From 30a7c6eb6dae961392db9e78a994904754091729 Mon Sep 17 00:00:00 2001 From: Marcello Seri Date: Mon, 9 Jan 2023 15:57:35 +0100 Subject: [PATCH] Fix a multitude of typos Signed-off-by: Marcello Seri --- 2-tangentbdl.tex | 6 +++--- 2b-submanifolds.tex | 6 +++--- 2c-vectorbdl.tex | 4 ++-- 3-vectorfields.tex | 6 +++--- 4-cotangentbdl.tex | 6 +++--- 5-tensors.tex | 6 +++--- 6 files changed, 17 insertions(+), 17 deletions(-) diff --git a/2-tangentbdl.tex b/2-tangentbdl.tex index 497b363..dd9fc80 100644 --- a/2-tangentbdl.tex +++ b/2-tangentbdl.tex @@ -698,7 +698,7 @@ \section{Tangent vectors as tangents to curves} \end{figure*} But how can this mapping between curves and tangent vector be well--defined? -Surely, there must be multiple curves with the same speed at a point which differ outside a neighbourhood of the point. +Surely, there must be multiple curves with the same velocity at a point which differ outside a neighbourhood of the point. \begin{lemma}\label{lem:equiv_tg_curves} Let $M$ be a smooth manifold and $\gamma, \delta : (-\epsilon, \epsilon) \to M$ two smooth curves with $\gamma(0) = \delta(0)$. Then, $\gamma'(0) = \delta'(0)$ as elements of $T_{\gamma(0)}M$ if and only if for some (and thus any) chart $\varphi:U\to\varphi(U)$, $\gamma(0)\in U$, we have $(\varphi\circ \gamma)'(0) = (\varphi\circ\delta)'(0)$. @@ -707,8 +707,8 @@ \section{Tangent vectors as tangents to curves} Let $(x^i)$ denote the coordinates of $\varphi$. The condition $(\varphi\circ \gamma)'(0) = (\varphi\circ\delta)'(0)$ is equivalent as stating that $(\gamma^i)'(0) = (\delta^i)'(0)$, where $\gamma^i = x^i\circ\gamma$ and $\delta^i=x^i\circ\delta$. Then, the claim follows from~\eqref{eq:tg_curve_vec} and the fact that $\left\{\frac{\partial}{\partial x^i}\big|_{\gamma(0)}\right\}$ is a basis of $T_{\gamma(0)}M$. \end{proof} -This seems to follow a pattern: until now, all the definitions of tangent vectors where in terms of classes of equivalence. -And it would seem reasonable to identify curves that that have the same tangent vector at $0$. +This seems to follow a pattern: until now, all the definitions of tangent vectors were in terms of classes of equivalence. +And it would seem reasonable to identify curves that have the same tangent vector at $0$. There is still a potential problem, though: we don't yet know if \emph{every} tangent vector can be written as the velocity vector of a curve. \begin{theorem} diff --git a/2b-submanifolds.tex b/2b-submanifolds.tex index fda33ba..1528f61 100644 --- a/2b-submanifolds.tex +++ b/2b-submanifolds.tex @@ -2,7 +2,7 @@ We have already seen an example at the begininng of the course. In Exercise~\ref{exe:subsetsmanifolds}, we proved that any open subset $U\subseteq M$ can be made into a smooth manifold with a differentiable structure induced by the one of $M$. These, somehow trivial, submanifolds are called \emph{open submanifolds}. -Bu there are many other examples beyond these ones. +But there are many other examples beyond these ones. In fact, you may have already seen them in multivariable analysis. In this chapter we will briefly explore what subsets of manifolds are still manifolds of their own rights, how their topologies are related and how their smooth structures are related. @@ -57,7 +57,7 @@ \end{pmatrix} \end{equation} has full rank (equal to $n$) and is therefore surjective. - Hence, $i$ is a submersion. + Hence, $\pi$ is a submersion. \item Let $m=1$, $n > 1$ and $\gamma:\R\to\R^n$ a smooth curve. The map $\gamma$ is an immersion if and only if its velocity vector satisfies $\gamma'(t)\neq0$ for all $t\in\R$. @@ -143,7 +143,7 @@ \end{enumerate} \end{proposition} -If $F$ is an embedding, and this $M$ is an embedded submanifold, then the set $F(U)$ above can be written as $F(U) = F(M)\cap W$ for some open set $W\subset N$. +If $F$ is an embedding, and thus $F(M)\subset N$ is an embedded submanifold, then the set $F(U)$ above can be written as $F(U) = F(M)\cap W$ for some open set $W\subset N$. By replacing $V$ in~\eqref{eq:slice_chart} with $V\cap W$, one gets \begin{equation} F(M)\cap V = \left\{ q \in V \;\mid\; y^{m+1}(q)=\cdots=y^n(q)=0\right\}. diff --git a/2c-vectorbdl.tex b/2c-vectorbdl.tex index f7e60d2..6d8d211 100644 --- a/2c-vectorbdl.tex +++ b/2c-vectorbdl.tex @@ -51,7 +51,7 @@ To compare vector bundles it is useful to define the following concept. \begin{definition} - An \emph{isomorphism} between two vector bundles $\pi_i: E_i \to M$, $i=1,2$, over the same base space $M$ is a diffeomorphism $h:E_1 \to E_2$ which maps every fiber $\pi_1^{-1}(p)$ to the corresponding fiber $\pi_2^{-1}(p)$ by a linear isomorphism. + An \emph{isomorphism} between two vector bundles $\pi_i: E_i \to M$, $i=1,2$, over the same base space $M$ is a homeomorphism $h:E_1 \to E_2$ which maps every fiber $\pi_1^{-1}(p)$ to the corresponding fiber $\pi_2^{-1}(p)$ by a linear isomorphism. \end{definition} Since an isomorphism preserves all the structure of a vector bundle, isomorphic bundles are often regarded as the same. @@ -85,7 +85,7 @@ \label{fig:vectorfield-rn} \end{marginfigure} -If the bundles are differentiable manifolds, then the definition of isomorphism nicely generalizes: they are \emph{diffeomorphic} if fibres are mapped to fibres diffeomorphically. +If the bundles are differentiable manifolds, then the definition of isomorphism nicely generalizes: they are \emph{diffeomorphic} if $h$ is a diffeomorphism such that fibres are mapped to fibres isomorphically. Even though, as we have seen, \emph{locally} $TM$ is diffeomorphic to $M\times\R^n$, this is not true in general with one exception. \begin{exercise}\label{ex:trivialisable} diff --git a/3-vectorfields.tex b/3-vectorfields.tex index 9b06121..a497172 100644 --- a/3-vectorfields.tex +++ b/3-vectorfields.tex @@ -1,5 +1,5 @@ We continue with our quest of generalizing multivariable calculus. -The next familiar object waiting to be questioned are vector fields. +The next familiar objects waiting to be questioned are vector fields. In the euclidean settings these are simply continuous maps that attach a vector to each point in their domain. \section{Vector fields} @@ -31,7 +31,7 @@ \section{Vector fields} \begin{example} If $(U, (x^i))$ is a smooth coordinate chart for a $n$-manifold $M$, the assignment $p \mapsto \frac{\partial}{\partial x^i}\big|_p$ determines a vector field on $U$, called the \emph{$i$th-coordinate vector field} and commonly denoted $\partial_{i}$, $\partial_{x^i}$ or $\frac{\partial}{\partial x^i}$. - The set $\{\frac{\partial}{\partial x^1}, \ldots, \frac{\partial}{\partial x^n}\}$ is then a local frame for $TM$. + The set $\{\frac{\partial}{\partial x^1}, \ldots, \frac{\partial}{\partial x^n}\}$ of vector fields that are pointwise a basis for the corresponding tangent spaces is called a local frame for $TM$. \end{example} The space of smooth vector fields is a vector space under pointwise addition and scalar multiplication: for all $p\in M$, $X,Y\in\fX(M)$, $\alpha, \beta \in \R$, we have @@ -51,7 +51,7 @@ \section{Vector fields} Let $M$ be a smooth manifold with or without boundary. \begin{enumerate} \item If $X,Y \in \fX(M)$ and $f,g\in C^\infty(M)$, then $fX+gY \in\fX(M)$. - \item $\fX(M)$ is a module over the ring $C^\infty(M)$. + \item $\fX(M)$ is a module\footnote{A module over a unital ring $R$ is a generalization of the concept of vector space where the field of scalars is replaced by the ring $R$.} over the ring $C^\infty(M)$. \end{enumerate} \end{proposition} diff --git a/4-cotangentbdl.tex b/4-cotangentbdl.tex index 3160cf5..3c9bdbc 100644 --- a/4-cotangentbdl.tex +++ b/4-cotangentbdl.tex @@ -7,9 +7,9 @@ \section{The cotangent space} Its \emph{dual space} $V^* := \cL(V, \R)$ is the $n$-dimensional real vector space of linear maps $\omega:V \to R$. The elements of $V^*$ are usually called \emph{linear functionals} and for $\omega\in V^*$ and $v\in V$ it is common to write \begin{equation} - \omega(v) =: (\omega, v) =: (\omega \mid v), + \omega(v) =: (\omega, v) =: (\omega \mid v). \end{equation} - even if the \emph{dual pairing} $(\omega \mid v)$ is \emph{not} a scalar product. + The notation can be deceiving: the \emph{dual pairing} $(\omega \mid v)$ is \emph{not} a scalar product. \end{definition} \begin{remark}\label{rmk:identification} @@ -507,7 +507,7 @@ \section{Line integrals} dE = \frac{\partial E}{\partial S}dS + \frac{\partial E}{\partial V} dV =: T dS - p dV. \end{equation} Here $T$ and $p$ are the two functions denoting respectively the temperature of the system and its pressure. - The $1$-form $TdS$ os called the \emph{heat} absorbed by the system while $-p dV$ os the \emph{work} performed by the system. + The $1$-form $TdS$ is called the \emph{heat} absorbed by the system while $-p dV$ is the \emph{work} performed by the system. Differently from the other properties of the system, these are not functions and thanks to this it makes sense to ask how much heat has been transferred or how much work has been performed: these are just the integrals of those one-forms over curves in the space of equilibrium states. diff --git a/5-tensors.tex b/5-tensors.tex index 111d69c..cea0c13 100644 --- a/5-tensors.tex +++ b/5-tensors.tex @@ -93,7 +93,7 @@ \section{Tensors} \begin{proposition}\label{prop:tensorbasis} Let $V$ be an $n$-dimensional vector space. Let $\{e_j\}$ and $\{\varepsilon^i\}$ respectively denote the bases of $V=T_0^1(V)$ and $V^*=T_1^0(V)$ respectively. - Then, every $\tau\in V_s^r$ can be uniquely written as the linear combination\marginnote{\textit{Exercise}: expand Einstein's notation to write the full sum on the left with the relevant indices.}% In the expression, all indices $j_1,\ldots,j_r$, $i_1,\ldots,i_s$ run from $1$ to $n$.} + Then, every $\tau\in T_s^r(V)$ can be uniquely written as the linear combination\marginnote{\textit{Exercise}: expand Einstein's notation to write the full sum on the left with the relevant indices.}% In the expression, all indices $j_1,\ldots,j_r$, $i_1,\ldots,i_s$ run from $1$ to $n$.} \begin{equation}\label{eq:tensor:decomposition} \tau = \tau^{j_1\cdots j_r}_{i_1\cdots i_s} \, e_{j_1}\otimes\cdots\otimes e_{j_r}\otimes \varepsilon^{i_1}\otimes \cdots\otimes \varepsilon^{i_s}, \end{equation} @@ -183,7 +183,7 @@ \section{Tensors} Note that, in general, $e^\flat_i\neq e^i$: indeed, by definition $e^\flat_i = g_{ij}e^j$. It turns out that these operators can be applied to tensors to produce new tensors. - For example, if $\tau$ is a $(0,2)$-tensor we can define an associated tensor $\tau'$ of type $(1,1)$ by $\tau(\omega, v) = \tau(\omega^\sharp, v)$. + For example, if $\tau$ is a $(0,2)$-tensor we can define an associated tensor $\tau'$ of type $(1,1)$ by $\tau'(\omega, v) = \tau(\omega^\sharp, v)$. Its components are $(\tau')_i^j = g^{jk}\tau_{ik}$. \end{example} @@ -211,7 +211,7 @@ \section{Tensors} It is convenient, at this point, to introduce the \emph{dual map} $L^* : V^* \to V^*$ to $L$. This is defined as the map such that for all $v\in V$ and for all $w\in V^*$ \begin{equation} - (w | Lv) = (L^* w | V). + (w | Lv) = (L^* w | v). \end{equation} But we already know how to define such kind of map! This is like a pullback: $L^*(w) := w \circ L$ for $w\in V^*$. \emph{As matrices}, this means that $w^T L v = (L^T w)^T v$, so the matrix of $L^*$ is the transpose of the matrix of $L$.