ is

$\frac{1}{\rho\sin\psi}\begin{pmatrix}C&D\\\ -\bar{D}&-C\end{pmatrix},$

where

$\begin{split}C&=-\frac{iY^{-1}\partial_{r}Y}{2}\\\
D&=-\frac{i}{2}\left[\cos^{2}\beta
e^{-i\theta}Y^{-1}\partial_{\bar{z}}\Sigma+\sin^{2}\beta
e^{i\theta}(Y^{-1}\partial_{z}\Sigma+\csc\beta
Y^{-1}z^{k})\right].\end{split}$ (3.13)

These calculations imply that

$A_{\rho}=\begin{pmatrix}0&A\\\ -\bar{A}&0\end{pmatrix},\quad
A_{\psi}=\rho\begin{pmatrix}0&B\\\ -\bar{B}&0\end{pmatrix},\qquad
A_{\theta}=r\begin{pmatrix}C&D\\\ -\bar{D}&-C\end{pmatrix}.$ (3.14)

###  3.3. The ansatz for the linearized problem

As we have discussed above, the hope is that the perturbation arguments of
this section will be able to produce a solution that is in the form (2.12).
One way to do this is to show that the linearized operator (3.8) is invertible
between appropriate Banach spaces, and then, since the solution we would get
is unique, we could argue that it must satisfy the required symmetries.
Although this is arguably the more natural approach, we employ a different one
here. We require that the section $s$ in unitary gauge have the form

$s^{U}=\begin{pmatrix}\gamma&e^{i(k+1)\theta}\delta\\\
e^{-i(k+1)\theta}\delta&-\gamma\end{pmatrix},$ (3.15)

where $\gamma$ and $\delta$ are real functions of $\psi$. We use the upper $U$
notation to keep track of when we work with the section in unitary gauge. The
reason we seek for a solution with this particular form is because it
preserves the ansatz (2.12). Below, we briefly demonstrate why this is the
case. We calculate

$e^{s^{U}}=\sum\limits_{n=0}^{\infty}\frac{(\gamma^{2}+\delta^{2})^{n}}{(2n)!}\text{Id}+\sum\limits_{n=0}^{\infty}\frac{(\gamma^{2}+\delta^{2})^{n}}{(2n+1)!}s^{U}.$

In order to go back to holomorphic gauge and compute the new hermitian metric
$H=H_{0}e^{s}$, we need to understand how $s$ transforms when we move from
unitary to holomorphic gauge. Since unitarity is preserved, $s$ must be equal
to $g^{-1}s^{U}g$ where

$g=\begin{pmatrix}Y^{-1/2}&-Y^{-1/2}\Sigma\\\ 0&Y^{1/2}\end{pmatrix},$

is such that $H_{0}=g^{\dagger}g$ as in section $2$. Therefore, the new metric
is going to be

$H_{0}e^{s}=g^{\dagger}e^{s^{U}}g.$

From this we get that

$\begin{split}Y&=Y_{0}\left(\sum\limits_{n=0}^{\infty}\frac{(\gamma^{2}+\delta^{2})^{n}}{(2n)!}+\sum\limits_{n=0}^{\infty}\frac{(\gamma^{2}+\delta^{2})^{n}}{(2n+1)!}\right)^{-1}\\\
\Sigma&=\Sigma_{0}-Y_{0}e^{i(k+1)\theta}\delta\sum\limits_{n=0}^{\infty}\frac{(\gamma^{2}+\delta^{2})^{n}}{(2n+1)!}\left(\sum\limits_{n=0}^{\infty}\frac{(\gamma^{2}+\delta^{2})^{n}}{(2n)!}+\sum\limits_{n=0}^{\infty}\frac{(\gamma^{2}+\delta^{2})^{n}}{(2n+1)!}\right)^{-1}.\end{split}$

From these formulas it is clear that $Y$ is of the form (2.12). As for
$\Sigma$, it is enough to notice that $Y_{0}=r^{k+1}F(\psi)$ and therefore
$Y_{0}e^{i(k+1)\theta}=z^{k+1}F(\psi)$ so this term is also of the form
(2.12).

######  Remark 3.4.

This calculation, on top of proving that sections of this form preserve
(2.12), shows the exact relation between the section $s^{U}$ and the functions
$u$ and $v$, which are the main objects of study in the next section, thus
providing a bridge between sections $3$ and $4$ of this paper.

In order to be able to consider a section of this form, the first thing we
need to check is that the linearized equation is algebraically consistent. By
this we mean that the error term should have the same algebraic form as
$\mathcal{L}_{H}s$. Thus, a good place to start, is to calculate the error
term that is produced when we plug in an exact solution for $\beta_{0}$ into
the equations (2.4) for $\beta_{1}$ close to $\beta_{0}$. Working in unitary
gauge, we write

$\displaystyle\mathcal{D}_{1}^{\beta_{1}}$
$\displaystyle=D_{1}^{\beta_{0}}-(\tan\beta_{1}-\tan\beta_{0})\phi_{\bar{z}}$
$\displaystyle\mathcal{D}_{2}^{\beta_{1}}$
$\displaystyle=D_{2}^{\beta_{0}}-(\tan\beta_{1}-\tan\beta_{0})\phi_{z}$
$\displaystyle\mathcal{D}_{3}^{\beta_{1}}$
$\displaystyle=D_{3}^{\beta_{0}}-i(\frac{1}{\cos\beta_{1}}-\frac{1}{\cos\beta_{0}})\phi_{1}.$

We will analyze only the error coming from the $\mathcal{D}_{1}$ term since
the others lead to similar expressions. We want to show that the error term
has the form (3.15) times $\rho^{-2}$. Defining
$\epsilon=\tan\beta_{1}-\tan\beta_{0}$, we write

$\begin{split}[\mathcal{D}_{1}^{\beta_{1}},\mathcal{D}_{1}^{\beta_{1},\dagger}]&=[\mathcal{D}_{1}^{\beta_{0}},\mathcal{D}_{1}^{\beta_{0},\dagger}]+\epsilon\left([\mathcal{D}_{1}^{\beta_{0},\dagger},\phi_{\bar{z}}]-[\mathcal{D}_{1}^{\beta_{0}},\phi_{z}]\right)+\epsilon^{2}[\phi_{\bar{z}},\phi_{z}]\\\
&=[\mathcal{D}_{1}^{\beta_{0}},\mathcal{D}_{1}^{\beta_{0},\dagger}]+\epsilon\left(\partial_{\bar{z}}\phi_{z}-\partial_{z}\phi_{\bar{z}}+[A_{\bar{z}},\phi_{z}]-[A_{z},\phi_{\bar{z}}]\right)+\epsilon^{2}[\phi_{\bar{z}},\phi_{z}].\end{split}$

At this point, notice that the matrices $\phi_{z}$ and $A_{z}$ given in (3.7)
are of the form

$\frac{e^{-i\theta}}{\rho}\begin{pmatrix}a&e^{i(k+1)\theta}b\\\
e^{-i(k+1)\theta}c&-a\end{pmatrix},$

where $a,b,c$ are real functions of $\psi$, while $\phi_{\bar{z}}$ and
$A_{\bar{z}}$ are of the form

$\frac{e^{-i\theta}}{\rho}\begin{pmatrix}a&e^{i(k+1)\theta}b\\\
e^{-i(k+1)\theta}c&-a\end{pmatrix}.$

Combining these expressions with the formulas
$\partial_{z}=\frac{e^{-i\theta}}{2}\left(\partial_{r}-\frac{i}{r}\partial_{\theta}\right)$
and
$\partial_{\bar{z}}=\frac{e^{i\theta}}{2}\left(\partial_{r}+\frac{i}{r}\partial_{\theta}\right)$,
a standard calculation gives the desired result.

Now that we know that the error term has this form and we require that $s$
have the same form, we need to show that the linearized operator preserves
this form as well. The key is that the commutator of two matrices of the form
(3.15) is of the same form. Since the connection coefficients $A_{\rho}$ and
$A_{\psi}$ are of this form, the following part of the linearized operator
preserves the required form

$D_{\rho}^{2}+\frac{2}{\rho}D_{\rho}+\frac{1}{\rho^{2}}(D_{\psi}^{2}+\cot\psi
D_{\psi}).$

Next, we see that for $s$ as in (3.15),

$\displaystyle D_{\theta}s$
$\displaystyle=\begin{pmatrix}0&i(k+1)e^{i(k+1)\theta}y\\\
-i(k+1)e^{-i(k+1)\theta}y&0\end{pmatrix}+A_{\theta}s$
$\displaystyle=\left(\frac{1}{2}\begin{pmatrix}i(k+1)&0\\\
0&-i(k+1)\end{pmatrix}+A_{\theta}\right)s$ $\displaystyle=:\hat{A}_{\theta}s,$

where $\hat{A}_{\theta}$ is has the required form up to a factor of $i$.
Therefore $D_{\theta}^{2}s$ also has the required form. $\phi_{1}$ is $i$
times a matrix of the form (3.15) and therefore the only term left to check is

$[\phi_{2},[\phi_{2},\star]]+[\phi_{3},[\phi_{3},\star]].$

We can re-write this sum as

$[\phi_{2},[\phi_{2},\star]]+[\phi_{3},[\phi_{3},\star]]=-2\left([\phi_{z},[\phi_{z}^{\dagger},\star]]+[\phi_{z}^{\dagger},[\phi_{z},\star]]\right)$

and use the expressions appearing in (3.7) to get the required result.
Therefore, we have showed that the linearized operator should at least in
principle admit a solution of the form (3.15). Proving that this is indeed the
case, is the subject of the next subsection.

###  3.4. Mapping properties

Finally, we are in position to prove the openness part for the continuity
argument. The linearized operator whose mapping properties we want to study is

$\Phi:=D_{\psi}^{2}+\cot\psi
D_{\psi}+N+\hat{A}_{\rho}^{2}+2\hat{A}_{\rho}+\frac{1}{\sin^{2}\psi}\hat{A}_{\theta}^{2},$
(3.16)

where $\hat{A}_{\rho}:=\rho A_{\rho}$ is independent of $\rho$ and
$\hat{A}_{\theta}$ was defined in the previous subsection. This operator turns
out to be a regular-singular operator acting on sections of the form (3.15).
Therefore, in order to understand its mapping properties we need to calculate
its indicial roots at the singular regions and introduce appropriate Banach
spaces adapted to the degeneracy of this operator.

We start with the calculation of the indicial roots of (3.16). We should
expect singular behavior at $\psi=0$ and $\psi=:\pi/2-\omega=\pi/2$. This
differs from the study of the indicial behavior of similar operators as in [5]
where they only have indicial behavior at $\omega=0$. In order to produce
solutions of the form (2.12), we considered $s$ of the form (3.15) and as a
result, we traded the dependence in $\theta$ with the appearance of indicial
behavior at $\psi=0$.

######  Lemma 3.5.

The set of indicial roots of $\Phi$ at $\omega=0$ is $\\{-1,2\\}$, in
accordance with the Nahm pole boundary condition, and the set of indicial
roots at $\psi=0$ is $\\{-k-1,0,0,k+1\\}$.

######  Remark 3.6.

The reason that there are four indicial roots at $\psi=0$ instead of two as
one would expect from a second order equation, is because we work with a
matrix-valued function $s$ and therefore we are actually dealing with a system
of equations. There are only two indicial roots at $\omega=0$ because the
system asymptotically decouples in that region.

###### Proof.

We start with the indicial behavior at $\omega=0$. Plugging the asymptotics
(2.13) into (3.14) we find that the indicial operator is given by

$\partial_{\omega}^{2}-\frac{2\text{Id}}{\omega^{2}}$

and therefore the indicial roots are $-1$ and $2$. This calculation should
come as no surprise. These are the indicial roots that characterize the Nahm
pole solutions for $SU(2)$ away from the knot and therefore, they should not
depend on $\beta$.

Next, we study the indicial behavior at $\psi=0$. In this region, the indicial
operator takes the form

$\partial_{\psi}^{2}+\frac{1}{\psi}\partial_{\psi}-\frac{(k+1)^{2}\text{Pr}_{\text{Ant}}}{\psi^{2}},$

where $\text{Pr}_{\text{Ant}}$ is projection to the anti-diagonal

$\text{Pr}_{\text{Ant}}\begin{pmatrix}x&y\\\
\bar{y}&-x\end{pmatrix}=\begin{pmatrix}0&y\\\ \bar{y}&0\end{pmatrix}.$

In order to see why this is the case, notice that as $\psi\to 0$ the
asymptotics (2.13) imply that all of the matrices (3.14) and all of the fields
$\phi_{i}$ are regular. Therefore, the part of the operator that contributes
to the indicial behavior is

$\partial_{\psi}^{2}+\cot\psi\partial_{\psi}+\frac{1}{4}\begin{pmatrix}i(k+1)&0\\\
0&-i(k+1)\end{pmatrix}^{2}.$

The last term should be interpreted as the operation of conjugating with the
matrix

$\frac{1}{2}\begin{pmatrix}i(k+1)&0\\\ 0&-i(k+1)\end{pmatrix}$

twice. Keeping only the lowest terms in $\psi$ we indeed get the required
formula for the indicial operator. As a result, the indicial roots are
$\pm(k+1)$ and $0$, the latter with double multiplicity.

∎

Before we state the main theorem of this section, let us define the
appropriate function spaces that will appear below. These are designed to be
sensitive to the singular behavior of the linearized operator as we will see.

######  Definition 3.7.

We will work with spaces of the form

$\hat{\Gamma}(S_{+}^{2}):=\Gamma_{\text{inv}}(S_{+}^{2},i\mathfrak{su}(E,H)|_{S_{+}^{2}}).$

The $\Gamma$ refers to the function space in which the norm of the section is
going to belong, the inv subscript means that we consider sections of the form
(3.15) and $i\mathfrak{su}(E,H)|_{S_{+}^{2}}$ is the restriction of the bundle
on the half sphere. The first type of spaces we want to consider is the
function spaces of the form

$\hat{H}^{k}_{0}(S_{+}^{2}),$

where $f\in\hat{H}^{k}_{0}(S_{+}^{2})$ if
$\omega\psi\partial_{\psi}f\in\hat{H}^{k-1}_{0}(S_{+}^{2})$, with
$\hat{H}^{0}_{0}(S_{+}^{2}):=\hat{L}^{2}(S_{+}^{2})$. The operator
$\omega\psi\partial_{\psi}$ is a zero operator at both ends in the sense of
[7]. The second type of spaces we will need is

$\hat{C}^{k,a}_{0}(S_{+}^{2}),$

so if $u$ is in such a space, then

$\|u\|_{L^{\infty}}+\sup\limits_{i\leq
k}[(\omega\psi\partial_{\psi})^{i}u]_{0;0,a}<\infty$

where

$[u]_{0;0,a}=\sup\limits_{\psi\neq\psi^{\prime}}\frac{|u(\psi)-u(\psi^{\prime})|(\psi+\psi^{\prime})^{a}}{|\psi-\psi^{\prime}|^{a}}.$

######  Theorem 3.8.

There is a unique self-adjoint realization

$\Phi:\hat{L}^{2}(S_{+}^{2})\rightarrow\hat{L}^{2}(S_{+}^{2})$

which has domain $\omega^{2}\psi^{2}\hat{H}^{2}_{0}(S_{+}^{2})$. The operators

$\begin{split}\Phi:\omega^{\gamma}\psi^{\delta}\hat{H}^{2}_{0}(S_{+}^{2})&\rightarrow\omega^{\gamma-2}\psi^{\delta-2}\hat{L}^{2}(S_{+}^{2})\\\
\Phi:\omega^{\mu}\psi^{\nu}\hat{C}^{2,a}_{0}(S_{+}^{2})&\rightarrow\omega^{\mu-2}\psi^{\nu-2}\hat{C}^{0,a}_{0}(S_{+}^{2})\end{split}$
(3.17)

are invertible provided that $-1/2<\gamma<5/2$, $1/2<\delta<k+3/2$ and
$-1<\mu<2$, $0<\nu<k+1$ respectively.

###### Proof.

The fact that the operator $\Phi$ is essentially self-adjoint can be checked
manually by an integration by parts argument. The fact that its domain in this
case is $\omega^{2}\psi^{2}\hat{H}^{2}_{0}(S_{+}^{2})$ follows from standard
techniques developed in [7]. The crucial property to check is that the kernel
of $\Phi$ mapping between these function spaces is zero. Then, since the
operator is self-adjoint, the index and as a result the co-kernel too is zero
and the operator is invertible. Applying the results of [7] give the
invertibility of $\Phi$ between the more general spaces as given in (3.17).
Assume on the contrary that the kernel is not empty. Let $s$ be an element of
the kernel. Then, from the calculation of the indicial roots above, such a
section must vanish at least quadratically near $\omega=0$ and at least as
fast as $\psi^{k+1}$ near $\psi=0$. Therefore, we can perform an integration
by parts since the boundary terms vanish. Remembering that we work in unitary
gauge, and that all the fields are anti-hermitian, the term $\langle
s,\hat{A}_{\rho}s\rangle$ vanishes and then

$\begin{split}0&=\langle s,\Phi s\rangle\\\ &=-|\sin\psi
D_{\psi}s|^{2}-|\Phi_{1}s|^{2}-|\Phi_{2}s|^{2}-|\Phi_{3}s|^{2}-|\hat{A}_{\rho}s|^{2}-|\frac{\hat{A}_{\theta}}{\sin\psi}s|^{2},\end{split}$

where $\Phi_{1}:=\frac{\rho\phi_{1}}{\cos\beta}$, $\Phi_{2}:=\rho\phi_{2}$ and
$\Phi_{3}:=\rho\phi_{3}$. In particular this implies that $D_{\psi}s=0$. Since
$s=0$ for $\psi=0$, this implies that $s$ is identically zero. Therefore there
is no kernel and we are done. ∎

###  3.5. Existence of solution

Following the development of the linear analysis in the last subsection, it is
now a simple matter to prove existence. Recall that from equation (3.2) we can
write $\Omega$ as

$\Omega_{H}=\Omega_{H_{0}}+\gamma(-s)\mathcal{L}_{H_{0}}s+Q(s),$

where

$Q_{H_{0}}(s)=2i\Lambda(\cos^{2}\beta(\mathcal{D}_{1}\gamma(-s))\mathcal{D}_{1}^{\dagger_{H_{0}}}-\sin^{2}\beta(\mathcal{D}_{2}\gamma(-s))\mathcal{D}_{2}^{\dagger_{H_{0}}})s+(\mathcal{D}_{3}\gamma(-s))\mathcal{D}_{3}^{\dagger_{H_{0}}}s.$

If we denote by $\mathcal{G}_{H_{0}}$ the inverse of $\mathcal{L}_{H_{0}}$,
then to rephrase our problem as a fixed point problem

$s=\mathcal{G}_{H_{0}}(\Omega_{H_{0}}+Q_{H_{0}}(s)),$

it is enough to prove a bound of the form

$\|\omega^{2-\mu}\psi^{2-\nu}(Q(s_{1})-Q(s_{2}))\|_{0;0,a}\leq
C(\|\omega^{-\mu}\psi^{-\nu}s_{1}\|_{2;0,a}+\|\omega^{-\mu}\psi^{-\nu}s_{1}\|_{2;0,a})\|\omega^{-\mu}\psi^{-\nu}(s_{1}-s_{2})\|_{2;0,a}$

as long as
$\|\omega^{-\mu}\psi^{-\nu}s_{1}\|_{2;0,a}+\|\omega^{-\mu}\psi^{-\nu}s_{1}\|_{2;0,a}\leq\epsilon$
for some appropriately small $\epsilon>0$. At this point, we re-write

$Q_{H_{0}}(s)=2i\Lambda(\cos^{2}\beta(\delta(-s)\mathcal{D}_{1}s)\gamma(-s)\mathcal{D}_{1}^{\dagger_{H_{0}}}s-\sin^{2}\beta(\delta(-s)\mathcal{D}_{2}s)\gamma(-s)\mathcal{D}_{2}^{\dagger_{H_{0}}}s)+(\delta(-s)\mathcal{D}_{3}s)\gamma(-s)\mathcal{D}_{3}^{\dagger_{H_{0}}}s,$

where

$\delta(-s):=\frac{1-\text{ad}s-e^{-\text{ad}s}}{(\text{ad}s)^{2}}.$

Clearly both $\gamma(-s)$ and $\delta(-s)$ are controlled by $s$. Let us show
how to bound the term

$\|\omega^{2-\mu}\psi^{2-\nu}(\delta(-s_{1})\mathcal{D}_{1}s_{1})\gamma(-s_{1})\mathcal{D}_{1}^{\dagger_{H_{0}}}s_{1}-\delta(-s_{2})\mathcal{D}_{1}s_{2})\gamma(-s_{2})\mathcal{D}_{1}^{\dagger_{H_{0}}}s_{2})\|_{0;0,a}$

since the other terms are bounded similarly. In order for the following
argument to work we need that both $\mu$ and $\nu$ be nonnegative. In fact,
$\nu$ must be strictly positive since $v=0$ corresponds to an indicial root.
It is a straightforward computation that the error term belongs to the space
$\omega^{-2}\hat{C}_{0}^{0,a}(S_{+}^{2})$. This corresponds to $\mu=0$ and
$\nu=2$. Therefore, since the error term is in the required region we can
proceed with the following argument. Using the triangle inequality the term we
wish to bound is less than or equal to

$\displaystyle\|\omega^{2}\delta(-s_{1})\mathcal{D}_{1}s_{1}\left(\gamma(-s_{1})\mathcal{D}_{1}^{H_{0}^{\dagger}}(s_{1}-s_{2})+(\gamma(-s_{1})-\gamma(-s_{2}))\mathcal{D}_{1}^{H_{0}^{\dagger}}s_{2}\right)\|_{0;0,a}+$
$\displaystyle\|\omega^{2}\gamma(-s_{2})\mathcal{D}_{1}^{H_{0}^{\dagger}}s_{2}\Bigl{(}\delta(-s_{2})\mathcal{D}_{1}(s_{1}-s_{2})+(\delta(-s_{1})-\delta(-s_{2}))\mathcal{D}_{1}s_{1}\Bigr{)}\|_{0;0,a}$

Let us focus on the first line of this bound since the second line can be
bounded in a similar way. The condition
$\|\psi^{-2}s_{1}\|_{2;0,a}+\|\psi^{-2}s_{1}\|_{2;0,a}\leq\epsilon$ implies
that both $\gamma(-s_{i})$ and $\delta(-s_{i})$ are bounded in the supremum
norm by a constant that only depends on $\epsilon$. Now,

$\displaystyle\|\omega\mathcal{D}_{1}^{H_{0}^{\dagger}}(s_{1}-s_{2})\|_{0;0,a}$
$\displaystyle\leq\|\omega\psi^{-1}\mathcal{D}_{1}^{H_{0}^{\dagger}}(s_{1}-s_{2})\|_{0;0,a}$
$\displaystyle\leq\|\psi^{-2}(s_{1}-s_{2})\|_{2;0,a}.$

Similarly
$\|\omega\mathcal{D}_{1}^{H_{0}^{\dagger}}s_{2}\|_{0;0,a}\leq\|\psi^{-2}s_{2}\|_{2;0,a}$
and $\|\omega\mathcal{D}_{1}s_{1}\|_{0;0,a}\leq\|\psi^{-2}s_{1}\|_{2;0,a}$.
Finally,

$\displaystyle|\gamma(-s_{1})-\gamma(-s_{2})|$ $\displaystyle\leq
C|s_{1}-s_{2}|$ $\displaystyle\leq C\|s_{1}-s_{2}\|_{2;0,a}$
$\displaystyle\leq C\|\psi^{-2}(s_{1}-s_{2})\|_{2;0,a},$

where the constant $C$ can change from line to line but is dependent only on
$\epsilon$. Putting all these bounds together, it is straightforward to obtain
the required inequality and therefore we are done.

##  4\. A priori estimates

By the results of the previous section, given a solution to the TEBE (2.4) of
the form (2.12) for some $\beta_{0}$, there is an open interval
$(\beta_{0}-\varepsilon,\beta_{0}+\varepsilon)$ such that (2.4) admits a
solution of this specified form for any $\beta$ in this interval. To prove
that solutions exist for every $\beta\in(-\pi/6,\pi/6)$, or equivalently
$\zeta\in[0,1/2)$, we invoke a standard continuity argument. It suffices to
show that if $\beta_{j}$ converges to some value in the appropriate open
interval, then the solutions $(u_{j}(\sigma),v_{j}(\sigma))$ associated to
these parameters have a subsequence which converges to a solution for the
limiting value of $\beta$. Given that this is an ODE system, it suffices to
deduce uniform $C^{1}$ bounds for $(u_{j},v_{j})$ along with bounds which
allow us to conclude that the asymptotic conditions (2.13) are preserved in
the limit. This is the goal of this section.

###  4.1. Preliminary bounds and a first integral

We begin by rewriting (2.14) in a slightly nicer form. Observe that the linear
part of these equations can be rewritten as

$(\sigma^{2}+1)u^{\prime\prime}+\sigma
u^{\prime}=((\sigma^{2}+1)^{1/2}\partial_{\sigma})^{2}u.$

This suggests the change of variables

$\frac{\,d\sigma}{\,d\tau}=(\sigma^{2}+1)^{1/2},\ \ \mbox{i.e.,}\ \
\sigma=\sinh\tau.$

Setting $\zeta=\sin\beta$, we also define the quantities

$V(\tau):=1-\zeta^{2}(k+1)v,\quad T(\tau):=\tanh\tau.$ (4.1)

In terms of these, (2.14) now takes the following simpler form

$\begin{split}u^{\prime\prime}&=-e^{-2u}(4V^{2}+\zeta^{2}v^{\prime
2}+4\zeta^{2}Vv^{\prime}T)\\\
v^{\prime\prime}&=u^{\prime}(2v^{\prime}+4VT)-4(k+1)V.\end{split}$ (4.2)

The reader should beware that in (4.2), the derivatives
$u^{\prime},v^{\prime}$, etc. are taken with respect to $\tau$, while in
(2.14) they are with respect to $\sigma$. Henceforth we only use the variable
$\tau$, so hopefully this should not cause confusion. The function $T(\tau)$
is not intended as a new independent variable.

We also state the asymptotic conditions:

$\begin{split}&u(\tau)\sim\log\tau,\ \tau\searrow 0,\ \ u(\tau)\sim(k+1)\tau,\
\tau\nearrow\infty,\\\ &v(\tau)\sim 2k\tau^{2},\tau\searrow 0,\ \
|v(\tau)|\leq Ce^{(k+1)(1-\epsilon)\tau}\ \ \tau\nearrow\infty.\end{split}$
(4.3)

######  Lemma 4.1.

If $(u,v)$ is a solution of (4.2) and satisfies the asymptotic conditions
(4.3), then

$u^{\prime 2}=(4V^{2}-\zeta^{2}v^{\prime 2})e^{-2u}+(k+1)^{2}.$ (4.4)

###### Proof.

Multiply the first equation by $u^{\prime}$ and the second equation by
$\zeta^{2}v^{\prime}e^{-2u}$ and add the two equations. After some calculaiton
this leads to the identity

$\begin{split}u^{\prime}u^{\prime\prime}&=-u^{\prime}e^{-2u}(4V^{2}-\zeta^{2}v^{\prime
2})-\zeta^{2}e^{-2u}v^{\prime}(v^{\prime\prime}+4(k+1)V)\\\
&=\frac{1}{2}((4V^{2}-\zeta^{2}v^{\prime 2})e^{-2u})^{\prime}.\end{split}$

Integrating this, we obtain the desired formula, at least up to some additive
constant.

To calculate the constant, we integrate from $1$ to $\tau$. This gives

$u^{\prime}(\tau)^{2}=\left(4(1-\zeta^{2}(k+1)v(\tau)^{2})^{2}-\zeta^{2}v^{\prime}(\tau)^{2}\right)e^{-2u(\tau)}+C_{1}$

for every $\tau>1$, where $C_{1}$ arises from the boundary terms at $\tau=1$.
Since $v(\tau)$ and $v^{\prime}(\tau)$ grow at some exponential rate less than
$k+1$ as $\tau\to\infty$, the first quantity on the right converges to $0$.
Since $u^{\prime}(\tau)\to k+1$, we conclude that $C_{1}=(k+1)^{2}$, as
claimed. ∎

We can deduce some immediate useful consequences from this.

######  Lemma 4.2.

If $(u,v)$ solves (4.2), then $u^{\prime}$ is monotone decreasing and

$4V^{2}-\zeta^{2}v^{\prime 2}\geq 0$ (4.5)

for $\tau>0$.

###### Proof.

We first show that $u^{\prime\prime}=-e^{-2u}(4V^{2}+\zeta^{2}v^{\prime
2}+4\zeta^{2}Vv^{\prime}T)\leq 0$. To do this, note that

$\displaystyle 4V^{2}+\zeta^{2}v^{\prime 2}+4\zeta^{2}Vv^{\prime}T$
$\displaystyle\geq 4V^{2}(1-\zeta^{2})+\zeta^{2}(4V^{2}+v^{\prime
2}-4|Vv^{\prime}|)$
$\displaystyle=4V^{2}(1-\zeta^{2})+\zeta^{2}(2|V|-|v^{\prime}|)^{2}$
$\displaystyle\geq 4V^{2}(1-\zeta^{2})$

Since $\zeta<1/2$, this implies that $u^{\prime\prime}\leq 0$ everywhere, so
$u^{\prime}$ is decreasing to its limit $k+1$ as $\tau$ grows. The identity
(4.4) then gives that $4V^{2}-\zeta^{2}v^{\prime 2}\geq 0$. ∎

######  Corollary 4.3.

The component $v(\tau)$ of the solution satisfies the following preliminary
uniform a priori bounds:

$e^{-2\zeta(k+1)\tau}\leq~{}V\leq
e^{2\zeta(k+1)\tau},\qquad|v^{\prime}(\tau)|\leq
2\zeta^{-1}e^{2\zeta(k+1)\tau}.$ (4.6)

###### Proof.

By the preceding lemma,

$-2\leq\frac{\zeta v^{\prime}}{1-\zeta^{2}(k+1)v}\leq 2,$

so integrating from $0$ to $\tau$ and using that $V(0)=1$, we obtain

$-2\zeta(k+1)\tau\leq\log V(\tau)\leq 2\zeta(k+1)\tau.$

This gives the first part of (4.6). The second part follows from this together
with (4.5). ∎

###  4.2. Improved bounds near zero

The next step is to derive uniform estimates for $u$ near $\tau=0$. This uses
the initial bounds of the last subsection and an improved bound for
$v^{\prime}$.

To simplify notation further, write $b:=k+1$ and $f:=e^{u}$.

######  Lemma 4.4.

There exists a positive constant $C$ and an interval $[0,\tau_{0}]$ both
independent of of $\zeta<1/2$ such that for any solution $(u,v)$ of equations
(4.2), $u$ satisfies the upper bound

$u(\tau)\leq\log(2\tau)+C\tau\ \ 0<\tau<\tau_{0}.$ (4.7)

###### Proof.

By (4.4) and the first part of (4.6) we get

$\displaystyle u^{\prime 2}$ $\displaystyle=(4V^{2}-\zeta^{2}v^{\prime
2})e^{-2u}+b^{2}$ $\displaystyle\leq 4V^{2}e^{-2u}+b^{2}$ $\displaystyle\leq
4e^{4\zeta(k+1)\tau}e^{-2u}+b^{2}.$

Multiplying both sides by $e^{2u}$ and dividing by $4+b^{2}f^{2}$,

$\frac{f^{\prime
2}}{b^{2}f^{2}+4}\leq\frac{4e^{4\zeta(k+1)\tau}+b^{2}f^{2}}{4+b^{2}f^{2}}\leq
e^{4\zeta(k+1)\tau}$

or equivalently

$\frac{f^{\prime}}{\sqrt{b^{2}f^{2}+4}}\leq e^{2\zeta(k+1)\tau}.$

Since $f(0)=0$, we can integrate this to get

$\frac{1}{b}\tanh^{-1}\left(\frac{bf}{\sqrt{b^{2}f^{2}+4}}\right)\leq\frac{e^{2\zeta(k+1)\tau}-1}{2\zeta(k+1)},$

which in turn implies that

$\frac{bf}{\sqrt{b^{2}f^{2}+4}}\leq\tanh\left(\frac{b(e^{2\zeta(k+1)\tau}-1)}{2\zeta(k+1)}\right)\leq\frac{b(e^{2\zeta(k+1)\tau}-1)}{2\zeta(k+1)}.$

Since $\zeta\leq 1$, there exist $C>0$ and $\tau_{0}>0$ independent of $\zeta$
such that

$\frac{b(e^{2\zeta(k+1)\tau}-1)}{2\zeta(k+1)}\leq b\tau+C\tau^{2}$

for $\tau<\tau_{0}$. Hence for such $\tau$, $f\leq 2\tau+C^{\prime}\tau^{2}$.
The statement of the lemma follows immediately from this last inequality. ∎

The bound from below for $u$ near $\tau=0$ is more subtle, and requires better
control of $v^{\prime}$ near $\tau=0$.

######  Lemma 4.5.

With $\tau_{0}$ as in the previous lemma, there exists $C>0$ independent of
both $\tau\in[0,\tau_{0}/2]$ and $\zeta$ such that

$v^{\prime}(\tau)\leq C\tau.$ (4.8)

###### Proof.

We look more carefully at the equation for $v$:

$v^{\prime\prime}=u^{\prime}(2v^{\prime}+4VT)-4(k+1)V.$

By (4.6), $4(k+1)V\leq C^{\prime}$ for $\tau\leq\tau_{0}$. By the same bound,
we also know that $V\geq 0$. Since $T\geq 0$, we obtain

$v^{\prime\prime}-2u^{\prime}v^{\prime}\geq-C.$

Multiplying both sides by $e^{-2u}$, this is equivalent to

$-Ce^{-2u}\leq(v^{\prime}e^{-2u})^{\prime}.$

On the other hand, we already showed in the proof of (4.2) that

$u^{\prime\prime}\leq-4V^{2}e^{-2u}(1-\zeta^{2})\leq-Ce^{-2u}.$ (4.9)

Combining this with the previous inequality, and using that $V\geq C>0$ for
$\tau\leq\tau_{0}$, we conclude that

$Cu^{\prime\prime}\leq(v^{\prime}e^{-2u})^{\prime}.$

Integrating this from any $\tau\leq\tau_{0}/2$ to $\tau_{0}$ gives

$C(u^{\prime}(\tau_{0})-u^{\prime}(\tau))\leq
v^{\prime}(\tau_{0})e^{-2u(\tau_{0})}-v^{\prime}(\tau)e^{-2u(\tau)}.$

Since $u^{\prime}(\tau_{0})>0$, we can drop it. We also have a uniform bound
$|v^{\prime}|\leq C$ for $\tau\leq\tau_{0}$. Therefore,

$v^{\prime}(\tau)\leq
C(u^{\prime}(\tau)e^{2u(\tau)}+e^{2(u(\tau)-u(\tau_{0}))}).$ (4.10)

We must show that there is a uniform bound for the right side.

Let us start by bounding $u^{\prime}(\tau)e^{2u(\tau)}=f^{\prime}f$. From the
preceding analysis, we have $f\leq C\tau$ for $\tau\leq\tau_{0}$, uniformly.
It is also straightforward to deduce from the proof of Lemma (4.4) that there
is some $C>0$ such that $f^{\prime}\leq C$. Taken together, this gives a bound
of $C\tau$ for the first term on the right.

The second term is harder to deal with for the following reason. The factor
$e^{-2u(\tau_{0})}$ might diverge to infinity as $\zeta\to\zeta_{0}<1/2$. We
show below that this is not the case, but unfortunately that proof relies on
the current lemma. We proceed instead in the following way. Since
$u^{\prime}\geq k+1$, $u(\tau)$ increases monotonically, and thus
$e^{-2u(\tau_{0})}\leq e^{-2u(\tau)}$ when $\tau\leq\tau_{0}$. Combining this
with (4.9), we see that

$u^{\prime\prime}\leq-Ce^{-2u(\tau_{0})}.$

Integrating from $\tau_{0}/2$ to $\tau_{0}$ and using that $u$ is increasing,
we get that if $\tau\leq\tau_{0}/2$,

$u(\tau)\leq
u(\tau_{0}/2)\leq-\frac{C}{4}\tau_{0}^{2}e^{-2u(\tau_{0})}-\frac{\tau_{0}}{2}u^{\prime}(\tau_{0})+u(\tau_{0})\leq-
C_{1}e^{-2u(\tau_{0})}+u(\tau_{0}),$

where $C_{1}=-\frac{1}{4}C\tau^{2}$. The last inequality holds since
$u^{\prime}(\tau_{0})>0$.

Finally,

$-C_{1}e^{-2u(\tau_{0})}\leq u(\tau_{0})\ \ \mbox{provided}\ \
u(\tau_{0})\leq-C.$

This is simply because the left hand side decreases exponentially with
$u(\tau_{0})$ while the right side only decreases linearly. In any case, this
now gives

$u(\tau)\leq 2u(\tau_{0})$

and hence, using Lemma (4.4),

$e^{2(u(\tau)-u(\tau_{0}))}=e^{u(\tau)+(u(\tau)-2u(\tau_{0})}\leq
e^{u(\tau)}\leq C\tau.$

This last inequality gives a contradiction if $u_{j}(\tau_{0})\to-\infty$ as
$\zeta_{j}\to\zeta_{0}$. On the other hand, if $u_{j}(\tau_{0})\geq-C$, then

$e^{2(u(\tau)-u(\tau_{0}))}\leq Ce^{2u(\tau)}\leq C\tau^{2}\leq
C^{\prime}\tau.$

Using this in (4.10) finishes the proof. ∎

We now prove the lower bound for $v^{\prime}(\tau)$.

######  Lemma 4.6.

There exists $C>0$ such that for any $0\leq\tau\leq\tau_{0}/2$,

$v^{\prime}(\tau)\geq-C\tau.$ (4.11)

###### Proof.

We again use the second equation in (4.2). Since $0\leq~{}V\leq C$ for
$\tau\leq\tau_{0}/2$ and $T\leq\tau$ for all $\tau$, we see that

$v^{\prime\prime}-2u^{\prime}v^{\prime}\leq C\tau u^{\prime}.$

Multiplying both sides by $e^{-2u}$ and integrating from
$\tau\in[0,\tau_{0}/2]$ to $\tau_{0}$, we obtain

$v^{\prime}(\tau_{0})e^{-2u(\tau_{0})}-v^{\prime}(\tau)e^{-2u(\tau)}\leq-C(\tau_{0}e^{-2u(\tau_{0})}-\tau
e^{-2u(\tau)})+C\int_{\tau}^{\tau_{0}}e^{-2u(\xi)}\,d\xi\leq
C\int_{\tau}^{\tau_{0}}e^{-2u(\xi)}\,d\xi;$

the first term on the right has been dropped since it is negative. By (4.6),
which implies that $v^{\prime}\geq-C$ and $u^{\prime\prime}\leq-
Ce^{-2u(\tau_{0})}$, we obtain

$\displaystyle-v^{\prime}(\tau)$ $\displaystyle\leq
C\tau+Ce^{2(u(\tau)-u(\tau_{0}))}-Ce^{2u(\tau)}\int_{\tau}^{\tau_{0}}u^{\prime\prime}(\xi)\,d\xi$
$\displaystyle\leq
C\tau+Ce^{2(u(\tau)-u(\tau_{0}))}+Ce^{2u(\tau)}(u^{\prime}(\tau)-u^{\prime}(\tau_{0}))$
$\displaystyle\leq
C\tau+Ce^{2(u(\tau)-u(\tau_{0}))}+Ce^{2u(\tau)}u^{\prime}(\tau)$
$\displaystyle\leq C\tau+Ce^{2(u(\tau)-u(\tau_{0}))}$ $\displaystyle\leq
C\tau,$

since we showed earlier that $Ce^{2u(\tau)}u^{\prime}(\tau)\leq C\tau$. From
the proof of Lemma (4.5) we can also bound $Ce^{2(u(\tau)-u(\tau_{0}))}$ by
$C\tau$. This gives the desired inequality. ∎

We have now established, in (4.8) and (4.11), the uniform estimate
$|v^{\prime}(\tau)|\leq C\tau$ for $0\leq\tau\leq\tau_{0}$, where $C$ and
$\tau_{0}$ are independent of $\zeta$. The next goal is to obtain uniform
bounds for $u$.

######  Lemma 4.7.

There exist constants $C>0$ and $\tau_{0}>0$, both independent of $\zeta$,
such that if $(u,v)$ solves (4.2) and satisfies the specified asymptotics,
then

$|u(\tau)-\log(2\tau)|\leq C\tau^{2},\qquad|u^{\prime}(\tau)-\tau^{-1}|\leq
C\tau$ (4.12)

for $\tau\leq\tau_{0}$.

###### Proof.

As before, we establish upper and lower bounds separately. The proof is
similar to that of Lemma (4.4).

Starting from (4.4), and using $\zeta^{2}v^{\prime 2}\leq C\tau^{2}$ for
$\tau\leq\tau_{0}$, we also get $V\geq 4-C\tau^{2}$ on this same interval.
This gives

$u^{\prime 2}\geq(4-C\tau^{2})e^{-2u}+b^{2},$

so that

$\frac{f^{\prime
2}}{b^{2}f^{2}+4}\geq\frac{4-C\tau^{2}+b^{2}f^{2}}{b^{2}f^{2}+4}\geq
1-C\tau^{2},\ \ \mbox{or more simply}\ \ \
\frac{f^{\prime}}{\sqrt{b^{2}f^{2}+4}}\geq 1-C\tau^{2},$

since $f^{\prime}=u^{\prime}e^{u}>0$. Integrating from $0$ to $\tau$, noting
that $f(0)=0$, we obtain

$\frac{bf}{\sqrt{b^{2}f^{2}+4}}\geq\tanh(b\tau-C\tau^{3})\geq
b\tau-C^{\prime}\tau^{3}.$

We can rewrite this as

$bf\geq\frac{b\tau-C\tau^{3}}{\sqrt{1-(b\tau-C\tau^{3})^{2}}}\geq
2b\tau-C\tau^{2},$

whence the desired inequality.

For the opposite inequality, $|v^{\prime}(\tau)|\leq C\tau^{2}$ implies that
$V\leq 4+C\tau^{2}$, so repeating the steps above, we get

$\frac{bf}{\sqrt{b^{2}f^{2}+4}}\leq\tanh(b\tau+C\tau^{3})\leq
b\tau+C\tau^{3},$

from which the required inequality is straightforward.

For the bounds on $u^{\prime}-\tau^{-1}$, we prove only the lower bound since
the upper bound is similar. We combine

$f^{\prime}\geq(1-C\tau^{2})\sqrt{b^{2}f^{2}+4}>2(1-C\tau^{2}),\ \ \mbox{and}\
\ \ f^{\prime}=u^{\prime}f\leq u^{\prime}(2\tau+C\tau^{3})$

to get

$u^{\prime}\geq\frac{2-C\tau^{2}}{2\tau+C\tau^{3}}\geq\frac{1}{\tau}-C\tau.$

∎

###  4.3. Bounds away from zero

We have now proved bounds on the components $u$, $v$ in some uniform interval
$[0,\tau_{0}]$. It remains to establish appropriate uniform bounds on
$[\tau_{0},\infty)$.

As a first step, decompose $u$ as $u=u_{0}+w$, where $u_{0}$ is the model
charge $k$ knot solution for the group $G=\mbox{SU(2)}$ when $\zeta=0$.
Explicitly,

$u_{0}(\sigma)=\log\left(\frac{((\sigma^{2}+1)^{1/2}+\sigma)^{k+1}-((\sigma^{2}+1)^{1/2}-\sigma)^{k+1}}{2(k+1)}\right)\\\
\Rightarrow
u_{0}(\tau)=\log\left((e^{(k+1)\tau}-e^{-(k+1)\tau})/2(k+1)\right)=\log\sinh((k+1)\tau)-\log(k+1).$

This function has precisely the same leading asymptotic behavior as the
solution $u$, both as $\tau\to 0$ and as $\tau\to\infty$.

######  Lemma 4.8.

Choosing $\tau_{0}>0$ as in the previous subsection, if $\zeta<1/2$, so
$d_{\zeta}:=1-2\zeta>0$, then

$\displaystyle-C\leq$ $\displaystyle\,w\leq C_{\zeta}$ (4.13) $\displaystyle-
Ce^{-2(k+1)\tau}\leq$ $\displaystyle\,w^{\prime}\leq
Ce^{-d_{\zeta}(k+1)\tau}.$ (4.14)

###### Proof.

Given that $e^{-2u_{0}}\leq Ce^{-2(k+1)\tau}$, where $C$ depends only on
$\tau_{0}$, then using (4.6), we see that

$\displaystyle(u_{0}^{\prime}+w^{\prime})^{2}$
$\displaystyle=(4V^{2}-\zeta^{2}v^{\prime 2})e^{-2u_{0}}e^{-2w}+b^{2}$
$\displaystyle\leq 4V^{2}e^{-2u_{0}}e^{-2w}+b^{2}$ $\displaystyle\leq
Ce^{2(2\zeta-1)(k+1)\tau}e^{-2w}+b^{2}\leq(Ce^{-d_{\zeta}b\tau}e^{-w}+b)^{2}.$

Since $b\leq u^{\prime}=u_{0}^{\prime}+w^{\prime}$ and $b\leq u_{0}^{\prime}$
as well, we have

$b-u^{\prime}_{0}\leq w^{\prime}\leq Ce^{-d_{\zeta}b\tau}e^{-w}.$ (4.15)

We analyze the left inequality first. We know that

$(u_{0}^{\prime})^{2}=4e^{-2u_{0}}+b^{2}\Rightarrow(u_{0}^{\prime})^{2}-b^{2}=4e^{-2u_{0}}$

hence

$w^{\prime}\geq b-u_{0}^{\prime}\geq-Ce^{-2b\tau}$

for $\tau\geq\tau_{0}$. The constant $C$ is uniform in $\zeta$ since
$e^{-2u_{0}}$ is independent of $\zeta$. We have established uniform control
of $w(\tau_{0})$ in (4.7), so integrating from $\tau_{0}$ to $\tau$ proves the
lower bounds for both $w$ and $w^{\prime}$.

As for the upper bounds, multiply both sides of the right inequality in (4.15)
by $e^{w}$ to get

$(e^{w})^{\prime}\leq Ce^{-d_{\zeta}b\tau},\ \ \ \tau\geq\tau_{0}.$

Integrating from $\tau_{0}$ to $\tau$ yields

$w(\tau)-w(\tau_{0})\leq\frac{C}{d_{\zeta}b}e^{-d_{\zeta}b\tau}.$

Since, as above, we have a bound on $w(\tau_{0})$, we get the upper bound in
(4.13). The lower bound in (4.13), which is already established, gives
$e^{-w}\leq C$, and hence

$w^{\prime}\leq Ce^{-d_{\zeta}b\tau}e^{-w}\leq Ce^{-d_{\zeta}b\tau}.$

This concludes the proof. ∎

We finally prove global bounds for $V$ and $v^{\prime}$:

######  Lemma 4.9.

There exists $C=C(\zeta_{0})>0$ such that for $\zeta\leq\zeta_{0}<1/2$ and
$\tau\geq\tau_{0}$,

$|V|\leq C,\quad-Ce^{-2\tau}\leq v^{\prime}\leq Ce^{-d_{\zeta}b\tau}.$

###### Proof.

Rerite the second equation in (4.2) as

$(v^{\prime}e^{-2u})^{\prime}=4e^{-2u}V(u^{\prime}T-(k+1)).$

By (4.14), we have

$b\leq u^{\prime}\leq b+Ce^{-d_{\zeta}b\tau}.$

Combining this with

$1-2e^{-2\tau}\leq T\leq 1,$

leads to

$-Ce^{-2u-2\tau}V\leq(v^{\prime}e^{-2u})^{\prime}\leq
Ce^{-2u-d_{\zeta}b\tau}V,\ \ \ \tau\geq\tau_{0}/$ (4.16)

Then (4.6) and the bounds for $u$ give

$-Ce^{-((2-2\zeta)b+2)\tau}\leq(v^{\prime}e^{-2u})^{\prime}\leq
Ce^{-(d_{\zeta}+2-2\zeta)b\tau}.$

Integrating from $\tau\geq\tau_{0}$ to $\infty$ and using (4.6) to get the
exponential decay of $v^{\prime}e^{-2u}$, we arrive at

$-Ce^{(2\zeta b-2)\tau}\leq v^{\prime}\leq Ce^{(2\zeta-d_{\zeta})b\tau}.$

This is an improvement compared to the second bound in (4.6): the exponent in
the lower bound has dropped from $2\zeta b$ to $2\zeta b-2$, while on the
right hand side it has dropped by $d_{\zeta}b$. Repeating this argument
iteratively, then so long as $d_{\zeta}>0$, we eventually obtain exponentially
decaying functions both above and below, as desired. Consider only the upper
bound for simplicity. Let $n_{\zeta}$ be the largest integer for which
$2\zeta-n_{\zeta}d_{\zeta}>0$. Integrating from $\tau_{0}$ to $\tau$ and using
(4.8) we get that

$V(\tau)\leq C+\frac{C}{(2\zeta-d_{\zeta})b}e^{(2\zeta-
d_{\zeta})b\tau}\leq\frac{C}{(2\zeta-d_{\zeta})b}e^{(2\zeta-d_{\zeta})b\tau}$

for some possibly different constant $C$. Plugging this improvement into
(4.16) and repeating the steps, we obtain inductively

$V(\tau)\leq Ce^{(2\zeta-
n_{\zeta}d_{\zeta})b\tau}\big{/}\prod\limits_{n=1}^{n_{\zeta}}(2\zeta-
nd_{\zeta})b).$

We must distinguish between two cases. If $2\zeta-(n_{\zeta}+1)d_{\zeta}<0$,
then using this final estimate in (4.16) gives the exponentially decaying
upper bound for $v^{\prime}$. Repeating the steps above, we also see that
$V\leq C$. Using this bound (4.16) one last time, we obtain the required upper
bound for $v^{\prime}$.

The other case is when $2\zeta-(n_{\zeta}+1)d_{\zeta}=0$. We then must repeat
the steps one additional time; at the first pass, we deduce that
$v^{\prime}\leq C$, so $V\leq C\tau$, but this implies that $v^{\prime}$ is
bounded above by an exponentially decreasing function, and hence that $V\leq
C$. ∎

###  4.4. Conclusion of the continuity argument

Now that have established uniform bounds for $u$ and $v$ for $\zeta$ in any
interval $[0,\zeta_{0}]$, $\zeta_{0}<1/2$, we can complete the proof of
existence of model knot solutions for every $\zeta\in[0,1/2)$.

######  Theorem 4.10.

Suppose that $\zeta_{j}\nearrow\bar{\zeta}<1/2$ and that $(u_{j},v_{j})$ is a
corresponding sequence of solutions to (4.2) with parameter $\zeta_{j}$. Then
there exists a subsequence $(u_{j^{\prime}},v_{j^{\prime}})$ which converges
smoothly on every compact interval $0<a\leq b<\infty$, and such that the
limiting solution $(\bar{u},\bar{v})$ satisfies the same asymptotics as
$\tau\to 0$ and $\tau\to\infty$.

###### Proof.

Write $u_{j}=u_{0}+w_{j}$ as before. By Lemmas (4.7) and (4.8), $w_{j}$ is
uniformly bounded on $\mathbb{R}^{+}$ and $|w_{j}^{\prime}(\tau)|$ decays at a
uniform exponential rate as $\tau\to\infty$ and is uniformly bounded near
$\tau=0$. Lemmas (4.5) , (4.6) and (4.9) provide the analogous uniform bounds
for $v_{j}$ and $v_{j}^{\prime}$. The equation itself now gives uniform bounds
on $u_{j}^{\prime\prime}$ and $v_{j}^{\prime\prime}$ on any compact interval,
and this may be bootstrapped to bounds for any derivative. By taking a
diagonal subsequence over increasing intervals, we obtain limiting functions
$(\bar{u},\bar{v})$, and by our uniform bounds near $0$ and $\infty$, this
limiting solution also has the same asymptotics.

Of course, $(\bar{u},\bar{v})$ is real analytic as well. Indeed, rewrite (4.2)
as

$\begin{split}w^{\prime\prime}&=-e^{-2u_{0}}(e^{-2w}(4V^{2}+\zeta^{2}v^{\prime
2}+4\zeta^{2}Vv^{\prime}T)-4)\\\
v^{\prime\prime}&=(u_{0}^{\prime}+w^{\prime})(2v^{\prime}+4VT)-4(k+1)V.\end{split}$
(4.17)

This system can be written in the form

$\tau U^{\prime}=F(U,\tau),$

where $\vec{U}:=(u,u^{\prime},v,v^{\prime})$ and $F$ is analytic in its
arguments, This is a regular singular equation, and classical theory gives a
convergent series expansion. Analyticity near $\tau=0$ then follows since the
indicial roots of this system at $\tau=0$ are positive integers. ∎

## References

  * [1] Davide Gaiotto and Edward Witten  “Knot invariants from four-dimensional gauge theory”  In _Advances in Theoretical and Mathematical Physics_ 16.3 International Press of Boston, 2012, pp. 935–1086
  * [2] Siqi He and Rafe Mazzeo  “Opers and the twisted Bogomolny equations”  In _Transactions of the AMS_ , To appear
  * [3] Siqi He and Rafe Mazzeo  “The extended Bogomolny equations with generalized Nahm pole boundary conditions, II”  In _Duke Mathematical Journal_ 169.12 Duke University Press, 2020, pp. 2281–2335
  * [4] Anton Kapustin and Edward Witten  “Electric-magnetic duality and the geometric Langlands program”  In _arXiv preprint hep-th/0604151_ , 2006
  * [5] Rafe Mazzeo and Edward Witten  “The KW equations and the Nahm pole boundary condition with knots”  In _Communications in Analysis and Geometry_ 28.4 International Press, 2020, pp. 871–942
  * [6] Rafe Mazzeo and Edward Witten  “The Nahm pole boundary condition”  In _The influence of Solomon Lefschetz in geometry and topology_ 50 Am. Math. Soc., 2014, pp. 171
  * [7] Mazzeo Rafe  “Elliptic theory of differential edge operators I”  In _Communications in Partial Differential Equations_ 16.10 Taylor & Francis, 1991, pp. 1615–1664
  * [8] Carlos T Simpson  “Constructing variations of Hodge structure using Yang-Mills theory and applications to uniformization”  In _Journal of the American Mathematical Society_ JSTOR, 1988, pp. 867–918
  * [9] Edward Witten  “Fivebranes and knots”  In _Quantum Topology_ 3.1, 2011, pp. 1–137