Commit 0ef72395 authored by Felix Hilsky's avatar Felix Hilsky
Browse files

various spellchecking

also some improvements but I get the impression that the whole things needs rewriting from scratch to match other parts of the work
parent 59bc3ed6
......@@ -16,13 +16,14 @@ Therefore we will use two specialised density results for the simply-connected c
\textinput{sobolev-lift-simply-connected}
\subsection{Orientability on surfaces} \label{sec:sobolev_orientability_on_surface} {
As mentioned in the \nameref{sec:introduction} liquid crystals exist in some cases like \parencite{q5-q3-vesicles} in very thin films that are best modelled as two-dimensional surfaces.
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 two-dimensional manifold.
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}.
......@@ -33,83 +34,75 @@ Therefore we will use two specialised density results for the simply-connected c
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 restriced $Q$-tensor field.
If $M$ is simply-connected, $Q$ is orientable by \cref{thm:orientability_on_simply_connected_manifolds} with $Q = P(n)$.
Then $(X, n^)$ is a global orthonormal frame.
If $M$ is not simply-connected we have to check if $Q$ is orientable along every generator of $π_1(M)$. (\cref{thm:orientability_of_continuous_line_fields})
Let $γ ⫶ [0,1] → M$ a loop.
By \cref{thm:orientability_on_a_path} $Q$ is orientable along $\rest{γ}{[0,1)}$ with $Q=P(n)$, so is orientable if $n(0) = \lim_{t → 1} n(t)$.
Since $M$ is orientable we can choose an orientation and choose $n$ such that $(X, n^)$ is oriented positively.
Since at every chart domain change, the orientation is preserved and within a chart the orientation is continuous, hence constant, $\lim_{t→1} (X, n(t)^)$ is also oriented positively.
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 $\rest{γ}{[0)}$ 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.
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
\begin{note}[Simply connected 2-manifolds are orientable] \label{note:simply_connected_2_manifolds_are_orientable}
The previous \cref{thm:triviality_of_two_dimensions} could be simplified by noting that simply-connected two-dimensional manifolds are always orientable.
\footnote{todo source? \url{https://math.stackexchange.com/q/3625351/1023694} has several proofs}
In three dimensions a similar statement is true: every closed (\ie compact without boundary) orientable three-dimensional manifold is parallizable. (See \autocite{q32-parallizable-3-manifolds}.)
\end{note} % end note Simply connected 2-manifolds are orientable
\textcite{q4-Ball2011} shows that for flat two-dimensional domains with holes orientability is reduced to checking the parity of the degree of the line field at the boundaries.
The degree describes how often it wraps around the circle along a path. If this is even you can find a covector field that wraps around half as often.
\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 parallizable 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 parallizable this covers important cases.\footnote{But it makes most of the previous work totally useless. tadum}
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 parallizable manifold with the global orthonormal frame $(X_1, X_2)$.
For $Q ∈ \secsob(𝒬^{𝕊}M)$ the components are given by $Q_{ij} = Q(X_i, X_j)$ and for $n ∈ \secsob(𝕊M)$ respectively $n_i = n(X_i)$ ($j ∈ \{1,2\}$). Then
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
\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
\footnote{If we choose the other orientation of $M$ ($(\tilde X_1, X_2) = (-X_1, X_2)$ we get $\tilde A(Q) = \cconj Q = Q^{-1}$ and $\cconj Z(n)\{-\cconj{Z(n)}, \cconj{Z(n)}\} = \{-Z(n)^{-1}, Z(n)^{-1}\}$, so also $\tilde A(P(n)) = Z(n)^2$.}
From \textcite[][120--130]{q4-q24-differential-topology} and \textcite[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 $$)
\footnote{todo: glossary $𝕊^1$ = sphere}
From \parencite[][120--130]{q4-q24-differential-topology} and \parencite[Theorem A.3][]{q4-q31-degree} we get that there exists a \newTerm{degree} for $f ∈ W^{\frac12, 2}(𝕊^1, 𝕊^1)$ ($𝕊^1$ as a subset of $$)
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 sufficiantly small pertubations of $f$.
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 transfered to surfaces:
\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 parallizable two-dimensional compact%
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
\footnote{the first = is supposed to be understood as a local abbreviation. Is that clear?}%
$\Tr Q = \rest{\Tr}{γ(𝕊^1)} Q ∈ \secsob(\rest{𝒬^{𝕊}M}{γ(𝕊^1)})$%
\footnote{todo: define and discuss Trace operator onto submanifolds like paths}
$\Tr Q := \rest{\Tr}{γ(𝕊^1)} Q ∈ \secsob(\rest{𝒬^{𝕊}M}{γ(𝕊^1)})$%
\todo{todo: define and discuss Trace operator onto submanifolds like paths}
is orientable if and only if $\deg(A(\Tr Q))2$.
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 $𝕊^1$.%
\footnote{todo: comment on how the norms are equivalent}
\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 generationg the fundamental group.
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 $L^2_1$.\footnote{todo: after formatting?}
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
\footnote{todo: proof that this also applies to my norm for vector fields}
\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}
\footnote{todo make title analogue to continuous theorm}
Let $M$ be a two-dimensional parallizable compact manifold.
\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 $\deg ⸨ A ⸨ \rest{\Tr}{γ(𝕊^1)} Q ⸩ ⸩ ∈ 2$ for all $γ⫶𝕊^1 → M$ in $G$.
......@@ -129,7 +122,7 @@ Therefore we will use two specialised density results for the simply-connected c
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.
\footnote{todo}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.
\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
}
......
Supports Markdown
0% or .
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment