Fix typos

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

6-differentiaforms.tex

@@ -733,7 +733,7 @@ \section{Lie derivative}
 \section{De Rham cohomology and Poincar\'e lemma}
 
 \begin{definition}
-We say that a smooth differential form $\omega\in\Omega^k(M)$ is \emph{closed} if $d\omega = 0$, and \emph{exact} if there exists a smooth $(k-1)$-form $\nu$ on $M$ such that $\omega = d\eta$.
+We say that a smooth differential form $\omega\in\Omega^k(M)$ is \emph{closed} if $d\omega = 0$, and \emph{exact} if there exists a smooth $(k-1)$-form $\eta$ on $M$ such that $\omega = d\eta$.
 
 The fact that $d\circ d = 0$ implies that every exact form is closed.
 \end{definition}
@@ -744,7 +744,7 @@ \section{De Rham cohomology and Poincar\'e lemma}
 We are going to prove it in two slightly different flavours: its classical version and a slight generalization.
 
 \begin{exercise}[\textit{[homework 4]}]
-  Let $M=\R^2\setminus\{0\}$ and $\omega$ the one-form on $M$ from Example~\ref{ex:li} given by
+  Let $M=\R^2\setminus\{0\}$, $\{x,y\}$ denote the standard euclidean coordinates in $\R^2$ and $\omega$ be the one-form on $M$ from Example~\ref{ex:li} given by
   \begin{equation}
     \omega = \frac{xdy - ydx}{x^2+y^2}.
   \end{equation}

7-integration.tex

@@ -108,17 +108,21 @@ \section{Orientation}
 \begin{proof}
   Let $\varphi$ and $\psi$ be two different charts with overlapping domains (otherwise there is nothing to check) and with local coordinates $(x^i)$ and $(y^i)$ respectively.
   Define the transition map $\sigma := \psi\circ\varphi^{-1}$, so that $(y^1,\ldots,y^n) = \phi(x)$.
-  %Since $d y^j = (D\sigma)_i^j dx^i$, we have that
-  Locally,
+  Since $d y^j = (D\sigma)_i^j dx^i$, we have that locally,
   \begin{align}
-    \omega
-    &= \omega(x) dx^{1}\wedge\cdots\wedge dx^{n} \\
+    \omega &= \widetilde \omega(y) dy^1\wedge\cdots\wedge dy^n \\
+    &= (\widetilde\omega \circ \sigma)(x) (D\sigma|_x)^1_{j_1}\cdots (D\sigma|_x)^n_{j_n} dx^{j_1}\wedge\cdots\wedge dx^{j_n} \\
+    &= (\widetilde\omega \circ \sigma)(x) \det(D\sigma|_x) dx^{1}\wedge\cdots\wedge dx^{n} \\
     &= \sigma^*(\widetilde \omega(y) dy^1\wedge\cdots\wedge dy^n) \\
-    &= (\widetilde\omega \circ \sigma)(x) \det(D\sigma|_x) dx^{1}\wedge\cdots\wedge dx^{n}
+    &= \omega(x) dx^{1}\wedge\cdots\wedge dx^{n} \\
     ,
   \end{align}
   where we used Proposition~\ref{prop:wedgeToJDet} and Theorem~\ref{thm:pullbacksdifferentialforms}.
-  Thus, $\omega(x)$ and $\widetilde\omega(y)$ have the same sign if and only if $\det(D\sigma|_x) > 0$.
+  Thus,
+  \begin{equation}\label{form:cov}
+    \omega(x) = (\widetilde\omega \circ \sigma)(x) \det(D\sigma|_x),
+  \end{equation}
+  and it has the same sign of $\widetilde\omega(y)$ if and only if $\det(D\sigma|_x) > 0$.
 \end{proof}
 
 \begin{definition}
@@ -181,9 +185,9 @@ \section{Orientation}
   The M\"obius strip inherits the differentiable structure from $\R^2$, so we need to show that there is no orientable atlas which is also compatible with the differentiable structure on $M$.
   \begin{enumerate}
     \item Define the map $\sigma:\R\times(-1,1)\to\R\times(-1,1)$ by $\sigma(x,y) = (x+1, -y)$ and show that $\pi\circ\sigma = \sigma$.
-    \item If $\nu\in\Omega^2(M)$ define $f$ by $\pi^* \nu = f \omega$ where $\omega$ is an area\footnote{I.e. a volume $2$-form.} form on $\R\times(-1,1)$.
+    \item If $\eta\in\Omega^2(M)$ defines\footnote{Cf. Exercise~\ref{ex:vfshape}.} $f$ by $\pi^* \eta = f \omega$ where $\omega$ is an area\footnote{I.e. a volume $2$-form.} form on $\R\times(-1,1)$.
     Show that $f(x+1, y) = - f(x,y)$.
-    \item Conclude that $f$ must vanish at some point of $\R\times(-1,1)$, which implies that $M$ is non-orientable \footnote{See also Exercise~\ref{ex:vfshape}}.
+    \item Conclude that $f$ must vanish at some point of $\R\times(-1,1)$, which implies that $M$ is non-orientable.
   \end{enumerate}
 \end{exercise}
 
@@ -195,7 +199,7 @@ \section{Orientation}
 
 \begin{exercise}[\textit{[homework 4]}]\label{ex:TMorient}
   Let $M$ be a smooth manifold without boundary and $\pi: TM \to M$ its tangent bundle.
-  Show that if $\{U,\phi\}$ is any atlas on $M$, then the corresponding\footnote{Remember Theorem~\ref{thm:tgbdlsmoothmfld}.} atlas on $TU$ is oriented.
+  Show that if $\{(U_\alpha,\phi_\alpha)\}$ is any atlas on $M$, then the corresponding\footnote{Remember Theorem~\ref{thm:tgbdlsmoothmfld}.} atlas $\{(TU_\alpha, \widetilde\phi_\alpha)\}$ on $TM$ is oriented.
   This, in particular, proves that the total space $TM$ of the tangent bundle is always orientable, regardless of the orientability of $M$.
 \end{exercise}
 
@@ -250,7 +254,7 @@ \section{Orientation}
   \end{enumerate}
 \end{exercise}
 
-Note that if $(U_i, \varphi_i)$ is a positively oriented atlas on $M$, then $(U_i|_{\partial M}, \varphi_i|_{\partial M})$ can be negatively oriented. 
+Note that if $\{(U_i, \varphi_i)\}$ is a positively oriented atlas on $M$, then $\{(U_i|_{\partial M}, \varphi_i|_{\partial M})\}$ can be negatively oriented. 
 Let $\omega = dx^1\wedge\cdots\wedge dx^n$ be a positive volume form on $M$ on one of the charts, then $-\frac{\partial}{\partial x^n}$ is an outward pointing on $\partial \cH^n$ and we have\footnote{Recall Lemma~\ref{lemma:intprod}.}
 \begin{align}
   \iota_{-{\partial}/\!{\partial x^n}} (dx^1\wedge\cdots\wedge dx^n)
@@ -269,7 +273,7 @@ \section{Orientation}
   Orientability is common but there are many examples of nonorientable manifolds.
   \begin{enumerate}
     \item Prove that $\bS^n$ is orientable.\\
-      \textit{\small Hint: there is a small exercise above that can help a lot here.}
+      \textit{\small Hint: above there is a small exercise that can help a lot here.}
     \item Prove that any Lie group is orientable.
     \item Prove that $\RP^n$ is orientable if and only if $n$ is odd. \\
       \textit{\small Hint: the antipodal map $x\mapsto -x$ on $\bS^n$ can help.}
@@ -284,19 +288,19 @@ \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 $M$.
-  If $\omega\in\Omega^n(M)$ be a $n$-form with compact support in $U$, we define the integral of $\omega$ as
+  If $\omega\in\Omega^n(M)$ be a $n$-form, $n > 0$, with compact support in $U$, we define the integral of $\omega$ as
   \begin{equation}
-    \int_M \omega = \int_U \omega := \int_{\varphi(U)} \varphi_*\omega := \int_{\R^n} \omega(x) d x^1\cdots dx^n,
+    \int_M \omega = \int_U \omega := \int_{\varphi(U)} \varphi_*\omega := \int_{\cH^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,
   \begin{equation}
-    \varphi_*\omega = \omega(x) e^{1}\wedge \cdots\wedge e^{n}\in\Omega^n(\cH^n).
+    \varphi_*\omega = \omega(x)\; e^{1}\wedge \cdots\wedge e^{n}\in\Omega^n(\cH^n).
   \end{equation}
-  For convenience we may write $d^n x := dx^1 \cdot dx^n$.
+  For convenience we may write $d^n x := dx^1 \cdots dx^n$.
 
-  If $M$ is an oriented $0$-dimensional manifold and $f$ is a $0$-form (that is, a smooth function) than we defined the integral to be the sum
+  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}
-    \int_M f = \sum_{p\in M} \pm f(p),
+    \int_M f := \sum_{p\in M} \pm f(p),
   \end{equation}
   where we take the positive sign at points where the orientation is positive and the negative sign at points where it is negative.
   The compactness assumption here implies that there are only finitely many nonzero terms in the sum.
@@ -309,14 +313,14 @@ \section{Integrals on manifolds}
   Then, the value of the integral $\int_M\omega$ is independent on the chosen chart.
 \end{lemma}
 \begin{proof}
-  Let $\varphi$ and $\psi$ be two charts on $U$ with the same orientation and local coordinates $x$ and $y$, let $\sigma = \psi\circ\varphi^{-1}$ be the corresponding transition map.
+  Let $\varphi$ and $\psi$ be two charts on $W = U\cap V$ with the same orientation and local coordinates $x$ and $y$, let $\sigma = \psi\circ\varphi^{-1}$ be the corresponding transition map.
   Then,
   \begin{align}
-    \int_{\varphi(U)} \varphi_*\omega &= \int \omega(x)\, d^n x \\
-    &= \int \sigma^*(\widetilde\omega(y) d^n y) \\
-    &\overset{(\bigstar)}{=} \int (\widetilde\omega \circ \sigma)(x) \det(D\sigma|_x) d^n x \\
+    \int_{\varphi(W)} \varphi_*\omega &= \int \omega(x)\, d^n x \\
+    &\overset{\eqref{form:cov}}{=} \int (\widetilde\omega \circ \sigma)(x) \det(D\sigma|_x) d^n x \\
+    &\overset{(\bigstar)}{=} \int \sigma^*(\widetilde\omega(y) d^n y) \\
     &\overset{(\clubsuit)}{=} \int \widetilde\omega(y) d^n y\\
-    &= \int_{\psi(V)} \psi_*\omega,
+    &= \int_{\psi(W)} \psi_*\omega,
   \end{align}
   where $\omega(p)$ and $\widetilde\omega(q)$ are the local expressions for $\omega$ in the two coordinate charts, in $(\bigstar)$ we applied Proposition~\ref{prop:wedgeToJDet} and in $(\clubsuit)$ we used the classical euclidean change of variables.
 \end{proof}
@@ -377,7 +381,7 @@ \section{Integrals on manifolds}
   which shows the commutativity of the following diagram
   \begin{equation}
     \begin{tikzcd}
-      \Omega^n(M) \arrow[rr, "F^*", yshift=0.5ex] \arrow[dr, "\int_M"] & & \Omega^n(N)\arrow[ll, "F_*", yshift=0.5ex] \arrow[dl, "\int_N"] \\
+      \Omega^n(M) \arrow[rr, "F^*", shift right = 0.5] \arrow[dr, "\int_M"] & & \Omega^n(N)\arrow[ll, "F_*", shift right = 0.5] \arrow[dl, "\int_N"] \\
       & \R &
     \end{tikzcd}
   \end{equation}
@@ -410,7 +414,7 @@ \section{Integrals on manifolds}
 \begin{exercise}[Fubini's theorem \textit{[homework 4]}]\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 $\pi_M^* \omega \wedge \pi_N^* \eta$.
+  \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 $\pi_M^* \omega \wedge \pi_N^* \eta$.
 
   If $\alpha\in\Omega^m(M)$ and $\beta\in\Omega^n(N)$ have compact support, show that
   \begin{equation}
@@ -631,7 +635,7 @@ \section{Stokes' Theorem}
 \begin{exercise}
   Let $D^n := \{x=(x^1, \ldots, x^n)\in\R^n \mid \|x\| \leq 1\}$ denote the unit disk in $\R^n$ centred at $0$. Recall that $\partial D^n = \bS^{n-1}$.
   \begin{enumerate}
-    \item Compute $\int_{\bS^1} \nu$ where $\nu$ is the following 1-form on $\R^2$: $\nu = -x^2 dx^1 + x^1 dx^2$.
+    \item Compute $\int_{\bS^1} \eta$ where $\eta$ is the following 1-form on $\R^2$: $\eta = -x^2 dx^1 + x^1 dx^2$.
     \item Compute $\int_{\bS^2} \omega$ where $\omega$ is the following 2-form on $\R^3$: $\omega = -x^1 dx^1\wedge dx^3 - x^2 dx^1\wedge dx^3 + x^3 dx^1\wedge dx^2$.
     \item Show that $\eta$ and $\omega$ above are closed but not exact (as differential forms on $\bS^1$ and $\bS^2$ respectively).
   \end{enumerate}
@@ -650,9 +654,9 @@ \section{Stokes' Theorem}
     &= 2\pi - 2\pi = 0.
   \end{align}
 
-  An important consequence of this is that while locally $\omega$ is the differential of the angle function $\theta$, this cannot be exact on all $M$: indeed, if $\omega = d\nu$, we would have
+  An important consequence of this is that while locally $\omega$ is the differential of the angle function $\theta$, this cannot be exact on all $M$: indeed, if $\omega = d\eta$, we would have
   \begin{equation}
-    2\pi = \int_{\bS^1}\omega = \int_{\bS^1}d\nu = \int_{\partial\bS^1}\nu = 0.
+    2\pi = \int_{\bS^1}\omega = \int_{\bS^1}d\eta = \int_{\partial\bS^1}\eta = 0.
   \end{equation}
 
   Moreover, since $2\pi = \int_{\bS^1}\omega$, Stokes' theorem also implies that $\bS^1$ is not the boundary of a compact regular domain in $\R^2\setminus\{0\}$.

aom.tex

@@ -32,7 +32,7 @@
 \usepackage{multicol} % multiple column layout facilities
 \usepackage{fancyvrb,xcolor} % extended verbatim environments
   \fvset{fontsize=\normalsize}% default font size for fancy-verbatim environments
-\usepackage{tikz-cd,bbm,mathbbol}
+\usepackage{tikz,tikz-cd,bbm,mathbbol}
 \DeclareSymbolFontAlphabet{\mathbbl}{bbold} %let's you use \mathbbl{k} for a field k
 \hypersetup{colorlinks} %puts color to hyperlinks
 \setcounter{secnumdepth}{2} 
@@ -111,7 +111,8 @@
     version={4.0},
 ]{doclicense}
 
-\usepackage{tcolorbox}
+\usepackage[dvipsnames]{xcolor}
+\usepackage[many]{tcolorbox}
 
 %\usepackage[numbers, sort]{natbib}
 \usepackage[style=alphabetic,natbib=true,sorting=anyt]{biblatex}
@@ -207,7 +208,7 @@
     \setlength{\parskip}{\baselineskip}
     Copyright \copyright\ \the\year\ \thanklessauthor
 
-    \par Version 0.9.3 -- \today
+    \par Version 0.9.5 -- \today
 
     \vfill
     \small{\doclicenseThis}
@@ -266,7 +267,7 @@ \chapter*{Introduction}
 In some sense I would like this course to provide the introduction to geometric analysis that I wish was there when I prepared my \href{https://www.mseri.me/lecture-notes-hamiltonian-mechanics/}{first edition} of the Hamiltonian mechanics course.
 
 I am extremely grateful to Martijn Kluitenberg for his careful reading of the notes and his useful comments and corrections.
-Many thanks also to Huub Bouwkamp, Bram Brongers, Mollie Jagoe Brown, Nicol\'as Moro, Luuk de Ridder, Jordan van Ekelenburg, Hanneke van Harten and Dave Verweg for reporting a number of misprints and corrections.
+Many thanks also to Huub Bouwkamp, Bram Brongers, Mollie Jagoe Brown, Nicol\'as Moro, Luuk de Ridder, Lisanne Sibma, Jordan van Ekelenburg, Hanneke van Harten and Dave Verweg for reporting a number of misprints and corrections.
 
 \mainmatter