Fix typos, add clarifications and exercises

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

Author: mseri

2-tangentbdl.tex

@@ -255,7 +255,7 @@ \section{Germs and derivations}
   For any $p\in U$, we can define a derivation of $C^\infty(U)$ at $p$ as
   \begin{equation}
     \frac{\partial}{\partial x^i}\Big|_p : C^\infty(U) \to \R, \quad
-    \frac{\partial}{\partial x^i}\Big|_p (f) := D_i(f\circ\varphi^{-1})(\varphi(p)).
+    \frac{\partial}{\partial x^i}\Big|_p (f) := \frac{\partial f}{\partial x^i}(p) := D_i(f\circ\varphi^{-1})(\varphi(p)).
   \end{equation}
   From now on, it will get more and more convenient to draw commutative diagrams to see ``how things are moving around'':
   \begin{equation}
@@ -343,6 +343,7 @@ \section{Germs and derivations}
     f = f(p) + x^i (g_i \circ \varphi),
     \quad g_i(0) = D_i (f \circ \varphi^{-1})(0) = \frac{\partial}{\partial x_i}\Big|_p(f).
   \end{equation}
+  \marginnote{If we are careful with the meaning of our notation, we could write more succintly $\frac{\partial f}{\partial x_i}(p)$ in place of $\frac{\partial}{\partial x_i}\big|_p(f)$ in the same fashion as in Example~\ref{ex:partialderivative}.}
   Thus, for any derivation $v$, we obtain
   \begin{equation}
     v(f) = v(f(p)) + v(x^i)g_i(0) + x^i(p) v(g_i\circ\varphi) = v(x^i)  \frac{\partial}{\partial x_i}\Big|_p(f).
@@ -523,9 +524,8 @@ \section{The differential of a smooth map}\label{sec:diffsmooth}
       \frac{\partial F^m}{\partial x^1} (p) & \cdot & \frac{\partial F^m}{\partial x^n} (p)
     \end{pmatrix}
   \end{equation}
-  without using the Proposition above.
-
-  \noindent\textit{\small Hint: show that $d F_p \left(\frac{\partial}{\partial x^i}\big|_p\right) (f) = \left(\frac{\partial F^j}{\partial x^i} (p) \frac{\partial}{\partial y^j}\big|_{F(p)}\right) (f)$.}
+  without using the Proposition above. Here $\frac{\partial F^i}{\partial x^j} (p) = \frac{\partial}{\partial x^j}\big|_p (F^i)$, where $F^i$ is the $i$th component of $F$ with respect to the chart with coordinates $y^j$.\\
+  \textit{\small Hint: show that $d F_p \left(\frac{\partial}{\partial x^i}\big|_p\right) (f) = \left(\frac{\partial F^j}{\partial x^i} (p) \frac{\partial}{\partial y^j}\big|_{F(p)}\right) (f)$.}
 \end{exercise}
 
 \newthought{A particularly important consequence} of this theorem is that our definition coincides with our old euclidean notions if we set $M=\R^m$ and $N=\R^n$.
@@ -565,7 +565,7 @@ \section{The differential of a smooth map}\label{sec:diffsmooth}
 The construction that we employed forced us to fix a basis for the spaces, if this was truly necessary it would defeat the purpose of this whole chapter.
 Fortunately for us, the following exercise shows that, at any given point, the tangent space to a vector space is \emph{canonically}\footnote{That is, independently of the choice of basis.} identified with the vector space itself.
 
-\begin{exercise}\label{ex:tg_curve_iso}
+\begin{exercise}[\textit{[homework 1]}]\label{ex:tg_curve_iso}
   Let $V$ and $W$ be finite-dimensional vector spaces, endowed with their standard smooth structure (see Exercise~\ref{exe:subsetsmanifolds}).
   \begin{enumerate}
     \item Fix $a\in V$. For any vector $v\in V$ define a map $\cT_a(v) : C^\infty(V) \to \R$ by
@@ -590,15 +590,13 @@ \section{The differential of a smooth map}\label{sec:diffsmooth}
 For example, since $GL_n(\R)$ is an open submanifolds of the vector space $M_n(\R)$, we can identify its tangent space at each point $X\in GL_n(\R)$ with the full space of matrices $M_n(\R)$.
 
 \begin{exercise}[Tangent space of a product manifold]
-  Let $M_1, \ldots, M_k$ be smooth manifolds, and for each $j$ let $\pi_j:M_1\times\cdots\times M_k \to M_j$ be the projection onto the $M_j$ factor.
+  Let $M_1, \ldots, M_k$ be smooth manifolds (without boundary\footnote{The statement is true also if one (only one!) of the $M_i$ spaces is a smooth manifold with boundary. If there is more than one manifold with boundary, the product space will have ``corners'' that cannot be mapped to half spaces and thus is not a smooth manifold, as a simple example you can consider the closed square $[0,1]\times [0,1]$.}), and for each $j$ let $\pi_j:M_1\times\cdots\times M_k \to M_j$ be the projection onto the $M_j$ factor.
   For any point $p=(p_1,\ldots,p_k)\in M_1\times\cdots\times M_k$, the map
   \begin{align}
     \sigma &: T_p(M_1\times\cdots\times M_k) \to T_p M_1\times\cdots\times T_p M_k\\
     \sigma &: v \mapsto \left(d(\pi_1)_p(v), \ldots, d(\pi_k)_p(v)\right)
   \end{align}
   is an isomorphism.
-
-  The same is true if some of the $M_i$ spaces are smooth manifolds with boundary (optional).
 \end{exercise}
 
 \begin{remark}
@@ -714,15 +712,15 @@ \section{Tangent vectors as tangents to curves}
   Let $F:M\to N$ b a smooth map between smooth manifolds and $\gamma:I\to M$ a smooth curve in $M$.
   Then
   \begin{equation}
-    d F_{\varphi(t)} (\varphi'(t)) = (F\circ\gamma)'(t).
+    d F_{\gamma(t)} (\gamma'(t)) = (F\circ\gamma)'(t).
   \end{equation}
 \end{proposition}
 \begin{proof}
   We are going to use~\eqref{eq:tg_curve_diff} as definition of $\gamma'(t)$.
   Applying the chain rule we obtain:
   \begin{align}
-    d F_{\varphi(t)} (\varphi'(t))
-    &= d F_{\varphi(t)} \circ d\varphi_t \left(\frac{\partial}{\partial t}\Big|_t\right) \\
+    d F_{\gamma(t)} (\gamma'(t))
+    &= d F_{\gamma(t)} \circ d\gamma_t\left(\frac{\partial}{\partial t}\Big|_t\right) \\
     &= d (F\circ\gamma)_t \left(\frac{\partial}{\partial t}\Big|_t\right) \\
     &= (F\circ\gamma)'(t).
   \end{align}
@@ -1208,6 +1206,24 @@ \section{Submanifolds}
   \end{enumerate}
 \end{exercise}
 
+Of course, we can also define subbundles.
+\begin{definition}
+  Let $\pi:E \to M$ be a rank-$n$ vector bundle and $F\subset E$ a submanifold.
+  If for all $p\in M$, the intersection $F_p := F\cap E_p$ is a $k$-dimensional subspace of the vector space $E_p$ and $\pi|_F : F \to M$ defines a rank-$k$ vector bundle, then $\pi|F: F \to M$ is called a \emph{subbundle} of $E$.
+\end{definition}
+
+\begin{exercise}[\textit{[homework 1]}]
+  Let $M$ be a smooth $m$-manifold and $N$ a smooth $n$-manifold.
+  Let $F:M\to N$ be an embedding and denote $\widetilde M = F(M)\subset N$.
+  \begin{enumerate}[(a)]
+    \item Show that the tangent bundle of $M$ in $N$, given by $T\widetilde M := dF(TM) \subset TN\Big|_{\widetilde M}$, is a subbundle of $TN\Big|_{\widetilde M}$ by providing explicit local trivialisations in terms of the charts $(U, \varphi)$ for $M$.
+    \item Assume that there exist a smooth function $\Phi:N\to\R^{n-k}$ such that $\widetilde M := \{p\in N \mid \Phi(p) = 0\}$ and $d\Phi_p$ has full rank for all $p\in\widetilde M$. Prove that
+    \begin{equation}
+      T\widetilde{M} = \{(p,v)\in TN|_{\widetilde{M}} \mid v\in\ker(d\Phi_p)\}.
+    \end{equation}
+  \end{enumerate}
+\end{exercise}
+
 We still have a question pending since the beginning of the chapter.
 Is the tangent space to a sphere the one that we naively imagine (see Figure~\ref{fig:tan-embedded-sphere})?
 To finally answer the question, we will prove one last proposition.
@@ -1258,7 +1274,7 @@ \section{Submanifolds}
   Show that $g$ is a smooth embedding and, therefore, that $g(U)$ is a smooth embedded $n$-dimensional submanifold\footnote{$g(U)$ is the the \emph{graph} of $f$!} of $\R^{n+1}$.
 \end{exercise}
 
-\begin{exercise}\label{exe:onsubmanifold}
+\begin{exercise}[\textit{[homework 1]}]\label{exe:onsubmanifold}
   Show that the orthogonal matrices $O(n) := \{ Q\in GL(n) \mid Q^TQ=\id \}$ form a $n(n-1)/2$-dimensional submanifold of the $n^2$-manifold $\mathrm{Mat}(n)$ of $n\times n$-matrices.
 
   Show also that

3-vectorfields.tex

@@ -29,7 +29,7 @@ \section{Vector fields}
 \end{equation}
 This defines $n$ functions $X^i: U\to\R$, called \emph{component functions of $X$} in the chart.
 
-\begin{exercise}
+\begin{exercise}[\textit{[homework 1]}]
   Show that, in the notation above, the restriction of $X$ to $U$ is smooth if and only if its component functions with respect to the chart are smooth.
 \end{exercise}
 
@@ -261,7 +261,7 @@ \section{Lie brackets}
         [X,Y] = \left(X^i\frac{\partial Y^j}{\partial x^i} - Y^i\frac{\partial X^j}{y^i}\right)\frac{\partial}{\partial x^j}.
     \end{equation}
 \end{proposition}
-\begin{exercise}
+\begin{exercise}[\textit{[homework 1]}]
   Prove the proposition.
 \end{exercise}
 

5-tensors.tex

@@ -151,6 +151,7 @@ \section{Tensors}
   \end{equation}
 
   A \emph{metric tensor} or \emph{scalar product} is a non-degenerate pseudo-metric tensor.
+  The Riemannian metric is a metric tensor on the tangent bundle of a manifold.
   An example of non-degenerate tensor which is not a metric is the so-called \emph{symplectic form}: a skew-symmetric non-degenerate $(0,2)$-tensor which is central in classical mechanics and the study of Hamiltonian systems.
 \end{definition}
 

aom.tex

@@ -207,7 +207,7 @@
     \setlength{\parskip}{\baselineskip}
     Copyright \copyright\ \the\year\ \thanklessauthor
 
-    \par Version 0.4.4 -- \today
+    \par Version 0.5 -- \today
 
     \vfill
     \small{\doclicenseThis}