continuous-Q-lift.tex 19.5 KB
 Felix Hilsky committed Mar 17, 2022 1 2 3 4 5 %! TEX program = lualatex \input{.maindir/tex/header/preamble-section} % inputs the preamble only if necessary \docStart Felix Hilsky committed Jul 28, 2022 6 \section{Orientability of continuous line fields} \label{sec:continuous_q_lift} Felix Hilsky committed Mar 17, 2022 7 Felix Hilsky committed Jul 25, 2022 8 9 10 11 12 13 14 15 16 17 18 19 20 { %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 $𝕊^{m-1}$. Felix Hilsky committed Mar 17, 2022 21 Felix Hilsky committed Jul 25, 2022 22 23 24 25 26 27 28 29 30 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. } Felix Hilsky committed Jul 25, 2022 31 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. Felix Hilsky committed Mar 17, 2022 32 \begin{lemma}[Projection as a covering map] \label{thm:projection_as_a_covering_map} Felix Hilsky committed Jul 22, 2022 33 The projection operators $P ⫶ 𝕊M → 𝒬^{𝕊}M$ and $P_N⫶𝕊^{N-1} → 𝒬^{𝕊'}ℝ^N$ ($N ∈ ℕ$) are covering maps with two sheets. Felix Hilsky committed Mar 17, 2022 34 35 \end{lemma} % end lemma Projection as a covering map \begin{proof} % of lemma Projection as a covering map Felix Hilsky committed Jul 22, 2022 36 37 38 $P$ 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 $T^{(2,0)}ℝ^N$. For $n ∈ 𝕊M$ or $𝕊^{N-1}$ choose a neighborhood $U$ that has a diameter of less than $2$ (measured in $TM$ or $ℝ^N$ respectively). Felix committed Jun 04, 2022 39 Then $P^{-1}(P(U)) = U ∪ -U$ (same for $P_N$) and $U$ and $-U$ are disjoint because the distance between opposite points is $2$. Felix Hilsky committed Jul 22, 2022 40 Since $P$ and $P_N$ are quadratic, they have continuous inverses if they are bijective. Hence they map $U$ and $-U$ homeomorphically onto $P(U)$. Felix Hilsky committed Mar 17, 2022 41 42 \end{proof} % of lemma Projection as a covering map Felix Hilsky committed Jul 22, 2022 43 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 Felix Hilsky committed Aug 10, 2022 44 \begin{externalresult}[Lifting of continuous maps] \label{thm:lifting_of_continuous_maps} Felix Hilsky committed Jul 22, 2022 45 46 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 $f_* ⸨ π_1(Y, y_0) ⸩ ⊆ p_* ⸨ π_1(\tilde X, \tilde x_0) ⸩$. Felix Hilsky committed Aug 10, 2022 47 \end{externalresult} % end proposition Lifting of continuous maps Felix Hilsky committed Jul 25, 2022 48 49 50 51 52 53 { % 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 $α$.% Felix Hilsky committed Mar 17, 2022 54 Felix Hilsky committed Jul 25, 2022 55 56 57 58 59 60 61 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. $Γ_R(\rest Eγ)$% \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 $γ ⊂ M$. } Felix Hilsky committed Jul 27, 2022 62 \begin{theorem}[Orientability of continuous line fields] \label{thm:orientability_of_continuous_line_fields} Felix Hilsky committed Jul 22, 2022 63 Consider a line field $Q ∈ Γ_C(𝒬^{𝕊}M)$. Felix Hilsky committed Mar 17, 2022 64 Let $G$ be a set of loops at a common point $p_0$ that generate the fundamental group $π_1(M, p_0)$. Felix Hilsky committed Jul 22, 2022 65 66 $Q$ 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 Felix Hilsky committed Jul 25, 2022 67 \begin{proof} % of theorem Orientability of continuous line fields Felix Hilsky committed Jul 22, 2022 68 As \cref{thm:projection_as_a_covering_map} shows, $P ⫶ 𝕊M → 𝒬^{𝕊}M$ is a covering map. Felix Hilsky committed Jul 25, 2022 69 70 Use $Q_{p_0}$ and a $n_0 ∈ 𝕊M$ with $P(n_0) = Q_{p_0}$ as the distinguished points. Felix Hilsky committed Jul 22, 2022 71 72 73 74 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 $P_* ⸨ π_1(𝕊M, n_0) ⸩$. Felix Hilsky committed Mar 17, 2022 75 For the \enquote{if} part Felix Hilsky committed Jul 22, 2022 76 let $γ ∈ G$, \ie $γ ⫶[0,1] → M$ continuous with $γ(0) = γ(1) = p_0$. Felix Hilsky committed Jul 25, 2022 77 $Q$ is orientable along $γ$. That means that there exists $n ∈ Γ_C(\rest{𝕊M}{γ([0,1])})$ with $P(n) = Q$. Felix Hilsky committed Jul 22, 2022 78 If the orientation of $Q$ along this path has $n(γ(0)) = -n_0$, use $-n$ instead. Felix Hilsky committed Jul 25, 2022 79 80 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 $Q_*(π_1(M, p_0)) ⊆ P_*(π_1(𝕊M, n_0))$. Felix Hilsky committed Jul 25, 2022 81 Hence there exists $n ∈ C(M, 𝕊M)$ with $P(n) = Q$ by \cref{thm:lifting_of_continuous_maps}, \ie $Q$ is orientable and $n ∈ Γ_C(𝕊M)$. Felix Hilsky committed Mar 17, 2022 82 83 For the \enquote{and only if} part restrict an orientation for $Q$ on the paths in $G$. Felix Hilsky committed Jul 25, 2022 84 \end{proof} % of theorem Orientability of continuous line fields Felix Hilsky committed Mar 17, 2022 85 Felix committed Jun 03, 2022 86 We formulate the same in the language of \cref{sec:embedding_M} and allow fields that are not tangent to $M$. Felix Hilsky committed Jul 27, 2022 87 \begin{lemma}[Orientability of continuous line fields via embedding of \texorpdfstring{$M$}{M}] \label{thm:orientability_of_line_fields_via_embedding_of_M} Felix Hilsky committed Jul 25, 2022 88 89 Let $ι$ be an isometric embedding of $M$ into $ℝ^N$ as in \cref{sec:embedding_M} and let $Q ∈ C(M, 𝒬^{𝕊'}ℝ^N)$ be a line field on but not necessarily tangent to $M$. Felix committed Jun 03, 2022 90 Let $G$ be a set of loops at a common point $p_0$ that generate the fundamental group $π_1(M, p_0)$. Felix Hilsky committed Jul 25, 2022 91 92 93 94 95 96 97 98 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, $P_N ⫶ 𝕊^{N-1} → 𝒬^{𝕊'}ℝ^N$ 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} Felix committed Jun 04, 2022 99 Felix committed Jun 03, 2022 100 For the \enquote{if} part Felix Hilsky committed Jul 25, 2022 101 102 103 104 105 106 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. Felix committed Jun 03, 2022 107 108 For the \enquote{and only if} part restrict an orientation for $Q$ on the paths in $G$. Felix Hilsky committed Jul 25, 2022 109 \end{proof} % of theorem Orientability of line fields on embedded M Felix committed Jun 03, 2022 110 Felix Hilsky committed Jul 25, 2022 111 {\textbf{Here another version with background colors instead of [].} Felix Hilsky committed Aug 07, 2022 112 \newcommand*{\ver}[2]{\sethlcolor{Lavender}\hl{#1}\sethlcolor{Yellow}\hl{#2}} Felix Hilsky committed Jul 28, 2022 113 \cref{thm:orientability_of_continuous_line_fields} and \cref{thm:orientability_of_line_fields_via_embedding_of_M} are very similar. Felix Hilsky committed Jul 25, 2022 114 Therefore the proofs are combined. Felix Hilsky committed Aug 07, 2022 115 \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}}}.% Felix Hilsky committed Jul 25, 2022 116 117 \textinput{proof_continuous_lift} } Felix Hilsky committed Jul 27, 2022 118 119 120 121 122 123 124 125 126 127 128 129 {\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} } Felix Hilsky committed Aug 07, 2022 130 131 132 133 134 135 136 137 138 139 140 {\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} } Felix Hilsky committed Mar 17, 2022 141 \begin{corollary}[Orientability on simply connected manifolds] \label{thm:orientability_on_simply_connected_manifolds} Felix Hilsky committed Jul 25, 2022 142 If $M$ is simply connected every continuous line field $Q ∈ Γ_C(𝒬^{𝕊}M)$ or $Q ∈ C(M, 𝒬^{𝕊'}ℝ^N)$ is orientable. Felix Hilsky committed Mar 17, 2022 143 144 \end{corollary} % end corollary Orientability on simply connected manifolds \begin{proof} % of corollary Orientability on simply connected manifolds Felix Hilsky committed Jul 25, 2022 145 $M$ 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. Felix Hilsky committed Jul 28, 2022 146 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. Felix Hilsky committed Jul 25, 2022 147 \end{proof} % of corollary Orientability on simply connected manifolds Felix Hilsky committed Mar 17, 2022 148 Felix Hilsky committed Apr 04, 2022 149 \begin{note}[Topological restriction on the manifold] \label{note:topological_restriction_on_the_manifold} 150 151 152 153 154 155 \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. Felix Hilsky committed Apr 04, 2022 156 \end{note} % end note Topological restriction on the manifold Felix Hilsky committed Mar 17, 2022 157 \begin{example}[Sphere] \label{ex:sphere} Felix Hilsky committed Jul 25, 2022 158 159 160 161 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}. Felix Hilsky committed Mar 17, 2022 162 163 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. Felix Hilsky committed Jul 25, 2022 164 165 166 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. Felix Hilsky committed Mar 17, 2022 167 168 169 \end{example} % end example Sphere \begin{example}[Torus] \label{ex:torus} Felix Hilsky committed Jul 25, 2022 170 171 \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 Felix Hilsky committed Mar 17, 2022 172 173 174 175 176 177 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. Felix Hilsky committed Jul 28, 2022 178 179 180 Define the (not continuous) $n(x, y) = \sin(xπ/2) \D x + \cos(xπ/2) \D y$. 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 $γ(0) = γ(1)$. Felix Hilsky committed Apr 04, 2022 181 \footnote{tikz picture. Maybe copy from \textcite[Chapter 4][20]{q37-index_formula_VMO}} Felix Hilsky committed Mar 17, 2022 182 183 184 \end{example} % end example Torus \begin{example}[Circle] \label{ex:circle} Felix Hilsky committed Jul 28, 2022 185 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. Felix Hilsky committed Mar 17, 2022 186 187 In this simple form the converse is not true. Felix Hilsky committed Jul 25, 2022 188 189 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. Felix Hilsky committed Mar 17, 2022 190 191 192 193 \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 Felix Hilsky committed Aug 11, 2022 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 \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 \abs {n^± - P^γ_{τt} \overline n}_g ≤ ε \text{ for all } t ∈ [t_1, t_2] and the other $n^- = - n^+$ satisfies \abs {n^± + P^γ_{τt} \overline n}_g ≤ ε \text{ for all } t ∈ [t_1, t_2] 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} Felix Hilsky committed Mar 17, 2022 244 \docEnd