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%! TEX program = lualatex

\input{.maindir/tex/header/preamble-section}
% inputs the preamble only if necessary
\docStart
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\section{Orientability of continuous line fields} \label{sec:continuous_q_lift}
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{ %intro
  The projection operator $P$ maps two opposite points on the sphere onto the corresponding line.
  Compare this with the construction of the real projective space $ℝℙ^d$ ($d ∈ ℕ$): We start with the sphere $𝕊^d$ and identify opposite points.
  This construction also shows that the sphere is a two-sheet covering space of $ℝℙ^d$.
  This similarity shows that $𝒬^{𝕊}M$ is a fiber bundle with fiber $ℝℙ^{m-1}$ and the projection $P$ is a covering map.
  \begin{definition}[Covering map] \label{def:covering_map}
    (from \textcite[1.3 Covering Spaces][56]{q4-q23-algebraic-topology})
    A \newTerm{covering space} of a topological space $X$ is a topological space $\tilde X$ together with a map $p ⫶ \tilde X → X$ satisfying the following condition:
    Each point $x ∈ X$ has an open neighborhood $U$ in $X$ such that $p^{-1}(U)$ is a union of disjoint open sets in $\tilde X$, each of which is mapped homeomorphically onto $U$ by $p$. %Such an $U$ is \newTerm{evenly covered} and
    The disjoint open sets in $\tilde X$ that project homeomorphically to $U$ by $p$ are called \newTerm{sheets} of $\tilde X$ over $U$.
  \end{definition} % end definition of Covering map
  This way algebraic geometry can answer the question of orientability.
  In the language of algebraic topology an orientation of a line field is a \newTerm{lift} from a map into $ℝℙ^{m-1}$ to a map into $𝕊^{m-1}$.
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  In this chapter we use algebraic topology to give a characterization of orientability of continuous line fields which generalises a result from \textcite[Theorem 1][506]{q4-Ball2011}.
  %
  % \textcite[3. Orientability issues][504]{q4-Ball2011} has a very good introduction as to why use algebraic topology.%
  % \footnote{Currently I cannot write it better. But at some point I probably should rephrase it.}
  %
  % $𝒬^{𝕊}_pM$ can be constructed as a subspace of $T^{(0,2)}T_pM$ but also from $𝕊_pM$  by identifying opposite points.
  % This establishes $𝕊_pM$ as a covering space of $𝒬^{𝕊}M \equiv ℝP^{m-1}$.
  % Algebraic topology gives us tools to determine which continuous maps can be lifted to a covering space.
}
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Since we want to lift from $𝒬^{𝕊}M$ to $𝕊M$, we need to check if $P ⫶ 𝕊M → 𝒬^{𝕊}M$ is a covering map.
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\begin{lemma}[Projection as a covering map] \label{thm:projection_as_a_covering_map}
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  The projection operators $P ⫶ 𝕊M → 𝒬^{𝕊}M$ and $P_N⫶𝕊^{N-1} → 𝒬^{𝕊'}^N$ ($N ∈ ℕ$) are covering maps with two sheets.
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\end{lemma} % end lemma Projection as a covering map
\begin{proof} % of lemma Projection as a covering map
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  $P$ is continuous since $n ↦ s ⸨ n ⊗ n - \frac1m g^{♯♯}$ ($m = \dim M$) is continuous as a map from $TM$ to $T^{(2,0)}TM$.
  $P_N$ is continuous since $n ↦ n ⊗ n$ is continuous as a map from $𝕊^{N-1}$ to $T^{(2,0)}^N$.
  For $n ∈ 𝕊M$ or $𝕊^{N-1}$ choose a neighborhood $U$ that has a diameter of less than $2$ (measured in $TM$ or $^N$ respectively).
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  Then $P^{-1}(P(U)) = U ∪ -U$ (same for $P_N$) and $U$ and $-U$ are disjoint because the distance between opposite points is $2$.
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  Since $P$ and $P_N$ are quadratic, they have continuous inverses if they are bijective. Hence they map $U$ and $-U$ homeomorphically onto $P(U)$.
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\end{proof} % of lemma Projection as a covering map

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We want to use the theorem about lifting of continuous map from algebraic topology from \autocite[Proposition 1.33][61]{q4-q23-algebraic-topology}): % page 70 in pdf
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\begin{externalresult}[Lifting of continuous maps] \label{thm:lifting_of_continuous_maps}
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  Suppose given a connected covering space $p ⫶(\tilde X, \tilde x_0)(X, x_0)$ and a map $f⫶(Y, y_0)(X,x_0)$ with $Y$ path-connected and locally path-connected.
  Then a lift $\tilde f ⫶ (Y, y_0)(\tilde X, \tilde x_0)$ of $f$ exists if and only if $f_* ⸨ π_1(Y, y_0) ⸩ ⊆ p_* ⸨ π_1(\tilde X, \tilde x_0)$.
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\end{externalresult} % end proposition Lifting of continuous maps
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{ % notation explanation
  The spaces in this theorem are \newTerm{pointed spaces}.
  They are equipped with one distinguished point which is used to define the fundamental group $π_1$.
  $π_1$ is the group of equivalence classes of loops through this point.
  The equivalence relation is being homotopic.
  The notation $f_*$ and $p_*$ denotes the induced homomorphism defined by $f_*([α]) = [f ∘ α]$ where $α$ is a loop in the domain of $f$ and $[α]$ is the homotopy equivalence class of $α$.%
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  We use \cref{thm:lifting_of_continuous_maps} to formulate a condition for the orientability of a continuous line field:
  we can check orientability on a selected number of loops.
  To define the field spaces we need notation to restrict functions to submanifolds.
  $Γ_R(\rest)$%
  \footnote{is this notation obvious enough? Currently it is just mentioned very shortly here without a number to refer to because I think it's very obvious and barely needs any explanation.}
  is the space of sections of regularity $R$ into the vector bundle $E$ on the submanifold $γ ⊂ M$.
}
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\begin{theorem}[Orientability of continuous line fields] \label{thm:orientability_of_continuous_line_fields}
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  Consider a line field $Q ∈ Γ_C(𝒬^{𝕊}M)$.
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  Let $G$ be a set of loops at a common point $p_0$ that generate the fundamental group $π_1(M, p_0)$.
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  $Q$ is orientable, \ie there exists $n ∈ Γ_C(𝕊M)$ such that $P(n) = Q$, if and only if $Q$ is orientable along every path $γ ∈ G$.
\end{theorem} % end theorem Orientability of continuous line fields
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\begin{proof} % of theorem Orientability of continuous line fields
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  As \cref{thm:projection_as_a_covering_map} shows, $P ⫶ 𝕊M → 𝒬^{𝕊}M$ is a covering map.
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  Use $Q_{p_0}$ and a $n_0 ∈ 𝕊M$ with $P(n_0) = Q_{p_0}$ as the distinguished points.

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  Connected manifolds are path-connected and locally path-connected.\footnote{it's correct, do I need to proof it? Not hard but needs some lines.}
  % $Q_* ⸨ π_1(M, p_0) ⸩ ⊆ P_* ⸨ π_1(𝕊M, n_0) ⸩$ (for $n_0 ∈ 𝕊_{p_0}M$ such that $Q_{p_0} = P(n_0)$)
  % is equivalent to all generators of $π_1(M, p_0)$ being in $P_* ⸨ π_1(𝕊M, n_0) ⸩$.

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  For the \enquote{if} part 
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  let $γ ∈ G$, \ie $γ ⫶[0,1] → M$ continuous with $γ(0) = γ(1) = p_0$.
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  $Q$ is orientable along $γ$. That means that there exists $n ∈ Γ_C(\rest{𝕊M}{γ([0,1])})$ with $P(n) = Q$.
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  If the orientation of $Q$ along this path has $n(γ(0)) = -n_0$, use $-n$ instead.
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  Then $n ∘ γ$ is a continuous loop at $n_0$ in $𝕊M$ with $P_* [n ∘ γ] = [P ∘ n ∘ γ] = [Q ∘ γ] = Q_* [γ]$.
  Since $γ$ is an arbitrary generator of $π_1(M, p_0)$ and $Q_*$ and $P_*$ are group homomorphisms, we get that $Q_*(π_1(M, p_0)) ⊆ P_*(π_1(𝕊M, n_0))$.
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  Hence there exists $n ∈ C(M, 𝕊M)$ with $P(n) = Q$ by \cref{thm:lifting_of_continuous_maps}, \ie $Q$ is orientable and $n ∈ Γ_C(𝕊M)$.
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  For the \enquote{and only if} part restrict an orientation for $Q$ on the paths in $G$. 
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\end{proof} % of theorem Orientability of continuous line fields
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We formulate the same in the language of \cref{sec:embedding_M} and allow fields that are not tangent to $M$.
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\begin{lemma}[Orientability of continuous line fields via embedding of \texorpdfstring{$M$}{M}] \label{thm:orientability_of_line_fields_via_embedding_of_M}
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  Let $ι$ be an isometric embedding of $M$ into $^N$ as in \cref{sec:embedding_M}
  and let $Q ∈ C(M, 𝒬^{𝕊'}^N)$ be a line field on but not necessarily tangent to $M$.
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  Let $G$ be a set of loops at a common point $p_0$ that generate the fundamental group $π_1(M, p_0)$.
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  Then $Q$ is orientable, \ie there exists $n ∈ C(𝕊^{N-1})$ such that $P_N(n) = Q$, if and only if $Q$ is orientable along every path $γ ∈ G$.
\end{lemma} % end theorem Orientability of line fields on embedded M
\begin{proof} % of theorem Orientability of line fields on embedded M
  As \cref{thm:projection_as_a_covering_map} showed, $P_N ⫶ 𝕊^{N-1} → 𝒬^{𝕊'}^N$ is a covering map.
  Use $Q(p_0)$ and a $n_0 ∈ 𝕊^{N-1}$ with $P(n_0) = Q(p_0)$ as the distinguished points.

  Connected manifolds are path-connected and locally path-connected.%
  \footnote{ggf. vorherigen Beweis zitieren}
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  For the \enquote{if} part 
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  let $γ ∈ G$, \ie $γ ⫶[0,1] → M$ continuous with $γ(0) = γ(1) = p_0$.
  $Q$ is orientable along $γ$. That means that there exists $n ∈ C(γ([0,1]), 𝕊^{N-1})$ with $P_N(n) = Q$.
  If the orientation of $Q$ along this path has $n(γ(0)) = -n_0$, use $-n$ instead.
  Then $n ∘ γ$ is a continuous loop at $n_0$ in $𝕊^{N-1}$ with $(P_N)_* [n ∘ γ] = [P_N ∘ n ∘ γ] = [Q ∘ γ] = Q_* [γ]$.
  Since $γ$ is an arbitrary generator of $π_1(M, p_0)$ and $Q_*$ and $P_*$ are group homomorphisms, we get that $Q_*(π_1(M, p_0))(P_N)_*(π_1(𝕊^{N-1}, n_0))$.
  Hence there exists $n ⫶ M → 𝕊^{N-1}$ with $P(n) = Q$ by \cref{thm:lifting_of_continuous_maps}, \ie $Q$ is orientable.
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  For the \enquote{and only if} part restrict an orientation for $Q$ on the paths in $G$. 
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\end{proof} % of theorem Orientability of line fields on embedded M
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{\textbf{Here another version with background colors instead of [].}
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  \newcommand*{\ver}[2]{\sethlcolor{Lavender}\hl{#1}\sethlcolor{Yellow}\hl{#2}}
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  \cref{thm:orientability_of_continuous_line_fields} and \cref{thm:orientability_of_line_fields_via_embedding_of_M} are very similar.
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  Therefore the proofs are combined.
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  \ver{Parts in pink are for \mbox{\cref{thm:orientability_of_continuous_line_fields}}}{and parts in yellow are for \mbox{\cref{thm:orientability_of_line_fields_via_embedding_of_M}}}.%
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  \textinput{proof_continuous_lift}
}
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{\textbf{Here another version with stacked text instead of [].}
  \newcommand*{\ver}[2]{$\begin{cases} \text{#1} \\ \text{#2} \end{cases}$}
  % use newcommand here instead of renewcommand because the first \newcommand
  % is inside { } and therefore local to this { }
  % hence \renewcommand at this place causes \ver not defined error
  \cref{thm:orientability_of_continuous_line_fields} and \cref{thm:orientability_of_line_fields_via_embedding_of_M} are very similar.
  Therefore the proofs are combined.
  \ver{The top line is for \cref{thm:orientability_of_continuous_line_fields}}{and the bottom line is for \cref{thm:orientability_of_line_fields_via_embedding_of_M}}.
  \textinput{proof_continuous_lift}
}

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{\textbf{Here another version with background colors and general notes.}
  \newcommand*{\ver}[2]{\sethlcolor{Lavender}\hl{#1}\sethlcolor{Yellow}\hl{#2}}
  \cref{thm:orientability_of_continuous_line_fields} and \cref{thm:orientability_of_line_fields_via_embedding_of_M} are very similar.
  Therefore the proofs are combined.
  \ver{Parts in pink are for \mbox{\cref{thm:orientability_of_continuous_line_fields}}}{and parts in yellow are for \mbox{\cref{thm:orientability_of_line_fields_via_embedding_of_M}}}.
  Additionally exchange $𝕊M$ with $𝕊^{N-1}$ and $P$ with $P_N$ for the proof of \cref{thm:orientability_of_line_fields_via_embedding_of_M}.
  \textinput{proof_continuous_lift_simplified}
}



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\begin{corollary}[Orientability on simply connected manifolds] \label{thm:orientability_on_simply_connected_manifolds}
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  If $M$ is simply connected every continuous line field $Q ∈ Γ_C(𝒬^{𝕊}M)$ or $Q ∈ C(M, 𝒬^{𝕊'}^N)$ is orientable.
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\end{corollary} % end corollary Orientability on simply connected manifolds
\begin{proof} % of corollary Orientability on simply connected manifolds
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  $M$ being simply connected means that $π_1(M, p_0)$ is the trivial group for any point $p_0 ∈ M$. Hence it has no generators.
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  Therefore the assumption in \cref{thm:orientability_of_continuous_line_fields} or \cref{thm:orientability_of_line_fields_via_embedding_of_M} is trivially fulfilled.
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\end{proof} % of corollary Orientability on simply connected manifolds
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\begin{note}[Topological restriction on the manifold] \label{note:topological_restriction_on_the_manifold}
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   \textcite{q52-poincare-hopf-theorem} showed that the existence of unit vector fields is a topological property of the manifold.
  There exists a unit vector fields on a smooth, compact manifold without boundary if and only if the Euler characteristic is zero.
  This result is called the \newTerm{Poincaré-Hopf Theorem}.
  % He furthermore showed that the Euler characteristic is equal to a sum of indices of isolated zeros of a vector field.
  For surfaces--two-dimensional manifolds--this implies that the only one with unit-vector fields is the torus.
  For all other compact boundaryless surfaces the question of orientability of continuous line fields is mute.
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\end{note} % end note Topological restriction on the manifold
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\begin{example}[Sphere] \label{ex:sphere}
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  The sphere $𝕊^2$ is simply connected but has Euler characteristic $2$.
  Therefore every continuous line field is orientable but there exist no continuous unit vector fields and hence no continuous line fields.

  On the $2$-sphere the Poincaré-Hopf Theorem is called \newTerm{Hairy Ball Theorem} \parencite{hairy-ball}.
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  Since the sphere and slightly distorted spheres are a very common object of study and real-life application many applications circumvent this problem in some way.
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  Usually the fields have \newTerm{point defects} which are points on the sphere where the line or unit vector field is not well-defined or $0$.
  For example \textcite{q53-application-shells-defects} studies tiny droplets with liquid crystals on the surface. Since the liquid crystal layer is so thin in comparison to the size of the droplet it can be modeled as two-dimensional.
  Then the liquid crystal must have defects that influence the properties of the entire droplet like bonds to neighboring particles.
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\end{example} % end example Sphere

\begin{example}[Torus] \label{ex:torus}
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  \footnote{todo: this example uses \enquote{orientability on path works} and \enquote{exactly two different orientations} which I have dropped so far. One might say they are of independent interest and should be proven anyway. If they are dropped, this section needs rewriting. Otherwise maybe as well.}
  The torus does have line and unit vector fields
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  but it is not simply connected.
  In order to construct an non-orientable line field we take a loop that is not homotopic equivalent to a point, \eg a circle around the hole. Call it $γ⫶[0,1]→M$.
  We know that any line field can be oriented along a path.
  So on $γ([0,1))$ we can find an orientation, so the problem must occur at $γ(0) = γ(1)$ and $Q$ must somehow force the oriented version $n$ to \enquote{turn around} along the curve.
  After defining $Q$ suitably on this loop we also have to check if we can extend it to all of $M$.
  For simple computation use the square $[-1, 1]^2$ with opposite sides identified.
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  Define the (not continuous) $n(x, y) = \sin(/2) \D x + \cos(/2) \D y$.
  It is continuous on $(-1, 1)^2$ and $n(0, y) = -n(1, y)$ and $n(x, 0) = n(x, 1)$.
  Hence $Q = P(n)$ is continuous on the entire torus but on the path $γ ⫶[0,1] → M$, $t ↦ (t, 0)$ an orientation of $Q$ it must be equal to $n$ (or $-n$) on $(0,1)$ but then has a discontinuity at $γ(0) = γ(1)$.
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  \footnote{tikz picture. Maybe copy from \textcite[Chapter 4][20]{q37-index_formula_VMO}}
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\end{example} % end example Torus

\begin{example}[Circle] \label{ex:circle}
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  The natural question that arises from \cref{thm:orientability_of_continuous_line_fields} is if the converse is also true, \ie if we can find a non-orientable line field on every non-simply connected manifold.
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  In this simple form the converse is not true.
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  For this consider the circle $𝕊^1$. $T_p𝕊^1$ is one-dimensional at every $p∈𝕊^1$.
  Therefore $𝕊^*𝕊^1$ consists of only two vectors at every base point and $𝒬^{𝕊}𝕊^1$ even only of one. Therefore there is only exactly one line field on the sphere and it is orientable.
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  \footnote{It appears plausible that the technique for the torus works for most manifolds of dimension 2 and higher since \enquote{turning around} on a loop seems easy and then we \enquote{only} need to extends this vector field to all of $M$. But this might not be trivial as the example of the sphere shows.
  }
\end{example} % end example Circle
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\todo{This paragraph about orientability on paths fits to continuous line lifting (this section) but is somehow more basic than the other theorems.
  So I would expect it to come at the beginning but the introductory text there does not fit with it because it's not necessary for the other results.
Now it's here at the end but a better thought-through structure of the section would be good.}
Before we consider lifting a restricted $Q$-tensor field on any manifold, we first look at paths.
In the flat case this was done in \textcite[Lemma 3][505]{q4-Ball2011} and we will reduce the manifold case to this.%
% it is actually used when showing that in 2-D we have a global ON-frame
% \footnote{it's quite likely that we will not need \cref{thm:orientability_on_a_path}. So remove it if not used}
\begin{proposition}[Orientability on a path] \label{thm:orientability_on_a_path}
  Let $M$ be a Riemannian $m$-dimensional manifold, $- ∞ < t_1 < t_2 < ∞$, $γ ⫶[t_1, t_2] → M$ be a continous injective path on $M$
  and $Q ∈ Γ_C(\rest{𝒬^{𝕊}M}{γ([t_1, t_2])})$ be a continous restricted $Q$-tensor field along the path $γ$.
  Then there exist exactly two continous maps (called \newTerm{liftings}\footnote{indeed new term? todo}) $n^+$ and $n^- ∈ Γ_C(𝕊M)$ so that
  \begin{equation*}
    Q_{γ(t)} = s ⸨ n^± ⊗ n^± - \frac1mg ⸩
  \end{equation*}
  and $n^± = - n^-$.
  Equivalently, given either of the two possible initial orientations at $γ(t_1)$, there exists a unique continous lifting with this initial orientation.

  Suppose in addition that $\overline{Q} = s ⸨ \overline n ⊗ \overline n - \frac1m g⸩ ∈ 𝒬_{γ(τ)}^{𝕊}M$, $n ∈ 𝕊_{γ(τ)}M$, $τ ∈ [t_1, t_2]$
  and that
  \begin{equation*}
    \text{there is some } 0 < ε < √2 \text{ such that } \abs { Q_{γ(t)} - P^γ_{τt} \overline Q }_g < \abs s ε \text{ for all } t ∈ [t_1, t_2]
  \end{equation*}
  Then one of the liftings, let us say $n^±$ satisfies
  \begin{equation}
    \abs {n^± - P^γ_{τt} \overline n}_g ≤ ε \text{ for all } t ∈ [t_1, t_2]
  \end{equation}
  and the other $n^- = - n^+$ satisfies
  \begin{equation}
    \abs {n^± + P^γ_{τt} \overline n}_g ≤ ε \text{ for all } t ∈ [t_1, t_2]
  \end{equation}
  Here $P^γ_{τt}$ is the parallel transport along $γ$ from $γ(τ)$ to $γ(t)$.
  (For definition see \textcite[Theorem 4.32][107]{q3-IntroRiemannLee}.)
\end{proposition} % end proposition Orientability on a path
\begin{proof} % of proposition Orientability on a path
  We use the parallel transport of $γ$ to transform $Q$ to a $Q$-tensor valued map agnostic of the manifold:
  \begin{equation*}
    \tilde Q (t) := P^γ_{tt_0} Q_{γ(t)}
  \end{equation*}
  Since $P^γ_{tt_0}$ is a linear isometry, it maps $𝒬_{γ(t)}^{𝕊}M$ to $𝒬_{γ(t_0)}^{𝕊}M$.
  Then $\tilde Q ⫶ [t_1, t_2] → 𝒬_{γ(t_0)}^{𝕊}M$.
  By choosing an orthonormal basis for $𝒬_{γ(t_0)}^{𝕊}M$ we can identify it with the space $Q$ as it is used in \textcite[Lemma 3][505]{q4-Ball2011}.
  Furthermore the parallel transport $P^γ_{tt_0}$ commutes with the projection $P$%
  \footnote{Please excuse the different uses of the letter $P$.}
  which is clear when writing $n$ and $P(n)$ in coordinates with respect to an orthonormal frame transported by $P^γ_{t_0t}$.

  With this construction the statement follows directly from \textcite[Lemma 3][505]{q4-Ball2011}.
\end{proof}


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\docEnd