Commit 52368c62 by Felix Hilsky

### move torus example after the necessary on path lifting proposition

also fix this proposition
parent d8157cd5
 ... ... @@ -357,7 +357,7 @@ publisher = {arXiv}, year = {2022} } @article{q37-index_formula_VMO, @article{q37-index-formula-VMO, doi = {10.48550/ARXIV.1407.1707}, doiurl = {https://arxiv.org/abs/1407.1707}, author = {Canevari, Giacomo and Segatti, Antonio and Veneroni, Marco}, ... ...
 ... ... @@ -166,21 +166,6 @@ We formulate the same in the language of \cref{sec:embedding_M} and allow fields Then the liquid crystal must have defects that influence the properties of the entire droplet like bonds to neighboring particles. \end{example} % end example Sphere \begin{example}[Torus] \label{ex:torus} \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 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. 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 $γ(0) = γ(1)$. \footnote{tikz picture. Maybe copy from \textcite[Chapter 4][20]{q37-index_formula_VMO}} \end{example} % end example Torus \begin{example}[Circle] \label{ex:circle} 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. ... ... @@ -192,53 +177,78 @@ We formulate the same in the language of \cref{sec:embedding_M} and allow fields } \end{example} % end example Circle \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.% We will also look at the one example of a boundaryless two-dimensional compact manifold with unit vector field and construct a non-orientable line field: a torus. In order to show that a line field is indeed non-orientable we have to understand \enquote{how it can fail}. Indeed if we look at paths that not loops but injective every line field can be oriented. We will also use this later to show in \cref{thm:triviality_of_two_dimensions} that orientable surfaces are parallelizable. % \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.} In the flat case this was done in Lemma~3 of \parencite{q4-Ball2011} %page 505 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 \newcommand{\Par}{\operatorname{Par}} \begin{proposition}[Orientability on a path] \label{thm:orientability_on_a_path} Let $γ ⫶ [0, 1] → M$ be a continuous injective path on $M$ and $Q ∈ Γ_C ⸨ \rest{𝒬^{𝕊}M}{γ([0,1])} ⸩$ be a continuous line field along the path $γ$. Then there exist exactly two orientations $n^+$ and $n^- ∈ Γ_C(\rest{𝕊M}{γ([0,1])})$ such 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] Q_{γ(t)} = P ⸨ n^±_{γ(t)} ⸩ \text{ for all } t ∈ [0,1] \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}.) and $n^- = - n^+$. Equivalently, given either of the two possible initial orientations at $γ(0)$, there exists a unique continuous orientation with this initial orientation. % add this if it is needed. But currently I don't think we need it % Suppose in addition that for some time $τ ∈ [0, 1]$ we have a reference line $‾Q = s ⸨ ‾n ⊗ ‾n - \frac1m g⸩ ∈ 𝒬_{γ(τ)}^{𝕊}M$ with $‾n ∈ 𝕊_{γ(τ)}M$, % such that % \begin{equation*} % \text{there is some } 0 < ε < √2 \text{ such that } \abs { Q_{γ(t)} - \Par^γ_{τt} \overline Q }_g < \abs s ε \text{ for all } t ∈ [0, 1] % \ . % \end{equation*} % Then one of the orientations, let us say $n^+$ satisfies % % \abs {n^+ - \Par^γ_{τt} \overline n}_g ≤ ε \text{ for all } t ∈ [0, 1] % % and the other $n^- = - n^+$ satisfies % % \abs {n^- + \Par^γ_{τt} \overline n}_g ≤ ε \text{ for all } t ∈ [0, 1] % \ . % \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)} \tilde Q (t) := \Par^γ_{t0} 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}. For a definition of the parallel transport along $γ$ from $γ(τ)$ to $γ(t)$ see Theorem 4.32 in \parencite{q3-IntroRiemannLee}. %page 107) Since $P^γ_{t0}$ is a linear isometry, it maps $𝒬_{γ(t)}^{𝕊}M$ to $𝒬_{γ(0)}^{𝕊}M$. Hence $\tilde Q$ maps from $[0, 1]$ to $𝒬_{γ(0)}^{𝕊}M$. By choosing an orthonormal basis for $𝒬_{γ(0)}^{𝕊}M$ we can identify it with the space $𝒬$ as it is used by \textcite{q4-Ball2011}. %page 505 Note that $g$ written in an orthonormal basis is the identity matrix used in the definition of $P$ used in \parencite{q4-Ball2011}. Furthermore the parallel transport $\Par^γ_{tt_0}$ commutes with the projection $P$% which is clear when writing $n$ and $P(n)$ in coordinates with respect to an orthonormal frame transported by $\Par^γ_{0t}$. With this construction the statement follows directly from Lemma 3 in \parencite{q4-Ball2011}. % page 505 Note that \parencite{q4-Ball2011} formally only considers the case $m = 3$ but the proof generalises trivially to any dimension. \end{proof} \begin{example}[Torus] \label{ex:torus} The torus does have line and unit vector fields and it is not simply connected. Therefore it is a candidate surface to exhibit non-orientable line fields. In order to construct an non-orientable line field we take a loop that is not homotopy equivalent to a point, \eg a circle around the hole. Call it $γ⫶[0,1]→M$. We know by \cref{thm:orientability_on_a_path} that any line field can be oriented along a path. Hence on $γ([0,1))$ we can find an orientation and thus 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 as a model for the torus. Define the discontinuous unit vector field $n(x, y) = \sin ⸨ \fracπ2 x ⸩ \tanvec x + \cos ⸨ \fracπ2 x ⸩ \tanvec y$. $n$ is continuous on $(-1, 1)^2$ and $n(0, y) = -n(1, y)$ and $n(x, 0) = n(x, 1)$ for all $x, y ∈ [0, 1]$. Hence $Q = P(n)$ is continuous on the entire torus but on the path $γ ⫶[0,1] → M$ with $γ(t) = (t, 0)$ an orientation of $Q$ it must be equal to $n$ (or $-n$) on $(0,1)$ but then has a discontinuity at $γ(0) = γ(1)$. \ask{Hi Joshua: kannst du ganz lieb sein und mir das tikzen?} %Something like that exists here: \textcite[Chapter 4][20]{q37-index-formula-VMO}} \end{example} % end example Torus \docEnd
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