global-analysis-and-the-banach-manifold-of-class-h1-curvers

diff --git a/sections/structure.tex b/sections/structure.tex
@@ -9,9 +9,9 @@ M)\)? Specifically, what is a class \(H^1\) curve in \(M\)?
 Given an interval \(I\), recall that a continuous curve \(\gamma : I \to
 \RR^n\) is called \emph{a class \(H^1\)} curve if \(\gamma\) is absolutely
 continuous, \(\dot \gamma(t)\) exists for almost all \(t \in I\) and
-\(\dot\gamma \in L^2(I, \RR^n)\). It is a well known fact that the so called
-\emph{Sobolev space \(H^1(I, \RR^n)\)} of all class \(H^1\) curves in \(\RR^n\)
-is a Hilbert space under the inner product given by
+\(\dot\gamma \in H^0(I, \RR^n) = L^2(I, \RR^n)\). It is a well known fact that
+the so called \emph{Sobolev space \(H^1(I, \RR^n)\)} of all class \(H^1\)
+curves in \(\RR^n\) is a Hilbert space under the inner product given by
 \[
   \langle \gamma, \eta \rangle_1
   = \int_0^1 \gamma(t) \cdot \eta(t) + \dot\gamma(t) \cdot \dot\eta(t) \; \dt
@@ -33,12 +33,12 @@ Finally, we may define\dots
   From now on we fix \(I = [0, 1]\).
 \end{note}
 
-We should note that every piece-wise smooth curve \(\gamma : I \to M\) is a
-class \(H^1\) curve. This answer raises and additional question though: why
+Notice in particular that every piece-wise smooth curve \(\gamma : I \to M\) is
+a class \(H^1\) curve. This answer raises and additional question though: why
 class \(H^1\) curves? The classical theory of the calculus of variations -- as
 described in \cite[ch.~5]{gorodski} for instance -- is usually exclusively
 concerned with the study of piece-wise smooth curves, so the fact that we are
-now interested a larger class of curves, highly non-smooth curves in fact,
+now interested a larger class of curves -- highly non-smooth curves, in fact --
 \emph{should} come as a surprise to the reader.
 
 To answer this second question we return to the case of \(M = \RR^n\). Denote
@@ -46,15 +46,18 @@ by \({C'}^\infty(I, \RR^n)\) the space of piece-wise curves in \(\RR^n\). As
 described in section~\ref{sec:introduction}, we would like \({C'}^\infty(I,
 \RR^n)\) to be a Banach manifold under which both the energy functional and the
 length functional are smooth maps. As most function spaces, \({C'}^\infty(I,
-\RR^n)\) admits several natural topologies. Perhaps the most obvious candidate
-is the \(L^2\) topology, which is to say, the topology induced by the norm
-\[
-  \norm{\gamma}_\infty = \sup_t \norm{\gamma(t)}
-\]
+\RR^n)\) admits several natural topologies. Some of the most obvious candidates
+are the uniform topology and the topology of \(H^0\) norm, which are the
+topologies induces by the norms
+\begin{align*}
+  \norm{\gamma}_\infty & = \sup_t \norm{\gamma(t)}            \\
+       \norm{\gamma}_0 & = \int_0^1 \norm{\gamma(t)}^2 \; \dt
+\end{align*}
+respectively.
 
-The problem with this choice is that \(\ell : {C'}^\infty(I, \RR^n) \to \RR\)
-is not a continuous map under the uniform topology. This can be readily seen by
-approximating the curve
+The problem with the first candidate is that \(\ell : {C'}^\infty(I, \RR^n) \to
+\RR\) is not a continuous map under the uniform topology. This can be readily
+seen by approximating the curve
 \begin{align*}
   \gamma : I & \to     \RR^2      \\
            t & \mapsto (t, 1 - t)
@@ -95,20 +98,21 @@ natural candidate for a norm in \({C'}^\infty(I, \RR^n)\) is
   \norm{\gamma}_1^2 = \norm{\gamma}_0^2 + \norm{\dot\gamma}_0^2,
 \]
 which is, of course, the norm induced by the inner product \(\langle \, ,
-\rangle_1\) -- here \(\norm{\cdot}_0\) denote the norm of \(H^0(I, \RR^n) =
-L^2(I, \RR^n)\).
+\rangle_1\) -- here \(\norm{\cdot}_0\) denote the norm of \(H^0(I, \RR^n)\).
 
 The other issue we face is one of completeness. Since \(\RR^n\) has a global
-chart, we to expect \({C'}^\infty(I, \RR^n)\) to be affine too. In other words,
-it is natural to expect \({C'}^\infty(I, \RR^n)\) to be Banach space. In
-particular, \({C'}^\infty(I, \RR^n)\) must be a normed Banach space. This is
-unfortunately not the case for \({C'}^\infty(I, \RR^n)\) the norm
-\(\norm\cdot_1\), but we can consider its completion, which, by a classical
-result by Lebesgue, just so happens to coincide with \(H^1(I, M)\).
+chart, we expect \({C'}^\infty(I, \RR^n)\) to be affine too. In other words, it
+is natural to expect \({C'}^\infty(I, \RR^n)\) to be Banach space. In
+particular, \({C'}^\infty(I, \RR^n)\) must be complete. This is unfortunately
+not the case for \({C'}^\infty(I, \RR^n)\) the norm \(\norm\cdot_1\), but we
+can consider its completion. Lo and behold, a classical result by Lebesgue
+establishes that this completion just so happens to coincide with \(H^1(I,
+\RR^n)\).
 
 It's also interesting to note that the completion of \({C'}^\infty(I, \RR^n)\)
-with respect to the norm \(\norm\cdot_\infty\) is the space \(C^0(I, \RR^n)\)
-of all continuous curves \(I \to \RR^n\), and that the natural inclusions
+with respect to the norms \(\norm\cdot_\infty\) and \(\norm\cdot_0\) are
+\(C^0(I, \RR^n)\) and \(H^0(I, \RR^n)\), respectively, and that the natural
+inclusions
 \begin{equation}\label{eq:continuous-inclusions-rn-curves}
   H^1(I, \RR^n)
   \longhookrightarrow C^0(I, \RR^n)
@@ -216,9 +220,9 @@ We begin with a technical lemma.
 
 \begin{lemma}\label{thm:section-in-open-is-open}
   Let \(W \subset TM\) be an open neighborhood of the zero section in \(TM\).
-  Given \(\gamma \in {C'}^\infty(I, M)\), denote by \(W_{\gamma, t}\) the set \(W
-  \cap T_\gamma(t) M\) and let \(W_\gamma = \bigcup_t W_{\gamma, t}\). Then
-  \(H^1(W_\gamma) = \{ X \in H^1(\gamma^* TM) : X_t \in W_{\gamma, t} \;
+  Given \(\gamma \in {C'}^\infty(I, M)\), denote by \(W_{\gamma, t}\) the set
+  \(W \cap T_{\gamma(t)} M\) and let \(W_\gamma = \bigcup_t W_{\gamma, t}\).
+  Then \(H^1(W_\gamma) = \{ X \in H^1(\gamma^* TM) : X_t \in W_{\gamma, t} \;
   \forall t \}\) is an open subset of \(H^1(\gamma^* TM)\).
 \end{lemma}
 
@@ -271,8 +275,11 @@ Finally, we find\dots
   U_\gamma \to H^1(\gamma^* TM))\}_{\gamma \in {C'}^\infty(I, M)}\) is an atlas
   for \(H^1(I, M)\) under the final topology of the maps \(\exp_\gamma\) --
   i.e. the coarsest topology such that such maps are continuous. This atlas
-  gives \(H^1(\gamma^* TM)\) the structure of a \emph{separable} Banach
-  manifold.
+  gives \(H^1(I, M)\) the structure of a \emph{separable} Banach manifold
+  modeled after separable Hilbert spaces, with typical representatives
+  \(H^1(\gamma^* TM) \cong H^1(I, \RR^n)\)\footnote{Any trivialization of
+  $\gamma^* TM$ induces an isomorphism $H^1(\gamma^* TM) \isoto H^1(I,
+  \RR^n)$.}.
 \end{theorem}
 
 The fact that \(\exp_\gamma\) is bijective should be clear from the definition
@@ -287,13 +294,8 @@ of this proof is showing that the transition maps
   \to H^1(\eta^* TM)
 \]
 are diffeomorphisms, as well as showing that \(H^1(I, M)\) is separable. We
-leave this details we leave as an exercise to the reader -- see theorem 2.3.12
-of \cite{klingenberg} for a full proof.
-
-The charts \(\exp_\gamma^{-1}\) are modeled after separable Hilbert spaces,
-with typical representatives \(H^1(\gamma^* TM) \cong H^1(I,
-\RR^n)\)\footnote{Any trivialization of $\gamma^* TM$ induces an isomorphism
-$H^1(\gamma^* TM) \isoto H^1(I, \RR^n)$.}.
+leave this details as an exercise to the reader -- see theorem 2.3.12 of
+\cite{klingenberg} for a full proof.
 
 It's interesting to note that this construction is functorial. More
 precisely\dots
@@ -426,7 +428,7 @@ words, we'll show\dots
 \end{theorem}
 
 \begin{proof}
-  Given \(\gamma \in {C'}^\infty(I, M)\) and \(X \in H^1(\gamma^*)\), let
+  Given \(\gamma \in {C'}^\infty(I, M)\) and \(X \in H^1(\gamma^* TM)\), let
   \begin{align*}
     g_X^\gamma : H^0(\gamma^* TM) \times H^0(\gamma^* TM) & \to \RR \\
     (Y, Z) &
@@ -449,7 +451,10 @@ words, we'll show\dots
   \[
     g_\gamma(X, Y)
     = g_0^\gamma(X, Y)
-    = \int_0^1 \langle (d \exp)_0 X_t, (d \exp)_0 Y_t \rangle \; \dt
+    = \int_0^1
+      \langle
+      (d \exp)_{0_{\gamma(t)}} X_t, (d \exp)_{0_{\gamma(t)}} Y_t
+      \rangle \; \dt
     = \int_0^1 \langle X_t, Y_t \rangle \; \dt
     = \langle X, Y \rangle_0
   \]