Update after AMS review

Signed-off-by: Marcello Seri <[email protected]>

1-manifolds.tex

@@ -21,7 +21,7 @@
     \begin{equation}\label{eq:diff}
       \lim_{\|h\|\to 0} \frac{\|f(x+h) - f(x) - T h\|}{\|h\|} = 0.
     \end{equation}
-    The map $Df(x) := T$ is the \emph{total derivative}\footnote{This is sometimes called (total) differential in multivariable analysis, but this terminology may become a source of confusion for us.} of $f$ and is nothing else than the Jacobian matrix of $f$ at the point $x$, that is
+    The map $Df(x) := T$ is the \emph{total derivative}\sidenote[][-2em]{This is sometimes called (total) differential in multivariable analysis, but this terminology may become a source of confusion for us.} of $f$ and is nothing else than the Jacobian matrix of $f$ at the point $x$, that is
     \begin{equation}\label{eq:jacobian}
       Df(x) = \begin{pmatrix}
         \frac{\partial f^1}{\partial x^1}(x) & \cdots & \frac{\partial f^1}{\partial x^n}(x) \\
@@ -61,7 +61,7 @@ \section{Topological manifolds}\label{sec:top_manifolds}
   \begin{enumerate}[(i)]
     \item $X$ and $\emptyset$ are open, i.e., $X\in \cT$ and $\emptyset\in\cT$;
     \item arbitrary unions of families of open subsets are open;
-    \item the intersection of finitely many\footnote{It is equivalent to require the intersection of any two open subsets to be open. (Why?)} open subsets is open.
+    \item the intersection of finitely many\sidenote[][-2em]{It is equivalent to require the intersection of any two open subsets to be open. (Why?)} open subsets is open.
   \end{enumerate}
 \end{definition}
 
@@ -107,8 +107,8 @@ \section{Topological manifolds}\label{sec:top_manifolds}
 \end{exercise}
 
 \begin{definition}[Topological manifold]
-  A topological space\footnote{From now on, if we say that $X$ is a topological space we are implying that there is a topology $\cT$ defined on $X$.} $M$ is a \emph{topological manifold} of dimension $n$, or topological $n$-manifold, if it has the following properties:
-  \marginnote[0.5em]{Note that the finite dimensionality is a somewhat artificial restriction: manifolds can be infinitely dimensional~\cite{book:lang:infinite}. For example, the space of continuous functions between manifolds is a so-called infinite-dimensional Banach manifold.\vspace{1em}}
+  A topological space\sidenote[][-1.5em]{From now on, if we say that $X$ is a topological space we are implying that there is a topology $\cT$ defined on $X$.} $M$ is a \emph{topological manifold} of dimension $n$, or topological $n$-manifold, if it has the following properties:
+  \marginnote[-0.5em]{Note that the finite dimensionality is a somewhat artificial restriction: manifolds can be infinitely dimensional~\cite{book:lang:infinite}. For example, the space of continuous functions between manifolds is a so-called infinite-dimensional Banach manifold.\vspace{1em}}
   \begin{enumerate}[(i)]
     \item $M$ is a Hausdorff space;
     \item $M$ is second countable;
@@ -141,7 +141,7 @@ \section{Topological manifolds}\label{sec:top_manifolds}
 \end{example}
 
 \begin{exercise}[The line with two origins]
-  Even though $\R^n$ with the euclidean topology is Hausdorff, being Hausdorff does not follow from being locally euclidean. A famous counterexample is the following\footnote{See also \cite[Problem 1-1]{book:lee} and \cite[Problem 5.1]{book:tu}.}.
+  Even though $\R^n$ with the euclidean topology is Hausdorff, being Hausdorff does not follow from being locally euclidean. A famous counterexample is the following\sidenote[][-2em]{See also \cite[Problem 1-1]{book:lee} and \cite[Problem 5.1]{book:tu}.}.
   \begin{marginfigure}
     \includegraphics{1_ex_1_0_11.pdf}
     \label{fig:hausdorff-not-locally-euclidean}
@@ -150,7 +150,7 @@ \section{Topological manifolds}\label{sec:top_manifolds}
   Let $A_1, A_2$ be two points not on the real line $\R$ and define $M:= (\R\setminus\{0\})\cup\{A_1,A_2\}$.
   Induce a topology on $M$ by taking as basis the collection of all open intervals in $\R$ that do not contain $0$, along with all the sets of the form $(-a, 0)\cup\{A_1\}\cup(0,a)$ and $(-a, 0)\cup\{A_2\}\cup(0,a)$, for $a>0$.
   \begin{enumerate}
-    \item Check that this forms a basis\footnote{That is, the basis elements cover $M$ and for any $B_1, B_2$ on the basis, for all $x \in I = B_1\cap B_2$, there is an element $B_3$ of the basis such that $x\in B_3$ and $B_3\subset I$.} for a topology on $M$.
+    \item Check that this forms a basis\sidenote[][-1.5em]{That is, the basis elements cover $M$ and for any $B_1, B_2$ on the basis, for all $x \in I = B_1\cap B_2$, there is an element $B_3$ of the basis such that $x\in B_3$ and $B_3\subset I$.} for a topology on $M$.
     \item Define the two charts
       \begin{equation}
         \varphi_j:(\R\setminus\{0\})\cup\{A_j\} \to \R, \quad

3-vectorfields.tex

@@ -134,11 +134,10 @@ \section{Vector fields}
     X_{F^{-1}(u,v)} = \frac{u^2}{v^2}\frac{\partial}{\partial x}\Big|_{F^{-1}(u,v)},
   \end{equation}
   and, thus, by~\eqref{eq:pfvf} (with $p=(u,v)$) we get
+  \marginnote[-5em]{For the computation, keep in mind that the Jacobian is from a change of coordinates between the two euclidean spaces $T_{F^{-1}(u,v)}M$ and $T_{(u,v)}N$, where on the first we are using the coordinate basis $\left\{\frac{\partial}{\partial x}\Big|_{F^{-1}(u,v)},\frac{\partial}{\partial y}\Big|_{F^{-1}(u,v)}\right\}$ and on the second we are using the coordinate basis $\left\{\frac{\partial}{\partial u}\Big|_{(u,v)},\frac{\partial}{\partial v}\Big|_{(u,v)}\right\}$.}
   \begin{equation}
     (F_*X)_{(u,v)} = \frac{u^2}{v^2}\frac{\partial}{\partial u}\Big|_{(u,v)} + \frac{u}{v}\frac{\partial}{\partial v}\Big|_{(u,v)}.
   \end{equation}
-  \marginnote[-5em]{
-  For this computation, keep in mind that the Jacobian here is a change of coordinate between the two euclidean spaces $T_{F^{-1}(u,v)}M$ and $T_{(u,v)}N$, where on the first we are using the coordinate basis $\left\{\frac{\partial}{\partial x}\Big|_{F^{-1}(u,v)},\frac{\partial}{\partial y}\Big|_{F^{-1}(u,v)}\right\}$ and on the second we are using the coordinate basis $\left\{\frac{\partial}{\partial x}\Big|_{(u,v)},\frac{\partial}{\partial y}\Big|_{(u,v)}\right\}$.}
 \end{example}
 
 \begin{exercise}
@@ -399,7 +398,7 @@ \section{Flows and integral curves}
 
 \begin{example}\label{ex:rotation}
   Let $(x,y)$ be standard coordinates on $\R^2$ and $Z = x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x}$ on $\R^2$.
-  If $\gamma:\R\to\R^2$ is a smooth curve, written in standard coordinates\footnote{In other words, $x(t)$ and $y(t)$ are the components of the function $\gamma : \R \to \R^2$ seen as functions $\R\to\R$, that is, ${\gamma(t) = (x\circ \gamma(t), y\circ \gamma(t)) =: (x(t), y(t))}$} as $\gamma(t) = (x(t), y(t))$, then we have
+  If $\gamma:\R\to\R^2$ is a smooth curve, written in standard coordinates\sidenote[][-2em]{In other words, $x(t)$ and $y(t)$ are the components of the function $\gamma : \R \to \R^2$ seen as functions $\R\to\R$, that is, ${\gamma(t) = (x\circ \gamma(t), y\circ \gamma(t)) =: (x(t), y(t))}$} as $\gamma(t) = (x(t), y(t))$, then we have
   \begin{align}
     \gamma'(t)
     &= d\gamma_t\left(\frac{d}{dt}\Big|_t\right) \\

6-differentiaforms.tex

@@ -951,10 +951,10 @@ \section{De Rham cohomology and Poincar\'e lemma}
     F^* = (K\circ i_0)^* = i_0^*\circ K^* = i_1^*\circ K^* = (K\circ i_1)^* = G^*.
   \end{equation}
 
-  In fact this is the case, thanks to the following theorem\footnote{This is a deep result related to the Whitney Embedding Theorem from Remark~\ref{rmk:WhitneyET} and is out of the scope of our course, for more details refer to~\cite[Chapter 6 and Theorems 6.26 and 9.27]{book:lee}.}.
+  In fact this is the case, thanks to the following theorem\sidenote[][-1em]{This is a deep result related to the Whitney Embedding Theorem from Remark~\ref{rmk:WhitneyET} and is out of the scope of our course, for more details refer to~\cite[Chapter 6 and Theorems 6.26 and 9.27]{book:lee}.}.
   %
   \begin{theorem}[Whitney Approximation Theorem for continuous maps]\label{thm:WhitneyApproxCont}
-    Given any continuous mapping $G \in C^0(M,N)$, there exists $F \in C^\infty(M,N)$ which is homotopic to $G$. Moreover, if $G$ is smooth\footnote{Note that a function $f : M \to N$ is defined to be smooth on a subset $A \subset M$ if there is some smooth function $g: U \to N$, defined on an open $U\supset A$ such that $g = f$ on $A$.} on a closed subset $A\subset M$, then one can choose $F$ so that $F=G$ on $A$.
+    Given any continuous mapping $G \in C^0(M,N)$, there exists $F \in C^\infty(M,N)$ which is homotopic to $G$. Moreover, if $G$ is smooth\sidenote[][1em]{Note that a function $f : M \to N$ is defined to be smooth on a subset $A \subset M$ if there is some smooth function $g: U \to N$, defined on an open $U\supset A$ such that $g = f$ on $A$.} on a closed subset $A\subset M$, then one can choose $F$ so that $F=G$ on $A$.
   \end{theorem}
   %
   In particular, if two smooth maps are homotopic then they are also smoothly homotopic: we can assume the map $K$ to be smooth.

7-integration.tex

@@ -212,7 +212,7 @@ \section{Orientation on manifolds}
     \quad\mbox{and}\quad
     \varphi_2(p) = \frac{2p^1}{1+p^2}.
   \end{equation}
-  Let's pick a pointwise orientation by choosing as basis $X_p\in T_pM$ given by\footnote{We are not making up anything, if you look carefully this is just $X_p = \partial_\theta$.} $X_p = -p^2 \frac{\partial}{\partial p^1} + p^1 \frac{\partial}{\partial p^2}$.
+  Let's pick a pointwise orientation by choosing as basis $X_p\in T_pM$ given by\sidenote[][-1em]{We are not making up anything, if you look carefully this is just $X_p = \partial_\theta$.} $X_p = -p^2 \frac{\partial}{\partial p^1} + p^1 \frac{\partial}{\partial p^2}$.
   Then, on $U_1$,
   \begin{align}
     (\varphi_1)_*(X) &= (d\varphi_1)_p(X) \\
@@ -361,16 +361,16 @@ \section{Integrals on manifolds}
 
 \begin{definition}\label{def:intnform:chart}
   Let $M$ be a smooth $n$-manifold and $(U,\varphi)$ be a chart from an oriented atlas of $M$ with coordinates $(x^i)$.
-  If $\omega\in\Omega^n(M)$ be a $n$-form, $n > 0$, with compact support in $U$, we define the integral of $\omega$ as\footnote{Recall that for a diffeomorphism $\phi$, $\phi_* = (\phi^{-1})^*$.}
+  If $\omega\in\Omega^n(M)$ be a $n$-form, $n > 0$, with compact support in $U$, we define the integral of $\omega$ as\sidenote[][-1em]{Recall that for a diffeomorphism $\phi$, $\phi_* = (\phi^{-1})^*$.}
   \begin{equation}
     \int_M \omega = \int_U \omega := \int_{\varphi(U)} \varphi_*\omega := \int_{\R^n} \omega(x) d x^1\cdots dx^n,
   \end{equation}
-  where the last is the usual Riemannian integral on $\R^n$ and, on the chart\footnote{In general, locally $\omega = \frac1{n!}\omega_{i_1,\ldots,i_n}(x) dx^{i_1}\wedge\cdots\wedge dx^{i_n}$, but, by reordering the basis element for each term in the sum, we can always write it in the more compact form presented here.}
+  where the last is the usual Riemannian integral on $\R^n$ and, on the chart\sidenote[][-2em]{In general, locally $\omega = \frac1{n!}\omega_{i_1,\ldots,i_n}(x) dx^{i_1}\wedge\cdots\wedge dx^{i_n}$, but, by reordering the basis element for each term in the sum, we can always write it in the more compact form presented here.}
   \begin{equation}
     \varphi_*\omega = \omega(x)\; dx^{1}\wedge \cdots\wedge dx^{n}\in\Omega^n(\R^n).
   \end{equation}
-  For convenience we may write $d^n x := dx^1 \cdots dx^n$.
-  \marginnote[-5em]{Everything remains valid if $\R^n$ is replaced by $\cH^n$.}
+  For convenience we may sometimes write $d^n x := dx^1 \cdots dx^n$.
+  \marginnote{Everything in Definition~\ref{def:intnform:chart} remains valid if $\R^n$ is replaced by $\cH^n$.}
 
   If $M$ is an oriented $0$-dimensional manifold and $f\in\Omega^0(M) = C^\infty(M)$, than we define the integral to be the sum
   \begin{equation}
@@ -493,9 +493,12 @@ \section{Integrals on manifolds}
 \end{example}
 
 \begin{exercise}[Fubini's theorem]\label{exe:fubini}
-  Let $M^m$ and $N^n$ be oriented manifolds. Endow $M\times N$ with the product orientation, that is\footnote{An equivalent way is to say that if $v_1,\ldots,v_m\in T_pM$ and $w_1, \ldots, w_n\in T_q N$ are positively oriented bases in the respective spaces, then \begin{equation}
-    (v_1,0),\ldots,(v_n,0),(0,w_1), \ldots, (0,w_n)\in T_{(p,q)}(M\times N)
-  \end{equation}is defined to be a positively oriented basis in the product.}, if $\pi_M : M\times N \to M$ and $\pi_N: M\times N \to N$ are the canonical projections on the elements of the product, and $\omega$ and $\eta$ respectively define orientations on $M$ and $N$, then the orientation on $M\times N$ is defined to be the orientation defined by $\pi_M^* \omega \wedge \pi_N^* \eta$.
+  Let $M^m$ and $N^n$ be oriented manifolds. Endow $M\times N$ with the product orientation, that is\footnote{An equivalent way is to say that if $v_1,\ldots,v_m\in T_pM$ and $w_1, \ldots, w_n\in T_q N$ are positively oriented bases in the respective spaces, then \begin{align}
+    &(v_1,0),\ldots,(v_n,0),\\
+    &(0,w_1), \ldots, (0,w_n)\\
+    &\quad\in T_{(p,q)}(M\times N)
+\end{align}
+is defined to be a positively oriented basis in the product.}, if $\pi_M : M\times N \to M$ and $\pi_N: M\times N \to N$ are the canonical projections on the elements of the product, and $\omega$ and $\eta$ respectively define orientations on $M$ and $N$, then the orientation on $M\times N$ is defined to be the orientation defined by $\pi_M^* \omega \wedge \pi_N^* \eta$.
 
   If $\alpha\in\Omega^m(M)$ and $\beta\in\Omega^n(N)$ have compact support, show that
   \begin{equation}
@@ -740,13 +743,13 @@ \section{Stokes' Theorem}
 \end{exercise}
 
 \begin{example}
+  \begin{marginfigure}
+    \includegraphics{8_3_8-annulus.pdf}
+  \end{marginfigure}
   Consider the annulus $M=\{(x,y)\in\R^2 \mid 1/2\leq x^2+y^2 \leq 1\}$ and the $1$-form $\omega = \frac{-y dx + x dy}{x^2 + y^2} = d\theta$ where $(x,y) = (\rho\cos\theta, \rho\sin\theta)$.
 
   Then $d\omega = 0$ and therefore $\int_M d\omega = 0$.
   Furthermore,
-  \begin{marginfigure}
-    \includegraphics{8_3_8-annulus.pdf}
-  \end{marginfigure}
   \begin{align}
     \int_{\partial M}\omega
     &= \int_{x^2 + y^2 =1} \omega + \int_{x^2+y^2 =1/2}\omega \\ 

aom.tex

@@ -213,7 +213,7 @@
   \setlength{\parskip}{\baselineskip}
   Copyright \copyright\ \the\year\ \thanklessauthor
 
-  \par Version 0.31 -- \today
+  \par Version 0.32 -- \today
 
   \vfill
   \small{\doclicenseThis}
@@ -264,7 +264,7 @@ \chapter*{Introduction}
 You should have access to both books via the University library and, in addition, Lee's ebook can be downloaded via the University proxy on \href{https://link.springer.com/book/10.1007/978-1-4419-9982-5}{SpringerLink}.
 
 The book~\cite{book:McInerney} is a nice compact companion that develops most of the concepts of the course in the specific case of $\R^n$ and could provide further examples and food for thoughts.
-The books~\cite{book:nicolaescu}\footnote{Beware of typos, there are many.},~\cite{book:crane} and~\cite{lectures:nanda}, freely available from the authors' website, are not really suitable as references for this courses but provides fantastic resources for the readers that want to dig further and see where the material discussed in the course can lead.
+The books \cite{book:nicolaescu}\footnote{Beware of typos, there are many.}, \cite{book:crane} and \cite{lectures:nanda}, freely available from the authors' website, are not really suitable as references for this courses but provides fantastic resources for the readers that want to dig further and see where the material discussed in the course can lead.
 Finally, a colleague mentioned~\cite{book:lang}. I don't have experience with this book but from a brief look it seems to follow a similar path as these lecture notes, so it might provide yet an alternative reference after all.
 
 The idea for the cut that I want to give to this course was inspired by the online \href{https://www.video.uni-erlangen.de/course/id/242}{Lectures on the Geometric Anatomy of Theoretical Physics} by Frederic Schuller, by the lecture notes of Stefan Teufel's Classical Mechanics course~\cite{lectures:teufel} (in German), by the classical mechanics book by Arnold~\cite{book:arnold} and by the Analysis of Manifold chapter in~\cite{book:thirring}.