text:
\section{Introduction}
\label{sec:introduction}
Let $(M,g)$ be a compact, two dimensional, oriented manifold
equipped with a smooth metric $g$ and
$k:M \to \rz$ a smooth positive function. We consider the following two equations
for curves $\gamma$ on $M$:
\begin{align}
\label{eq:magnetic_geodesic}
D_{t,g} \dot \gamma = k(\gamma) J_{g}(\gamma) \dot \gamma,
\end{align}
and
\begin{align}
\label{eq:1}
D_{t,g} \dot \gamma = |\dot \gamma|_{g} k(\gamma) J_{g}(\gamma) \dot \gamma,
\end{align}
where $D_{t,g}$ is the covariant derivative with respect to $g$,
and $J_g(x)$ is the rotation by $\pi/2$ in $T_xM$ measured with $g$ and
the given orientation.\\
Equation (\ref{eq:magnetic_geodesic}) describes the motion
of a charge in a magnetic field corresponding to the magnetic form
$k dV_g$
and solutions to \eqref{eq:magnetic_geodesic}
will be called {\em ($k$-)magnetic geodesics}
(see \cite{MR890489,MR676612}).
Equation (\ref{eq:1}) corresponds to the problem of {\em prescribing geodesic curvature},
as its solutions $\gamma$
are constant speed curves with geodesic curvature $k_g(\gamma,t)$ given by $k(\gamma(t))$
(see \cite{arXiv:0808.4038}).\\
It is easy to see that a nonconstant magnetic geodesic $\gamma$ lies in a fixed energy level
$E_c$, i.e. there is $c>0$, such that
\begin{align*}
(\gamma,\dot\gamma) \in E_c := \{(x,V) \in TM\where |V|_g=c\}.
\end{align*}
For fixed $k$ and $c>0$
the equations \eqref{eq:magnetic_geodesic} and \eqref{eq:1} are equivalent in the following
sense: If $\gamma$ is a nonconstant solution of \eqref{eq:1} with $k$ replaced by
$k/c$, then the curve
$\gamma_c(t):=\gamma(ct/|\dot \gamma|_g)$
is a $k$-magnetic geodesic in $E_c$, and a $k$-magnetic geodesic in $E_c$
solves \eqref{eq:1} with $k$ replaced by $k/c$. We emphazise that $k$-magnetic geodesics
on different energy levels are not reparameterizations of each other.\\
We study the existence of closed curves with prescribed geodesic curvature
or equivalently the existence of periodic magnetic geodesics on prescribed energy levels
$E_c$.\\
There is a vast literature on the existence of closed magnetic geodesics.
We limit ourselves to quote \cite{MR730159,MR1185286}
for the approach via Morse-Novikov theory
for (possibly multi-valued) variational functionals,
\cite{MR890489,MR902290} for the
application of the theory of dynamical
systems and symplectic geometry, \cite{MR2036336} concerning Aubry-Mather's theory,
and \cite{arXiv:0808.4038}, where the theory
of vector fields on infinite dimensional manifolds is applied to \eqref{eq:1}.
We refer to \cite{MR2036336,MR2534483,MR1432462,MR2593232} for a survey
and additional references.\\
From the example of the horocycle flow below, closed magnetic geodesics need not
exist on a fixed energy level in general. However, from \cite{MR1185286,MR1133303,MR1432462},
there are always closed magnetic geodesics
in high and low energy levels, i.e. in $E_c$ with $c\ge c_0$ and $c\le (c_0)^{-1}$,
where $c_0>0$ depends on $(M,g)$ and $k$ (in case of a flat torus and high energy levels
$k$ is assumed
not to vanish).
If the magnetic form is exact, i.e. $[k dV_g]=0$ in $H_{dR}^2(M)$,
then there is a periodic magnetic geodesic in every energy level (see \cite{MR2036336}).
Concerning non exact magnetic forms positive functions $k$ are of special
interest, since the magnetic form is symplectic in this case.
For $k>0$ a closed magnetic geodesic exists
in every energy level, if $(M,g)$ is a flat torus \cite{MR970068,MR786086}
or if $(M,g)$ is a sphere $S^2$ with nonnegative curvature
\cite{arXiv:0903.1128}. The (essentially) only nonexistence result
for closed magnetic geodesics is based on an old result of
Hedlund \cite{MR1545946}.
\begin{example*}[Horocycle flow \cite{MR1432462}]
Let $(M,g)$ be a compact hyperbolic surface of constant curvature $K_g\equiv -1$
and $k \equiv 1$.
\begin{enumerate}
\item  If $0<c<1$, then $E_c$ contains a contractible closed magnetic geodesic.
\item  There are no closed magnetic geodesics in $E_1$.
\item If $c>1$, there are no contractible closed magnetic geodesics in $E_c$,
but any non trivial free homotopy class of closed curves can be represented by one.
\end{enumerate}
\end{example*}
The existence question for closed magnetic geodesics on hyperbolic surfaces
for non constant functions $k$ is poorly understood. We shall show: If
$(M,g)$ is a compact hyperbolic surface with Gaussian curvature $K_g \ge -1$
and $k\ge 1$ a positive function, then there is a
contractible closed magnetic geodesics in $E_c$ for all $0<c<1$.
The example of the horocycle flow shows that this existence result is sharp.\\
We consider curves, that
are Alexandrov embedded.
\begin{definition}{(oriented Alexandrov embedded)}
\label{def:alexandrov}
Let $B \subset \rz^2$ denote the open ball of radius $1$ centered at $0\in \rz^2$.
An immersion $\gamma\in C^1(\rand B,M)$ will be called {\em oriented Alexandrov embedded},
if there is an immersion $F\in C^1(\overline{B},M)$, such that
$F\eval_{\rand B}=\gamma$ and $F$ is orientation preserving in the sense that
\begin{align*}
\langle DF\eval_{x}x, J_g(\gamma(x))\dot\gamma(x)\rangle_{T_{\gamma(x)}S^2,g}>0
\end{align*}
for all $x \in \rand B$.
\end{definition}
We shall prove
\begin{theorem}
\label{thm_existence}
Let $(M,g)$ be a smooth, compact, orientable surface with negative Euler characteristic
and $k \in C^\infty(M)$ a positive function. Assume there is $K_0>0$ such that
$k$ and the Gaussian curvature $K_g$ of $(M,g)$ satisfy
\begin{align*}
k>(K_0)^\frac12 \text{ and } K_g\ge -K_0.
\end{align*}
Then there is an oriented Alexandrov embedded curve $\gamma \in C^2(S^1,M)$ that
solves \eqref{eq:1} and
the number of such solutions is at least
$-\chi(M)$
provided they are all nondegenerate.
\end{theorem}
The equivalence between (\ref{eq:magnetic_geodesic}) and (\ref{eq:1}) leads to
\begin{corollary}
\label{cor_existence}
Let $(M,g)$ be a smooth, compact, orientable surface with negative Euler characteristic
and $k \in C^\infty(M)$ a positive function. Assume there is $K_0>0$ such that
$k$ and the Gaussian curvature $K_g$ of $(M,g)$ satisfy
\begin{align*}
k\ge(K_0)^\frac12 \text{ and } K_g\ge -K_0.
\end{align*}
Then every energy level $E_c$ with $0<c<1$
contains an oriented Alexandrov embedded closed magnetic geodesic
and the number of such closed magnetic geodesics in $E_c$ is at least $-\chi(M)$
provided they are all nondegenerate.
\end{corollary}
The proof of our existence results is organized as follows.
We consider solutions to \eqref{eq:1} as zeros of
the vector field $X_{k,g}$ defined on the Sobolev space $H^{2,2}(S^1,M)$:
For $\gamma \in H^{2,2}(S^1,M)$ we let $X_{k,g}(\gamma)$ be the unique weak solution of
\begin{align}
\label{eq:def_vector_field}
\big(-D_{t,g}^{2} + 1\big)X_{k,g}(\gamma)=
-D_{t,g} \dot\gamma + |\dot \gamma|_{g} k(\gamma)J_{g}(\gamma)\dot\gamma
\end{align}
in $T_\gamma H^{2,2}(S^1,M)$.
The uniqueness implies that any zero of $X_{k,g}$ is a weak solution of \eqref{eq:1}
which is a classical solution in $C^{2}(S^1,M)$ applying standard regularity
theory.\\
After setting up notation in Section \ref{s:preliminiaries} and introducing the classes of maps
and spaces needed for our analysis
we recall in Section \ref{sec:s1_euler}
the definition and properties of the $S^1$-equivariant Poincar\'{e}-Hopf index
defined in \cite{arXiv:0808.4038},
$$\chi_{S^1}(X_{k,g},\mathscr{M}_A) \in \zz,$$
where
$\mathscr{M}_A$ is the set of oriented Alexandrov embedded regular curves
in $H^{2,2}(S^1,M)$.\\
From the uniformization theorem
$(M,g)$
is isometric to $(\hz/\Gamma,e^\phi g_0)$,
where $\Gamma$ is a group
of isometries of the standard hyperbolic plane $(\hz,g_0)$ acting
freely and properly discontinuously and $\phi$ is a function in
$C^\infty(\hz/\Gamma,\rz)$.
Since the problem of prescribing
geodesic curvature is invariant under isometries we may assume
without loss of generality that
\begin{align*}
(M,g) = (\hz/\Gamma,e^\phi g_0).
\end{align*}
In Section \ref{sec:unperturbed} we analyze the {\em unperturbed problem}
with $k\equiv k_0>0$ and $g=g_0$:
We compute the set of oriented Alexandrov embedded zeros of $X_{k_0,g_0}$
and the image and kernel of the corresponding linearizations.
The perturbative analysis in Section \ref{sec:perturbed}, which carries
over from \cite{arXiv:0808.4038}, is used to compute the degree
of the unperturbed problem in Section \ref{sec:computation_degree}:
For large positive constants $k_0$ and the standard metric $g_0$
we shall show that
\begin{align*}
\chi_{S^1}(X_{k_0,g_0},\mathscr{M}_A)=-\chi(M),
\end{align*}
where $\chi(M)$ denotes the Euler characteristic of $M$.\\
Section \ref{sec:apriori-estimate} contains the apriori estimate which implies
that under the assumptions of Theorem \ref{thm_existence}
the set of solutions to \eqref{eq:1} is compact in $\mathscr{M}_A$.
The homotopy invariance of the $S^{1}$-equivariant Poincar\'{e}-Hopf index then leads
to the identity
\begin{align*}
\chi_{S^1}(X_{k,g},\mathscr{M}_A)=\chi_{S^1}(X_{k_0,g_{can}},\mathscr{M}_A)=-\chi(M).
\end{align*}
The resulting proof of Theorem \ref{thm_existence} is given in Section \ref{sec:existence}.

\section{Preliminaries}
\label{s:preliminiaries}
It is convenient for the functional analytic setting to assume that $M$ is embedded
in some $\rz^{q_M}$.
We consider for $m\in \nz_0$ the set of Sobolev functions
\begin{align*}
H^{m,2}(S^{1},M) := \{\gamma \in H^{m,2}(S^{1},\rz^{q_M}) \where \gamma(t) \in M
\text{ for a.e. } t \in S^{1}.\}
\end{align*}
For $m \ge 1$ the set
$H^{m,2}(S^{1},M)$ is a sub-manifold of the Hilbert space $H^{m,2}(S^{1},\rz^{q_M})$
and is contained in $C^{m-1}(S^{1},\rz^{q_M})$.
Hence, if $m\ge 1$ then $\gamma \in H^{m,2}(S^{1}, M)$
satisfies
$\gamma(t) \in M$ for all $t\in S^{1}$.
In this case the tangent space
at $\gamma \in H^{m,2}(S^{1}, M)$ is given by
\begin{align*}
T_\gamma H^{m,2}(S^{1}, M) :=
\{V \in H^{m,2}(S^{1},\rz^{q_M})\where V(t) \in T_{\gamma(t)}M \text{ for all }t \in S^{1}\}.
\end{align*}
For $m=0$ the set $H^{0,2}(S^{1},M)=L^{2}(S^{1},M)$ fails to be a manifold.
We define for $\gamma \in H^{1,2}(S^{1},M)$ the space
$T_\gamma L^{2}(S^{1},M)$ by
\begin{align*}
T_\gamma L^{2}(S^{1},M):=
\{V \in L^{2}(S^{1},\rz^{q_M})\where V(t) \in T_{\gamma(t)}M \text{ for a.e. }t \in S^{1}\}.
\end{align*}
A metric $g$ on $M$ induces a metric on $H^{m,2}(S^{1},M)$ for $m\ge 1$
by setting for $\gamma \in H^{m,2}(S^{1},M)$
and $V,\,W \in T_\gamma H^{m,2}(S^{1}, S^{2})$
\begin{align*}
\langle W,V\rangle_{T_\gamma H^{m,2}(S^{1}, S^{2}),g}
:= \int_{S^{1}}\Big\langle &\big((-1)^{\llcorner \frac{m}{2}\lrcorner}(D_{t,g})^{m}+1\big)V(t),\\
&\big((-1)^{\llcorner \frac{m}{2}\lrcorner}(D_{t,g})^{m}+1\big)W(t)\Big\rangle_{\gamma(t),g}\, dt,
\end{align*}
where $\llcorner m/2\lrcorner$ denotes the largest integer that does not exceed $m/2$.\\
Since $g$ and $k$ are smooth, $X_{k,g}$ is a smooth vector field (see
\cite[Sec. 6]{arXiv:0808.4038,MR0464304})
on the set $H^{2,2}_{reg}(S^{1},M)$ of regular curves,
\begin{align*}
H^{2,2}_{reg}(S^{1},M)
:= \{\gamma \in H^{2,2}(S^{1},M)\where \dot\gamma(t) \neq 0 \text{ for all }t\in S^{1}\}.
\end{align*}
From \cite{arXiv:0808.4038} there holds
\begin{align}
\label{eq:dg_x_g_formula}
\big(-&D_{t,g}^2+1\big)D_{g} X_{k,g}\eval_\gamma(V) \notag\\
&= -D_{t,g}^2 V - R_g\big(V,\dot\gamma \big)\dot\gamma
+ |\dot\gamma|_g^{-1}\langle D_{t,g}V,\dot\gamma\rangle_g k(\gamma)J_{g}(\gamma)\dot\gamma
 \notag \\
&\quad +|\dot\gamma|_g \big(k'(\gamma)V\big)J_{g}(\gamma)\dot\gamma
+ |\dot\gamma|_g k(\gamma)
\Big(\big(D_gJ_{g}\eval_{\gamma}V\big)\dot\gamma
+J_{g}(\gamma) D_{t,g}V\Big).
\end{align}
We note that (see also \cite[Thm. 6.1]{MR493919})
\begin{align*}
\big(-D_{t,g}^2+1\big)D_{g} X_{k,g}\eval_\gamma(V) = (-D_{t,g}^2 +1)V +T(V),
\end{align*}
where $T$ is a linear map from $T_\gamma H^{2,2}(S^{1},M)$
to $T_\gamma L^{2}(S^{1},M)$ that
depends only on the first derivatives of $V$ and is therefore compact.
Taking the inverse $(-D_{t,g}^2+1)^{-1}$ we deduce that $D_{g} X_{k,g}\eval_\gamma$
is the form $identity + compact$ and thus a Rothe map (see \cite{arXiv:0808.4038}).\\
The vector field $X_{k,g}$ as well as the set of solutions to \eqref{eq:1}
is invariant under a circle action:
For $\theta \in S^{1}=\rz/\zz$ and $\gamma \in H^{2,2}(S^1,M)$
we define $\theta*\gamma \in H^{2,2}(S^1,M)$
by
\begin{align*}
\theta*\gamma(t) = \gamma(t+\theta).
\end{align*}
Moreover, for $V \in T_\gamma H^{2,2}(S^1,M)$ we let
\begin{align*}
\theta*V := V(\cdot+\theta) \in T_{\theta*\gamma} H^{2,2}(S^1,M).
\end{align*}
Then $X_{k,g}(\theta * \gamma) = \theta*X_{k,g}(\gamma)$ for any $\gamma \in H^{2,2}(S^1,M)$
and $\theta\in S^{1}$. Thus, any zero gives rise to a $S^{1}$-orbit of zeros.
We call $\gamma$ a {\em prime} curve, if
the isotropy group $\{\theta \in S^1\where \theta*\gamma=\gamma\}$ of $\gamma$ is trivial.\\
For $m\ge 1$ the exponential map $Exp_{g}: TH^{m,2}(S^{1},M) \to H^{m,2}(S^{1},M)$
is defined for $\gamma \in H^{m,2}(S^{1},M)$ and $V \in T_\gamma H^{m,2}(S^{1},M)$ by
\begin{align*}
Exp_{\gamma,g}(V)(t) := Exp_{\gamma(t),g}(V(t)),
\end{align*}
where $Exp_{z,g}$ denotes the exponential map on $(M,g)$ at $z \in M$.
Due to its pointwise definition
\begin{align*}
\theta*Exp_{\gamma,g}(V)(t) = Exp_{\theta*\gamma,g}(\theta*V)(t).
\end{align*}
We shall find solutions to \eqref{eq:1} in the class of oriented Alexandrov
embedded curves. Let $\gamma \in H^{2,2}(S^1,M)$ be an oriented Alexandrov
embedded curves with corresponding oriented immersion $F$ from $B$ to $M$.
If we equip $B$ with the metric $F^*g$ induced by $F$, then the outer normal $N_B(x)$
at $x \in \rand B$ with respect to $F^*g$ satisfies
\begin{align*}
DF\eval_x N_B(x) = N_\gamma(x)
\end{align*}
where $N_\gamma(x)$ denotes the normal to the curve $\gamma$ at $x \in \rand B$
defined by
\begin{align*}
N_\gamma(x):= |\dot\gamma(x)|^{-1}J_g(\gamma(x))\dot\gamma(x).
\end{align*}
In \cite{arXiv:0903.1128} the following two basic properties
of oriented Alexandrov embedded curves are shown.
\begin{lemma}
\label{lem:basis_alexandrov}
$ $
\begin{enumerate}
\item Let $(\gamma_n)$ in $C^2(\rand B,M)$ be a sequence of immersions,
which are oriented Alexandrov embedded, such that $(\gamma_n)$ converges
to an immersion $\gamma_0$ in $C^2(\rand B, M)$ with strictly positive
geodesic curvature.
Then $\gamma_0$ is oriented Alexandrov embedded.
\item The set of regular, oriented Alexandrov embedded curves is open in
$H^{2,2}(S^1,M)$.
\end{enumerate}
\end{lemma}
Property $(1)$ and $(2)$ are given in \cite{arXiv:0903.1128} for closed
curves in $S^2$. Since the analysis in the
proof of $(1)$ and $(2)$ is done in tubular neighborhoods
of closed curves, properties $(1)$ and $(2)$ continue to hold
if $S^2$ is replaced by a general surface $M$.

\section{The $S^{1}$-Poincar\'{e}-Hopf index}
\label{sec:s1_euler}
In \cite{arXiv:0808.4038} a $S^1$-equivariant Poincar\'{e}-Hopf index
or $S^{1}$-degree is introduced for equivariant vector fields on subsets
of $H^{2,2}(S^{1},S^{2})$. The $S^{1}$-degree is based on an
equivariant version of the Sard-Smale lemma \cite[Lem 3.9]{arXiv:0808.4038},
which depends on an appropriate change of a vector field locally around its critical orbits.
It's merely a matter of form to extend this local argument, when $S^{2}$ is replaced
by a general surface $M$. We give a short account of the definition
and properties of the $S^{1}$-degree for equivariant vector fields on subsets
of $H^{2,2}(S^{1},M)$.\\
We define a $C^2$ equivariant vector field $W_g$ on $H^{2,2}(S^1,M)$ by
\begin{align*}
W_{g}(\gamma) = (-(D_{t,g})^2 + 1)^{-1} \dot \gamma, \text{ for } \gamma \in H^{2,2}(S^1,M).
\end{align*}
We will compute the $S^1$-Poincar\'{e}-Hopf index for the following class of vector fields.
\begin{definition}
Let $\mathscr{M}$ be an open
$S^1$-invariant subset of prime curves in $H^{2,2}(S^{1},M)$.
A $C^2$ vector field $X$ on $\mathscr{M}$ is called $(\mathscr{M},g,S^{1})$-admissible, if
\begin{itemize}
\item[(1)] $X$ is $S^1$-equivariant, i.e. $X(\theta*\gamma)=\theta*X(\gamma)$ for all
$(\theta,\gamma)\in S^1\times \mathscr{M}$.
\item[(2)] $X$ is proper in $\mathscr{M}$, i.e. the set
$\{\gamma \in \mathscr{M} \where X(\gamma)=0 \}$ is compact,
\item[(3)] $X$ is orthogonal to $W_{g}$, i.e.
$\langle X(\gamma), W_g(\gamma)\rangle_{T_\gamma H^{2,2}(S^1,M)}=0$
for all $\gamma \in \mathscr{M}$.
\item[(4)] $X$ is a Rothe field, i.e.
if $X(S^{1}*\gamma)=0$ then $D_{g}X\eval_\gamma$ and
$\text{Proj}_{\langle W_g(\gamma)\rangle^\perp} \circ D_{g}X\eval_\gamma$
are Rothe maps in $\mathscr{L}(T_\gamma H^{2,2}(S^1,M))$ and
$\mathscr{L}(\langle W_g(\gamma)\rangle^\perp)$, respectively.
\item[(5)] $X$ is elliptic, i.e. there is $\eps>0$ such that
for all finite sets of charts
\begin{align*}
\{(Exp_{\gamma_i,g},B_{2\delta_i}(0))\where \gamma_i\in H^{4,2}(S^1,M)
\text{ for }1\le i\le n\},
\end{align*}
and finite sets
\begin{align*}
\{W_i \in T_{\gamma_i}H^{4,2}(S^1,M)\where \|W_i\|_{T_{\gamma_i}H^{4,2}(S^1,M)}<\eps
\text{ for }1\le i\le n\},
\end{align*}
there holds: If
$\alpha \in \capl_{i=1}^n Exp_{\gamma_i,g}(B_{\delta_i}(0))\subset H^{2,2}(S^1,M)$
satisfies
\begin{align*}
X(\alpha) = \sum_{i=1}^n \text{Proj}_{\langle W_g(\alpha)\rangle^\perp}\circ
D Exp_{\gamma_i,g}\eval_{Exp_{\gamma_i,g}^{-1}(\alpha)}(W_i)
\end{align*}
then $\alpha$ is in $H^{4,2}(S^1,M)$.
\end{itemize}
\end{definition}
It is shown in \cite{arXiv:0808.4038} that $X_{k,g}$ satisfies properties
$(3)-(4)$. Hence, $X_{k,g}$ is $(\mathscr{M},g,S^{1})$-admissible if and only
if $X_{k,g}$ is proper in $\mathscr{M}$.
Note that the regularity property $(5)$, taking $W_i=0$, shows that
any zero of $X$ belongs to $H^{4,2}(S^1,M)$.
Furthermore, for $\gamma \in H^{4,2}(S^1,M)$ the map
$\theta \mapsto \theta*\gamma$ is $C^2$ from $S^1$ to $H^{2,2}(S^{1},M)$. Hence, if
$X(\gamma)=0$ then
\begin{align*}
0 &= D_\theta(X(\theta*\gamma))\eval_{\theta=0} = D_{g}X\eval_\gamma (\dot\gamma),
\end{align*}
such that the kernel of $D_{g}X\eval_\gamma$ is nontrivial.
If $X$ is a vector field orthogonal to $W_{g}$
and $X(\gamma)=0$, then
\begin{align*}
0 &= D\big(\langle X(\alpha),W_g(\alpha)\rangle_{T_\alpha H^{2,2}(S^{1},M),g}\big)\eval_\gamma
= \langle D_{g}X\eval_\gamma,W_g(\gamma)\rangle_{T_\gamma H^{2,2}(S^{1},M),g}
\end{align*}
where the various curvature terms and terms containing derivatives
of $W_g$ vanish as $X(\gamma)=0$.
Thus, $X(\gamma)=0$ implies
\begin{align}
\label{eq:dx_to_wg_perp}
D_{g}X\eval_\gamma:\:T_\gamma H^{2,2}(S^{1},M) \to \langle W_g(\gamma)\rangle^{\perp},
\end{align}
and the projection $\text{Proj}_{\langle W_g(\gamma)\rangle^{\perp}}$ in $(4)$ is unnecessary.
\begin{definition}
\label{d:critical_orbit}
Let $\mathscr{M}$ be an open
$S^1$-invariant subset of prime curves in $H^{2,2}(S^{1},M)$, $S^{1}*\gamma \subset \mathscr{M}$,
and $X$ a $(\mathscr{M},g,S^{1})$-admissible vector field on $\mathscr{M}$.\\
The orbit $S^{1}*\gamma$ is called a {\em critical
orbit} of $X$, if $X(\gamma)=0$.\\
The orbit $S^{1}*\gamma$ is called a {\em nondegenerate critical
orbit} of $X$, if $X(\gamma)=0$ and
\begin{align*}
D_{g}X\eval_\gamma:\: \langle W_{g}(\gamma)\rangle^{\perp}
\xpfeil{}{} \langle W_{g}(\gamma)\rangle^{\perp}
\end{align*}
is an isomorphism.
\end{definition}
Note that if $\gamma \in H^{4,2}(S^1,M)\subset H^{2,2}(S^1,M) $ then
$\dot\gamma \not\in \langle W_{g}(\gamma)\rangle^{\perp}$.
\begin{definition}
\label{d:M_g_t_s_1_homotopy}
Let $\{g_t \where t \in [0,1]\}$ be a family of smooth metrics on $M$, which induces
a corresponding family of metrics on $H^{2,2}(S^1,M)$, still denoted by $g_t$.
Let $\mathscr{M}$ be an open
$S^1$-invariant subset of prime curves in $H^{2,2}(S^{1},M)$ and $X_0$, $X_1$
two vector-fields on $\mathscr{M}$ such that
$X_i$ is $(\mathscr{M},g_i,S^1)$-admissible for $i=0,1$.
A $C^2$ family of vector-fields $X(t,\cdot)$ on $\mathscr{M}$ for $t\in [0,1]$ is called
a $(\mathscr{M},g_t,S^1)$-homotopy between $X_0$ and $X_1$, if
\begin{itemize}
\item $X(0,\cdot)=X_0$ and $X(1,\cdot)=X_1$,
\item $\{(t,\gamma)\in[0,1]\times \mathscr{M} \where X(t,\gamma)=0\}$ is compact,
\item $X_t:= X(t,\cdot)$ is $(\mathscr{M},g_t,S^1)$-admissible for all $t\in [0,1]$.
\end{itemize}
We write $(\mathscr{M},g,S^{1})$-homotopy,
if the family of metrics $\{g_t\}$ is constant.
\end{definition}
Note that,
if $\{k_t \in C^\infty(M,\rz)\where t \in [0,1]\}$ is a $C^2$ family of smooth function,
then $t \mapsto X_{k_t,g_t}$ is a $(\mathscr{M},g_t,S^1)$-homotopy, if and only if
the set
\begin{align*}
\{(t,\gamma)\in[0,1]\times \mathscr{M} \where X_{k_t,g_t}(\gamma)=0\}
\end{align*}
is compact.\\
We let $\mathscr{M}$ be an open
$S^1$-invariant subset of prime curves in $H^{2,2}(S^{1},M)$ and $X$
a $(\mathscr{M},g,S^{1})$-admissible vector field on $\mathscr{M}$.
The local $S^1$-degree of an isolated, nondegenerate critical orbit $S^1*\gamma_0$
is defined by
\begin{align*}
\deg_{loc,S^1}(X,S^1*\gamma_0):=\sgn D_g X\eval_{\gamma_0},
\end{align*}
where $\sgn D_g X\eval_{\gamma_0}$ is the sign of the Rothe map $D_g X\eval_{\gamma_0}$
in $\mathscr{L}(\langle W_{g}(\gamma)\rangle^{\perp})$. Since $D_g X_{k,g}\eval_{\gamma_0}$
is of the form $identity+compact$, in the above situation $\sgn D_g X_{k,g}\eval_{\gamma_0}$
is given by the usual Leray-Schauder degree.\\
Using an equivariant version of the Sard-Smale lemma a $S^1$-equivariant Poincar\'{e}-Hopf index
\begin{align*}
\chi(X,\mathscr{M}) \in \zz
\end{align*}
is defined in \cite{arXiv:0808.4038} with the following properties.
\begin{lemma}
\label{lem:degree_properties}
$ $
\begin{enumerate}
\item If $X$ is $(\mathscr{M},g,S^{1})$-admissible with only finitely many critical orbits,
that are all nondegenerate, then
\begin{align*}
\chi_{S^1}(X,\mathscr{M}) := \sum_{\{S^{1}*\gamma\subset \mathscr{M} \where X(S^{1}*\gamma)=0\}}
\deg_{loc,S^1}(X,S^{1}*\gamma).
\end{align*}
\item If $X_0$ and $X_1$ are $(\mathscr{M},g_t,S^1)$-homotop, then
$\chi(X_0,\mathscr{M})=\chi(X_1,\mathscr{M})$.
\end{enumerate}
\end{lemma}

\section{The Unperturbed Problem}
\label{sec:unperturbed}
Let $\hz \subset \rz^{3}$ be the standard hyperbolic plane
\begin{align*}
\hz:= \{(\xi^1,\xi^2,\tau) \in \rz^3 \where \tau^2-|\xi|^2=1 \text{ and }\tau >0\}
\end{align*}
with metric $g_{0}$ induced by the Minkowski metric $g_m$,
\begin{align*}
g_m := (d\xi^1)^2+(d\xi^2)^2-(d\tau)^2=\langle \cdot,\cdot\rangle_m.
\end{align*}
We choose the orientation on $\hz$
such that $J_{g_0}(y)$ is given for $y \in \hz$ by
\begin{align*}
J_{g_0}(y)(v) := y \cross_m v \text{ for all }v \in T_y \hz,
\end{align*}
where $\cross_m$ denotes the twisted cross product in $\rz^{3}$,
\begin{align*}
\begin{pmatrix}
v^1\\v^2\\v^3
\end{pmatrix}
\cross_m
\begin{pmatrix}
w^1\\w^2\\w^3
\end{pmatrix}
:=
\begin{pmatrix}
v^3w^2-v^2w^3\\v^1w^3-v^3w^1\\v^1w^2-v^2w^1
\end{pmatrix}.
\end{align*}
The twisted cross product $\cross_m$ is related
to the usual cross product $\cross$ in $\rz^3$ by
$v \cross_m w = I_{2,1}v \cross I_{2,1}w$, where
$I_{2,1}$ is given by
\begin{align*}
I_{2,1}:=
\begin{pmatrix}
1 &0 &0\\
0 &1 &0\\
0&0&-1
\end{pmatrix},
\end{align*}
and satisfies for $a,\,b,\,c,\,d \in \rz^3$
\begin{align*}
\langle (a\cross_m b),a\rangle_m=0=\langle (a\cross_m b),b\rangle_m,\\
a\cross_m(b\cross_m c)= -b\langle a,c\rangle_m + c\langle a,b\rangle_m,\\
\langle (a\cross_m b),(c \cross_m d)\rangle_m
= -\langle a,c\rangle_m \langle b,d\rangle_m +
\langle b,c\rangle_m\langle a,d\rangle_m
\end{align*}
We fix a compact, orientable Riemannian surface $(M,g_0)$,
\begin{align*}
M := \hz/\Gamma,
\end{align*}
where $\Gamma \subset SO(2,1)_+$ is a group
of oriented isometries acting
freely and properly discontinuously on $\hz$. Concerning the metric we will
be sloppy and denote by $g_0$ the
metric on $\hz$ as well as the induced metric on $\hz/\Gamma$.
The unperturbed problem on $M$ is given by
\begin{align}
\label{eq:unperturbed_M}
D_{t,g_0} \dot \gamma = |\dot \gamma|_{g_0} k_0 J_{g_0}(\gamma) \dot \gamma,
\end{align}
where $k_0$ is a positive constant.\\
We shall compute the $S^1$-degree of the unperturbed equation \eqref{eq:unperturbed_M}
in three steps. {\em Step 1:} We compute explicitly the set $\mathcal{Z}_M$
of Alexandrov embedded solutions in $H^{2,2}(S^1,M)$ to \eqref{eq:unperturbed_M}
and show that $\mathcal{Z}_M$ is a finite dimensional, nondegenerate manifold,
in the sense that we have for all $\tilde{\alpha} \in \mathcal{Z}_M$
\begin{align*}
T_{\tilde{\alpha}} \mathcal{Z}_M = \text{kernel}(D_{g_0}X_{k_0,g_0}\eval_{\tilde{\alpha}}),\\
T_{\tilde{\alpha}} H^{2,2}(S^1,M) =
T_{\tilde{\alpha}} \mathcal{Z}_M \oplus R(D_{g_0}X_{k_0,g_0}\eval_{\tilde{\alpha}}).
\end{align*}
{\em Step 2:} In Section \ref{sec:perturbed}
we perform a finite dimensional reduction of a slightly perturbed problem:
We consider for $k_1\in C^\infty(M,\rz)$, which will be chosen later,
and $\eps \in \rz$, which is assumed to be very small,
the perturbed vector field $X_{g_0,\eps}$ defined by
\begin{align*}
X_{g_0,\eps}(\gamma)&:=
(-D_{t,g_0}^{2} + 1)^{-1}
\big(-D_{t,g_0} \dot\gamma + |\dot \gamma|_{g_0}
(k_0+\eps k_1(\gamma))J_{g_0}(\gamma)\dot\gamma\big)\\
&= X_{k_0,g_0}(\gamma)+\eps K_1(\gamma),
\end{align*}
where the vector field $K_1$ is given by
\begin{align*}
K_1(\gamma):= (-D_{t,g_0}^{2} + 1)^{-1} |\dot \gamma|_{g_0}
\big(k_1(\gamma)J_{g_0}(\gamma)\dot\gamma \big).
\end{align*}
We show that if $S^1*\tilde{\alpha}_0 \subset \mathcal{Z}_M$ is a nondegenerate
critical orbit
of the vector field $\tilde{\alpha} \mapsto P_1(\tilde{\alpha}) \circ K_1(\tilde{\alpha})$
on $\mathcal{Z}_M$,
where $P_1(\tilde{\alpha})$ is a projection onto $T_{\tilde{\alpha}} \mathcal{Z}_M$ defined below,
then for any $0<\eps <<1$
there is a unique nondegenerate critical orbit $S^1*\tilde{\gamma}(\eps)$
of $X_{g_0,\eps}$
 such that $\tilde{\gamma}(\eps)$ converges to $\tilde{\alpha}_0$ as $\eps \to 0^+$
and
\begin{align*}
\deg_{loc, S^{1}}(X_{g_0,\eps},S^1*\gamma(\eps))
&= -\deg_{loc}(P_1(\cdot)\circ K_1(\cdot),S^1*\tilde{\alpha}_0).
\end{align*}
{\em Step 3:} In Section \ref{sec:computation_degree}
we choose a Morse function $k_1 \in C^\infty(M,\rz)$ with critical points
\begin{align*}
\{\tilde{w}_i \in M \where 1 \le i \le n\}.
\end{align*}
We show that if $k_0 >>1 $ is large, then $P_1(\cdot)\circ K_1(\cdot)$ has exactly
$n$ critical orbits $\{S^1*\tilde{\alpha}_{i,k_0} \where 1 \le i \le n\}$ such that
for $1 \le i \le n$
\begin{align*}
\deg_{loc}(P_1(\cdot)\circ K_1(\cdot),S^1*\tilde{\alpha}_{i,k_0})=
\deg_{loc}(\nabla k_1, \tilde{w}_i).
\end{align*}
This yields the formula
$\chi_{S^1}(X_{k_0,g_0},\mathscr{M}_A)=-\chi(M)$, where $\mathscr{M}_A$ is the subset of
$H^{2,2}(S^{1},M)$ consisting of Alexandrov embedded, regular curves.\\
{\em Step 1:}
The prescribed geodesic curvature equation with $k\equiv k_0$ on $(\hz,g_0)$ is given by
\begin{align}
\label{eq:unpert:1}
Proj_{\gamma^\perp,g_m}\ddot \gamma = |\dot \gamma|_m k_0 \gamma \cross_m \dot \gamma,
\end{align}
where $\gamma \in H^{2,2}(S^1,\hz)$, $\dot \gamma$ and $\ddot \gamma$ are the usual derivatives
of $\gamma$ considered as a curve in $\rz^3$, $|\dot \gamma|_m$ is the Minkowski norm
of $\dot \gamma$
in $(\rz^3,g_m)$.\\
If $k_0>1$ then there is a unique $r=r(k_0)>0$ such that
\begin{align*}
k_0= \frac{\sqrt{1+r^2}}{r}.
\end{align*}
We call a triple of vectors $\{v_0,v_1,w\}$ in $\rz^3$ a positive oriented orthonormal system
with respect to $g_m$, if
\begin{align*}
\langle v_0,v_1\rangle_m = \langle v_0,w\rangle_m= \langle v_1,w\rangle_m=0,\\
\langle v_0,v_0\rangle_m =\langle v_1,v_1\rangle_m = -\langle w,w\rangle_m=1,\\
v_0 \cross_m v_1 = w.
\end{align*}
We define for $\lambda>0$ and a positive oriented orthonormal system $\{v_0,v_1,w\}$
the function $\alpha \in C^\infty(\rz,\hz)$ by
\begin{align}
\label{eq:soln_alpha}
\alpha(t,\lambda,v_0,v_1,w) := \sqrt{1+r^2}w+r\cos(\lambda r^{-1}t)v_1+r\sin(\lambda r^{-1}t)v_0
\end{align}
A direct calculation shows that $\alpha(\cdot,\lambda,v_0,v_1,w)$ solves (\ref{eq:unpert:1}).
We fix $(\gamma_0,\tilde{v}_0) \in T\hz$ with $\tilde{v}_0 \neq 0$
and define
the parameter $\lambda := |\tilde{v}_0|_m$ and
the positive oriented orthonormal system $(v_0,v_1,w)$ by
\begin{align*}
v_0 &:= \lambda^{-1} \tilde{v}_0,\, v_1 := -r\gamma_0-\sqrt{1+r^2}(v_0\cross_m \gamma_0),\,
w := v_0\cross_m v_1.
\end{align*}
Then $\alpha(\cdot,\lambda,v_0,v_1,w)$ satisfies the initial conditions
\begin{align*}
\alpha(0,\lambda,v_0,v_1,w)=\gamma_0,\; \dot\alpha(0,\lambda,v_0,v_1,w)= \tilde{v}_0,
\end{align*}
and we deduce that all non constant solutions of \eqref{eq:unpert:1} are obtained in this way.
Since we are only interested in solutions in $H^{2,2}(S^1,\hz)$ we get an extra condition
on $\lambda$, i.e. the $1$-periodicity leads to
\begin{align*}
\lambda \in 2\pi \nz r.
\end{align*}
\begin{lemma}
\label{l:alexandrov_soln_hz}
The oriented Alexandrov embedded solutions in
$H^{2,2}(S^1,\hz)$ of equation (\ref{eq:unpert:1}) are
given by the set of simple solutions
\begin{align*}
\mathcal{Z}_{\hz} :=
\big\{&\alpha(\cdot, 2\pi r,v_0,v_1,w )\where \\
&\{v_0,v_1,w\}
\text{ is a pos. orth. system in }(\rz^3,g_m)\big\}.
\end{align*}
\end{lemma}
\begin{proof}
From the analysis above the periodic solutions to (\ref{eq:unpert:1}) are given by
\begin{align*}
\big\{\alpha(\cdot, 2\pi n r,v_0,v_1,w )\where
&n \in \nz \text{ and }\\
&\{v_0,v_1,w\}
\text{ is a pos. orth. system in }(\rz^3,g_m)\big\}.
\end{align*}
We fix $n \in \nz$ and a positive orthonormal system $\{v_0,v_1,w\}$ and write
\begin{align*}
\gamma_n:= \alpha(\cdot, 2\pi n r,v_0,v_1,w).
\end{align*}
Assume $\gamma_n$ is oriented Alexandrov embedded and let $F_n$ be the
corresponding immersion.
Since $\hz$ is diffeomorphic to $\rz^2$
we may assume
that $\gamma_1$ is a simple curve in the plane $(\rz^2,\delta)$ with standard metric $\delta$.
If we apply the Gau{\ss}-Bonnet formula to $(B,F_n^*\delta)$ and the embedded
curve $\gamma_1$ in the plane, we obtain
\begin{align*}
2\pi &= \int_{\rand B} k_{F_n^*\delta}\, dS_{F_n^*\delta} + \int_B K_{F_n^*\delta}\, dA_{F_n^*\delta}\\
&= \int_{\gamma_n} k_{\delta} \, dS_{\delta}
= n\int_{\gamma_1} k_{\delta} \, dS_{\delta} = n2\pi,
\end{align*}
which is only possible for $n=1$.\\
The curve $\gamma_1$ is oriented Alexandrov embedded using polar coordinates and
\begin{align*}
[0,2\pi]\times [0,1] \ni (t,s) \mapsto
\sqrt{1+s^2r^2}w+sr\cos(t)v_1+sr\sin(t)v_0.
\end{align*}
\end{proof}
The Lorentz transformations
$S0(2,1)_+$ of $(\rz^3,g_m)$,
\begin{align*}
SO(2,1)_+ := \{A \in O(2,1) \where A(\hz)\subset \hz \text{ and } \det A = 1\},
\end{align*}
correspond to the oriented isometries of $(\hz,g_0)$ and act on solutions:
if $\gamma$ solves (\ref{eq:unpert:1}) so does $A\circ\gamma$
for any $A\in SO(2,1)_+$. We have
\begin{align*}
A\circ \alpha(\cdot, \lambda,v_0,v_1,w ) = \alpha(\cdot, \lambda,A(v_0),A(v_1),A(w)).
\end{align*}
Moreover, there holds,
\begin{align}
\label{eq:param_crit_orbits_hz}
\alpha(\cdot, 2\pi r,v_0,v_1,w ) = \theta*\alpha(\cdot, 2\pi r,v_0',v_1',w')
\end{align}
for some $\theta \in S^1$ if and only if $w=w'$.
Consequently, the critical orbits of \eqref{eq:unpert:1} in $\hz$,
$
\{S^{1}*\gamma \where \gamma \in \mathcal{Z}_{\hz}\},
$
are parametrized by $w \in \hz$ and correspond to ``circles'' with radius $r$
around the center $w$ in $\hz$.\\
We let $\pi_M$ be the natural projection, $\pi_M: \hz \to \hz/\Gamma$.
Any point $z \in \hz$ admits a neighborhood $U=B_{\delta}(z)$
such that $\pi_M\eval_U: U \to  \pi_M(U)$
is an isometry. From \eqref{eq:soln_alpha} there is $C_{k_0}>1$ such that
if $k_0\ge C_{k_0}$ then any solution to \eqref{eq:unpert:1} on $\hz$ passing through $z$
remains in $U$. For $M$ is compact $C_{k_0}= C_{k_0}(\Gamma)$ and $\delta>0$ may be chosen
independently of $z$. Equation \eqref{eq:1} is invariant under isometries,
hence the set of solutions to \eqref{eq:unperturbed_M} with $k_0 \ge C_{k_0}$
is given by
\begin{align*}
\big\{&\pi_M\circ \alpha(\cdot, 2\pi r,v_0,v_1,w )\where \\
&\{v_0,v_1,w\}
\text{ is a pos. orth. system in }(\rz^3,g_m)\big\}.
\end{align*}
Moreover, we have
\begin{lemma}
\label{l:alexandrov_soln_M}
If $k_0 \ge C_{k_0}$, then
the oriented Alexandrov embedded solutions in
$H^{2,2}(S^1,M)$ of equation \eqref{eq:unperturbed_M} are
given by the set of simple solutions
\begin{align*}
\mathcal{Z}_{M} :=
\big\{&\tilde{\alpha}=\pi_M \circ \alpha(\cdot, 2\pi r,v_0,v_1,w )\where \\
&\{v_0,v_1,w\}
\text{ is a pos. orth. system in }(\rz^3,g_m)\big\}.
\end{align*}
\end{lemma}
\begin{proof}
We fix $n \in \nz$ and a positive orthonormal system $\{v_0,v_1,w\}$ and write
\begin{align*}
\gamma_n:= \pi_M \circ \alpha(\cdot, 2\pi n r,v_0,v_1,w)
\end{align*}
From the above analysis any periodic solution to \eqref{eq:unperturbed_M} on $(M,g_0)$ is
of this form.
Hence, it is enough
to show that
$\gamma_n$ is oriented Alexandrov embedded, if and only if $n=1$.\\
Concatenating the immersion in the proof of Lemma \ref{l:alexandrov_soln_hz}
with $\pi_M$ we deduce that $\gamma_1$ is oriented Alexandrov embedded.
Suppose $\gamma_n$ is oriented Alexandrov embedded with an immersion $F_n:B \to M$.
From the homotopy lifting property of the covering $\pi_M:\hz \to M$ we may lift $F_n$
to see that $\alpha(\cdot, 2\pi n r,v_0,v_1,w)$ is oriented Alexandrov embedded
in $\hz$. From Lemma \ref{l:alexandrov_soln_hz} this is only possible for $n=1$.
\end{proof}
From \eqref{eq:param_crit_orbits_hz} we find
\begin{align*}
\pi_M\circ \alpha(\cdot, 2\pi r,v_0,v_1,w ) = \theta* \pi_M\circ \alpha(\cdot, 2\pi r,v_0',v_1',w')
\end{align*}
for some $\theta \in S^1$ if and only if $\pi_M(w)=\pi_M(w')$,
such that the critical orbits of \eqref{eq:unpert:1} in $M$
are parametrized by $w \in M$
and correspond to projections on $M$ of ``circles'' in $\hz$.\\
In the following we always assume that
\begin{align*}
k_0 \ge C_{k_0}.
\end{align*}
We denote by $X_{k_0,g_0,\hz}$ the vector field on $H^{2,2}(S^1,\hz)$ corresponding
to equation \eqref{eq:unpert:1}.
We fix a solution $\alpha=\alpha(\cdot,2\pi r,v_0,v_1,w)$ of \eqref{eq:unpert:1}
and note that for $V \in  T_{\alpha}H^{2,2}(S^1,\hz)$
\begin{align*}
R_{g_0}(V,\dot\alpha)\dot\alpha = -V|\dot \alpha|_m^{2}+\langle V,\dot\alpha\rangle_m \dot\alpha.
\end{align*}
By \eqref{eq:dg_x_g_formula} a vector field $W$ is contained in
the kernel of $D_{g_0}X_{k_0,g_0,\hz}\eval_{\alpha}$ if and only if $W$ is a periodic solution
of
\begin{align}
\label{eq:linearization}
0 &=
-D_{t,g_0}^2W + W|\dot \alpha|_m^{2}-\langle W,\dot\alpha\rangle_m \dot\alpha \notag \\
&\quad + |\dot\alpha|_m^{-1}\langle D_{t,g_0}W,\dot\alpha\rangle_m k_0 (\alpha\cross_m \dot\alpha)
+|\dot\alpha|_m  k_0 (\alpha \cross_m D_{t,g_0}W).
\end{align}
Due to the geometric origin of equation (\ref{eq:unpert:1}) and the
$SO(2,1)_+$ invariance we find that
\begin{align}
\label{eq:12}
W_0(t,v_0,v_1,w) &:= t\dot\alpha,\\
W_1(t,v_0,v_1,w) &:=  \dot\alpha= 2\pi r(-\sin(2\pi t)v_1 + \cos(2\pi t)v_0), \notag \\
W_2(t,v_0,v_1,w) &:=  (1+r^2)^{\frac12} v_1+r\cos(2\pi t)w, \notag \\
W_3(t,v_0,v_1,w) &:= (1+r^2)^{\frac12} v_0+r\sin(2\pi t)w,\notag
\end{align}
solve \eqref{eq:linearization}. In the sequel,
we will omit the dependence of $W_i$ on $(v_0,v_1,w)$, if there is no possibility of confusion.
The initial values of $W_0,\dots, W_3$
\begin{align*}
W_0(0,v_0,v_1,w)& =0 ,\,
D_{t,g_0} W_0(0,v_0,v_1,w)= 2\pi r v_0,\\
W_1(0,v_0,v_1,w) &=2\pi r v_0,\, D_{t,g_0} W_1(0,v_0,v_1,w)=
-4\pi^2 r^3k_0(k_0v_1+w),\\
W_2(0,v_0,v_1,w) &= rk_0 v_1+rw,\, D_{t,g_0} W_2(0,v_0,v_1,w)= 0,\notag\\
W_3(0,v_0,v_1,w) &= r k_0 v_0,\, D_{t,g_0} W_3(0,v_0,v_1,w)= -2\pi r^{3}(k_0v_1+w).
\end{align*}
are a basis of $\big(T_{\alpha(0)}\hz\big)^2$, such that any solution to \eqref{eq:linearization}
is a linear combination of $W_0,\dots, W_3$.
As only $W_1,\dots, W_3$ are periodic, we obtain
\begin{align}
\label{eq:kernel_DX_0}
\text{kernel}(D_{g_0}X_{k_0,g_0,\hz}\eval_{\alpha})=
\langle W_1,\,W_2,\,W_3 \rangle.
\end{align}
We fix a neighborhood $U$ of $\alpha(0)$ as above, where $\pi_M:\: U \to \pi_M(U)$
is an isometry. Then $\alpha \in H^{2,2}(S^1,U)$ and $\pi_M$ induces isomorphisms
\begin{align*}
\pi_M:\: H^{2,2}(S^1,U) \to H^{2,2}(S^1,\pi_M(U)), \, \alpha \mapsto \pi_M \circ \alpha,\\
(\pi_M)_*:\: T_\alpha H^{2,2}(S^1,\hz) \to T_{\pi_M \circ \alpha} H^{2,2}(S^1,M),
\, V \mapsto d\pi_M\eval_\alpha V,
\end{align*}
where $(\pi_M)_*$ is an isometry. Moreover, there holds on $H^{2,2}(S^1,U)$
\begin{align}
\label{eq:3}
(\pi_M)_* \circ X_{g_0,k_0,\hz} = X_{g_0,k_0}\circ \pi_M, \notag \\
(\pi_M)_* \circ D_{g_0}X_{g_0,k_0,\hz}\eval_\alpha
=D_{g_0} X_{g_0,k_0}\eval_{\pi_M \circ \alpha} \circ (\pi_M)_*.
\end{align}
Since
$\mathcal{Z}_{\hz}$
and $\mathcal{Z}_{M}$ are three dimensional submanifolds of $H^{2,2}(S^{1},\hz)$
and $H^{2,2}(S^{1},M)$, respectively, we have for $\alpha \in \mathcal{Z}_{\hz}$
and $\tilde{\alpha}=\pi_M \circ \alpha \in \mathcal{Z}_{M}$
\begin{align*}
T_\alpha \mathcal{Z}_{\hz} &=
\text{kernel}(D_{g_0}X_{k_0,g_0,\hz}\eval_{\alpha})=\langle W_1,\,W_2,\,W_3 \rangle,\\
T_{\tilde{\alpha}} \mathcal{Z}_{M} &=
\text{kernel}(D_{g_0}X_{k_0,g_0}\eval_{\tilde{\alpha}})\\
&=\langle \tilde{W}_i:= (\pi_M)_* \circ W_i \where 1 \le i \le 3 \rangle.
\end{align*}
To compute the image of $D_{g_0}X_{k_0,g_0,\hz}\eval_{\alpha}$ we note
that
$\{\dot\alpha,\alpha\cross_m\dot\alpha\}$
is an orthogonal system in $T_{\alpha}\hz$ for any $t\in S^{1}$.
Thus any $V \in T_\alpha H^{2,2}(S^1,\hz)$ may be written
as
\begin{align*}
V= \lambda_1 \dot\alpha + \lambda_2 (\alpha\cross_m \dot\alpha)
\end{align*}
for some functions $\lambda_1,\,\lambda_2 \in H^{2,2}(S^{1},\rz)$.
Using the fact that
\begin{align*}
D_{t,g_0} \dot\alpha &= |\dot \alpha|_m k_0(\alpha \cross_m \dot\alpha) \text{ and }
D_{t,g_0} (\alpha \cross_m \dot\alpha) = -|\dot \alpha|_m k_0 \dot\alpha,
\end{align*}
we obtain
\begin{align}
\label{eq:d_x_g0_alpha}
D_{g_0}X_{k_0,g_0,\hz}\eval_{\alpha}(V) &= (-D_{t,g_0}^2+1)^{-1}
\big((-\lambda_1''+2\pi\sqrt{1+r^{2}} \lambda_2') \dot\alpha \notag\\
&\qquad +(-\lambda_2''-(2\pi)^{2}\lambda_2)(\alpha\cross_m \dot\alpha)\big).
\end{align}
Concerning $W_1,\dots, W_3$ and $W_{g_0}$ we find
\begin{align}
W_1(t) &= \dot\alpha(t),\notag \\
W_2(t) &= -\frac{1}{2\pi r}\big(\sqrt{1+r^{2}} \sin(2\pi t)\dot\alpha(t) + \cos(2\pi t)(\alpha\cross_m
\dot\alpha)\big),\notag \\
W_3(t) &= -\frac{1}{2\pi r}\big(-\sqrt{1+r^{2}} \cos(2\pi t)\dot\alpha(t)
+ \sin(2\pi t)(\alpha\cross_m \dot\alpha)\big)\notag \\
\label{eq:8}
W_{g_0}(\alpha) &= (1+|\dot\alpha|_m^{2}k_0^{2})^{-1}\dot\alpha
=  (1+4\pi^2(1+r^2))^{-1} W_1.
\end{align}

\begin{lemma}
\label{l:kernel_not_in_range}
If $r\neq (2\pi)^{-1}$, then we have
for $\alpha \in \mathcal{Z}_{\hz}$
\begin{align*}
\{0\}&=\langle W_1,W_2,W_3\rangle \cap R\big(D_{g_0}X_{k_0,g_0,\hz}\eval_{\alpha}\big) ,\\
\langle W_1\rangle^{\perp}&=
\langle W_2,W_3\rangle \oplus R\big(D_{g_0}X_{k_0,g_0,\hz}\eval_{\alpha}\big)
\end{align*}
\end{lemma}
\begin{proof}
For $\lambda_1,\lambda_2 \in H^{2,2}(S^{1},\rz)$ we have
\begin{align*}
 (-D_{t,g_0}^2+1)&\big(\lambda_1 \dot\alpha + \lambda_2(\alpha \cross_m \dot\alpha)\big)\\
&= \big(-\lambda_1''+4\pi\sqrt{1+r^{2}} \lambda_2' +(4\pi^{2}(1+r^{2})+1) \lambda_1\big)\dot\alpha\\
&\quad +\big(-\lambda_2''-4\pi\sqrt{1+r^{2}} \lambda_1' +(4\pi^{2}(1+r^{2})+1) \lambda_2\big)
\alpha \cross_m \dot\alpha
\end{align*}
Hence we get by direct calculations
\begin{align}
(-D_{t,g_0}^2+1)(W_1) &= (4\pi^{2}(1+r^{2})+1) \dot\alpha,\notag \\
(-D_{t,g_0}^2+1)(-2\pi r W_2) &= \sqrt{1+r^{2}}(4\pi^{2}r^{2} +1) \sin(2\pi t) \dot\alpha \notag \\
\label{eq:D2_1_w2}
&\quad + (-4\pi^{2} r^{2}+1)\cos(2\pi t) (\alpha \cross_m \dot\alpha),\\
(-D_{t,g_0}^2+1)(-2\pi r W_3) &= -\sqrt{1+r^{2}}(4\pi^{2}r^{2} +1) \cos(2\pi t) \dot\alpha\notag\\
\label{eq:D2_1_w3}
&\quad + (-4\pi^{2} r^{2}+1)\sin(2\pi t) (\alpha \cross_m \dot\alpha).
\end{align}
Consequently, by (\ref{eq:dx_to_wg_perp}), \eqref{eq:8}, and the above computations
$W_1$ is orthogonal
to $\langle W_2,W_3\rangle$ and to $R\big(D_{g_0}X_{k_0,g_0,\hz}\eval_{\alpha}\big)$
in $T_\alpha H^{2,2}(S^1,\hz)$.
As in $L^{2}(S^{1},\rz)$
\begin{align*}
\lambda_2''+(2\pi)^{2}\lambda_2 \perp_{L^2} \langle \cos(2\pi t),\sin(2\pi t)\rangle,\;
\langle \lambda_1'',\lambda_2'\rangle  \perp_{L^2} {\rm const},
\end{align*}
we get from \eqref{eq:d_x_g0_alpha} and the fact that $1-4\pi^2r^2\neq 0$
\begin{align*}
\{0\} &=  (-D_{t,g_0}^2+1)\big(\langle W_1,W_2,W_3\rangle\big)\\
&\qquad \cap
(-D_{t,g_0}^2+1)D_{g_0}X_{k_0,g_0,\hz}\eval_{\alpha}(T_{\alpha}H^{2,2}(S^{1},\hz))
\end{align*}
and the claim follows for $D_{g_0}X_{k_0,g_0,\hz}\eval_{\alpha}$ is a Fredholm operator of index $0$.
\end{proof}
Moreover, we see for $\alpha \in \mathcal{Z}_{\hz}$
\begin{align}
\label{eq:4}
R(D_{g_0}X_{k_0,g_0,\hz}\eval_{\alpha}) &=
\big\{
(-D_{t,g_0}^2+1)^{-1}\big(
(-\lambda_1''+2\pi\sqrt{1+r^2}\lambda_2')\dot{\alpha} \notag\\
&\qquad -(\lambda_2''+(2\pi)^2\lambda_2)(\alpha \cross_m \dot{\alpha})\big)
\where \lambda_1,\,\lambda_2 \in H^{2,2}(S^1,\rz)
\big\} \notag \\
&=
\big\{
(-D_{t,g_0}^2+1)^{-1}\big(
\lambda_1\dot{\alpha}+\lambda_2(\alpha \cross_m \dot{\alpha})\big)
\where
\lambda_i \in L^2(S^1,\rz), \notag \\
&\qquad  \lambda_1 \perp_{L^2} 1,\,
\lambda_2 \perp_{L^2} \langle \cos(2\pi t),\sin(2\pi t)\rangle
\big\}\notag \\
&= \langle (\alpha \cross_m \dot{\alpha}) \rangle \oplus E_{+},
\end{align}
where $E_{+}$ is given by
\begin{align*}
E_{+} &=
\big\{
(-D_{t,g_0}^2+1)^{-1}\big(
\lambda_1\dot{\alpha}+\lambda_2(\alpha \cross_m \dot{\alpha})\big)
\where\\
&\qquad \lambda_i\in L^2(S^1,\rz),\, \lambda_1 \perp_{L^2}1,\,
\lambda_2 \perp_{L^2} \langle 1, \cos(2\pi t),\sin(2\pi t)\rangle
\big\}
\end{align*}
We have for $V=\lambda_1\dot{\alpha}+\lambda_2 (\alpha \cross_m \dot{\alpha})$ in $T_\alpha H^{2,2}(S^{1},\hz)$
\begin{align*}
D_{g_0}X_{k_0,g_0,\hz}\eval_{\alpha}(V) \in E_{+} \aequi \lambda_2 \perp_{L^{2}} 1
\aequi V \perp_{L^{2}} (\alpha \cross_m \dot{\alpha}).
\end{align*}
We fix
\begin{align*}
V=(-D_{t,g_0}^{2}+1)^{-1}(\lambda_1\dot{\alpha}+\lambda_2 (\alpha \cross_m \dot{\alpha}))
\in E_{+}.
\end{align*}
Then
\begin{align*}
\int_{S^{1}} \langle V,&\alpha \cross_m \dot{\alpha}\rangle_m\\
&=
\int_{S^{1}} \langle
(-D_{t,g_0}^{2}+1)^{-1}(\lambda_1\dot{\alpha}+\lambda_2 (\alpha \cross_m \dot{\alpha})),
\alpha \cross_m \dot{\alpha}\rangle_m\\
&= \int_{S^{1}} \langle\lambda_1\dot{\alpha}+\lambda_2 (\alpha \cross_m \dot{\alpha})
,(-D_{t,g_0}^{2}+1)^{-1}(\alpha \cross_m \dot{\alpha})\rangle_m\\
&= (4\pi^{2}(1+r^{2})+1)^{-1}\int_{S^{1}}
\langle \lambda_1\dot{\alpha}+\lambda_2 (\alpha \cross_m \dot{\alpha}),
\alpha \cross_m \dot{\alpha}\rangle_m=0.
\end{align*}
Consequently, $D_{g_0}X_{k_0,g_0,\hz}\eval_{\alpha}(E_{+})=E_{+}$.\\
$E_+$ is $L^{2}$-orthogonal to $\alpha \times_m \dot\alpha$ and $\dot\alpha$, we may thus write
\begin{align*}
V &= (\nu_1+f_1)\dot{\alpha}+(\nu_2+f_2)(\alpha \cross_m \dot{\alpha}),
\end{align*}
where
\begin{align*}
\nu_1,\nu_2 \perp_{L^2} \langle 1, \sin(2\pi \cdot),\cos(2\pi \cdot)\rangle
\text{ and }
f_1,f_2 \in \langle \sin(2\pi \cdot),\cos(2\pi \cdot)\rangle.
\end{align*}
Then
\begin{align}
\label{eq:5}
\langle (-D_{t,g_0}^{2}&+1) D_{g_0}X_{k_0,g_0,\hz}\eval_{\alpha}(V),V\rangle_{L^{2}} \notag \\
&=
\int_{S^{1}} (\nu_1')^2-2\pi \sqrt{1+r^2}\nu_1'\nu_2+(\nu_2')^2-4\pi^2(\nu_2)^2 \notag \\
&\quad \int_{S^{1}} (f_1')^2-2\pi \sqrt{1+r^2}f_1'f_2.
\end{align}
Since $\nu_2 \perp_{L^2} \langle 1,\, \cos(2\pi\cdot),\sin(2\pi\cdot)\rangle$
we have
\begin{align*}
\int_{S^{1}}(\nu_2')^{2}-4\pi^{2}(\nu_2)^{2}\ge \int_{S^{1}}16\pi^{2}(\nu_2)^{2}
\end{align*}
and for $0<r\le 1$
\begin{align*}
\int_{S^{1}} (\nu_1')^2&-2\pi \sqrt{1+r^2}\nu_1'\nu_2+(\nu_2')^2-4\pi^2(\nu_2)^2\\
&\ge
\int_{S^{1}} (\nu_1')^2-\frac{1}{4}(\nu_1')^2-4\pi^2(1+r^2)(\nu_2)^2
+(\nu_2')^2-4\pi^2(\nu_2)^2\\
&\ge
\int_{S^{1}} \frac{3}{4}(\nu_1')^2+4\pi^2(\nu_2)^2.
\end{align*}
Concerning the remaining term in \eqref{eq:5} we note that
as $(-D_{t,g_0}^{2}+1)$ maps
\begin{align*}
\big\{\lambda_1 \dot{\alpha}+ \lambda_2 (\alpha \cross_m \dot{\alpha}) \where
\lambda_1,\lambda_2 \in \langle \sin(2\pi \cdot),\cos(2\pi \cdot)\rangle
\big\}
\end{align*}
into itself and $V \in E_{+}$ there holds
\begin{align*}
f_1 \dot{\alpha}+f_2(\alpha \cross_m \dot{\alpha})
\in (-D_{t,g_0}^{2}+1)^{-1}
\big\langle \cos(2\pi \cdot) \dot\alpha,\sin(2\pi \cdot) \dot\alpha\big\rangle.
\end{align*}
Hence, by explicit computations there are $x,y \in \rz$ satisfying
\begin{align*}
f_1(t) &= x \cos(2\pi t)+ y \sin(2\pi t),\\
f_2(t) &= \frac{8\pi^2\sqrt{1+r^2}}{4\pi^2(2+r^2)+1}
\big(
y\cos(2\pi t)-x\sin(2\pi t)
\big),
\end{align*}
such that
\begin{align*}
\int_{S^{1}} (f_1')^2&-2\pi \sqrt{1+r^2}f_1'f_2
= \frac{2\pi^2(1-4\pi^2r^2)}{4\pi^2(2+r^2)+1}(x^2+y^2).
\end{align*}
This shows that if $r < (2\pi)^{-1}$, then
\begin{align*}
\langle (-D_{t,g_0}^{2}+1) D_{g_0}X_{k_0,g_0,\hz}\eval_{\alpha}(V),V\rangle_{L^{2}} >0
\text{ for all } V \in E_{+}\setminus \{0\},
\end{align*}
and the homotopy
\begin{align*}
[0,1] \ni s \mapsto (1-s) \big(D_{g_0}X_{k_0,g_0,\hz}\eval_{\alpha}\big)\eval_{E_{+}} +s\, id\eval_{E_{+}}
\end{align*}
is admissible.
We use the decomposition in \eqref{eq:4} and
\begin{align*}
D_{g_0}X_{k_0,g_0,\hz}\eval_{\alpha}(\alpha \cross_m \dot{\alpha}) = -\frac{4 \pi^2}{4\pi^2(1+r^2)+1}
(\alpha \cross_m \dot{\alpha})
\end{align*}
to see that under the assumption $r < (2\pi)^{-1}$
\begin{align*}
\big(D_{g_0}X_{k_0,g_0,\hz}\eval_{\alpha}\big)\eval_{R(D_{g_0}X_{k_0,g_0,\hz}\eval_{\alpha})}
\sim
\begin{pmatrix}
-1 & 0\\
0 &   id\eval_{E_{+}}
\end{pmatrix}.
\end{align*}
Consequently, for $r < (2\pi)^{-1}$
\begin{align}
\label{eq:sgn_dx_g0_range}
\sgn \big(D_{g_0}X_{k_0,g_0,\hz}\eval_{\alpha}\big)\eval_{R(D_{g_0}X_{k_0,g_0,\hz}\eval_{\alpha})} =-1.
\end{align}
We remark that
the formula for the degree continues to hold for $r > (2\pi)^{-1}$.\\
From \eqref{eq:3} and the fact that $(\pi_M)_*$ is an isometry we obtain
for $\tilde{\alpha} \in \mathcal{Z}_{M}$
\begin{align}
\label{eq:6}
\{0\}&=\langle \tilde{W}_1,\tilde{W}_2,\tilde{W}_3\rangle
\cap R\big(D_{g_0}X_{k_0,g_0}\eval_{\tilde{\alpha}}\big) ,\notag \\
\langle \tilde{W}_1\rangle^{\perp}&=
\langle \tilde{W}_2,\tilde{W}_3\rangle \oplus R\big(D_{g_0}X_{k_0,g_0}
\eval_{\tilde{\alpha}}\big),
\notag\\
-1 &= sgn \big(D_{g_0}X_{k_0,g_0}\eval_{\tilde{\alpha}}\big)
\eval_{R(D_{g_0}X_{k_0,g_0}\eval_{\tilde{\alpha}})}.
\end{align}
We fix $\tilde{\alpha}_0 \in \mathcal{Z}_{M}$ and a parametrization $\phi$ of $\mathcal{Z}_{M}$,
which maps an open neighborhood of
$0$ in $\langle \tilde{W}_1(\tilde{\alpha}_0),\tilde{W}_2(\tilde{\alpha}_0),\tilde{W}_3(\tilde{\alpha}_0)\rangle$ into $\mathcal{Z}_{M}$, such that
\begin{align*}
\phi(0)=\tilde{\alpha}_0 \text{ and } D\phi\eval_0=id.
\end{align*}
As $\mathcal{Z}_{M}$ consists of smooth functions,
$\mathcal{Z}_{M}$ is a sub-manifold of $H^{m,2}(S^{1},M)$ for $1\le m<\infty$.
We define a map $\Phi$ from an open neighborhood $\mathcal{U}$
of $0$ in
\begin{align*}
T_{\tilde{\alpha}_0}H^{2,2}(S^{1},M)=\langle \tilde{W}_1(\tilde{\alpha}_0),\tilde{W}_2(\tilde{\alpha}_0),\tilde{W}_3(\tilde{\alpha}_0)\rangle \oplus
\text{Range}(D_{g_0}X_{k_0,g_0}\eval_{\tilde{\alpha}_0})
\end{align*}
to $H^{2,2}(S^{1},M)$ by
\begin{align}
\label{eq:15}
\Phi(W,U):= Exp_{\tilde{\alpha}_0,g_0}\big(Exp_{\tilde{\alpha}_0,g_0}^{-1}(\phi(W))+U\big).
\end{align}
Then $(\Phi,\mathcal{U})$ is a chart
of $H^{2,2}(S^{1},M)$ around $\tilde{\alpha}_0$ such that
$\mathcal{U}$ is an open neighborhood of $0$ in $T_{\tilde{\alpha}_0}H^{2,2}(S^{1},M)$,
$\Phi(0)=\tilde{\alpha}_0$, and
\begin{align*}
 D\Phi\eval_0=id,\;
\Phi^{-1}\big(\mathcal{Z}_{M}\cap \Phi(\mathcal{U})\big)= \mathcal{U}\cap \langle \tilde{W}_1(\tilde{\alpha}_0),\tilde{W}_2(\tilde{\alpha}_0),\tilde{W}_3(\tilde{\alpha}_0)\rangle.
\end{align*}
From the properties of $Exp_{\tilde{\alpha}_0,g_0}$ the map $\Phi$ is a chart of $H^{k,2}(S^{1},M)$
around
$\tilde{\alpha}_0$ for any $1\le k \le 4$ and shrinking $\mathcal{U}$ we may assume that
\begin{align}
\label{eq:def_trans_1_phi}
T_{\Phi(V)}H^{1,2}(S^{1},M)
&= \langle \frac{d}{dt}\Phi(V)\rangle \oplus D\Phi\eval_V(\langle \dot\tilde{\alpha}_0\rangle^{\perp,H^{1,2}}),\\
\label{eq:def_trans_2_phi}
T_{\Phi(V)}H^{2,2}(S^{1},M)
&= \langle W_{g_0}(\Phi(V))\rangle \oplus D\Phi\eval_V(\langle W_{g_0}(\tilde{\alpha}_0)\rangle^{\perp}),\\
\label{eq:def_trans_3_phi}
\text{Proj}_{\langle W_{g_0}(\Phi(V)\rangle^\perp}\circ D \Phi\eval_{V}&:\,
\langle W_{g_0}(\tilde{\alpha}_0)\rangle^\perp \xpfeil{\iso}{} \langle W_{g_0}(\Phi(V)\rangle^\perp,
\end{align}
and the norm of the projections in (\ref{eq:def_trans_1_phi}) and (\ref{eq:def_trans_2_phi})
as well as the norm of the map in (\ref{eq:def_trans_3_phi})
and its inverse are uniformly bounded with respect to $V$.

\section{The perturbative analysis}
\label{sec:perturbed}
For $\tilde{\alpha}_0 \in \mathcal{Z}_{M}$
the vectors $\tilde{W}_1(\tilde{\alpha}_0)$ and $W_{g_0}(\tilde{\alpha}_0)$ are collinear
and we use $\langle \tilde{W}_1(\tilde{\alpha}_0)\rangle$
instead of $\langle W_{g_0}(\tilde{\alpha}_0)\rangle$
in the analysis below.\\
We define a $S^{1}$-invariant
vector bundle $SH^{2,2}(S^{1},M)$ by
\begin{align*}
SH^{2,2}(S^{1},M) := \{(\gamma,V)\in TH^{2,2}(S^{1},M)
\where \gamma \neq {\rm const},\,V\in \langle W_g(\gamma)\rangle^\perp\}.
\end{align*}
As in \cite[Sec. 4]{arXiv:0808.4038} we obtain
a chart $\Psi$ for the bundle $SH^{2,2}(S^{1},M)$ around $(\tilde{\alpha}_0,0)$ by,
\begin{align*}
\Psi: \mathcal{U}\times \mathcal{U}\cap \langle \tilde{W}_1(\tilde{\alpha}_0)\rangle^{\perp}
\to SH^{2,2}(S^{1},M),\\
\Psi(V,U):= \big(\Phi(V),Proj_{\langle W_{g_0}(\Phi(V))\rangle^{\perp}}\circ D\Phi\eval_V(U)\big).
\end{align*}
We define
\begin{align*}
X_{g_0,\eps}^{\Phi}:\: \mathcal{U}\cap \langle \tilde{W}_1(\tilde{\alpha}_0)\rangle^{\perp}
\to \langle \tilde{W}_1(\tilde{\alpha}_0)\rangle^{\perp}
\end{align*}
by
\begin{align*}
X_{g_0,\eps}^{\Phi}(V) := Proj_2 \circ \Psi^{-1}\big(\Phi(V),X_{g_0,\eps}(\Phi(V))\big).
\end{align*}
As in \cite[Lem. 3.5]{arXiv:0808.4038} it is easy to see that
\begin{align}
\label{eq:nondegenerate_phi}
V \in \mathcal{U}\cap \langle \tilde{W}_1(\tilde{\alpha}_0)\rangle^{\perp}
\text{ is a (nondegenerate) zero of } X_{g_0,\eps}^{\Phi} \text{ if and only if } \notag \\
S^{1}*\Phi(V) \text{ is a (nondegenerate) critical orbit of } X_{g_0,\eps},
\end{align}
and
if $X_{g_0,\eps}^{\Phi}(V)=0$, then
\begin{align}
\label{eq:d_x_phi}
D_{g_0}X_{g_0,\eps}^{\Phi}\eval_V = A_V^{-1}\circ D_{g_0}X_{g_0,\eps}\eval_{\Phi(V)} \circ D\Phi\eval_V,
\end{align}
where the isomorphism $A_V:\: \langle \tilde{W}_1(\tilde{\alpha}_0)\rangle^{\perp} \to \langle W_{g_0}(\Phi(V))\rangle^{\perp}$
is given by
\begin{align*}
A_V = Proj_{\langle W_{g_0}(\Phi(V))\rangle^{\perp}}\circ D\Phi\eval_V.
\end{align*}
From Lemma \ref{l:kernel_not_in_range} we may assume
\begin{align*}
\mathcal{U}\cap \langle \tilde{W}_1(\tilde{\alpha}_0)\rangle^{\perp} = \mathcal{U}_1\times \mathcal{U}_2,
\end{align*}
where $\mathcal{U}_1$ and $\mathcal{U}_2$ are open neighborhoods of $0$
in $\langle \tilde{W}_2(\tilde{\alpha}_0),\tilde{W}_3(\tilde{\alpha}_0)\rangle$
and $R\big(D_{g_0}X_{k_0,g_0}\eval_{\tilde{\alpha}_0}\big)$.
We denote for $\tilde{\alpha} \in \mathcal{Z}_{M}$ by $P_2(\tilde{\alpha})$ the projection onto
$R(D_{g_0}X_{g_0,0}\eval_{\tilde{\alpha}})$ with respect to the decomposition
\begin{align*}
\langle \tilde{W}_1(\tilde{\alpha})\rangle^{\perp}&=
\langle \tilde{W}_2(\tilde{\alpha}),\tilde{W}_3(\tilde{\alpha})\rangle \oplus R\big(D_{g_0}X_{k_0,g_0}\eval_{\tilde{\alpha}}\big),
\end{align*}
and by $P_1(\tilde{\alpha})$ the projection onto $\langle \tilde{W}_2(\tilde{\alpha}),\tilde{W}_3(\tilde{\alpha})\rangle$.
Moreover, for $W \in \mathcal{U}_1$ we define for $i=1,2$
\begin{align*}
P_i^{\Phi}(W):= (A_W)^{-1}\circ P_i(\Phi(W)) \circ A_W.
\end{align*}
The projections $P_1^{\Phi}(W)$ and $P_2^{\Phi}(W)$ correspond to the decomposition
\begin{align}
\label{eq:decomp_phi_w}
\langle \tilde{W}_1(\tilde{\alpha}_0)\rangle^{\perp}&=
\langle \tilde{W}_2(\tilde{\alpha}_0),\tilde{W}_3(\tilde{\alpha}_0)\rangle \oplus R\big(D_{g_0}X_{g_0,0}^{\Phi}\eval_{W}\big),
\end{align}
as we have for $W \in \mathcal{U}_1$
\begin{align*}
D_{g_0}X_{g_0,0}^{\Phi}\eval_W = A_W^{-1}\circ D_{g_0}X_{g_0,0}\eval_{\Phi(W)} \circ A_W.
\end{align*}
Moreover, for $\tilde{\alpha} \in \mathcal{Z}_{M}$ the vector field $K_1(\tilde{\alpha})$ is
orthogonal to $\tilde{W}_1(\tilde{\alpha})$ and we may define a vector field on $\mathcal{Z}_{M}$ by
\begin{align*}
\mathcal{Z}_{M} \ni \tilde{\alpha} \mapsto P_1(\tilde{\alpha}) \circ K_1(\tilde{\alpha}) \in
\langle \tilde{W}_2(\tilde{\alpha}),\tilde{W}_3(\tilde{\alpha})\rangle.
\end{align*}
Note that $P_1(\cdot) \circ K_1(\cdot)$ is $S^1$-equivariant, i.e.
\begin{align*}
\theta * \big(P_1(\tilde{\alpha}) \circ K_1(\tilde{\alpha})\big)
= P_1(\theta*\tilde{\alpha}) \circ K_1(\theta*\tilde{\alpha}) \text{ for all }(\theta,\tilde{\alpha})
\in S^1\times \mathcal{Z}_{M}.
\end{align*}
If $P_1(\tilde{\alpha}_0) \circ K_1(\tilde{\alpha}_0) =0$ for some $\tilde{\alpha}_0 \in \mathcal{Z}_{M}$
differentiating the identity
\begin{align*}
0 \equiv
\langle P_1(\tilde{\alpha}) \circ K_1(\tilde{\alpha}), \tilde{W}_1(\tilde{\alpha})\rangle
\end{align*}
we find that the covariant derivative
$D_{\mathcal{Z}_{M}}\big(P_1(\cdot) \circ K_1(\cdot)\big)\eval_{\tilde{\alpha}_0}$
maps
\begin{align*}
T_{\tilde{\alpha}_0}\mathcal{Z}_{M}
= \langle \tilde{W}_1(\tilde{\alpha}_0),\tilde{W}_2(\tilde{\alpha}_0),
\tilde{W}_3(\tilde{\alpha}_0)\rangle
\end{align*}
to $\langle \tilde{W}_2(\tilde{\alpha}_0),\tilde{W}_3(\tilde{\alpha}_0)\rangle$
and the $S^1$ equivariance leads to
\begin{align*}
D_{\mathcal{Z}_{M}}\big(P_1(\cdot) \circ K_1(\cdot)\big)\eval_{\tilde{\alpha}_0}
\big(\tilde{W}_1(\tilde{\alpha}_0)\big) =0.
\end{align*}
Consequently, we say that $S^1*\tilde{\alpha}_0 \in \mathcal{Z}_{M}$ is a nondegenerate zero orbit
of $P_1(\cdot) \circ K_1(\cdot)$, if $P_1(\tilde{\alpha}_0) \circ K_1(\tilde{\alpha}_0)=0$ and
\begin{align*}
D_{\mathcal{Z}_{M}}\big(P_1(\cdot) \circ K_1(\cdot)\big)\eval_{\tilde{\alpha}_0}:\:
\langle \tilde{W}_2(\tilde{\alpha}_0),\tilde{W}_3(\tilde{\alpha}_0)\rangle \to \langle \tilde{W}_2(\tilde{\alpha}_0),\tilde{W}_3(\tilde{\alpha}_0)\rangle
\end{align*}
is invertible.\\
Using the above notation the perturbative analysis done in \cite{arXiv:0808.4038}
carries over and we state the following four results without proof
(see \cite[Lem. 5.2-5.5]{arXiv:0808.4038}).
\begin{lemma}
\label{l:implicit_function}
For $\tilde{\alpha}_0\in \mathcal{Z}$ after possibly shrinking $\mathcal{U}$
there are $\eps_0>0$ and
\begin{align*}
U &\in C^2([-\eps_0,\eps_0]\times \mathcal{U}_1,\langle \tilde{W}_1(\tilde{\alpha}_0)\rangle^\perp),\\
R &\in C^2([-\eps_0,\eps_0]\times \mathcal{U}_1,\langle \tilde{W}_2(\tilde{\alpha}_0),\tilde{W}_3(\tilde{\alpha}_0)\rangle),
\end{align*}
such that for all $(\eps,W) \in [-\eps_0,\eps_0]\times \mathcal{U}_1$
\begin{align*}
R(\eps,W) &=X_{g_0,\eps}^{\Phi}(W+U(\eps,W)),\\
0&= P_1^\Phi(W)\circ U(\eps,W),\\
O(\eps)_{\eps \to 0} &=\|U(\eps,W)\|+\|D_W U(\eps,W)\|+\|R(\eps,W)\|+\|D_W R(\eps,W)\|,\\
R(\eps,W)&= \eps P_1^\Phi(W)\circ K_1^{\Phi}(W)+o(\eps)_{\eps \to 0},\\
U(\eps,W) &= -\eps (D_{g_0}X_{g_0,0}^\Phi\eval_{W})^{-1}\circ  P_2^\Phi(W)\circ K_1^{\Phi}(W)+o(\eps)_{\eps \to 0}.
\end{align*}
Moreover, the functions $U(\eps, W)$ and $R(\eps,W)$ are unique, in the sense that,
if $(\eps,W,U,R)$ in $[-\eps_0,\eps_0]\times \mathcal{U}_1\times \mathcal{U}\cap\langle \tilde{W}_1(\tilde{\alpha}_0)\rangle^\perp
\times \mathcal{U}_1$
satisfies
\begin{align*}
X_{g_0,\eps}^{\Phi}(W+U)=R \text{ and }
P_1^\Phi(W)\big(U\big) =0,
\end{align*}
then $U=U(\eps,W)$ and $R=R(\eps,W)$.
\end{lemma}

\begin{lemma}
\label{l:expansion_eps_0}
Under the assumptions of Lemma \ref{l:implicit_function} we have
as $\eps \to 0$
\begin{align*}
X_{g_0,\eps}^{\Phi}(W+U(\eps,W))
&= \eps P_1^\Phi(W)\circ K_1^{\Phi}(W) + O(\eps^{2})_{\eps \to 0},
\end{align*}
where $K_1^{\Phi}$ is the vector-field $K_1$ in the coordinates $\Phi$, i.e.
\begin{align*}
K_1^{\Phi} = X_{g_0,1}^\Phi-X_{g_0,0}^\Phi.
\end{align*}
\end{lemma}

\begin{lemma}
\label{l:nondeg_zero}
Under the assumptions of Lemma \ref{l:implicit_function}
suppose $0$ is a nondegenerate zero of the vector-field $P_1^\Phi(\cdot)\circ K_1^\Phi(\cdot)$,
in the sense that $P_1^\Phi(0)\circ K_1^\Phi(0)=0$
and
\begin{align*}
D_W(P_1^\Phi(\cdot)\circ K_1^\Phi(\cdot))\eval_0
\in \mathcal{L}(\langle {W}_2(\tilde{\alpha}_0),{W}_3(\tilde{\alpha}_0)\rangle)
\end{align*}
is an isomorphism. Then, after possibly shrinking $\eps_0$ and $\mathcal{U}$,
for any $0<\eps\le \eps_0 $ there is a unique $W(\eps)\in \mathcal{U}_1$
such that
\begin{align*}
X_{g_0,\eps}^{\Phi}(W(\eps)+U(\eps,W(\eps)))=0,\\
W(\eps) \to 0 \text{ as } \eps \to 0.
\end{align*}
Moreover, $V(\eps):= W(\eps)+U(\eps,W(\eps))$ is the only zero of
$X_{g_0,\eps}^{\Phi}$ in $\mathcal{U}\cap \langle \tilde{W}_1(\tilde{\alpha}_0)\rangle^{\perp}$
and is nondegenerate with
\begin{align*}
\sgn(D X_{g_0,\eps}^{\Phi}\eval_{V(\eps)})
&= -det(D_W(P_1^\Phi(\cdot)\circ K_1^\Phi(\cdot))\eval_0).
\end{align*}
\end{lemma}

\begin{lemma}
\label{l:nondeg_zero_gamma}
Under the assumptions of Lemma \ref{l:implicit_function}
suppose $\tilde{\alpha}_0$ is a nondegenerate zero of the vector-field $P_1(\cdot)\circ K_1(\cdot)$ on $\mathcal{Z}_{M}$,
in the sense that $P_1(\tilde{\alpha}_0)\circ K_1(\tilde{\alpha}_0)=0$
and
\begin{align*}
D_{\mathcal{Z}}(P_1(\cdot)\circ K_1(\cdot))\eval_{\tilde{\alpha}_0}
\in \mathcal{L}(\langle {W}_2(\tilde{\alpha}_0),{W}_3(\tilde{\alpha}_0)\rangle)
\end{align*}
is an isomorphism. Then
for any $0<\eps<\eps_0$ there is $\tilde{\gamma}(\eps) \in \Phi(\mathcal{U})$ satisfying
\begin{align*}
X_{g_0,\eps}(\tilde{\gamma}(\eps))=0 \text{ and }
\tilde{\gamma}(\eps) \to \tilde{\alpha}_0 \text{ as } \eps \to 0.
\end{align*}
Moreover, $S^1*\tilde{\gamma}(\eps)$ is the unique critical orbit of $X_{g_0,\eps}$ in $\Phi(\mathcal{U})$
and is nondegenerate with
\begin{align*}
\deg_{loc, S^{1}}(X_{g_0,\eps},S^1*\tilde{\gamma}(\eps)) &= -\det(D_{\mathcal{Z}_{M}}(P_1(\cdot)\circ K_1(\cdot))\eval_{\tilde{\alpha}_0}).
\end{align*}
\end{lemma}

\section{The computation of the degree}
\label{sec:computation_degree}
In order to compute the $S^{1}$-degree of $X_{g_0,\eps}$ we choose a smooth Morse function
$k_1 \in C^{\infty}(M,\rz)$.
The corresponding vector-field $K_1$ on $H^{2,2}(S^{1},M)$ is given by
\begin{align*}
K_1(\gamma) = (-D_{t,g_0}^{2}+1)^{-1}(|\dot \gamma|_{g_0} k_1(\tilde{\gamma}) J_{g_0}(\gamma)\dot\gamma).
\end{align*}
We note that for $\tilde{\alpha} = \pi_M \circ \alpha(\cdot,2\pi|r|,v_0,v_1,w) \in \mathcal{Z}_{M}$
and $r>0$ small enough we have
\begin{align*}
K_1(\tilde{\alpha}) = 2\pi r
(-D_{t,g_0}^{2}+1)^{-1}
\big(
k_1(\tilde{\alpha}) (\phi_M)_*(\alpha \times_m \dot\alpha)
\big).
\end{align*}
Consequently, from \eqref{eq:D2_1_w2}, \eqref{eq:D2_1_w3}, and \eqref{eq:4}
\begin{align}
\label{eq:11}
P_1(\tilde{\alpha})\circ K_1(\tilde{\alpha}) =
\sigma_2(\tilde{\alpha}) \tilde{W}_2(\tilde{\alpha})
+ \sigma_3(\tilde{\alpha}) \tilde{W}_3(\tilde{\alpha}),
\end{align}
where $\sigma_2(\tilde{\alpha}),\, \sigma_2(\tilde{\alpha}) \in \rz$ are defined by the condition that
\begin{align*}
2\pi r k_1(\tilde{\alpha})
-\frac{\sigma_2(\tilde{\alpha})}{2\pi r}
(1-4\pi^{2}r^{2})\cos(2\pi \cdot)
-\frac{\sigma_3(\tilde{\alpha})}{2\pi r}
(1-4\pi^{2}r^{2})\sin(2\pi \cdot)
\end{align*}
is $L^{2}$-orthogonal to $\langle \cos(2\pi \cdot), \sin(2\pi \cdot)\rangle$.
Hence,
\begin{align*}
\sigma_2(\tilde{\alpha}) &= \frac{8\pi^{2} r^{2}}{1-4\pi^{2}r^{2}}
\int^{1}_{0} k_1\circ \tilde{\alpha}(t) \cos(2\pi t) \, dt,\\
\sigma_3(\tilde{\alpha}) &= \frac{8\pi^{2} r^{2}}{1-4\pi^{2}r^{2}}
\int^{1}_{0} k_1\circ \tilde{\alpha}(t) \sin(2\pi t) \, dt.
\end{align*}
In the following we are interested in the asympotics of $\sigma_2$ and
$\sigma_3$ as $r \to 0^+$ or equivalently as $k_0 \to \infty$.
There holds
\begin{align}
\label{eq:7}
&\frac{1-4\pi^{2}r^{2}}{8\pi^{2} r^{2}}\sigma_2(\tilde{\alpha}) \notag \\
&\quad = \int^{1}_{0}
\Big( k_1\circ \pi_M(w)
+ r dk_1\eval_{\pi_M(w)} \cos(2\pi t)(\pi_M)_* v_1 \notag \\
&\qquad +r dk_1\eval_{\pi_M(w)}\sin(2\pi t)(\pi_M)_* v_0 + O(r^2)\Big)
\cos(2\pi t) \, dt \notag\\
&\quad = \frac{1}{2} r dk_1\eval_{\pi_M(w)} (\pi_M)_* v_1 + O(r^2),
\end{align}
and analogously we find
\begin{align}
\label{eq:9}
\frac{1-4\pi^{2}r^{2}}{8\pi^{2} r^{2}}\sigma_3(\tilde{\alpha})
&=  \frac{1}{2} r dk_1\eval_{\pi_M(w)} (\pi_M)_* v_0 + O(r^2).
\end{align}
From the above expansion we easily deduce
\begin{lemma}
\label{l:k_1_1}
For all $\delta>0$ there is $r_0>0$ such that for all $0<r\le r_0$ and
\begin{align*}
\tilde{\alpha}= \pi_M(\sqrt{1+r^2}w+r\cos(2\pi t)v_1+r\sin(2\pi t)v_0) \in \mathcal{Z}_{M}
\end{align*}
satisfying $P_1(\tilde{\alpha})\circ K_1(\tilde{\alpha}) =0$ there holds
\begin{align*}
\pi_M(w) \in \cupl_{i=1}^n B_{\delta}(\tilde{w}_i),
\end{align*}
where $\{\tilde{w}_i\where 1\le i \le n\}$ denotes the set of critical points of $k_1$
in $M$.
\end{lemma}
Fix $w_0 \in \hz$ and a positive orthonormal system $\{v_0,v_1,w_0\}$ in $(\rz^3,m)$ such that
$\pi_M(w_0)$ is a critical point of $k_1$ in $M$. We choose $\delta>0$, a parametrization
\begin{align*}
w:\: B_1(0) \subset \rz^2 \to B_{\delta}(w_0) \subset \hz,\;
(x,y) \mapsto w(x,y),
\end{align*}
and smooth maps $v_0, v_1:\: B_1(0) \to \rz^3$ such that
$\{v_0(x,y),v_1(x,y),w(x,y)\}$ is orthonormal for all $(x,y) \in B_1(0)$
and
\begin{align}
\label{eq:10}
(v_0(0,0),v_1(0,0),w(0,0))=(v_0,v_1,w_0),\,
\frac{\rand w}{\rand x}\eval_{(0,0)}=v_1,\, \frac{\rand w}{\rand y}\eval_{(0,0)}=v_0.
\end{align}
Shrinking $\delta>0$ we may assume that
$\pi_M \circ \phi_w$ parametrizes $M$ and that
$(x,y) \mapsto \tilde{\alpha}(x,y)$
is an injective immersion from $B_1(0)$ to $\mathcal{Z}_M$, where
\begin{align*}
\tilde{\alpha}(x,y) :=
\pi_M \big(\sqrt{1+r^2} w(x,y)+r \cos(2\pi \cdot) v_1(x,y) +r \sin(2\pi \cdot) v_0(x,y)\big).
\end{align*}
From \eqref{eq:12} and \eqref{eq:10} we get as $r,\delta \to 0^+$
\begin{align}
\label{eq:13}
\frac{\rand}{\rand x} \tilde{\alpha}\eval_{(x,y)}
&= \tilde{W}_2(\tilde{\alpha}(x,y)) + O(r)+O(\delta) \notag\\
\frac{\rand}{\rand y} \tilde{\alpha}\eval_{(x,y)}
&= \tilde{W}_3(\tilde{\alpha}(x,y)) + O(r)+O(\delta).
\end{align}
Define $H:\: B_1(0) \to \rz^2$ by
\begin{align*}
H(x,y) := \big(\sigma_2(\tilde{\alpha}(x,y)),\sigma_3(\tilde{\alpha}(x,y))\big).
\end{align*}
By \eqref{eq:7} and \eqref{eq:9} we have as $r \to 0^+$
\begin{align*}
\frac{H(x,y)}{8\pi^2r^3} &:=
\big(dk_1\eval_{\pi_M(w(x,y))} (\pi_M)_* v_1(x,y),
dk_1\eval_{\pi_M(w(x,y))} (\pi_M)_* v_0(x,y)\big)\\
&\quad +O(r).
\end{align*}
Since
\begin{align*}
\frac{d}{dx} \pi_M \circ w\eval_{0,0} = (\pi_M)_* v_1(0,0),\;
\frac{d}{dy} \pi_M \circ w\eval_{0,0} = (\pi_M)_* v_0(0,0)
\end{align*}
we find for small values of $\delta>0$ and $r>0$
\begin{align*}
\deg(H,B_1(0),0)&= \deg(\nabla (k_1\circ \pi_M \circ w),B_1(0),0)\\
&= \deg(\nabla k_1,B_{\delta}(\pi_M(w_0)),0)
= \deg_{loc}(\nabla k_1,\pi_M(w_0)),
\end{align*}
and the set of zeros of $H$ in $B_1(0)$ is non-empty. Fix a zero $(x_0,y_0) \in B_1(0)$ of $H$.
Then
\begin{align*}
dH\eval_{(x_0,y_0)} =
\begin{pmatrix}
\frac{\rand}{\rand x}(\sigma_2 \circ \tilde{\alpha})\eval_{(x_0,y_0)} &
\frac{\rand}{\rand y}(\sigma_2 \circ \tilde{\alpha})\eval_{(x_0,y_0)}\\
\frac{\rand}{\rand x}(\sigma_3 \circ \tilde{\alpha})\eval_{(x_0,y_0)} &
\frac{\rand}{\rand y}(\sigma_3 \circ \tilde{\alpha})\eval_{(x_0,y_0)}\\
\end{pmatrix}.
\end{align*}
From \eqref{eq:7}, \eqref{eq:9}, and the fact that $H(x_0,y_0)=0$ we get
\begin{align*}
dk_1\eval_{\pi_M(w(x_0,y_0))} = O(r).
\end{align*}
Thus, we have as $r \to 0^+$
\begin{align*}
dk_1&\eval_{\tilde{\alpha}(x_0,y_0)(t)} \frac{\rand}{\rand x}\tilde{\alpha}\eval_{(x_0,y_0)}(t)\\
&= dk_1\eval_{\pi_M(w(x_0,y_0))} (\pi_M)_* \frac{\rand }{\rand x}w\eval_{(x_0,y_0)}\\
&\quad + r \Big(\nabla_{\frac{\rand}{\rand r} \tilde{\alpha}(x_0,y_0)(t)\eval_{r=0}}
dk_1\eval_{\pi_M(w(x_0,y_0))} (\pi_M)_*\frac{\rand }{\rand x}w\eval_{(x_0,y_0)}\\
&\quad +dk_1\eval_{\pi_M(w(x_0,y_0))}
\nabla_{\frac{\rand}{\rand r} \tilde{\alpha}(x_0,y_0)(t)\eval_{r=0}}
\frac{\rand}{\rand x}\tilde{\alpha}(x_0,y_0)(t)\eval_{r=0}\Big)
+O(r^2)\\
&= dk_1\eval_{\pi_M(w(x_0,y_0))} (\pi_M)_* \frac{\rand }{\rand x}w\eval_{(x_0,y_0)}\\
&\quad + r (\nabla dk_1)\eval_{\pi_M(w(x_0,y_0))}
\Big((\pi_M)_* \frac{\rand }{\rand x}w\eval_{(x_0,y_0)},\\
&\qquad\cos(2\pi t) (\pi_M)_* v_1(x_0,y_0)+\sin(2\pi t) (\pi_M)_* v_0(x_0,y_0)\Big)
+O(r^2).
\end{align*}
Using \eqref{eq:10}, this leads to, as $r,\delta \to 0^{+}$
\begin{align*}
\frac{\rand}{\rand x}&(\sigma_2 \circ \tilde{\alpha})\eval_{(x_0,y_0)}\\
&= \int_0^1 dk_1\eval_{\tilde{\alpha}(x_0,y_0)(t)}
\frac{\rand}{\rand x}\tilde{\alpha}\eval_{(x_0,y_0)}(t)
\cos(2\pi t) dt\\
&=  \frac{r}{2} (\nabla dk_1)\eval_{\pi_M(w(x_0,y_0))}
((\pi_M)_* \frac{\rand }{\rand x}w\eval_{(x_0,y_0)}, (\pi_M)_* v_1(x_0,y_0))
+O(r^2)\\
&=  \frac{r}{2} (\nabla dk_1)\eval_{\pi_M(w_0)}
\Big((\pi_M)_* v_1, (\pi_M)_* v_1\Big)
+O(r^2)+O(r\delta).
\end{align*}
Analogously, we may compute the remaining partial derivatives of $H$
and we find for small values of $\delta>0$ and $r>0$
\begin{align}
\label{eq:14}
\sgn \det(dH\eval_{(x_0,y_0)}) =
\sgn \det(\nabla dk_1\eval_{\pi_M(w_0)})=\deg_{loc}(\nabla k_1,\pi_M(w_0)),
\end{align}
such that $(x_0,y_0)$ is the unique zero of $H$ in $B_1(0)$.
From \eqref{eq:11} we see that
\begin{align*}
P_1(\tilde{\alpha}(x_0,y_0)) \circ K_1(\tilde{\alpha}(x_0,y_0)) =0,
\end{align*}
by \eqref{eq:13} we obtain as $r,\delta \to 0^+$
\begin{align*}
&\nabla_{\tilde{W}_2}\big(P_1(\cdot) \circ K_1(\cdot)\big)\eval_{\tilde{\alpha}(x_0,y_0)}\\
&\quad= \Big(d\sigma_2\eval_{\tilde{\alpha}(x_0,y_0)} \tilde{W}_2(\tilde{\alpha}(x_0,y_0))\Big)
\tilde{W}_2(\tilde{\alpha}(x_0,y_0))\\
&\qquad +
\Big(d\sigma_3\eval_{\tilde{\alpha}(x_0,y_0)} \tilde{W}_2(\tilde{\alpha}(x_0,y_0))\Big)
\tilde{W}_3(\tilde{\alpha}(x_0,y_0))\\
&\quad = \frac{\rand}{\rand x}(\sigma_2 \circ \tilde{\alpha})\eval_{(x_0,y_0)}
\tilde{W}_2(\tilde{\alpha}(x_0,y_0))\\
&\qquad +
\frac{\rand}{\rand x}(\sigma_3 \circ \tilde{\alpha})\eval_{(x_0,y_0)}
\tilde{W}_3(\tilde{\alpha}(x_0,y_0)) + O(r)+O(\delta).
\end{align*}
Concerning the covariant derivative of $P_1(\cdot) \circ K_1(\cdot)$ in
direction $\tilde{W}_3$ we have to replace $\frac{\rand}{\rand x}$
by $\frac{\rand}{\rand y}$ in the above formula. Consequently,
from \eqref{eq:14}
\begin{align*}
\sgn \det(D_{\mathcal{Z}_{M}}(P_1(\cdot)\circ K_1(\cdot))\eval_{\tilde{\alpha}(x_0,y_0)})
= \deg_{loc}(\nabla k_1,\pi_M(w_0)).
\end{align*}
Thus we arrive at the following
\begin{lemma}
\label{l:k_1_2}
Let $\{\tilde{w}_i\where 1\le i \le n\}$ denote the set of critical points of $k_1$
in $M$. Then there is $r_0>0$ such that
for all $0<r\le r_0$ the set of critical orbits of
$P_1(\cdot )\circ K_1(\cdot)$ is given by $\{S^1*\tilde{\alpha}_{i,r}\where 1\le i \le n\}$,
where
\begin{align*}
\tilde{\alpha}_{i,r}= \pi_M(\sqrt{1+r^2}w_{i,r}+r\cos(2\pi t)v_{1,i,r}
+r\sin(2\pi t)v_{0,i,r}) \in \mathcal{Z}_{M}.
\end{align*}
Moreover, we have for $1\le i\le n$
\begin{align*}
\pi_M(w_{i,r}) \to  \tilde{w}_i \text{ as }r \to 0^+,\\
\sgn \det(D_{\mathcal{Z}_{M}}(P_1(\cdot)\circ K_1(\cdot))\eval_{\tilde{\alpha}_{i,r}})
= \deg_{loc}(\nabla k_1,\tilde{w}_i).
\end{align*}
\end{lemma}
\begin{proof}
From Lemma \ref{l:k_1_1} and the analysis of $H$ we may choose $\delta>0$ and $r_0>0$
such that the union $\cup_{i=1}^n B_{\delta}(\tilde{w}_i)$ is disjoint and
for every $i$ and $0<r\le r_0$
there is a unique $\pi_M(w_{i,r}) \in B_{\delta}(\tilde{w}_i)$ corresponding
to a critical orbit $S^1*\tilde{\alpha}_{i,r}$. Moreover,
if $r \to 0^+$ we may shrink $\delta>0$, which yields together with the uniqueness
of $\pi_M(w_{i,r})$ the claimed asymptotic.
\end{proof}

\begin{lemma}
\label{l:euler_charak_unperturbed}
Let $\mathscr{M}_A$ be the set of oriented Alexandrov embedded regular curves
in $H^{2,2}(S^1,M)$. There is $C_{k_0}>0$ such that for all $k_0\ge C_{k_0}$
we have
\begin{align*}
\chi_{S^1}(X_{k_0,g_0},\mathscr{M}_A)=-\chi(M),
\end{align*}
where $\chi(M)$ denotes the Euler characteristic of $M$.
\end{lemma}
\begin{proof}
We choose a Morse function $k_1$ on $M$ with nondegenerate critical points
$\{\tilde{w}_i\where 1\le i \le n\}$.
From Lemma \ref{l:k_1_2} we obtain $C_{k_0}>0$ such that for all
$k_0\ge C_{k_0}$ the critical orbits of $P_1(\cdot)\circ K_1(\cdot)$
are given by $\{S^1*\tilde{\alpha}_{i,k_0}\where 1\le i \le n\}$ satisfying
\begin{align*}
\sgn \det(D_{\mathcal{Z}_{M}}(P_1(\cdot)\circ K_1(\cdot))\eval_{\tilde{\alpha}_{i,k_0}})
= \deg_{loc}(\nabla k_1,\tilde{w}_i).
\end{align*}
We fix $k_0\ge C_{k_0}$.
By Lemma \ref{l:nondeg_zero} there is $\eps>0$ such that
for any $0<\eps<\eps_0$ and $1\le i\le n$
there is $\tilde{\gamma}_{i}(\eps) \in \Phi(\mathcal{U}_i)$ satisfying
\begin{align*}
X_{g_0,\eps}(\tilde{\gamma}_{i}(\eps))=0 \text{ and }
\tilde{\gamma}(\eps) \to \tilde{\alpha}_{i,k_0} \text{ as } \eps \to 0.
\end{align*}
Moreover, $S^1*\tilde{\gamma}_{i}(\eps)$ is the unique critical orbit of $X_{g_0,\eps}$
in $\Phi(\mathcal{U}_i)$ and is nondegenerate with
\begin{align}
\label{eq:16}
\deg_{loc, S^{1}}(X_{g_0,\eps},S^1*\tilde{\gamma}(\eps)) &= -\deg_{loc}(\nabla k_1,\tilde{w}_i).
\end{align}
To show that there is an open neighborhood
$\mathcal{U}$ of $\mathcal{Z}_M$ and $\eps_0>0$
such that for all $0<\eps<\eps_0$ the critical orbits
of $X_{g_0,\eps}$ in $\mathcal{U}$ are given exactly
by $\{\tilde{\gamma}_i(\eps)\where 1 \le i \le n\}$
we argue by contradiction. Suppose there are $\eps_n \to 0^+$ and
a sequence $(\tilde{\alpha}_n)$ of zeros of $X_{g_0,\eps_n}$ that converges to
$\mathcal{Z}_M$ but $\tilde{\alpha}_n \notin \{\tilde{\gamma}_i(\eps)\where 1 \le i \le n\}$.
Up to a subsequence we may assume
\begin{align*}
\tilde{\alpha}_n \to \tilde{\alpha}_0 \in \mathcal{Z}_M
\end{align*}
as $n \to \infty$. For large $n$ we use the chart $\Phi$ around $\tilde{\alpha}_0$
given in \eqref{eq:15}.
From the existence of a slice of the $S^{1}$-action (see \cite[Lem. 3.1]{arXiv:0808.4038})
we get sequences $\theta_n \in \rz/\zz$ and
$V_n \in \langle \tilde{W}_1(\tilde{\alpha}_0)\rangle^\perp$
converging to $0$ such that
\begin{align*}
\theta_n*\tilde{\alpha}_n = \Phi(V_n).
\end{align*}
Note that from the $S^1$-invariance and by construction
\begin{align*}
X_{g_0,\eps_n}(\theta_n*\tilde{\alpha}_n)=0 \text{ and }
X_{g_0,\eps_n}^{\Phi}(V_n)=0.
\end{align*}
We consider the map
\begin{align*}
\Lambda:\:
\langle \tilde{W}_1(\tilde{\alpha}_0)\rangle^{\perp}=
\langle \tilde{W}_2(\tilde{\alpha}_0),\tilde{W}_3(\tilde{\alpha}_0)\rangle
\oplus R\big(D_{g_0}X_{g_0,0}^{\Phi}\eval_{0}\big)
\to \langle \tilde{W}_1(\tilde{\alpha}_0)\rangle^{\perp},
\end{align*}
defined by
\begin{align*}
\Lambda(W,V) := W + D_{g_0}X_{g_0,0}^{\Phi}\eval_{W}(V).
\end{align*}
From \eqref{eq:decomp_phi_w} the map $\Lambda$ is a diffeomorphism
locally around $(0,0)$, hence
we may decompose
\begin{align*}
V_n= \Phi^{-1}(\theta_n*\alpha_n)= W_n+U_n,
\end{align*}
where $W_n \in \langle \tilde{W}_1(\alpha_0)\rangle^\perp$ and
$U_n \in R(D_{g_0}X_{k_0,g_0}^\Phi\eval_{W_n})$ converge to $0$
as $n \to \infty$.
From the uniqueness part of Lemma \ref{l:implicit_function},
as $X_{g_0,\eps_n}^{\Phi}(W_n+U_n)=0$,
we get $U_n= U(\eps_n,W_n)$. By Lemma \ref{l:expansion_eps_0} we see that
\begin{align*}
P_1(\tilde{\alpha}_0)\circ K_1(\tilde{\alpha}_0)=0.
\end{align*}
Consequently, $S^1*\alpha_0 \in \{S^1*\tilde{\alpha}_{i,k_0}\where 1\le i \le n\}$.
From the uniqueness part in Lemma \ref{l:nondeg_zero} we finally arrive at the contradiction
\begin{align*}
S^1*\tilde{\alpha}_n \in \{S^1*\tilde{\gamma}_{i}(\eps_n)\where 1 \le i \le n\}.
\end{align*}
From the definition of the $S^1$-equivariant Poincar\'{e}-Hopf index,
the classification of Alexandrov embedded zeros of $X_{k_0,g_0}$, and \eqref{eq:16}
there holds for small $\eps>0$
\begin{align*}
\chi_{S^1}(X_{k_0,g_0},\mathscr{M}_A) &= \chi_{S^1}(X_{k_0,g_0},\mathcal{U})
= \chi_{S^1}(X_{g_0,\eps},\mathcal{U})\\
&= \sum_{i=1}^n \deg_{loc, S^{1}}(X_{g_0,\eps},S^1*\tilde{\gamma}_i(\eps))\\
&= -\sum_{i=1}^n \deg_{loc}(\nabla k_1,\tilde{w}_i) = -\chi(M).
\end{align*}
\end{proof}

\section{The apriori estimate}
\label{sec:apriori-estimate}
We fix a continuous family of metrics $\{g_t\where t\in [0,1]\}$ on $M$
and a continuous family of positive continuous function $\{k_t \where t\in [0,1]\}$ on $M$.
We assume that there is $K_0>0$,
such that the Gaussian curvature $K_{g_t}$ of each metric $g_t$ on $M$
and the functions $\{k_t\}$ satisfy
\begin{align}
\label{eq:17}
K_{g_t} \ge -K_0,\\
\label{eq:18}
k_{inf}:= \inf\{k_t(x)\where (x,t)\in M\times [0,1]\}>(K_0)^\frac12.
\end{align}
We let $X_{t}$ be the vector field on $H^{2,2}(S^1,M)$ defined by
\begin{align*}
X_t:= X_{k_t,g_t}.
\end{align*}
We denote by $\mathscr{M}_A\subset H^{2,2}(S^1,M)$ the set
\begin{align*}
\mathscr{M}_A:= \{\gamma \in H^{2,2}_{reg}(S^1,M)
\where \gamma \text{ is prime and oriented Alexandrov embedded.}\}.
\end{align*}
We shall show that the set
\begin{align*}
X^{-1}(0) := \{(\gamma,t) \in \mathscr{M}_A\times [0,1] \where X_t(\gamma)=0\}
\end{align*}
is compact in $\mathscr{M}_A\times [0,1]$.
Fix $(\gamma,t)\in X^{-1}(0)$. Then there is an oriented immersion $F:\overline{B}\to M$
with $F\eval_{\rand B}=\gamma$. We denote by $F^*g_t$ the induced metric on $B$.
\begin{lemma}
\label{lem_soln_prescribed_gauss}
For any $(\gamma,t)\in X^{-1}(0)$
there is $\phi \in C^2(\overline{B},\rz)$ satisfying
\begin{align}
\label{eq:19}
-\laplace_{F^*g_t}\phi + K_{F^*g_t}+K_0 e^\phi = 0 \text{ in } B, \notag \\
\rand_{\nu} \phi = 0 \text{ on } \rand B,
\end{align}
where $\nu$ denotes the unit normal oriented to the outside.\\
Moreover, there is $C_0>0$, which may be chosen independently of $(\gamma,t)\in X^{-1}(0)$,
such that
\begin{align*}
0\ge \phi \ge -C_0.
\end{align*}
\end{lemma}
\begin{proof}
To show the existence of a solution $\phi$ we use the method of upper and lower solutions
(see also \cite{MR0343205}).
The function $\phi_+ \equiv 0$ satisfies
\begin{align*}
-\laplace_{F^*g_t}\phi_+ + K_{F^*g_t}+K_0 e^{\phi_+} = K_{F^*g_t}+K_0 \ge 0,
\end{align*}
from \eqref{eq:17} and the fact that $F$ is a local isometry. Hence, $\phi_+$
is a supersolution of \eqref{eq:19}. To find a subsolution, we let $\phi_1 \in C^\infty(M,\rz)$
be defined as the solution to the linear equation
\begin{align*}
-\laplace_{g_t}\phi_1 + K_{g_t}-2\pi \chi(M) \text{vol}(M,g_t)^{-1}= 0 \text{ in }M,\;
\int_M \phi_1 dg_t =0.
\end{align*}
By standard elliptic estimates using a Green's function on $(M,g_t)$
(see \cite[Thm 4.13]{MR1636569}) we have
\begin{align*}
\sup_{M}|\phi_1| &\le C(g_t)
\big(\sup_{M}|K_{g_t}|- 2\pi \chi(M) \text{vol}(M,g_t)^{-1}\big)\\
&\le C_1,
\end{align*}
because $\{g_t\where t\in [0,1]\}$ is a compact set of smooth metrics.
We may choose $C_2>1$ such that we have for all $t\in [0,1]$
\begin{align*}
-C_2 \le \ln\big(-2\pi \chi(M)K_0 \text{vol}(M,g_t)^{-1}\big).
\end{align*}
Since $F$ is a local isometry, there holds
\begin{align*}
\laplace_{F^*g_t} (\phi_1 \circ F)= (\laplace_{g_t} \phi_1) \circ F.
\end{align*}
We define $\phi_- \in C^2(B,\rz)$ by
\begin{align*}
\phi_-:= \phi_1\circ F -C_1-C_2
\end{align*}
and get
\begin{align*}
-\laplace_{F^*g_t}\phi_- + K_{F^*g_t}+K_0 e^{\phi_-} =
2\pi \chi(M) \text{vol}(M,g_t)^{-1} +K_0 e^{\phi_1 \circ F-C_1-C_2}
\le 0.
\end{align*}
Hence $\phi_-$ is a subsolution of \eqref{eq:19} satisfying
\begin{align*}
-C_0 := -(2C_1+C_2) \le \phi_-  < \phi_+.
\end{align*}
Using a version of the method of upper and lower solutions
given in \cite{MR2316146} we find a solution $\phi$ to \eqref{eq:19}
satisfying $\phi_- \le \phi \le \phi_+$.
\end{proof}
We consider $B$ equipped with the metric  $h_t:= e^\phi F^* g_t$.
Then the Gaussian curvature $K_{h_t}$ and the geodesic
curvature $k_{h_t}$ of $\rand B$ with respect to $(B,h_t)$ are given by
(see \cite[Sec 5.8.2]{MR1636569})
\begin{align*}
K_{h_t} \equiv -K_0 \text { and } k_{h_t} = k_{F^*g_t} e^{-\frac{\phi}{2}}.
\end{align*}
Consequently, since $0\le \phi$,
\begin{align*}
\inf_{\rand B} k_{h_t} \ge \inf_{\rand B} k_{F^*g_t}
\ge k_{\inf}.
\end{align*}
The Gauss-Bonnet formula applied to $(B,h_t)$ gives
\begin{align*}
2\pi &= -\int_{B} K_0 \,dh_t +
\int_{\rand B}k_{h_t} \, dS_{h_t}
&\ge
-K_0 A(B,h_t)+k_{inf} L(\rand B, h_t),
\end{align*}
where $A(B,h_t)$ denotes the area of $B$ and $L(\rand B, h_t)$ the length of $\rand B$
with respect to $h_t$.
The isoperimetric inequality (see \cite[Thm 4.3]{MR0500557}) yields
\begin{align*}
L(\rand B, h_t)^2 \ge 4\pi A(B,h_t) + K_0 A(B,h_t)^2 \ge K_0 A(B,h_t)^2.
\end{align*}
Thus we arrive at
\begin{align*}
2\pi \ge -(K_0)^{\frac12} L(\rand B, h_t) + k_{inf} L(\rand B, h_t).
\end{align*}
This yields
\begin{align*}
L(\gamma,g_t) &= L(\rand B,F^* g_t) \le e^{C_0} L(\rand B,h_t)
\le e^{C_0} \frac{2\pi}{k_{inf}- (K_0)^{\frac12}}.
\end{align*}
Using again the Gauss-Bonnet formula we see
\begin{align*}
2\pi &= -\int_{B} K_0 \,dh_t +
\int_{\rand B}k_{h_t} \, dS_{h_t}\\
&\le e^{\frac{C_0}{2}}\big(\sup\{k_t(x)\where (x,t)\in M\times [0,1]\}\big)
L(\rand B,h_t)\\
&\le e^{2C_0}\big(\sup\{k_t(x)\where (x,t)\in M\times [0,1]\}\big)
L(\gamma,g_t).
\end{align*}
Consequently, there is $C>0$, such that
\begin{align}
\label{eq:21}
C \le L(\gamma,g_t) \le C^{-1},
\end{align}
for all $(\gamma,t) \in X^{-1}(0)$.\\
Fix a sequence $(\gamma_n,t_n)_{n\in \nz}$ in $X^{-1}(0)$.
As a solution each $\gamma_n$ is parameterized proportional
to its arc-length. From \eqref{eq:21},
$(\gamma_n)$ is uniformly bounded in $C^1(S^1,M)$. Using the equation
\eqref{eq:1} we obtain a uniform bound of $(\gamma_n)$
in $C^3(S^1,M)$, such that we may extract a subsequence, still denoted
by $(\gamma_n,t_n)_{n \in \nz}$, which
converges in $C^2(S^1,M)\times [0,1]$ to $(\gamma_0,t_0)$.
The convergence in $C^2(S^1,M)$ and the lower bound in \eqref{eq:21} imply that
$X_{t_0}(\gamma_0)=0$ and that $\gamma_0$ is an immersion.
By Lemma \ref{lem:basis_alexandrov}
the curve $\gamma_0$ is oriented Alexandrov embedded and hence
$(\gamma_0,t_0)\in X^{-1}(0)$. This shows that
\begin{lemma}
\label{l:apriori-estimate}
Under the assumptions \eqref{eq:17} and \eqref{eq:18}
the set $X^{-1}(0)$ is compact.
\end{lemma}

\section{Existence results}
\label{sec:existence}
We give the proof of our main existence result.
\begin{proof}[{Proof of Theorem \ref{thm_existence}}]
From the uniformization theorem
$(M,g)$
is isometric to $(\hz/\Gamma,e^\phi g_0)$,
where $\Gamma \subset O(2,1)_+$ is a group
of isometries acting
freely and properly discontinuously and $\phi$ is a function in
$C^\infty(\hz/\Gamma,\rz)$.
Due to the invariance of \eqref{eq:1} under isometries we may assume
without loss of generality that
\begin{align*}
(M,g) = (\hz/\Gamma,e^\phi g_0).
\end{align*}
We consider the family of metrics $\{g_t:= e^{t\phi} g_{0} \where t\in [0,1]\}$
and choose a large constant $k_0>>1$, such that
\begin{align*}
k_0 > \big(-\inf\{K_{g_t}(x) \where (x,t) \in M\times [0,1]\}\big)^\frac12
+\inf_{M}k + C_{k_0},
\end{align*}
where $K_{g_t}$ denotes the Gaussian curvature of the metric $g_t$ given by
\begin{align*}
K_{g_t} &= e^{-t\phi}\big(-t\laplace_{g_0}(\phi)+2\big).
\end{align*}
From Lemma \ref{l:apriori-estimate} the homotopy
\begin{align*}
[0,1]\ni t\mapsto X_{k_0,g_t}
\end{align*}
is $(\mathscr{M}_A,g_t,S^1)$-admissible. By Lemma \ref{l:euler_charak_unperturbed}
and the homotopy invariance of the $S^{1}$-equivariant Poincar\'{e}-Hopf index
we obtain
\begin{align*}
-\chi(M)= \chi_{S^1}(X_{k_0,g_0},\mathscr{M}_A)=\chi_{S^1}(X_{k_0,g},\mathscr{M}_A).
\end{align*}
For $t\in [0,1]$ we define $k_t \in C^\infty(M,\rz)$ by
\begin{align*}
k_t(x) := (1-t)k_0 + t k(x).
\end{align*}
Then
\begin{align*}
\inf\{k_t(x)\where (x,t)\in M\times [0,1]\}= \inf_{M}k >
\big(-\inf_{M}K_{g}\big)^\frac12.
\end{align*}
From Lemma \ref{l:apriori-estimate} the homotopy
\begin{align*}
[0,1]\ni t\mapsto X_{k_t,g}
\end{align*}
is $(\mathscr{M}_A,g,S^1)$-admissible and there holds
\begin{align*}
\chi_{S^1}(X_{k,g},\mathscr{M}_A)=\chi_{S^1}(X_{k_0,g},\mathscr{M}_A)= -\chi(M).
\end{align*}
This gives the claim.
\end{proof}

\bibliographystyle{plain}

redpajama_set_name: RedPajamaArXiv


text:
The hidden world of Somalia's abandoned pharmaceutical industry
My parents had left Somalia before the start of the civil war, and they only had good memories of the place they had called home all their lives. This changed very quickly , as the civil war completely devastated Somalia, a country with a capital that was once hailed the 'pearl of the Indian Ocean'. I received a Bachelor's degree in Medical pharmacology from Coventry University where I gained an appreciation for the methodical and rigorous guidelines pharmaceutical companies in the UK must adhere to, thus ensuring consumer safety. Subsequently, I then completed a master's degree in Global Health and Development at the University College London, where I was able to use my prior knowledge of pharmacology to understand the significance that access to safe, quality medicines have on health systems. My background gave me a unique perspective on the burgeoning health issues around the world. As a first-generation child of immigrants, my Somali heritage and culture were always very close to me , thus it was important for me to write my Master's thesis on Somalia.
I had heard countless stories from my parents and family friends about people in Somalia who had been fallen ill and used medication they purchased from unlicensed pharmacies and subsequently became worse or even died. This deeply disturbed me and inspired me to delve into Somalia's pharmaceutical industry. My initial thesis topic was looking at how the unregulated pharmaceutical industry in Somali impacted key health indicators in the country. However, after a preliminary bibliographic search, it was apparent that there was a lack of research and data on this topic. It was very disheartening for me to see , however, it also made me understand that since there was not enough research to base a thesis on, there definitely was not enough to base well-informed health policies on. Consequently, I then revised my research question to look at health system in failed and fragile states, with Somalia following the 1991 civil war as a case study. My research yielded a lot of harsh truths about the current state of Somalia's ailing pharmaceutical industry and of the health system as a whole.
The current private, unregulated health system has resulted in widespread imports of cheap counterfeit and substandard medicines. Local pharmacies and laboratory services in Somalia have been rapidly expanding as a cheaper alternative to seeking medical services at licenced and certified health centres and pharmacies. UNICEF conducted interviews with healthcare providers in Somalia in 2012 and reported that the majority of pharmacies did not check the quality of pharmaceutical products they had procured. Furthermore, serial numbers and batch numbers were not regularly referenced to ensure that they had received the correct shipment. If quality assurance checks were carried out, the most that would happen would be pharmacy workers confirming that brand names were spelled correctly, and the overall general impressions were 'good'. A combination of low literacy rates, poor health literacy and low health seeking behaviour levels, made patients more susceptible to ignoring medical advice given by their healthcare workers or to not seek advice on the administration and dosage regiments entirely.
Over the counter medication that patients receive from local pharmacies cannot be regulated and controlled and therefore are not always effective or safe. The usage of these unregulated medicines can be very detrimental and, in some cases fatal. For patients, the lack of regulation allows importers and their international distributers to use Somalia as a pharmaceutical dumpsite for their expired and falsified medications. Dr Dufle, a physician who runs one of the largest private hospitals in Mogadishu said in a recent article that falsified medicines are "worse than bombs, because a bomb kills 10 or 20 people. But these drugs can kill hundreds of people and no one will hear about it". This profound statement emphasized the magnitude of despair caused by falsified and counterfeit medications in Somalia.
Somali health officials have acknowledged that there is a wide scale problem in accessing safe medicines, however without firstly strengthening the healthcare system, it would be near impossible to provide safer alternative methods to procuring medical products. However, due to a myriad of contributing factors that stem from a poorly funded and weakened healthcare system, it is not feasible to attempt to change this without implementing regulatory policies and investing in the necessary technology and personnel.
This is why campaigns like Fight The Fakes are particularly important in low resource settings, as they are able to raise awareness on this detrimental issue that is very often overlooked. They are able to focus on advocacy and bringing in multi-sectoral stakeholders to the discussion without being anchored down by the never-ending tides of politics in healthcare. My work at Fight The Fakes has allowed me to bring my passion of Global Health and my unwavering hope of a better Somalia to fruition.
Nabina Ibrahim, Fight the Fakes Secretariat
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Categories:	STORIESUPDATES

redpajama_set_name: RedPajamaCommonCrawl


Gregor Meritis knows he's inviting trouble when he helps a wounded shape-shifter—especially when the law of the land is to kill them on sight—but he can't pass up the opportunity to study one before they become extinct. She disturbs the quiet order of his scholarly existence, vexes him in more ways than he can count, and encourages him to break enough of the kingdom's laws so that not even being the king's cousin will save his head. When she challenges him to join her on a quest to find the legendary Tears of Elios, he discovers magic far more powerful than he's ever imagined. But once he unlocks their secrets, will he be able to use their power for good? Or will he let them fall into the hands of a tyrant who's out to destroy all non-humans, including the shape-shifter who's captured his heart?
E-Book available at iBooks, B&N, Google Play, Kobo, and Amazon .
An epilogue was written for my readers who asked what happens to Kira and Galen.


text:
Good afternoon, Ursinus!
Complete Archive of the Lantern Available in Digital Commons
Explore the History of the Ursinus College Literary Magazine
A digital archive of the Lantern Literary Magazines is available to browse, search and download in the Ursinus College Digital Commons. Dating back to the first magazine published in May 1933, over 180 issues are available. The magazine has changed in many ways over the past 85 years, but one will always find a wonderful blend of student fiction, nonfiction, poetry and art. Some of the earliest issues also feature book reviews and nonfiction pieces related to campus history and the local community. As explained in the first issue, the "Lantern" name was chosen "because it represents a distinctive feature of campus architecture, and because it symbolizes the light shed by creative work." Reading this creative work can provide a fascinating glimpse into the past of Ursinus College as well as the historic events which shaped previous generations of students here. Physical copies of some past issues are also available to read on the main floor of Myrin Library and all issues can be viewed by appointment in the College Archives.
The Lantern: Ursinus's Literary Magazine
Andy Prock
Scholarly Communications & Metadata Librarian
aprock@ursinus.edu
Carolyn Weigel
College Archivist
cweigel@ursinus.edu


We are collecting toys and clothes for children in need in our town to brighten their holiday!
Your contribution would be greatly appreciated by selecting an ornament from the tree.
Click Tree of Giving Flyer for more details.
Thank you for all your support and Happy Holidays!


Chewing is a behavior in which all dogs engage. And it's not just when they are eating that they like to chew. It's used in play, hunting, and eating. Did you know, dog dentition is shaped differently than our own? The majority of dog teeth are conical in shape with the exception of the back molar teeth. Additionally, dogs have 5-10 times less enamel than we have covering the surface of our teeth.
Over the past number of years, you may have heard about groomers, pet shops and even some veterinary providers offering a procedure referred to as Anesthesia Free Dentistry (AFD) or No Anesthesia Dentistry (NAD). These procedures involve the scaling of a dog or cat's teeth with an instrument without the use of anesthesia. In these procedures, the animal must still be restrained while only calculus above the gingival margin (gum line) is removed.
Similar to people, dogs and cats can develop tumors within their oral cavity. A tumor is an abnormal growth of tissue that can be benign (non-cancerous) or malignant (cancerous). Oral tumors in cats and dogs may arise from dental tissue (odontogenic), inflammation (periodontal disease), or can be malignant (cancerous). Oral neoplasia—an uncontrolled growth of cells that is often cancerous—accounts for up to 10% of all feline tumors and 6% of canine tumors.
Dogs frequently fracture teeth. Studies have shown that up to 10% of dogs have some type of fractured tooth. All fractured teeth should be evaluated by a veterinarian and be imaged with a dental radiograph. If the fracture does not expose the tooth's pulp and the radiograph is normal, it can be treated with a procedure that smooths the fracture and seals the exposed dentin. This kind of procedure will decrease any possible pain and inhibit infection.
What is a COHAT? Why We Don't Just Call it a "Pet Dental Cleaning"


text:
(In)Convenient Choices and Queer Affective Lives: The Story of My Mother and Me
By Sohini Chatterjee Categories Choice and Sexuality Voices March 5, 2019
Making choices is akin to constructing, stumbling upon, or re-signifying meanings by ourselves and/or those with whom we share affective ties. Choices are never absolutely infused with agency, and this is especially true for queer people. Even when we demonstrate agency, choices are strategic and critical in the moment of their making, revealing our situation even through our hesitations while making them or not making them at all. Choice making, active or passive, is then reflective of the nature of our familial and social relationships and the heteropatriarchal conditions that inform it.
I came out of the closet a few years ago, for the first time, by unleashing on my mother the difficult truth of my queerness – a revelation she was ill-equipped to handle at the time, despite my family's liberal pretences. For a while she was sustained by the comfort of denial, evidenced by her casual remark that I was queer because of my cerebral curiosity in gender and sexuality studies. At 21, I lacked the intellectual and emotional wherewithal to establish my identity through astute, and enervating, contestations. I chose silence over resistance, realising strategic self-censorship would be the pragmatic thing to do, given the circumstances. My queerness was not discussed thereafter, as I moved out of home for two years and dated a cisgender heterosexual man during that time. My mother probably did not think my queerness would ever be a point of discussion or contention between us. For all intents and purposes, I had become straight to her. Ironically however, those two years paved the way for my mother's evolved understanding of feminism and sexuality.
After the relationship with the man ended and I returned home, I started coming out to her with renewed conviction, I realised a lot of groundwork had already been done and she had learnt and chosen to be quietly accommodating instead of being blithely dismissive of my queerness. Our queer journey has so far been an expedition involving myriad quiet agreements, respectful exchanges, gentle understanding, and discreet but sound support despite the inconvenience of convenient choices leading her to tolerance, and helping her inch towards, often unspoken, acceptance.
My relationship with my mother was improved by the distance between us – physical and intellectual. I introduced into her vocabulary the language of rights, justice, bodily autonomy and agency as I was beginning to develop my own nuanced understanding of these concepts in and out of university. This was a world of critical exploration of being and becoming that she had not had access to in her time. But much of this resonated with her as someone who had performed her own infractions by marrying my father against socio-cultural norms and norms of morality. I realised for the first time that she was willing and able to recognise my agency over my body when I told her, after coming back home last year, at the risk of a discord between us, that I was travelling to see my then boyfriend and would be staying at his place for a few days. "That's up to you", my mother had said. It was not helpless withdrawal but recognition of my agency. That was the day I realised my mother was willing to recognise the choices I was capable of, and will be, making as an adult individual. Her choices fostered my evolution and strengthened my own choices.
Conversations about my queerness began slowly and casually the second time around. I chose to not furnish this information with the arrogance and entitlement I had demonstrated earlier, without realising that my mother is from a generation where conversations about gender and sexuality were not had openly, uninhibitedly, critically, and were not part of the public discourse. Yet, she was amenable to these conversations, was willing to address her own prejudices and recognise the violence of erasure. It was perhaps because she had had to fight a similar fight, albeit differently, with her family, that had also entailed questions of romantic choice, cohabitation, familial piety, and the flouting of patriarchal diktats, that she could attempt to cultivate an understanding of my queerness. She was no stranger to the world of romantic dissidence, but her heterosexual privilege had made her acquaintance with it short-lived. The fundamental difference between us is that her dissidence was disciplined and sanitised through marriage, whereas mine will never be because the law of the land does not permit it, and even if it did, I remain doubtful if I would ever have enough faith in the institution of marriage to commit to it. However, our transgressions are not disparate and we have now begun to be united by them.
The first time I fell in love with a girl, I had nobody to confide in. Now, whether they last a day, a week, or a few months, my mother is aware of all my infatuations. However, they are not easy admissions, no matter how necessary they are. She is learning a new way of loving and living through me, and I have to be patient while she comes to terms with it. Patience is a conscious choice, as are constant negotiations in the lived realities of queer people within and outside of the family. Those with the latter are more intimate and hence all the more challenging. The road from understanding to acceptance is a long one, and often bumpy. There are multiple choices – from the well-thought out to the organic – that are made for this journey to sustain itself.
A few days ago, I was telling my mother about the impossibility of a queer future with someone who lived in a different city. "But I really like her", I found myself switching from our native Bangla to a foreign tongue– a language we hardly speak at home. Hiding behind this language was a choice that acted as an armour, helping me in the articulation of desires that hitherto had no space in my home and the language that was spoken in it. Home is always associated with a language that carries with it the burden of its prejudices. While on our way to undoing prejudicial inhibitions, we often choose to dissociate the language from it, not entirely, but only in critical moments to re-introduce it into the language of desire. The intercession of English helps us move on and brokers a truce with my discomfiture.
I talk and my mother listens to stories of my love and longing in foreign and familiar tongues. She listens most of the time without saying anything in return. For a long time, I identified her silence with sorrow. Now, a year later, I have found language and reassurance in it. Silence is often a choice she makes in trying to navigate her role as a single parent of a queer child. Her silence had seemed torturous to me for a long time. I felt unheard when she would look down at her phone while I told her of a budding queer romance or of queer conversations with a long lost friend. But then there are moments like this that nod towards the progress we have made: I told her recently I was going to Mumbai to meet someone, and in reply my mother said something along the lines of, "But doesn't your love interest live in Delhi?"
My mother and I have both made certain choices, sometimes inconvenient for me, sometimes difficult for her, but those choices have revealed to me the strength of our relationship and alternative possibilities that she and I can imagine together. My choices may be more vocal but hers are sounder in their headway against prejudices she is choosing to unlearn, everyday, bit by bit. Our choices, even though now compatible, are performed in keeping with our vastly different lives. Yet, they are more complimentary than conflicting.
ChoiceconversationsFamilymotherQueerSexuality
Sohini Chatterjee / About Author
Sohini Chatterjee identifies as a queer feminist and writes primarily on gender, culture and politics. She holds an MA in International Relations.
More posts by Sohini Chatterjee


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'BLUE chip' companies, SMEs and not-for profit organisations will have the opportunity of learning how to create and develop winning brands from some of the country's top marketing and brand building experts.
Promotional merchandise marketing experts, Orb Group, has announced the lineup of speakers at their showcase at Glasgow's Marriott hotel on Tuesday March 12.
Managing director, Mike Smith, revealed that the event will feature seminars from marketing and brand-building experts who have been involved in the creation and development of top brands.
Mike said: "Laura Vernett from Harper Macleod and Miles Duncan from Leverage Business Performance will be giving their insight on marketing for professional services.
"We have Graeme Atha from the Marketing Society talking about how to inspire bolder leadership in marketing.
Seminars throughout the day include, expert speakers on current best practices, campaign case studies, cutting edge marketing strategies, how to win new business and integrating social media with your campaign.
Follow #orbevents for more information.
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