Big update to cleanup some more todos and typos

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

1-manifolds.tex

@@ -285,9 +285,9 @@ \section{Differentiable manifolds}
   Why is this independent of the choice of the isomorphism $T$?
 
   This fact has a very interesting consequence.
-  The space $\mathrm{Mat}(n)$ of $n\times n$-matrices can be identified with $\R^{n^2}$ by writing the elements of the matrix as a $n^2$-vector.
-  This gives to $\mathrm{Mat}(n)$ a structure of differentiable manifold.
-  The subset of invertible matrices $GL(n) = \{ A \in \mathrm{Mat}(n) \;\mid\; \det A \neq 0\}$, widely known as the \emph{general linear group}, being an open subset of $\mathrm{Mat}(n)$ (why?) is itself a differentiable manifold.
+  The space $\mathrm{Mat}(n, \R)$ of real $n\times n$-matrices can be identified with $\R^{n^2}$ by writing the elements of the matrix as a $n^2$-vector.
+  This gives to $\mathrm{Mat}(n, \R)$ a structure of differentiable manifold.
+  The subset of invertible matrices $GL(n) := GL(n, \R) = \{ A \in \mathrm{Mat}(n, \R) \;\mid\; \det A \neq 0\}$, widely known as the \emph{general linear group}, being an open subset of $\mathrm{Mat}(n, \R)$ (why?) is itself a differentiable manifold.
   Is such manifold connected? Why?
 \end{exercise}
 
@@ -382,41 +382,6 @@ \section{Differentiable manifolds}
 However, they are \emph{metrizable}\footnote{In fact, all the topological manifolds are metrizable. This property is far more general and harder to prove~\cite[Theorem 34.1 and Exercise 1 of Chapter 4.36]{book:munkres:topology} or \cite{nlab:urysohn_metrization_theorem}. Note that not all topological spaces are metrizable, for example a space with more than one point endowed with the discrete topology is not. And even if a topological space is metrizable, the metric will be far from unique: for example, proportional metrics generate the same collection of open sets.}: there exists some metric on the manifold that induces the given topology on it.
 This allows to always view manifolds as metric spaces.
 
-\begin{example}[A different smooth structure on $\R$]
-  Consider the homeomorphism $\psi:\R\to\R$, $\psi(x) = x^3$.
-  The atlas consisting of the global chart $(\R, \psi)$ defines a smooth structure on $\R$.
-  This chart is not smoothly compatible with the standard smooth structure on $\R$ since $\id_\R \circ \psi^{-1} (y) = y^{1/3}$ is not smooth at $y=0$.
-  Therefore, the smooth structure defined on $\R$ by $\psi$ is different from the standard one.
-  You can adapt this idea to construct many different smooth structures on topological manifolds provided that they at least have one smooth structure.
-\end{example}
-
-\begin{exercise}
-  Show that there exists a diffeomorphism between the smooth structures $(\R, \id_\R)$ and $(\R, \psi)$ from the previous example.
-\end{exercise}
-
-\begin{exercise}[\textit{[homework 1]}]
-  For $r>0$, let $\phi_r:\R\to\R$ be the map given by
-  \begin{equation}
-    \phi_r(t) := \begin{cases}
-      t, & \mbox{if } t<0,\\
-      rt, & \mbox{if } t\geq0.
-    \end{cases}
-  \end{equation}
-  Let $\cA_r$ denote the maximal atlas on $\R$ containing the chart $(\R, \phi_r)$.
-  \begin{enumerate}
-    \item Show that the differentiable structures on $\R$ defined by $\cA_r$ and $\cA_s$, $0<r<s$, are different.
-    This shows that there are uncountably many families of different differential structures on $\R$.
-    \item Let $M_r$ be the manifold $\R$ equipped with the atlas $\cA_r$.
-    Show that $M_r$ and $M_s$ are diffeomorphic for $r,s >0$.
-  \end{enumerate}
-\end{exercise}
-
-\begin{remark}
-	There exist examples of topological manifolds without smooth structures.
-	It is also known that smooth manifolds of dimension $n < 4$ have exactly one smooth structure (up to diffeomorphisms) while ones of dimension $n > 4$ have finitely many\footnote{A beautiful example of this is the $7$-sphere $\bS^7$ which is known to have 28 non-diffeomorphic smooth structures.}.
-	The case $n = 4$ is unknown: if you prove that there is only one smooth structure, you will have shown the smooth Poincar\'e conjecture.
-\end{remark}
-
 Instead of always constructing a topological manifold and then specify a smooth structure, it is often convenient to combine these steps into a single construction.
 This is especially useful when the initial set is not equipped with a topology.
 In this respect, the following lemma provides a welcome shortcut: in brief it says that given a set with suitable ``charts'' that overlap smoothly, we can use those to define both a topology and a smooth structure on the set.
@@ -722,6 +687,41 @@ \section{Smooth maps and differentiability}
 From now on, when we write manifold, chart, atlas, etc. we always mean smooth manifold, smooth chart, smooth atlas, etc..
 \end{tcolorbox}
 
+\begin{example}[A different smooth structure on $\R$]
+  Consider the homeomorphism $\psi:\R\to\R$, $\psi(x) = x^3$.
+  The atlas consisting of the global chart $(\R, \psi)$ defines a smooth structure on $\R$.
+  This chart is not smoothly compatible with the standard smooth structure on $\R$ since $\id_\R \circ \psi^{-1} (y) = y^{1/3}$ is not smooth at $y=0$.
+  Therefore, the smooth structure defined on $\R$ by $\psi$ is different from the standard one.
+  You can adapt this idea to construct many different smooth structures on topological manifolds provided that they at least have one smooth structure.
+\end{example}
+
+\begin{exercise}
+  Show that there exists a diffeomorphism between the smooth structures $(\R, \id_\R)$ and $(\R, \psi)$ from the previous example.
+\end{exercise}
+
+\begin{exercise}[\textit{[homework 1]}]
+  For $r>0$, let $\phi_r:\R\to\R$ be the map given by
+  \begin{equation}
+    \phi_r(t) := \begin{cases}
+      t, & \mbox{if } t<0,\\
+      rt, & \mbox{if } t\geq0.
+    \end{cases}
+  \end{equation}
+  Let $\cA_r$ denote the maximal atlas on $\R$ containing the chart $(\R, \phi_r)$.
+  \begin{enumerate}
+    \item Show that the differentiable structures on $\R$ defined by $\cA_r$ and $\cA_s$, $0<r<s$, are different.
+    This shows that there are uncountably many families of different differential structures on $\R$.
+    \item Let $M_r$ be the manifold $\R$ equipped with the atlas $\cA_r$.
+    Show that $M_r$ and $M_s$ are diffeomorphic for $r,s >0$.
+  \end{enumerate}
+\end{exercise}
+
+\begin{remark}
+	There exist examples of topological manifolds without smooth structures.
+	It is also known that smooth manifolds of dimension $n < 4$ have exactly one smooth structure (up to diffeomorphisms) while ones of dimension $n > 4$ have finitely many\footnote{A beautiful example of this is the $7$-sphere $\bS^7$ which is known to have 28 non-diffeomorphic smooth structures.}.
+	The case $n = 4$ is unknown: if you prove that there is only one smooth structure, you will have shown the smooth Poincar\'e conjecture.
+\end{remark}
+
 \section{Partitions of unity}
 
 \newthought{Cutoff functions} are a class of smooth functions that will be of crucial importance throughout the course, and whose existence cannot be given for granted.

2-tangentbdl.tex

@@ -596,7 +596,7 @@ \section{The differential of a smooth map}\label{sec:diffsmooth}
 
 An important consequence of what we have seen so far is that we can routinely \emph{identify} tangent vectors to a finite-dimensional vector space with elements of the space itself.
 More generally, if $M$ is an open submanifold of a vector space $V$, we can combine the identifications $T_p M \simeq T_p V \simeq V$ to obtain a canonical identification of each tangent space to $M$ with $V$.
-For example, since $GL_n(\R)$ is an open submanifold of the vector space $\mathrm{Mat}(n)$, we can identify its tangent space at each point $X\in GL_n(\R)$ with the full space of matrices $\mathrm{Mat}(n)$.
+For example, since $GL_n(\R)$ is an open submanifold of the vector space $\mathrm{Mat}(n, \R)$, we can identify its tangent space at each point $X\in GL_n(\R)$ with the full space of matrices $\mathrm{Mat}(n, \R)$.
 
 \begin{exercise}[Tangent space of a product manifold]
   Let $M_1, \ldots, M_k$ be smooth manifolds (without boundary\sidenote[][-8em]{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.
@@ -1218,24 +1218,6 @@ \section{Submanifolds}
   \end{enumerate}
 \end{exercise}
 
-\newthought{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 2]}]
-  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}
-    \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 $\Phi:N\to\R^{n-m}$ 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}
-
 \newthought{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.
@@ -1287,16 +1269,38 @@ \section{Submanifolds}
 \end{exercise}
 
 \begin{exercise}[\textit{[homework 2]}]\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 that the orthogonal matrices
+  \begin{equation}
+    O(n) := O(n, \R) = \{ Q\in \mathrm{Mat}(n, \R) \mid Q^TQ=\id \}
+  \end{equation}
+  form a $n(n-1)/2$-dimensional submanifold of the $n^2$-manifold $\mathrm{Mat}(n, \R)$ of $n\times n$-matrices.
 
   Show also that
   \begin{equation}
-    T_Q O(n) = \left\lbrace B \in \mathrm{Mat}(n) \mid (Q^{-1} B)^T = -Q^{-1}B \right\rbrace,
+    T_Q O(n) = \left\lbrace B \in \mathrm{Mat}(n, \R) \mid (Q^{-1} B)^T = -Q^{-1}B \right\rbrace,
   \end{equation}
   and, thus, that $T_{\id} O(n)$ is the space of skew-symmetric matrices
   \begin{equation}
-    T_{\id} O(n) = \left\{ B \in \mathrm{Mat}(n) \mid B^T = -B \right\}.
+    T_{\id} O(n) = \left\{ B \in \mathrm{Mat}(n, \R) \mid B^T = -B \right\}.
   \end{equation}
-  \textit{\small Hint: Find a suitable map $F: \mathrm{Mat}(n) \to \mathrm{Sym}(n)$ such that $F^{-1}(\{p\}) = O(n)$ for some point $p$ in the image, e.g. $0$ or $\id_n$.
+  \textit{\small Hint: Find a suitable map $F: \mathrm{Mat}(n, \R) \to \mathrm{Sym}(n)$ such that $F^{-1}(\{p\}) = O(n)$ for some point $p$ in the image, e.g. $0$ or $\id_n$.
   Here $\mathrm{Sym}(n)$ denotes the space of symmetric matrices.}
 \end{exercise}
+
+\newthought{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 2]}]
+  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}
+    \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 $\Phi:N\to\R^{n-m}$ 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}

3-vectorfields.tex

@@ -344,7 +344,7 @@ \section{Lie brackets}
   This may seem an alien concept at first, however there are many simple examples of Lie algebras. To name a few:
   \begin{enumerate}
     \item $\R^3$ with the cross product $[x,y]:=x\times y$ is a $3$-dimensional Lie algebra;
-    \item the set $\mathrm{Mat}(n)$ of $n\times n$-matrices with the matrix commutator $[A,B] = AB-BA$ is a $n^2$-dimensional Lie algebra, usually denoted $\mathfrak{gl}(n, \R)$;
+    \item the set $\mathrm{Mat}(n, \R)$ of $n\times n$-matrices with the matrix commutator $[A,B] = AB-BA$ is a $n^2$-dimensional Lie algebra, usually denoted $\mathfrak{gl}(n, \R)$;
     \item any vector space $V$ turns into an (abelian) Lie algebra by defining $[v,w]=0$;
     \item if $V$ is a vector space, the vector space of all linear maps from $V$ to itself becomes a Lie algebra, denoted $\mathfrak{gl}(V)$, with the brackets defined by $[A,B] = A\circ B-B\circ A$. Note that with the usual identification of $n\times n$ matrices with linear maps from $\R^n$ to itself, $\mathfrak{gl}(\R^n)$ coincides with $\mathfrak{gl}(n, \R)$.
   \end{enumerate}

3b-liegroups.tex

@@ -133,6 +133,27 @@ \section{Lie groups}
   You have proven this when you solved Exercise~\ref{exe:onsubmanifold}.
 \end{example}
 
+\begin{exercise}\label{ex:SL2LGA}
+  For this exercise is useful to remember that we can identify the space $\mathrm{Mat}(2,\R)$ of $2\times 2$-matrices with $\R^4$ by associating the matrix $X = \begin{pmatrix}x_{11} & x_{12}\\ x_{21} & x_{22}\end{pmatrix}$ with the point $(x_{11}, x_{12}, x_{21}, x_{22})\in\R^4$.
+
+  \begin{enumerate}
+    \item Show that the set
+    \begin{equation*}
+      \mathrm{SL}(2) := \mathrm{SL}(2,\R) = \{ A\in \mathrm{Mat}(2,\R) \;\mid\; \det A = 1 \}
+    \end{equation*}
+    is a 3-dimensional smooth submanifold of $\mathrm{Mat}(2,\R)$.
+    \item Let $e\in\mathrm{Mat}(2,\R)$ denote the identity matrix. Show that
+    \begin{equation*}
+      T_e \mathrm{SL}(2) = \{ A\in \mathrm{Mat}(2,\R) \;\mid\; \mathrm{tr}\, A = 0 \},
+    \end{equation*}
+    where $\mathrm{tr} A$ denotes the matrix trace, i.e., the sum of the diagonal entries of $A$.
+    \item Let $\iota: \mathrm{SL}(2)\to \mathrm{SL}(2)$ be the map $\iota(A) = A^{-1}$. Show that $\iota$ is smooth.
+    \item Show that $d\iota_e: T_e\mathrm{SL}(2)\to T_e\mathrm{SL}(2)$ is given by $d\iota_e(A) = -A$.
+  \end{enumerate}
+\end{exercise}
+
+In fact, some parts of the Exercise~\ref{ex:SL2LGA} above are instances of a them more general statement of Exercise~\ref{ex:DiffGroupMaps}.
+
 \begin{exercise}\label{ex:DiffGroupMaps}
   Let $G$ be a Lie group.
   \begin{enumerate}
@@ -163,13 +184,14 @@ \section{Lie algebras}
 
 \begin{example}
   \begin{enumerate}
-    \item The Lie algebra of $GL(n)$ is $\mathfrak{gl}(n)\simeq \mathrm{Mat}(n)$.
+    \item The Lie algebra of $GL(n)$ is $\mathfrak{gl}(n)\simeq \mathrm{Mat}(n, \R)$.
     \item The Lie algebra of $O(n)$ is $\mathfrak{o}(n) = \{A \in \mathfrak{gl}(n) \mid A+A^T = 0\}$. You have shown it in Exercise~\ref{exe:onsubmanifold}.
+    \item Can you guess what is the Lie algebra of $SL(2)$ from Exercise~\ref{ex:SL2LGA}? 
   \end{enumerate}
 \end{example}
 
 \begin{exercise}
-  The Lie algebra of $\bT^n$ is $\R^n$.\\
+  Show that the Lie algebra of $\bT^n$ is $\R^n$.\\
   \textit{\small Hint: using the fact that $T(M\times N) \simeq T(M)\times T(N)$ and look at what happens in the case $n=1$.}
 \end{exercise}
 
@@ -327,7 +349,7 @@ \section{Lie algebras}
   Use the inclusion $i:H\hookrightarrow G$ as the homomorphism, then $di_e:\fh = T_eH\to \fg = T_eG$ is the Lie algebra homomorphism.
 \end{proof}
 
-If we go back to the example of $GL(n)$, now we have two possibly different Lie brackets on $\mathfrak{gl}(n)=\mathrm{Mat}(n)$: the one coming from the previous corollary and the matrix commutator.
+If we go back to the example of $GL(n)$, now we have two possibly different Lie brackets on $\mathfrak{gl}(n)=\mathrm{Mat}(n, \R)$: the one coming from the previous corollary and the matrix commutator.
 The next result, which we will not prove, shows that they coincide.
 
 \marginnote{See~\cite[Proposition 8.41]{book:lee} for reference.}

4-cotangentbdl.tex

@@ -245,11 +245,11 @@ \section{One-forms and the cotangent bundle}
 The cotangent bundle is a vector bundle of rank $n$ with projection $\pi:T^*M\to M$, $(p,\omega)\mapsto p$.
 The cotangent spaces are the fibres of the cotangent bundle. 
 
-\begin{theorem}
+\begin{theorem}\label{thm:starmbld}
   Let $M$ be a smooth $n$-manifold.
-  The smooth structure on $M$ naturally induces a smooth structure on $T^*M$, making $T^*M$ into a smooth manifold of dimension $2n$ for which all coordinate covector fields are smooth local sections.
+  The smooth structure on $M$ naturally induces a smooth structure on $T^*M$, making $T^*M$ into a smooth manifold of dimension $2n$.
 \end{theorem}
-\begin{proof}
+%\begin{proof}
   % The proof is analogous to that of Theorem~\ref{thm:tgbdlsmoothmfld}, so we will not do it again.
   % The atlas is obtained by an atlas $\{(U_i, \varphi_i)\}$ of $M$ by defining the new atlas
   % \begin{equation}
@@ -260,21 +260,25 @@ \section{One-forms and the cotangent bundle}
   %   \left(\varphi_i, (\varphi_i^{-1})^*\right) : T^* U_i &\to T^*\varphi(U_i),\\
   %   (p, \omega) &\mapsto \left(\varphi_i(p), d(\varphi_i^{-1})^*)_p\omega \right).
   % \end{align}
-  \begin{exercise}\textit{[homework 3]}
-    Mimicking what we did for Theorem~\ref{thm:tgbdlsmoothmfld}, complete this proof.
-  \end{exercise}
-\end{proof}
+\begin{exercise}\label{exe:prooftstarmbld}\textit{[homework 3]}
+  Mimicking what we did for Theorem~\ref{thm:tgbdlsmoothmfld}, complete the proof of Theorem~\ref{thm:starmbld}.
+\end{exercise}
+%\end{proof}
 
-\begin{definition}
+\begin{definition}\label{def:covfield}
   A \emph{covector field} or a \emph{(differential) $1$-form} on $M$ is a smooth section of $T^*M$.
   That is, a $1$-form $\omega\in\Gamma(T^*M)$ is a smooth map $\omega: p \to \omega_p \in T_p^*M$ that assigns to each point $p\in M$ a cotangent vector at $p$.
   We denote the space of all smooth covector fields on $M$ by $\fX^*(M)$.
   \marginnote[-1em]{For reasons related to tensor fields that we will understand soon, this is sometimes denoted $\cT_1^0(M)$.}
 
-  As for vector fields, we can define \emph{$C^p$-covector fields} as the $C^p$-maps $\omega:M\to T^*M$ such that $\pi\circ\omega = \id_M$.
+  Of course, one can also define \emph{$C^p$-covector fields} as the $C^p$-maps $\omega:M\to T^*M$ such that $\pi\circ\omega = \id_M$.
 \end{definition}
 
-Also in this case, we will often identify for a covector field $\omega\in\fX^*(M)$ its value $\omega(p) = \omega_p \in \{p\}\times T^*_p M$ at $p\in M$ with its part in $T_p^*M$ without necessarily making this explicit in the notation by projecting on the second factor.
+\begin{remark}
+  In Exercise~\ref{exe:prooftstarmbld} you have shown that coordinate covector fields are smooth local sections for the cotangent bundle.
+\end{remark}
+
+In analogy with the previous chapters, for covector fields $\omega\in\fX^*(M)$ we will often make the identification of its value $\omega(p) = \omega_p \in \{p\}\times T^*_p M$ at $p\in M$ with its part in $T_p^*M$ without necessarily making this explicit in the notation by projecting on the second factor.
 
 \begin{example}
   Let $f\in C^\infty(M)$, then the map