Added some more exercises

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

6-differentiaforms.tex

@@ -248,7 +248,7 @@ \section{The interior product}
     \end{equation}
   \end{enumerate}
 \end{lemma}
-\begin{exercise}
+\begin{exercise}[\textit{[homework 4]}]
   Prove the Lemma.
 \end{exercise}
 
@@ -298,7 +298,7 @@ \section{Differential forms on manifolds}
     F^*\left(\omega_J dx^J\right) = (\omega_{j_1,\ldots, j_k}\circ F) d(x^{j_1}\circ F)\wedge\cdots\wedge d(x^{j_k}\circ F).
   \end{equation}
 \end{theorem}
-\begin{exercise}
+\begin{exercise}%[\textit{[homework 4]}]
   Prove the theorem.
 \end{exercise}
 
@@ -350,7 +350,7 @@ \section{Differential forms on manifolds}
   where $DF$ represents the Jacobian matrix of $F$ in these coordinates.
 \end{proposition}
 
-\begin{exercise}
+\begin{exercise}\textit{[homework 4]}
   Prove the Proposition~\ref{prop:wedgeToJDet}.\\
   \textit{\small Hint: look at Theorem~\ref{thm:pullbacksdifferentialforms}.}
 \end{exercise}
@@ -451,7 +451,12 @@ \section{Exterior derivative}
   \end{align}
 \end{proof}
 
-In particular, this means that for any diffeomorphism $F:M\to N$ and $\omega\in\Omega^k(M)$, $F^*d\omega = d F^*\omega$.
+\begin{exercise}\label{ex:smoothpushforward}
+  Let $F:M\to N$ be a smooth map between smooth manifolds and $\omega\in\Omega^k(N)$, then
+  \begin{equation}
+    F^* (d\omega) = d(F^* \omega).
+  \end{equation}
+\end{exercise}
 
 \begin{lemma}
   The exterior derivative satisfies the following properties.
@@ -475,16 +480,9 @@ \section{Exterior derivative}
   \end{align}
 \end{proof}
 
-\begin{exercise}\label{ex:smoothpushforward}
-  Let $F:M\to N$ be a smooth map between smooth manifolds and $\omega\in\Omega^k(N)$, then
-  \begin{equation}
-    F^* d\omega = d(F^* \omega).
-  \end{equation}
-\end{exercise}
-
 Let $N\subset M$ a submanifold and $i:N\hookrightarrow M$ the corresponding injection.
 For $\omega\in\Omega^k(M)$, we call $i^*\omega \in \Omega^k(N)$ the restriction of $\omega$ to $N$.
-The previous exercise, then, implies that restriction and exterior derivative commute, that is, $i^*d\omega = d(i^*\omega)$.
+Exercise~\ref{ex:smoothpushforward}, then, implies that restriction and exterior derivative commute, that is, $i^*d\omega = d(i^*\omega)$.
 
 \begin{example}[Exterior derivatives and vector calculus in $\R^3$]
   Let $M=\R^3$. Any smooth $1$-form $\omega\in\Omega^1(\R^3)$ can be written as
@@ -517,9 +515,9 @@ \section{Exterior derivative}
 
   Moreover, the interior multiplication yields another map $\beta: \fX(\R^3)\to\Omega^2(\R^3)$ defined by $\beta(X) = \iota_X (dx\wedge dy\wedge dz)$, which is linear over $C^\infty(\R^3)$ (why?) and, thus, corresponds to a smooth bundle homomorphism from $T\R^3$ to $\Lambda^2(\R^3)$ (why?).
 
-  In a similar fashion, we can also define a smooth bundle isomorphism $(\bigstar): C^\infty(\R^3) \to \Omega^3(\R^3)$ via
+  In a similar fashion, we can also define a smooth bundle isomorphism $\bigstar: C^\infty(\R^3) \to \Omega^3(\R^3)$ via
   \begin{equation}
-    (\bigstar)(f) = f dx\wedge dy\wedge dz.
+    \bigstar(f) = f dx\wedge dy\wedge dz.
   \end{equation}
 
   If $f\in C^\infty(\R^3)$ and $v\in\cT_0^1(\R^3)$, we can use the exterior derivatives to observe that the following diagram commutes
@@ -528,7 +526,7 @@ \section{Exterior derivative}
       C^\infty(\R^3) \arrow[d, "\id"] \arrow[r, "\nabla"] &
       \fX(\R^3) \arrow[d, "{}^\flat"] \arrow[r, "\nabla\times"] &
       \fX(\R^3) \arrow[d, "\beta"] \arrow[r, "\nabla\cdot"] &
-      C^\infty(\R^3) \arrow[d, "(\bigstar)"] \\
+      C^\infty(\R^3) \arrow[d, "\bigstar"] \\
       \Omega^0(\R^3) \arrow[r, "d"] &
       \Omega^1(\R^3) \arrow[r, "d"] &
       \Omega^2(\R^3) \arrow[r, "d"] &
@@ -541,7 +539,7 @@ \section{Exterior derivative}
 \end{example}
 
 \begin{exercise}
-  Show that the diagram~\eqref{diag:comm:r3ops} commutes and use it to give a quick proof that $\nabla\times\circ \nabla \equiv 0$ and that $\nabla\cdot \circ \nabla\times \equiv 0$.
+  Show that the diagram~\eqref{diag:comm:r3ops} commutes and use it to give a quick proof that $(\nabla\times)\circ \nabla \equiv 0$ and that $(\nabla\cdot) \circ (\nabla\times) \equiv 0$.
   For example,
   \begin{equation}
     df = \frac{\partial f}{\partial x^i} = (\nabla f)_i dx^i = (\nabla f)^\flat.
@@ -567,7 +565,7 @@ \section{Exterior derivative}
     Show that $\omega := d\lambda$ is a symplectic form on $T^* M$, that is, every cotangent bundle is a symplectic manifold.
   \end{enumerate}
 
-  For example, $\omega = \sum_{i=1}^n \alpha^i\wedge \alpha^{i+n}\in\Omega^2(\R^{2n})$ is a symplectic form and plays a central role in classical mechanics. There one usually calls $(\alpha^{n+1},\ldots,\alpha^{2n})$ the \emph{position coordinates} and $(\alpha^{1},\ldots,\alpha^{n})$ the \emph{momentum coordinates}.
+  For example, $\omega = \sum_{i=1}^n \alpha^i\wedge \alpha^{i+n}\in\Omega^2(\R^{2n})$ is a symplectic form and plays a central role in classical mechanics. There, one usually calls $(\alpha^{n+1},\ldots,\alpha^{2n})$ the \emph{position coordinates} and $(\alpha^{1},\ldots,\alpha^{n})$ the \emph{momentum coordinates}.
   \begin{enumerate}
     \item[c)] Show that for $n=2$, $\omega \wedge \omega = -2 \alpha^1\wedge\alpha^2\wedge\alpha^3\wedge\alpha^4$.
 
@@ -707,7 +705,7 @@ \section{De Rham cohomology and Poincar\'e lemma}
 This is the statement of the so-called Poincar\'e lemma.
 We are going to prove it in two slightly different flavours: its classical version and a slight generalization.
 
-\begin{exercise}
+\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
   \begin{equation}
     \omega = \frac{xdy - ydx}{x^2+y^2}.

7-integration.tex

@@ -38,7 +38,7 @@ \section{Orientation}
 \end{definition}
 
 \begin{example}
-  If $e_i$ is the standard $i$th basis vector in $\R^n$, the standard orientation of $\R^n$ is given by declaring that $e_1\wedge\cdots e_n$ is a positive basis of $\Lambda^n(\R^n)$ and thus that $\{e_1,\ldots,e_n\}$ is a positive basis of $\R^n$.
+  If $e_i$ is the standard $i$th basis vector in $\R^n$, the standard orientation of $\R^n$ is given by declaring that $e_1\wedge\cdots\wedge e_n$ is a positive basis of $\Lambda^n(\R^n)$ and thus that $\{e_1,\ldots,e_n\}$ is a positive basis of $\R^n$.
 \end{example}
 
 The key in the preservation of orientation now resides only in the way different bases are transformed by $n$-forms, as the following lemma shows.
@@ -174,8 +174,19 @@ \section{Orientation}
   \end{enumerate}
 \end{exercise}
 
-What about orientation on the boundaries?
+\begin{exercise}
+  Let $f\in C^\infty(\R^{n+1})$ with $0$ as a regular value.
+  Show that $f^{-1}(0)$ is an orientable submanifold of $\R^{n+1}$.
+  %In particular, the unit $n$-sphere $\bS^n\subset\R^{n+1}$ is orientable.
+\end{exercise}
+
+\begin{exercise}[\textit{[homework 4]}]
+  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.
+  This, in particular, proves that the total space $TM$ of the tangent bundle is always orientable, regardless of the orientability of $M$.
+\end{exercise}
 
+\newthought{What about orientation on the boundaries?}
 Let's first look at the tangent space. 
 
 Let $M$ be a smooth $n$-manifold with boundary and $p\in \partial M$.
@@ -286,11 +297,11 @@ \section{Integrals on manifolds}
   Then,
   \begin{align}
     \int_{\varphi(U)} \varphi_*\omega &= \int \omega(p)\, d^n p \\
-    &\overset{((\bigstar))}{=} \int (\widetilde\omega \circ \varphi)(p) \det(D\varphi|_p) d^n p \\
-    &\overset{((\clubsuit))}{=} \int \widetilde\omega(q) d^n q\\
+    &\overset{(\bigstar)}{=} \int (\widetilde\omega \circ \varphi)(p) \det(D\varphi|_p) d^n p \\
+    &\overset{(\clubsuit)}{=} \int \widetilde\omega(q) d^n q\\
     &= \int_{\psi(U)} \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.
+  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}
 
 To be able to integrate charts which are not supported in the domain of a single chart, we now need the help of a partition of unity.
@@ -318,11 +329,11 @@ \section{Integrals on manifolds}
     \sum_j \int_{\varphi_j(U_j)} (\varphi_j)_*\left(\rho_j \omega\right)
     &= \sum_j \int_{\varphi_j(U_j)} (\varphi_j)_*\left(\rho_j \sum_k \widetilde\rho_k\omega\right) \\
     &= \sum_{j,k} \int_{\phi_j(U_j\cap V_k)} (\varphi_j)_* \left(\rho_j \widetilde\rho_k\omega\right) \\
-    &\overset{((\spadesuit))}{=} \sum_{j,k} \int_{\psi_k(U_j\cap V_k)} (\psi_k)_* \left(\rho_j \widetilde\rho_k\omega\right) \\
+    &\overset{(\spadesuit)}{=} \sum_{j,k} \int_{\psi_k(U_j\cap V_k)} (\psi_k)_* \left(\rho_j \widetilde\rho_k\omega\right) \\
     &= \sum_k \int_{\psi_k(U_j\cap V_k)} (\psi_k)_*\left(\rho_j \widetilde\rho_k \sum_j\rho_j \omega\right) \\
     &= \sum_k \int_{\psi_k(V_k)} (\psi_k)_*\left( \widetilde\rho_k \omega\right),
   \end{align}
-  where in $((\spadesuit))$ we used Lemma~\ref{lemma:intindep:chart}.
+  where in $(\spadesuit)$ we used Lemma~\ref{lemma:intindep:chart}.
 \end{proof}
 
 This result can be nicely formalized as follows.
@@ -379,7 +390,7 @@ \section{Integrals on manifolds}
   \end{equation}
 \end{example}
 
-\begin{exercise}[Fubini's theorem]\label{exe:fubini}
+\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 $\pi_M^* \omega \wedge \pi_N^* \eta$.
@@ -577,10 +588,10 @@ \section{Stokes' Theorem}
     \int_M d\omega
     &= \int_{\cH^n}(\phi^{-1})^* dw \\
     &= \int_{\cH^n} d\left((\phi^{-1})^* \omega\right) \\
-    &\overset{((\spadesuit))}{=} \int_{\partial \cH^n}(\phi^{-1})^* w \\
+    &\overset{(\spadesuit)}{=} \int_{\partial \cH^n}(\phi^{-1})^* w \\
     &= \int_{\partial M} w , 
   \end{align}
-  where in the $((\spadesuit))$ step we applied the computations above and where $\partial \cH^n$ has the orientation induced by $\cH^n$.
+  where in the $(\spadesuit)$ step we applied the computations above and where $\partial \cH^n$ has the orientation induced by $\cH^n$.
   For a negatively oriented smooth boundary chart, the computation applies with an extra negative sign on each side of the equation.
   For an interior chart, we get the same computations with $\R^n$ in place of $\cH^n$.
 

aom.tex

@@ -207,7 +207,7 @@
     \setlength{\parskip}{\baselineskip}
     Copyright \copyright\ \the\year\ \thanklessauthor
 
-    \par Version 0.7.7 -- \today
+    \par Version 0.8 -- \today
 
     \vfill
     \small{\doclicenseThis}