sobolev-Q-lift.tex

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\section{Orientability of Sobolev line fields} \label{sec:orientability_of_sobolev_line_fields}
% \footnote{\textcite{q43-sobolev-liftings} solves a lot of it - but for $0<s<1$ but I have $s = 1$. And the referenced work only looks at lifting $𝕊^1$-valued map to $ℝ$ (angle function) and the other referenced work is supposed to show sth about arbitrary coverings but only works with $ℝ → 𝕊^1$. Weird on-first-sight-wrong-citation.}

Since algebraic topology is only concerned with continuous maps it does not give us tools to directly study the orientability of Sobolev line fields.
Instead we use approximation results that help us to reduce the question to continuous fields.
% \cref{sec:sobolev_spaces} shows that we can view
Smooth functions between manifolds are in general not dense in the Sobolev function space though.
Therefore we will use two specialised density results for the simply-connected case (\cref{thm:sequentially_weak_density_of_smooth_manifold_maps}) and for surfaces (\cref{thm:density_of_smooth_maps_on_2_manifolds}).

\textinput{sobolev-lift-simply-connected}

\subsection{Orientability on surfaces} \label{sec:sobolev_orientability_on_surface} {
  As mentioned in the introduction\todo{rename if the introduction section is renamed. nameref doesn't work here since I don't want to have it capitalised}
  liquid crystals exist in some cases like \parencite{q5-q3-vesicles} in very thin films that are best modelled as two-dimensional surfaces.
  Therefore the case $\dim M = 2$ is of special interest.
  Section 4 of \parencite{q4-Ball2011} gives a roadmap how to transfer the orientability condition from continuous to Sobolev line fields.
  In order to use those ideas we notice that the tangent bundle is trivial in the case of oriented surfaces with unit vector fields.\todo{somehow mention unorientable case}
  This allows us to consider unit vector fields as maps into $𝕊^1$ as discussed in \cref{sec:sobolev_field_spaces_on_parallelizable_manifolds}.
  \begin{lemma}[Triviality of Two Dimensions] \label{thm:triviality_of_two_dimensions}
    Let $M$ be a compact orientable surface.
    Then either $M$ admits no smooth unit tangent field or has a global orthonormal frame.
  \end{lemma} % end lemma Triviality of Two Dimensions
  The prominent example of a manifold without unit tangent field is the sphere by the Hairy Ball theorem \autocite{hairy-ball}.
  A manifold with a global frame is called \newTerm{parallelizable}.
  Any global frame can be orthonormalized to an orthonormal global frame with the Gram-Schmidt process.
  \begin{proof} % of lemma Triviality of Two Dimensions
    If $M$ admits no smooth unit tangent field, we are done.
    Otherwise let $X ∈ Γ_{C^∞}(𝕊M)$.
    Then in each point $p ∈M$, the orthogonal complement $X_p^⊥$ is a one-dimensional subspace with exactly two unit vectors.
    Call them $X_p^{⊥+}$ and $X_p^{⊥-}$.
    Then $Q := P(X_p^{⊥+}) = P(X_p^{⊥-})$ and since the two orientations can be chosen smoothly locally, $Q$ is a smooth line field.

    Now we have to check if $Q$ is orientable along every generator of $π_1(M)$. (\cref{thm:orientability_of_continuous_line_fields})
    Let $γ ⫶ [0,1] → M$ be a loop.
    By \cref{thm:orientability_on_a_path} $Q$ is orientable along $\rest{γ}{[0,τ)}$ with $Q=P(n)$ with any $τ < 1$ and hence also along $\restγ{[0,1)}$.
    Therefore $Q$ is orientable along the entire loop $γ$ if $n(0) = \lim_{t → 1} n(t)$.
    Since $M$ is orientable we can choose an orientation of $M$ and choose $n$ such that $(X, n)$ is oriented positively.
    Since at every chart domain change, the orientation of a frame is preserved and within a chart the orientation is continuous, hence constant, $\lim_{t→1} (X, n(t))$ is also oriented positively.
    Therefore $Q$ is orientable along the loop $γ$.
    By \cref{thm:orientability_of_continuous_line_fields} $Q$ is orientable with $Q = P(n)$ and then $(X, n)$ is a global orthonormal frame.
  \end{proof} % of lemma Triviality of Two Dimensions

  \textcite{q4-Ball2011} show that for flat two-dimensional domains with holes orientability can be checked with the parity of the winding number of an auxiliary unit vector field representing a line field at the boundaries.
  The winding number describes how often a function wraps around the circle along a loop.
  If this is even you can find a unit vector field that wraps around half as often.
  The same argument for manifolds is in general more complicated since it is not clear what \enquote{wrapping around the circle} is supposed to mean if the circle is a different one in each point.
  On the Möbius strip the analogue statement is even false since a line field has to \enquote{turn around} once while moving around the strip because the strip \enquote{turns around} as well.\footnote{Back this up with drawing and calculation.}
  This complication does not appear in the case of parallelizable two-dimensional manifolds because there exists a global orthonormal frame that lets us identify all tangent spaces, thus making the tangent bundle trivial.
  Since at least the orientable two-dimensional manifolds that have any unit tangent fields are parallelizable this covers important cases.\footnote{But it makes most of the previous work totally useless. tadum}
  \footnote{For non-orientable manifolds, there exists a \newTerm{mod 2 degree} according to \autocite[1.6 Theorem][125 (136)]{q4-q24-differential-topology}. Maybe that is enough for the Möbius strip?}
  \begin{definition}[Complex version of line field] \label{def:complex_version_of_line_field}
    Let $M$ be a two-dimensional parallelizable manifold with the global orthonormal frame $(X_1, X_2)$.
    We write a line field $Q ∈ \secsob(𝒬^{𝕊}M)$ in these coordinates: $Q = Q^{ij} X_i ⊗ X_j$  and $n ∈ \secsob(𝕊M)$ respectively as $n = n^i X_i$ ($i, j ∈ \{1,2\}$). Then
    \begin{align}
      \label{eq:def:complex-Q} A(Q) = \frac2s (Q^{11} + \im Q^{12}) ∈ 𝕊^1 ⊂ ℂ \\
      \label{eq:def:complex-n} Z(n) = n^1 + \im n^2 ∈ 𝕊^1 ⊂ ℂ
    \end{align}
    Note that $A(P(n)) = Z(n)^2$.
  \end{definition} % end definition of Complex version of line field

  From \parencite[][120--130]{q4-q24-differential-topology} and \parencite[Theorem A.3][]{q4-q31-degree} we get that there exists a \newTerm{degree} for $f ∈ W^{\frac12, 2}(𝕊^1, 𝕊^1)$ ($𝕊^1$ as a subset of $ℂ$)
  computed by
  \begin{equation} \label{eq:def:degree}
    \deg f = \frac1{2πi} ∫_{𝕊^1} f^{-1} \frac{∂f}{∂θ} \D θ
    \ .
  \end{equation}
  This degree is an integer, it is invariant under sufficiently small perturbations of $f$.
  The integral in \eqref{eq:def:degree} is to be understood in the sense of distributions since $f, f^{-1} ∈ W^{\frac12, 2}(𝕊^1, 𝕊^1)$ and $\frac{∂f}{∂θ} ∈ W^{-\frac12,2}(S^1, 𝕊^1)$.
  For some more discussion on the degree, see \textcite[Section 4][516]{q4-Ball2011}.
  \textcite[Propositon 6][516]{q4-Ball2011} can then be directly transferred to surfaces:
  \begin{proposition}[Orientability on loops] \label{thm:orientability_on_loops}
    Let $M$ be a parallelizable two-dimensional compact%
    \footnote{do I need it here?}
    manifold. Let $Q ∈ \secsob(𝒬^{𝕊}M)$ and let $γ⫶𝕊^1→ M$ be a loop in\footnote{on?} $M$.
    Then
    $\Tr Q := \rest{\Tr}{γ(𝕊^1)} Q ∈ \secsob(\rest{𝒬^{𝕊}M}{γ(𝕊^1)})$%
    \todo{todo: define and discuss Trace operator onto submanifolds like paths}
    is orientable if and only if $\deg(A(\Tr Q)) ∈ 2 ℤ$.
    Moreover if it is orientable with $n ∈ \secsob(\rest{𝕊M}{γ(𝕊^1)})$ with $\Tr Q = P(n)$,
    then $2 \deg n = \deg (\Tr Q)$.
  \end{proposition} % end proposition Orientability on loops
  \begin{proof} % of proposition Orientability on loops
    By identifying $𝕊^1$ with $γ(𝕊^1) ⊂ M$ and identifying the tangent spaces at $M$ via the orthonormal frame we can regard $\Tr Q$ and $n$ as $W^{\frac12, 2}$ map from $𝕊^1$ to $𝕊^1$.%
    \todo{comment on how the norms are equivalent}
    Then the same proof as for \textcite[Proposition 6][517]{q4-Ball2011} holds.
  \end{proof} % of proposition

  For continuous line fields we already know that orientability can be checked on loops generating the fundamental group.
  In order to use this for Sobolev line fields it is useful if we can approximate them with smooth maps.
  \begin{externalresult}[Density of smooth maps on 2-manifolds] \label{thm:density_of_smooth_maps_on_2_manifolds} (from \textcite[Proposition in section 4][267]{q4-q34-approximation} ) Let $M$ be a 2-dimensional compact manifold. Let $N$ be a compact manifold without boundary. Then $C^∞(M, N)$ is dense in $W^{1,2}(M, N)$.
    Note that the authors denote $W^{1,2}$ by $L^2_1$.\todo{after formatting?}
  \end{externalresult} % end proposition Density of smooth maps on 2-manifolds
  \todo{proof that this also applies to my norm for vector fields}
  As the authors show this is in general not true for higher dimensions of the domain $M$.

  The transfer of \textcite[Propositon 7][517]{q4-Ball2011} to our surfaces is straight-forward:
  \begin{theorem}[Characterisation of Orientability of Sobolev Line Fields] \label{thm:characterisation_of_orientability_of_sobolev_line_fields}
    \todo{make title analogue to continuous theorem}
    Let $M$ be a two-dimensional parallelizable compact manifold.
    Let $G$ be a set of loops at a common point $p_0$ that generate the fundamental group $π_1(M, p_0)$.
    Let $Q ∈ \secsob(𝒬^{𝕊}M)$.
    Then $Q$ is orientable if and only if $\deg ⸨ A ⸨ \rest{\Tr}{γ(𝕊^1)} Q ⸩ ⸩ ∈ 2 ℤ$ for all $γ⫶𝕊^1 → M$ in $G$.
  \end{theorem} % end theorem Characterisation of Orientability of Sobolev Line Fields
  \begin{proof} % of theorem Characterisation of Orientability of Sobolev Line Fields
    Essentially the same proof as for \textcite[Propositon 7][517]{q4-Ball2011}
    with some remarks.

    The condition that the degree at the outer boundary fits is correct but unnecessary. It also does not appear in the continuous case. But it follows from the orientability on the whole domain, so it is not wrong.

    The citation for [34] should be more precise: \textcite[Proposition][267]{q4-q34-approximation}. Note that there $W^{1,2}$ is called $L^2_1$. That is very confusing for readers that just skim the paper nowadays.
    This theorem is for any compact surface (so only two-dimensional!) into a compact manifold without boundary. So it also does not work for $M → TM$ space.

    In \autocite[Lemma 10][518]{q4-Ball2011} the big part of the proof is used for the direction that is not needed in the proof of Proposition 7 but later for construction of a counterexample. So it is actually much simpler and should not be part of the proof.

    The stability of the degree for VMO maps is found in \textcite[Theorem 1][213]{q4-q9-VMO-BMO-degree}.
    The relation of $W^{\frac12, 2}$ and BMO-norm is found in \textcite[Example 2 Section I.2][210]{q4-q9-VMO-BMO-degree}.

    \textcite[Lemma 9][514]{q4-Ball2011} can be used directly.
    \todo{to do}Instead of \autocite[Lemma 10][518]{q4-Ball2011} I have to redo parts of Lemma 1. The argument from $n$ continuous + $P(n) = Q ∈ \secsob(𝒬^{𝕊}M)$ implies $n ∈ \secsob(𝕊M)$ is missing so far.
  \end{proof} % of theorem Characterisation of Orientability of Sobolev Line Fields

}

\begin{note}[Non-orientable manifold] \label{note:non_orientable_manifold}
  For non-orientable manifolds we need some more thoughts, guided by \textcite{stackex_orientation_cover}.
  The double (or orienting) cover gives us a manifold $\tilde M$ that is orientable and a double covering of the original one $M$.
  A loop on $M$ lifts to a loop on $\tilde M$ which in some cases \enquote{loops around twice}.
  $Q$ lifts to $\tilde Q$ on $\tilde M$ via $\D π^{-1}$ (which is a pointwise/local isometry).
  The problem is that $\deg (\tilde Q) = 0$ since $A(\tilde Q)(z) = A(\tilde Q)(-z) ^{-1}$, so the first half cancels out the second half.
  Therefore the degree does not tell us anything. Details in handwritten notes.

  On the other hand, for continuous line fields the same condition for orientability as for orientable manifolds holds and the approximation result also holds.
  So it is a question of finding the right reason why orientability on loops transfers from $Q$ to approximations $Q_k$.

  Some more rambling thoughts:

  \begin{itemize}
    \item Choose a unit vector field $X$ on $M$. Then $X^⊥$ is a line field on $M$.
      Given a loop $γ ⫶ [0,1] → M$, we can choose an orientation of $X^⊥$ on $γ([0,1))$.
      Call this vector field $Y ⫶γ([0,1)) → TM$, $Y_p ∈ T_pM$, $Y_p ⊥ X_p$ for all $p ∈ γ([0,1)]$.
      If $Y ∘ γ(t) → Y ∘ γ(0)$ for $t → 1$ we have an orientation along $γ$, so we can use the technique for orientable manifolds.
      Otherwise $Y ∘ γ(t) → - Y ∘ γ(0)$, so the orientation \enquote{turns around} along $γ$. That's what we assume in the following. Think of the central circle of the Möbius strip, $X$ pointing along the circle, $Y$ perpendicular to it.
    \item Write $Y_γ(t) = \begin{cases} Y(γ(t)) & \text{for } t ∈ [0,1) \\ -Y(γ(0)) & \text{for } t = 1 \end{cases}$. $X_γ = X ∘ γ$.
    \item We can write $A(Q)$ and $Z(n)$ according to this orthonormal frame. Similarly to $Y_γ$ write $A_γ(Q) = \frac 1s ⸨ Q(X_γ,X_γ) + \im Q(X_γ, Y_γ) ⸩$ and
      $Z_γ(n) = n(X_γ) + \im n(Y_γ)$. Then $A_γ(Q)(1) = \cconj{A_γ(Q)(0)}$.
    \item After playing around with different $A_γ(Q)(0)$ one sees that the condition for orientability is that $A_γ(Q)$ passes the $1 ∈ 𝕊^1 ⊂ ℂ$ an odd number of times.
      I could not find a good way yet how to encode this condition into something that makes sense for $W^{\frac12,2}$ functions.
    \item The degree term is not defined for $A_γ(Q)$ since start and endpoint are not equal. To close the circle we need the arc from $A_γ(Q)(1)$ to $A_γ(Q)(0)$.
      To compute the integral expression for this arc: Let $A_γ(Q)(0) = e^{\im α}$, $α ∈ [0, 2π)$.
      Let
      \begin{align*}
        g &⫶ [0,1] → 𝕊^1 ⊂ ℂ \\
        g(θ) &= e^{\im α (2θ - 1)} \\
        \text{Then } g(0) &= e^{- \im α} = A_γ(Q)(1) \text{ and } g(1) = e^{\im α} = A_γ(Q)(0) \\
        \frac1{2π\im} ∫_0^1 g(θ)^{-1} \frac{∂g}{∂θ} \D θ
                          &= \frac1{2π\im} ∫_0^1 e^{-\im α(2θ-1)} 2 \im α e^{\im α (2θ-1)} \D θ = \fracαπ
      \end{align*}
    \item So, if we append $g$ to $A_γ(Q)$ we add another traversal of $1$ and get a function with an integer degree.
      Since then the traversal number is equal to the degree ($\mod 2$),
      we get the condition for orientability (here $α = \arg(A_γ(Q)(0))$)
      \begin{equation} \label{eq:moebius_orientability_condition}
        \frac1{2π\im} ∫_0^1 A_γ(Q)^{-1} \frac{∂A(Q)}{∂θ} \D θ
        = 2 k - \frac1π \arg (A_γ(Q)(0)) \qquad (k ∈ ℤ)
      \end{equation}
      Unfortunately this $α$ changes under small changes of $Q$ and is not well-defined for $W^{\frac12, 2}$-functions.
      At the jump points of $\arg$, it \enquote{jumps} by a value of $2π$, thus not changing the condition ($\mod 2$). Maybe that's good.
    \item Hence it would be nice if we can show that
      \begin{align*}
        Q
        &↦ \frac1{2π\im} ∫_0^1 A_γ(Q)^{-1} \frac{∂A_γ(Q)}{∂θ} \D θ
        + \frac1π \arg A_γ(Q)(0) \\
        Γ_{W^{\frac12, 2}∩C}(\rest{𝒬^{𝕊}M}{γ([0,1])})
        &→ ℤ_2 \\
        \text{is continuous with respect to }
        &\norm·_{W^{\frac12,2}} \text{ and the discrete metric on } ℤ_2
      \end{align*}
      Note that $A_γ(Q) ⫶ [0,1] → 𝕊^1$ has no degree definition since \autocite[][122]{q4-q24-differential-topology} has degree definition only for maps from and to compact manifolds without boundary. (So called closed manifold.)
      But $[0,1]$ has a boundary and $(0,1)$ is not compact.
    \item The function that should be continuous depends on the point value of $A(Q)$ in $0$.
      Taking point values of $H^{\frac12,2}$-functions is not well-defined,
      so taking point values of continuous functions should not be a continuous operator with respect to the $H^{\frac12,2}$-norm.
      Note that the trace theorem would naively imply that $T ⫶ W^{\frac12,2}([0,1]) → W^{\frac12 - \frac12, 2}(\{0,1\}) = L^2(\{0,1\})$ is continuous.
      But check \textcite[Proposition 3.8][19]{q31-hitch-frac-Sob} assumes the differentiation degree to be $s > \frac12$.
      So this theorem is not applicable here.
  \end{itemize}
\end{note} % end note Non-orientable manifold
\docEnd