# The Loewner Energy via the Renormalised Energy of
Moving Frames

Alexis Michelat***EPFL B, Station 8, CH-1015 Lausanne, Switzerland
<EMAIL_ADDRESS><EMAIL_ADDRESS>Yilin Wang†††Institut
des Hautes Études Scientifiques, Bures-sur-Yvette, France<EMAIL_ADDRESS>
(August 27, 2024)

###### Abstract

We obtain a new formula for the Loewner energy of Jordan curves on the sphere,
which is a Kähler potential for the essentially unique Kähler metric on the
Weil-Petersson universal Teichmüller space, as the renormalised energy of
moving frames on the two domains of the sphere delimited by the given curve.

###### Contents

  1. 1 Introduction
    1. 1.1 Background on Weil-Petersson quasicircles
    2. 1.2 Moving Frames and the Ginzburg-Landau Equations
    3. 1.3 Main Results
  2. 2 Moving Frame Energy via Zeta-Regularised Determinants for Smooth Curves
  3. 3 Construction of Harmonic Moving Frames for Weil-Petersson Curves
  4. 4 Proof of the Main Theorems for Non-Smooth Curves
  5. 5 Appendix
    1. 5.1 Properties of the Geodesic Curvature for Weil-Petersson Quasicircles
    2. 5.2 A consequence of Theorem 4.5

Mathematical subject classification : 53C42, 30C35.

##  1 Introduction

###  1.1 Background on Weil-Petersson quasicircles

In [40, 30], the second author and Steffen Rohde introduced the Möbius-
invariant Loewner energy to measure the roundness of Jordan curves on the
Riemann sphere $\mathbb{C}\cup\\{\infty\\}$ using the Loewner transform [25].
The original motivation comes from the probabilistic theory of Schramm-Loewner
evolutions, see, e.g., [42] for an overview. It was proved in [41] that the
Loewner energy is proportional to the _universal Liouville action_ introduced
by Takhtajan and Teo [14]. In particular, the class of finite energy curves
corresponds exactly to the _Weil-Petersson_ class of quasicircles which has
already been studied extensively by both physicists and mathematicians since
the eighties, see, _e.g._ , [8, 43, 28, 27, 31, 11, 14, 32, 15, 33, 16, 4, 37,
38, 20], and is still an active research area. See the introduction of [4]
(see also the companion papers [5] and [6] for more on this topic) for a
summary and a list of equivalent definitions of very different nature.

In this article, we sometimes view Jordan curves as curves on
$S^{2}\subset\mathbb{R}^{3}$ and give new characterisations of the Loewner
energy in terms of the moving frames on $S^{2}$. Note that in this article,
$S^{2}$ refers to the sphere of radius $1$ centred at the origin in
$\mathbb{R}^{3}$ equipped with the induced round metric $g_{0}$ from its
embedding into $\mathbb{R}^{3}$. Therefore, $S^{2}$ is isometric to
$\widehat{\mathbb{C}}=\mathbb{C}\cup\left\\{\infty\right\\}$ endowed with the
metric

$\displaystyle g_{\widehat{\mathbb{C}}}=\frac{4|dz|^{2}}{(1+|z|^{2})^{2}}$

by the stereographic projection. To distinguish the two setups, we will let
$\gamma$ denote a Jordan curve in $\widehat{\mathbb{C}}$ and let $\Gamma$
denote a Jordan curve in $S^{2}$. Let us first list a few equivalent
definitions of Weil-Petersson quasicircles that are relevant to this work.

######  Theorem 1.1 (Cui, [11], Tahktajan-Teo, [14], Shen, [33], Bishop, [4]).

Let $\gamma\subset\mathbb{C}$ be a Jordan curve, $\Omega$ be the bounded
connected component of $\mathbb{C}\setminus\gamma$, and let
$f:\mathbb{D}\rightarrow\Omega$ and
$g:\mathbb{C}\setminus\overline{\mathbb{D}}\rightarrow\mathbb{C}\setminus\overline{\Omega}$
be biholomorphic maps such that $g(\infty)=\infty$. The following conditions
are equivalent:

  1. _(1)_

There exists a quasiconformal extension of $g$ to $\mathbb{C}$ such that the
Beltrami coefficient
$\mu=\dfrac{\partial_{\overline{z}}g}{\partial_{z}g}:\mathbb{D}\rightarrow\mathbb{D}$
of $g|_{\mathbb{D}}$ satisfies
$\displaystyle\int_{\mathbb{D}}|\mu(z)|^{2}\frac{|dz|^{2}}{(1-|z|^{2})^{2}}<\infty$.

  2. _(2)_

$\displaystyle\int_{\mathbb{D}}|\nabla\log|f^{\prime}(z)||^{2}|dz|^{2}=\int_{\mathbb{D}}\left|\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}\right|^{2}|dz|^{2}<\infty$.

  3. _(3)_

$\displaystyle\int_{\mathbb{C}\setminus\overline{\mathbb{D}}}\left|\frac{g^{\prime\prime}(z)}{g^{\prime}(z)}\right|^{2}|dz|^{2}<\infty$.

  4. _(4)_

The _(conformal) welding function_ $\varphi=g^{-1}\circ f|_{S^{1}}$ satisfies
$\log\varphi^{\prime}$ belongs to the Sobolev space $H^{1/2}(S^{1})$.

  5. _(5)_

The curve $\gamma$ is chord-arc and the unit tangent $\tau:\gamma\rightarrow
S^{1}$ belongs to $H^{1/2}(\gamma)$.

  6. _(6)_

Every minimal surface
$\Sigma\subset\mathbb{H}^{3}\simeq\mathbb{C}\times\mathbb{R}_{+}^{\ast}$ with
asymptotic boundary $\gamma$ has finite renormalised area, _i.e._ ,

$\displaystyle{\mathcal{RA}(\Sigma)=\lim\limits_{\varepsilon\rightarrow
0}\left(\mathrm{Area}(\Sigma_{\varepsilon})-\mathrm{Length}(\partial\Sigma_{\varepsilon})\right)=-2\pi\chi(\Sigma)-\int_{\Sigma}|\mathring{A}|^{2}d\mathrm{vol}_{\Sigma}>-\infty,}$
(1.1)

where for all $\varepsilon>0$,
$\Sigma_{\varepsilon}=\Sigma\cap\left\\{(z,t):t>\varepsilon\right\\}$ and
$\partial\Sigma_{\varepsilon}=\Sigma\cap\left\\{(z,t):t=\varepsilon\right\\}$.

If $\gamma$ satisfies any of those conditions, $\gamma$ is called a _Weil-
Petersson quasicircle_.

The equivalences $(1)$ and $(2)$ are due to Cui, and independently to
Takhtajan and Teo who proved the equivalences $(1),(2),(3)$. In $(4)$, the
continuous extension of $f,g$ to $S^{1}$ is well-defined by a classical
theorem of Carathéodory [9]. The equivalence between $(1)$ and $(4)$ is proved
by Shen. The second condition is perhaps the simplest one since it corresponds
to the condition $\log|f^{\prime}|\in W^{1,2}(\mathbb{D})$, the Sobolev space
of functions with squared-integrable weak derivatives.

For $(5)$, we recall that a Jordan curve is _chord-arc_ if there exists
$K<\infty$ such that for all $x,y\in\gamma$, we have $\ell(x,y)\leq K|x-y|$,
where $\ell(x,y)$ is the length of the shortest arc joining $x$ to $y$. We
mention that Weil-Petersson quasicircles are not only chord-arc but even
_asymptotically smooth_ , namely, the ratio $\ell(x,y)/|x-y|$ tends to $1$ as
$x$ tends to $y$. These curves are not necessarily $C^{1}$ for they allow
certain types of infinite spirals. See Section 5.1 for an explicit
construction of such spirals. Recall that for any Jordan chord-arc curve
$\gamma$, a function $u:\gamma\rightarrow\mathbb{C}$ belongs to the Sobolev
space $H^{1/2}(\gamma)$ if and only if

$\displaystyle\int_{\gamma}\int_{\gamma}\left|\frac{u(z)-u(w)}{z-w}\right|^{2}|dz||dw|<\infty,$
(1.2)

where $|dz|$ is the arc-length measure.

The equivalence between ($1$) and ($5$) was proven by Y. Shen and L. Wu ([36];
see also [18, 33, 34, 35]), and also by Christopher Bishop [4]. The last
characterisation ($6$) due to Bishop [4] using the notion of renormalised area
was first investigated for Willmore surfaces by S. Alexakis and R. Mazzeo
([1], [2]) which has strong motivations arising from string theory [17]. The
integral of the squared trace-free second fundamental form $\mathring{A}$ in
$(6)$ is the Willmore energy of $\Sigma$ which is of particular interest for
being conformally invariant. Amongst the important previous contribution that
inspired this work, we should mention Epstein’s work ([12], [13]).

Not only we can characterize this class of curves qualitatively, as listed
above, there is an important quantity associated with each element of the
class. Indeed, after appropriate normalisation, the class of Weil-Petersson
quasicircles can be identified with the Weil-Petersson universal Teichmüller
space $T_{0}(1)$ via conformal welding. Takhtajan and Teo [14] showed that
$T_{0}(1)$ carries an essentially unique homogeneous Kähler metric and
introduced the _universal Liouville action_ $S_{1}$. They showed that $S_{1}$
is a Kähler potential on $T_{0}(1)$ which is of critical importance for the
Kähler geometry. We take an analytic instead of a Teichmüller theoretic
viewpoint, so we will consider $S_{1}$ as defined for Weil-Petersson
quasicircles instead of their welding functions. Explicitly, for a Weil-
Petersson quasicircle $\gamma$,

$\displaystyle
S_{1}(\gamma)=\int_{\mathbb{D}}\left|\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}\right|^{2}|dz|^{2}+\int_{\mathbb{C}\setminus\overline{\mathbb{D}}}\left|\frac{g^{\prime\prime}(z)}{g^{\prime}(z)}\right|^{2}|dz|^{2}+4\pi\log|f^{\prime}(0)|-4\pi\log|g^{\prime}(\infty)|.$
(1.3)

######  Theorem 1.2 (Y. Wang, [41]).

A Jordan curve $\gamma$ has finite Loewner energy $I^{L}(\gamma)$ if and only
if $\gamma$ is a Weil-Petersson quasicircle. Furthermore, we have

$\displaystyle I^{L}(\gamma)=\frac{1}{\pi}S_{1}(\gamma).$ (1.4)

We will therefore use interchangeably the terms “Jordan curve of finite
Loewner energy,” “Weil-Petersson quasicircle,” or simply “Weil-Petersson
curve.” As we did not define explicitly the Loewner energy $I^{L}(\gamma)$,
readers may consider (1.4) as its definition. It may not be obvious from the
expression of $S_{1}$ that it is invariant under Möbius transformations, such
as the inversion ${\mathfrak{i}}:z\mapsto 1/z$. However, it would follow
directly from the definition using Loewner transform in [30]. Provided that
$\gamma$ separates $0$ from $\infty$, we may choose the biholomorphic
functions $f$ and $g$ as in Theorem 1.1 and assume further that $f(0)=0$.
Applying the invariance of the Loewner energy under ${\mathfrak{i}}$, we get

$\displaystyle I^{L}(\gamma)=I^{L}({\mathfrak{i}}(\gamma))$ (1.5)
$\displaystyle=\frac{1}{\pi}\int_{\mathbb{D}}\left|\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}-2\frac{f^{\prime}(z)}{f(z)}+\frac{2}{z}\right|^{2}|dz|^{2}+\frac{1}{\pi}\int_{\mathbb{C}\setminus\overline{\mathbb{D}}}\left|\frac{g^{\prime\prime}(z)}{g^{\prime}(z)}-2\frac{g^{\prime}(z)}{g(z)}+\frac{2}{z}\right|^{2}|dz|^{2}+4\log|f^{\prime}(0)|-4\log|g^{\prime}(\infty)|.$

###  1.2 Moving Frames and the Ginzburg-Landau Equations

Moving frames, first introduced by Darboux in the late $19$th century to study
curves and surfaces, were later generalised by Élie Cartan and permit one to
reformulate astutely a wide class of differential-geometric problems. One of
the rather recent such use of this theory is found in the work of Frédéric
Hélein on harmonic maps ([19]), where the moving frame pave the way towards
new regularity results.

In [23], Paul Laurain and Romain Petrides suggest a new approach to relate the
Loewner energy to the renormalised energy of moving frames using the Ginzburg-
Landau energy in a minimal regularity setting (which is of independent
interest). Although the Ginzburg-Landau is normally used to construct harmonic
maps with values into $S^{1}$ under topological constraints where no smooth
solutions exist ([3]), it should be seen—although we will not use this
functional here—more generally as a way to construct (singular) moving frames
on surfaces. Through this approach, one may hope to link quantatively the
Loewner energy and the Willmore energy that can also be written in terms of
moving frames ([26]).

Let $\Omega\subset\mathbb{C}$ be a simply connected domain, and
$\gamma=\partial\Omega$. In [23], they show that the Bethuel-Brezis-Hélein
([3]) analysis carries on for general chorc-arc curves and $H^{1/2}$ boundary
data. Using this delicate analysis, they obtain the following result, which is
the most relevant one in this article.

######  Theorem 1.3 (Laurain-Petrides, [23], Theorem $0.2$, Theorem $0.3$).

Let $\Omega\subset\mathbb{C}$ be a bounded simply connected domain such that
$\gamma=\partial\Omega$ is a Weil-Petersson quasicircle. Then, there exists a
harmonic map $\vec{u}:\Omega\setminus\left\\{p\right\\}\rightarrow S^{1}$ with
boundary data $\tau:\Gamma\rightarrow S^{1}$ which is the unit tangent vector
of $\partial\Omega=\Gamma$. Let $\vec{v}=-i\,\vec{u}$ and
$\omega=\langle\vec{u},d\vec{v}\,\rangle$, then there exists a harmonic
function $\mu:\Omega\rightarrow\mathbb{R}$ such that
$\omega=\ast\,d\left(G_{\Omega}+\mu\right)$, and a conformal map
$f:\mathbb{D}\rightarrow\Omega$ such that $f(0)=p$, and

$\displaystyle\left\\{\begin{aligned}
\frac{1}{r}\partial_{\theta}f&=e^{\mu\circ f}\vec{u}\circ f\\\
\partial_{r}f&=e^{\mu\circ f}\vec{v}\circ f.\end{aligned}\right.$ (1.6)

Furthermore, we have

$\displaystyle\int_{\Omega}|\omega-\ast\,dG_{\Omega}|^{2}dx=\int_{\Omega}|\nabla\mu|^{2}dx=\int_{\mathbb{D}}\left|\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}\right|^{2}|dz|^{2},$
(1.7)

where $G_{\Omega}$ is a Green’s function with Dirichlet boundary condition on
$\partial\Omega$.

The other main result of [23] is to identify the renormalised energy in the
sense of Bethuel-Brezis-Hélein as an explicit term involving (1.7).

######  Remark 1.4.

The harmonic function $\mu$ is explicitly given by
$\mu=\log|\partial_{r}f|\circ
f^{-1}=\log\left|\frac{1}{r}\partial_{\theta}f\right|\circ
f^{-1}=\log|f^{\prime}|\circ f^{-1}$. The last identities follow from the
conformality of $f$. We note that in [23], the point $p$ is a special point
such that any biholomorphic map $f$ with $f(0)=p$ maximizes $|f^{\prime}(0)|$
amongst all biholomorphic maps $\mathbb{D}\to\Omega$.

We see that the frame energy (1.7) coincides with the first term in (1.3). To
obtain the second half of the Loewner energy involving

$\displaystyle\int_{\mathbb{C}\setminus\overline{\mathbb{D}}}\left|\frac{g^{\prime\prime}(z)}{g^{\prime}(z)}\right|^{2}|dz|^{2},$
(1.8)

we cannot easily use the Ginzburg-Landau equation to construct the moving
frames since that would force us to work on the non-compact domain
$\mathbb{C}\setminus\overline{\Omega}$. Using the inversion ${\mathfrak{i}}$
will not suffice either. If we choose the biholomorphic map
$\widetilde{g}:\mathbb{D}\rightarrow{\mathfrak{i}}(\mathbb{C}\setminus\overline{\Omega})$
so that $\widetilde{g}={\mathfrak{i}}\circ g\circ{\mathfrak{i}}$, we have

$\displaystyle\int_{\mathbb{D}}\left|\frac{\widetilde{g}^{\prime\prime}(z)}{\widetilde{g}^{\prime}(z)}\right|^{2}|dz|^{2}=\int_{\mathbb{C}\setminus\overline{\mathbb{D}}}\left|\frac{g^{\prime\prime}(z)}{g^{\prime}(z)}-2\frac{g^{\prime}(z)}{g(z)}+\frac{2}{z}\right|^{2}|dz|^{2}$

which is in general different from (1.8). To overcome this technicality, we
work directly on $S^{2}$ to obtain a formula of the Loewner energy in terms of
moving frames.

###  1.3 Main Results

######  Theorem A.

Let $\Gamma\subset S^{2}\subset\mathbb{R}^{3}$ be a Weil-Petersson
quasicircle, $\Omega_{1},\Omega_{2}\subset S^{2}$ be the two disjoint open
connected components of $S^{2}\setminus\Gamma$. Fix some $j=1,2$. Then, for
any $p_{j}\in\Omega_{j}$, there exists harmonic moving frames
$(\vec{u}_{j},\vec{v}_{j}):\Omega_{j}\setminus\left\\{p_{j}\right\\}\rightarrow
U\Omega_{j}\times U\Omega_{j}$ such that the Cartan form
$\omega_{j}=\langle\vec{u}_{j},d\vec{v}_{j}\rangle$ admits the decomposition

$\displaystyle\omega_{j}=\ast\,d\left(G_{\Omega_{j}}+\mu_{j}\right),$ (1.9)

where
$G_{\Omega_{j}}:\Omega_{j}\setminus\left\\{p_{j}\right\\}\rightarrow\mathbb{R}$
is the Green’s function of the Laplacian $\Delta_{g_{0}}$ on $\Omega_{j}$ with
Dirichlet boundary condition, and $\mu_{j}\in C^{\infty}(\Omega_{j})$
satisfies

$\displaystyle\left\\{\begin{aligned}
-\Delta_{g_{0}}\mu_{j}&=1\qquad&&\text{in}\;\,\Omega_{j}\\\
\partial_{\nu}\mu_{j}&=k_{g_{0}}-\partial_{\nu}G_{\Omega_{j}}\qquad&&\text{on}\;\,\partial\Omega_{j},\end{aligned}\right.$
(1.10)

where $k_{g_{0}}$ is the geodesic curvature on $\Gamma=\partial\Omega_{j}$.
Define the functional $\mathscr{E}$ _(_ that we call the renormalised energy
associated to the frames $(\vec{u}_{1},\vec{v}_{1})$ and
$(\vec{u}_{2},\vec{v}_{2})$_)_ by

$\displaystyle\mathscr{E}(\Gamma)=\int_{\Omega_{1}}|d\mu_{1}|^{2}_{g_{0}}d\mathrm{vol}_{g_{0}}+\int_{\Omega_{2}}|d\mu_{2}|^{2}_{g_{0}}d\mathrm{vol}_{g_{0}}+2\int_{\Omega_{1}}G_{\Omega_{1}}K_{g_{0}}d\mathrm{vol}_{g_{0}}+2\int_{\Omega_{2}}G_{\Omega_{2}}K_{g_{0}}d\mathrm{vol}_{g_{0}}+4\pi.$
(1.11)

Then there exists conformal maps $f_{1}:\mathbb{D}\rightarrow\Omega_{1}$ and
$f_{2}:\mathbb{D}\rightarrow\Omega_{2}$ such that $f_{1}(0)=p_{1}$,
$f_{2}(0)=p_{2}$ and

$\displaystyle I^{L}(\Gamma)=\frac{1}{\pi}\mathscr{E}(\Gamma)+4\log|\nabla
f_{1}(0)|+4\log|\nabla
f_{2}(0)|-12\log(2)=\frac{1}{\pi}\mathscr{E}_{0}(\Gamma).$ (1.12)

Figure 1:  Harmonic moving frames on the sphere associated to a Weil-Petersson
quasicircle.

######  Remark 1.10.

  1. (1)

In the theorem above, we wrote $U\Omega_{j}$ ($j=1,2$) for the unit tangent
bundle. The function $\mu_{j}$, explicitly given by

$\displaystyle\mu_{j}=\frac{1}{2}\log\left(\frac{|\nabla
f_{j}|^{2}}{2}\right)=\log|\nabla f_{j}|-\frac{1}{2}\log(2){,}$ (1.13)

correspond to the conformal parameter of the conformal maps
$f_{1},f_{2}:\mathbb{D}\rightarrow S^{2}\subset\mathbb{R}^{3}$.

  2. (2)

The constant term $4\pi$ in the definition of $\mathscr{E}$ is arranged so
that $\mathscr{E}(S^{1})=0$ (see Remark 2.8). Furthermore, the name
renormalised energy is justified by the following identity

$\displaystyle\mathscr{E}(\Gamma)=\int_{\Omega_{1}}\left(|d\vec{u}_{1}|^{2}_{g_{0}}+|d\vec{v}_{1}|_{g_{0}}^{2}-2|dG_{\Omega_{1}}|^{2}_{g_{0}}\right)d\mathrm{vol}_{g_{0}}+\int_{\Omega_{2}}\left(|d\vec{u}_{2}|^{2}_{g_{0}}+|d\vec{v}_{2}|_{g_{0}}^{2}-2|dG_{\Omega_{2}}|^{2}_{g_{0}}\right)d\mathrm{vol}_{g_{0}},$

where no constant term is involved.

  3. (3)

The solution (1.10) to the Dirichlet problem is unique, and so are the moving
frames once the singularities $(p_{1},p_{2})\in\Omega_{1}\times\Omega_{2}$ are
fixed. See Theorem 3.2 and 3.5. Notice that the geodesic curvature is
understood in the distributional sense here (see Section 5.1 from the appendix
for more details).

This theorem corresponds to Theorem 2.6 in the article for smooth curves and
to Theorem 4.5 for general Weil-Petersson quasicircles. The general case
follows essentially from the following result which can also be viewed as a
restatement of Theorem A without any mention of moving frames.

######  Theorem B (See Theorem 4.5).

Let $\Gamma\subset S^{2}$ be a Weil-Petersson quasicircle and
$\Omega_{1},\Omega_{2}\subset S^{2}\setminus\Gamma$ be the two connected
components of $S^{2}\setminus\Gamma$. For all conformal maps
$f_{1}:\mathbb{D}\rightarrow\Omega_{1}$ and
$f_{2}:\mathbb{D}\rightarrow\Omega_{2}$, we have

$\displaystyle I^{L}(\Gamma)$
$\displaystyle=\frac{1}{\pi}\sum_{j=1}^{2}\left(\int_{\mathbb{D}}|\nabla\log|\nabla
f_{j}||^{2}|dz|^{2}+\int_{\mathbb{D}}\log|z||\nabla
f_{j}|^{2}|dz|^{2}+\mathrm{Area}(\Omega_{j})+4\pi\log|\nabla
f_{j}(0)|\right)-12\log(2).$ (1.14)

Acknowledgements. This paper is part of a common project between Paul Laurain
and Romain Petrides and the two authors on the various characterisations of
Weil-Petersson quasicircles. We thank Paul Laurain and Romain Petrides for
useful discussions and kindly sharing their manuscript with us. We also thank
Christopher Bishop for allowing us into his topic class which helped us
understand his recent work [4]. A. M. is supported by the Early
Postdoc.Mobility _Variational Methods in Geometric Analysis_
P$2$EZP$2$_$191893$. Y. W. is partially supported by NSF grant DMS-$1953945$.

##  2 Moving Frame Energy via Zeta-Regularised Determinants for Smooth Curves

The following expression of the Loewner energy will prove crucial in this
section.

######  Theorem 2.1 (Y. Wang [41]).

Let $\alpha\in C^{\infty}(S^{2},\mathbb{R})$, $g=e^{2\alpha}g_{0}$ be a metric
conformally equivalent to the spherical metric $g_{0}$ of $S^{2}$, and
$\Gamma\subset S^{2}$ be a simple smooth curve. Let
$\Omega_{1},\Omega_{2}\subset S^{2}$ be the two disjoint open connected
components of $S^{2}\setminus\Gamma$. Then we have

$\displaystyle
I^{L}(\Gamma)=12\,\log\frac{\det_{\zeta}(-\Delta_{S^{2}_{-},g})\det_{\zeta}(-\Delta_{S^{2}_{+},g})}{\det_{\zeta}(-\Delta_{\Omega_{1},g})\det_{\zeta}(-\Delta_{\Omega_{2},g})},$
(2.1)

where $S_{-}^{2}$ _(_ resp. $S^{2}_{+}$_)_ is the southern hemisphere _(_
resp. the northern hemisphere _)_.

We now use the formula (2.1) expressing the Loewner energy in terms of zeta-
regularised determinants to link the Loewner energy to the renormalised energy
of moving frames on $S^{2}$. First, let $g_{0}=g_{S^{2}}$ be the standard
round metric on $S^{2}$. Let $\Gamma\subset S^{2}$ be a simple _smooth_ ‡‡‡It
is necessary to assume that the curve is smooth for one will need to recurse
to the Froebenius theorem below. Furthermore, the formula for the Loewner
energy using the zeta-regularised determinants ([41]) only works for smooth
(or at least $C^{3}$) curves ([14], Corollary $3.12$). curve, and let
$\Omega_{1},\Omega_{2}\subset S^{2}$ the two disjoint open connected
components of $S^{2}\setminus\Gamma$. Since we are working on a curved
manifold, we cannot directly use the result of [23] to construct moving frames
with the Ginzburg-Landau method. However, we will construct them directly in
Section 3 (see Theorem 3.2). Therefore, let us assume that
$(\vec{u}_{1},\vec{v}_{1}):\Omega_{1}\setminus\left\\{p_{1}\right\\}\rightarrow
US^{2}\times US^{2}$ are harmonic vector fields such that $\vec{u}_{1}=\tau$
on $\partial\Omega_{1}=\Gamma$ (where $\tau$ is the unit tangent on $\Gamma$),
and the $1$-form $\omega=\langle\vec{u}_{1},d\vec{v}_{1}\rangle$ satisfies

$\displaystyle\omega=\ast\,d\,\left(G_{\Omega_{1}}+\mu_{1}\right)\qquad\text{in}\;\,\mathscr{D}^{\prime}(\Omega_{1})$
(2.2)

where
$G_{\Omega_{1}}:\Omega_{1}\setminus\left\\{p_{1}\right\\}\rightarrow\mathbb{R}$
is the Green’s function for the Laplacian on
$\Omega_{1}\setminus\left\\{p_{1}\right\\}$ with Dirichlet boundary condition.
Namely, $G_{\Omega_{1}}$ satisfies

$\displaystyle\left\\{\begin{aligned}
\Delta_{g_{0}}G_{\Omega_{1}}&=2\pi\delta_{p_{1}}\qquad&&\text{in}\;\,\mathscr{D}^{\prime}(\Omega_{1})\\\
G_{\Omega_{1}}&=0\qquad&&\text{on}\;\,\partial\Omega_{1},\end{aligned}\right.$
(2.3)

and $\mu_{1}:\Omega_{1}\rightarrow\mathbb{R}$ is a smooth function satisfying

$\displaystyle\left\\{\begin{aligned}
-\Delta_{g_{0}}\mu_{1}&=1\qquad&&\text{in}\;\,\Omega_{1}\\\
\partial_{\nu}\mu_{1}&=k_{g_{0}}-\partial_{\nu}G_{\Omega_{1}}\qquad&&\text{on}\;\,\partial\Omega_{1}.\end{aligned}\right.$
(2.4)

where $k_{g_{0}}$ is the geodesic curvature with respect to the round metric
$g_{0}$, and the normal derivative is taken with respect to the $g_{0}$.

To fix notations, we recall the following result.

######  Theorem 2.2 ([22], see also, [21, 39]).

Let $\Omega\subset\mathbb{C}$ _(_ resp. $\Omega\subset S^{2}$_)_ be a bounded
simply connected domain such that $\partial\Omega$ is chord-arc, and let
$g_{0}$ be the flat metric on $\Omega$ _(_ resp. $g_{0}$ be the round metric
on $S^{2}$_)_. Then for all $p\in\Omega$, there exists a unique Green’s
function $G_{\Omega,p}\in
C^{\infty}(\Omega\setminus\left\\{p\right\\},\mathbb{R})$ with Dirichlet
boundary condition. Furthermore, for every $h\in
H^{1/2}(\partial\Omega,\mathbb{R})$, there exists a unique function $u\in
W^{1,2}(\Omega,\mathbb{R})$ such that

$\displaystyle\left\\{\begin{aligned}
\Delta_{g_{0}}u&=0\qquad&&\text{in}\;\,\Omega\\\
u&=h\qquad&&\text{on}\;\,\partial\Omega.\end{aligned}\right.$

Whenever it is clear from context, we will write $G_{\Omega_{1}}$ for
$G_{\Omega_{1},p_{1}}$.

######  Remark 2.3.

The existence of a Green’s function follows from its conformal invariance and
the uniformisation theorem. Indeed, if $\Omega$ is a Jordan domain, and
$f:\mathbb{D}\rightarrow\Omega$ is a biholomorphic map such that $f(0)=p$, and
$G_{\mathbb{D},0}=\log|z|$, then $G_{\Omega,p}=G_{\mathbb{D},0}\circ f^{-1}$.
We assume that $\partial\Omega$ is chord-arc so that the trace theorems apply
as in [22, 39]. The passage from $\mathbb{C}$ to $S^{2}$ is easy using a
stereographic projection and the conformal invariance of Green’s functions.

Now, following Proposition $5.1$ of [23], it is not hard to see that their
proof using the Froebenius theorem also works for domains of the sphere, and
we get a conformal diffeomorphism ${\varphi}:(-\infty,0)\times\partial
B(0,\rho)\rightarrow\Omega_{1}\setminus\left\\{p_{1}\right\\}$ for some
$\rho>0$ such that

$\displaystyle\begin{aligned}
\partial_{s}{\varphi}(s,\theta)&=e^{G_{\Omega_{1}}\circ{\varphi}+\mu_{1}\circ{\varphi}}\vec{v}_{1}\circ{\varphi}\\\
\partial_{\theta}{\varphi}(s,\theta)&=e^{G_{\Omega_{1}}\circ{\varphi}+\mu_{1}\circ{\varphi}}\vec{u}_{1}\circ{\varphi}.\end{aligned}$

Notice that the Proposition $5.1$ of [23] gives a privileged
$p_{1}\in\Omega_{1}$, but we will show in Theorem 3.2 that $p_{1}$ can be
taken arbitrarily (see also Theorem 4.5). However, the proof works for an
arbitrary harmonic moving frame whose Cartan form admits an expansion as in
(2.2) where $\mu_{1}$ solves (2.4). Since $\mu_{1}$ is defined up to an
additive constant, we can assume that $\rho=1$ in the following. We define the
conformal map $f_{1}:\mathbb{D}\rightarrow\Omega_{1}$ using the polar
coordinates by

$\displaystyle f_{1}(r,\theta)={\varphi}(\log(r),\theta),$

we can continuously extend $f_{1}$ at $z=0$ such that $f_{1}(0)=p_{1}$.

Now we relate $\mu_{1}$ to $f_{1}$. Since $f_{1}$ is conformal, the function
$G=G_{\Omega_{1}}\circ
f_{1}:\mathbb{D}\setminus\left\\{0\right\\}\rightarrow\mathbb{R}$ is harmonic
on $\mathbb{D}\setminus\left\\{0\right\\}$, satisfies $G=0$ on
$\partial\mathbb{D}$, so by (2.3), we deduce that

$\displaystyle G=G_{\mathbb{D},{0}}=\log|z|.$

Therefore, we have

$\displaystyle\left\\{\begin{aligned}
\partial_{r}f_{1}&=\frac{1}{r}\partial_{s}{\varphi}(\log(r),\theta)=\frac{1}{r}e^{\log(r)+\mu_{1}\circ
f_{1}}{\vec{v}}_{1}\circ f_{1}=e^{\mu_{1}\circ f_{1}}\vec{v}_{1}\circ f_{1}\\\
\frac{1}{r}\partial_{\theta}f_{1}&=\frac{1}{r}\partial_{\theta}{\varphi}(\log(r),\theta)=e^{\mu_{1}\circ
f_{1}}\vec{u}_{1}\circ f_{1}.\end{aligned}\right.$ (2.5)

Since $|\vec{u}_{1}|=|\vec{v}_{1}|=1$, and
$\langle\vec{u}_{1},\vec{v}_{1}\rangle=0$, we deduce that

$\displaystyle|\partial_{r}f_{1}|^{2}=\frac{1}{r^{2}}\left|\partial_{\theta}f_{1}\right|^{2}=e^{2\mu_{1}\circ
f_{1}}$
$\displaystyle\langle\partial_{r}f_{1},\partial_{\theta}f_{1}\rangle=0,$

which shows that the conformal parameter of $f$ is

$\displaystyle\frac{1}{2}|\nabla
f_{1}|^{2}={\frac{1}{2}\left(|\partial_{r}f_{1}|^{2}+\frac{1}{r^{2}}\left|\partial_{\theta}f_{1}\right|^{2}\right)}=e^{2\mu_{1}\circ
f_{1}},$

which implies that

$\displaystyle\mu_{1}=\log|\nabla f_{1}|\circ f_{1}^{-1}-\frac{1}{2}\log(2).$
(2.6)

In particular, we have

$\displaystyle\mu_{1}(p_{1})=\log|\nabla f_{1}(0)|-\frac{1}{2}\log(2),$ (2.7)

where $p_{1}\in\Omega_{1}$ is the singularity of the moving frame
$(\vec{u}_{1},\vec{v}_{1}):\Omega_{1}\setminus\left\\{p_{1}\right\\}\rightarrow
US^{2}\times US^{2}$.

We can relate the change of metric by $f_{1}$ to $\mu_{1}$ as follows. If
$\iota:\Omega_{1}\subset S^{2}\hookrightarrow\mathbb{R}^{3}$ is the inclusion
map, we have ${{g_{0}}_{|\Omega_{1}}}=\iota^{\ast}g_{\mathbb{R}^{3}}$.

As $f_{1}$ is conformal, we have

$\displaystyle f_{1}^{\ast}{g_{0}}_{|\Omega_{1}}$
$\displaystyle=f_{1}^{\ast}\iota^{\ast}g_{\mathbb{R}^{3}}=(\iota\circ
f_{1})^{\ast}g_{\mathbb{R}^{3}}={\frac{1}{2}}|\nabla
f_{1}(z)|^{2}|dz|^{2}=e^{2\mu_{1}\circ f_{1}(z)}|dz|^{2}=e^{2\mu_{1}\circ
f_{1}(z)-2\psi(z)}\frac{4|dz|^{2}}{(1+|z|^{2})^{2}}$
$\displaystyle=e^{2\mu_{1}\circ
f_{1}-2\psi}(({\pi^{-1}})^{\ast}g_{0})|_{\mathbb{D}},$ (2.8)

where

$\displaystyle\psi(z)=\log\left(\frac{2}{1+|z|^{2}}\right),$

and $\pi^{-1}:\mathbb{C}\rightarrow S^{2}\setminus\left\\{N\right\\}$ is the
inverse stereographic projection. Writing for simplicity

$\displaystyle
g_{\mathbb{D}}=\frac{4|dz|^{2}}{(1+|z|^{2})^{2}}=e^{2\psi(z)}|dz|^{2}=(\pi^{-1})^{\ast}g_{0}|_{\mathbb{D}},$

we deduce by (2) that

$\displaystyle{g_{0}}_{|\Omega_{1}}=(f_{1}\circ
f_{1}^{-1})^{\ast}(g_{0|\Omega_{1}})=(f_{1}^{-1})^{\ast}f_{1}^{\ast}{g_{0}}_{|\Omega_{1}}=e^{2\mu_{1}-2\psi\circ
f_{1}^{-1}}(f_{1}^{-1})^{\ast}(g_{\mathbb{D}}),$

so that (by an abuse of notation for the last identity)

$\displaystyle(f_{1}^{-1})^{\ast}(g_{\mathbb{D}})=e^{-2\mu_{1}+2\psi\circ
f_{1}^{-1}}{g_{0}}_{|\Omega_{1}}=e^{2\alpha_{1}}{g_{0}}_{|\Omega_{1}}$ (2.9)

where

$\displaystyle\alpha_{1}(z)=-\mu_{1}(z)+\psi{\circ}f_{1}^{-1}(z).$

######  Remark 2.4.

To summarize, the above discussion shows that the moving frame
$(\vec{u}_{1},\vec{v}_{1})$ satisfying the boundary condition
$\vec{u}_{1}=\tau$ on $\Gamma$, (2.2), and (2.4) is tightly related to a
conformal map $f_{1}:\mathbb{D}\to\Omega_{1}$ using Froebenius theorem as in
[23], in the way that the moving frame satisfies (2.5). However, we can start
directly with any conformal map $f_{1}$ and (2.5) gives a moving frame
$(\vec{u}_{1},\vec{v}_{1})$ which satisfies (2.2) and (2.4). This is the
approach we take in Section 3 which allows us to relax the regularity
assumption of $\partial\Omega_{1}=\Gamma$.

######  Definition 2.5.

Define the open subsets $S_{+}^{2},S^{2}_{-}\subset S^{2}$ by

$\displaystyle S_{+}^{2}$
$\displaystyle=S^{2}\cap\left\\{(x,y,z){\in\mathbb{R}^{3}}:z>0\right\\}$
$\displaystyle S_{-}^{2}$
$\displaystyle=S^{2}\cap\left\\{(x,y,z){\in\mathbb{R}^{3}}:z<0\right\\}.$

######  Theorem 2.6.

Let $\Gamma\subset S^{2}$ be a smooth Jordan curve, and let
$\Omega_{1},\Omega_{2}\subset S^{2}$ the two disjoint open connected
components of $S^{2}\setminus\Gamma$. Fix some $j=1,2$. Then, for all
$p_{j}\in\Omega_{j}$ and for all harmonic moving frames
$(\vec{u}_{j},\vec{v}_{j}):\Omega_{j}\setminus\left\\{p_{j}\right\\}\rightarrow
U\Omega_{j}\times U\Omega_{j}$ such that the Cartan form
$\omega_{j}=\langle\vec{u}_{j},d\vec{v}_{j}\rangle$ admits the decomposition

$\displaystyle\omega_{j}=\ast\,d\left(G_{\Omega_{j}}+\mu_{j}\right),$

where
$G_{\Omega_{j}}:\Omega_{j}\setminus\left\\{p_{j}\right\\}\rightarrow\mathbb{R}$
is the Green’s function of the Laplacian $\Delta_{g_{0}}$ on $\Omega_{j}$ with
Dirichlet boundary condition, and $\mu_{j}\in C^{\infty}(\Omega_{j})$
satisfies

$\displaystyle\left\\{\begin{aligned}
-\Delta_{g_{0}}\mu_{j}&=1\qquad&&\text{in}\;\,\Omega_{j}\\\
\partial_{\nu}\mu_{j}&=k_{g_{0}}-\partial_{\nu}G_{\Omega_{j}}\qquad&&\text{on}\;\,\partial\Omega_{j},\end{aligned}\right.$
(2.10)

where $k_{g_{0}}$ is the geodesic curvature on $\Gamma=\partial\Omega_{j}$.
Define the functional $\mathscr{E}$ _(_ that we call the renormalised energy
associated to the frames $(\vec{u}_{1},\vec{v}_{1})$ and
$(\vec{u}_{2},\vec{v}_{2})$_)_ by

$\displaystyle\mathscr{E}(\Gamma)=\int_{\Omega_{1}}|d\mu_{1}|^{2}_{g_{0}}d\mathrm{vol}_{g_{0}}+\int_{\Omega_{2}}|d\mu_{2}|^{2}_{g_{0}}d\mathrm{vol}_{g_{0}}+2\int_{\Omega_{1}}G_{\Omega_{1}}K_{g_{0}}d\mathrm{vol}_{g_{0}}+2\int_{\Omega_{2}}G_{\Omega_{2}}K_{g_{0}}d\mathrm{vol}_{g_{0}}+4\pi.$

Then there exists conformal maps $f_{1}:\mathbb{D}\rightarrow\Omega_{1}$ and
$f_{2}:\mathbb{D}\rightarrow\Omega_{2}$ such that $f_{1}(0)=p_{1}$,
$f_{2}(0)=p_{2}$ and

$\displaystyle I^{L}(\Gamma)=\frac{1}{\pi}\mathscr{E}(\Gamma)+4\log|\nabla
f_{1}(0)|+4\log|\nabla
f_{2}(0)|-12\log(2)=\frac{1}{\pi}\mathscr{E}_{0}(\Gamma).$

######  Remark 2.7.

If $\mathscr{W}_{j}$ ($j=1,2$) is the renormalised energy in the sense of
Bethuel-Brezis-Hélein associated to the moving frame
$(\vec{u}_{j},\vec{v}_{j})$—or more precisely, to its boundary data—(see [3]
and [23]), we have

$\displaystyle\mathscr{W}_{1}+\mathscr{W}_{2}=\mathscr{E}(\Gamma)+2\pi\log|\nabla
f_{1}(0)|+2\pi\log|\nabla f_{2}(0)|,$

since $\omega_{1}-\ast\,dG_{\Omega_{1}}=\ast\,d\mu_{1}$.

###### Proof of Theorem 2.6.

If $\pi^{-1}:\mathbb{C}\rightarrow S^{2}\setminus\left\\{N\right\\}$ is the
inverse stereographic projection,

$\displaystyle
g_{\mathbb{D}}=\frac{4|dz|^{2}}{(1+|z|^{2})^{2}}=e^{2\psi(z)}|dz|^{2},$

and $S^{2}_{-}$ is the southern hemisphere, we deduce that

$\displaystyle\det\hphantom{}_{\kern-1.42262pt\zeta}(-\Delta_{S^{2}_{-},g_{0}})$
$\displaystyle=\det\hphantom{}_{\kern-1.42262pt\zeta}(-\Delta_{\mathbb{D},\pi^{\ast}g_{0}})=\det\hphantom{}_{\kern-1.42262pt\zeta}(-\Delta_{\mathbb{D},g_{\mathbb{D}}})=\det\hphantom{}_{\kern-1.42262pt\zeta}(-\Delta_{\Omega_{1},\varphi^{\ast}g_{\mathbb{D}}})=\det\hphantom{}_{\kern-1.42262pt\zeta}(-\Delta_{\Omega_{1},g_{1}}),$

and by the Alvarez-Polyakov formula (see (1.17) of [29]) and (2.9), we have

$\displaystyle\det\hphantom{}_{\kern-1.42262pt\zeta}(-\Delta_{S^{2}_{-},g_{0}})-\det\hphantom{}_{\kern-1.42262pt\zeta}(-\Delta_{\Omega_{1},g_{0}})=\det\hphantom{}_{\kern-1.42262pt\zeta}(-\Delta_{\Omega_{1},g_{1}})-\det\hphantom{}_{\kern-1.42262pt\zeta}(-\Delta_{\Omega_{1},g_{0}})$
$\displaystyle=-\frac{1}{12\pi}\bigg{\\{}\int_{\Omega_{1}}|d\alpha_{1}|_{g_{0}}^{2}d\mathrm{vol}_{g_{0}}+2\int_{\Omega_{1}}K_{g_{0}}\alpha_{1}\,d\mathrm{vol}_{g_{0}}+2\int_{\Gamma}k_{g_{0}}\alpha_{1}\,d\mathscr{H}^{1}_{g_{0}}+3\int_{\Gamma}\partial_{\nu}\alpha_{1}\,d\mathscr{H}^{1}_{g_{0}}\bigg{\\}}$
$\displaystyle=-\frac{1}{12\pi}\bigg{\\{}\int_{\Omega_{1}}|d(-\mu_{1}+\theta_{1})|_{g_{0}}^{2}d\mathrm{vol}_{g_{0}}+2\int_{\Omega_{1}}K_{g_{0}}(-\mu_{1}+\theta_{1})d\mathrm{vol}_{g_{0}}+2\int_{\Gamma}k_{g_{0}}(-\mu_{1}+\theta_{1})d\mathscr{H}^{1}_{g_{0}}$
$\displaystyle+3\int_{\Gamma}\partial_{\nu}(-\mu_{1}+\theta_{1})d\mathscr{H}^{1}_{g_{0}}\bigg{\\}},$

if we choose $\Gamma$ to be oriented as $\partial\Omega_{1}$, and

$\displaystyle\theta_{1}=\psi\circ f^{-1}_{1}.$

Therefore, using subscripts with evident notations, we deduce by Theorem 2.1
with $g=g_{0}$ that

$\displaystyle-\pi\,I^{L}(\Gamma)=-12\pi\,\log\frac{\det_{\zeta}(-\Delta_{S^{2}_{-},g_{0}})\det_{\zeta}(-\Delta_{S^{2}_{+},g_{0}})}{\det_{\zeta}(-\Delta_{\Omega_{1},g_{0}})\det_{\zeta}(-\Delta_{\Omega_{2},g_{0}})}=-12\pi\,\log\frac{\det_{\zeta}(-\Delta_{\Omega_{1},g_{1}})\det_{\zeta}(-\Delta_{\Omega_{2},g_{2}})}{\det_{\zeta}(-\Delta_{\Omega_{1},g_{0}})\det_{\zeta}(-\Delta_{\Omega_{2},g_{0}})}$
$\displaystyle=\int_{\Omega_{1}}|d(-\mu_{1}+\theta_{1})|_{g_{0}}^{2}d\mathrm{vol}_{g_{0}}+2\int_{\Omega_{1}}K_{g_{0}}(-\mu_{1}+\theta_{1})d\mathrm{vol}_{g_{0}}+2\int_{\partial\Omega_{1}}k_{g_{0}}(-\mu_{1}+\theta_{1})d\mathscr{H}^{1}_{g_{0}}$
$\displaystyle+3\int_{\partial\Omega_{1}}\partial_{\nu}(-\mu_{1}+\theta_{1})d\mathscr{H}^{1}_{g_{0}}$
$\displaystyle+\int_{\Omega_{2}}|d(-\mu_{2}+\theta_{2})|_{g_{0}}^{2}d\mathrm{vol}_{g_{0}}+2\int_{\Omega_{2}}K_{g_{0}}(-\mu_{2}+\theta_{2})d\mathrm{vol}_{g_{0}}+2\int_{\partial\Omega_{2}}k_{g_{0}}(-\mu_{2}+\theta_{2})d\mathscr{H}^{1}_{g_{0}}$
$\displaystyle+3\int_{\partial\Omega_{2}}\partial_{\nu}(-\mu_{2}+\theta_{2})d\mathscr{H}^{1}_{g_{0}}.$
(2.11)

Notice that provided that $\Gamma$ be given with the same orientation of
$\partial\Omega_{1}$, we have

$\displaystyle
2\int_{\partial\Omega_{1}}k_{g_{0}}(-\mu_{1}+\theta_{1})d\mathscr{H}^{1}_{g_{0}}+2\int_{\partial\Omega_{2}}k_{g_{0}}(-\mu_{2}+\theta_{2})d\mathscr{H}^{1}_{g_{0}}=2\int_{\Gamma}k_{g_{0}}(-\mu_{1}+\theta_{1}+\mu_{2}-\theta_{2})\mathscr{H}^{1}_{g_{0}}.$
(2.12)

Since $K_{g_{0}}=1=-\Delta_{g_{0}}\mu_{1}$ on $\Omega_{1}$, we have

$\displaystyle\int_{\Omega_{1}}K_{g_{0}}(-\mu_{1})d\mathrm{vol}_{g_{0}}=\int_{\Omega_{1}}\mu_{1}\,\Delta_{g_{0}}\mu_{1}d\mathrm{vol}_{g_{0}}$
$\displaystyle=-\int_{\Omega_{1}}|d\mu_{1}|_{g_{0}}^{2}d\mathrm{vol}_{g_{0}}+\int_{\Gamma}\mu_{1}\partial_{\nu}\mu_{1}d\mathscr{H}^{1}_{g_{0}}$
$\displaystyle=-\int_{\Omega_{1}}|d\mu_{1}|_{g_{0}}^{2}d\mathrm{vol}_{g_{0}}+\int_{\Gamma}\left(k_{g_{0}}-\partial_{\nu}G_{\Omega_{1}}\right)\mu_{1}d\mathscr{H}^{1}_{g_{0}}$
(2.13)

by (2.4). Therefore, we deduce that

$\displaystyle
2\int_{\Omega_{1}}K_{g_{0}}(-\mu_{1}+\theta_{1})d\mathrm{vol}_{g_{0}}+2\int_{\partial\Omega_{1}}k_{g_{0}}(-\mu_{1}+\theta_{1})d\mathscr{H}^{1}_{g_{0}}$
$\displaystyle=-2\int_{\Omega_{1}}|d\mu_{1}|_{g_{0}}^{2}d\mathrm{vol}_{g_{0}}-2\int_{\Gamma}\partial_{\nu}G_{\Omega_{1}}\mu_{1}d\mathscr{H}^{1}_{g_{0}}+2\int_{\Omega_{1}}\theta_{1}d\mathrm{vol}_{g_{0}}+2\int_{\Gamma}k_{g_{0}}\theta_{1}d\mathscr{H}^{1}_{g_{0}}.$
(2.14)

Now, we have

$\displaystyle\int_{\Omega_{1}}|d(-\mu_{1}+\theta_{1})|_{g_{0}}^{2}d\mathrm{vol}_{g_{0}}=\int_{\Omega_{1}}|d(-\mu_{1}+\psi\circ
f_{1}^{-1})|^{2}_{g_{0}}d\mathrm{vol}_{g_{0}}$
$\displaystyle=\int_{\Omega_{1}}|d\mu_{1}|_{g_{0}}^{2}d\mathrm{vol}_{g_{0}}-2\int_{\Omega_{1}}\langle
d\mu_{1},d(\psi\circ
f_{1}^{-1})\rangle_{g_{0}}d\mathrm{vol}_{g_{0}}+\int_{\Omega_{1}}|d(\psi\circ
f_{1}^{-1})|_{g_{0}}^{2}d\mathrm{vol}_{g_{0}}.$ (2.15)

Since $-\Delta_{g_{0}}\mu_{1}=1$, we deduce that

$\displaystyle-2\int_{\Omega_{1}}\langle d\mu_{1},d(\psi\circ
f^{-1})\rangle_{g_{0}}d\mathrm{vol}_{g_{0}}$
$\displaystyle=-2\int_{\Omega_{1}}\langle
d\mu_{1},d\theta_{1}\rangle_{g_{0}}d\mathrm{vol}_{g_{0}}=2\int_{\Omega_{1}}\theta_{1}\Delta_{g_{0}}\mu_{1}d\mathrm{vol}_{g_{0}}-2\int_{\Gamma}\theta_{1}\partial_{\nu}\mu_{1}d\mathscr{H}^{1}_{g_{0}}$
$\displaystyle=-2\int_{\Omega_{1}}\theta_{1}d\mathrm{vol}_{g_{0}}-2\int_{\Gamma}k_{g_{0}}\theta_{1}d\mathscr{H}^{1}_{g_{0}}+2\int_{\Gamma}\partial_{\nu}G_{\Omega_{1}}\theta_{1}d\mathscr{H}^{1}_{g_{0}}.$
(2.16)

Now, by conformal invariance of the Dirichlet energy, we have

$\displaystyle\int_{\Omega_{1}}|d(\psi\circ
f^{-1})|^{2}_{g_{0}}d\mathrm{vol}_{g_{0}}=\int_{\mathbb{D}}|\nabla\psi|^{2}|dz|^{2}$

Since $\psi(z)=\log(2)-\log(1+|z|^{2})$ and $\psi$ is real, we have

$\displaystyle\int_{\mathbb{D}}|\nabla\psi|^{2}|dz|^{2}$
$\displaystyle=4\int_{\mathbb{D}}|\partial_{z}\psi|^{2}|dz|^{2}=4\int_{\mathbb{D}}\frac{|z|^{2}|dz|^{2}}{(1+|z|^{2})^{2}}=4\pi\log(2)-2\pi.$
(2.17)

Therefore, we get (2), (2), (2) and (2.17)

$\displaystyle\int_{\Omega_{1}}|d(-\mu_{1}+\theta_{1})|_{g_{0}}^{2}d\mathrm{vol}_{g_{0}}+2\int_{\Omega_{1}}K_{g_{0}}(-\mu_{1}+\theta_{1})d\mathrm{vol}_{g_{0}}+2\int_{\partial\Omega_{1}}k_{g_{0}}(-\mu_{1}+\theta_{1})d\mathscr{H}^{1}_{g_{0}}$
$\displaystyle=\int_{\Omega_{1}}|d\mu_{1}|_{g_{0}}^{2}d\mathrm{vol}_{g_{0}}-2\int_{\Omega_{1}}\langle
d\mu_{1},d\theta_{1}\rangle_{g_{0}}d\mathrm{vol}_{g_{0}}+\int_{\Omega_{1}}|d\theta_{1}|_{g_{0}}^{2}d\mathrm{vol}_{g_{0}}$
$\displaystyle-2\int_{\Omega_{1}}\theta_{1}d\mathrm{vol}_{g_{0}}-2\int_{\Gamma}k_{g_{0}}\theta_{1}d\mathscr{H}^{1}_{g_{0}}+2\int_{\Gamma}\partial_{\nu}G_{\Omega_{1}}\theta_{1}d\mathscr{H}^{1}_{g_{0}}$
$\displaystyle=\int_{\Omega_{1}}|d\mu_{1}|_{g_{0}}^{2}d\mathrm{vol}_{g_{0}}-\cancel{2\int_{\Omega_{1}}\theta_{1}d\mathrm{vol}_{g_{0}}}-\cancel{2\int_{\Gamma}k_{g_{0}}\theta_{1}d\mathscr{H}^{1}_{g_{0}}}+2\int_{\Gamma}\partial_{\nu}G_{\Omega_{1}}\theta_{1}d\mathscr{H}^{1}_{g_{0}}+\int_{\Omega_{1}}|d\theta_{1}|_{g_{0}}^{2}d\mathrm{vol}_{g_{0}}$
$\displaystyle-2\int_{\Omega_{1}}|d\mu_{1}|_{g_{0}}^{2}d\mathrm{vol}_{g_{0}}-2\int_{\Gamma}\partial_{\nu}G_{\Omega_{1}}\mu_{1}d\mathscr{H}^{1}_{g_{0}}+\cancel{2\int_{\Omega_{1}}\theta_{1}d\mathrm{vol}_{g_{0}}}+\cancel{2\int_{\Gamma}k_{g_{0}}\theta_{1}d\mathscr{H}^{1}_{g_{0}}}$
$\displaystyle=-\int_{\Omega_{1}}|d\mu_{1}|_{g_{0}}^{2}d\mathrm{vol}_{g_{0}}+\int_{\Omega_{1}}|d\theta_{1}|_{g_{0}}^{2}d\mathrm{vol}_{g_{0}}+2\int_{\Gamma}\partial_{\nu}G_{\Omega_{1}}(-\mu_{1}+\theta_{1})d\mathscr{H}^{1}_{g_{0}}$
$\displaystyle=-\int_{\Omega_{1}}|d\mu_{1}|_{g_{0}}^{2}d\mathrm{vol}_{g_{0}}+2\int_{\Gamma}\partial_{\nu}G_{\Omega_{1}}(-\mu_{1}+\theta_{1})d\mathscr{H}^{1}_{g_{0}}+4\pi\log(2)-2\pi.$
(2.18)

Now, since $\theta_{1}=\psi\circ f^{-1}$, and $\psi(z)=0$ for all $z\in
S^{1}$, we have $\theta_{1}=0$ on $\Gamma$. Therefore, we have

$\displaystyle
2\int_{\Gamma}\partial_{\nu}G_{\Omega_{1}}(-\mu_{1}+\theta_{1})d\mathscr{H}^{1}_{g_{0}}$
$\displaystyle=-2\int_{\Gamma}\partial_{\nu}G_{\Omega_{1}}\mu_{1}d\mathscr{H}^{1}_{g_{0}}.$
(2.19)

Now, since $-\Delta_{g_{0}}\mu_{1}=1$ and $K_{g_{0}}=1$, we have

$\displaystyle\int_{\Omega_{1}}G_{\Omega_{1}}K_{g_{0}}d\mathrm{vol}_{g_{0}}$
$\displaystyle=-\int_{\Omega_{1}}G_{\Omega_{1}}\Delta_{g_{0}}\mu_{1}d\mathrm{vol}_{g_{0}}$
$\displaystyle=-\int_{\Omega_{1}}\mu_{1}\Delta_{g_{0}}G_{\Omega_{1}}d\mathrm{vol}_{g_{0}}-\int_{\Omega_{1}}\left(G_{\Omega_{1}}\partial_{\nu}\mu_{1}-\mu_{1}\partial_{\nu}G_{\Omega_{1}}\right)d\mathscr{H}^{1}_{g_{0}}$
$\displaystyle=-2\pi\mu_{1}(p_{1})+\int_{\partial\Omega_{1}}\mu_{1}\partial_{\nu}G_{\Omega_{1}}d\mathscr{H}^{1}_{g_{0}},$
(2.20)

where we used the Dirichlet condition $G_{\Omega_{1}}=0$ on
$\partial\Omega_{1}=\Gamma$. Therefore, (2.19) and (2) imply that

$\displaystyle
2\int_{\Gamma}\partial_{\nu}G_{\Omega_{1}}(-\mu_{1}+\theta_{1})d\mathscr{H}^{1}_{g_{0}}=-2\int_{\Omega_{1}}G_{\Omega_{1}}K_{g_{0}}d\mathrm{vol}_{g_{0}}-4\pi\mu_{1}(p_{1}).$
(2.21)

Gathering (2) and (2.21) yields

$\displaystyle\int_{\Omega_{1}}|d(-\mu_{1}+\theta_{1})|_{g_{0}}^{2}d\mathrm{vol}_{g_{0}}+2\int_{\Omega_{1}}K_{g_{0}}(-\mu_{1}+\theta_{1})d\mathrm{vol}_{g_{0}}+2\int_{\partial\Omega_{1}}k_{g_{0}}(-\mu_{1}+\theta_{1})d\mathscr{H}^{1}_{g_{0}}$
$\displaystyle=-\int_{\Omega_{1}}|d\mu_{1}|_{g_{0}}^{2}d\mathrm{vol}_{g_{0}}-2\int_{\Omega_{1}}G_{\Omega_{1}}K_{g_{0}}d\mathrm{vol}_{g_{0}}-4\pi\mu_{1}(p_{1})+4\pi\log(2)-2\pi.$
(2.22)

We also have

$\displaystyle\int_{\partial\Omega_{1}}\partial_{\nu}\left(-\mu_{1}+\theta_{1}\right)d\mathscr{H}^{1}_{g_{0}}+\int_{\partial\Omega_{2}}\partial_{\nu}\left(-\mu_{2}+\theta_{2}\right)d\mathscr{H}^{1}_{g_{0}}=0.$
(2.23)

Indeed, we have by the boundary conditions (2.10)

$\displaystyle\int_{\partial\Omega_{1}}\partial_{\nu}\mu_{1}d\mathscr{H}^{1}_{g_{0}}$
$\displaystyle=\int_{\Gamma}k_{g_{0}}d\mathscr{H}^{1}_{g_{0}}-\int_{\partial\Omega}\partial_{\nu}G_{\Omega_{j}}d\mathscr{H}^{1}_{g_{0}}=\int_{\Gamma}k_{g_{0}}d\mathscr{H}^{1}_{g_{0}}-\int_{\Omega_{1}}\Delta_{g_{0}}G_{\Omega_{1}}d\mathrm{vol}_{g_{0}}$
$\displaystyle=\int_{\Gamma}k_{g_{0}}d\mathrm{vol}_{g_{0}}-2\pi$
$\displaystyle\int_{\partial\Omega_{2}}\partial_{\nu}\mu_{2}d\mathscr{H}^{1}_{g_{0}}$
$\displaystyle=-\int_{\Gamma}k_{g_{0}}d\mathrm{vol}_{g_{0}}-2\pi.$ (2.24)

We also have by the conformal invariance of the Dirichlet energy

$\displaystyle\int_{\partial\Omega_{1}}\partial_{\nu}\theta_{1}d\mathscr{H}^{1}_{g_{0}}=\int_{\Omega_{1}}\Delta_{g_{0}}\theta_{1}d\mathrm{vol}_{g_{0}}=\int_{\mathbb{D}}\Delta\psi|dz|^{2}=\int_{S^{1}}\partial_{\nu}\psi\,d\mathscr{H}^{1}=-2\pi.$
(2.25)

Therefore, we finally get by (2) and (2.25)

$\displaystyle\int_{\partial\Omega_{1}}\partial_{\nu}\left(-\mu_{1}+\theta_{1}\right)d\mathscr{H}^{1}_{g_{0}}+\int_{\partial\Omega_{2}}\partial_{\nu}\left(-\mu_{2}+\theta_{2}\right)d\mathscr{H}^{1}_{g_{0}}$
$\displaystyle=-\left(\int_{\Gamma}k_{g_{0}}d\mathrm{vol}_{g_{0}}-2\pi\right)-2\pi-\left(-\int_{\Gamma}k_{g_{0}}d\mathrm{vol}_{g_{0}}-2\pi\right)-2\pi=0$

which proves (2.23).

Finally, we deduce by (2), (2) and (2.23) that

$\displaystyle-\pi\,I^{L}(\Gamma)=-\int_{\Omega_{1}}|d\mu_{1}|_{g_{0}}^{2}d\mathrm{vol}_{g_{0}}-2\int_{\Omega_{1}}G_{\Omega_{1}}K_{g_{0}}d\mathrm{vol}_{g_{0}}-4\pi\mu_{1}(p_{1})+4\pi\log(2)-2\pi$
$\displaystyle-\int_{\Omega_{2}}|d\mu_{2}|_{g_{0}}^{2}d\mathrm{vol}_{g_{0}}-2\int_{\Omega_{1}}G_{\Omega_{2}}K_{g_{0}}d\mathrm{vol}_{g_{0}}-4\pi\mu_{2}(p_{2})+4\pi\log(2)-2\pi$
$\displaystyle=-\int_{\Omega_{1}}|d\mu_{1}|_{g_{0}}^{2}d\mathrm{vol}_{g_{0}}-\int_{\Omega_{2}}|d\mu_{2}|_{g_{0}}^{2}d\mathrm{vol}_{g_{0}}-2\int_{\Omega_{1}}G_{\Omega_{1}}K_{g_{0}}d\mathrm{vol}_{g_{0}}-2\int_{\Omega_{1}}G_{\Omega_{2}}K_{g_{0}}d\mathrm{vol}_{g_{0}}$
$\displaystyle-4\pi\mu_{1}(p_{1})-4\pi\mu_{2}(p_{2})+8\pi\log(2)-4\pi.$ (2.26)

Recalling the identity (2.7), we finally deduce that

$\displaystyle\pi\,I^{L}(\Gamma)$
$\displaystyle=\int_{\Omega_{1}}|d\mu_{1}|_{g_{0}}^{2}d\mathrm{vol}_{g_{0}}+\int_{\Omega_{2}}|d\mu_{2}|_{g_{0}}^{2}d\mathrm{vol}_{g_{0}}+2\int_{\Omega_{1}}G_{\Omega_{1}}K_{g_{0}}d\mathrm{vol}_{g_{0}}+2\int_{\Omega_{1}}G_{\Omega_{2}}K_{g_{0}}d\mathrm{vol}_{g_{0}}+4\pi$
$\displaystyle+4\pi\log|\nabla f_{1}(0)|+4\pi\log|\nabla
f_{2}(0)|-12\pi\log(2).$ (2.27)

Now we introduce the functional

$\displaystyle\mathscr{E}(\Gamma)$
$\displaystyle=\int_{\Omega_{1}}|d\mu_{1}|^{2}_{g_{0}}d\mathrm{vol}_{g_{0}}+\int_{\Omega_{2}}|d\mu_{2}|^{2}_{g_{0}}d\mathrm{vol}_{g_{0}}+2\int_{\Omega_{1}}G_{\Omega_{1}}K_{g_{0}}d\mathrm{vol}_{g_{0}}+2\int_{\Omega_{2}}G_{\Omega_{2}}K_{g_{0}}d\mathrm{vol}_{g_{0}}+4\pi.$
(2.28)

We deduce that

$\displaystyle I^{L}(\Gamma)=\frac{1}{\pi}\mathscr{E}(\Gamma)+4\log|\nabla
f_{1}(0)|+4\log|\nabla f_{2}(0)|-12\log(2).$ (2.29)

This concludes the proof of the theorem. ∎

######  Remark 2.8.

We check that equality (2.29) holds for the equator $S^{1}$. Using the
definition (1.3) with the conformal maps $f,g$ being the identity maps, we see
that $I^{L}(S^{1})$ vanishes.

Let us first check that

$\displaystyle\mathscr{E}(S^{1})=0$

with the marked points $p_{1}=S=(0,0,-1)$ and $p_{2}=N=(0,0,1)$. This identity
justifies the term $4\pi$ in the definition of $\mathscr{E}$ as we remarked
earlier.

For this, since $K_{g_{0}}=1$ on $\Omega_{1}=S_{-}^{2}$, after making a
stereographic projection $\pi:S^{2}\setminus\\{N\\}\to\mathbb{C}$ sending $S$
to $0$, we find

$\displaystyle\int_{S^{2}_{-}}G_{S^{2}_{-}}K_{g_{0}}d\mathrm{vol}_{g_{0}}$
$\displaystyle=\int_{\mathbb{D}}G_{\mathbb{D}}(z)\frac{4|dz|^{2}}{(1+|z|^{2})^{2}}=\int_{\mathbb{D}}\frac{4\log|z|}{(1+|z|^{2})^{2}}|dz|^{2}=8\pi\int_{0}^{1}\frac{r\log
r}{(1+r^{2})^{2}}dr$
$\displaystyle=8\pi\left[-\frac{1}{2(1+r^{2})}\log(r)+\frac{1}{2}\log(r)-\frac{1}{4}\log(1+r^{2})\right]_{0}^{1}=-2\pi\log(2).$
(2.30)

We take $f_{1}=\pi^{-1}|_{\mathbb{D}}$ which is consistent with
$f_{1}(0)=p_{1}=S$. By (2.4) we have $\partial_{\nu}\mu=-1$ on $S^{1}$ and
$-\Delta_{g_{0}}\mu=1$ in $S^{2}_{-}$ which translates to

$\displaystyle-\Delta\mu(z)=\frac{4}{(1+|z|^{2})^{2}}\qquad\text{in}\;\,\mathbb{D}.$

We deduce by a direct verification that $\mu(z)=-\log(1+|z|^{2})$. (This is
easy to guess since by (2.6), $\mu$ can be computed from the conformal factor
of $f_{1}$.) Therefore, we have by the conformal invariance of the Dirichlet
energy

$\displaystyle\int_{S^{2}_{-}}\left|\omega-\ast
dG_{S^{2}_{-}}\right|^{2}_{g_{0}}d\mathrm{vol}_{g_{0}}$
$\displaystyle=\int_{S^{2}_{-}}|d\mu|_{g_{0}}^{2}d\mathrm{vol}_{g_{0}}=\int_{\mathbb{D}}|\nabla\mu(x)|^{2}dx=\int_{\mathbb{D}}\left|\frac{2x}{1+|x|^{2}}\right|^{2}dx=8\pi\int_{0}^{1}\frac{r^{3}}{(1+r^{2})^{2}}dr$
$\displaystyle=8\pi\int_{0}^{1}\left(\frac{r}{1+r^{2}}-\frac{r}{(1+r^{2})^{2}}\right)dr=8\pi\left[\frac{1}{2}\log(1+r^{2})+\frac{1}{2}\frac{1}{1+r^{2}}\right]_{0}^{1}$
$\displaystyle=8\pi\left(\frac{1}{2}\log(2)-\frac{1}{4}\right)=4\pi\log(2)-2\pi.$
(2.31)

Finally, by (2.8) and (2.8), we have

$\displaystyle\int_{S^{2}_{-}}|\omega-\ast\,dG_{S^{2}_{-}}|_{g_{0}}^{2}d\mathrm{vol}_{g_{0}}+2\int_{S^{2}_{-}}G_{S^{2}_{-}}\,K_{g_{0}}d\mathrm{vol}_{g_{0}}+2\pi$
$\displaystyle=\left(4\pi\log(2)-2\pi\right)-4\pi\log(2)+2\pi=0.$

Applying the same computation to $\Omega_{2}=S^{2}_{+}$ with
$f_{2}=-f_{1}:\mathbb{D}\to S^{2}_{+}$ which is consistent with the choice
$p_{2}=N=f_{2}(0)$, we obtain the claimed identity $\mathscr{E}(S^{1})=0$.

Now we show that

$\displaystyle 4\log|\nabla f_{1}(0)|+4\log|\nabla f_{2}(0)|-12\log(2)=0.$
(2.32)

Since the inverse stereographic projection is given by

$\displaystyle
f_{1}(z)=\pi^{-1}(z)=\left(\frac{2\,\mathrm{Re}\,(z)}{1+|z|^{2}},\frac{2\,\mathrm{Im}\,(z)}{1+|z|^{2}},\frac{-1+|z|^{2}}{1+|z|^{2}}\right),$

we compute directly that

$\displaystyle|\nabla f_{1}(0)|=|\nabla f_{2}(0)|=2\sqrt{2}$

which concludes the proof of (2.32) and shows the identity (2.29) for the
circle $S^{1}$.

##  3 Construction of Harmonic Moving Frames for Weil-Petersson Curves

In the previous section, we showed that in the case of smooth curves, the
Loewner energy was equal to a renormalised Dirichlet energy of a specific
harmonic moving frame. In this section, we will directly construct harmonic
moving frames satisfying appropriate boundary conditions for arbitrary Weil-
Petersson quasicircles. In the next section, we will show that Theorem 2.6
holds for non-smooth curves.

Before stating the main theorem of this section, recall an easy lemma on
harmonic vector fields.

######  Lemma 3.1.

Let $\Sigma\subset\mathbb{R}^{3}$ be a smooth surface,
$\vec{n}:\Sigma\rightarrow S^{2}$ its unit normal, and
$g=\iota^{\ast}g_{\mathbb{R}^{3}}$ be the induced metric. Assume that
$\vec{u}:\Sigma\rightarrow S^{2}$ is a smooth critical point of the Dirichlet
energy amongst $S^{2}$-valued maps such that
$\langle{\vec{u}},\vec{n}\rangle=0$. Then $\vec{u}$ satisfies the following
Euler-Lagrange equation:

$\displaystyle-\Delta_{g}\vec{u}=|d\vec{u}|_{g}^{2}\vec{u}+\left(\langle
d\vec{u},d\vec{n}\rangle_{g}+\langle\vec{u},\Delta_{g}\vec{n}\rangle\right)\vec{n}.$
(3.1)

###### Proof.

We proceed as in [19] in Lemme ($1.4.10$), taking variations $\vec{X}$ that
also satisfy $\langle\vec{X},\vec{n}\rangle=0$. ∎

The following result is the same as Theorem 2.6, but for a general Weil-
Petersson quasicircle.

######  Theorem 3.2.

Let $\Gamma\subset S^{2}$ a Weil-Petersson quasicircle and let
$\Omega_{1},\Omega_{2}$ be the two open connected components of
$S^{2}\setminus\Gamma$. For $j=1,2$ and for all $p_{j}\in\Omega_{j}$, there
exists a harmonic moving frame
$(\vec{u}_{j},\vec{v}_{j}):\Omega_{j}\setminus\left\\{p_{j}\right\\}\rightarrow
U\Omega_{j}\times U\Omega_{j}$ such that the Cartan form
$\omega_{j}=\langle\vec{u}_{j},d\vec{v}_{j}\rangle$ admits the decomposition

$\displaystyle\omega_{j}=\ast\,d\left(G_{\Omega_{j}}+\mu_{j}\right),$ (3.2)

where
$G_{\Omega_{j}}=G_{\Omega_{j},p_{j}}:\Omega_{j}\setminus\left\\{p_{j}\right\\}\rightarrow\mathbb{R}$
is the Green’s function of the Laplacian $\Delta_{g_{0}}$ on $\Omega_{j}$ with
Dirichlet boundary condition and singularity $p_{j}\in\Omega_{j}$, and
$\mu_{j}\in C^{\infty}(\Omega_{j})$ satisfies

$\displaystyle\left\\{\begin{aligned}
-\Delta_{g_{0}}\mu_{j}&=1\qquad&&\text{in}\;\,\Omega_{j}\\\
\partial_{\nu}\mu_{j}&=k_{g_{0}}-\partial_{\nu}G_{\Omega_{j}}\qquad&&\text{on}\;\,\partial\Omega_{j},\end{aligned}\right.$
(3.3)

where $k_{g_{0}}$ is the geodesic curvature on $\Gamma=\partial\Omega_{j}$.

######  Remark 3.3.

The Neumann condition for $\mu_{j}$ ($1\leq j\leq 2$) is understood in the
sense of distributions, since the geodesic curvature is only in
$H^{-1/2}(\Gamma)$ in general (see the appendix for more details).

###### Proof.

Rather than using the moving frame that comes from a Ginzburg-Landau type
minimisation as in [23]—that would have had to be carried in the geometric
setting of domains of $S^{2}$—we directly use the uniformisation theorem and
the geometric formula of [41] (that does not require any regularity on the
curve $\Gamma$) to construct the relevant moving frame. We now construct the
moving frame on $\Omega_{1}$. The construction for $\Omega_{2}$ is similar.

Step 1. Definition of $(\vec{u}_{1},\vec{v}_{1})$ and $\mu_{1}$.

Let $\pi:S^{2}\setminus\left\\{N\right\\}\rightarrow\mathbb{C}$ be the
standard stereographic projection and assume without loss of generality that
$N\in\Omega_{2}$. Let $\Omega=\pi(\Omega_{1})\subset\mathbb{C}$ be the image
domain and $\gamma=\pi(\Gamma)\subset\mathbb{C}$ be the image curve. Thanks to
the Uniformisation Theorem, there exists a univalent holomorphic map
$f:\mathbb{D}\rightarrow\Omega$ such that $f(0)=\pi(p_{1})$.

Now, let $f_{1}=\pi^{-1}\circ f:\mathbb{D}\rightarrow\Omega_{1}$. Notice that
$f_{1}(0)=p_{1}$. Explicitly, we have

$\displaystyle
f_{1}(z)=\pi^{-1}(f(z))=\left(\frac{2\,\mathrm{Re}\,(f(z))}{1+|f(z)|^{2}},\frac{2\,\mathrm{Im}\,(f(z))}{1+|f(z)|^{2}},\frac{-1+|f(z)|^{2}}{1+|f(z)|^{2}}\right).$

A direct computation show that

$\displaystyle\partial_{z}f_{1}=f^{\prime}\left(\frac{(1-\overline{f}^{2})}{(1+|f|^{2})^{2}},\frac{-i(1+\overline{f}^{2})}{(1+|f|^{2})^{2}},\frac{2\overline{f}}{(1+|f|^{2})^{2}}\right).$

Now, by analogy with the construction in Section 2 (see also [23], Proposition
$5.1$), define $\mu_{1}:\Omega_{1}\rightarrow\mathbb{R}$ and
$\vec{u}_{1}:\Omega_{1}\rightarrow US^{2}$ and
$\vec{v}_{1}:\Omega_{1}\rightarrow US^{2}$ by

$\displaystyle\left\\{\begin{aligned} \partial_{r}f_{1}&=e^{\mu_{1}\circ
f_{1}}\vec{v}_{1}\circ f_{1}\\\
\frac{1}{r}\partial_{\theta}f_{1}&=e^{\mu_{1}\circ f_{1}}\vec{u}_{1}\circ
f_{1}.\end{aligned}\right.$

Then, we have from direct computations

$\displaystyle e^{2\mu_{1}\circ
f_{1}}=|\partial_{r}f_{1}|^{2}=\frac{1}{r^{2}}|\partial_{\theta}f_{1}|^{2}=2|\partial_{z}f_{1}|^{2}=\frac{4|f^{\prime}(z)|^{2}}{(1+|f(z)|^{2})^{2}}.$

Therefore, we deduce if $\mu=\mu_{1}\circ f_{1}$ that

$\displaystyle\mu(z)=\log|f^{\prime}(z)|-\log\left(1+|f(z)|^{2}\right)+\log(2).$
(3.4)

Since $\partial_{z}=\dfrac{1}{2}(\partial_{x}-i\,\partial_{y})$, we have

$\displaystyle\left\\{\begin{aligned}
\partial_{r}f_{1}&=\cos(\theta)\partial_{x}f_{1}+\sin(\theta)\partial_{y}f_{1}=\mathrm{Re}\,\left(\frac{z}{|z|}\right)\mathrm{Re}\,(\partial_{z}f_{1})-\mathrm{Im}\,\left(\frac{z}{|z|}\right)\mathrm{Im}\,\left(\partial_{z}f_{1}\right)\\\
\frac{1}{r}\partial_{\theta}f_{1}&=-\sin(\theta)\partial_{x}f_{1}+\cos(\theta)\partial_{y}f_{1}=-\mathrm{Im}\,\left(\frac{z}{|z|}\right)\mathrm{Re}\,(\partial_{z}f_{1})-\mathrm{Re}\,\left(\frac{z}{|z|}\right)\mathrm{Im}\,\left(\partial_{z}f_{1}\right).\end{aligned}\right.$

By the elementary identities for all $a,b\in\mathbb{C}$

$\displaystyle\left\\{\begin{aligned}
\mathrm{Re}\,(a)\mathrm{Re}\,(b)+\mathrm{Im}\,(a)\mathrm{Im}\,(b)&=\mathrm{Re}\,(ab)\\\
\mathrm{Re}\,(a)\mathrm{Im}\,(b)+\mathrm{Im}\,(a)\mathrm{Re}\,(b)&=\mathrm{Im}\,(ab),\end{aligned}\right.$

we deduce that

$\displaystyle\left\\{\begin{aligned}
\partial_{r}f_{1}&=\mathrm{Re}\,\left(\frac{z}{|z|}\partial_{z}f_{1}\right)=\mathrm{Re}\,\left(\frac{\overline{z}}{|z|}\overline{f^{\prime}(z)}\left(\frac{(1-f(z)^{2})}{(1+|f(z)|^{2})^{2}},\frac{i(1+f(z)^{2})}{(1+|f(z)|^{2})^{2}},\frac{2f(z)}{(1+|f(z)|^{2})^{2}}\right)\right)\\\
\frac{1}{r}\partial_{\theta}f_{1}&=-\mathrm{Im}\,\left(\frac{z}{|z|}\partial_{z}f_{1}\right)=\mathrm{Im}\,\left(\frac{\overline{z}}{|z|}\overline{f^{\prime}(z)}\left(\frac{(1-f(z)^{2})}{(1+|f(z)|^{2})^{2}},\frac{i(1+f(z)^{2})}{(1+|f(z)|^{2})^{2}},\frac{2f(z)}{(1+|f(z)|^{2})^{2}}\right)\right).\end{aligned}\right.$

More generally, if $\varphi:\mathbb{C}\rightarrow\mathbb{C}$ is a smooth
complex function, we have

$\displaystyle\partial_{\theta}\varphi=-\mathrm{Im}\,\left(z\right)\left(\partial_{z}+\partial_{\overline{z}}\right)\varphi+\mathrm{Re}\,\left({z}\right)i\left(\partial_{z}-\partial_{\overline{z}}\right)\varphi=i\left(z\,\partial_{z}\varphi-\overline{z}\,\partial_{\overline{z}}\varphi\right).$
(3.5)

Since

$\displaystyle|\partial_{z}f_{1}|=\frac{1}{r}|\partial_{\theta}f_{1}|=\frac{2|f^{\prime}(z)|}{1+|f(z)|^{2}},$

we deduce that

$\displaystyle\left\\{\begin{aligned} \vec{v}_{1}\circ
f_{1}&=\mathrm{Re}\,\left(\overline{\frac{zf^{\prime}(z)}{|zf^{\prime}(z)|}}\left(\frac{(1-f(z)^{2})}{(1+|f(z)|^{2})},\frac{i(1+f(z)^{2})}{(1+|f(z)|^{2})},\frac{2f(z)}{(1+|f(z)|^{2})}\right)\right)\\\
\vec{u}_{1}\circ
f_{1}&=\mathrm{Im}\,\left(\overline{\frac{zf^{\prime}(z)}{|zf^{\prime}(z)|}}\left(\frac{(1-f(z)^{2})}{(1+|f(z)|^{2})},\frac{i(1+f(z)^{2})}{(1+|f(z)|^{2})},\frac{2f(z)}{(1+|f(z)|^{2})}\right)\right).\end{aligned}\right.$
(3.6)

Notice that

$\displaystyle F(z)=\left((1-f(z)^{2}),i(1+f(z)^{2}),2f(z)\right)$

is a holomorphic null vector, _i.e._ $\langle F(z),F(z)\rangle=0$, so we see
directly since $|\vec{u}_{1}|=|\vec{v}_{1}|=1$ that

$\displaystyle\langle\vec{u}_{1},\vec{v}_{1}\rangle=0.$

Step 2. Verification of the system (3.3).

Part 1. Equation on $\Omega_{1}$ for $\mu_{1}$.

Since $\mu=\mu_{1}\circ f_{1}$, the equation $-\Delta_{g_{0}}\mu_{1}=1$ is
equivalent to

$\displaystyle-\Delta\mu=e^{2\mu}$ (3.7)

Thanks to the explicit expression in (3.4), and by harmonicity of
$\log|f^{\prime}|$, we have

$\displaystyle\Delta\mu=4\,\partial_{\overline{z}}\left(\frac{f^{\prime}(z)\overline{f^{\prime}(z)}}{1+|f(z)|^{2}}\right)=4\left(\frac{|f^{\prime}(z)|^{2}}{1+|f(z)|^{2}}-\frac{|f^{\prime}(z)|^{2}|f(z)|^{2}}{(1+|f(z)|^{2})^{2}}\right)=\frac{4|f^{\prime}(z)|^{2}}{(1+|f(z)|^{2})^{2}}=e^{2\mu}.$

Recalling that

$\displaystyle
g_{\widehat{\mathbb{C}}}=\frac{4|dz|^{2}}{(1+|z|^{2})^{2}}=(\pi^{-1})^{\ast}g_{0},$

we deduce that

$\displaystyle\frac{4|f^{\prime}(z)|^{2}|dz|^{2}}{(1+|f(z)|^{2})^{2}}=f^{\ast}g_{\widehat{\mathbb{C}}}=f^{\ast}((\pi^{-1})^{\ast}g_{0})=(\pi^{-1}\circ
f)^{\ast}g_{0}=f_{1}^{\ast}g_{0}.$

Therefore, (3.7) can be rewritten as

$\displaystyle-\Delta_{f_{1}^{\ast}g_{0}}(\mu_{1}\circ f_{1})=1$

or by conformal invariance of the Dirichlet energy

$\displaystyle-\Delta_{g_{0}}\mu_{1}=1.$ (3.8)

Part 2. Boundary conditions.

If ${h}:\mathbb{C}\rightarrow\mathbb{R}$ is a smooth function, we have

$\displaystyle\partial_{\nu}{h}$
$\displaystyle=\frac{x}{\sqrt{x^{2}+y^{2}}}\partial_{x}{h}+\frac{y}{\sqrt{x^{2}+y^{2}}}\partial_{y}{h}=\frac{\mathrm{Re}\,(z)}{|z|}\left(\partial_{z}+\partial_{\overline{z}}\right){h}+\frac{\mathrm{Im}\,(z)}{|z|}i\left(\partial_{z}-\partial_{\overline{z}}\right){h}=2\,\mathrm{Re}\,\left(\frac{z}{|z|}\partial_{z}{h}\right).$

This implies since $\mu(z)=\log|f^{\prime}(z)|-\log(1+|f(z)|^{2})+\log(2)$ by
(3.4) that

$\displaystyle\partial_{z}\mu(z)=\frac{1}{2}\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}-\frac{f^{\prime}(z)}{f(z)}\frac{|f(z)|^{2}}{1+|f(z)|^{2}},$

and

$\displaystyle\partial_{\nu}\mu=\mathrm{Re}\,\left(z\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}-2z\frac{f^{\prime}(z)}{f(z)}\frac{|f(z)|^{2}}{1+|f(z)|^{2}}\right)$

on $\partial\mathbb{D}$ in the distributional sense. We will comment on it in
Remark 4.6. Recall that the geodesic curvature on $\partial\Omega_{1}$ is
given (see [10]) by

$\displaystyle
k_{g_{0}}=\langle\vec{u}_{1},\partial_{\theta}\vec{v}_{1}\rangle.$ (3.9)

From (3.6) it is natural to define

$\displaystyle\varphi(z)=\overline{\frac{zf^{\prime}(z)}{|zf^{\prime}(z)|}}\left(\frac{(1-f(z)^{2})}{(1+|f(z)|^{2})},\frac{i(1+f(z)^{2})}{(1+|f(z)|^{2})},\frac{2f(z)}{(1+|f(z)|^{2})}\right)=\chi(z)\psi(z),$

where

$\displaystyle\left\\{\begin{aligned}
\chi(z)&=\frac{\overline{zf^{\prime}(z)}}{|zf^{\prime}(z)|}=\exp\left(\frac{1}{2}\log\left(\overline{zf^{\prime}(z)}\right)-\frac{1}{2}\log\left(zf^{\prime}(z)\right)\right)\\\
\psi(z)&=\left(\frac{1-f(z)^{2}}{1+|f(z)|^{2}},\frac{i(1+f(z)^{2})}{1+|f(z)|^{2}},\frac{2f(z)}{1+|f(z)|^{2}}\right),\end{aligned}\right.$
(3.10)

so that $\vec{v}_{1}\circ f_{1}=\mathrm{Re}\,(\varphi)$ and $\vec{u}_{1}\circ
f_{1}=\mathrm{Im}\,(\varphi)$. Then, we compute

$\displaystyle\left\\{\begin{aligned}
\partial_{z}\chi&=-\frac{1}{2}\left(\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}+\frac{1}{z}\right)\chi\\\
\partial_{\overline{z}}\chi&=\frac{1}{2}\overline{\left(\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}+\frac{1}{z}\right)}\chi.\end{aligned}\right.$
(3.11)

We also get

$\displaystyle\left\\{\begin{aligned}
\partial_{z}\psi&=-\frac{f^{\prime}(z)\overline{f(z)}}{1+|f(z)|^{2}}\psi+\frac{2f^{\prime}(z)}{1+|f(z)|^{2}}\left(-f(z),i\,f(z),1\right)\\\
\partial_{\overline{z}}\psi&=-\frac{\overline{f^{\prime}(z)}f(z)}{1+|f(z)|^{2}}\psi.\end{aligned}\right.$
(3.12)

Since $\langle\psi,\psi\rangle=0$, we have
$\langle\partial_{z}\psi,\psi\rangle=\langle\partial_{\overline{z}}\psi,\psi\rangle=0$.
In particular, we have

$\displaystyle\langle(-f(z),i\,f(z),1),\psi\rangle=\frac{1}{1+|f(z)|^{2}}\langle(-f(z),i\,f(z),1),(1-f(z)^{2},i(1+f(z)^{2}),2f(z))\rangle=0,$
(3.13)

while

$\displaystyle\left\langle(-f(z),i\,f(z),1),\left(1-\overline{f(z)}^{2},-i\left(1+\overline{f(z)}^{2}\right),2\overline{f(z)}\right)\right\rangle$
$\displaystyle=-f(z)+\overline{f(z)}|f(z)|^{2}+f(z)+\overline{f(z)}|f(z)|^{2}+2{\overline{f(z)}}=2\overline{f(z)}(1+|f(z)|^{2}),$

so that

$\displaystyle\langle(-f(z),i\,f(z),1),{\overline{\psi}}\rangle=2\overline{f(z)}.$
(3.14)

Therefore, we deduce by (3.12), (3.13) and (3.14) that

$\displaystyle\left\\{\begin{aligned}
&\langle\varphi,\varphi\rangle=\langle\partial_{z}\varphi,\varphi\rangle=\langle\partial_{\overline{z}}\varphi,\varphi\rangle=\langle\psi,\psi\rangle=\langle\partial_{z}\psi,\psi\rangle=\langle\partial_{\overline{z}}\psi,\psi\rangle=0\\\
&\langle\partial_{z}\psi,\overline{\psi}\rangle=\frac{2f^{\prime}(z)\overline{f(z)}}{1+|f(z)|^{2}}\\\
&\langle\partial_{\overline{z}}\psi,\overline{\psi}\rangle=-\frac{2\overline{f^{\prime}(z)}f(z)}{1+|f(z)|^{2}}.\end{aligned}\right.$
(3.15)

The identities (3.11) and (3.15) imply that

$\displaystyle
z\,\partial_{z}\varphi-\overline{z}\,\partial_{z}\varphi=-\left(\mathrm{Re}\,\left(z\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}\right)+1\right)\varphi+\chi(z)\left(z\,\partial_{z}\psi-\overline{z}\,\partial_{\overline{z}}\psi\right),$

and since $|\chi|^{2}=1$ and $|\psi|^{2}=2$, we have

$\displaystyle\langle
z\,\partial_{z}\varphi-\overline{z}\,\partial_{\overline{z}}\varphi,\varphi+\overline{\varphi}\rangle=-2\,\left(\mathrm{Re}\,\left(z\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}\right)+1\right)$
$\displaystyle+\chi(z)\left(z\left(\chi(z)\langle\partial_{z}\psi,\psi\rangle+\overline{\chi(z)}\langle\partial_{z}\psi,\overline{\psi}\rangle\right)-\overline{z}\left(\chi(z)\langle\partial_{\overline{z}}\varphi,\varphi\rangle+\overline{\chi(z)}\langle\partial_{\overline{z}}\psi,\overline{\psi}\rangle\right)\right)$
$\displaystyle=-2\,\left(\mathrm{Re}\,\left(z\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}\right)+1\right)+2z\frac{f^{\prime}(z)}{f(z)}\frac{|f(z)|^{2}}{1+|f(z)|^{2}}+2\overline{z}\overline{\frac{f^{\prime}(z)}{f(z)}}\frac{|f(z)|^{2}}{1+|f(z)|^{2}}$
$\displaystyle=-2-2\,\mathrm{Re}\,\left(z\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}-2z\frac{f^{\prime}(z)}{f(z)}\frac{|f(z)|^{2}}{1+|f(z)|^{2}}\right),$

so that

$\displaystyle k_{g_{0}}$
$\displaystyle=\langle\vec{u},\partial_{\theta}\vec{v}\rangle=-\langle\partial_{\theta}\vec{u},\vec{v}\rangle=-\langle\partial_{\theta}\mathrm{Im}\,(\varphi),\mathrm{Re}\,(\varphi)\rangle=-\mathrm{Im}\,\left(\langle\partial_{\theta}\varphi,\mathrm{Re}\,(\varphi)\rangle\right)$
$\displaystyle=-\frac{1}{2}\mathrm{Im}\,\left(\langle
i\left(z\partial_{z}\varphi-\overline{z}\partial_{\overline{z}}\right),\varphi+\overline{\varphi}\rangle\right)=-\frac{1}{2}\mathrm{Re}\,\left(\langle
z\partial_{z}\varphi-\overline{z}\partial_{\overline{z}}\varphi,\varphi+\overline{\varphi}\rangle\right)$
$\displaystyle=\mathrm{Re}\,\left(z\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}-2z\frac{f^{\prime}(z)}{f(z)}\frac{|f(z)|^{2}}{1+|f(z)|^{2}}\right)+1=\partial_{\nu}\mu+1=\partial_{\nu}\mu+\partial_{\nu}G_{\mathbb{D}},$
(3.16)

which concludes the proof of the system (3.3) by the conformal invariance of
the Green’s function (we denoted for simplicity
$G_{\mathbb{D}}=G_{\mathbb{D},0}=\log|\,\cdot\,|$).

Step 3. Verification that $(\vec{u}_{1},\vec{v}_{1})$ is a harmonic moving
frame.

Now, thanks to Lemma 3.1 and (3.1), the maps $\vec{u}_{1}$ and $\vec{v}_{1}$
are unit harmonic moving frames if and only if they satisfy in the
distributional sense (see Theorem 3.5) the system (writing
$\vec{u}=\vec{u}_{1}\circ f_{1}$ and $\vec{v}=\vec{v}_{1}\circ f_{1}$ for
simplicity)

$\displaystyle\left\\{\begin{aligned}
-\Delta\vec{u}&=|\nabla\vec{u}|^{2}\vec{u}+\left(2\langle\nabla\vec{u},\nabla\vec{n}\rangle+\langle\vec{u},\Delta\vec{n}\rangle\right)\vec{n}\\\
-\Delta\vec{v}&=|\nabla\vec{v}|^{2}\vec{v}+\left(2\langle\nabla\vec{v},\nabla\vec{n}\rangle+\langle\vec{v},\Delta\vec{n}\rangle\right)\vec{n}.\end{aligned}\right.$
(3.17)

where $\vec{n}:\mathbb{D}\rightarrow S^{2}$ is the same map as $f_{1}$ but
viewed as the Gauss map associated to the branched minimal immersion of the
disk from $\mathbb{D}$ into $\mathbb{R}^{3}$ with Weierstrass data $(f,dz)$.
It is given by

$\displaystyle\vec{n}(z)=\left(\frac{2\,\mathrm{Re}\,(f(z))}{1+|f(z)|^{2}},\frac{2\,\mathrm{Im}\,(f(z))}{1+|f(z)|^{2}},\frac{-1+|f(z)|^{2}}{1+|f(z)|^{2}}\right).$

By a direct computation, we see that the Gauss map satisfies the following
equations

$\displaystyle|\nabla\vec{n}(z)|^{2}$
$\displaystyle=\frac{8|f^{\prime}(z)|^{2}}{(1+|f(z)|^{2})^{2}}$
$\displaystyle-\Delta\vec{n}$ $\displaystyle=|\nabla\vec{n}|^{2}\vec{n}.$

In particular, the previous equation (3.17) must reduce to

$\displaystyle\left\\{\begin{aligned}
-\Delta\vec{u}&=|\nabla\vec{u}|^{2}\vec{u}+2\langle\nabla\vec{u},\nabla\vec{n}\rangle\vec{n}\\\
-\Delta\vec{v}&=|\nabla\vec{v}|^{2}\vec{v}+2\langle\nabla\vec{v},\nabla\vec{n}\rangle\vec{n}.\end{aligned}\right.$
(3.18)

However, since
$\langle\vec{u},\Delta\vec{n}\rangle=|\nabla\vec{n}|^{2}\langle\vec{u},\vec{n}\rangle=0$,
we deduce that
$-\langle\Delta\vec{u},\vec{n}\rangle=2\langle\nabla\vec{u},\nabla\vec{n}\rangle$,
and since $|\nabla\vec{u}|^{2}=1$, we also get (by taking the Laplacian of
$|\vec{u}|^{2}=1$) that
$-\langle\Delta\vec{u},\vec{u}\rangle=|\nabla\vec{u}|^{2}\vec{u}$. Therefore,
we need only check that

$\displaystyle\langle\Delta\vec{u},\vec{v}\,\rangle=\langle\Delta\vec{v},\vec{u}\,\rangle=0$
(3.19)

to show that $\vec{u}$ and $\vec{v}$ satisfy the equations (3.18). Recall from
(3.10) that $\vec{u}=\mathrm{Re}\,(\varphi)$ and
$\vec{v}=\mathrm{Im}\,(\varphi)$, we deduce that

$\displaystyle\Delta\vec{u}=\mathrm{Re}\,(\Delta\varphi),\qquad\Delta\vec{v}=\mathrm{Im}\,(\Delta\varphi)$

and we have

$\displaystyle\left\\{\begin{aligned}
\langle\mathrm{Re}\,(\Delta\varphi),\mathrm{Im}\,(\varphi)\rangle&=\frac{1}{2}\mathrm{Im}\,\left(\langle\Delta\varphi,\varphi\rangle\right)-\frac{1}{2}\mathrm{Im}\,(\langle\Delta\varphi,\overline{\varphi}\rangle)\\\
\langle\mathrm{Im}\,(\Delta\varphi),\mathrm{Re}\,(\varphi)\rangle&=\frac{1}{2}\mathrm{Im}\,\left(\langle\Delta\varphi,\varphi\rangle\right)+\frac{1}{2}\mathrm{Im}\,(\langle\Delta\varphi,\overline{\varphi}\rangle).\end{aligned}\right.$
(3.20)

Therefore, the equations (3.19) are equivalent to

$\displaystyle\mathrm{Im}\,(\langle\Delta\varphi,\varphi\rangle)=\mathrm{Im}\,(\langle\Delta\varphi,\overline{\varphi}\rangle)=0.$
(3.21)

Using (3.12), (3.13) and (3.15), we get

$\displaystyle\langle\Delta\varphi,\varphi\rangle=-4\langle\partial_{z}\varphi,\partial_{\overline{z}}\varphi\rangle$
(3.22)
$\displaystyle=-4\left\langle-\frac{1}{2}\left(\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}+\frac{1}{z}\right)\varphi-\frac{f^{\prime}(z)\overline{f(z)}}{1+|f(z)|^{2}}\left(-f(z),i\,f(z),1\right),\left(\frac{1}{2}\overline{\left(\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}+\frac{1}{z}\right)}-\frac{\overline{f^{\prime}(z)}f(z)}{1+|f(z)|^{2}}\right)\varphi\right\rangle=0,$

which implies in particular that
$\mathrm{Im}\,\left(\langle\Delta\varphi,\varphi\rangle\right)=0$. Then, we
compute

$\displaystyle\langle\partial_{\overline{z}}\varphi,\overline{\varphi}\rangle=\left\langle\left(\frac{1}{2}\overline{\left(\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}+\frac{1}{z}\right)}-\frac{\overline{f^{\prime}(z)}f(z)}{1+|f(z)|^{2}}\right)\varphi,\overline{\varphi}\right\rangle=\left(\overline{\left(\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}+\frac{1}{z}\right)}-2\frac{\overline{f^{\prime}(z)}f(z)}{1+|f(z)|^{2}}\right).$
(3.23)

Therefore, we have

$\displaystyle\frac{1}{4}\langle\Delta\varphi,\overline{\varphi}\rangle=\partial_{z}\langle\partial_{\overline{z}}\varphi,\overline{\varphi}\rangle-\langle\partial_{\overline{z}}\varphi,\partial_{z}\overline{\varphi}\rangle=\partial_{z}\langle\partial_{\overline{z}}\varphi,\overline{\varphi}\rangle-|\partial_{\overline{z}}\varphi|^{2},$

where we used
$\partial_{z}\overline{\varphi}=\overline{\partial_{\overline{z}}\varphi}$. By
(3.23), we deduce that

$\displaystyle\partial_{z}\langle\partial_{\overline{z}}\varphi,\overline{\varphi}\rangle=-2\frac{|f^{\prime}(z)|^{2}}{1+|f(z)|^{2}}+2\frac{|f^{\prime}(z)|^{2}|f(z)|^{2}}{(1+|f(z)|^{2})^{2}}=-\frac{2|f^{\prime}(z)|^{2}}{(1+|f(z)|^{2})^{2}},$

so that

$\displaystyle\langle\Delta\varphi,\overline{\varphi}\rangle=-\frac{8|f^{\prime}(z)|^{2}}{(1+|f(z)|^{2})^{2}}-4|\partial_{\overline{z}}\varphi|^{2}\in\mathbb{R},$

which implies that

$\displaystyle\mathrm{Im}\,\left(\langle\Delta\varphi,\overline{\varphi}\rangle\right)=0.$
(3.24)

Therefore, we deduce that (3.21) holds, which implies that $\vec{u}$ and
$\vec{v}$ solve the equations (3.18).

Step 4. Proof of the decomposition
$\omega_{1}=\ast\,d(G_{\Omega_{1}}+\mu_{1})$.

Recall that $\vec{u}=\vec{u}_{1}$ and $\vec{v}=\vec{v}_{1}$, and let

$\displaystyle\omega=\langle\vec{u},d\vec{v}\,\rangle=\langle\vec{u},\partial\vec{v}\,\rangle+\langle\vec{u},\overline{\partial}\vec{v}\,\rangle.$

Recall that since $\ast\,dx=dy$ and $\ast\,dy=-dx$, we have

$\displaystyle\ast\,dz$
$\displaystyle=\ast\left(dx+i\,dy\right)=dy-i\,dx=-i(dx+i\,dy)=-i\,dz$
$\displaystyle\ast\,d\overline{z}$ $\displaystyle=i\,d\overline{z}.$

Therefore, $\omega=\ast\,d\left(\mu+G\right)$ (where we write for simplicity
$G=G_{\mathbb{D}}=\log|\,\cdot\,|$) if and only if

$\displaystyle\langle\vec{u},\partial\vec{v}\,\rangle+\langle\vec{u},\overline{\partial}\vec{v}\,\rangle=\ast\left(\partial\left(\mu+G\right)+\overline{\partial}\left(\mu+G\right)\right)=-i\,\partial(\mu+G)+i\,\overline{\partial}(\mu+G),$

which is equivalent to the identity

$\displaystyle\langle\vec{u},\partial\vec{v}\,\rangle=-i\,\partial\left(\mu+G\right).$
(3.25)

We have by (3.11) and (3.12)

$\displaystyle\partial_{z}\vec{v}=\partial_{z}\mathrm{Re}\,(\varphi)=\frac{1}{2}\left(\partial_{z}\varphi+\overline{\partial_{\overline{z}}\varphi}\right)=\frac{1}{2}\left(-\frac{1}{2}\left(\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}+\frac{1}{z}\right)\varphi-\frac{f^{\prime}(z)\overline{f(z)}}{1+|f(z)|^{2}}\varphi\right.$
$\displaystyle\left.+\frac{2f^{\prime}(z)}{1+|f(z)|^{2}}\chi(z)(-f(z),i\,f(z),1)+\frac{1}{2}{\left(\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}+\frac{1}{z}\right)}\overline{\varphi}-\frac{f^{\prime}(z)\overline{f(z)}}{1+|f(z)|^{2}}\overline{\varphi}\right)=-\frac{i}{2}\left(\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}-\frac{2f^{\prime}(z)\overline{f(z)}}{1+|f(z)|^{2}}\right).$

Therefore, using (3.13), (3.14), (3.15), and
$\langle\vec{u},\vec{v}\rangle=0$, we deduce that

$\displaystyle\langle\vec{u},\partial_{z}\vec{v}\,\rangle$
$\displaystyle=\langle\mathrm{Im}\,(\varphi),\partial_{z}\mathrm{Re}\,(\varphi)\rangle=-\frac{i}{2}\left(\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}+\frac{1}{z}\right)+\frac{i}{2}\frac{f^{\prime}(z)}{1+|f(z)|^{2}}\langle(-f(z),i\,f(z),1),\overline{\psi}\rangle$
$\displaystyle=-\frac{i}{2}\left(\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}-\frac{2f^{\prime}(z)\overline{f(z)}}{1+|f(z)|^{2}}+\frac{1}{z}\right),$

and this concludes the proof of (3.25) since by (3.4)

$\displaystyle\partial_{z}(\mu(z)+\log|z|)$
$\displaystyle=\partial_{z}\left(\log|f^{\prime}(z)|-\log(1+|f(z)|^{2})-\frac{1}{2}\log(2)+\log|z|\right)$
$\displaystyle=\frac{1}{2}\left(\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}-\frac{2f^{\prime}(z)\overline{f(z)}}{1+|f(z)|^{2}}+\frac{1}{z}\right).$

This last identity concludes the proof of the theorem. ∎

Finally, we will establish the uniqueness of distributional solutions of the
system (3.2) with appropriate boundary conditions (3.3). This is the exact
analogous of Remark I.$1$ of [3]. First, we need to define explicit maps that
yield trivialisations of vector fields on simply connected domains of the
sphere. Let $\Omega_{1}\subset S^{2}$ be as Theorem 3.2. Using the
stereographic projection $\pi:S^{2}\setminus\left\\{N\right\\}\to\mathbb{C}$,
we have one holomorphic chart $z$ on $S^{2}\setminus\left\\{N\right\\}$, and
for a domain $\Omega_{1}\subset S^{2}\setminus\left\\{N\right\\}$, it yields a
trivialisation $T\Omega_{1}\rightarrow\Omega\times\mathbb{C}$ where
$\Omega=\pi(\Omega_{1})$.

More explicitly, let $X:\Omega\rightarrow\mathbb{R}^{3}$ be a vector field
such that $\langle X,\vec{n}\rangle=0$, where $\vec{n}:\Omega\rightarrow
S^{2}$ is the unit normal given by

$\displaystyle\pi^{-1}(z)=\vec{n}(z)=\left(\frac{2\,\mathrm{Re}\,(z)}{1+|z|^{2}},\frac{2\,\mathrm{Im}\,(z)}{1+|z|^{2}},\frac{-1+|z|^{2}}{1+|z|^{2}}\right).$

Now, we introduce the function $\psi:\mathbb{C}\rightarrow\mathbb{C}^{3}$,
given by

$\displaystyle\psi(z)=\left(\frac{1-z^{2}}{1+|z|^{2}},\frac{i(1+z^{2})}{1+|z|^{2}},\frac{2z}{1+|z|^{2}}\right),$

and we easily check that

$\displaystyle\langle\psi,\psi\rangle=0\qquad|\psi|^{2}=\langle\psi,\overline{\psi}\rangle=2.$
(3.26)

Therefore, we deduce that $(\vec{u}_{1},\vec{u}_{2})$ defined as follows is a
tangent unit moving frame (orthogonal to $\vec{n}$ )

$\displaystyle\vec{u}_{1}(z)=\mathrm{Re}\,(\psi(z)),\quad\vec{u}_{2}(z)=\mathrm{Im}\,(\psi(z)).$

The trivialisation map on $\Omega_{1}\subset S^{2}\setminus\left\\{N\right\\}$
is then given by

$\displaystyle T\Omega_{1}$ $\displaystyle\rightarrow\Omega\times\mathbb{C}$
$\displaystyle(z,v)$ $\displaystyle\mapsto(z,\langle
v,\vec{u}_{1}(z)\rangle+i\,\langle v,\vec{u}_{2}(z)\rangle),$ (3.27)

while the trivialisation map of sections is given by

$\displaystyle\Psi_{\Omega_{1}}:\Gamma(T\Omega_{1})$ $\displaystyle\rightarrow
C^{\infty}(\Omega_{1},\mathbb{C})$ $\displaystyle X$
$\displaystyle\mapsto\langle X,\vec{u}_{1}\rangle+i\,\langle
X,\vec{u}_{2}\rangle.$ (3.28)

Notice that for all tangent vector field $X$, we have $\langle
X,\vec{n}\rangle=0$, which implies that there exists real functions
$\lambda_{1},\lambda_{2}:\Omega_{1}\rightarrow\mathbb{R}$ such that

$\displaystyle X=\lambda_{1}\vec{u}_{1}+\lambda_{2}\vec{u}_{2}.$

######  Remark 3.4.

Using the next Theorem 3.5, it is easy to check that
$(\vec{u}_{j},\vec{v}_{j})$ ($j=1,2$) are harmonic vector fields since by
(3.10) and (3.11), we have

$\displaystyle-\Delta\chi=\left|\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}+\frac{1}{z}\right|^{2}\chi=|\nabla\chi|^{2}\chi,$

_i.e._ $\chi:\mathbb{D}\rightarrow S^{1}$ is a harmonic map with values into
$S^{1}$.

######  Theorem 3.5.

Under the conditions of Theorem 3.2, let
$\Omega_{1}^{\ast}=\Omega_{1}\setminus\left\\{p_{1}\right\\}$ and $\vec{u}\in
W^{1,2}_{\mathrm{loc}}(\Omega_{1}^{\ast},U\Omega_{1})\cap
W^{1,1}(\Omega_{1},U\Omega_{1})$ be a unit vector field in $\Omega_{1}$, and
let $\vec{u}_{0}=\Psi_{\Omega_{1}}(\vec{u}):\Omega_{1}\rightarrow S^{1}$. Then
$\vec{u}$ is a harmonic vector field on $\Omega_{1}$, _i.e._

$\displaystyle-\Delta_{g_{0}}\vec{u}=|d\vec{u}|_{g_{0}}^{2}{\vec{u}}+\left(2\langle
d\vec{u},d\vec{n}\rangle_{g_{0}}+\langle\vec{u},\Delta_{g_{0}}\vec{n}\rangle\right)\vec{n}$

if and only if $\vec{u}_{0}$ is a harmonic map with values into $S^{1}$,
_i.e._

$\displaystyle-\Delta_{g_{0}}\vec{u}_{0}=|d\vec{u}_{0}|^{2}_{g_{0}}\vec{u}_{0}.$

In particular, for all degree $1$ boundary data $h\in
H^{1/2}(\partial\Omega_{1},S^{1})$ and $p\in\Omega_{1}$, there exists a unique
unit vector-field $\vec{u}\in
W^{1,2}_{\mathrm{loc}}(\Omega_{1}^{\ast},U\Omega_{1})\cap
W^{1,1}(\Omega_{1},U\Omega_{1})$ such that $\vec{u}=\Psi_{\Omega_{1}}^{-1}(h)$
on $\partial\Omega_{1}$ and such that $\vec{u}_{0}=\Psi_{\Omega_{1}}(\vec{u})$
satisfies in the distributional sense

$\displaystyle\operatorname{div}\left(\vec{u}_{0}\times\nabla\vec{u}_{0}\right)=0\qquad\text{in}\;\,\mathscr{D}^{\prime}(\Omega_{1}).$

######  Remark 3.6.

If $\vec{u}_{0}:\Omega\rightarrow S^{1}$, writing locally
$\vec{u}_{0}=e^{i\varphi}$ for some real-valued function $\varphi$, we deduce
that $\vec{u}_{0}$ is harmonic if and only if

$\displaystyle-\Delta\vec{u}_{0}=\left(|\nabla\varphi|^{2}-i\,(\Delta\varphi)\right)\vec{u}_{0}=|\nabla\vec{u}_{0}|^{2}\vec{u}_{0}.$

Therefore, $\vec{u}_{0}$ is harmonic as a map with values into $S^{1}$ if and
only if $\varphi$ is harmonic, _i.e._ $\Delta\varphi=0$.

###### Proof of Theorem 3.5.

By making a stereographic projection, thanks to the conformal invariance of
the harmonic equation, we deduce that for all unit vector-field
$\vec{u}\in\Gamma(T\Omega_{1}^{\ast})$ is given in
$\Omega=\pi_{N}(\Omega_{1})$ as

$\displaystyle\vec{u}=\lambda_{1}\,\mathrm{Re}\,(\psi)+\lambda_{2}\,\mathrm{Im}\,(\psi),$
(3.29)

where

$\displaystyle\psi(z)=\left(\frac{1-z^{2}}{1+|z|^{2}},\frac{i(1+z^{2})}{1+|z|^{2}},\frac{2z}{1+|z|^{2}}\right).$

Furthermore, we have $\lambda_{1}^{2}+\lambda_{2}^{2}=1$, which implies that
there exists a measurable function $\varphi$ such that
$\lambda_{1}+i\lambda_{2}=e^{-i\varphi}$. In particular, we can rewrite (3.29)
as

$\displaystyle\vec{u}=\cos(\varphi)\,\mathrm{Re}\,(\psi)-\sin(\varphi)\,\mathrm{Im}\,(\psi)=\mathrm{Re}\,(e^{-i\varphi})\,\mathrm{Re}\,(\psi)+\mathrm{Im}\,(e^{-i\varphi})\,\mathrm{Im}\,(\psi)=\mathrm{Re}\,\left(e^{i\varphi}\psi\right),$

where we used the identity
$\mathrm{Re}\,(a)\,\mathrm{Re}\,(b)+\mathrm{Im}\,(a)\,\mathrm{Im}\,(b)=\mathrm{Re}\,(\overline{a}\,b)$
valid for all $a,b\in\mathbb{C}$. If

$\displaystyle\vec{v}=\sin(\varphi)\mathrm{Re}\,(\psi)+\cos(\varphi)\mathrm{Im}\,(\psi)=\mathrm{Im}\,(e^{i\varphi}\psi),$

we immediately have $\langle\vec{u},\vec{v}\,\rangle=0$, and since
$|\vec{u}\,|^{2}=|\vec{v}\,|^{2}=1$, while
$-\Delta\vec{n}=|\nabla\vec{n}|^{2}\vec{n}$, we get

$\displaystyle\langle\Delta\vec{u},\vec{u}\,\rangle=-|\nabla\vec{u}\,|^{2}$
$\displaystyle\langle\Delta\vec{u},\vec{n}\,\rangle=-2\langle\nabla\vec{u},\nabla\vec{n}\,\rangle-\langle\vec{u},\Delta\vec{n}\,\rangle=-2\langle\nabla\vec{u},\nabla\vec{n}\,\rangle,$

and similar formulae for $\vec{v}$. Therefore, we deduce that
$(\vec{u},\vec{v})$ solves the system

$\displaystyle\left\\{\begin{aligned}
-\Delta\vec{u}&=|\nabla\vec{u}\,|^{2}\vec{u}+2\langle\nabla\vec{u},\nabla\vec{n}\,\rangle\vec{n}\qquad&&\text{in}\;\,\Omega\\\
-\Delta\vec{v}&=|\nabla\vec{v}\,|^{2}\vec{v}+2\langle\nabla\vec{v},\nabla\vec{n}\,\rangle\vec{n}\qquad&&\text{in}\;\,\Omega,\end{aligned}\right.$

if and only if

$\displaystyle\langle\Delta\vec{u},\vec{v}\,\rangle=\langle\Delta\vec{v},\vec{u}\,\rangle=0.$

Now, we compute

$\displaystyle\Delta\vec{u}$
$\displaystyle=\mathrm{Re}\,\left((i\,\Delta\varphi-|\nabla\varphi|^{2})e^{i\varphi}\psi\right)+2\,\mathrm{Re}\,\left(ie^{i\varphi}\nabla\varphi\cdot\nabla\psi\right)+\mathrm{Re}\,\left(e^{i\varphi}\Delta\psi\right)$
$\displaystyle=-(\Delta\varphi)\mathrm{Im}\,(e^{i\varphi}\psi)-|\nabla\varphi|^{2}\mathrm{Re}\,(e^{i\varphi}\psi)+\mathrm{Re}\,\left(e^{i\varphi}\Delta\psi\right)$
$\displaystyle=-(\Delta\varphi)\,\vec{v}-|\nabla\varphi|^{2}\vec{u}+2\,\mathrm{Re}\,\left(ie^{i\varphi}\nabla\varphi\cdot\nabla\psi\right)+\mathrm{Re}\,(e^{i\varphi}\Delta\psi)$
$\displaystyle\Delta\vec{v}$
$\displaystyle=(\Delta\varphi)\,\vec{u}-|\nabla\varphi|^{2}\vec{u}+2\,\mathrm{Im}\,\left(ie^{i\varphi}\nabla\varphi\cdot\nabla\psi\right)+\mathrm{Im}\,(e^{i\varphi}\Delta\psi).$

We have since $\langle\nabla\varphi,\varphi\rangle=0$ the identity

$\displaystyle\left\langle\mathrm{Re}\,(ie^{i\varphi}\nabla\varphi\cdot\nabla\psi),\vec{v}\right\rangle$
$\displaystyle=\mathrm{Re}\,\left\langle
ie^{i\varphi}\nabla\varphi\cdot\nabla\psi,\frac{e^{i\varphi}\psi-e^{-i\varphi}\overline{\psi}}{2i}\right\rangle=-\frac{1}{2}\mathrm{Re}\,\left(\nabla\varphi\cdot\langle\nabla\psi,\overline{\psi}\rangle\right)$
$\displaystyle\left\langle\mathrm{Im}\,(ie^{i\varphi}\nabla\varphi\cdot\nabla\psi),\vec{u}\right\rangle$
$\displaystyle=\mathrm{Im}\,\left\langle
ie^{i\varphi}\nabla\varphi\cdot\nabla\psi,\frac{e^{i\varphi}\psi+e^{-i\varphi}\overline{\psi}}{2}\right\rangle=\frac{1}{2}\mathrm{Re}\,\left(\nabla\varphi\cdot\langle\nabla\psi,\overline{\psi}\rangle\right)$
$\displaystyle\langle\mathrm{Re}\,(e^{i\varphi}\Delta\psi),\vec{v}\rangle$
$\displaystyle=\mathrm{Re}\,\left\langle
e^{i\varphi}\Delta\psi,\frac{e^{i\varphi}\psi-e^{-i\varphi}\psi}{2i}\right\rangle=\frac{1}{2}\mathrm{Im}\,\left(e^{2i\varphi}\langle\Delta\psi,\psi\rangle\right)-\frac{1}{2}\mathrm{Im}\,\langle\Delta\psi,\psi\rangle$
$\displaystyle\langle\mathrm{Im}\,(e^{i\varphi}\Delta\psi),\vec{u}\rangle$
$\displaystyle=\frac{1}{2}\mathrm{Im}\,\left(e^{2i\varphi}\langle\Delta\psi,\psi\rangle\right)+\frac{1}{2}\mathrm{Im}\,\langle\Delta\psi,\overline{\psi}\rangle.$

In particular, we have

$\displaystyle\langle\Delta\vec{u},\vec{v}\rangle=-(\Delta\varphi)-\mathrm{Re}\,\left(\nabla\varphi\cdot\langle\nabla\psi,\overline{\psi}\rangle\right)+\frac{1}{2}\mathrm{Im}\,\left(e^{2i\varphi}\langle\Delta\psi,\psi\rangle\right)-\frac{1}{2}\mathrm{Im}\,\langle\Delta\psi,\psi\rangle$
$\displaystyle\langle\Delta\vec{v},\vec{u}\rangle=(\Delta\varphi)+\mathrm{Re}\,\left(\nabla\varphi\cdot\langle\nabla\psi,\overline{\psi}\rangle\right)+\frac{1}{2}\mathrm{Im}\,\left(e^{2i\varphi}\langle\Delta\psi,\psi\rangle\right)+\frac{1}{2}\mathrm{Im}\,\langle\Delta\psi,\overline{\psi}\rangle.$

Summing those equations and substracting the first one to the second one
yields the system

$\displaystyle\left\\{\begin{aligned}
&\mathrm{Im}\,\left(e^{2i\varphi}\langle\Delta\psi,\psi\rangle\right)=0\\\
&2(\Delta\varphi)+2\,\mathrm{Re}\,\left(\nabla\varphi\cdot\langle\nabla\psi,\overline{\psi}\rangle\right)+\mathrm{Im}\,\langle\Delta\psi,\overline{\psi}\rangle=0.\end{aligned}\right.$
(3.30)

We will show that for all smooth _real-valued_ function
$\varphi:\Omega\rightarrow\mathbb{R}$

$\displaystyle\mathrm{Re}\,\left(\nabla\varphi\cdot\langle\nabla\psi,\overline{\psi}\rangle\right)=\langle\Delta\psi,\psi\rangle=\mathrm{Im}\,\langle\Delta\psi,\overline{\psi}\rangle=0,$
(3.31)

which will imply that $(\vec{u},\vec{v})$ solves the system (3.30) if and only
if $\Delta\varphi=0$, or $\varphi$ is harmonic.

Now, we compute

$\displaystyle\partial_{z}\psi$
$\displaystyle=-\frac{\overline{z}}{1+|z|^{2}}\psi+\frac{2}{1+|z|^{2}}\left(-z,i\,z,1\right)$
$\displaystyle\partial_{\overline{z}}\psi$
$\displaystyle=-\frac{z}{1+|z|^{2}}\psi$

We have

$\displaystyle\nabla\varphi\cdot\nabla\psi$
$\displaystyle=2\,\partial_{z}\varphi\cdot\partial_{\overline{z}}\psi+2\,\partial_{\overline{z}}\varphi\cdot\partial_{z}\psi$
$\displaystyle=-2\frac{\overline{z}\partial_{z}\varphi}{1+|z|^{2}}\psi+\frac{4}{1+|z|^{2}}(-z\partial_{z}\varphi,iz\partial_{z}\varphi,\partial_{z}\varphi)-2\frac{z\partial_{\overline{z}}\varphi}{1+|z|^{2}}\psi$
$\displaystyle=-4\,\mathrm{Re}\,\left(\frac{\overline{z}\partial_{z}\varphi}{1+|z|^{2}}\right)\psi+\frac{4}{1+|z|^{2}}(-z\partial_{z}\varphi,iz\partial_{z}\varphi,\partial_{z}\varphi).$

Then we have

$\displaystyle\frac{1}{4}\Delta\psi=\partial_{z\overline{z}}^{2}\psi=\frac{-1+|z|^{2}}{(1+|z|^{2})^{2}}\psi-\frac{2}{(1+|z|^{2})^{2}}(-z^{2},iz^{2},z)$
(3.32)

Now, notice that

$\displaystyle\langle(-z^{2},iz^{2},z),\psi\rangle$
$\displaystyle=\frac{1}{1+|z|^{2}}\langle(-z^{2},iz^{2},z),\left(1-z^{2},i(1+z^{2}),2z\right)\rangle$
$\displaystyle=\frac{1}{1+|z|^{2}}\left(-z^{2}(1-\overline{z}^{2})+z^{2}(1+\overline{z}^{2})+2z^{2}\right)=0,$

which implies as $\langle\psi,\psi\rangle=0$ and by (3.32) that

$\displaystyle\langle\Delta\psi,\psi\rangle=0.$ (3.33)

Now, we have

$\displaystyle\langle(-z^{2},iz^{2},z),\overline{\psi}\rangle=\frac{1}{1+|z|^{2}}\left(-z^{2}(1-\overline{z}^{2})+z^{2}(1+\overline{z}^{2})+2|z|^{2}\right)=2|z|^{2}.$

Since $|\psi|^{2}=\langle\psi,\overline{\psi}\rangle=2$, we deduce that

$\displaystyle\langle\Delta\psi,\overline{\psi}\rangle=4\left(\frac{2(-1+|z|^{2})}{(1+|z|^{2})^{2}}-\frac{4|z|^{2}}{(1+|z|^{2})^{2}}\right)=-\frac{8}{1+|z|^{2}}\in\mathbb{R}.$
(3.34)

We now compute

$\displaystyle\langle(-z,iz,1),\overline{\psi}\rangle=\frac{1}{1+|z|^{2}}\left(-z(1-\overline{z}^{2})+z(1+\overline{z}^{2})+2\overline{z}\right)=2\overline{z}.$

which shows since $|\psi|^{2}=2$ that

$\displaystyle\langle\partial_{z}\psi,\psi\rangle$
$\displaystyle=-\frac{2\overline{z}}{1+|z|^{2}}+\frac{4\overline{z}}{1+|z|^{2}}=\frac{2\overline{z}}{1+|z|^{2}}$
$\displaystyle\langle\partial_{\overline{z}}\psi,\psi\rangle$
$\displaystyle=-\frac{2z}{1+|z|^{2}}.$

Therefore, we have

$\displaystyle\nabla\varphi\cdot\langle\nabla\psi,\psi\rangle=2\,\partial_{\overline{z}}\varphi\cdot\langle\partial_{z}\psi,\psi\rangle+2\,\partial_{z}\varphi\cdot\langle\partial_{\overline{z}}\psi,\psi\rangle=\frac{4\,\overline{z}\,\partial_{\overline{z}}\varphi}{1+|z|^{2}}-\frac{4\,z\,\partial_{z}\varphi}{1+|z|^{2}}=8i\,\mathrm{Im}\,\left(\frac{\overline{z}\,\partial_{\overline{z}}\varphi}{1+|z|^{2}}\right)\in
i\mathbb{R},$

and this immediately implies that

$\displaystyle\mathrm{Re}\,\left(\nabla\varphi\cdot\langle\nabla\psi,\psi\rangle\right)=0.$
(3.35)

Finally, we deduce by (3.33), (3.34) and (3.35) that (3.31) holds and that the
system (3.30) holds if and only if $\Delta\varphi=0$. If
$\vec{u}=g_{\Omega_{1}}=\Psi_{\Omega_{1}}(g)$ for some
$g:\partial\Omega_{1}=\Gamma\rightarrow S^{1}$, then we have

$\displaystyle\lambda_{1}+i\lambda_{2}=g,$

or

$\displaystyle e^{-i\varphi}=g\qquad\text{on}\;\,\Gamma.$

In particular, the function
$\vec{u}_{0}=e^{-i\varphi}:\Omega_{1}\setminus\left\\{p_{1}\right\\}\rightarrow
S^{1}$ is a harmonic map on $\Omega_{1}\setminus\left\\{p_{1}\right\\}$
satisfying $\vec{u}_{0}=h$ on $\Gamma$. Now, notice that provided $\vec{u}\in
W^{1,1}(\Omega_{1})$, one can rewrite the equation distributionally as

$\displaystyle\operatorname{div}\left(\vec{u}\times\nabla\vec{u}\right)=2\langle\nabla\vec{u},\nabla\vec{n}\rangle\vec{n}\times\vec{u}.$

In particular, we deduce as $u_{0}$ is harmonic that

$\displaystyle\operatorname{div}\left(\vec{u}_{0}\times\nabla\vec{u}_{0}\right)=\frac{\partial}{\partial
x_{1}}\left(\vec{u}_{0}\times\frac{\partial\vec{u}_{0}}{\partial
x_{1}}\right)+\frac{\partial}{\partial
x_{2}}\left(\vec{u}_{0}\times\frac{\partial\vec{u}_{0}}{\partial
x_{2}}\right)=0.$

By Theorem I.$5$ and Remark I.$1$ of [3], we deduce that $\vec{u}_{0}$ is the
unique harmonic function with a singularity at $p_{1}$ such that
$\vec{u}_{0}=h$ on $\partial\Omega_{1}$. This concludes the proof of the
theorem. ∎

##  4 Proof of the Main Theorems for Non-Smooth Curves

In order to extend Theorem 2.6 to the non-smooth setting, we will obtain
another formula for $\mathscr{E}_{0}$ in terms of conformal maps and that
holds true for any closed simple curve of finite Loewner energy. Using this
additional formula, the convergence result will be easily obtained.

Under the preceding notations, if $\Gamma\subset S^{2}$ Weil-Petersson
quasicircle, from Remark 2.4, thanks to Theorem 3.2, there exists harmonic
moving frames $(\vec{u}_{1},\vec{v}_{1})$ and $(\vec{u}_{2},\vec{v}_{2})$ on
$\Omega_{1}$ and $\Omega_{2}$ with arbitrary singularities $p_{1}$ and $p_{2}$
respectively, such that

$\displaystyle\mathscr{E}(\Gamma)=\int_{\Omega_{1}}|d\mu_{1}|^{2}_{g_{0}}d\mathrm{vol}_{g_{0}}+\int_{\Omega_{2}}|d\mu_{2}|^{2}_{g_{0}}d\mathrm{vol}_{g_{0}}+2\int_{\Omega_{1}}G_{\Omega_{1}}K_{g_{0}}d\mathrm{vol}_{g_{0}}+2\int_{\Omega_{2}}G_{\Omega_{2}}K_{g_{0}}d\mathrm{vol}_{g_{0}}+4\pi.$

where
$\omega_{j}=\langle\vec{u}_{j},d\vec{v}_{j}\rangle=\ast\,d(G_{\Omega_{j}}+\mu_{j})$
in $\mathscr{D}^{\prime}(\Omega_{j})$ for $j=1,2$, and $\mu_{j}$ satisfies
(2.10). We saw in Theorem 2.6 that in the case of smooth curves, there exists
conformal maps $f_{1}:\mathbb{D}\rightarrow\Omega_{1}$ and
$f_{2}:\mathbb{D}\rightarrow\Omega_{2}$ such that

$\displaystyle I^{L}(\Gamma)=\frac{1}{\pi}\mathscr{E}(\Gamma)+4\log|\nabla
f_{1}(0)|+4\log|\nabla
f_{2}(0)|-12\log(2)=\frac{1}{\pi}\mathscr{E}_{0}(\Gamma).$

In this section, we generalise this result for curves of finite Loewner
energy. Now, if
$\pi:S^{2}\setminus\left\\{p_{2}\right\\}\rightarrow\mathbb{C}$ is a
stereographic projection, since $f_{j}:\mathbb{D}\rightarrow\Omega_{j}$ is
conformal and $\pi$ is also conformal, we deduce that $\pi\circ
f_{j}:\mathbb{D}\rightarrow\pi(\Omega_{j})\subset\mathbb{R}$ is also
conformal. Therefore, these maps are biholomorphic or anti-biholomorphic, so
up to a complex conjugate (which is an isometry), we can assume that they are
holomorphic. Notice that $\Omega=\pi(\Omega_{1})$ is bounded, while
$\pi(\Omega_{2})=\mathbb{C}\setminus\overline{\Omega}$ is unbounded.
Therefore, if
${\mathfrak{i}}:\mathbb{C}\setminus\left\\{0\right\\}\rightarrow\mathbb{C}\setminus\left\\{0\right\\}$
is the inversion, we let $g=\pi\circ
f_{2}\circ{\mathfrak{i}}:\mathbb{C}\setminus\overline{\mathbb{D}}\rightarrow\mathbb{C}\setminus\overline{\Omega}$
and $f={\pi\circ f_{1}}:\mathbb{D}\rightarrow\Omega$. From (1.3) and (1.4), if
$\gamma=\pi(\Gamma)$, we have

$\displaystyle
I^{L}(\Gamma)=I^{L}(\gamma)=\int_{\mathbb{D}}\left|\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}\right|^{2}|dz|^{2}+\int_{\mathbb{C}\setminus\overline{\mathbb{D}}}\left|\frac{g^{\prime\prime}(z)}{g^{\prime}(z)}\right|^{2}|dz|^{2}+4\pi\log|f^{\prime}(0)|-4\pi\log|g^{\prime}(\infty)|.$
(4.1)

Indeed, since $f_{2}(0)=p_{2}$, we have $g(\infty)=\infty$, so that the
functions $f$, $g$ satisfy the needed conditions to apply Theorem 1.1.

Now, with the previous notations, define the functional

$\displaystyle\mathscr{E}_{0}({\Gamma})=\mathscr{E}({\Gamma})+4\pi\log|\nabla
f_{1}(0)|+4\pi\log|\nabla f_{2}(0)|-12\pi\log(2).$

######  Definition 4.1.

Let $\gamma$ be a Jordan curve with finite Loewner energy. Let
$f:\mathbb{D}\rightarrow\Omega$,
$g:\mathbb{C}\setminus\overline{\mathbb{D}}\rightarrow\mathbb{C}\setminus\overline{\Omega}$
be biholomorphic maps such that $g(\infty)=\infty$, we define the third
universal Liouville action $S_{3}$ by

$\displaystyle S_{3}(\gamma)$
$\displaystyle=\int_{\mathbb{D}}\left|\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}-2\frac{f^{\prime}(z)}{f(z)}\frac{|f(z)|^{2}}{1+|f(z)|^{2}}\right|^{2}|dz|^{2}+\int_{\mathbb{C}\setminus\overline{\mathbb{\mathbb{D}}}}\left|\frac{g^{\prime\prime}(z)}{g^{\prime}(z)}-2\frac{g^{\prime}(z)}{g(z)}\frac{|g(z)|^{2}}{1+|g(z)|^{2}}+\frac{2}{z}\right|^{2}|dz|^{2}$
$\displaystyle+2\int_{\mathbb{D}}\log|z|\frac{4|f^{\prime}(z)|^{2}|dz|^{2}}{(1+|f(z)|^{2})^{2}}-2\int_{\mathbb{C}\setminus\overline{\mathbb{D}}}\log|z|\frac{4|g^{\prime}(z)|^{2}|dz|^{2}}{(1+|g(z)|^{2})^{2}}+4\pi$
$\displaystyle+4\pi\log|f^{\prime}(0)|-4\pi\log|g^{\prime}(\infty)|-4\pi\log(1+|f(0)|^{2}).$

######  Remark 4.2.

  1. ($1$)

One may wonder from where this definition comes from. It will be made clear in
the proof of the next theorem where we explicitly rewrite $\mathscr{E}_{0}$
with the help of the conformal maps $f$ and $g$ defined above.

  2. ($2$)

We call this quantity $S_{3}$ since a functional called $S_{2}$ was defined in
[14] as the log-determinant of the Grunsky operator associated with the curve
$\gamma$ (up to a factor $-\frac{1}{12}$).

The goal of this section is to show the identity

${\pi}I^{L}=S_{1}=S_{3}=\mathscr{E}_{0}.$ (4.2)

The third equality is straightforward and is proved in Theorem 4.3, and the
proof of the whole identity is completed in Theorem 4.5.

######  Theorem 4.3.

Let $\Gamma\subset S^{2}$ be a simple curve of finite Loewner energy. Then we
have

$\displaystyle\mathscr{E}_{0}(\Gamma)=S_{3}(\Gamma)$

###### Proof.

If $\Gamma\subset S^{2}$ is a curve of finite Loewner energy $\Omega_{1}$ and
$\Omega_{2}$ the two connected components of $S^{2}\setminus\Gamma$, and
$f_{1}:\mathbb{D}\rightarrow\Omega_{1}$,
$f_{2}:\mathbb{D}\rightarrow\Omega_{2}$ are the conformal maps associated to
$\Gamma$ in the definition of $\mathscr{E}$ with $f_{j}(0)=p_{j}$ for $j=1,2$.
Now, recall from (2.6) that

$\displaystyle\log|\nabla f_{1}|=\frac{1}{2}\log(2)+\mu_{1}\circ f_{1}.$

We have by conformal invariance of the Dirichlet energy

$\displaystyle\int_{\mathbb{D}}\left|\nabla\log|\nabla
f_{1}|\right|^{2}|dz|^{2}=\int_{\mathbb{D}}\left|\nabla(\mu_{1}\circ
f_{1})\right|^{2}|dz|^{2}=\int_{\Omega_{1}}|d\mu_{1}|_{g_{0}}^{2}d\mathrm{vol}_{g_{0}}.$
(4.3)

Since $f_{1}$ is conformal and $f_{1}(0)=p_{1}$, we have

$\displaystyle G_{\Omega_{1}}\circ f_{1}(z)=G_{\Omega_{1},p_{1}}\circ
f_{1}(z)=G_{\mathbb{D},0}(z)=\log|z|.$

A change of variable gives

$\displaystyle
2\int_{\Omega_{1}}G_{\Omega_{1}}d\mathrm{vol}_{g_{0}}=\int_{\mathbb{D}}\log|z||\nabla
f_{1}|^{2}|dz|^{2}.$ (4.4)

Finally, we deduce by (4.3) and (4.4) that

$\displaystyle\mathscr{E}({\Gamma})$
$\displaystyle=\int_{\Omega_{1}}|d\mu_{1}|^{2}_{g_{0}}d\mathrm{vol}_{g_{0}}+\int_{\Omega_{2}}|d\mu_{2}|^{2}_{g_{0}}d\mathrm{vol}_{g_{0}}+2\int_{\Omega_{1}}G_{\Omega_{1}}K_{g_{0}}d\mathrm{vol}_{g_{0}}+2\int_{\Omega_{2}}G_{\Omega_{2}}K_{g_{0}}d\mathrm{vol}_{g_{0}}+4\pi$
$\displaystyle+4\pi\log|\nabla f_{1}(0)|+4\pi\log|\nabla
f_{2}(0)|-12\pi\log(2)$ $\displaystyle=\int_{\mathbb{D}}|\nabla\log|\nabla
f_{1}||^{2}|dz|^{2}+\int_{\mathbb{D}}|\nabla\log|\nabla
f_{2}||^{2}|dz|^{2}+\int_{\mathbb{D}}\log|z||\nabla
f_{1}|^{2}|dz|^{2}+\int_{\mathbb{D}}\log|z||\nabla f_{2}|^{2}|dz|^{2}+4\pi$
$\displaystyle+4\pi\log|\nabla f_{1}(0)|+4\pi\log|\nabla
f_{2}(0)|-12\pi\log(2).$ (4.5)

Up to a rotation of $S^{2}$, we can assume that $p_{2}=N$ and if
$\pi:S^{2}\setminus\left\\{N\right\\}\rightarrow\mathbb{C}$ is the standard
stereographic projection, let

$\displaystyle f$ $\displaystyle=\pi\circ f_{1}:\mathbb{D}\rightarrow\Omega$
$\displaystyle\widetilde{g}$ $\displaystyle=\pi\circ
f_{2}:\mathbb{D}\rightarrow\mathbb{C}\setminus\overline{\Omega}$

which we assume without loss of generality to be biholomorphic (up to a
complex conjugation). Now, since

$\displaystyle
f_{1}(z)=\pi^{-1}(f(z))=\left(\frac{2\,\mathrm{Re}\,(f(z))}{1+|f(z)|^{2}},\frac{2\,\mathrm{Im}\,(f(z))}{1+|f(z)|^{2}},\frac{-1+|f(z)|^{2}}{1+|f(z)|^{2}}\right),$

a computation shows that

$\displaystyle\partial_{z}f_{1}=f^{\prime}\left(\frac{(1-\overline{f}^{2})}{(1+|f|^{2})^{2}},\frac{-i(1+\overline{f}^{2})}{(1+|f|^{2})^{2}},\frac{2\overline{f}}{(1+|f|^{2})^{2}}\right),$

which implies that

$\displaystyle|\partial_{z}f_{1}|^{2}=\frac{|f^{\prime}|^{2}}{(1+|f|^{2})^{4}}\left(|1-f^{2}|^{2}+|1+f^{2}|^{2}+4|f|^{2}\right)=\frac{2|f^{\prime}|^{2}}{(1+|f|^{2})^{2}}.$

We deduce that

$\displaystyle|\nabla
f_{1}|^{2}=4|\partial_{z}f_{1}|^{2}=\frac{8|f^{\prime}|^{2}}{(1+|f|^{2})^{2}}.$
(4.6)

Therefore, we have

$\displaystyle\log|\nabla
f_{1}|=\log|f^{\prime}|-\log(1+|f|^{2})+\frac{3}{2}\log(2),$

so that

$\displaystyle 4\pi\log|\nabla
f_{1}(0)|=4\pi\log|f^{\prime}(0)|-4\pi\log(1+|f(0)|^{2})+6\pi\log(2)$ (4.7)

Since $\Omega=f(\mathbb{D})$ is compact, we have

$\displaystyle\int_{\mathbb{D}}|f^{\prime}(z)|^{2}\frac{|f(z)|^{2}|dz|^{2}}{(1+|f(z)|^{2})^{2}}\leq\frac{1}{4}\int_{\mathbb{D}}|f^{\prime}(z)|^{2}|dz|^{2}=\frac{1}{4}\mathrm{Area}(\Omega)<\infty.$
(4.8)

Therefore, (4.8) implies that $\nabla\log|\nabla f_{1}|\in L^{2}(\mathbb{D})$
and

$\displaystyle\int_{\mathbb{D}}|\nabla\log|\nabla
f_{1}||^{2}|dz|^{2}=\int_{\mathbb{D}}\left|\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}-2\frac{f^{\prime}(z)}{f(z)}\frac{|f(z)|^{2}}{1+|f(z)|^{2}}\right|^{2}|dz|^{2}<\infty,$
(4.9)

while (4.6) implies that

$\displaystyle\int_{\mathbb{D}}\log|z||\nabla
f_{1}|^{2}|dz|^{2}=\int_{\mathbb{D}}\log|z|\frac{8|f^{\prime}(z)|^{2}|dz|^{2}}{(1+|f(z)|^{2})^{2}}<\infty,$
(4.10)

which is finite by (4.8) and the smoothness of $f$ in $\mathbb{D}$. Since the
function
$\widetilde{g}:\mathbb{D}\rightarrow\mathbb{C}\setminus\overline{\Omega}$ is
unbounded at $0$, we do not see trivially that

$\displaystyle\int_{\mathbb{D}}\left|\frac{\widetilde{g}^{\prime\prime}(z)}{\widetilde{g}^{\prime}(z)}-2\frac{\widetilde{g}^{\prime}(z)\overline{\widetilde{g}(z)}}{1+|\widetilde{g}(z)|^{2}}\right|^{2}|dz|^{2}=\int_{\mathbb{D}}\left|\frac{\widetilde{g}^{\prime\prime}(z)}{\widetilde{g}^{\prime}(z)}-2\frac{\widetilde{g}^{\prime}(z)}{g(z)}\frac{|\widetilde{g}(z)|^{2}}{1+|\widetilde{g}(z)|^{2}}\right|^{2}|dz|^{2}<\infty.$

For this, as $\widetilde{g}$ is univalent and $\widetilde{g}(0)=\infty$, we
deduce that $\widetilde{g}$ admits the following meromorphic expansion at
$z=0$ for some $a\in\mathbb{C}\setminus\left\\{0\right\\}$ and
$a_{0},a_{1}\in\mathbb{C}$

$\displaystyle\widetilde{g}(z)=\frac{a}{z}+a_{0}+a_{1}z+O(|z|^{2}).$ (4.11)

Therefore, we have by a direct computation

$\displaystyle\frac{\widetilde{g}^{\prime\prime}(z)}{\widetilde{g}^{\prime}(z)}-2\frac{\widetilde{g}^{\prime}(z)}{\widetilde{g}(z)}\frac{|\widetilde{g}(z)|^{2}}{1+|\widetilde{g}(z)|^{2}}$
$\displaystyle=-\frac{a_{0}}{a}+\left(\frac{a_{0}^{2}}{a^{2}}-\frac{4a_{1}}{a}\right)z-\frac{\overline{z}}{|a|^{2}}+O(|z|^{2})\in
L^{\infty}_{\mathrm{loc}}(\mathbb{D}).$ (4.12)

Since $\Gamma$ is a Weil-Petersson quasicircle, we deduce by estimates similar
to (4.8) and (4.9) that $\nabla\log|\widetilde{g}^{\prime}|\in
L^{2}(\mathbb{D}\setminus\overline{\mathbb{D}}(0,\varepsilon))$ and
$\widetilde{g}^{\prime}\in
L^{2}(\mathbb{D}\setminus\overline{\mathbb{D}}(0,\varepsilon))$ for all
$\varepsilon>0$ and we finally deduce that

$\displaystyle\int_{\mathbb{D}}\left|\frac{\widetilde{g}^{\prime\prime}(z)}{\widetilde{g}^{\prime}(z)}-2\frac{\widetilde{g}^{\prime}(z)}{\widetilde{g}(z)}\frac{|\widetilde{g}(z)|^{2}}{1+|\widetilde{g}(z)|^{2}}\right|^{2}|dz|^{2}<\infty.$

Now, if
${g}=\widetilde{g}\circ{\mathfrak{i}}:\mathbb{C}\setminus\overline{\mathbb{D}}\rightarrow\mathbb{C}\setminus\overline{\Omega}$,
we compute and

$\displaystyle\frac{\widetilde{g}^{\prime\prime}(z)}{\widetilde{g}^{\prime}(z)}-2\frac{\widetilde{g}^{\prime}(z)}{\widetilde{g}(z)}\frac{|\widetilde{g}(z)|^{2}}{1+|\widetilde{g}(z)|^{2}}$
$\displaystyle=-\frac{1}{z^{2}}\left(\frac{{g}^{\prime\prime}(1/z)}{{g}^{\prime}(1/z)}-2\frac{{g}^{\prime}(1/z)}{{g}(1/z)}\frac{|{g}(1/z)|^{2}}{1+|{g}(1/z)|^{2}}+2z\right).$

A change of variable shows that

$\displaystyle\int_{\mathbb{D}}\left|\frac{\widetilde{g}^{\prime\prime}(z)}{\widetilde{g}^{\prime}(z)}-2\frac{\widetilde{g}^{\prime}(z)}{\widetilde{g}(z)}\frac{|\widetilde{g}(z)|^{2}}{1+|\widetilde{g}(z)|^{2}}\right|^{2}|dz|^{2}=\int_{\mathbb{C}\setminus\overline{\mathbb{D}}}\left|\frac{{g}^{\prime\prime}(z)}{{g}^{\prime}(z)}-2\frac{{g}^{\prime}(z)}{{g}(z)}\frac{|{g}(z)|^{2}}{1+|{g}(z)|^{2}}+\frac{2}{z}\right|^{2}|dz|^{2}$
(4.13)

Furthermore, we directly get

$\displaystyle\int_{\mathbb{D}}\log|z|\frac{|\widetilde{g}^{\prime}(z)|^{2}|dz|^{2}}{(1+|\widetilde{g}(z)|^{2})^{2}}$
$\displaystyle=\int_{\mathbb{D}}\log|z|\frac{|{g}^{\prime}(1/z)|^{2}}{(1+|{g}(1/z)|^{2})^{2}}\frac{|dz|^{2}}{|z|^{4}}=\int_{\mathbb{C}\setminus\overline{\mathbb{D}}}\log\left(\frac{1}{|z|}\right)\frac{|{g}^{\prime}(z)|^{2}|dz|^{2}}{(1+|{g}(z)|^{2})^{2}}$
$\displaystyle=-\int_{\mathbb{C}\setminus\overline{\mathbb{D}}}\log|z|\frac{|g^{\prime}(z)|^{2}|dz|^{2}}{(1+|g(z)|^{2})^{2}}.$
(4.14)

Now, notice that

$\displaystyle|\nabla
f_{2}|^{2}=\frac{8|g^{\prime}(z)|^{2}}{(1+|g(z)|^{2})^{2}}=\frac{8}{|a|^{2}}+O(|z|),$

which implies that

$\displaystyle\log|\nabla f_{2}|=\frac{3}{2}\log(2)-\log|a|.$

Furthermore, the expansion (4.11) shows that as $|z|\rightarrow\infty$, we
have

$\displaystyle{g}(z)=az+O(1),$

so that $|a|=|{g}^{\prime}(\infty)|$, and

$\displaystyle 4\pi\log|\nabla
f_{2}(0)|=-4\pi\log|{g}^{\prime}(\infty)|+6\pi\log(2).$ (4.15)

Finally, we deduce by (4.7) and (4.15) that

$\displaystyle 4\pi\log|\nabla f_{1}(0)|+4\pi\log|\nabla
f_{2}(0)|-12\pi\log(2)=4\pi\log|f^{\prime}(0)|-4\pi\log{g}^{\prime}(\infty)|-4\pi\log(1+|f(0)|^{2}).$
(4.16)

Gathering (4), (4.9), (4.10), (4.13), (4), and (4.16), we finally deduce that

$\displaystyle\mathscr{E}(\Gamma)$
$\displaystyle=\int_{\mathbb{D}}|\nabla\log|\nabla
f_{1}||^{2}|dz|^{2}+\int_{\mathbb{D}}|\nabla\log|\nabla
f_{2}||^{2}|dz|^{2}+\int_{\mathbb{D}}\log|z||\nabla
f_{1}|^{2}|dz|^{2}+\int_{\mathbb{D}}\log|z||\nabla f_{2}|^{2}|dz|^{2}+4\pi$
$\displaystyle+4\pi\log|\nabla f_{1}(0)|+4\pi\log|\nabla
f_{2}(0)|-12\pi\log(2)$
$\displaystyle=\int_{\mathbb{D}}\left|\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}-2\frac{f^{\prime}(z)}{f(z)}\frac{|f(z)|^{2}}{1+|f(z)|^{2}}\right|^{2}|dz|^{2}+\int_{\mathbb{C}\setminus\overline{\mathbb{\mathbb{D}}}}\left|\frac{{g}^{\prime\prime}(z)}{{g}^{\prime}(z)}-2\frac{{g}^{\prime}(z)}{{g}(z)}\frac{|{g}(z)|^{2}}{1+|{g}(z)|^{2}}+\frac{2}{z}\right|^{2}|dz|^{2}$
$\displaystyle+2\int_{\mathbb{D}}\log|z|\frac{4|f^{\prime}(z)|^{2}|dz|^{2}}{(1+|f(z)|^{2})^{2}}-2\int_{\mathbb{C}\setminus\overline{\mathbb{D}}}\log|z|\frac{4|{g}^{\prime}(z)|^{2}|dz|^{2}}{(1+|{g}(z)|^{2})^{2}}+4\pi$
$\displaystyle+4\pi\log|f^{\prime}(0)|-4\pi\log|\widetilde{g}^{\prime}(\infty)|-4\pi\log(1+|f(0)|^{2})$
$\displaystyle=S_{3}(\Gamma)$

which concludes the proof of the theorem. ∎

######  Remark 4.4.

If $\Gamma=S^{1}$, then we can take $f=\mathrm{Id}_{\mathbb{D}}$ and
$g=\mathrm{Id}_{\mathbb{C}\setminus\overline{\mathbb{D}}}$, and we compute

$\displaystyle S_{3}(\Gamma)$
$\displaystyle=\int_{\mathbb{D}}\left|\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}-2\frac{f^{\prime}(z)}{f(z)}\frac{|f(z)|^{2}}{1+|f(z)|^{2}}\right|^{2}|dz|^{2}+\int_{\mathbb{C}\setminus\overline{\mathbb{\mathbb{D}}}}\left|\frac{g^{\prime\prime}(z)}{g^{\prime}(z)}-2\frac{g^{\prime}(z)}{g(z)}\frac{|g(z)|^{2}}{1+|g(z)|^{2}}+\frac{2}{z}\right|^{2}|dz|^{2}$
$\displaystyle+2\int_{\mathbb{D}}\log|z|\frac{4|f^{\prime}(z)|^{2}|dz|^{2}}{(1+|f(z)|^{2})^{2}}-2\int_{\mathbb{C}\setminus\overline{\mathbb{D}}}\log|z|\frac{4|g^{\prime}(z)|^{2}|dz|^{2}}{(1+|g(z)|^{2})^{2}}+4\pi$
$\displaystyle+4\pi\log|f^{\prime}(0)|-4\pi\log|g^{\prime}(\infty)|-4\pi\log(1+|f(0)|^{2})$
$\displaystyle=8\int_{\mathbb{D}}\frac{|z|^{2}|dz|^{2}}{(1+|z|^{2})^{2}}+16\int_{\mathbb{D}}\log|z|\frac{|z|^{2}|dz|^{2}}{(1+|z|^{2})^{2}}+4\pi$
$\displaystyle=16\pi\int_{0}^{1}\frac{r^{3}dr}{(1+r^{2})^{2}}+32\pi\int_{0}^{1}\frac{r\log(r)dr}{(1+r^{2})^{2}}+4\pi$
$\displaystyle=16\pi\left(\frac{1}{4}(2\log(2)-1)\right)+32\pi\left(-\frac{1}{4}\log(2)\right)+4\pi$
$\displaystyle=0$

as expected.

In the next theorem, we finally complete the proof of (4.2) by showing that
$\pi I^{L}(\Gamma)=S_{3}(\Gamma)$.

######  Theorem 4.5.

Let $\Gamma\subset S^{2}$ be a closed simple curve of finite Loewner energy.
Then we have

$\displaystyle I^{L}(\Gamma)=\frac{1}{\pi}\mathscr{E}_{0}(\Gamma),$

where $\mathscr{E}_{0}$ is defined in (1.12). Furthermore, if
$\Omega_{1},\Omega_{2}\subset S^{2}\setminus\Gamma$ are the two connected
components of $S^{2}\setminus\Gamma$, for all conformal maps
$f_{1}:\mathbb{D}\rightarrow\Omega_{1}$ and
$f_{2}:\mathbb{D}\rightarrow\Omega_{2}$, we have

$\displaystyle I^{L}(\Gamma)$
$\displaystyle=\frac{1}{\pi}\sum_{j=1}^{2}\left(\int_{\mathbb{D}}|\nabla\log|\nabla
f_{j}||^{2}|dz|^{2}+\int_{\mathbb{D}}\log|z||\nabla
f_{j}|^{2}|dz|^{2}+\mathrm{Area}(\Omega_{j})+4\pi\log|\nabla
f_{j}(0)|\right)-12\log(2)$

Figure 2: Spherical formula for the Loewner energy with respect to conformal
maps.

###### Proof.

By Theorem 2.6, we have the identity
$I^{L}(\Gamma)=\frac{1}{\pi}\mathscr{E}_{0}(\Gamma)$ for all smooth $\Gamma$,
and by the preceding Theorem 4.3, we have
$\mathscr{E}_{0}(\Gamma)=S_{3}(\Gamma)$ for any Jordan curve $\Gamma$ of
finite Loewner energy. Therefore, we will prove that
$I^{L}=\frac{1}{\pi}S_{3}$ which will imply our result.

We now let $\Omega_{1},\Omega_{2}\subset S^{2}\setminus{\Gamma}$ be the two
connected components of $S^{2}\setminus\Gamma$, and
$f_{1}:\mathbb{D}\rightarrow\Omega_{1}$,
$f_{2}:\mathbb{D}\rightarrow\Omega_{2}$ be the two conformal maps associated
to $\Omega_{1}$ and $\Omega_{2}$, and let $p_{1}=f_{1}(0)$ and
$p_{2}=f_{2}(0)$. Up to a rotation on $S^{2}$ (which does not change any of
the energies considered), we can assume that $p_{2}=N$. If
$\pi:S^{2}\setminus\left\\{N\right\\}\rightarrow\mathbb{C}$ is the standard
stereographic projection, let $\gamma=\pi(\Gamma)$, and $\Omega$ the bounded
component of $\mathbb{C}\setminus\gamma$ and define $f=\pi\circ
f_{1}:\mathbb{D}\rightarrow\pi(\Omega_{1})=\Omega$ and $g=\pi\circ
f_{2}\circ{\mathfrak{i}}:\mathbb{C}\setminus\overline{\mathbb{D}}\rightarrow\mathbb{C}\setminus\overline{\Omega}$
such that (using Theorem 4.3)

$\displaystyle\mathscr{E}_{0}(\Gamma)$
$\displaystyle=S_{3}(\gamma)=\int_{\mathbb{D}}\left|\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}-2\frac{f^{\prime}(z)}{f(z)}\frac{|f(z)|^{2}}{1+|f(z)|^{2}}\right|^{2}|dz|^{2}+\int_{\mathbb{C}\setminus\overline{\mathbb{\mathbb{D}}}}\left|\frac{g^{\prime\prime}(z)}{g^{\prime}(z)}-2\frac{g^{\prime}(z)}{g(z)}\frac{|g(z)|^{2}}{1+|g(z)|^{2}}+\frac{2}{z}\right|^{2}|dz|^{2}$
$\displaystyle+2\int_{\mathbb{D}}\log|z|\frac{4|f^{\prime}(z)|^{2}|dz|^{2}}{(1+|f(z)|^{2})^{2}}-2\int_{\mathbb{C}\setminus\overline{\mathbb{D}}}\log|z|\frac{4|g^{\prime}(z)|^{2}|dz|^{2}}{(1+|g(z)|^{2})^{2}}+4\pi$
$\displaystyle+4\pi\log|f^{\prime}(0)|-4\pi\log|g^{\prime}(\infty)|-4\pi\log(1+|f(0)|^{2}).$
(4.17)

Now, by Corollary A.$4$ of [14] and Theorem $8.1$ [41], if
$\left\\{\gamma_{n}\right\\}_{n\in\mathbb{N}}$ is a sequence of smooth curves
converging uniformly to a simple curve $\gamma$ and such that for a sequence
of maps $f_{n}:\mathbb{D}\rightarrow\mathbb{C}$ such that $f_{n}(0)=0$,
$f_{n}^{\prime}(0)=1$ and $f_{n}(\mathbb{D})=\Omega_{n}$, where $\Omega_{n}$
is the bounded component of $\mathbb{C}\setminus\gamma_{n}$, and satisfies

$\displaystyle\lim\limits_{n\rightarrow\infty}\int_{\mathbb{D}}\left|\frac{f_{n}^{\prime\prime}(z)}{f_{n}^{\prime}(z)}-\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}\right|^{2}|dz|^{2}=0,$

where $f:\mathbb{D}\rightarrow\Omega$ is a univalent function such that
$f(0)=0$ and $f^{\prime}(0)=1$, we have

$\displaystyle
I^{L}(\Gamma_{n})\underset{n\rightarrow\infty}{\longrightarrow}I^{L}(\Gamma).$
(4.18)

In particular, for any sequence of holomorphic maps
$g_{n}:\mathbb{C}\setminus\overline{\mathbb{D}}\rightarrow\mathbb{C}\setminus\overline{\Omega_{n}}$
such that $g_{n}(\infty)=\infty$, since

$\displaystyle S_{1}(\Gamma_{n})=\pi\,I^{L}(\Gamma_{n})$
$\displaystyle=\int_{\mathbb{D}}\left|\frac{f^{\prime\prime}_{n}(z)}{f_{n}^{\prime}(z)}\right|^{2}|dz|^{2}+\int_{\mathbb{C}\setminus\overline{\mathbb{D}}}\left|\frac{g_{n}^{\prime\prime}(z)}{g_{n}^{\prime}(z)}\right|^{2}|dz|^{2}+4\pi\log|f_{n}^{\prime}(0)|-4\pi\log|g_{n}^{\prime}(\infty)|$
$\displaystyle=\int_{\mathbb{D}}\left|\frac{f^{\prime\prime}_{n}(z)}{f_{n}^{\prime}(z)}\right|^{2}|dz|^{2}+\int_{\mathbb{C}\setminus\overline{\mathbb{D}}}\left|\frac{g_{n}^{\prime\prime}(z)}{g_{n}^{\prime}(z)}\right|^{2}|dz|^{2}-4\pi\log|g_{n}^{\prime}(\infty)|$

we deduce that

$\displaystyle\int_{\mathbb{C}\setminus\overline{\mathbb{D}}}\left|\frac{g_{n}^{\prime\prime}(z)}{g_{n}^{\prime}(z)}\right|^{2}|dz|^{2}-4\pi\log|g_{n}^{\prime}(\infty)|\underset{n\rightarrow\infty}{\longrightarrow}\int_{\mathbb{C}\setminus\overline{\mathbb{D}}}\left|\frac{g^{\prime\prime}(z)}{g^{\prime}(z)}\right|^{2}|dz|^{2}-4\pi\log|g^{\prime}(\infty)|$
(4.19)

for all univalent function
$g:\mathbb{C}\setminus\overline{\mathbb{D}}\rightarrow\mathbb{C}\setminus\overline{\Omega}$
such that $g(\infty)=\infty$. Now, if $\gamma=\pi(\Gamma)\subset\mathbb{C}$,
let $\left\\{\varepsilon_{n}\right\\}_{n\in\mathbb{N}}\subset(0,\infty)$ such
that $\varepsilon_{n}\underset{n\rightarrow\infty}{\longrightarrow}0$, and
define

$\displaystyle f_{n}:\mathbb{D}$ $\displaystyle\rightarrow\mathbb{C}$
$\displaystyle z$ $\displaystyle\mapsto
f((1-\varepsilon_{n})z)/{(1-\varepsilon_{n})}.$

Then $\gamma_{n}=f_{n}(S^{1})$ is smooth and uniformly converges to $\gamma$.
Furthermore, we have

$\displaystyle\int_{\mathbb{D}}\left|\frac{f^{\prime\prime}_{n}(z)}{f^{\prime}_{n}(z)}-\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}\right|^{2}|dz|^{2}\underset{n\rightarrow\infty}{\longrightarrow}0.$
(4.20)

which implies that

$\displaystyle
I^{L}(\gamma_{n})=\frac{1}{\pi}S_{1}(\gamma_{n})\underset{n\rightarrow\infty}{\longrightarrow}\frac{1}{\pi}S_{1}(\gamma)=I^{L}(\gamma).$

Now, we need to show the result
$f_{n}^{\prime}\underset{n\rightarrow{\infty}}{\longrightarrow}f^{\prime}$ in
$L^{2}(\mathbb{D})$ strongly. Notice that since $f^{\prime}$ is smooth in
$\mathbb{D}$, we have by construction
$f_{n}^{\prime}\underset{n\rightarrow\infty}{\longrightarrow}f^{\prime}$
almost everywhere. Furthermore, a linear change of variable shows that

$\displaystyle\int_{\mathbb{D}}|f_{n}^{\prime}(z)|^{2}|dz|^{2}=\int_{\mathbb{D}}|f^{\prime}((1-\varepsilon_{n})z))|^{2}|dz|^{2}$
$\displaystyle=\frac{1}{(1-\varepsilon_{n})^{2}}\int_{\mathbb{D}(0,1-\varepsilon_{n})}|f^{\prime}(w)|^{2}|dw|^{2}\underset{n\rightarrow\infty}{\longrightarrow}\int_{\mathbb{D}}|f^{\prime}(w)|^{2}|dw|^{2}.$

By Brezis-Lieb lemma ([24]), since
$f_{n}^{\prime}\underset{n\rightarrow\infty}{\longrightarrow}f^{\prime}$
almost everywhere and
$\left\|f_{n}^{\prime}\right\|_{\mathrm{L}^{2}(\mathbb{D})}\underset{n\rightarrow\infty}{\longrightarrow}\left\|f^{\prime}\right\|_{\mathrm{L}^{2}(\mathbb{D})}$,
we deduce that

$\displaystyle
f_{n}^{\prime}\underset{n\rightarrow\infty}{\longrightarrow}f^{\prime}\quad\text{strongly
in}\;\,L^{2}(\mathbb{D}).$ (4.21)

Therefore, we also get the convergence

$\displaystyle\begin{aligned}
\frac{f_{n}^{\prime}}{f_{n}}\frac{|f_{n}|^{2}}{1+|f_{n}|^{2}}&\underset{n\rightarrow\infty}{\longrightarrow}\frac{f^{\prime}}{f}\frac{|f|^{2}}{1+|f|^{2}}\quad&&\text{in}\;\,L^{2}(\mathbb{D})\end{aligned}$

which finally shows by (4.20) that

$\displaystyle\int_{\mathbb{D}}\left|\frac{f_{n}^{\prime\prime}(z)}{f_{n}^{\prime}(z)}-2\frac{f_{n}^{\prime}(z)}{f_{n}(z)}\frac{|f_{n}(z)|^{2}}{1+|f_{n}(z)|^{2}}\right|^{2}|dz|^{2}$
$\displaystyle\underset{n\rightarrow\infty}{\longrightarrow}\int_{\mathbb{D}}\left|\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}-2\frac{f^{\prime}(z)}{f(z)}\frac{|f(z)|^{2}}{1+|f(z)|^{2}}\right|^{2}|dz|^{2}$
$\displaystyle\int_{\mathbb{D}}\log|z|\frac{4|f_{n}^{\prime}(z)|^{2}|dz|^{2}}{(1+|f_{n}(z)|^{2})^{2}}$
$\displaystyle\underset{n\rightarrow\infty}{\longrightarrow}\int_{\mathbb{D}}\log|z|\frac{4|f^{\prime}(z)|^{2}|dz|^{2}}{(1+|f(z)|^{2})^{2}}.$
(4.22)

Finally, we also have $f_{n}(0)=f(0)$ and

$\displaystyle
4\pi\log|f_{n}^{\prime}(0)|=4\pi\log|f^{\prime}(0)|+4\pi\log(1-\varepsilon_{n})\underset{n\rightarrow\infty}{\longrightarrow}4\pi\log|f^{\prime}(0)|.$
(4.23)

Therefore, if $\Omega_{n}=f_{n}(\mathbb{D})$, and
$g_{n}:\mathbb{C}\setminus\overline{\mathbb{D}}\rightarrow\mathbb{C}\setminus\overline{\Omega_{n}}$
is any univalent map such that $g_{n}(\infty)=\infty$, since
$\gamma_{n}\underset{n\rightarrow\infty}{\longrightarrow}\gamma$ uniformly, we
can assume without loss of generality that
$g_{n}^{\prime}(\infty)\underset{n\rightarrow\infty}{\longrightarrow}g^{\prime}(\infty)$.
Furthermore, by Corollary A.$4$ of [14], we also get

$\displaystyle\lim\limits_{n\rightarrow\infty}\int_{\mathbb{C}\setminus\overline{\mathbb{D}}}\left|\frac{g_{n}^{\prime\prime}(z)}{g_{n}^{\prime}(z)}-\frac{g^{\prime\prime}(z)}{g(z)}\right|^{2}=0$
(4.24)
$\displaystyle\lim\limits_{n\rightarrow\infty}\int_{\mathbb{C}\setminus\overline{\mathbb{D}}}\left|\left(\frac{g_{n}^{\prime\prime}(z)}{g_{n}^{\prime}(z)}-2\frac{g_{n}^{\prime}(z)}{g_{n}(z)}+\frac{2}{z}\right)-\left(\frac{g^{\prime\prime}(z)}{g^{\prime}(z)}-2\frac{g^{\prime}(z)}{g(z)}+\frac{2}{z}\right)\right|^{2}|dz|^{2}=0.$
(4.25)

As previously, we have

$\displaystyle\frac{g_{n}^{\prime}}{g_{n}}\frac{1}{1+|g_{n}|^{2}}$
$\displaystyle\underset{n\rightarrow\infty}{\longrightarrow}\frac{g^{\prime}}{g}\frac{1}{1+|g|^{2}}\quad\text{in}\;\,L^{2}(\mathbb{C}\setminus\overline{\mathbb{D}})$
$\displaystyle{g_{n}^{\prime}}\frac{1}{1+|g_{n}|^{2}}$
$\displaystyle\underset{n\rightarrow\infty}{\longrightarrow}{g^{\prime}}\frac{1}{1+|g|^{2}}\quad\text{in}\;\,L^{2}(\mathbb{C}\setminus\overline{\mathbb{D}})$
(4.26)

Therefore, (4.24) and (4) imply that

$\displaystyle\int_{\mathbb{C}\setminus\overline{\mathbb{\mathbb{D}}}}\left|\frac{g_{n}^{\prime\prime}(z)}{g_{n}^{\prime}(z)}-2\frac{g_{n}^{\prime}(z)}{g_{n}(z)}\frac{|g_{n}(z)|^{2}}{1+|g_{n}(z)|^{2}}+\frac{2}{z}\right|^{2}|dz|^{2}$
$\displaystyle\underset{n\rightarrow\infty}{\longrightarrow}\int_{\mathbb{C}\setminus\overline{\mathbb{\mathbb{D}}}}\left|\frac{g^{\prime\prime}(z)}{g^{\prime}(z)}-2\frac{g^{\prime}(z)}{g(z)}\frac{|g(z)|^{2}}{1+|g(z)|^{2}}+\frac{2}{z}\right|^{2}|dz|^{2}$
$\displaystyle\int_{\mathbb{C}\setminus\overline{\mathbb{\mathbb{D}}}}\log|z|\frac{4|g_{n}^{\prime}(z)|^{2}|dz|^{2}}{(1+|g_{n}(z)|^{2})^{2}}$
$\displaystyle\underset{n\rightarrow\infty}{\longrightarrow}\int_{\mathbb{C}\setminus\overline{\mathbb{\mathbb{D}}}}\log|z|\frac{4|g^{\prime}(z)|^{2}|dz|^{2}}{(1+|g(z)|^{2})^{2}}.$
(4.27)

Finally, we deduce by (4), (4), (4.23) and (4) that

$\displaystyle
S_{3}(\gamma_{n})\underset{n\rightarrow\infty}{\longrightarrow}S_{3}(\gamma)$

which concludes the proof of the theorem by (4.18), Theorem 2.6 and Theorem
4.3. ∎

######  Remarks 4.6.

Notice that we can also directly express the Loewner energy using moving
frames. First, we trivially have

$\displaystyle I^{L}(\Gamma)$
$\displaystyle=\frac{1}{\pi}\left\\{\sum_{i=1}^{2}\int_{\Omega_{i}}|\omega_{i}-\ast\,dG_{\Omega_{i}}|_{g_{0}}^{2}+2\int_{\Omega_{i}}G_{\Omega_{i}}K_{g_{0}}d\mathrm{vol}_{g_{0}}+\mathrm{Area}(\Omega_{i})\right\\}$
$\displaystyle+4\log|\nabla f_{1}(0)|+4\log|\nabla f_{2}(0)|-12\,\log(2).$

Alternatively, we have

$\displaystyle I^{L}(\Gamma)$
$\displaystyle=\frac{1}{2{\pi}}\int_{\Omega_{1}}\left(|d\vec{u}_{1}|^{2}_{g_{0}}+|d\vec{v}_{1}|^{2}_{g_{0}}-2|dG_{\Omega_{1}}|^{2}_{g_{0}}\right)d\mathrm{vol}_{g_{0}}+\frac{1}{2{\pi}}\int_{\Omega_{2}}\left(|d\vec{u}_{2}|^{2}_{g_{0}}+|d\vec{v}_{2}|^{2}_{g_{0}}-2|dG_{\Omega_{2}}|^{2}_{g_{0}}\right)d\mathrm{vol}_{g_{0}}$
$\displaystyle+4\log|\nabla f_{1}(0)|+4\log|\nabla f_{2}(0)|-12\log(2),$

which is (up to the second line involving the conformal maps $f_{1}$ and
$f_{2}$) very reminiscent of the Ginzburg-Landau renormalised energy ([3,
Chapter VIII]).

To see this equality, since $\vec{u}_{1}$, $\vec{v}_{1}$ and $\vec{n}$ are
unitary, we have

$\displaystyle|d\vec{u}_{1}|^{2}_{g_{0}}$ $\displaystyle=|\langle
d\vec{u}_{1},\vec{v}_{1}\rangle|_{g_{0}}^{2}+|\langle
d\vec{u}_{1},\vec{n}\rangle|_{g_{0}}^{2}=|\omega_{1}|^{2}_{g_{0}}+|\langle
d\vec{n},\vec{u}_{1}\rangle|_{g_{0}}^{2}$
$\displaystyle|d\vec{v}_{1}|^{2}_{g_{0}}$
$\displaystyle=|\omega_{1}|^{2}_{g_{0}}+|\langle
d\vec{n},\vec{v}_{1}\rangle|_{g_{0}}^{2}$
$\displaystyle|d\vec{u}_{1}|^{2}_{g_{0}}+|d\vec{v}_{1}|_{g_{0}}^{2}$
$\displaystyle=2|\omega_{1}|^{2}_{g_{0}}+|d\vec{n}|_{g_{0}}^{2}=2|\omega_{1}|_{g_{0}}^{2}+2.$

Then, integrating by parts and using that $G_{\Omega_{1}}=0$ on
$\partial\Omega_{1}$, we deduce by Stokes theorem—and the equation (that
follows from (3.8))

$\displaystyle
d\left(\omega_{1}-\ast\,dG_{\Omega_{1}}\right)=-K_{g_{0}}d\mathrm{Area}_{g_{0}},$

where $K_{g_{0}}=1$ is the Gauss curvature of the sphere—that

$\displaystyle\frac{1}{2}\int_{\Omega_{1}}\left(|d\vec{u}_{1}|^{2}_{g_{0}}+|d\vec{v}_{1}|_{g_{0}}^{2}-2|dG_{\Omega_{1}}|_{g_{0}}^{2}\right)d\mathrm{vol}_{g_{0}}=\int_{\Omega_{1}}\left(|\omega_{1}|_{g_{0}}^{2}-|dG_{\Omega_{1}}|^{2}_{g_{0}}\right)d\mathrm{vol}_{g_{0}}+\mathrm{Area}_{g_{0}}(\Omega_{1})$
$\displaystyle=\int_{\Omega_{1}}\langle\omega_{1}-\ast\,dG_{\Omega_{1}},\omega_{1}+\ast\,dG_{\Omega_{1}}\rangle_{g_{0}}d\mathrm{vol}_{g_{0}}+\mathrm{Area}_{g_{0}}(\Omega_{1})$
$\displaystyle=\int_{\Omega_{1}}|\omega_{1}-\ast\,dG_{\Omega_{1}}|_{g_{0}}^{2}d\mathrm{vol}_{g_{0}}+2\int_{\Omega_{1}}(\omega_{1}-\ast\,dG_{\Omega_{1}})\wedge
dG_{\Omega_{1}}+\mathrm{Area}_{g_{0}}(\Omega_{1})$
$\displaystyle=\int_{\Omega_{1}}|\omega_{1}-\ast\,dG_{\Omega_{1}}|_{g_{0}}^{2}{d\mathrm{vol}_{g_{0}}}-2\int_{\Omega_{1}}G_{\Omega_{1}}d(\omega_{1}-\ast\,dG_{\Omega_{1}})+\mathrm{Area}_{g_{0}}(\Omega_{1})$
$\displaystyle=\int_{\Omega_{1}}|d\mu_{1}|_{g_{0}}^{2}d\mathrm{vol}_{g_{0}}+2\int_{\Omega_{1}}G_{\Omega_{1}}K_{g_{0}}d\mathrm{vol}_{g_{0}}+\mathrm{Area}_{g_{0}}(\Omega_{1})$

which implies since $\mathrm{Area}_{g_{0}}(S^{2})=4\pi$ that

$\displaystyle\frac{1}{2}\int_{\Omega_{1}}\left(|d\vec{u}_{1}|^{2}_{g_{0}}+|d\vec{v}_{1}|^{2}_{g_{0}}-2|dG_{\Omega_{1}}|^{2}_{g_{0}}\right)d\mathrm{vol}_{g_{0}}+\frac{1}{2}\int_{\Omega_{1}}\left(|d\vec{u}_{2}|^{2}_{g_{0}}+|d\vec{v}_{2}|^{2}_{g_{0}}-2|dG_{\Omega_{2}}|^{2}_{g_{0}}\right)d\mathrm{vol}_{g_{0}}$
$\displaystyle=\int_{\Omega_{1}}|d\mu_{1}|_{g_{0}}^{2}d\mathrm{vol}_{g_{0}}+\int_{\Omega_{2}}|d\mu_{2}|_{g_{0}}^{2}+2\int_{\Omega_{1}}G_{\Omega_{1}}K_{g_{0}}d\mathrm{vol}_{g_{0}}+2\int_{\Omega_{2}}G_{\Omega_{2}}K_{g_{0}}d\mathrm{vol}_{g_{0}}+4\pi.$

Notice that it gives another explanation for the factor $4\pi$ in the
definition of $\mathscr{E}$.

##  5 Appendix

In this appendix, we provide more details on the geodesic curvature for Weil-
Petersson quasicircles and show a consequence of Theorem 4.5 which is an
identity on univalent functions associated to a Weil-Petersson quasicircle.

###  5.1 Properties of the Geodesic Curvature for Weil-Petersson Quasicircles

######  Lemma 5.1.

Let $\mathbb{H}=\mathbb{C}\cap\left\\{z:\mathrm{Im}\,(z)>0\right\\}$ be the
Poincaré half-plane, and $f:\mathbb{H}\rightarrow\mathbb{C}$ a univalent
holomorphic map, $\Omega=f(\mathbb{H})$, and assume that
$\gamma=\partial\Omega$ is a simple curve of finite Loewner energy. Then the
geodesic curvature $k_{g_{0}}$ of $\gamma$ is given in the distributional
sense by

$\displaystyle
k_{g_{0}}=\mathrm{Im}\,\left(\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}\right)\qquad\text{for
all}\;\,z\in\partial_{\infty}\mathbb{H}=\mathbb{R}.$ (5.1)

###### Proof.

The geodesic curvature is given by

$\displaystyle k_{g_{0}}=\langle\partial_{x}\vec{u},\vec{v}\rangle,$

if $(\vec{u},\vec{v})$ is the Cartesian frame given by (in the following
formulae, $f$ is seen as a $\mathbb{R}^{2}$-valued function)

$\displaystyle\left\\{\begin{aligned}
\vec{u}&=\frac{\partial_{x}f}{|\partial_{x}f|}=\left(\mathrm{Re}\,\left(\frac{f^{\prime}(z)}{|f^{\prime}(z)|}\right),\mathrm{Im}\,\left(\frac{f^{\prime}(z)}{|f^{\prime}(z)|}\right)\right)=\frac{f^{\prime}(z)}{|f^{\prime}(z)|}\\\
\vec{v}&=\frac{\partial_{y}f}{|\partial_{y}f|}=\left(-\mathrm{Im}\,\left(\frac{f^{\prime}(z)}{|f^{\prime}(z)|}\right),\mathrm{Re}\,\left(\frac{f^{\prime}(z)}{|f^{\prime}(z)|}\right)\right)=i\,\frac{f^{\prime}(z)}{|f^{\prime}(z)|}.\end{aligned}\right.$

Define $\vec{u}_{z}=\dfrac{f^{\prime}(z)}{|f^{\prime}(z)|}$. Then we have

$\displaystyle\partial_{z}\vec{u}_{z}$
$\displaystyle=\partial_{z}\left(\frac{f^{\prime}(z)}{|f^{\prime}(z)|}\right)=\frac{f^{\prime\prime}(z)}{|f^{\prime}(z)|}-\frac{1}{2}\frac{f^{\prime\prime}(z)}{|f^{\prime}(z)|}=\frac{1}{2}\frac{f^{\prime\prime}(z)}{|f^{\prime}(z)|}=\frac{1}{2}\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}\vec{u}_{z}$
$\displaystyle\partial_{\overline{z}}\vec{u}_{z}$
$\displaystyle=\partial_{\overline{z}}\left(\frac{f^{\prime}(z)}{|f^{\prime}(z)|}\right)=-\frac{1}{2}\frac{f^{\prime}(z)^{2}\overline{f^{\prime\prime}(z)}}{|f^{\prime}(z)|^{3}}=-\frac{1}{2}\overline{\left(\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}\right)}\vec{u}_{z}.$

Therefore, we deduce that

$\displaystyle\partial_{z}\mathrm{Re}\,\left(\frac{f^{\prime}(z)}{|f^{\prime}(z)|}\right)$
$\displaystyle=\frac{1}{2}\left(\partial_{z}\left(\frac{f^{\prime}(z)}{|f^{\prime}(z)|}\right)+\overline{\partial_{\overline{z}}\left(\frac{{f^{\prime}(z)}}{|f^{\prime}(z)|}\right)}\right)=\frac{1}{4}\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}\left(\vec{u}_{z}-\vec{u}_{\overline{z}}\right)=\frac{i}{2}\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}\mathrm{Im}\,\left(\frac{f^{\prime}(z)}{|f^{\prime}(z)|}\right)$
$\displaystyle\partial_{z}\mathrm{Im}\,\left(\frac{f^{\prime}(z)}{|f^{\prime}(z)|}\right)$
$\displaystyle=-\frac{i}{2}\left(\frac{1}{2}\frac{f^{\prime\prime}(z)}{|f^{\prime}(z)|}+\frac{1}{2}\overline{\frac{f^{\prime\prime}(z)}{|f^{\prime}(z)|}}\right)=-\frac{i}{4}\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}\left(\vec{u}_{z}+\vec{u}_{\overline{z}}\right)=-\frac{i}{2}\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}\mathrm{Re}\,\left(\frac{f^{\prime}(z)}{|f^{\prime}(z)|}\right)$

where we used

Therefore, we have

$\displaystyle\partial_{x}\mathrm{Re}\,\left(\frac{f^{\prime}(z)}{|f^{\prime}(z)|}\right)$
$\displaystyle=2\,\mathrm{Re}\,\left(\partial_{z}\mathrm{Re}\,\left(\frac{f^{\prime}(z)}{|f^{\prime}(z)|}\right)\right)=-\mathrm{Im}\,\left(\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}\right)\mathrm{Im}\,\left(\frac{f^{\prime}(z)}{|f^{\prime}(z)|}\right)$
$\displaystyle\partial_{x}\mathrm{Im}\,\left(\frac{f^{\prime}(z)}{|f^{\prime}(z)|}\right)$
$\displaystyle=\mathrm{Im}\,\left(\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}\right)\mathrm{Re}\,\left(\frac{f^{\prime}(z)}{f^{\prime}(z)}\right),$

so that
$\partial_{x}\vec{u}=\mathrm{Im}\,\left(\dfrac{f^{\prime\prime}(z)}{f^{\prime}(z)}\right)\vec{v},$
and

$\displaystyle
k_{g_{0}}=\langle\partial_{x}\vec{u},\vec{v}\rangle=\mathrm{Im}\,\left(\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}\right),$

which concludes the proof of the lemma. ∎

######  Lemma 5.2.

Let $\gamma\subset\mathbb{C}$ be a Weil-Petersson quasicircle. Then the
geodesic curvature $k_{g_{0}}:S^{1}\rightarrow\mathbb{R}$ is a tempered
distribution of order at most $2$. More precisely, we have $k_{g_{0}}\in
H^{-1/2}(S^{1})$.

###### Proof.

Either using the Poincaré half-plane $\mathbb{H}$ and the formula (5.1) or
(3), we get

$\displaystyle
k_{g_{0}}=\mathrm{Re}\,\left(z\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}\right)+1.$
(5.2)

Now, if $0<\varepsilon<1$ and $f_{\varepsilon}:\mathbb{D}\rightarrow\Omega$ is
defined by

$\displaystyle
f_{\varepsilon}(z)=\frac{1}{1-\varepsilon}f((1-\varepsilon)z)\qquad
z\in\mathbb{D},$

we have (see [41], Lemma $8.2$)

$\displaystyle\lim\limits_{\varepsilon\rightarrow
0}\int_{\mathbb{D}}\left|\frac{f_{\varepsilon}^{\prime\prime}(z)}{f_{\varepsilon}^{\prime}(z)}-\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}\right|^{2}|dz|^{2}=0,$

which is equivalent by trace theory to

$\displaystyle\lim\limits_{\varepsilon\rightarrow
0}\left\|\log|f^{\prime}_{\varepsilon}|-\log|f^{\prime}|\right\|_{\mathrm{H}^{{1}/{2}}(S^{1})}=0,$
$\displaystyle\lim\limits_{\varepsilon\rightarrow
0}\left\|\arg(f^{\prime}_{\varepsilon})-\arg(f^{\prime})\right\|_{\mathrm{H}^{{1}/{2}}(S^{1})}=0.$

and using the equivalent norm for $H^{s}$ spaces given by

$\displaystyle\left\|u\right\|_{\mathrm{H}^{s}(S^{1})}=\sum_{n\in\mathbb{Z}}|n|^{2s}|a_{n}|^{2},\qquad\text{if}\;\,u(z)=\sum_{n\in\mathbb{Z}}a_{n}z^{n}.$

we deduce that

$\displaystyle\lim\limits_{\varepsilon\rightarrow
0}\left\|\partial_{\theta}\log(f_{\varepsilon}^{\prime})-\partial_{\theta}\log(f^{\prime})\right\|_{\mathrm{H}^{-{1}/{2}}(S^{1})}=0.$

Since

$\displaystyle\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}=\frac{1}{ie^{i\theta}}\partial_{\theta}\log(f^{\prime}(z)),$

we deduce that $\dfrac{f^{\prime\prime}}{f^{\prime}}\in H^{-1/2}(S^{1})$ and
that concludes the proof of the lemma by (5.2). ∎

######  Remark 5.3.

For other considerations related to trace spaces, see the Definition $5$ of
[7] and [5].

######  Proposition 5.4.

For all $\varepsilon>0$, let
$\mathbb{D}_{+}(0,\varepsilon)=\mathbb{H}\cap\left\\{z:|z|<\varepsilon\right\\}$,
where $\mathbb{H}=\mathbb{C}\cap\left\\{z:\mathrm{Im}\,(z)>0\right\\}$ is the
Poincaré half-plane. For $\varepsilon>0$ small enough, the map
$f:\mathbb{D}_{+}(0,\varepsilon)\rightarrow\mathbb{C}$ defined by

$\displaystyle f(z)=z\,e^{i\log\log(z)},$

where $\log(z)$ is the principal value of the logarithm on $\mathbb{H}$, is an
immersion and $\log|f^{\prime}|\in W^{1,2}(\mathbb{D}_{+}(0,\varepsilon))$. In
particular, the curve $\gamma:(-\varepsilon,\varepsilon)\rightarrow\mathbb{C}$
such that $\gamma(t)=te^{i\log\log(t)}$ for all
$t\in(-\varepsilon,\varepsilon)$ is a part of a Weil-Petersson
quasicircle.***Beware that the $\log$ function here is defined as the trace of
our continuous determination of the logarithm on the upper-half plane and is
not the standard $\log$ function on $(0,\infty)$. Furthermore, its geodesic
curvature $k_{g_{0}}$ is given by

$\displaystyle
k_{g_{0}}=-\frac{1}{t(1+\log^{2}(t))}+\mathrm{p.v.}\int_{-\varepsilon}^{\varepsilon}\frac{dt}{t\log(t)}.$

###### Proof.

We compute

$\displaystyle f^{\prime}(z)$
$\displaystyle=e^{i\log\log(z)}+\frac{i}{\log(z)}e^{i\log\log(z)}=\left(1+\frac{i}{\log(z)}\right)e^{i\log\log(z)}$
$\displaystyle f^{\prime\prime}(z)$
$\displaystyle=-\frac{i}{z\log^{2}(z)}e^{i\log\log(z)}+\frac{i}{z\log(z)}\left(1+\frac{i}{\log(z)}\right)e^{i\log\log(z)}.$

Therefore, we have

$\displaystyle\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}=-\frac{i}{z\log(z)(i+\log(z))}+\frac{i}{z\log(z)}=i\frac{-1+i+\log(z)}{z\log(z)(i+\log(z))}.$

Notice that we have

$\displaystyle|i+\log(z)|^{2}=|\log|z|+i(1+\arg(z))|^{2}=\log^{2}|z|+(1+\arg(z))^{2}$

Therefore, we have

$\displaystyle\left|\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}\right|^{2}=\frac{(\log|z|-1)^{2}+(1+\arg(z))^{2}}{|z|^{2}\log^{2}|z|(\log^{2}|z|+(1+\arg(z))^{2}}\leq\frac{1}{\log^{2}|z|}+\frac{-2\log|z|+1}{\log^{4}|z|}\leq\frac{2}{|z|^{2}\log^{4}|z|}+\frac{2}{|z|^{2}\log^{2}|z|},$

and

$\displaystyle\int_{\mathbb{D}(0,\frac{1}{2})}\left|\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}\right|^{2}|dz|^{2}\leq
4\pi\int_{0}^{\frac{1}{2}}\frac{dr}{r\log^{4}(r)}+4\pi\int_{0}^{\frac{1}{2}}\frac{dr}{r\log^{2}(r)}=\frac{4\pi}{\log(2)}+\frac{4\pi}{3\log^{2}(2)}<\infty,$

which shows that $\gamma$ is a Weil-Petersson quasicircle. Then, we have by
Lemma 5.1 for $z\in\mathbb{R}$

$\displaystyle
k_{g_{0}}=\mathrm{Im}\,\left(\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}\right)=\frac{1-\log|z|+\log^{2}|z|}{|z|\log|z|(1+\log^{2}|z|)}=-\frac{1}{|z|(1+\log^{2}|z|)}+\frac{1}{|z|\log|z|},$

which concludes the proof of the proposition. ∎

######  Remark 5.5.

In particular, we see that there exists curves whose geodesic curvature is a
distribution of order $1$. This curve is an example of spiral mentioned
earlier in the introduction.

###  5.2 A consequence of Theorem 4.5

The new identity of $\pi\,I^{L}=S_{3}$ from Theorem 4.3 and Theorem 4.5
provides a new identity about holomorphic univalent maps of the plane.

######  Lemma 5.6.

Let $\gamma\subset\mathbb{C}$ be a closed simple curve with finite Loewner
energy. We have

$\displaystyle
4\,\mathrm{Re}\,\int_{\mathbb{D}}\left(\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}-2\frac{f^{\prime}(z)}{f(z)}+\frac{2}{z}\right)\overline{\left(\frac{f^{\prime}(z)}{f(z)}\frac{1}{1+|f(z)|^{2}}-\frac{2}{z}\right)}|dz|^{2}+4\int_{\mathbb{D}}\left|\frac{f^{\prime}(z)}{f(z)}\frac{1}{1+|f(z)|^{2}}-\frac{1}{z}\right|^{2}|dz|^{2}$
$\displaystyle+4\,\mathrm{Re}\,\int_{\mathbb{C}\setminus\overline{\mathbb{D}}}\left(\frac{g^{\prime\prime}(z)}{g^{\prime}(z)}-2\frac{g^{\prime}(z)}{g(z)}+\frac{2}{z}\right)\overline{\left(\frac{g^{\prime}(z)}{g(z)}\frac{1}{1+|g(z)|^{2}}\right)}|dz|^{2}+4\int_{\mathbb{C}\setminus\overline{\mathbb{D}}}\left|\frac{g^{\prime}(z)}{g(z)}\frac{1}{1+|g(z)|^{2}}\right|^{2}|dz|^{2}$
$\displaystyle+2\int_{\mathbb{D}}\log|z|\frac{4|f^{\prime}(z)|^{2}|dz|^{2}}{(1+|f(z)|^{2})^{2}}-2\int_{\mathbb{C}\setminus\overline{\mathbb{D}}}\log|z|\frac{4|g^{\prime}(z)|^{2}|dz|^{2}}{(1+|g(z)|^{2})^{2}}+4\pi=0,$

where $f$ and $g$ are univalent maps as in Definition 4.1.

###### Proof.

Recall the definition of $S_{3}(\gamma)$ in Definition 4.1:

$\displaystyle S_{3}(\gamma)$
$\displaystyle=\int_{\mathbb{D}}\left|\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}-2\frac{f^{\prime}(z)}{f(z)}\frac{|f(z)|^{2}}{1+|f(z)|^{2}}\right|^{2}|dz|^{2}+\int_{\mathbb{C}\setminus\overline{\mathbb{\mathbb{D}}}}\left|\frac{g^{\prime\prime}(z)}{g^{\prime}(z)}-2\frac{g^{\prime}(z)}{g(z)}\frac{|g(z)|^{2}}{1+|g(z)|^{2}}+\frac{2}{z}\right|^{2}|dz|^{2}$
$\displaystyle+2\int_{\mathbb{D}}\log|z|\frac{4|f^{\prime}(z)|^{2}|dz|^{2}}{(1+|f(z)|^{2})^{2}}-2\int_{\mathbb{C}\setminus\overline{\mathbb{D}}}\log|z|\frac{4|g^{\prime}(z)|^{2}|dz|^{2}}{(1+|g(z)|^{2})^{2}}+4\pi$
$\displaystyle+4\pi\log|f^{\prime}(0)|-4\pi\log|g^{\prime}(\infty)|$ (5.3)

and that $\pi I^{L}(\gamma)=S_{1}(\gamma)=S_{3}(\gamma)$ from Theorem 4.5.
Using the identity (1.5), we obtain

$\displaystyle S_{3}(\gamma)$ $\displaystyle=S_{3}(\mathfrak{i}(\gamma))$
(5.4)
$\displaystyle=\int_{\mathbb{D}}\left|\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}-2\frac{f^{\prime}(z)}{f(z)}+\frac{2}{z}+2\frac{f^{\prime}(z)}{f(z)}\frac{1}{1+|f(z)|^{2}}-\frac{2}{z}\right|^{2}|dz|^{2}$
$\displaystyle\quad+\int_{\mathbb{C}\setminus\overline{\mathbb{D}}}\left|\frac{g^{\prime\prime}(z)}{g^{\prime}(z)}-2\frac{g^{\prime}(z)}{g(z)}+\frac{2}{z}+2\frac{g^{\prime}(z)}{g(z)}\frac{1}{1+|g(z)|^{2}}\right|^{2}|dz|^{2}+4\pi\log\left\lvert\frac{f^{\prime}(0)}{g^{\prime}(\infty)}\right\rvert$
$\displaystyle\quad+2\int_{\mathbb{D}}\log|z|\frac{4|f^{\prime}(z)|^{2}|dz|^{2}}{(1+|f(z)|^{2})^{2}}-2\int_{\mathbb{C}\setminus\overline{\mathbb{D}}}\log|z|\frac{4|g^{\prime}(z)|^{2}|dz|^{2}}{(1+|g(z)|^{2})^{2}}+4\pi$
$\displaystyle=\int_{\mathbb{D}}\left|\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}-2\frac{f^{\prime}(z)}{f(z)}+\frac{2}{z}\right|^{2}|dz|^{2}+\int_{\mathbb{C}\setminus\overline{\mathbb{D}}}\left|\frac{g^{\prime\prime}(z)}{g^{\prime}(z)}-2\frac{g^{\prime}(z)}{g(z)}+\frac{2}{z}\right|^{2}|dz|^{2}+4\pi\log\left\lvert\frac{f^{\prime}(0)}{g^{\prime}(\infty)}\right\rvert$
$\displaystyle\quad+4\,\mathrm{Re}\,\int_{\mathbb{D}}\left(\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}-2\frac{f^{\prime}(z)}{f(z)}+\frac{2}{z}\right)\overline{\left(\frac{f^{\prime}(z)}{f(z)}\frac{1}{1+|f(z)|^{2}}-\frac{2}{z}\right)}|dz|^{2}$