sobolev-lift-simply-connected.tex 8.08 KB
 1 2 3 4 5 6 7 %! TEX program = lualatex \input{.maindir/tex/header/preamble-section} % inputs the preamble only if necessary \docStart \subsection{Orientability on simply-connected manifolds} \label{sec:sobolev_orientability_on_simply_connected_manifolds} If $M$ is simply-connected we do not have density of smooth maps in the norm sense but in a weak sense.  Felix Hilsky committed Aug 10, 2022 8 \begin{externalresult}[Sequentially Weak Density of Smooth Manifold Maps] \label{thm:sequentially_weak_density_of_smooth_manifold_maps}  9 10 11 12 13  (\autocite[Theorem I][225]{q4-q33-density-of-manifold-to-manifold-maps}) Let $M$, $N$ be compact smooth manifolds with $M$ simply connected. The Sobolev space $W^{1,2}(M, N)$ is defined as $\setDef{f ∈ W^{1,2}(M, ℝ^d)}{f ∈ N \text{ \ae}}$ where $N$ is embedded isometrically into $ℝ^d$. Then for $u ∈ W^{1,2}(M, N)$ there exists a sequence $(u^{(k)})_k$ with $u^{(k)} ∈ C^∞(M, N)$ for all $k ∈ ℕ$ so that $u^{(k)}$ converges weakly to $u$.  Felix Hilsky committed Aug 10, 2022 14 \end{externalresult} % end theorem Sequentially Weak Density of Smooth Manifold Maps  15 16 17 18 19 20 21 22 23 24 25 26  To use \cref{thm:sequentially_weak_density_of_smooth_manifold_maps} we need to show that orientability is preserved by weak convergence to transfer orientability of the continuous approximations to the approximated Sobolev function. \begin{proposition}[Orientability preserved by weak convergence] \label{thm:orientability_preserved_by_weak_convergence_M_embedding} Let $M$ be embedded into $ℝ^N$ by $ι$ as in \cref{sec:embedding_M}. Let $q ∈ [1, ∞]$\footnote{which $q$ are actually OK or which need special attention?} Let $(Q^{(k)})_k$ be a sequence of orientable functions in $W^{1,p}(M, 𝒬^{𝕊'}ℝ^N)$ weakly converging to $Q ∈ W^{1,p}(M, 𝒬^{𝕊'}ℝ^N)$. Call the orientations $n^{(k)} ∈ W^{1,p}(M, 𝕊^{N-1})$ with $P_N(n^{(k)}) = Q^{(k)}$ and for all $k ∈ ℕ$. Then $(n^{(k)})_k$ weakly converges in $W^{1,p}(M, 𝕊^{N-1})$ to some $n ∈ W^{1,p}(M, 𝕊^{N-1})$ with $P_N(n) = Q$. If $Q ∈ Γ_{W^{1,p}}(𝒬^{𝕊'}M)$, \ie $Q$ is tangent to $M$ \ae, then $n∈Γ_{W^{1,p}}(𝕊M)$ is a tangent \ae as well.  Felix Hilsky committed Aug 10, 2022 27  \textbf{Now again with formulas. Which is easier to comprehend?}\ask{Which one should I use?}  28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66  Let $ι ⫶ M \hookrightarrow ℝ^N$ be an isometric embedding as in \cref{sec:embedding_M}, $q ∈ [1, ∞]$. Let \begin{align*} Q_{(k)} &∈ W^{1,q}(M, 𝒬^{𝕊'}ℝ^N), & (k ∈ ℕ) \\ Q_{(k)} &⇀ Q ∈ W^{1,q}(M, 𝒬^{𝕊'}ℝ^N), & (k → ∞) \\ Q_{(k)} &= P_N(n_{(k)}) \text{ with } n_{(k)} ∈ W^{1,q}(M, 𝕊^{N-1}) \ . & (k ∈ ℕ) \\ \text{Then } n_{(k)} &⇀ n ∈ n ∈ W^{1,q}(M, 𝕊^{N-1}) \text{ with } P_N(n) = Q(n) \ . & (k → ∞) \\ \text{If } Q &∈ Γ_{W^{1,q}}(𝒬^{𝕊'}M) \text{, then } n ∈ Γ_{W^{1,q}}(𝕊M) \ . \end{align*} \end{proposition} % end proposition Orientability preserved by weak convergence \begin{proof} % of proposition Orientability preserved by weak convergence In order to get a weak limit of the $n_{(k)}$'s we need to show that the sequence is bounded. For this, calculate the Euclidean derivative $‾∇_v \hat n$ for any $\hat Q = P_N(\hat n)$ and $\hat n$ weakly differentiable and the direction $v ∈ TM$. Note that in $ℝ^N$ the musical operators $♯$ and $♭$ are given by transposition $·^T$. The contravariant tensor $\hat Q$ and vector $\hat n$ are here interpreted as a bilinear maps on two and one covectors respectively. As a preparation note that $\hat n · \hat n = \abs {\hat n}^2 = 1$ and therefore \begin{align} \label{eq:‾∇n(n)=0} 0 = \frac12 ‾∇_v (\hat n · \hat n) = \frac12 (‾∇_v \hat n) · \hat n + \frac12 \hat n · (‾∇_v \hat n) = ‾∇_v \hat n · \hat n = ‾∇_v \hat n (\hat n^T) \ . \end{align} \begin{align} ‾∇_v \hat Q(\hat n^T, ·) &= ‾∇_v ⸨ \hat n ⊗ \hat n ⸩ (\hat n^T) \nonumber \\ &= (‾∇_v \hat n) (\hat n^T) ⊗ \hat n + \hat n(\hat n^T) ⊗ ‾∇_v \hat n \nonumber \\ &= ‾∇_v \hat n & (\text{\eqref{eq:‾∇n(n)=0} and } \hat n(\hat n^T) = \hat n · \hat n = 1) \nonumber \\ ⇒ ‾∇_v \hat n &= (‾∇_v \hat Q)(\hat n^T, ·) \label{eq:∇ninQterms} \end{align}  Felix Hilsky committed Aug 10, 2022 67  Hence $‾∇\hat n$ is in $L^p$. Since also $\abs{\hat n} = 1$, $\hat n ∈ W^{1,p}(M, 𝕊^{N-1})$, the norm of $\hat n$ is bounded by the norm of $\hat Q$.  68 69 70 71 72 73 74 75 76 77 78 79 80 81  Since $(Q_{(k)})_k$ converges weakly, it is a bounded sequence by the Uniform Boundedness Principle. Therefore the previous calculation \eqref{eq:∇ninQterms} shows that $(n_{(k)})_k$ is bounded as well. The Banach-Alaoglu Theorem further gives a subsequence $(n_{(k_l)})_l$ of $(n_{(k)})_k$ that weakly converges to some $n ∈ W^{1,p}(M, ℝ^N)$. By the Kondrakov Theorem \autocite[§11 2.34 Theorem]{q33-aubin-laplace-equation} $W^{1,p}(M, ℝ^N)$ is compactly embedded in $L^p(M, ℝ^N)$. Therefore we can find a further subsequence $(n_{(k_s)})_s$ that converges in $L^p$-norm and therefore pointwise \ae. The same reason shows that $( Q_{(k_s)} ) _s$ has a subsequence that converges pointwise \ae and hence $P_N(n) = Q$, \ie $Q$ is orientable. If $n_p · η = n_p(η^T) \neq 0$ for some $p ∈ M$ and $η ⊥ T_pM$, then $P_N(n) (η^T, η^T) = Q(η^T, η^T) = (n ⊗ n)(η^T, η^T) = (n(η^T))^2 \neq 0$. In the tangent case, \ie $Q ∈ Γ_{W^{1,p}}(𝒬^{𝕊'}M)$, this is false \ae. Hence $n ∈ Γ_{W^{1,p}}(𝕊M)$ is tangent to $M$. \end{proof} % of proposition Orientability preserved by weak convergence  Felix Hilsky committed Aug 10, 2022 82 83 84 85 86 87 88 \begin{note}[Choice of codomain] \label{note:choice_of_codomain} If we only used the intrinsic definition of the tensor fields it would be unclear how to use \cref{thm:sequentially_weak_density_of_smooth_manifold_maps}. If we used it with $N = 𝒬M$, we would not have the guarantee that $u^{(k)}_p ∈ T_pM$. If we used it with $N = 𝒬_pM$ we have to somehow identify all tangent spaces which is -- in the general case -- not possible in a smooth way. By embedding $M$ and considering non-tangent tensor fields we solve this problem but need \cref{thm:orientability_on_simply_connected_manifolds} and \cref{thm:orientability_preserved_by_weak_convergence_M_embedding} for $𝒬^{𝕊'}ℝ^N$- and $𝕊^{N-1}$-valued functions. \end{note} % end note Choice of codomain  89 90  \begin{theorem}[Sobolev orientability on simply connected manifolds] \label{thm:sobolev_orientability_on_simply_connected_manifolds}  Felix Hilsky committed Aug 10, 2022 91  Let $M$ be a simply-connected and $ι ⫶ M \hookrightarrow ℝ^N$ an isometric embedding.  92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111  Let $q ∈ [2, ∞)$\footnote{$q=∞$ braucht irgendeine Sonderbehandlung?} Let $Q ∈ \secsob(𝒬^{𝕊}M)$. Then $Q$ is orientable, \ie there exists $n ∈ \secsob(𝕊M)$ with $P(n) = Q$. \end{theorem} % end theorem Sobolev orientability on simply connected manifolds \begin{proof} % of theorem Sobolev orientability on simply connected manifolds Since $M$ is compact and $2 ≤ q$, $Q ∈ \secsob[2](𝒬^{𝕊})$. By \cref{thm:standard_and_tangential_norm} $ι_𝒬(Q) ∈ W^{1,2}(M, 𝒬^{𝕊}ℝ^N)$. $𝒬^{𝕊'}ℝ^N$ is a model for $ℝℙ^{N-1}$ as mentioned in \cref{sec:continuous_q_lift} and hence a smooth compact manifold. Therefore by \cref{thm:sequentially_weak_density_of_smooth_manifold_maps} there exists a sequence $(Q_{(k)})_k$ in $C^∞(M, 𝒬^{𝕊'}ℝ^N)$ converging weakly to $ι_𝒬(Q)$. By \cref{thm:orientability_on_simply_connected_manifolds} $Q_{(k)}$ is orientable for all $k ∈ ℕ$ with $P(n_{(k)}) = Q_{(k)}$ and $n ∈ C(M, 𝕊^{N-1})$. By \cref{thm:image_of_embedding_based_projection} $n_{(k)} ∈ W^{1,2}(M, 𝕊^{N-1})$. \cref{thm:orientability_preserved_by_weak_convergence_M_embedding} now shows that $ι_𝒬(Q)$ is orientable with $P_N(n) = ι_𝒬(Q)$ and $n ∈ W^{1,2}(𝕊M)$. \eqref{eq:∇ninQterms} gives a formula for $∇n$ that shows that $∇n$ is also in $L^q$ since $∇ι_*(Q)$ is in $L^q$. % Since $ι_𝒬(Q) ∈ W^{1,p}(𝒬^{𝕊'}M)$ formular \eqref{eq:∇ninQterms} further shows that $n ∈ W^{1,p}(𝕊^*M)$. Transferring $n$ back with $ι_*^{-1}$, we get $P(ι_*^{-1}n) = Q$, \ie $Q$ is orientable. \end{proof} % of theorem Sobolev orientability on simply connected manifolds \docEnd