Move metrizability and add proof for the Riemannian case

Add better structure to the sections of some chapters

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

1-manifolds.tex

@@ -391,9 +391,6 @@ \section{Differentiable manifolds}
   Given two manifolds $(M_1, \cA_1)$ and $(M_2, \cA_2)$, we can define the \emph{product manifold} $M_1 \times M_2$ by equipping $M_1 \times M_2$ with the product topology\footnote{Open sets in the product are products of open sets from the respective topological spaces.} and covering the space with the atlas $\{ (U_1\times U_2, (\varphi_1, \varphi_2)) \;\mid\; (U_1, \varphi_1)\in\cA_1, (U_2, \varphi_2)\in \cA_2\}$.
 \end{example}
 
-Note that smooth manifolds do not yet have a metric structure: distances between the points are not defined.
-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.
 
 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.
@@ -799,7 +796,7 @@ \section{Smooth maps and differentiability}\label{sec:smoothfn}
   Note that this, in particular, holds for $N=\R^n$.
 \end{exercise}
 
-\section{Partitions of unity}
+\section{Partitions of unity}\label{sec:partition_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.
 Since their definition and construction does not require more than what we have just seen, let's talk about them now.
@@ -910,7 +907,7 @@ \section{Partitions of unity}
 
 \begin{proof}[Proof of Proposition~\ref{prop:cutoff}.]
   Consider the open cover of $M$ given by $\cC:=\{M\setminus K, U\}$.
-  Then Theorem~\ref{thm:partitionof1} implies that there exists a partition of unity $\{\rho_U, \rho_{M\setminus K}\}$ adapted to $\cC$. The function $\chi := \rho_U$ is our cutoff function.
+  Then Theorem~\ref{thm:partitionof1} implies that there exists a partition of unity $\{\rho_U, \rho_{M\setminus K}\}$ subordinate to $\cC$. The function $\chi := \rho_U$ is our cutoff function.
 \end{proof}
 
 \section{Manifolds with boundary}\label{sec:mbnd}

2c-vectorbdl.tex

@@ -1,5 +1,7 @@
 What we have seen in the previous chapter is our first example of vector bundle, which is just a way to call a vector space depending continuously (or smoothly) on some parameters, for example points on a manifold.
 
+\section{Vector bundles}
+
 \begin{definition}\label{def:vector_bundle}
   A \emph{vector bundle of rank $r$} on a manifold $M$ is a manifold $E$ together with a smooth surjective map $\pi : E \to M$ such that, for all $p\in M$, the following properties hold:
   \begin{enumerate}[(i)]

5-tensors.tex

@@ -149,9 +149,9 @@ \section{Tensors}
   \end{equation}
 
   A \emph{metric tensor} or \emph{scalar product} is a non-degenerate pseudo-metric tensor, that is a symmetric, positive definite $(0,2)$-tensor.
-  The Riemannian metric is a metric tensor on the tangent bundle of a manifold.
+  We will briefly see later that a Riemannian metric provides a metric tensor on the tangent spaces 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 fundamental in classical mechanics and the study of Hamiltonian systems.
+  An example of non-degenerate tensor which is not a metric is the so-called \emph{symplectic tensor}: a skew-symmetric non-degenerate $(0,2)$-tensor related to the symplectic form, a fundamental object in topology, classical mechanics and the study of Hamiltonian systems.
 \end{definition}
 
 \begin{example}\label{ex:musicaliso}
@@ -252,7 +252,7 @@ \section{Tensors}
   \end{align}
   % TODO: improve remark with coordinate representation
   % In coordinates the one maping (r,s)-tensors to (r+s, 0)-tensors is $I_g(\tau) = \tau_{i_1\ldots i_r}^{j_1\ldots j_s} g_{j_1 j_{r+1}} \ldots g_{j_s j_{r+s}}}$
-  In general, one can use the metric to raise or lower arbitrary indices, changing the tensor type from $(r,s)$ to $(r+1, s-1)$ or $(r-1, s+1)$.
+  In general, one can use the metric tensor to raise or lower arbitrary indices, changing the tensor type from $(r,s)$ to $(r+1, s-1)$ or $(r-1, s+1)$.
 
   A neat application of this is showing that a non-degenerate bilinear map $g\in T_2^0(V)$ can be lifted to a non-degenerate bilinear map on arbitrary tensors, that is
   \begin{equation}
@@ -261,7 +261,7 @@ \section{Tensors}
     G(\tau, \widetilde\tau) := (\cI_g(\tau), \widetilde\tau),
   \end{equation}
   where the scalar product is defined via the requirement that tensor products of basis elements of $V$ are orthinormal and that $G$ is invariant under the musical isomorphism.
-  In particular, if $g$ is a metric on $V$, then $G$ is a metric on $T_s^r(V)$.
+  In particular, if $g$ is a metric tensor on $V$, then $G$ is a metric tensor on $T_s^r(V)$.
 
 % TODO: improve remark with coordinate representation
 % If \((g_{ij})\) and \((g^{ij})\) are the matrix element of the matrices representing metric tensor and its inverse resp, then \[ G(\sigma, \tau) = \langle\sigma, \tau\rangle_g := g^{k_1 l_1}\cdots g^{k_rl_r} g_{i_1j_1} \cdots g_{i_sj_s} \sigma_{k_1,\ldots,k_r}^{i_1,\ldots,i_s} \tau_{l_1,\ldots,l_r}^{j_1,\ldots,j_s} \]
@@ -361,17 +361,6 @@ \section{Tensor bundles}
 \end{equation}
 where $\tau^{j_1\cdots j_r}_{i_1\cdots i_s}\in C^\infty(M)$.
 
-\begin{example}
-  A non-degenerate symmetric bilinear form $g\in \cT_2^0(M)$ is a \emph{pseudo-Riemannian metric} and the pair $(M,g)$ a \emph{pseudo-Riemannian manifold}\footnote{Also called \emph{semi-Riemannian manifold}.}.
-  If $g$ is also fibre-wise positive definite, then $g$ is a \emph{Riemannian metric} and $(M,g)$ is a \emph{Riemannian manifold}.
-  From this you see that the Riemannian metric is just an inner product on the tangent bundle of the manifold.
-
-  \begin{enumerate}
-    \item The euclidean space $\R^n$ is a Riemannian manifold with the usual scalar product, which we can represent as $g = \sum_{i=1}^n dx^i\otimes dx^i$ (What is its matrix form?).
-    \item If $M=\R^4$, an example of pseudo-Riemannian metric is the Minkowski metric $g = g_{ij} dx^i\otimes dx^j$ where $[g_{ij}] = {\left(\begin{smallmatrix} -1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{smallmatrix}\right)}$. The pseudo-Riemannian manifold $(M, g)$ is the space-time manifold of special relativity, with $x^1 = t$ is the time and $(x^2, x^3, x^4) = (x,y,z)$ is the space.
-  \end{enumerate}
-\end{example}
-
 \begin{definition}
   The \emph{support} of a tensor field $\tau\in\cT_s^r(M)$ is defined as the set
   \begin{equation}
@@ -534,3 +523,60 @@ \section{Tensor bundles}
 \end{equation}
 Can you show why?
 
+\newthought{If you work on Riemannian geometry or in Relativit}y, you cannot avoid hearing about tensors.
+Let's briefly look at the reason.
+
+\begin{definition}
+  A non-degenerate symmetric bilinear form $g\in \cT_2^0(M)$ is a \emph{pseudo-Riemannian metric} and the pair $(M,g)$ a \emph{pseudo-Riemannian} (or \emph{semi-Riemannian}) \emph{manifold}.
+  If $g$ is also fibre-wise positive definite, then $g$ is a \emph{Riemannian metric} and $(M,g)$ is a \emph{Riemannian manifold}.
+  From this you see that the Riemannian metric is just an inner product on the tangent bundle of the manifold.
+\end{definition}
+
+\begin{example}
+  \begin{enumerate}
+    \item The euclidean space $\R^n$ is a Riemannian manifold with the usual scalar product, which we can represent as $g = \sum_{i=1}^n dx^i\otimes dx^i$ (What is its matrix form?).
+    \item If $M=\R^4$, an example of pseudo-Riemannian metric is the Minkowski metric $g = g_{ij} dx^i\otimes dx^j$ where $[g_{ij}] = {\left(\begin{smallmatrix} -1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{smallmatrix}\right)}$. The pseudo-Riemannian manifold $(M, g)$ is the space-time manifold of special relativity, with $x^1 = t$ is the time and $(x^2, x^3, x^4) = (x,y,z)$ is the space.
+  \end{enumerate}
+\end{example}
+
+A metric on a manifold $M$ provides an inner product at each point of the tangent bundle: this allows to compute lengths of curves and angles between vectors. We can then induce a distance function $d: M \times M \to [0, \infty)$ by defining
+\begin{equation}
+  d(p,q) = \inf_{\gamma \in \cC(p,q)} \ell(\gamma),
+\end{equation}
+where
+\begin{equation}
+  \cC(p,q) = \left\{\gamma : [0,1] \to M \mbox{ piecewise smooth} \;\mid\; \gamma(0)=p, \gamma(1)=q\right\}
+\end{equation}
+and
+\begin{equation}
+  \ell(\gamma) := \int_0^1 g(\gamma'(t), \gamma'(t))\, dt.
+\end{equation}
+This will make $(M, d)$ also a metric space whose metric topology is the same as the original manifold topology.
+We will not go further into proving these claims or discussing their many interesting consequences.
+If you are interested or curious you can look at any good book in Riemannian geometry,
+a good and concise one is \cite[Chapter 6]{book:lee:riemannian}.
+
+The relation between manifolds and metric spaces does not end here. As it happens, all smooth manifolds are \emph{metrizable}\footnote{In fact, all the topological manifolds are metrizable. This more general statement is far 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 distance on the manifold that induces the given topology on it.
+
+We will show here the proof in the case of smooth manifolds since it is relatively simple consequence of the existence of partitions of unity.
+
+\begin{theorem}
+  Every smooth manifold admits a Riemannian metric.
+\end{theorem}
+\begin{proof}
+  Let $M$ be a smooth $m$-dimensional manifold, let $\{(U_i, \varphi_i)\}_{i\in I}$ be a countable atlas for the manifold and let $\{\rho_i\}_{i\in I}$ be a partition of unity adapted to it. See Section~\ref{sec:partition_of_unity}.
+  Denote $g_{\R^m}$ the Euclidean metric on $\R^n$, that is, for any $x \in \R^m$ and any $v,w \in T_x\R^m\simeq\R^m$, $g_{\R^m}(v, w) = v \cdot w$.
+
+  We define the metric\footnote{Exercise: check that it is actually a metric} $g$ on $M$ by setting
+  \begin{align}
+    g := \sum_{i\in I} \rho_i \hat{g}_i,
+    \quad\mbox{where}\quad
+    \hat{g}_i := \begin{cases}
+      \varphi_i^* g_{\R^m} & \mbox{on } U_i, \\
+      0 & \mbox{otherwise}
+    \end{cases},
+  \end{align}
+  concluding the proof.
+\end{proof}
+
+Once we have a Riemannian metric, we can define the distance as described above.

6b-cohomology.tex

@@ -1,3 +1,4 @@
+\section{De Rham cohomology}
 \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 $\eta$ on $M$ such that $\omega = d\eta$.
 
@@ -38,6 +39,7 @@
   If $F:M\to N$ is a smooth map, then $F^*$ induces a well-defined map $F^*:H_{\mathrm{dR}}^k(N) \to H_{\mathrm{dR}}^k(M)$ (denoted with the same symbol) via $[\omega]\mapsto[F^*\omega]$.
 \end{corollary}
 
+\section{Poincar\'e lemma}
 Without further ado, let's look at a first version of Poincar\'e lemma on manifolds.
 As for all the local concepts we have seen so far, the proof will reduce the problem to a euclidean statement to which we will apply the Poincar\'e lemma that you have seen in multivariable calculus.
 

aom.bib

@@ -111,11 +111,24 @@ @book{book:lee
   url       = {https://link.springer.com/book/10.1007/978-1-4419-9982-5}
 }
 
+@book{book:lee:riemannian,
+    doi       = {10.1007/978-3-319-91755-9},
+    url       = {https://link.springer.com/book/10.1007/978-3-319-91755-9},
+    year      = 2018,
+    publisher = {Springer, New York},
+    author    = {John M. Lee},
+    volume    = {176}, 
+    series    = {Graduate Texts in Mathematics},
+    title     = {Introduction to Riemannian Manifolds}
+}
+
 @book{book:lee:topology,
   doi       = {10.1007/978-1-4419-7940-7},
   year      = 2011,
-  publisher = {Springer New York},
+  publisher = {Springer, New York},
   author    = {John M. Lee},
+  volume    = {202}, 
+  series    = {Graduate Texts in Mathematics},
   title     = {Introduction to Topological Manifolds},
   url       = {https://link.springer.com/book/10.1007/978-1-4419-7940-7}
 }

aom.tex

@@ -213,7 +213,7 @@
   \setlength{\parskip}{\baselineskip}
   Copyright \copyright\ \the\year\ \thanklessauthor
 
-  \par Version 1.4.1 -- \today
+  \par Version 1.5 -- \today
 
   \vfill
   \small{\doclicenseThis}