├── ib-geometry ├── geometry.pdf ├── geo-chap-a1.tex ├── geo-chap-03.tex ├── geo-chap-07.tex ├── geo-chap-05.tex ├── geo-chap-04.tex └── geo-chap-06.tex ├── notebooks ├── tripos-dotted.pdf └── tripos-dotted.tex ├── ib-linear-algebra ├── linear-algebra.pdf ├── linear-algebra.tex └── linalg-chap-04.tex ├── ib-met-top-spaces ├── met-top-spaces.pdf ├── met-top-spaces.tex ├── mettop-chap-03.tex ├── mettop-chap-04.tex ├── mettop-chap-02.tex └── mettop-chap-01.tex └── README.md /ib-geometry/geometry.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/alexwlchan/maths-courses/master/ib-geometry/geometry.pdf -------------------------------------------------------------------------------- /notebooks/tripos-dotted.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/alexwlchan/maths-courses/master/notebooks/tripos-dotted.pdf -------------------------------------------------------------------------------- /ib-linear-algebra/linear-algebra.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/alexwlchan/maths-courses/master/ib-linear-algebra/linear-algebra.pdf -------------------------------------------------------------------------------- /ib-met-top-spaces/met-top-spaces.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/alexwlchan/maths-courses/master/ib-met-top-spaces/met-top-spaces.pdf -------------------------------------------------------------------------------- /notebooks/tripos-dotted.tex: -------------------------------------------------------------------------------- 1 | \documentclass[a4paper]{article} 2 | 3 | \usepackage[margin=2cm]{geometry} 4 | 5 | \usepackage{tikz} 6 | 7 | \definecolor{dotblue}{RGB}{189, 204, 224} 8 | 9 | \tikzset{ 10 | mark coordinate/.style={ 11 | inner sep=0pt, 12 | outer sep=0pt, 13 | minimum size=2pt, 14 | fill=dotblue, 15 | circle} 16 | } 17 | 18 | \begin{document} 19 | 20 | \pagestyle{empty} 21 | 22 | \begin{center} 23 | \begin{tikzpicture} 24 | 25 | \draw (0,0) rectangle (16.2,-0.9); 26 | \foreach \s in {1.8,14.4,12.6,10.8} {\draw (\s,0) -- (\s,-0.9);} 27 | 28 | \foreach \s in {0,...,18} { 29 | \foreach \t in {2,...,28} 30 | { 31 | \coordinate [mark coordinate] (\s \t) at (\s*0.9,-\t*0.9); 32 | } 33 | } 34 | 35 | \end{tikzpicture} 36 | \end{center} 37 | 38 | \pagebreak 39 | 40 | \begin{center} 41 | \begin{tikzpicture} 42 | 43 | \foreach \s in {0,...,18} { 44 | \foreach \t in {0,...,28} 45 | { 46 | \coordinate [mark coordinate] (\s \t) at (\s*0.9,-\t*0.9); 47 | } 48 | } 49 | 50 | \end{tikzpicture} 51 | \end{center} 52 | 53 | \end{document} -------------------------------------------------------------------------------- /README.md: -------------------------------------------------------------------------------- 1 | maths-courses 2 | ============= 3 | 4 | A collection of notes for courses in the Cambridge Maths Tripos, typeset by me and released with permission from lecturers. Intended for students to use; please do not use them for teaching purposes without the explicit permission of both myself and the associated lecturer. 5 | 6 | Currently available: 7 | 8 | * IB *Geometry*, lectured in Lent 2013 by Prof. Rasmussen. 9 | * IB *Linear Algebra*, lectured in Michaelmas 2012 by Prof. Grojnowski. 10 | * IB *Metric and Topological Spaces*, lectured in Easter 2012 by Prof. Wilson. 11 | * Some custom notebook styles that I use for maths. 12 | 13 | LaTeX files require XeLaTeX and style files `drangreport` and `drangsymbols`. In addition, certain courses require `awlc-algebra` and `awlc-statistics` (which provide subject-specific shortcuts and commands). This can be downloaded [from my `drangreport` GitHub repo][1]. 14 | 15 | I use a lot of packages in my work, so you may need to install new packages before the LaTeX files will compile. I explicitly *do not provide* assistance in getting these files to compile; they are provided "as is". I am, however, willing to discuss specific sections of the code if you have a particular question about how something works. 16 | 17 | Please send comments and corrections to [awlc2@cam.ac.uk][3]. 18 | 19 | This work is licensed under a [Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License][2]. 20 | 21 | These resources are *not* endorsed by the University of Cambridge. 22 | 23 | [1]: https://github.com/alexwlchan/drangreport 24 | [2]: http://creativecommons.org/licenses/by-nc-sa/3.0/ 25 | [3]: mailto:awlc2@cam.ac.uk -------------------------------------------------------------------------------- /ib-linear-algebra/linear-algebra.tex: -------------------------------------------------------------------------------- 1 | \documentclass[twoside]{scrartcl} 2 | 3 | \usepackage[blacklinks]{drangreport} 4 | % \usepackage[caecilia]{drangreport} 5 | 6 | \usepackage{stmaryrd} 7 | 8 | \usepackage{awlc-algebra} 9 | 10 | \title{IB Linear Algebra} 11 | 12 | \let\Im\relax 13 | \DeclareMathOperator*{\Im}{Im} 14 | 15 | \newcommand{\herm}[1]{\overline{#1}^\Trans} 16 | 17 | \DeclareMathOperator*{\ch}{ch} 18 | \renewcommand{\chi}{\textstyle\ch} 19 | 20 | \begin{document} 21 | 22 | \NotesTitle{IB}{Michaelmas 2012}{Linear Algebra}{Prof.~I.}{Grojnowski} 23 | { 24 | {\Large\bfseries\sffamily{Course schedule}} 25 | 26 | Definition of a vector space (over $\R$ or $\C$), subspaces, the space spanned by a subset. Linear independence, bases, dimension. Direct sums and complementary subspaces. \hfill [3] 27 | 28 | Linear maps, isomorphisms. Relation between rank and nullity. The space of linear maps from $U$ to $V$, representation by matrices. Change of basis. Row rank and column rank. \hfill [4] 29 | 30 | Determinant and trace of a square matrix. Determinant of a product of two matrices and of the inverse matrix. Determinant of an endomorphism. The adjugate matrix. \hfill [3] 31 | 32 | Eigenvalues and eigenvectors. Diagonal and triangular forms. Characteristic and minimal polynomials. Cayley-Hamilton Theorem over $\C$. Algebraic and geometric multiplicity of eigenvalues. Statement and illustration of Jordan normal form. \hfill [4] 33 | 34 | Dual of a finite-dimensional vector space, dual bases and maps. Matrix representation, rank and determinant of dual map. \hfill [2] 35 | 36 | Bilinear forms. Matrix representation, change of basis. Symmetric forms and their link with quadratic forms. Diagonalisation of quadratic forms. Law of inertia, classification by rank and signature. Complex Hermitian forms. \hfill [4] 37 | 38 | Inner product spaces, orthonormal sets, orthogonal projection, $V=W\oplus W^\perp$. Gram-Schmidt orthogonalisation. Adjoints. Diagonalisation of Hermitian matrices. Orthogonality of eigenvectors and properties of eigenvalues. \hfill [4] 39 | 40 | \subsection*{Appropriate books} 41 | 42 | {\shortskip 43 | C.W.~Curtis Linear \emph{Algebra: an Introductory Approach}. Springer 1984 (£38.50 hardback) 44 | 45 | P.R.~Halmos \emph{Finite-Dimensional Vector Spaces}. Springer 1974 (£31.50 hardback) 46 | 47 | K.~Hoffman and R.~Kunze \emph{Linear Algebra}. Prentice-Hall 1971 (£72.99 hardback)}} 48 | 49 | \TableofContents 50 | 51 | \input{linalg-chap-01} % Vector spaces 52 | \input{linalg-chap-02} % Endomorhisms 53 | \input{linalg-chap-03} % Jordan normal form 54 | \input{linalg-chap-04} % Continuous random variables 55 | \input{linalg-chap-05} % Inequalities 56 | \input{linalg-chap-06} % Geometrical probability 57 | 58 | \end{document} -------------------------------------------------------------------------------- /ib-met-top-spaces/met-top-spaces.tex: -------------------------------------------------------------------------------- 1 | \documentclass[twoside, numbers=noendperiod]{scrartcl} 2 | 3 | \usepackage[blacklinks]{drangreport} 4 | 5 | \usepackage{awlc-algebra} 6 | \usepackage{multirow} 7 | 8 | \newcommand{\xx}{\vec{x}} 9 | \newcommand{\yy}{\vec{y}} 10 | 11 | \newcommand{\closure}[1]{\overline{#1}} 12 | 13 | \DeclareMathOperator{\Int}{Int} 14 | \DeclareMathOperator{\Cl}{Cl} 15 | 16 | \newcommand{\Eucl}{\text{Eucl}} 17 | 18 | \newcommand{\Abar}{\overline{A}} 19 | \newcommand{\fbar}{\overline{f}} 20 | 21 | \newcommand{\xbar}{\overline{x}} 22 | \newcommand{\ybar}{\overline{y}} 23 | 24 | \newcommand{\boundary}{\del} 25 | 26 | \newcommand{\base}{\cal{B}} 27 | \newcommand{\cover}{\cal{U}} 28 | \newcommand{\subcover}{\cal{V}} 29 | 30 | \title{IB Metric \& Topological Spaces} 31 | 32 | \begin{document} 33 | 34 | \NotesTitle{IB}{Easter 2012}{Metric \& Topological Spaces}{Prof.~P.~M.~H.}{Wilson} 35 | { 36 | \section*{Course schedule} % (fold) 37 | \label{sub:course_schedule} 38 | 39 | \subsubsection*{Metrics} % (fold) 40 | \label{ssub:metrics} 41 | 42 | Definition and examples. Limits and continuity. Open sets and neighbourhoods. Characterizing limits and continuity using neighbourhoods and open sets. \hfill [3] 43 | 44 | % subsubsection metrics (end) 45 | 46 | \subsubsection*{Topology} % (fold) 47 | \label{ssub:topology} 48 | 49 | Definition of a topology. Metric topologies. Further examples. Neighbourhoods, closed sets, convergence and continuity. Hausdorff spaces. Homeomorphisms. Topological and non-topological properties. Completeness. Subspace, quotient and product topologies. \\ 50 | \mbox{ } \hfill [3] 51 | 52 | % subsubsection topology (end) 53 | 54 | \subsubsection*{Connectedness} % (fold) 55 | \label{ssub:connectedness} 56 | 57 | Definition using open sets and integer-valued functions. Examples, including intervals. Components. The continuous image of a connected space is connected. Path-connectedness. Path-connected spaces are connected but not conversely. Connected open sets in Euclidean space are path-connected. \hfill [3] 58 | 59 | % subsubsection connectedness (end) 60 | 61 | \subsubsection*{Compactness} % (fold) 62 | \label{ssub:compactness} 63 | 64 | Definition using open covers. Examples: finite sets and $[0,1]$. Closed subsets of compact spaces are compact. Compact subsets of a Hausdorff space must be closed. The compact subsets of the real line. Continuous images of compact sets are compact. Quotient spaces. Continuous real-valued functions on a compact space are bounded and attain their bounds. The product of two compact spaces is compact. The compact subsets of Euclidean space. Sequential compactness. \hfill [3] 65 | 66 | % subsubsection compactness (end) 67 | 68 | % subsubsection modules (end) 69 | 70 | % subsection course_schedule (end) 71 | 72 | \subsection*{Appropriate books} % (fold) 73 | \label{sub:literature} 74 | 75 | {\shortskip 76 | W.A.~Sutherland \emph{Introduction to Metric and Topological Spaces}. Clarendon 1975 (£21.00 paperback). 77 | 78 | A.J.~White \emph{Real Analysis: an Introduction}. Addison-Wesley 1968 (out of print) 79 | 80 | B.~Mendelson \emph{Introduction to Topology}. Dover, 1990 (£5.27 paperback)} 81 | 82 | % subsection literature (end) 83 | } 84 | 85 | \TableofContents 86 | 87 | \input{mettop-chap-01} % Metric spaces 88 | \input{mettop-chap-02} % Topological spaces 89 | \input{mettop-chap-03} % Connectedness 90 | \input{mettop-chap-04} % Compactness 91 | 92 | \lecturenotesend 93 | 94 | \end{document} -------------------------------------------------------------------------------- /ib-geometry/geo-chap-a1.tex: -------------------------------------------------------------------------------- 1 | %!TEX root = geometry.tex 2 | \stepcounter{lecture} 3 | \setcounter{lecture}{1} 4 | \appendix 5 | \sektion{Appendix: Review sheets} 6 | \label{app:review-sheets} 7 | 8 | \subsection{Euclidean geometry} % (fold) 9 | \label{sub:euclidean_geometry} 10 | 11 | Lines: 12 | \begin{itemize} 13 | \shortskip 14 | \item A line is the shortest path between two points. 15 | \item Plane separation: the complement of a line is a disconnected topological space. 16 | \item There is a unique line passing through two distinct points. 17 | \item Two distinct lines intersect in at most one point. 18 | \item Given a point $\xx$ and a line $L$ not containing $\xx$, there is a unique line passing through $\xx$ and parallel to $L$. % 19 | \item Given a point $\xx$ and a line $L$ not containing $\xx$, there is a unique line passing through $\xx$ and perpendicular to $L$. % 20 | \end{itemize} 21 | 22 | Circles: 23 | \begin{itemize} 24 | \shortskip 25 | \item A line and a circle intersect in at most two points. 26 | \item Two distinct circles intersect in at most two points. 27 | \item The perimeter of a circle of radius $R$ is $2\pi R$. 28 | \end{itemize} 29 | 30 | Isometries: 31 | \begin{itemize} 32 | \shortskip 33 | \item If $F_1, F_2$ are orthogonal frames, then there is a unique isometry taking $F_1$ to $F_2$. 34 | \item Any isometry which fixes three non-colinear points is the identity. 35 | \item Any isometry can be written as the composition of at most three reflections. 36 | \end{itemize} 37 | 38 | Triangles: 39 | \begin{itemize} 40 | \shortskip 41 | \item The sum of the interior angles in a triangle is $\pi$. 42 | \item If $A_1,A_2,A_3$ and $A_1\p,A_2\p,A_3\p$ are two sets of non-colinear points with $d(A_i,A_j) = d(A_i\p,A_j\p)$, then there is a unique $\phi\in\Isom(\R^2)$ with $\phi(A_i) = A_i\p$. % 43 | \item If instead we have $d(A_1,A_j) = d(A_1\p,A_j\p)$ and $\angle A_2 A_1 A_3 = \angle A_2\p A_1\p A_3\p$, then there is a unique $\phi\in\Isom(\R^2)$ with $\phi(A_i) = A_i\p$. % 44 | \end{itemize} 45 | 46 | Trigonometry: 47 | \begin{itemize} 48 | \item If $\triangle ABC$ has sides $a,b,c$ and opposite angles $\alpha,\beta,\gamma$, then 49 | \begin{equation*} 50 | \f{\sin \alpha}{a} = \f{\sin \beta}{b} = \f{\sin\gamma}{c}, \qquad 51 | c^2 = a^2 + b^2 - 2ab \cos \gamma. 52 | \end{equation*} 53 | \end{itemize} 54 | 55 | % subsection euclidean_geometry (end) 56 | 57 | \pagebreak 58 | 59 | \subsection{Spherical/projective geometry} % (fold) 60 | \label{sub:spherical_p} 61 | 62 | Spherical lines: 63 | \begin{itemize} 64 | \shortskip 65 | \item A line is the shortest path between two points. 66 | \item Plane separation: the complement of a line is a disconnected topological space. 67 | \item There is a unique line passing through two distinct, non-antipodal points. 68 | \item Two distinct lines intersect in two points. 69 | \item Given a point $\xx$ and a line $L$ not containing $\xx$, there is a line passing through $\xx$ and perpendicular to $L$. 70 | \end{itemize} 71 | 72 | Projective lines: 73 | \begin{itemize} 74 | \shortskip 75 | \item A line is the shortest path between two points. 76 | \item The complement of a line is connected. 77 | \item There is a unique line passing through two distinct, points. 78 | \item Two distinct lines intersect in exactly one point. 79 | \item Given a point $\xx$ and a line $L$ not containing $\xx$, there is a line passing through $\xx$ and perpendicular to $L$. 80 | \end{itemize} 81 | 82 | Circles: 83 | \begin{itemize} 84 | \shortskip 85 | \item A line and a circle which is distinct from it intersect in at most two points. 86 | \item Two distinct circles intersect in at most two points. 87 | \item The perimeter of a circle of radius $R$ is $2\pi \sin R$. 88 | \end{itemize} 89 | 90 | Isometries: 91 | \begin{itemize} 92 | \shortskip 93 | \item If $F_1, F_2$ are orthogonal frames, then there is a unique isometry taking $F_1$ to $F_2$. 94 | \item Any isometry which fixes three non-colinear points is the identity. 95 | \item Any isometry can be written as the composition of at most three reflections. 96 | \end{itemize} 97 | 98 | Triangles: 99 | \begin{itemize} 100 | \shortskip 101 | \item The sum of the interior angles in a $\triangle ABC$ is $\pi+\Area(ABC)$. 102 | \item If $A_1,A_2,A_3$ and $A_1\p,A_2\p,A_3\p$ are two sets of non-colinear points with $d(A_i,A_j) = d(A_i\p,A_j\p)$, then there is a unique $\phi\in\Isom(\R^2)$ with $\phi(A_i) = A_i\p$. % 103 | \item If instead we have $d(A_1,A_j) = d(A_1\p,A_j\p)$ and $\angle A_2 A_1 A_3 = \angle A_2\p A_1\p A_3\p$, then there is a unique $\phi\in\Isom(\R^2)$ with $\phi(A_i) = A_i\p$. % 104 | \end{itemize} 105 | 106 | Trigonometry: 107 | \begin{itemize} 108 | \item If $\triangle ABC$ has sides $a,b,c$ and opposite angles $\alpha,\beta,\gamma$, then 109 | \begin{equation*} 110 | \f{\sin \alpha}{\sin a} = \f{\sin \beta}{\sin b} = \f{\sin\gamma}{\sin c}, \qquad 111 | \begin{array}{l} 112 | \cos a = \ph \cos b \cos c + \sin \alpha \sin b \sin c, \\[3pt] 113 | \cos\alpha = -\cos\beta \cos \gamma + \sin a \sin \beta\sin\gamma. 114 | \end{array} 115 | \end{equation*} 116 | \end{itemize} 117 | 118 | % subsection spherical_p (end) 119 | 120 | \pagebreak 121 | 122 | \subsection{Hyperbolic geometry} % (fold) 123 | \label{sub:hyperbolic_geometry} 124 | 125 | Models: 126 | \begin{itemize} 127 | \shortskip 128 | \item Hyperboloid: % Originally "hyperboloid model" 129 | $S = \{(x,y,z): x^2+y^2-z^2=-1, z<0\}$, with the Minkowski metric $\dif{x^2} + \dif{y^2} - \dif{z^2}$ on $\R^3$. Lines are intersections of $S$ with planes through the origin. 130 | \item Unit disk model: $D = \left\{z \iC: \left\vert z \right\vert < 1\right\}$ with metric 131 | \begin{equation*} 132 | g^D = \f{4\left( \dif{x^2} + \dif{y^2} \right)}{\left( 1-x^2-y^2 \right)^2}. 133 | \end{equation*} 134 | Lines are Euclidean lines/circles perpendicular to $\boundary{D}$. 135 | \item Upper half plane model: $H = \left\{z \iC: \Re (z)>0\right\}$ with metric 136 | \begin{equation*} 137 | g^H = \f{\dif{x^2} + \dif{y^2}}{y^2}. 138 | \end{equation*} 139 | Lines are Euclidean lines/circles perpendicular to $\boundary{H}$. 140 | \end{itemize} 141 | 142 | Lines: 143 | \begin{itemize} 144 | \shortskip 145 | \item A line is the shortest path between two points. 146 | \item Plane separation: the complement of a line is a disconnected topogical space. 147 | \item There is a unique line passing through two distinct points. 148 | \item Two distinct lines intersect in at most one point. 149 | \item Given $\xx$ and $L$ as previously, there is a unique line passing through $\xx$ perpendicular to $L$, and infinitely many lines passing through $\xx$ which do not intersect $L$. 150 | \end{itemize} 151 | 152 | Circles: 153 | \begin{itemize} 154 | \shortskip 155 | \item In either the upper half-plane or the unit disk models, circles are Euclidean circles (but their centres are not the Euclidean centres.) 156 | \item A line and a circle, or two distinct circles, intersect in at most two points. 157 | \item The perimeter of a circle of radius $R$ is $2\pi\sinh R$. 158 | \end{itemize} 159 | 160 | Isometries: 161 | \begin{itemize} 162 | \shortskip 163 | \item If $F_1, F_2$ are orthogonal frames, then there is a unique isometry taking $F_1$ to $F_2$. 164 | \item Any isometry which fixes three non-colinear points is the identity. 165 | \item Any isometry can be written as the composition of at most three reflections. 166 | \end{itemize} 167 | 168 | Triangles: 169 | \begin{itemize} 170 | \shortskip 171 | \item The sum of the interior angles in a $\triangle ABC$ is $\pi-\Area(ABC)$. 172 | \item If $A_1,A_2,A_3$ and $A_1\p,A_2\p,A_3\p$ are two sets of non-colinear points with $d(A_i,A_j) = d(A_i\p,A_j\p)$, then there is a unique $\phi\in\Isom(\R^2)$ with $\phi(A_i) = A_i\p$. % 173 | \item If instead we have $d(A_1,A_j) = d(A_1\p,A_j\p)$ and $\angle A_2 A_1 A_3 = \angle A_2\p A_1\p A_3\p$, then there is a unique $\phi\in\Isom(\R^2)$ with $\phi(A_i) = A_i\p$. % 174 | \end{itemize} 175 | 176 | Trigonometry: 177 | \begin{itemize} 178 | \item If $\triangle ABC$ has sides $a,b,c$ and opposite angles $\alpha,\beta,\gamma$, then 179 | \begin{equation*} 180 | \f{\sin \alpha}{\sinh a} = \f{\sin \beta}{\sinh b} = \f{\sin\gamma}{\sinh c}, \qquad 181 | \begin{array}{l} 182 | \cosh a = \cosh b \cosh c - \cos \alpha \sinh b \sinh c, \\[3pt] 183 | \cos\alpha = -\cos\beta \cos \gamma + \cosh a \sinh \beta\sinh\gamma. 184 | \end{array} 185 | \end{equation*} 186 | \end{itemize} 187 | 188 | \vfill 189 | 190 | % subsection hyperbolic_geometry (end) -------------------------------------------------------------------------------- /ib-linear-algebra/linalg-chap-04.tex: -------------------------------------------------------------------------------- 1 | %!TEX root = linear-algebra.tex 2 | \stepcounter{lecture} 3 | \setcounter{lecture}{4} 4 | \sektion{Duals} 5 | This chapter really belongs after chapter 1 -- it's just definitions and intepretations of row reduction. 6 | 7 | \lecturemarker{16}{9 Nov} 8 | \begin{definition} 9 | Let $V$ be a vector space over a field $\F$. Then % 10 | \begin{equation*} 11 | V^* = \Lin(V,\F) = \left\{\text{linear functions } V\to\F\right\} 12 | \end{equation*} 13 | is the \emph{dual space} of $V$. 14 | \end{definition} 15 | 16 | \begin{examples} 17 | \mbox{} 18 | \begin{enumerate} 19 | \shortskip 20 | \item Let $V=\R^3$. Then $(x,y,z) \mapsto x-y$ is in $ V^*$. 21 | \item If $V=C([0,1])=\left\langle \text{continuous functions } [0,1]\to\R \right\rangle$, then $f\mapsto\int_0^1 f(t)\dif{t}$ is in $C([0,1])^*$. % 22 | \end{enumerate} 23 | \end{examples} 24 | 25 | \begin{definition} 26 | Let $V$ be a finite dimensional vector space over $\F$, and $v_1,\ldots,v_n$ be a basis of $V$. Then define $v_i^*\in V^*$ by % 27 | \begin{equation*} 28 | v_i^*(v_j) = \delta_{ij} = 29 | \begin{cases} 30 | 0 & \text{if } i\neq j, \\ % 31 | 1 & \text{if } i=j, 32 | \end{cases} 33 | \end{equation*} 34 | and extend linearly. That is, $v_i^*\left( \sum_j \lambda_j\,v_j \right) = \lambda_i$. 35 | \end{definition} 36 | 37 | \begin{lemma} 38 | The set $v_1^*,\ldots,v_n^*$ is a basis for $V^*$, called the \emph{basis dual to} or \emph{dual basis for} $v_1,\ldots,v_n$. In particular, $\dim V^*=\dim V$. % 39 | \end{lemma} 40 | 41 | \begin{proof} 42 | Linear independence: if $\sum \lambda_i\,v_i^*=0$, then $0=\left( \sum \lambda_i \, v_i^* \right) (v_j) = \lambda_j$, so $\lambda_j=0$ for all $j$. Span: if $\varphi\in V^*$, then we claim % 43 | \begin{equation*} 44 | \varphi = \sum_{j=1}^n \varphi(v_j) \cdot v_j^*. 45 | \end{equation*} 46 | As $\varphi$ is linear, it is enough to check that the right hand side % of the above 47 | applied to $v_k$ is $\varphi(v_k)$. But % 48 | \begin{equation*} 49 | \textstyle \sum_j \varphi(v_j) \, v_j^* (v_k) = \sum_j \varphi(v_j) \, \delta_{jk} = \varphi(v_k). \qedhere % 50 | \end{equation*} 51 | \end{proof} 52 | 53 | \begin{remarks} 54 | \mbox{} 55 | \begin{enumerate} 56 | \item We know in general that $\dim \Lin(V,W)=\dim V \dim W$. 57 | \item If $V$ is finite dimensional, then this shows that $V\isom V^*$, as any two vector spaces of dimension $n$ are isomorphic. But they are not \emph{canonically} isomorphic (there is no natural choice of isomorphism). % 58 | 59 | If the vector space $V$ has more structure (for example, a group $G$ acts upon it), then $V$ and $V^*$ are not usually isomorphic in a way that respects this structure. % 60 | 61 | \item If $V=F[x]$, then $V^* \isomto \F^\N$ by the isomorphism $\theta\in V^* \mapsto (\theta(1),\theta(x),\theta(x^2),\ldots)$, and conversely, if $\lambda_i\iF$, $i=0,1,2,\ldots$ is any sequence of elements of $\F$, we get an element of $V^*$ by sending $\sum a_i \, x^i \mapsto \sum a_i\,\lambda_i$ (notice this is a finite sum). % 62 | 63 | Thus $V$ and $V^*$ are not isomorphic, since $\dim V$ is countable, but $\dim \F^\N$ is uncountable. 64 | \end{enumerate} 65 | \end{remarks} 66 | 67 | \begin{definition} 68 | Let $V$ and $W$ be vector space over $\F$, and $\alpha$ a linear map $V\to W$, $\alpha\in \Lin(V,W)$. Then we define $\alpha^*:W^*\to V^* \in \Lin(W^*,V^*)$, by setting $\alpha^*(\theta) = \theta\alpha:V\to\F$. % 69 | 70 | (Note: $\alpha$ linear, $\theta$ linear implies $\theta\alpha$ linear, and so $\alpha^*(\theta)\in V^*$ as claimed, if $\theta\in W^*$.) % 71 | \end{definition} 72 | 73 | \begin{lemma} 74 | Let $V,W$ be finite dimensional vector spaces, with % 75 | \begin{itemize} 76 | \shortskip 77 | \item [] $v_1,\ldots,v_n$ a basis of $V$, and $w_1,\ldots,w_m$ a basis for $W$; 78 | \item [] $v_1^*,\ldots,v_n^*$ the dual basis of $V^*$, and $w_1^*,\ldots,w_m^*$ the dual basis for $W^*$; % 79 | \end{itemize} 80 | If $\alpha$ is a linear map $V\to W$, and $A$ is the matrix of $\alpha$ with respect to $v_i,w_j$, then $A^\Trans$ is the matrix of $\alpha^*:W^*\to V^*$ with respect to $w_j^*,v_i^*$. 81 | \end{lemma} 82 | 83 | \begin{proof} 84 | Write $\alpha^*(w_i^*) = \sum_{j=1}^n c_{ji} \, v_j^*$, so $c_{ij}$ is a matrix of $\alpha^*$. Apply this to $v_k$: % 85 | \begin{align*} 86 | \LHS 87 | &= \left( \alpha^* (w_i^*) \right)(v_k) 88 | = w_i^*(\alpha(v_k)) 89 | = w_i^*\left( \textstyle\sum_\ell a_{\ell k} \,w_\ell \right) = a_{ik} \\ 90 | \RHS 91 | &= c_{ki}, 92 | \end{align*} 93 | that is, $c_{ji}=a_{ij}$ for all $i,j$. 94 | \end{proof} 95 | This was the promised interpretation of $A^\Trans$. % 96 | 97 | 98 | \begin{corollary} 99 | \mbox{} 100 | \begin{enumerate} 101 | \shortskip 102 | \item $(\alpha\beta)^* = \beta^* \alpha^*$; 103 | \item $(\alpha+\beta)^* = \alpha^* + \beta^*$; 104 | \item $\det \alpha^* = \det\alpha$ 105 | \end{enumerate} 106 | \end{corollary} 107 | 108 | \begin{proof} 109 | (i) and (ii) are immediate from the definition, or use the result $(AB)^\Trans = B^\Trans A^\Trans$. (iii) we proved in the section on determinants where we showed that $\det A^\Trans=\det A$. % 110 | \end{proof} 111 | 112 | Now observe that $\left.(A^\Trans)\right.^\Trans = A$. What does this mean? 113 | 114 | \begin{proposition} 115 | \mbox{} 116 | \begin{enumerate} 117 | \item Consider the map $V\to V^{**} = (V^*)^*$ taking $v\mapsto \hat{\hat{v}}$, where $\hat{\hat{v}}(\theta) = \theta(v)$ if $\theta\in V^*$. Then $\hat{\hat{v}} \in V^{**}$, and the map $V\mapsto V^{**}$ is linear and injective. % 118 | \item Hence if $V$ is a finite dimensional vector space over $\F$, then this map is an isomorphism, so $V\isomto V^{**}$ canonically. % 119 | \end{enumerate} 120 | \end{proposition} 121 | 122 | \begin{proof} 123 | \mbox{} 124 | \begin{enumerate} 125 | \item We first show $\hat{\hat{v}}\in V^{**}$, that is $\hat{\hat{v}}:V^*\to\F$, is linear: % 126 | \begin{align*} 127 | \hat{\hat{v}} (a_1\theta_1+a_2\theta_2) 128 | &= (a_1\theta_1+a_2\theta_2)(v) = a_1\,\theta_1(v) + a_2 \,\theta_2(v) \\ 129 | &= a_1\,\hat{\hat{v}}(\theta_1) + a_2\,\hat{\hat{v}}(\theta_2). 130 | \end{align*} 131 | Next, the map $V\to V^{**}$ is linear. This is because 132 | \begin{align*} 133 | \left( \lambda_1 v_1 + \lambda_2 v_2 \right)^{\hat\hat}(\theta) 134 | &= \theta\left( \lambda_1 v_1 + \lambda_2 v_2 \right) \\ 135 | &= \lambda_1\,\theta(v_1) + \lambda_2\,\theta(v_2) \\ 136 | &= \left( \lambda_1 \hat{\hat{v_1}} + \lambda_2 \hat{\hat{v_2}} \right)(\theta). 137 | \end{align*} 138 | Finally, if $v\neq 0$, then there exists a linear function $\theta:V\to\F$ such that $\theta(v)\neq 0$. % 139 | 140 | (Proof: extend $v$ to a basis, and then define $\theta$ on this basis. We've only proved that this is okay when $V$ is finite dimensional, but it's always okay.) % 141 | 142 | Thus $\hat{\hat{v}}(\theta)\neq 0$, so $\hat{\hat{v}}\neq 0$, and $V\to V^{**}$ is injective. % 143 | \item Immediate. \qedhere 144 | \end{enumerate} 145 | \end{proof} 146 | 147 | \begin{definition} 148 | \mbox{} 149 | \begin{enumerate} 150 | \item If $U\leq V$, then define 151 | \begin{equation*} 152 | U^\circ = \left\{\theta\in V^* \mid \theta(U)=0\right\} 153 | = \left\{\theta\in V^* \mid \theta(u) = 0 \;\forall u\in U\right\} 154 | \leq V^*. 155 | \end{equation*} 156 | This is the \emph{annihilator} of $U$, a subspace of $V^*$, often denoted $U^\perp$. % 157 | \item If $W\leq V^*$, then define 158 | \begin{equation*} 159 | {}^\circ W = \left\{v\in V \mid \varphi(v)=0\;\forall\varphi\in W\right\}\leq V. 160 | \end{equation*} 161 | This is often denoted $^\perp W$. 162 | \end{enumerate} 163 | \end{definition} 164 | 165 | \begin{example} 166 | If $V=\R^3$, $U=\left\langle (1,2,1) \right\rangle$, then % 167 | \begin{equation*} 168 | U^\circ = \left\{\sum_{i=1}^3 a_i\,e_i^* \in V^* \mid a_1+2a_2+a_3 = 0 \right\} 169 | = \left\langle \mat{-2 \\ 1 \\ 0}, \mat{0 \\ 1 \\ -2} \right\rangle. 170 | \end{equation*} 171 | \end{example} 172 | 173 | \begin{remark} 174 | If $V$ is finite dimensional, and $W\leq V^*$, then under the canonical isomorphism $V\to V^{**}$, we have $^\circ W \mapsto W^\circ$, where $^\circ W\leq V$ and $(W^\circ)\leq (V^*)^*$. Proof is an exercise. % 175 | \end{remark} 176 | 177 | \begin{lemma} 178 | Let $V$ be a finite dimensional vector space with $U\leq V$. Then % 179 | \begin{equation*} 180 | \dim U+\dim U^\circ=\dim V. 181 | \end{equation*} 182 | \end{lemma} 183 | 184 | \begin{proof} 185 | Consider the restriction map $\Res:V^* \to U^*$ taking $\varphi\mapsto \varphi |_U$. (Note that $\Res=\iota^*$, where $\iota: U\injto V$ is the inclusion.) % 186 | 187 | Then $\ker\Res=U^\circ$, by definition, and $\Res$ is surjective (why?). 188 | 189 | So the rank-nullity theorem implies the result, as $\dim V^*=\dim V$. 190 | \end{proof} 191 | 192 | \begin{proposition} 193 | Let $V,W$ be a finite dimensional vector space over $\F$, with $\alpha\in \Lin(V,W)$. Then % 194 | \begin{enumerate} 195 | \shortskip 196 | \item $\ker(\alpha^*:W^*\to V^*)=\left( \Im\alpha \right)^\circ$ ($\leq W^*$); 197 | \item $\rk(\alpha^*) = \rk(\alpha)$; that is, $\rk A^\Trans = \rk A$, as promised; 198 | \item $\Im \alpha^* = \left( \ker\alpha \right)^\circ$. 199 | \end{enumerate} 200 | \end{proposition} 201 | 202 | \begin{proof} 203 | \mbox{} 204 | \begin{enumerate} 205 | \shortskip 206 | \item Let $\theta\in W^*$. Then $\theta\in\ker\alpha^* \iff \theta\alpha=0 \iff \theta\alpha(v)=0 \;\forall v\in V \iff \theta\in\left( \Im\alpha \right)^\circ$. % 207 | \item By rank-nullity, we have 208 | \begin{align*} 209 | \rk\alpha^* 210 | &= \dim W-\dim\ker\alpha^* \\ 211 | &= \dim W-\dim(\Im \alpha)^\circ, \text{by (i)}, \\ 212 | &= \dim \Im\alpha, \text{ by the previous lemma,} \\ 213 | &= \rk\alpha, \text{ by definition.} 214 | \end{align*} 215 | \item Let $\varphi\in\Im\alpha^*$, and then $\varphi=\theta\circ\alpha$ for some $\theta\in W^*$. Now, let $v\in\ker\alpha$. Then $\varphi(v)=\theta\alpha(v)=0$, so $\varphi\in\left( \ker\alpha \right)^\circ$; that is, $\Im\alpha^* \subseteq\left( \ker\alpha \right)^\circ$. % 216 | 217 | But by (ii), 218 | \begin{align*} 219 | \dim\Im\alpha^* = \rk(\alpha^*) 220 | &= \rk\alpha = \dim V - \dim\ker\alpha \\ 221 | &= \dim\left( \ker\alpha \right)^\circ 222 | \end{align*} 223 | by the previous lemma; that is, they both have the same dimension, so they are equal. \qedhere % 224 | \end{enumerate} 225 | \end{proof} 226 | 227 | \begin{lemma} 228 | Let $U_1,U_2\leq V$, and $V$ be finite dimensional. Then % 229 | \begin{enumerate} 230 | \shortskip 231 | \item $U_1^{\circ\circ} \isomto {}^\circ\!(U_1^\circ) \isomto U_1$ under the isomorphism $V\isomto V^{**}$. % 232 | \item $(U_1+U_2)^\circ = U_1^\circ \cap U_2^\circ$. 233 | \item $(U_1\cap U_2)^\circ = U_1^\circ + U_2^\circ$. 234 | \end{enumerate} 235 | \end{lemma} 236 | 237 | \begin{proof*} 238 | Exercise! 239 | \end{proof*} 240 | -------------------------------------------------------------------------------- /ib-geometry/geo-chap-03.tex: -------------------------------------------------------------------------------- 1 | %!TEX root = geometry.tex 2 | \stepcounter{lecture} 3 | \setcounter{lecture}{3} 4 | \sektion{Möbius transformations} 5 | \label{sec:m_bius_transformations} 6 | 7 | \subsection{Stereographic projection} % (fold) 8 | \label{sub:stereographic_projection} 9 | 10 | \lecturemarker{8}{13 Feb} 11 | We've encountered Möbius transformations before, in \emph{Groups}. These are transformations of the \emph{extended complex plane}, $\C\cup\{\infty\} = \C_\infty$. We'd like to consider how they apply to the sphere, $S^{\,2}$. To do this, first we need to map the sphere to the complex plane, which we do using \emph{stereographic projections}. 12 | 13 | \begin{definition} 14 | Let $\NN=(0,0,1) \in S^2$ (the ``north pole''). Then we define the \emph{stereographic projection map} $\pi:S^{\,2}\backslash\{\NN\} \to \R^2 = \C$ by 15 | \begin{equation*} 16 | \pi(\pp) = \text{intersection of the Euclidean ray $\pp-\NN$ with the $x,y$ plane.} 17 | \end{equation*} 18 | Let's consider a slightly flattened picture of this: if $\pp=(x,0,z)$: 19 | 20 | \begin{center} 21 | \begin{tikzpicture}[scale=2] 22 | 23 | \bigarrow (-1.3,0) -- (2.5,0) node [right=2pt] {$z$}; 24 | \bigarrow (0,-1.3) -- (0,1.6) node [right] {$x$}; 25 | 26 | \draw [thick] (0,0) circle (1); 27 | 28 | \draw (0,1) node [above left=2pt] {$\NN$} -- (1.732,0) node {$\bullet$}; 29 | 30 | \draw (1.732,0) node [below=2pt] {$\pi(\pp)$}; 31 | 32 | \draw (0,1) node {$\bullet$}; 33 | \draw (30:1) node {$\bullet$}; 34 | \draw (30:1) node [above right] {$\pp=(x,0,z)$}; 35 | 36 | \draw [dashed] (30:1) -- ++ (0,-0.5); 37 | \end{tikzpicture} 38 | \end{center} 39 | \end{definition} 40 | 41 | This picture gives us a way to compute the value of $\pi(\pp)$ for a given point $\pp$. By considering the two similar triangles, we see that 42 | \begin{equation*} 43 | \f{z}{1} = \f{\pi(\pp) - x}{\pi(\pp)} \implies \pi(\pp) = \f{x}{1-z}, 44 | \end{equation*} 45 | and so the $x$-coordinate of $\pi(\pp)$ is $x/(1-z)$. 46 | 47 | The projection is radially symmetric about the $z$-axis, and so by rotating, we have 48 | \begin{equation*} 49 | \pi\left( (x,y,z) \right) = \f{x+iy}{1-z}. 50 | \end{equation*} 51 | So now we naturally ask how to invert the projection. In other words, given $w\iC$, can we find $\pp\in S^{\,2}$ with $\pi(\pp) = w$. We consider $w\iC$ with 52 | \begin{equation*} 53 | w = \f{x+iy}{1-z}, \qquad x^2+y^2+z^2=1. 54 | \end{equation*} 55 | Squaring this equation gives 56 | \begin{equation*} 57 | \left\vert w \right\vert^2 58 | = \f{x^2+y^2}{\left( 1-z^2 \right)^2} 59 | = \f{1-z^2}{\left( 1-z \right)^2} 60 | = \f{1+z}{1-z}. 61 | \end{equation*} 62 | We can solve this for $z$ in terms of $\left\vert w \right\vert$, which gives 63 | \begin{equation*} 64 | z = \f{\left\vert w \right\vert^2-1}{\left\vert w \right\vert^2+1} 65 | \implies 1-z = \f{2}{\left\vert w \right\vert^2+1}. 66 | \end{equation*} 67 | Now we return to our definition of $w$. We have 68 | \begin{equation*} 69 | x = \left( 1-z \right) \Re(w) 70 | \qqand 71 | y = \left( 1-z \right) \Im(w), 72 | \end{equation*} 73 | and thus we can write 74 | \begin{equation*} 75 | \pi^{-1}(w) = \left( \f{2\,\Re(w)}{1+\left\vert w \right\vert^2}, \f{2\,\Im(w)}{1+\left\vert w \right\vert^2}, \f{\left\vert w \right\vert^2-1}{\left\vert w \right\vert^2+1} \right). 76 | \end{equation*} 77 | This will turn out to be a very useful formula. 78 | 79 | This projection map does most of the work of identifying $\C_\infty$ to $S^{\,2}$. There are just two points left unaccounted for: $\infty\in\C_\infty$, and $\NN\in S^{\,2}$. It naturally follows that we can identify $\C_\infty$ and $S^{\,2}$ using the map 80 | \begin{align*} 81 | w\iC & \longleftrightarrow \pi^{-1}(w) \in S^{\,2}, \\ 82 | \infty & \longleftrightarrow \NN \in S^{\,2}. 83 | \end{align*} 84 | This tells us that $\{w_n\} \to \infty$ in $\C_\infty$ if and only if $\left\vert w_n \right\vert \to \infty$ also. 85 | 86 | In this context, we call $\C_\infty$ the \emph{Riemann sphere}, and we will also encounter it in \emph{Complex Analysis} and \emph{Complex Methods}. 87 | 88 | % We will encounter this in \emph{Complex Analysis} or \emph{Complex Methods}. A mereomophic fn is analogous to a holoc map $\C\to\C_\infty$. 89 | 90 | % subsection stereographic_projection (end) 91 | 92 | \subsection{Möbius group} % (fold) 93 | \label{sub:m_bius_group} 94 | 95 | Consider an invertible matrix 96 | \begin{equation*} 97 | A=\mat{a & b \\c & d} \in \GL_2(\C). 98 | \end{equation*} 99 | This induces a Möbius map. We define 100 | \begin{equation*} 101 | \fullfunction{\phi_A}{\C_\infty}{\C_\infty}{w}{\f{aw+b}{cw+d}}, 102 | \end{equation*} 103 | with $\phi_A(-d/c)=\infty$ and $\phi_A(\infty) = a/c$. 104 | 105 | % Thinking of Möbius maps as induced by matrices gives us some useful results: 106 | 107 | \begin{lemma} 108 | \mbox{} 109 | \begin{enumerate} 110 | \shortskip 111 | \item $\phi_{\lambda A}(w)=\phi_A(w)$; 112 | \item $\phi_A(\phi_B(w)) = \phi_{AB}(w)$. 113 | \end{enumerate} 114 | \end{lemma} 115 | 116 | \begin{proof} 117 | Part (i) is easy and left as an exercise. For (ii), define 118 | \begin{equation*} 119 | X = \left\{\ww\iC^2 : \ww\neq\bf{0}\right\}/\sim, \qquad \ww\sim\lambda\ww, \qquad \lambda\iC^*. 120 | \end{equation*} 121 | We can define a map $P:X\to\C_\infty$ by $P(w_1,w_2) = w_1/w_2$. 122 | 123 | Then $\GL_2(\C)$ acts on $X$ by $A\cdot\ww = A\ww$ (by matrix multiplication) and 124 | \begin{equation*} 125 | P(A\ww) = \phi_A(P(\ww)). 126 | \end{equation*} 127 | Then we have 128 | \begin{equation*} 129 | \phi_A(\phi_B(P(\ww))) 130 | = P(A \cdot (B \cdot \ww)) 131 | = P(AB\ww) 132 | = \phi_{AB}(P(\ww)). \qedhere 133 | \end{equation*} 134 | \end{proof} 135 | 136 | It would have been easy to do this by simply plugging in matrices and turning the handle on some algebra, but this is a cleaner proof. It gives us some understanding of why the result is true. 137 | 138 | \pagebreak 139 | 140 | \begin{corollary} 141 | We define the \emph{projective general linear group} as 142 | \begin{equation*} 143 | \PGL_2(\C) := \f{\GL_2(\C)}{\left\{\lambda I: \lambda\iC\right\}}. 144 | \end{equation*} 145 | This acts on $\C_\infty$. 146 | \end{corollary} 147 | 148 | \vspace{-3pt} 149 | 150 | \begin{exercise} 151 | If $\SL_2(\C)$ is the special linear group, then show that 152 | \begin{equation*} 153 | \PGL_2(\C) = \f{\SL_2(\C)}{\left\{\pm I\right\}} =: \PSL_2(\C). 154 | \end{equation*} 155 | \end{exercise} 156 | 157 | \begin{definition} 158 | The \emph{Mobius group} is given by 159 | \begin{equation*} 160 | \Mobgp = \left\{\phi:\C_\infty \to \C_\infty: \phi(w)=\phi_A(w), A\in\GL_2(\C)\right\} \cong \PSL_2(\C). 161 | \end{equation*} 162 | Then $\phi\in\Mobgp$ is a \emph{Mobius transformation}. This is the group of all invertible holomorphic maps $\C_\infty \to \C_\infty$. 163 | \end{definition} 164 | 165 | \begin{lemma} 166 | \mbox{} 167 | \begin{enumerate} 168 | \item We can generate $\Mobgp$ with maps of the form 169 | \begin{itemize} 170 | \shortskip 171 | \item $z\mapsto az$, $a\iC^*$ (dilation); 172 | \item $z\mapsto z+b$, $b\iC$ (translation); 173 | \item $z\mapsto 1/z$ (inversion). 174 | \end{itemize} 175 | \item If $z_1,z_2,z_3$ and $w_1,w_2,w_3$ are two sets of distinct points in $\C_\infty$, then there is a unique $\phi\in\Mobgp$ with $\phi(z_i)=w_i$. 176 | \item \emph{Cross ratios.} If $z_1,z_2,z_3,z_4\iC_\infty$ are distinct and $\phi\in\Mobgp$ with $\phi(z_i)=w_i$, then cross ratios are preserved. That is, 177 | \begin{equation*} 178 | \f{\left( z_2-z_3 \right)\left( z_4-z_1 \right)}{\left( z_2-z_1 \right)\left( z_4-z_3 \right)} 179 | = \f{\left( w_2-w_3 \right)\left( w_4-w_1 \right)}{\left( w_2-w_1 \right)\left( w_4-w_3 \right)}. 180 | \end{equation*} 181 | \end{enumerate} 182 | \end{lemma} 183 | 184 | \vspace{-6pt} 185 | 186 | Proof is left as an exercise. 187 | 188 | \begin{definition} 189 | Let $\cal{C}$ be the set of Euclidean lines and circles in $\C$. 190 | \end{definition} 191 | 192 | \begin{lemma} 193 | Let $S\subset\C$. Then $S\in\cal{C}$ if and only if $S$ satisfies an equation of the form \label{lem:eq-lines-circles} 194 | \begin{equation*} 195 | az\zbar + bz+\overline{bz}+c=0, 196 | \end{equation*} 197 | for $a,c\iR$, $b\iC$, and not all zero. 198 | \end{lemma} 199 | 200 | \begin{proof} 201 | A line satifies $\alpha x+\beta y = \gamma$, for $\alpha,\beta,\gamma\iR$, so 202 | \begin{equation*} 203 | \alpha\left( \f{z+\zbar}{2} \right) + \beta\left( \f{z-\zbar}{2i} \right) = \gamma. 204 | \end{equation*} 205 | Rearranging this gives 206 | \begin{equation*} 207 | \left( \f{\alpha-i\beta}{2} \right)z + \left( \f{\alpha+i\beta}{2} \right)\zbar = \gamma, 208 | \end{equation*} 209 | which is what we want. 210 | 211 | A circle satisfies $\left\vert z-p \right\vert^2=r^2$, so 212 | \begin{equation*} 213 | z\zbar-p\zbar-\overline{p}z+\left\vert p \right\vert^2 = r^2 \qquad \text{or} \qquad 214 | z\zbar-p\zbar-\overline{p}z+\left( \left\vert p \right\vert^2-r^2 \right) = 0, 215 | \end{equation*} 216 | which is again the desired form. 217 | 218 | The converse is very similar: if $a\neq 0$, then divide by $a$ and complete the square. If $a=0$, then we have a line. 219 | \end{proof} 220 | 221 | \begin{corollary} 222 | If $S\in\cal{C}$, $\phi\in\Mobgp$, then $\phi(S)\in\cal{C}$. That is, Möbius maps takes lines and circles to lines and circles. 223 | \end{corollary} 224 | 225 | \begin{proof} 226 | It sufficies to check that this is true for the generators of $\Mobgp$. 227 | 228 | Take $w=\phi(z)=\alpha z$. If the equation of $S$ is $az\zbar + bz+b\zbar + c = 0$, then 229 | \begin{equation*} 230 | z = \alpha^{-1} w \implies 231 | \f{a}{\left\vert \alpha \right\vert^2}\,w\wbar + \f{b}{\alpha}\,w + \f{\overline{b}}{\overline{\alpha}}\,\wbar + c = 0. 232 | \end{equation*} 233 | This equation is of the same form, and so elements of $\cal{C}$ map to other elements of $\cal{C}$. 234 | 235 | The cases $z\mapsto z+b$ and $z\mapsto 1/z$ are similar. For the latter, the new equation is 236 | \begin{equation*} 237 | \f{a}{w\wbar} + \f{b}{w} + \f{\overline{b}}{\wbar} + c = 0 \implies a + b\wbar + cw\wbar = 0, 238 | \end{equation*} 239 | which is again of the same form. 240 | \end{proof} 241 | 242 | \begin{corollary} 243 | There's a unique element of $\cal{C}$ passing through any three distinct points $z_1,z_2,z_3\iC_\infty$. 244 | \end{corollary} 245 | 246 | \begin{proof} 247 | Choose $\phi\in\Mobgp$ with $\phi(z_1)=0$, $\phi(z_2)=1$ and $\phi(z_3)=2$. There's a unique line $S\in\cal{C}$ passing through $0,1,2$ in $\R$. So $C=\phi^{-1}(\R)$ is the set that we want. 248 | \end{proof} 249 | 250 | \begin{corollary} 251 | The group $\Mobgp$ acts transitively on $\cal{C}$. 252 | \end{corollary} 253 | 254 | \begin{proof} 255 | Given $C_1,C_2$, pick $z_1,z_2,z_3$ on $C_1$, $w_1,w_2,w_3$ on $C_2$, and $\phi$ with $\phi(z_i) = w_i$. Then $\phi(C_1)$ passes though $w_1,w_2,w_3$, and so $\phi(C_1)=C_2$. 256 | \end{proof} 257 | 258 | \begin{examples} 259 | \mbox{} 260 | \begin{enumerate} 261 | \item If $\phi(z) = \df{z-i}{z+i}$, then $\phi(\R)=S^{\,1}\subset \C$. 262 | 263 | Consider: if $z\iR$, then $\left\vert z-i \right\vert = \left\vert z+i \right\vert = \sqrt{z^2+1}$. 264 | \item The stabiliser of the real line is given by 265 | \begin{equation*} 266 | A=\left\{\phi\in\Mobgp: \phi(\R)=\R\right\} = \left\{\phi_A: A\in\GL_2(\R)\right\}. 267 | \end{equation*} 268 | Similarly, the stabiliser of the circle has 269 | \begin{equation*} 270 | B 271 | = \left\{\phi\in\Mobgp: \phi(S^{\,1}) = S^{\,1}\right\} 272 | = \left\{\phi: \phi(z) = \lambda\,\f{z+\alpha}{\overline{\alpha}z+1}, \lambda\in S^{\,1}, \alpha\iC, |\alpha|^2\neq 1\right\}. 273 | \end{equation*} 274 | The idea of the proof is that $B=\phi A\phi^{-1}$, where $\phi$ is as in the previous example. 275 | \end{enumerate} 276 | \end{examples} 277 | 278 | % subsection m_bius_group (end) -------------------------------------------------------------------------------- /ib-geometry/geo-chap-07.tex: -------------------------------------------------------------------------------- 1 | %!TEX root = geometry.tex 2 | \stepcounter{lecture} 3 | \setcounter{lecture}{7} 4 | \sektion{Surfaces} 5 | 6 | \subsection{First fundamental form} % (fold) 7 | \label{sub:first_fundamental_form} 8 | 9 | Suppose $\sigma:U \to \R^3$ is a parameterised surface $S$ and let $g$ be the induced metric on $U$. 10 | 11 | \begin{definition} 12 | The \emph{first fundamental form} at a point $p\in U$ is the bilinear form on $\R^2$ given by 13 | \begin{equation*} 14 | B_{I,p} := g_p(\vv,\ww). 15 | \end{equation*} 16 | This is represented by the matrix 17 | \begin{equation*} 18 | \mat{\sigma_x \\ \sigma_y} \mat{\sigma_x & \sigma_y} = \mat{\bsig_x \cdot \bsig_x & \bsig_x \cdot \bsig_y \\ \bsig_y \cdot \bsig_x & \bsig_y \cdot \bsig_y} = \mat{E & F \\ F & G}. 19 | \end{equation*} 20 | \end{definition} 21 | 22 | % subsection first_fundamental_form (end) 23 | 24 | \subsection{Second fundamental form} % (fold) 25 | \label{sub:second_fundamental_form} 26 | 27 | The tangent space $T_{\sigma(p)} S$ is spanned by $\dif{\sigma_p}(1,0) = \bsig_x$ and $\dif{\sigma_p}(0,1) = \bsig_y$. 28 | 29 | The unit normal to $T_{\sigma(p)} S$ at $\sigma(p)$ is 30 | \begin{equation*} 31 | \nn(p) = \f{\bsig_x \cross \bsig_y}{\left\vert \bsig_x \cross \bsig_y \right\vert}. 32 | \end{equation*} 33 | The map $\nn:U\to S^2 \subset \R^3$ is called the \emph{Gauss map}. 34 | 35 | \begin{definition} 36 | The \emph{second fundamental form} of $\sigma$ at $\pp$ is the bilinear form on $\R^2$ defined by 37 | \begin{equation*} 38 | B_{\text{\emph{II}},p}(\vv,\ww) = -\dif{\sigma_p(\vv)} \cdot \dif{\nn_p}(\ww) 39 | \end{equation*} 40 | \end{definition} 41 | 42 | There's a useful procedure for computing it. Let 43 | \begin{itemize} 44 | \shortskip 45 | \item $S=\mat{\sigma_x & \sigma_y}$ be the $3\times 2$ matrix that represents $\dif{\sigma}$; and 46 | \item $N=\mat{\nn_x & \nn_y}$ be the matrix that represents $\dif{\nn}$. 47 | \end{itemize} 48 | Then we have 49 | \begin{equation*} 50 | B_{\text{\emph{II}}}(\vv,\ww) = -(S\vv)^\Trans (N\ww) 51 | = - v^\Trans S^\Trans N \ww 52 | = -\vv^\Trans \mat{\sigma_x \\ \sigma_y} \mat{\nn_x & \nn_y} \ww. 53 | \end{equation*} 54 | Thus $B_{\text{\emph{II}}}$ is given by the matrix 55 | \begin{equation*} 56 | -\mat{\sigma_x \\ \sigma_y} \mat{n_x & n_y} 57 | = -\mat{\sigma_x \cdot n_x & \sigma_x \cdot \nn_y \\ \sigma_y \cdot n_x & \sigma_y \cdot n_y} 58 | = \mat{L & M_1 \\ M_2 & N}. 59 | \end{equation*} 60 | 61 | \begin{lemma} 62 | \begin{equation*} 63 | \mat{L & M_1 \\ M_2 & N} = \mat{\sigma_{xx} \cdot \nn & \sigma_{xy} \cdot \nn \\ \sigma_{yx} \cdot \nn & \sigma_{yy} \cdot \nn} 64 | \end{equation*} 65 | \end{lemma} 66 | 67 | \begin{proof} 68 | If $\sigma_x \in T_{\sigma(p)} S$, then $\sigma_x \cdot \nn = 0$. Then $\sigma_{xx} \cdot \nn + \sigma_x \cdot \nn_x = 0$, so $-\sigma_x \cdot n_x = \sigma_{xx} \cdot \nn$. Thus $L=\sigma_{xx} \cdot \nn$. Other entries are similar. 69 | \end{proof} 70 | 71 | \begin{corollary} 72 | $B_{\text{\emph{II}}}$ is symmetric. 73 | \end{corollary} 74 | 75 | \begin{proof} 76 | We have $\sigma_{xy} = \sigma_{yx}$, so done. 77 | \end{proof} 78 | 79 | \pagebreak 80 | 81 | \begin{example} 82 | We have $\sigma(\theta,z) = (\cos \theta, \sin\theta, z)$; a cylinder of radius $1$. Thus 83 | \begin{equation*} 84 | \sigma_\theta=(-\sin \theta, \cos\theta, 0) \qqand 85 | \sigma_z= (0,0,1). 86 | \end{equation*} 87 | Then we have 88 | \begin{equation*} 89 | \mat{E & F \\ F & G} = \mat{1 & 0 \\ 0 & 1} \qqand 90 | g =\dif{z^2} + \dif{\theta^2}, 91 | \end{equation*} 92 | so this is locally Euclidean. Thus the normal is 93 | \begin{equation*} 94 | \nn 95 | = \f{\sigma_\theta \cross \sigma_z}{\left\vert \sigma_{\theta} \cross \sigma_z \right\vert} 96 | = (\cos\theta, \sin\theta, 0) 97 | \end{equation*} 98 | Taking second derivatives gives 99 | \begin{equation*} 100 | \sigma_{\theta\theta} = (-\cos\theta, -\sin\theta,0) \qqand 101 | \sigma_{\theta z} = \sigma_{zz} = 0. 102 | \end{equation*} 103 | Thus our matrix is given by 104 | \begin{equation*} 105 | \mat{L & M \\ M & N} = \mat{-1 & 0 \\ 0 & 0} \qqand 106 | B_{\text{\emph{II}}} = \dif{\theta^2}. 107 | \end{equation*} 108 | \end{example} 109 | 110 | \begin{theorem} 111 | [Gauss' theorema egregium] If $g$ is the metric induced by $\sigma$, then 112 | \begin{equation*} 113 | K_p(g) 114 | = \f{\det(B_{\text{\emph{II}}}(\sigma))}{\det(B_I(\sigma))} 115 | = \f{LN-M^2}{EG-F^2}. 116 | \end{equation*} 117 | \end{theorem} 118 | 119 | \emph{See handout.} 120 | 121 | % subsection second_fundamental_form (end) 122 | 123 | % This was labelled section 7.4, but it's only 7.3 in LaTeX. Am I missing something? 124 | 125 | \subsection{Closed surfaces and charts} % (fold) 126 | \label{sub:closed_surfaces_and_charts} 127 | 128 | We have the following basic problem: 129 | \lecturemarker{16}{13 Mar} 130 | 131 | \textbf{Problem.} A compact surface $S$ (such as the sphere $S^{\,2}$) cannot be written as the image of a single map $\sigma: U\to S$, where $U$ is open in $\R^2$. 132 | 133 | This is actually a theorem, which can be proved using \emph{Algebraic Topology}. 134 | 135 | \emph{Solution.} Cover $S$ with open sets, each of which is parameterised. This gives us something close to what we want. We require the following definition: 136 | 137 | \begin{definition} 138 | If $S \subset \R^3$, a \emph{chart} for $S$ is an open set $V \subset S$ and a bijective map $f:V \xrightarrow{\text{open}} \R^2$ such that $\sigma = f^{-1}$ is a parametrisation. 139 | 140 | It might seem strange to think of the inverse of the map, but later it will be more convenient to think of charts in this way. 141 | 142 | If $f_i:V_i \to U_i$, $i=1,2$ are two charts on $S$, then the \emph{transition function} $\phi_{12}: f_1(V_1 \cap V_2) \to f_2 (V_1 \cap V_2)$ is given by $\phi_{12} = f_2 \circ f_1^{-1}$. 143 | 144 | We say that $f_1$ and $f_2$ are \emph{compatible} if $\phi_{12}$ and $\phi_{21} = \phi_{12}^{-1}$ are both differentiable. 145 | % \missingfigure{Geo 16/1} 146 | \end{definition} 147 | 148 | This might seem like a strange statement to make, because after some algebra we can prove that it always holds for embedded surfaces (the only surfaces that we've been considering). But later, when we consider abstract surfaces, this will turn out to be very useful. 149 | 150 | \begin{definition} 151 | An \emph{atlas} for $S\subset \R^3$ is a set of compatible charts $f_i: V_i \to U_i$ such that the $V_i$ covers $S$. We say $S$ is an \emph{embedded surface} in $\R^3$ if it has an atlas. 152 | \end{definition} 153 | 154 | \begin{example} 155 | An atlas for $S^{\,2} \subset \R^3$ is 156 | \begin{itemize} 157 | \shortskip 158 | \item $\pi_1:S^2-\{N\} \to \R^2$ is stereographic projection from the north pole $N$; 159 | \item $\pi_2:S^2-\{S\} \to \R^2$ is stereographic projection from the north pole $S$. 160 | \end{itemize} 161 | We treat $\R^2-\{0\}$ as $\C^*$, and then our transition function is 162 | \begin{equation*} 163 | \fullfunction{\phi_{12}}{\C^*}{\C^*}{z}{1/\overline{z}}. 164 | \end{equation*} 165 | \end{example} 166 | 167 | \subsubsection*{Metrics} % (fold) 168 | \label{ssub:metrics} 169 | 170 | If $f_1,f_2$ are compatible charts on $S$, then $f_i^{-1}$ induces a Riemannian metric $g_i$ on $U_i$,, given by 171 | \begin{equation*} 172 | g_i(\vv,\ww) = (\dif{f_i})^{-1}(\vv) \cdot (\dif{f_i^{-1}})(\ww). 173 | \end{equation*} 174 | 175 | \begin{lemma} 176 | $\phi_{12}:(f_1(V_1 \cap V_2), g_1)) \to (f_2(V_1 \cap V_2), g_2)$ is an isometry. 177 | \end{lemma} 178 | 179 | \begin{proof} 180 | Working through the algebra: 181 | \begin{align*} 182 | g_2(\dif{\phi_{12}(\vv)}, \dif{\phi_{12}(\ww)}) 183 | &= (\dif{f_2})^{-1} (\dif{f_2} \circ (\dif{f_1})^{-1} (\vv)) \cdot \dif{f_2^{-1}}(\dif{f_2} \cdot (\dif{f_1})^{-1}(\ww)) \\ 184 | &= \dif{f_1^{-1}}(\vv) \cdot \dif{f_1^{-1}}(\ww) \\ 185 | &= g_1(\vv,\ww). \qedhere 186 | \end{align*} 187 | \end{proof} 188 | 189 | If $S\subset \R^3$ is a smoothly embedded surface, and $p\in S$, then the Gauss curvature is given by $K_p(S) := K_{f(p)}(g)$, where $f:V\to U$ is a chart defined in a neighbourhood of $p$ and $g$ is the metric induced on $f$. 190 | 191 | The lemma implies that this is well-defined. 192 | 193 | Similarly, $\gamma:(a,b) \to S$ is a geodesic if $f\circ \gamma$ is a geodesic with respect to the metric $g$ induced by $f$, where $f$ is any chart of $S$. 194 | 195 | % subsubsection metrics (end) 196 | 197 | % subsection closed_surfaces_and_charts (end) 198 | 199 | \subsection{Abstract surfaces} % (fold) 200 | \label{sub:abstract_surfaces} 201 | 202 | Suppose $S$ is a Hausdorff, second-countable topological space. A chart on $S$ is an open set $V\subset S$ and a bijective map $f:V\to U \subset \R^2$, with $U$ open. (Don't worry if some of these terms are unfamiliar; they will be introduced formally in \emph{Metric \& Topological Spaces}. They are cited here merely for completeness.) Many definitions are the same as with closed surfaces: 203 | 204 | If $f_1,f_2$ are charts on $S$, then the transition function $\phi_{12}:f_1(V_1 \cap V_2) \to f_2(V_1 \cap V_2)$ is given by $\phi_{12} = f_2 \circ f_1^{-1}$. 205 | 206 | We say that $f_1$ and $f_2$ are compatible if $\phi_{12}$ are differentiable. This definition is exactly the same as before, but now it has teeth. In the embedded case, we merely need to ask that $f_1^{-1}$ be differentiable for $f_1$ and $f_2$ to be compatible. In this case, it doesn't make sense to ask that $f_1^{-1}$ be differentiable, so this is actually a useful distinction to make. 207 | 208 | An atlas on $S$ is a set of compatible charts $f_i:V_i \to U_i$ such that the $V_i$ cover all of $S$. 209 | 210 | \begin{definition} 211 | An \emph{abstract smooth surface} is a space $S$ as above together with an atlas on $S$. 212 | \end{definition} 213 | 214 | In some sense, there's nothing special about two dimensions in this definition. We could similarly define an abstract smooth $n$-manifold. Some other properties aren't so nice though. There are some four-manifolds which don't admit any structure as a smooth manifold, whereas $\R^4$ can be made into a smooth manifold in uncountably many ways. 215 | 216 | In almost all cases, it is better to think about smooth manifolds, but these are not discussed in this course. We mention them here only for completeness, and will henceforth restrict our discussion to surfaces. 217 | 218 | \begin{example} 219 | Consider the torus $T^2=\R^2/\Z^2$. There is a projection map $\pi:\R^2\to T^2$. 220 | Charts on $T^2$ are inverses of maps $\pi_U:U \to T^2$, the restriction of $\pi$ to an open set $U=B_\epsilon(p)$, $\epsilon<1/2$. 221 | % 222 | % ensure that ball does not overlap itself when projected 223 | % inverse of map gives a triangle 224 | % 225 | Transition functions are translations by $(n,m)\iZ^2$. 226 | \end{example} 227 | 228 | \begin{definition} 229 | If $\left\{f_i:V_i \to U_i\right\}$ is an atlas on an abstract surface $S$, then a Riemannian metric on $S$ is a set of metrics $g_i$ on $U_i$ so that the transition functions $\phi_{ij}(f_i(V_i \cap V_j), g_i) \to (f_j(V_i \cap V_j), g_j)$ are all isometries. 230 | \end{definition} 231 | 232 | In an embedded surfaces, we get these as isometries for free. Here, we have to include it as part of the definition. 233 | 234 | \begin{example} 235 | The flat metric on $T^2$ is defined by taking the atlas in the previous example, and equipping each $U$ with the Euclidean metric $\dif{x^2} + \dif{y^2}$. The transition functions are all translations, so isometries under $g^E$. The Gauss curvature of $g$ is identically zero. 236 | 237 | However, there is no way to embed $T^2$ into $\R^3$ such that the Gauss curvature is identically zero (see examples sheet). 238 | 239 | In some sense, it is better to think of this embedding as treating $T^2$ as the quotient of $(\R^2, g^E)$, by the action of a group of isometries. 240 | \end{example} 241 | 242 | Here the phrase \emph{flat metric} is used to describe a surface (or manifold) with identically zero Gauss curvature. 243 | 244 | \begin{example} 245 | The Möbius strip also has a flat metric. The strip is given by $M=\R\times (-1,1)/G$, where $G\cong \Z$ and $K\cdot(X,Y) = (x+k, \left( -1 \right)^k + y)$. (Take the two sides of an infinite strip and glue them together with a strip, as we illustrated previously.) Again, this is an isometry of the Euclidean metric. 246 | \end{example} 247 | 248 | % subsection abstract_surfaces (end) 249 | 250 | \pagebreak 251 | 252 | \subsection{Global Gauss-Bonnet} % (fold) 253 | \label{sub:global_gauss_bonnet} 254 | 255 | This leads us to the final theorem of the course, generalising the local Gauss-Bonnet theorem we saw previously: 256 | 257 | \begin{theorem} 258 | If $(S,g)$ is a compact abstract surface equipped with a Riemannian metric $g$, then 259 | \begin{equation*} 260 | 2\pi \,\chi(S) = \int_S K(g) \dif{A_g}. 261 | \end{equation*} 262 | \end{theorem} 263 | 264 | There are all sorts of beautiful theorems like this, which relate global topological information to local properties. This is not an isolated example, although it is the only such theorem we study in this course. 265 | 266 | \begin{example} 267 | Take $S=S^{\,2}$ and let $g=g^{\,S}$ be the spherical metric. We know Gaussian curvature is $K \equiv 1$. Then 268 | \begin{equation*} 269 | 2\pi \cdot 2 = \int_{S^2} 1 \dif{A} = 4\pi, 270 | \end{equation*} 271 | and everything is consistent. 272 | \end{example} 273 | 274 | The idea behind the theorem is quite easy. Technical details are needed to make it into a complete proof; here we present the main ideas. 275 | 276 | \begin{proof} 277 | [Sketch proof] Find a geodesic triangulation of $S$ (that is, a triangulation where edges are geodesics), and so that each face is contained in a chart. 278 | % 279 | The idea is to start with any triangulation, and subdivide the edges, replacing small edges by geodesics. 280 | 281 | \begin{center} 282 | \begin{tikzpicture}[scale=2.5] 283 | \draw [thick] plot [smooth] coordinates {(-1,0) (-0.9, 0.171) (-0.8, 0.288) (-0.7, 0.357) (-0.6, 0.384) (-0.5, 0.375) (-0.4, 0.336) (-0.3, 0.273) (-0.2, 0.192) (-0.1, 0.099) (0,0) (0.1, -0.099) (0.2, -0.192) (0.3, -0.273) (0.4, -0.336) (0.5, -0.375) (0.6, -0.384) (0.7, -0.357) (0.8, -0.288) (0.9, -0.171) (1,0)}; % 284 | 285 | \foreach \s/\t/\u/\v in {-1/0/-0.7/0.357, -0.7/0.357/-0.4/0.336, -0.4/0.336/0.1/-0.099, 0.1/-0.099/0.7/-0.357, 0.7/-0.357/1/0} 286 | { 287 | \draw (\s,\t) -- (\u,\v); 288 | \draw (\s,\t) node {$\bullet$}; 289 | } 290 | 291 | \draw (1,0) node {$\bullet$}; 292 | 293 | \end{tikzpicture} 294 | \end{center} 295 | 296 | Importantly, this does not change the topology of the triangulation. 297 | 298 | Now suppose triangulation has $V$ vertices, $E$ edges and $F$ faces. We know that $E=\f{3}{2}F$ (recall our discussion of the Euler characteristic for the sphere). Then 299 | \begin{align*} 300 | \iint_S K \dif{A_g} 301 | &= \sum_{i=1}^F \iint_{f_i} K \dif{A_g} \\ 302 | \intertext{where $f_i$ is the $i$th face. Then we apply local Gauss-Bonnet, and letting $\alpha_{ij}$, $j=1,2,3$ be the angles in $f_i$:} 303 | &= \sum_{i=1}^F \defect(f_i) \\ 304 | &= \sum_{i=1}^F \left( \alpha_{i1} + \alpha_{i2} + \alpha_{i3} - \pi \right) \\ 305 | &= \sum_{i,j} \alpha_{ij} - \pi F 306 | = 2\pi V - \pi F 307 | = 2\pi\left( V-E+F \right). \qedhere 308 | \end{align*} 309 | \end{proof} 310 | 311 | % subsection global_gauss_bonnet (end) 312 | -------------------------------------------------------------------------------- /ib-met-top-spaces/mettop-chap-03.tex: -------------------------------------------------------------------------------- 1 | %!TEX root = met-top-spaces.tex 2 | \stepcounter{lecture} 3 | \setcounter{lecture}{3} 4 | \sektion{Connectedness} 5 | 6 | \subsection{Basic notions} % (fold) 7 | \label{sub:basic_notions} 8 | 9 | \begin{definition} 10 | A topological space $X$ is \emph{disconnected} if there are non-empty open subsets $U, V$ with $U\cap V = \emptyset$ and $X=U\cup V$. Otherwise $X$ is called \emph{connected}. 11 | 12 | In other words, $X$ is connected if and only if, given open sets $U,V$ with $U \cap V = \emptyset$ and $X=U\cup V$, either $U=\emptyset$ or $V=\emptyset$ (equivalently, either $U=X$ or $V=X$). 13 | \end{definition} 14 | 15 | Clearly connectedness is a topological property. We can extend this definition to subsets: 16 | 17 | \begin{definition} 18 | If $Y$ is a subset of the topological space $X$, then $Y$ is disconnected in the subspace topology if and only if there are $U, V$ open in $X$ such that $U\cap V=\emptyset$, $V\cap Y\neq \emptyset$ but $U\cap V\cap Y=\emptyset$ and $Y\subset U\cup V$. In this case, we say that $U$ and $V$ \emph{disconnect} $Y$. 19 | \end{definition} 20 | 21 | \begin{proposition} 22 | Let $X$ be a topological space. Then the following are equivalent: \label{prop:equivalent-properties} 23 | \begin{enumerate} 24 | \shortskip 25 | \item $X$ is connected; 26 | \item The only subsets of $X$ that are both open and closed are $\emptyset$ and $X$; 27 | \item Every continuous function $f:X\to\Z$ is constant. 28 | \end{enumerate} 29 | \end{proposition} 30 | 31 | \begin{proof} 32 | $\text{(i)} \iff \text{(ii)}$ is trivial, since for $U\subset X$, we consider that $U \cup (X\backslash U) = X$. 33 | 34 | $\text{(i)} \implies \text{(iii)}$. Suppose there is a non-constant map $f:X\to\Z$, then there exists $mm\})$ are open, non-empty sets disconnecting $X$; that is, $X$ is \emph{not} connected. 35 | 36 | $\text{(iii)} \implies \text{(i)}$. Suppose that $X$ is not connected; that is, there are non-empty, disjoint open $U$, $V$ such that $X=U\cup V$. Consider the map $f:X\to\Z$ defined by % 37 | \begin{equation*} 38 | f(x) = 39 | \begin{cases} 40 | 0 & \text{if } x\in U, \\ % 41 | 1 & \text{if } x\in V. 42 | \end{cases} 43 | \end{equation*} 44 | This is continuous (even locally constant) function, but not globally constant. 45 | \end{proof} 46 | 47 | \begin{proposition} 48 | A continuous image of a connected space is connected. 49 | \end{proposition} 50 | 51 | \begin{proof} 52 | If $f:X\to Y$ is a surjective continuous map of topological spaces and if $U, V$ disconnect $Y$, then their pre-images $f^{-1}(U), f^{-1}(V)$ disconnect $X$. Thus, if $X$ is connected, then so is $Y$. 53 | \end{proof} 54 | 55 | \pagebreak 56 | 57 | \subsubsection*{Connectedness in R} % (fold) 58 | \label{ssub:connectedness_in_r} 59 | 60 | \vspace{9pt} 61 | 62 | \begin{definition} 63 | A set $I\subseteq \R$ is called an \emph{interval} if, given $x,z\in I$ with $x\leq z$, we have $y \in I$ for all $y$ such that $x\leq y\leq z$. 64 | 65 | We have the cases $\inf I=a\iR$ and $a\in I$, or $a\not\in I$, or $\inf I = -\infty$. Similarly, we have the cases $\sup I=b\iR$ and $b\in I$, or $b\not\in I$, or $\sup I = \infty$. Thus any interval $I$ takes the form $[a,b]$, $[a,b)$, $(a,b]$, $[a,\infty)$, $(a,\infty)$, $(-\infty,b]$, $(-\infty,b)$ or $(-\infty,\infty)$. 66 | \end{definition} 67 | 68 | \begin{theorem} 69 | A subset of $\R$ is connected if and only if it is an interval. \label{thm:cnncted-R-interval} 70 | \end{theorem} 71 | 72 | \begin{proof} 73 | First suppose that $X\subseteq\R$ is not an interval. Then we can find $x0$ with $(s-\delta,s+\delta) \subset U$. Hence there is $s\p \in [u,v]\cap U$ with $s\p>s$, contradicting $s$ as an upper bound. 78 | 79 | If $s\in V$, then there exists $\delta$ such that $(s-\delta,s+\delta) \subset V$. Then $(s-\delta,s+\delta) \cap U = \emptyset$ implies $[u,v] \cap U \subset [u,s-\delta]$, which contradicts $s$ as a \emph{least} upper bound. Thus $I$ is connected. 80 | \end{proof} 81 | 82 | \begin{corollary} 83 | [Intermediate value theorem] Let $a0$ for all $t0$ such that $B(y,\delta) \subseteq X$. Then for all $z\in B(y,\delta)$, we have $y\sim z$, by taking the straight line segment. Transitivity implies that $x\sim z$ for all $z\in B(y,\delta)$, and thus $B(y,\delta) \subset U$, and $U$ is open. 306 | 307 | Similarly, $X\backslash U$ is open. Suppose $y\in X\backslash U$. Since $X$ is open, there exists $\delta>0$ such that $B(y,\delta) \subseteq X$. For $z\in B(y,\delta)$, we have $y\sim z$ as above, and then $x \not\sim z$, hence $B(y,\delta) \subsetneq X\backslash U$. 308 | 309 | Since $X$ is connected, we must have $X\backslash U=\emptyset$ and $U=X$. Hence $X$ is path-connected, since $x\sim y$ for all $y\in X$. 310 | \end{proof} 311 | 312 | % subsection path_connectedness (end) 313 | 314 | \pagebreak 315 | 316 | \subsection{Products of connected spaces} % (fold) 317 | \label{sub:products_of_connected_spaces} 318 | 319 | \begin{proposition} 320 | \label{prop:3-path-connected-product-topology} 321 | Let $X$ and $Y$ be topological spaces. If $X$ and $Y$ are path-connected, then so too is $X\times Y$, with the product topology. 322 | \end{proposition} 323 | 324 | \begin{proof} 325 | Given $(x_1,y_1)$ and $(x_2,y_2)\in X\times Y$, we know that there are paths $\gamma_1:[0,1]\to X$ and $\gamma_2:[0,1]\to Y$ with $\gamma_1(0)=x_1$, $\gamma_1(1)=x_2$, $\gamma_2(0)=y_1$ and $\gamma_2(1)=y_2$. 326 | 327 | Now define a map $\gamma:[0,1]\to X\times Y$ by $\gamma(t) = (\gamma_1(t),\gamma_2(t))$. % We claim that this $\gamma$ is continuous. 328 | % 329 | The base for the topology on $X\times Y$ consists of the open sets $U\times V$, with $U$ open in $X$ and $V$ open in $Y$. So it is sufficient to prove that $\gamma^{-1}(U\times V)$ us open for all such $U$ and $V$. But $\gamma^{-1}(U\times V)=\gamma^{-1}(U) \cap \gamma^{-1}(V)$ is clearly open. So $\gamma$ is continuous and defines a path from $(x_1,y_1)$ to $(x_2,y_2)$. 330 | \end{proof} 331 | 332 | \begin{proposition} 333 | If $X$ and $Y$ are connected, then so too is $X \times Y$ with the product topology. \label{prop:connected-products} 334 | \end{proposition} 335 | 336 | \vspace{-8pt} 337 | 338 | First we make some general comments about product topologies. Given $y\in Y$, the set $X\times\{y\}$ with the subspace topology is homeomorphic to $X$, using the projection map $\pi_1:X \times \{y\} \to X$. We already know that this map is a continuous bijection. 339 | 340 | However, a base for the topology on $X \times Y$ consists of open sets $U\times V$, with $U$ open in $X$ and $V$ open in $Y$. This implies that a base for the subspace topology on $X\times\{y\}$ consists of subsets $U\times\{y\}$, for $U$ open subsets of $X$. Thus, under $\eval[0]{\pi_1}_{X\times\{y\}}$, open sets do correspond, and hence $\pi_1:X\times \{y\} \to X$ is a homeomorphism. 341 | 342 | Similarly, for $x\in X$, $\{x\} \times X$ is homeomorphic to $Y$, and so $X \times \{y\}$ is connected for all $y\in Y$. Thus $\{x\} \times X$ is connected for all $x\in X$. 343 | 344 | \begin{proof} 345 | [Proof of proposition~\ref{prop:connected-products}] Given a continuous function $f:X \times Y \to \Z$, it is obvious that $f$ is constant on each slice $\{x\} \times Y$ and $X \times \{y\}$, by connectedness. 346 | 347 | \begin{center} 348 | \begin{tikzpicture} 349 | 350 | \bigarrow (0,0) -- (4.5,0) node [below=2pt] {$X$}; 351 | \bigarrow (0,0) -- (0,4.5) node [left=2pt] {$Y$}; 352 | 353 | \foreach \s in {1,2} { 354 | \draw [dashed] (\s*1.5,0) node [below=3pt] {$x_\s$} -- (\s*1.5,4.2); 355 | \draw [dashed] (0,\s*1.5) node [left=2pt] {$y_\s$} -- (4.2,\s*1.5); 356 | \draw (\s*1.5,\s*1.5) node {$\bullet$}; 357 | } 358 | 359 | \draw (1.5,1.5) node [below left] {$(x_1,y_1)$}; 360 | \draw (3,3) node [above right] {$(x_2,y_2)$}; 361 | 362 | \draw (0,0) node [below left=2pt] {$0$}; 363 | \end{tikzpicture} 364 | \end{center} 365 | 366 | Given arbitary points $(x_1,y_1), (x_2,y_2) \in X\times Y$, we deduce that $f(x_1,y_1) = f(x_1,y_2) = f(x_2,y_2)$ (see diagram). Hence $f$ is constant on $X\times Y$ and $X\times Y$ is connected. 367 | \end{proof} 368 | 369 | \vspace{3pt} 370 | 371 | \begin{remark} 372 | A similar argument also proves proposition \ref{prop:3-path-connected-product-topology}: there is a path joining $(x_1,y_1)$ to $(x_1,y_2)$ and a path joining $(x_1,y_2)$ to $(x_2,y-2)$ in $X\times Y$. 373 | \end{remark} 374 | 375 | % subsection products_of_connected_spaces (end) -------------------------------------------------------------------------------- /ib-met-top-spaces/mettop-chap-04.tex: -------------------------------------------------------------------------------- 1 | %!TEX root = met-top-spaces.tex 2 | \stepcounter{lecture} 3 | \setcounter{lecture}{4} 4 | \sektion{Compactness} 5 | 6 | \subsection{Basic notions} % (fold) 7 | \label{sub:compact:basic_notions} 8 | 9 | \begin{definition} 10 | Let $(X,\tau)$ be a topological space. An \emph{open cover of $X$} is a collection of open subsets $\cover=\{U_\gamma:\gamma\in\Gamma\}$ such that $X=\bigcup_{\gamma\in\Gamma} U_\gamma$. 11 | 12 | If $Y\subset X$, then an open cover of $Y$ is a collection of open subsets in $X$ $\cover=\{U_\gamma:\gamma\in\Gamma\}$ such that $Y\subset\bigcup_{\gamma\in\Gamma} U_\gamma$. 13 | \end{definition} 14 | 15 | \vspace{3pt} 16 | 17 | \begin{remark} 18 | Such an open cover of $Y$ provides a base of open sets $\cover = \{U_\gamma \cap Y: \gamma\in\Gamma\}$ for $Y$ with the subspace topology, and conversely. 19 | \end{remark} 20 | 21 | \vspace{-3pt} 22 | 23 | \begin{definition} 24 | A \emph{subcover} of an open cover $\cover$ is a subcollection $\subcover \subseteq \cover$ which is still an open cover of $Y$. 25 | \end{definition} 26 | 27 | \begin{example} 28 | The intervals $I_n=(-n,n)$, where $n=1,2,\ldots$, form an open cover of $\R$, and $I_{n^2}$ is a proper subcover. The intervals $J_n = (n-1,n+1)$, where $n\iZ$, form an open cover of $\R$ with no proper subcover. 29 | \end{example} 30 | 31 | \begin{definition} 32 | A topological space $(X,\tau)$ is \emph{compact} if every open cover has a finite subcover. 33 | \end{definition} 34 | 35 | \begin{examples} 36 | In this sense, $\R$ is not compact, as the open covers described above have no finite subcovers. Any finite topological space is compact, as is any set with the indiscrete topology (the only open subsets of $X$ being $\emptyset$ and $X$), or with the cofinite topology. 37 | \end{examples} 38 | 39 | With this definition, compactness is a topological property. 40 | 41 | \begin{lemma} 42 | Let $(X,\tau)$ be a topological space with $Y\subset X$. Then $Y$ is compact in the subspace topology if and only if every open cover $\{U_\gamma\}$ of $Y$ has a finite subcover. 43 | \end{lemma} 44 | 45 | \begin{proof} 46 | First let $\{U_\gamma:\gamma \in \Gamma\}$ be an open cover of $Y$, then $Y = \bigcup_{\gamma\in\Gamma} (U_\gamma \cap Y)$, where the $U_\gamma \cap Y$ are open in $Y$. Since $Y$ is compact, there exist $\gamma_1,\ldots,\gamma_n \in \Gamma$ such that $Y=\bigcup_{i=1}^n (U_{\gamma_i}\cap Y)$, and then $\{U_{\gamma_i}: i=1,\ldots,n\}$ covers $Y$. 47 | 48 | Now the converse. Suppose $Y=\bigcup_{\gamma\in\Gamma} V_\gamma$, where the $V_\gamma$ are open in $Y$. Write $V_\gamma = U_\gamma \cap Y$, where the $U_\gamma$ are open in $X$ and form an open cover of $Y$. Then there exist $\gamma_1,\ldots,\gamma_n \in \Gamma$ such that $Y \subseteq \bigcup_{i=1}^n U_{\gamma_i}$, and hence $Y=\bigcup_{i=1}^n V_{\gamma_i}$. 49 | \end{proof} 50 | 51 | \begin{example} 52 | The open interval $(0,1)$ is not compact: consider the open cover by intervals $(1/n,1-1/n$ for $n=3,4,\ldots$. 53 | \end{example} 54 | 55 | \pagebreak 56 | 57 | \begin{theorem} 58 | [Heine-Borel theorem] The closed interval $[a,b] \subset \R$ is compact. 59 | \end{theorem} 60 | 61 | \begin{proof} 62 | Let $[a,b] \subset \bigcup_{\gamma\in\Gamma} U_\gamma$, for $U_\gamma$ open in $\R$. Then set 63 | \begin{equation*} 64 | K = \left\{x\in[a,b]: \text{$[a,x]$ is contained in a finite union of the $U_\gamma$}\right\}. 65 | \end{equation*} 66 | Clearly $a\in K$, so $K\neq \emptyset$. Let $r=\sup K$. Then $r\in[a,b]$, and so $r\in U_{\gamma_1}$ for some $\gamma_1 \in \Gamma$. Since $U_{\gamma_1}$ is open, there exists $\delta>0$ such that $[r-\delta,r+\delta] \subseteq U_{\gamma_1}$. 67 | 68 | \begin{center} 69 | \begin{tikzpicture}[xscale=2,scale=3] 70 | 71 | \draw (-0.2,0) -- (1.2,0); 72 | \draw [very thick, |-|] (0,0) node [below=5pt] {$0$} -- (1,0) node [below=5pt] {$1$}; 73 | 74 | \draw [thick, |-|] (0,0) -- (0.6,0) node [below=7.2pt] {\small{$r$}}; 75 | 76 | \draw [thick, |-|] (0.45,0) node [below=5pt] {\small{$r-\delta$}} -- (0.75,0) node [below=5pt] {\small{$r+\delta$}}; 77 | 78 | \draw [ultra thick, (-] (0.4,0) -- (0.41,0); 79 | \draw [ultra thick, -)] (0.94,0) -- (0.95,0); 80 | 81 | \bigarrow [dashed, <->] (0.4,0.2) -- (0.95,0.2); 82 | \bigarrow [dashed] (0.95,0.2) -- (0.4,0.2); 83 | \draw (0.675,0.2) node [above] {$U_{\gamma_1}$}; 84 | 85 | \end{tikzpicture} 86 | \end{center} 87 | 88 | By the definition of $r$, there exists $c\in[r-\delta,r]$ such that $[a,c]$ is contained in a finite union of the $U_\gamma$. Hence, the same is true for $[a,r+\delta] \cap [a,b]$ (we just need to include $U_{\gamma_1}$ also). But this contradicts $r$ as an upper bound, unless $r=b$, in which case, the above argument tells us that $[a,b]$ is covered by finitely many of the $U_\gamma$. (There exists $c\in[b-\delta,b]$ such that $[a,c]$ is covered by finitely many $U_\gamma$, and include $U_{\gamma_1}$ also.) 89 | 90 | Thus $[a,b]$ is compact. 91 | \end{proof} 92 | 93 | \begin{proposition} 94 | A continuous image of a compact set is compact. \label{prop:cts-compact} 95 | \end{proposition} 96 | 97 | \begin{proof} 98 | Suppose $f:(X,\tau)\to(Y,\sigma)$ is a continuous map of topological spaces, and $K\subseteq X$ is compact. Then we wish to show that $f(K)$ is compact. 99 | 100 | Suppose $f(K)\subseteq\bigcup_{\gamma\in\Gamma} U_\gamma$, with $U_\gamma$ open in $Y$. Since $f$ is continuous, $K\subset\bigcup_{\gamma\in\Gamma} f^{-1}(U_\gamma)$, and each $f^{-1}(U_\gamma)$ is open in $X$. Since $K$ is compact, there exist $\gamma_1,\ldots,\gamma_n \in \Gamma$ such that $K\subseteq\bigcup_{i=1}^n f^{-1}(U_{\gamma_i})$, and hence $f(K)\subseteq\bigcup_{i=1}^n U_{\gamma_i}$. % 101 | \end{proof} 102 | 103 | \begin{proposition} 104 | A closed subset of a compact topological space $X$ is compact. \label{prop:closed-compact} 105 | \end{proposition} 106 | 107 | \begin{proof} 108 | Let $X$ be a compact topological space, and $K\subseteq X$ be closed. If $K=\emptyset$ then this is trivial, so assume not. Suppose $K\subseteq \bigcup_{\gamma\in\Gamma} U_\gamma$, where the $U_\gamma$ are open in $X$. Then $X=(X\backslash K)\cup\left( \bigcup_{\gamma\in\Gamma}U_\gamma \right)$, where $X\backslash K$ is also open. Since $X$ is compact, there is a finite subcover, and hence there exists $\gamma_1,\ldots,\gamma_n \in \Gamma$ such that $X=(X\backslash K) \cup \left( \bigcup_{i=1}^n U_{\gamma_i} \right)$. Thus $K\subseteq \bigcup_{i=1}^n U_{\gamma_i}$. 109 | \end{proof} 110 | 111 | \begin{proposition} 112 | Every compact subset of a Hausdorff topological space is closed. \label{prop:compact-hausdorff-closed} 113 | \end{proposition} 114 | 115 | \begin{proof} 116 | Let $X$ be a Hausdorff topological space, and $K \subseteq X$ be compact. If $K=X$, this is trivial, so suppose not. Then we show that $X\backslash K$ is open. 117 | 118 | 119 | Given $x\in X\backslash K$, for any $y\in K$< there are disjoint open sets with $U_y \ni x$ and $V_y \ni y$. Now $\{V_u: y\in K\}$ is an open cover of $K$. Since $K$ is compact, there exist $y_1,\ldots,y_n \in K$ such that $K \subseteq \bigcup_{i=1}^n V_{y_i}$. Then $U=\bigcap_{i=1}^n U_{y_i}$ is an open neighbourhood of $x$ with $U\cap K = \emptyset$, so $U\subseteq X\backslash K$, as required. 120 | \end{proof} 121 | 122 | \pagebreak 123 | 124 | \begin{corollary} 125 | A set $X\subset\R$ is compact if and only if it is closed and bounded. \label{cor:comp-iff-clsed-bded} 126 | \end{corollary} 127 | 128 | \begin{proof} 129 | If $X\subset\R$ is compact, then proposition~\ref{prop:compact-hausdorff-closed} tells us that $X$ is closed (since $\R$ is Hausdorff). It is also bounded: if not, the open sets $U_m=\left\{x\iR: \left\vert x \right\vert>}}}, 183 | postaction={decorate}, 184 | shorten >=0.4pt]} 185 | 186 | \vspace{3pt} 187 | 188 | \begin{center} 189 | \begin{minipage}{0.5\textwidth} 190 | \centering 191 | \begin{tikzpicture}[scale=2.25] 192 | 193 | \draw (0,0) rectangle (1,1); 194 | 195 | \foreach \s in {0,1} { 196 | \bigarrow (0,\s) -- (0.5,\s); 197 | \biggarrow (\s,0) -- (\s,0.6); 198 | } 199 | 200 | \end{tikzpicture} 201 | \end{minipage} 202 | \begin{minipage}{0.45\textwidth} 203 | \centering 204 | \begin{tikzpicture}[scale=0.2, yscale=0.5] 205 | 206 | \draw [thick] (0,0) circle (9.7 and 9); 207 | 208 | \draw (0,3.3) .. controls (3,3.3) and (5,2) .. (5.4,1.2); 209 | \draw (0,-0.5) .. controls (3,-0.5) and (5,0.5) .. (5.4,1.2); 210 | 211 | \draw (0,3.3) .. controls (-3,3.3) and (-5,2) .. (-5.4,1.2); 212 | \draw (0,-0.5) .. controls (-3,-0.5) and (-5,0.5) .. (-5.4,1.2); 213 | 214 | \draw [thin, gray] (5.4,1.2) .. controls (5.7,1.5) and (6.2,2.9) .. (7.5,3.3); 215 | \draw [thin, gray] (-5.4,1.2) .. controls (-5.7,1.5) and (-6.2,2.9) .. (-7.5,3.3); 216 | \end{tikzpicture} 217 | \end{minipage} 218 | \end{center} 219 | 220 | \begin{center} 221 | \begin{minipage}{0.5\textwidth} 222 | \centering 223 | the 2-D torus $\R^2/\Z^2$ 224 | \end{minipage} 225 | \begin{minipage}{0.45\textwidth} 226 | \centering 227 | the embedded torus $X\subseteq \R^3$ 228 | \end{minipage} 229 | \end{center} 230 | 231 | In \emph{Analysis I}, you learnt that continuous real-valued functions on $[a,b]\subseteq\R$ are bounded and attain their bounds. Here, we are using the compactness of $[a,b]$. This statement is still true, for instance, for continuous real-value functions on the Cantor set. 232 | \end{example} 233 | 234 | \begin{proposition} 235 | Continuous real-valued functions on a compact space $X$ are bounded and attain their bounds. 236 | \end{proposition} 237 | 238 | \begin{proof} 239 | Suppose $X$ is compact and $f:X\to\R$ is continuous. Then proposition~\ref{prop:cts-compact} tells us that $f(X)$ is compact, and so corollary~\ref{cor:compact-clsd-bded} tells us that $f(X)$ is bounded and closed. 240 | 241 | Since $f(X)$ is closed, it contains all of its accumulation points. But $\sup f(X)$ and $\inf f(X)$ (which exist because $f(X)$ is bounded) are accumilation points for $f(X)$, so $\sup f(X), \inf f(X) \in f(X)$, which gives the desire result. 242 | \end{proof} 243 | 244 | \begin{theorem} 245 | The product of two compact spaces is compact. 246 | \end{theorem} 247 | 248 | \begin{proof} 249 | Suppose $X$ and $Y$ are compact and that $X\times Y=\bigcup_{\gamma\in\Gamma} U_\gamma$, where the $U_\gamma$ are open in $X\times Y$. 250 | 251 | By the definition of the product topology, each $U_\gamma$ is the union of ``basic open sets'' of the form $V \times W$, where $V$ is open in $X$ and $W$ is open in $Y$. Thus 252 | \begin{equation*} 253 | X\times Y=\bigcup_{\delta\in\Delta} V_\delta \times W_\delta, 254 | \end{equation*} 255 | with $V_\delta$ open in $X$, $W_\delta$ open in $Y$ and $V_\delta \times W_\delta$ a subset of some $U_\gamma$. 256 | 257 | Now let $x\in X$, and then we have 258 | \begin{equation*} 259 | \{x\} \times Y \subseteq \bigcup_{\delta\in\Delta} V_\delta \times W_\delta, 260 | \end{equation*} 261 | such that $x\in V_\delta$. 262 | 263 | Now, since $Y$ is compact, there exist $\delta_1,\ldots,\delta_m$ such that $Y=\bigcup_{i=1}^m W_{\delta_i}$. 264 | 265 | Then let $V_x = \bigcap_{i=1}^n V_{\delta_i}$ be an open neighbourhood of $x$ such that $V_x \times Y \subseteq \bigcap_{i=1}^n V_{\delta_i} \times W_{\delta_i}$. The $V_x$ obtained in this way form an open cover of $X$, and so there exist $x_1,\ldots,x_n$ such that $X = \bigcup_{j=1}^n V_{x_j}$. 266 | 267 | Now $X\times Y=\bigcup_{j=1}^n V_{x_j} \times Y$ and each $V_{x_j} \times Y$ has a finite cover by $V_\delta\times W_\delta$'s. Thus $X\times Y$ has a finite cover by such sets. Since each $V_\delta\times W_\delta$ is a subset of some $U_\gamma$, $X\times Y$ has a finite cover by $U_\gamma$'s. Thus $X\times Y$ is compact. 268 | \end{proof} 269 | 270 | \pagebreak 271 | 272 | \begin{remarks} 273 | Given topological spaces $X,Y,Z$, the product $X\times Y \times Z$ is homeomorphic to $X \times \left( Y\times Z \right)$ under the obvious map (since a base for the topology of $X\times Y \times Z$ consists of subsets $U\times V \times W$, and a base for the topology of $X \times \left( Y\times Z \right)$ consists of subsets $U\times \left( V\times W \right)$, for $U$ open in $X$, $V$ open in $Y$ and $W$ open in $Z$. Hence open sets in $X \times Y \times Z$ correspond to open sets in $X\times \left( Y\times Z \right)$.) By induction, the above theorem implies that the product of finitely many compact spaces is compact. 274 | 275 | Now applying corollary~\ref{cor:compact-clsd-bded}, we see that $\left[ -M,M \right]^n$ is a compact subset of $\Rn$. The proof of the same corollary may be extended to show that $X\subseteq \Rn$ is compact if and only if $X$ is closed and bounded. 276 | \end{remarks} 277 | 278 | \begin{proposition} 279 | Let $X$ be a compact metric space, $Y$ a metric space and $f:X \to Y$ a continuous map. Then $f$ is \emph{uniformly continuous}; that is, given $\epsilon>0$, there exists $\delta>0$ such that for all $x,y\in X$, $d_X(x,y) < \delta$ implies $d_Y(f(x),f(y))<\epsilon$. 280 | \end{proposition} 281 | 282 | \begin{proof} 283 | Since $f$ is continuous, for all $x\in X$, there exists $\delta_x$ such that $d_X(x,y) < 2\delta_x$ implies $d_Y(f(x),f(y)) < \epsilon/2$. Now let 284 | \begin{equation*} 285 | U_x = \left\{y: d_X(x,y) < \delta_x \right\}. 286 | \end{equation*} 287 | The $U_x$ form an open cover of $X$, and so there exist $x_1,\ldots,x_n$ such that $X=\bigcup_{i=1}^n U_{x_i}$. Let $\delta=\min\{\delta_{x_i}\}$. 288 | 289 | Suppose now $d_X(y,z) < \delta$; since the $U_{x_i}$ form a cover, we can find $x_i$ such that $d(y,x_i) < \delta_{x_i}$. Since $d_X(y,z) < \delta<\delta_{x_i}$, we deduce that $d_X(z,x_i) < \delta + \delta_i < 2\delta_i$ (from the triangle inequality). Thus 290 | \begin{equation*} 291 | d_Y(f(y),f(z)) 292 | < d_Y(f(y),f(x_i)) + d_Y(f(x_i),f(z)) 293 | < \epsilon/2 + \epsilon/2 294 | = \epsilon. \qedhere 295 | \end{equation*} 296 | \end{proof} 297 | 298 | % subsection compact:basic_notions (end) 299 | 300 | \subsection{Sequential compactness} % (fold) 301 | \label{sub:sequential_compactness} 302 | 303 | \begin{definition} 304 | A topological space is \emph{sequentially compact} if every sequence in $X$ has a convergent subsequence. 305 | \end{definition} 306 | 307 | \vspace{3pt} 308 | 309 | \begin{remark} 310 | For general topological spaces, the property of compactness and sequential compactness are independent; neither implies the other. 311 | \end{remark} 312 | 313 | \begin{proposition} 314 | Any compact metric space is sequentially compact. 315 | \end{proposition} 316 | 317 | \vspace{-6pt} 318 | 319 | Notice that this can be reduced to the Bolzano-Weierstrass theorem: namely, that any closed bounded subset of $\Rn$ is sequentially compact. 320 | 321 | \begin{proof} 322 | Suppose $(X,d)$ is a metric space and $(x_n)_{n=1}^\infty$ is a sequence in $X$ with no convergent subsequence (in particular, there are infinitely many distinct $x_n$). We claim that for all $x\in X$, there exists $\delta>0$ such that $d(x,x_n) < \delta$ for at most finitely many $n$. 323 | 324 | If not, then there exists $x\in X$ such that for all $m>0$ in $\N$, $d(x,x_n)<1/m$ for infinitely many $n$, and hence there is a subsequence of $(x_n)$ converging to $x$, which is a contradiction. 325 | 326 | For each $x$, pick such a $\delta=\delta(x)$ and let $U_x = \{y:d(x,y) < \delta\}$. Each $U_x$ contains $x_n$ for only finitely many $n$. But $\{U_x:x\in X\}$ is an open cover for $X$, for which no finite subcover can exist. Hence $X$ is not compact. 327 | \end{proof} 328 | 329 | \begin{exercise} 330 | Show directly that if $X\subseteq\Rn$ is sequentially compact, then it is bounded, closed and hence compact. (Bounded, since otherwise we can find $(x_n)$ such that $d(x_n,x_0) > n$, which implies that there is no convergent subsequence.) 331 | \end{exercise} 332 | 333 | More generally, we have: 334 | 335 | \begin{theorem} 336 | Suppose $(X,d)$ is a sequentially compact metric space. Then 337 | \begin{enumerate} 338 | \shortskip 339 | \item Given any $\epsilon>0$, there exists $x_1,\ldots,x_n$ such that $X=\bigcup_{i=1}^n B(x_i,\epsilon)$; 340 | \item Given any open cover $\cover$ of $X$, there exists $\epsilon>0$ such that for all $x\in X$, $B(x,\epsilon)$ is contained in some element $U$ of $\cover$. 341 | \item $(X,d)$ is compact. 342 | \end{enumerate} 343 | \end{theorem} 344 | 345 | \begin{proof} 346 | \mbox{} 347 | \begin{enumerate} 348 | \item Suppose not. Then by induction, we can construct a sequence $(x_n)$ in $X$ such that $d(x_m,x_n) \geq \epsilon$ for all $m\neq n$. Clearly such a sequence has no convergent subsequence, since no subsequence can satisfy the Cauchy condition. Contradiction. 349 | 350 | \item Suppose not. Then there exists an open cover $\cover$ of $X$ such that for all $n$, there exists $x_n\in X$ such that $B(x_n,1/n) \not \subseteq U$, for all $U \in \cover$. But $(x_n)$ has a subsequence $(x_{n(r)})$ tending to $x\in X$. So let $x\in U_0$, for some $U_0 \in \cover$. Since $U_0$ is open, there exists $m>0$ such that $B(x,2/m) \subseteq U_0$. 351 | 352 | \begin{center} 353 | \begin{tikzpicture} 354 | 355 | \foreach \s/\t in {0/0, -0.25/0.2} {\fill [black] (\s,\t) circle (1.5pt);} 356 | 357 | \draw (0,0) node [right] {$x$}; 358 | \draw (-0.25,0.2) node [above] {$x_{n(r)}$}; 359 | 360 | \draw [thick] (0,0) circle (2); 361 | \draw [dashed] (0,0) circle (1.05); 362 | 363 | \draw (-0.25,0.2) circle (0.9); 364 | 365 | \draw (1.41,1.41) -- (2.4,1.41) node [right] {\small{$B(x,2/m)$}}; 366 | \draw (0.725,0.725) -- (2.4,0.725) node [right] {\small{$B(x,1/m)$}}; 367 | 368 | \draw (-1.15,0.2) -- (-2.4,0.2) node [left] {\small{$B(x_{n(r)},1/n(r))$}}; 369 | \end{tikzpicture} 370 | \end{center} 371 | 372 | Now, there exists $N$ such that $x_{n(r)} \in B(x,1/m)$ for all $r\geq N$. Additionally, if $n(r)>m$, and $y\in B(x_{n(r)},1/n(r))$, then $d(x,y) \leq d(x,x_{n(r)}) + d(x_{n(r)},y) < 2/m$. So for such $n(r)$, $B(x_{n(r)}, 1/n(r)) \subseteq B(x,2/m) \subseteq U_0$. Contradiction. 373 | 374 | \item Let $\cover{U}$ be an open cover of $X$. Choose $\epsilon>0$ as in (ii). For this $\epsilon$, using (i), there exists $x_1,\ldots,x_n \in X$ such that $X=\bigcup_{i=1}^n B(x_i,\epsilon)$. For each $i$, $B(x_i,\epsilon) \subseteq U_i$ for some $U_i \in \cover$, by (ii). Thus $X=\bigcup_{i=1}^n U_i$, and $X$ is compact. \qedhere 375 | \end{enumerate} 376 | \end{proof} 377 | 378 | % subsection sequential_compactness (end) -------------------------------------------------------------------------------- /ib-geometry/geo-chap-05.tex: -------------------------------------------------------------------------------- 1 | %!TEX root = geometry.tex 2 | \stepcounter{lecture} 3 | \setcounter{lecture}{5} 4 | \sektion{Hyperbolic geometry} 5 | 6 | \subsection{Hyperboloid model} % (fold) 7 | \label{sub:hyperboloid_model} 8 | 9 | \begin{definition} 10 | We consider the surface $S \subset \R^3$ given by $x^2+y^2-z^2=-1$, $z<0$. This \emph{hyperboloid sheet} gives us another way to think of points. The sketch below illustrates the sheet: it asymptotically approaches the planes $x=y$, $y=z$ and $z=x$; we take a cross section view. 11 | 12 | \begin{center} % Geo 11/1 13 | \begin{tikzpicture}[yscale=1.4, rotate=180] 14 | \bigarrow (2,0) -- (-2,0); 15 | \bigarrow (0,2.5) -- (0,-1); 16 | 17 | \begin{scope} [yscale=3, xscale=1.4] 18 | \draw [thick] (0,0.3) arc (-90:-55:1.8); 19 | \end{scope} 20 | 21 | \begin{scope} [yscale=3, rotate=180, xscale=1.4] 22 | \draw [thick] (0,-0.3) arc (90:55:1.8); 23 | \end{scope} 24 | 25 | \draw [thick] (-0.05,0.9) -- (0.05,0.9); 26 | 27 | \draw [dashed] (2,2) -- (0,0) -- (-2,2); 28 | \end{tikzpicture} 29 | \end{center} 30 | 31 | With this in mind, we give $\R^3$ the \emph{Minkowski metric} 32 | \begin{equation*} 33 | g^M = \dif{x^2} + \dif{y^2} - \dif{z^2}. 34 | \end{equation*} 35 | Formally, we've taken the surface $x^2+y^2+z^2=-1$, and replaced $z$ by $iz$. 36 | \end{definition} 37 | 38 | At first, these two definitions might seem unnatural, but in some sense, it's the most natural thing in the world. Note, however, that the Minkowski metric is \emph{not} a Riemannian metric. It is sometimes called the \emph{pseudo-Riemannian metric}. 39 | 40 | As before, we consider the stereographic projection $\pi:S\to\C$. And as before, we have a ``north pole'' $\NN=(0,0,1)$, and for any $\pi\in S$ that is not $\NN$, we define $\pi(\pp)$ to be the intersection of $NP$ with the $xy$ plane. Thus 41 | \begin{equation*} 42 | \pi\left( (x,y,z) \right) = \f{x+iy}{1-z} = w, 43 | \end{equation*} 44 | the same as the sphere. Inversion is similar; we first consider 45 | \begin{equation*} 46 | \left\vert w \right\vert^2 47 | = \f{x^2+y^2}{\left( 1-z \right)^2} 48 | = \f{z^2-1}{\left( 1-z \right)^2} 49 | = \f{z+1}{z-1} 50 | \implies 51 | z = \f{\left\vert w \right\vert^2+1}{\left\vert w \right\vert^2-1}. 52 | \end{equation*} 53 | Now, if $z<0$, then $\left\vert w \right\vert^2<1$, and so 54 | \begin{equation*} 55 | \Image(\pi) = D = \left\{w\iC: \left\vert w \right\vert<1\right\}. 56 | \end{equation*} 57 | So using the same process as the sphere, we have 58 | \begin{equation*} 59 | 1-z = \f{2}{1-\left\vert w \right\vert^2}, 60 | \end{equation*} 61 | and so the inverse is given by 62 | \begin{equation*} 63 | \pi^{-1}(w) = \left( \f{2\,\Re(w)}{1-\left\vert w \right\vert^2}, \f{2\,\Im(w)}{1-\left\vert w \right\vert^2}, \f{\left\vert w \right\vert^2+1}{\left\vert w \right\vert^2-1} \right). 64 | \end{equation*} 65 | 66 | % subsection hyperboloid_model (end) 67 | 68 | \subsection{Unit disc model} % (fold) 69 | \label{sub:unit_disc_model} 70 | 71 | Let $g^D$ be the metric induced on $g$ using $g^M$, that is, 72 | \begin{equation*} 73 | g^D = \dif{\sigma_1^2} + \dif{\sigma_2^2} - \dif{\sigma_3^2}, 74 | \end{equation*} 75 | where 76 | \begin{equation*} 77 | \sigma(x,y) = \left( \f{2x}{1-x^2-y^2}, \f{2y}{1-x^2-y^2}, \f{1+x^2+y^2}{1-x^2-y^2} \right). 78 | \end{equation*} 79 | Note that we've now switched to using $x$ and $y$ as coordinates on $D$, not on $\R^3$. This is essentially the same calculation as for $g^S$, and we obtain 80 | \begin{equation*} 81 | g^D = \f{4\left( \dif{x^2} + \dif{y^2} \right)}{\left.(1-x^2-y^2)\right.^2}. 82 | \end{equation*} 83 | Again, this is conformal to $g^E$. 84 | 85 | % subsection unit_disc_model (end) 86 | 87 | \subsection{Upper half-plane model} % (fold) 88 | \label{sub:upper_half_plane_model} 89 | 90 | This gives us another way to do hyperbolic geometry. Let $H=\{z\iC: \Im(z)>0\}$, the upper half-plane. Define 91 | \begin{equation*} 92 | \fullfunction{\varphi}{\C_\infty}{\C_\infty}{z}{(z-i)/(z+i)} 93 | \end{equation*} 94 | We saw in section~\ref{sec:m_bius_transformations} that $\phi(\R\cup\{\infty\}) = S^1$, and since $\phi(i)=0$, we have $\phi(H)=D$. 95 | 96 | \begin{definition} 97 | Let $g^H$ be the Riemannian metric on $H$ induced from $g^D$ using $\phi$: 98 | \begin{equation*} 99 | g_\pp^H(\va,\vb) = g^D_{\phi(\pp)}(\ddif{\phi}{\pp}(\va), \ddif{\phi}{\pp}(\vb)) 100 | \end{equation*} 101 | By definition, $\phi$ is a Riemann isometry from $g^H$ to $g^D$. 102 | \end{definition} 103 | 104 | To compute $g^H$, write $w=x+iy$. Then $\dif{x^2} \dif{y^2} = \dif{w} \dif{\wbar}$. For $w\in D$, we have 105 | \begin{equation*} 106 | w=\phi(z) = \f{z-i}{z+i} = 1-\f{zi}{z+i}. 107 | \end{equation*} 108 | By careful consideration, this gives us 109 | \begin{equation*} 110 | \dif{w} = \f{zi}{\left( z+i \right)^2} \qqand 111 | \dif{\wbar} = \f{-zi}{\left( \zbar-i \right)^2} \dif{\zbar}. 112 | \end{equation*} 113 | By substituting appropriately, and writing $z=u+iv$, we have 114 | \begin{align*} 115 | g^H 116 | = \f{4\left( \df{zi \dif{z}}{\left( z+i \right)^2} \right) \left( \df{-zi \dif{\zbar}}{\left( \zbar-i \right)^2} \right)}{\left( 1-\df{\left( z-i \right)\left( \zbar+i \right)}{\left( z+i \right)\left( \zbar-i \right)} \right)^2} 117 | &= \f{16 \dif{z} \dif{\zbar}}{\left[ \left( z+i \right)\left( \zbar-i \right) - \left( z-i \right)\left( \zbar+i \right) \right]^2} \\ 118 | &= \f{16 \dif{z} \dif{\zbar}}{\left[ zi\left( \zbar-z \right) \right]^2} \\ 119 | &= \f{16 \dif{z} \dif{\zbar}}{\left( 4v \right)^2} \\ 120 | &= \f{\dif{u^2} + \dif{v^2}}{v^2}. 121 | \end{align*} 122 | 123 | % $G^H = \PSL_2(\R)$ acts on $H$ $\leftrightarrow$ $G^D = {\phi: \phi(z) = e^{i\theta} \f{z-a}{az-1}, \theta\iR, a\in D}$ 124 | % $G^D = \phi G^H \phi^{-1}$. 125 | 126 | % subsection upper_half_plane_model (end) 127 | 128 | \subsection{Geometry of the hyperbolic plane} % (fold) 129 | \label{sub:geometry_of_the_hyperbolic_plane} 130 | 131 | \lecturemarker{12}{27 Feb} 132 | We've now seen two models of the hyperbolic plane: the upper half-plane model $H$, and the unit disc model $D$. 133 | 134 | In particular, recall that $G^H = \PSL_2(\R)$ acts on $H$, with 135 | \begin{equation*} 136 | G^D = \left\{\phi: \phi(z) = e^{i\theta} \f{z-a}{az-a}, \theta\iR, a\in D\right\} = \phi G^H \phi^{-1}. 137 | \end{equation*} 138 | This $\phi$ gives us a way to compare boundaries: $\boundary{H} = \R\cup\infty=\R_\infty$, and $\boundary{D} =S^{\,1}$. Then $\phi(\R_\infty) = S^{\,1}$. 139 | 140 | We may use both of these models, depending on which is more convenient at the time. We use $\HH$ to denote either one, without specifying which. With these two models in hand, we can discuss all the features of geometry that we've talked about before. 141 | 142 | Angles are simple. Both $g^D$ and $g^H$ are conformal, so angles between curves in $g^D$ or $g^H$ is the same as the Euclidean angle. 143 | 144 | Now let's consider isometries. Let $\Isom(H)$ be the group of Riemannian isometries of $(H,g^H)$, and similarly, $\Isom(D)$ be the group of Riemannian isometries of $(D,g^D)$. We come to our first result: 145 | 146 | \begin{proposition} 147 | $G^H \subset \Isom(H)$. 148 | \end{proposition} 149 | 150 | \vspace{-8pt} 151 | 152 | Note that $G^H$ isn't all isometries, as not all isometries are orientiation preserving. 153 | 154 | \begin{proof} 155 | Recall that $G^H = \PSL_2(\R)$ is generated by three kinds of maps: 156 | \begin{enumerate} 157 | \shortskip 158 | \item $z\mapsto z+b$, $b\iR$; 159 | \item $z \mapsto az$, $a\iR$; 160 | \item $z\mapsto -1/z$. 161 | \end{enumerate} 162 | It suffices to check that (i), (ii) and (iii) are in $\Isom(H)$. 163 | \begin{enumerate} 164 | \item If $z=\phi(z\p) = z\p+b$, then we have 165 | \begin{equation*} 166 | \begin{array}{lcl} 167 | x=x\p + b & & \dif{x} = \dif{x\p} \\ 168 | y=y\p & & \dif{y} = \dif{y\p}. 169 | \end{array} 170 | \end{equation*} 171 | Thus we have 172 | \begin{equation*} 173 | g^H 174 | = \f{\dif{x^2} + \dif{y^2}}{y^2} 175 | \xrightarrow[\text{induced by $\phi$}]{\text{metric}} 176 | \f{\left( \dif{x\p} \right)^2 + \left( \dif{y\p} \right)^2}{\left( y\p \right)^2} 177 | = g^H. 178 | \end{equation*} 179 | 180 | \item If $z=\phi(z\p)=az\p$, then $x=ax\p$, $y=ay\p$ and 181 | \begin{equation*} 182 | g^H 183 | = \f{\dif{x^2} + \dif{y^2}}{y^2} 184 | \longrightarrow 185 | \f{a^2\left( \dif{x\p} \right)^2 + a^2\left( \dif{y\p} \right)^2}{a^2\left( y\p \right)^2} = g^H. 186 | \end{equation*} 187 | 188 | \item If $z=\phi(w)=-1/w$, then 189 | \begin{equation*} 190 | \dif{z} = \f{\dif{w}}{w^2} \qqand 191 | \dif{\zbar} = \f{\dif{\wbar}}{\wbar^2}. 192 | \end{equation*} 193 | Then we have 194 | \begin{equation*} 195 | g^H 196 | = \f{\dif{x^2} + \dif{y^2}}{y^2} 197 | = \f{\dif z \dif{\zbar}}{\left( \f{1}{2}\left( \zbar-z \right) \right)^2} 198 | \longrightarrow 199 | \f{\left( -\f{\dif{w}}{w^2} \right)\left( -\f{\dif{\wbar}}{\wbar^2} \right)}{\left( \f{1}{2}\left( \f{1}{\wbar}-\f{1}{w} \right) \right)^2} 200 | = \f{\dif{w} \dif{\wbar}}{\left( \f{1}{2}\left( w-\wbar \right) \right)^2} = g^H. \qedhere 201 | \end{equation*} 202 | \end{enumerate} 203 | \end{proof} 204 | 205 | \begin{corollary} 206 | $G^D \subset \Isom(D)$ 207 | \end{corollary} 208 | 209 | \begin{proof} 210 | If $\psi \in G^D$, then $\psi=\phi_0 \chi \phi_0^{-1}$, where $\chi\in G^H$. Thus $\phi_0, \chi, \phi_0^{-1}$ are all isometries, and the composition of isometries is an isometry. Thus $\psi \in \Isom(D)$. 211 | \end{proof} 212 | 213 | Now we consider hyperbolic lines. These are defined in a very similar way to spherical lines. 214 | 215 | \begin{definition} 216 | A \emph{hyperbolic line} in $H$ is $L = H \cap C$, where $C$ is a Euclidean line or circle which is perpendicular to $\partial{H}$. A similar definition under the disc model comes by replacing $H$ by $D$. 217 | 218 | \begin{center} 219 | \begin{minipage}{0.4\textwidth} % Geo 12/1, 12/1b 220 | \centering 221 | \begin{tikzpicture} 222 | \draw (0,0) -- (5,0); 223 | \draw [very thick] (1,0) -- (1,2); 224 | \draw [very thick] (4,0) arc (0:180:1); 225 | \end{tikzpicture} 226 | \end{minipage} 227 | \hspace{0.2cm} 228 | \begin{minipage}{0.4\textwidth} 229 | \centering 230 | \begin{tikzpicture}[scale=0.6] 231 | \draw (2,2) circle (2); 232 | \clip (2,2) circle (2); 233 | \draw [very thick] (2,2) -- ++ (45:2) -- ++ (-135:4); 234 | \draw [very thick] (2,2) ++ (-45:2) circle (1); 235 | \end{tikzpicture} 236 | \end{minipage} 237 | 238 | \begin{minipage}{0.4\textwidth} 239 | \centering 240 | half-plane model 241 | \end{minipage} 242 | \hspace{0.2cm} 243 | \begin{minipage}{0.4\textwidth} 244 | \centering 245 | disc model 246 | \end{minipage} 247 | \end{center} 248 | \end{definition} 249 | 250 | For the rest of the chapter, when we say ``line'', we mean ``hyperbolic line'' unless otherwise specified. 251 | 252 | Once we have lines, then it's natural to define rays: 253 | 254 | \begin{definition} 255 | If $\gamma:\R\to H$ is a parameterisation of a line, then $R=\gamma([c,\infty))$ is a \emph{hyperbolic ray} starting at $\gamma(c)$ and with direction $\gamma\p(c)$. 256 | \end{definition} 257 | 258 | Now let's consider a few basic results involving lines: 259 | 260 | \begin{lemma} 261 | $L$ is a line in $H$ if and only if $\phi_0(L)$ is a line in $D$. 262 | \end{lemma} 263 | 264 | \begin{proof} 265 | As $\phi_0\in\Mobgp$, it preserves angles, and it takes Euclidean lines and circles to Euclidean lines and circles. Also, $\phi_0(\boundary{H}) = \boundary{D}$. Thus, if $L$ is a line in $H$, then $\phi_0(L)$ is a line in $D$. 266 | 267 | The converse is similar. 268 | \end{proof} 269 | 270 | \begin{lemma} 271 | Given $\va \neq \vec{0}$, there is a unique hyperbolic line through $\vec{0} \in D$ which is tangent to $\va$ at $\vec{0}$. 272 | \end{lemma} 273 | 274 | \begin{proof} 275 | First we show that any line through $\vec{0}$ is a diameter of $D$. Suppose $C$ is a Euclidean circle passing through $\vec{0}$, perpendicular to $\boundary{D}$. Let $B$ be its centre. 276 | 277 | % \begin{center} % Geo 12/3 278 | % \begin{tikzpicture} 279 | % \draw [thick, red] (0,0) circle (1.4); 280 | % \draw [thick, blue] (2,0) circle (2); 281 | % \draw (0,0) node [below left] {$O$} -- (2,0) node [below right] {$B$} -- (0.49,1.31) -- cycle; 282 | % \draw (0.42,1.31) node [above=2pt] {$A$}; 283 | 284 | % \foreach \s/\t in {0/0, 2/0, .49/1.31} {\draw (\s,\t) node {$\bullet$};} 285 | % \end{tikzpicture} 286 | % \end{center} 287 | 288 | Let $A$ be a point in $C \cap \boundary{D}$, then $\triangle OAB$ is isoceles. Thus $\angle OAB = \pi/2 = \angle AOB$, and so the sum of the angles is more than $\pi$. But this is ridiculous. 289 | 290 | The lemma now follows, sine there's a unique Euclidean line through $\vec{0}$ with direction vector $\va$. 291 | \end{proof} 292 | 293 | \begin{corollary} 294 | If $p\in\HH$ and $\va\neq 0$, then there's a unique ray stating at $\pp$ with direction vector $\va$. 295 | \end{corollary} 296 | 297 | \begin{proof*} 298 | [In $D$] Choose $\psi \in G^D$ with $\psi(\pp)=0$, such as $\psi(z) = \df{z-p}{\overline{p}z-1}$. 299 | 300 | Then there's a unique ray $R\p$ starting at $\vec{0}$ with direction $\dif{\psi_p(\va)} \neq 0$. Then the ray $R$ which we want is $R=\psi^{-1}(R\p)$. \qed 301 | \end{proof*} 302 | 303 | \begin{proposition} 304 | If $R_1,R_2$ are hyperbolic rays starting at $p_1,p_2 \iH$, then there is $\psi\in G$ with $\psi(p_1) = p_2$, $\psi(R_1)=R_2$. 305 | \end{proposition} 306 | 307 | \begin{proof*} 308 | [In $D$] Let $R_0$ be the positive real axis. Let 309 | \begin{equation*} 310 | \psi_1(z) = \f{z-p_1}{\overline{p_1}z-1}. 311 | \end{equation*} 312 | If $\psi_1(p_1)=0$, then $\psi_1(R_1)$ is a radius of $D$. 313 | \begin{center} 314 | \begin{tikzpicture}[scale=2] 315 | \draw (0,0) circle (1) node {$\bullet$}; 316 | 317 | \draw (1,0) -- (0,0) -- ++ (60:1); 318 | \draw (0.3,0) arc (0:60:0.3); 319 | \draw (0.25,0.2) node [right] {$\theta$}; 320 | 321 | \draw (0.5,0) node [below] {$R_0$}; 322 | \end{tikzpicture} 323 | \end{center} 324 | 325 | Let $\psi_R(z) = e^{-i\theta} \psi_1(z)$, where $\theta$ is the angle between $R_0$ and $\psi_1(R_1)$. Then $\psi_{R_1}(R_1) = R_0$. Construct $R_2$ similarly, and then take $\psi = \psi_{R_2}^{-1} \circ \psi_{R_1}$. \'.e' 326 | \end{proof*} 327 | 328 | \begin{proposition} 329 | There's a unique line containing two distinct points $p_1,p_2 \iH$. 330 | \end{proposition} 331 | 332 | \begin{proof*} 333 | [In $D$] Choose $\psi \in G^D$ with $\psi(p_1) = \vec{0}$. There's a unique hyperbolic line $L$ containing $\vec{0}$ and $\psi(p_2)$, namely through the diameter of $D$ through $\psi(\overline{p_2})$. Thus $\psi^{-1}(L)$ is the unique line containing $p_1$ and $p_2$. \qed 334 | \end{proof*} 335 | 336 | \begin{proposition} 337 | Two lines $L_1$ and $L_2$ intersect in at most one point in $\HH$. 338 | \end{proposition} 339 | 340 | \begin{proof*} 341 | [In $H$] After applying an element $\psi\in G^H$, we may assume that 342 | \begin{itemize} 343 | \shortskip 344 | \item $\psi(L_1)$ is the positive imaginary axis. 345 | \item $\psi(L_2)$ is (i) a circle centred on the real axis or (ii) a vertical line 346 | \end{itemize} 347 | Consider the two cases for $\psi(L_2)$: 348 | \begin{enumerate} 349 | \shortskip 350 | \item At most one intersection with $\psi(L_1)$ in $H$ (other is in the lower half-plane); 351 | \item Has none. \qed 352 | \end{enumerate} 353 | \end{proof*} 354 | 355 | This final proposition motivates the following definition: 356 | 357 | \begin{definition} 358 | We say that $L_1$ and $L_2$ are \emph{haroparallel} if they intersect in $\boundary{\HH}$ and are \emph{ultraparallel} if they do not intersect in $\boundary{\HH}$. 359 | 360 | \begin{center} 361 | \begin{minipage}{0.4\textwidth} % Geo 12/5 362 | \centering 363 | \begin{tikzpicture} 364 | \draw (0,0) -- (5,0); 365 | \draw [very thick] (1,0) -- (1,2); 366 | \draw [very thick] (3,0) arc (0:180:1); 367 | \end{tikzpicture} 368 | \end{minipage} 369 | \hspace{0.2cm} 370 | \begin{minipage}{0.4\textwidth} 371 | \centering 372 | \begin{tikzpicture} 373 | \draw (0,0) -- (5,0); 374 | \draw [very thick] (1,0) -- (1,2); 375 | \draw [very thick] (4,0) arc (0:180:1); 376 | \end{tikzpicture} 377 | \end{minipage} 378 | 379 | \begin{minipage}{0.4\textwidth} 380 | \centering 381 | haroparallel 382 | \end{minipage} 383 | \hspace{0.2cm} 384 | \begin{minipage}{0.4\textwidth} 385 | \centering 386 | ultraparallel 387 | \end{minipage} 388 | \end{center} 389 | 390 | If $L$ is a line, $p\not\in L$, then there are infinitely many ultraparallel lines to $L$ that pass through $p$. 391 | \end{definition} 392 | 393 | \pagebreak 394 | 395 | Now we consider distance; specifically, the shortest distance between two points. 396 | 397 | \begin{proposition} 398 | If $p,q\iH$, then the line segment from $\pp$ to $\qq$ is the shortest path from $\pp$ to $\qq$ in $H$. 399 | \end{proposition} 400 | 401 | \begin{proof*} 402 | [In $H$] Let $L$ be the unique line segment from $\pp$ to $\qq$. After composing with $\psi\in G$, we may assume that $L$ is the positive real axis. 403 | 404 | So we have $p=ia$, $q=ib$, $a,b\iR$. Let $\gamma(t)$ be a path from $\pp$ to $\qq$ in $H$. Then 405 | \begin{equation*} 406 | L_{g^H}(\gamma) 407 | = \int_0^1 \sqrt{\f{\gamma_1^2+\gamma_2^2}{\gamma_2^2}} \dif{t} 408 | \geq \int_0^1 \left\vert \f{\gamma_2\p(t)}{\gamma_2\p(t)} \right\vert \dif{t} 409 | \geq \left\vert \int_0^1 \f{\gamma\p(t)}{\gamma(t)} \dif{t} \right\vert 410 | = \left\vert \ln b - \ln a \right\vert 411 | = \left\vert \ln(b/a) \right\vert. 412 | \end{equation*} 413 | with equality if and only if $\gamma_1\p=0$ and $\gamma_2\p$ has constant sign; that is, if $\gamma$ is a vertical line segment. \qed 414 | \end{proof*} 415 | 416 | \begin{corollary} 417 | The distance from $ia$ to $ib$ in $H$ is $\left\vert \ln(b/a) \right\vert$. 418 | \end{corollary} 419 | 420 | \begin{corollary} 421 | The distance from $\vec{0}$ to $re^{i\theta}$ in $D$ is $\ln\left[ (1+r)/(1-r) \right] = 2\tanh^{-1} r$. 422 | \end{corollary} 423 | 424 | % subsection geometry_of_the_hyperbolic_plane (end) 425 | 426 | \subsection{Isometries of the hyperbolic plane} % (fold) 427 | \label{sub:isometries_of_the_hyperbolic_plane} 428 | 429 | \lecturemarker{13}{4 Mar} 430 | We extend complex conjugation to a map $c:\C_\infty \to \C_\infty$ by setting $c(\infty) = \infty$. Now, if $\phi_A$ is the Möbius transformation defined by the matrix $A\in\GL_2(\C)$, then we see that 431 | \begin{equation*} 432 | \phi_A \circ c = c \circ \phi_{\bar{A}}. 433 | \end{equation*} 434 | 435 | \begin{definition} 436 | The \emph{extended Möbius group} is given by 437 | \begin{equation*} 438 | \overline{\Mobgp} = \left\{ \phi:\C_\infty \to \C_\infty : \phi\in\Mobgp \text{or } \phi\circ c\in\Mobgp \right\}. 439 | \end{equation*} 440 | \end{definition} 441 | 442 | We observe that $c^2=\iota$, so the second condition is equivalent to saying that $\phi = \psi\circ c$, where $\psi\in\Mobgp$. It follows from our first equation that the extended Möbius group is closed under composition, and thus it indeed forms a group, containing the Möbius group as an index two subgroup. 443 | 444 | Then elements of $\Mobgp$ are \emph{orientation preserving}, while elements of $\overline{\Mobgp}$ not in $\Mobgp$ are \emph{orientation reversing}. We can compare this to rotations and reflections in Euclidean geometry. We already have our rotations (given by Möbius maps), so now let's consider reflections: 445 | 446 | \begin{definition} 447 | If $C\subset\C_\infty$ is a Euclidean line or circle, the reflection in $C$ is the extended Möbius transformation defined by 448 | \begin{equation*} 449 | R_C = \psi^{-1} \circ c \circ \psi, 450 | \end{equation*} 451 | where $\psi\in\Mobgp$ satisfies $\phi(C) = \R\cup\{\infty\}$. 452 | \end{definition} 453 | 454 | Hopefully this definition is reasonably intuitive; now we just need to check that it makes sense. We need to check that our choice of $\psi$ doesn't matter. So suppose we have $\psi\p(C)=\R\cup\{\infty\}$. Then $\psi\p \circ \psi^{-1}(\R) = \R$, and so $\psi\p \circ \psi^{-1} = \phi_A$, for some $A\in\GL_2(\R)$. (As in the Euclidean plane, two reflections form a rotation.) Then 455 | \begin{equation*} 456 | \psi^{\,\prime^{-1}} \circ c \circ \psi\p 457 | = \psi^{-1} \circ \phi^{-1}_A \circ c \circ \phi_A \circ \psi 458 | = \psi^{-1} \circ c \circ \psi, 459 | \end{equation*} 460 | and so $R_C$ is well-defined. 461 | 462 | \begin{example} 463 | If $C$ is the unit circle, then $R_C=\psi_0 \circ c \circ \psi_0^{-1}$, and so 464 | \begin{equation*} 465 | R_C(z) 466 | = \psi_0\left( -i\,\f{\zbar+1}{\zbar-1} \right) 467 | = \f{-i\f{\zbar+1}{\zbar-1} - i}{-i\f{\zbar+1}{\zbar-1} + i} 468 | = \f{1}{\zbar}. 469 | \end{equation*} 470 | More generally, if $C_r$ is a circle of radius $r$ centred at $0$, then $R_{C_r} = \psi_r \circ R_{C_1} \circ \psi_{1/r}$, where $\psi_a(z) = az$. Thus $R_{C_r} = r^2/\zbar$. 471 | \end{example} 472 | 473 | \begin{proposition} 474 | $\overline{\Mobgp}$ is generated by reflections. 475 | \end{proposition} 476 | 477 | \begin{proof} 478 | It is enough to check that the maps 479 | \begin{enumerate} 480 | \shortskip 481 | \item $z\mapsto z+b$, $b\iC$; 482 | \item $z\mapsto az$, $a\iC$, $a\neq 0$; 483 | \item $z\mapsto 1/z$; 484 | \end{enumerate} 485 | are compositions of reflections, since these maps generate $\Mobgp$. 486 | 487 | Map (i) is generated by $R_{L_1} \circ R_{L_2}$, where $L_1$ and $L_2$ are two Euclidean lines perpendicular to $b$ and separated by a distance $b/2$. 488 | 489 | For map (ii), multiplication by $a\iR$ is $R_{C_1} \circ R_{C_2}$, where $C_2$ is the unit circle and $C_1$ is a circle of radius $\sqrt{a}\,$ centred at the origin, while multiplication by $e^{i\theta}$ is $R_{L_1} \circ R_{L_2}$, where $L_1$ and $L_2$ are two lines which intersect in an angle $\theta/2$ at the origin. Using these two maps, we can compose for any $a\iC\backslash\{0\}$. 490 | 491 | Finally, map (iii) is the composition of reflection in the unit circle with reflection in $\R$. This completes the proof. 492 | \end{proof} 493 | 494 | We can view the groups $\Isom(S^2)$, $\Isom(\R^2)$ and $\Isom(D)$ as subgroups of the extended Möbius group, corresponding to the extension of the following subgroups of $\Mobgp$ by $c$: 495 | \begin{align*} 496 | \Isom^+(S^2) &= \left\{\phi_A : A = \mat{\alpha & \beta \\ -\overline{\beta} & \overline{\alpha}}, \det A = 1 \right\}, \\ 497 | \Isom^+(\R^2) &= \left\{\phi_A : A = \mat{\alpha & \beta \\ 0 & \overline{\alpha}}, \det A = 1 \right\}, \\ 498 | \Isom^+(D) &= \left\{\phi_A : A = \mat{\alpha & \beta \\ \overline{\beta} & \overline{\alpha}}, \det A = 1 \right\}, \\ 499 | \end{align*} 500 | 501 | % subsection isometries_of_the_hyperbolic_plane (end) -------------------------------------------------------------------------------- /ib-geometry/geo-chap-04.tex: -------------------------------------------------------------------------------- 1 | %!TEX root = geometry.tex 2 | \stepcounter{lecture} 3 | \setcounter{lecture}{4} 4 | \sektion{Riemannian geometry} 5 | 6 | \lecturemarker{9}{18 Feb} 7 | All the functions we will encounter in this chapter are smooth (that is, infinitely differentiable) unless otherwise stated. 8 | 9 | \subsection{Parameterised spaces} % (fold) 10 | \label{sub:parameterised_spaces} 11 | 12 | \begin{definition} 13 | A \emph{parametrised surface} $S\subset\R^3$ is a map $\sigma:U\to\R^3$, where $U$ is an open subset of $\R^2$ such that 14 | \begin{enumerate} 15 | \shortskip 16 | \item $\sigma$ is injective and $\Image(\sigma)=\sigma$; 17 | \item For each $\pp\in U$, $\ddif{\sigma}{\pp}$ is injective. 18 | \end{enumerate} 19 | This condition is actually slightly more restrictive than it needs to be. 20 | 21 | If $S$ satisfies (ii), then we say that it is \emph{smoothly embedded}. 22 | \end{definition} 23 | 24 | Recall that if $\sigma=(\sigma_1,\sigma_2,\sigma_3)$, then $\ddif{\sigma}{\pp}:\R^2\to\R^3$ has matrix representation 25 | $$\mat{\ssig{1x} & \ssig{1y} \\ 26 | \ssig{2x} & \ssig{2y} \\ 27 | \ssig{3x} & \ssig{3y}}, \qquad 28 | \text{where $\sigma_{ix} = \dpd{\sigma_i}{x}$.}$$ 29 | 30 | \begin{example} 31 | Consider the following two parametrisations of $S^{\,2}$. First, spherical coordinates: 32 | \begin{equation*} 33 | \fullfunction{\sigma}{(0,2\pi) \times (0,\pi)}{\R^3}{(\theta,\phi)}{(\cos\theta \sin\phi, \sin\theta\sin\phi, \cos\phi)}, 34 | \end{equation*} 35 | Alternatively, consider the inverse of stereographic projection: 36 | \begin{equation*} 37 | \fullfunction{\sigma}{\R^2}{\R^3}{(x,y)}{\left( \f{2x}{1+x^2+y^2}, \f{2y}{1+x^2+y^2}, \f{x^2+y^2-1}{1+x^2+y^2} \right)}. 38 | \end{equation*} 39 | \end{example} 40 | 41 | We can construct paths on parametrised surfaces in the obvious way: if $\gamma:[0,1] \to U$ is a path in $U$, then $\Gamma = \sigma \circ \gamma$ is a path in $S$. 42 | 43 | The chain rule holds as we would expect: 44 | \begin{equation*} 45 | \Gamma\p(t) = \ddif{\sigma}{\gamma(t)}[\gamma\p(t)]. 46 | \end{equation*} 47 | 48 | \begin{definition} 49 | The \emph{tangent space} to $S$ at $\sigma(\pp)$ is 50 | \begin{equation*} 51 | T_{\sigma(\pp)} S := \Image \ddif{\sigma}{\pp}, 52 | \end{equation*} 53 | which is a linear subspace of $\R^3$. 54 | \end{definition} 55 | 56 | As with spherical geometry, the derivative of a path at a point is in the tangent space: 57 | \begin{equation*} 58 | \Gamma\p(t) = \ddif{\sigma}{\gamma(t)} [\gamma\p(t)] \in T_{\Gamma(t)}. 59 | \end{equation*} 60 | This leads to the following fact, which we shall return to later: 61 | \begin{equation*} 62 | \left.\vert \Gamma\p(t) \vert\right.^2 = \ddif{\sigma}{\gamma(t)} [\gamma\p(t)] \cdot \ddif{\sigma}{\gamma(t)} [\gamma\p(t)]. 63 | \end{equation*} 64 | 65 | % subsection parameterised_spaces (end) 66 | 67 | \subsection{Riemannian metrics} % (fold) 68 | \label{sub:riemannian_metrics} 69 | 70 | \begin{definition} 71 | If $U\subset\R^2$ is open, then a \emph{Riemannian metric} $g$ on $U$ is a smooth map $g:U\to\Mat_2(\R)$ such that for each $\pp\in U$, $g_{\pp} := g(\pp)$ is symmetric and positive definite. That is, 72 | \begin{equation*} 73 | g_\pp = \mat{E(x,y) & F(x,y) \\ F(x,y) & G(x,y)} \text{ with } E(x,y)>0, EG-F^2>0. 74 | \end{equation*} 75 | \end{definition} 76 | 77 | We saw in \emph{Linear Algebra} that a symmetric, positive definite matrix is analogous to an inner product, and that's what we really care about. 78 | 79 | For each $\pp\in U$, $g_\pp$ defines an inner product on $\R^2$ 80 | \begin{equation*} 81 | \left\langle \va,\vb \right\rangle = \va^\Trans \mat{E & F \\ F & G} \vb =: g_\pp(\va,\vb). 82 | \end{equation*} 83 | If $\sigma:U\to\R^3$ is a parameterised surface, then define $g$ by 84 | \begin{equation*} 85 | g_\pp(\va,\vb) = \ddif{\sigma}{\pp}(\va) \cdot \ddif{\sigma}{\pp}(\vb), 86 | \end{equation*} 87 | which is the usual inner product on $\R^3$. 88 | 89 | \begin{note} 90 | If $\Gamma=\sigma\circ\gamma$, then 91 | \begin{equation*} 92 | \Gamma\p(t) \cdot \Gamma\p(t) = g_{\gamma(t)}(\gamma\p(t),\gamma\p(t)). 93 | \end{equation*} 94 | \end{note} 95 | 96 | Now if we have 97 | \begin{equation*} 98 | A 99 | = \ddif{\bsig}{\pp} 100 | = \mat{\sigma_{1x} & \sigma_{1y} \\ \sigma_{2x} & \sigma_{2y} \\ \sigma_{3x} & \sigma_{3y}} 101 | = \mat{\bsig_x & \bsig_y}, 102 | \end{equation*} 103 | then we can write 104 | \begin{equation*} 105 | \ddif{\sigma}{\pp}(\va) \cdot \ddif{\sigma}{\pp}(\vb) = A\va \cdot A\vb = \va^\Trans A^\Trans A \vb. 106 | \end{equation*} 107 | This gives us 108 | \begin{equation*} 109 | g 110 | = A^\Trans a 111 | = \mat{\bsig_x \\ \bsig_y} \mat{\bsig_x \bsig_y} 112 | = \mat{\bsig_x \cdot \bsig_x & \bsig_x \cdot \bsig_y \\ \bsig_y \cdot \bsig_x & \bsig_y \cdot \bsig_y}. 113 | \end{equation*} 114 | Now we must show that this really is a metric: 115 | 116 | \begin{lemma} 117 | As defined above, $g$ is a Riemannian metric. 118 | \end{lemma} 119 | 120 | \begin{proof} 121 | As $\bsig_x \cdot \bsig_y = \bsig_y \cdot \bsig_x$, the matrix is symmetric. 122 | 123 | To show that it is positive definite, write 124 | \begin{equation*} 125 | g_\pp(\va,\va) = \ddif{\sigma}{\pp}(\va) \cdot \ddif{\sigma}{\pp}(\va)\geq 0, 126 | \end{equation*} 127 | with equality if and only if $\ddif{\sigma}{\pp}(\va)=0$, which is true if and only if $\va=0$, since $\dif{\sigma}$ is injective. (This is where our embedding hypothesis comes in.) 128 | \end{proof} 129 | 130 | \emph{Notation.} We don't usually write $g$ as a $2\times 2$ matrix; instead we write $g=E\dif{x^2} + 2F \dif{x} \dif{y} + G \dif{y^2}$, where 131 | \begin{equation*} 132 | E = \bsig_x \cdot \bsig_x, \qquad F = \bsig_x \cdot \bsig_y, \qquad G = \bsig_y \cdot \bsig_y. 133 | \end{equation*} 134 | We say that this $g$ is the Riemannian metric on $U$ \emph{induced} by $\sigma$. 135 | 136 | Note that not every Riemannian metric arises in this way. 137 | 138 | \pagebreak 139 | 140 | \begin{example} 141 | The Euclidean metric on $\R^2$ is $\dif{x^2}+\dif{y^2}$. 142 | 143 | The Euclidean metric on $\R^3$ is $\dif{u_1^2} + \dif{u_2^2} + \dif{u_3^2}$, for coordinates $(u_1,u_2,u_3)$ on $\R^3$. We write 144 | \begin{equation*} 145 | \dif{u_i} = \pd{\sigma_i}{x} \dif{x} + \pd{\sigma_i}{y} \dif{y}, 146 | \end{equation*} 147 | then the metric induced by $\sigma$ is $\dif{u_1^2} + \dif{u_2^2} + \dif{u_3^2}$. 148 | \end{example} 149 | 150 | Now let's look at a more complicated example: 151 | 152 | \begin{example} 153 | Consider 154 | \begin{equation*} 155 | \sigma(x,y) = \left( \f{2x}{1+x^2+y^2}, \f{2y}{1+x^2+y^2}, \f{x^2+y^2-1}{1+x^2+y^2} \right), \qquad \alpha=1+x^2+y^2. 156 | \end{equation*} 157 | Then we have 158 | \begin{align*} 159 | \dif{\sigma_1} 160 | &= \left( \f{2}{\alpha} - \f{4x^2}{\alpha^2} \right) \dif{x} - \f{4xy}{\alpha^2} \dif{y} 161 | = 2 \left( \f{1+y^2-x^2}{\alpha^2} \dif{x} - \f{2xy}{\alpha^2} \dif{y} \right) \\ 162 | \dif{\sigma_2} 163 | &= 2 \left( \f{1+x^2-y^2}{\alpha^2} \dif{y} - \f{2xy}{\alpha^2} \dif{x} \right) \\ 164 | \dif{\sigma_3} 165 | &= \f{4x}{\alpha^2} \dif{x} + \f{4y}{\alpha^2} \dif{y}. 166 | \end{align*} 167 | So we have 168 | \begin{align*} 169 | g 170 | &= \left( \dif{\sigma_1} \right)^2 + \left( \dif{\sigma_2} \right)^2 + \left( \dif{\sigma_3} \right)^2 \\ 171 | &= \f{4}{\alpha^4} \left[ \left[ \left( 1+y^2-x^2 \right)^2 + 4x^2 y^2 + 4x^2 \right] \dif{x^2} \right. \\ 172 | & \qquad + \left[ -2xy-2xy+4xy \right] \dif{x} \dif{y} \\ 173 | & \qquad\qquad \left.+ \left[ \left( 1+x^2 - y^2 \right)^2 + 4x^2 y^2 + 4y^2 \right] \dif{y}^2 \right] \\ 174 | &= \f{4}{\alpha^4} \left( \alpha^2 \dif{x^2} + \alpha^2 \dif{y^2} \right) \\ 175 | &= \f{4\left( \dif{x^2} + \dif{y^2} \right)}{\left( 1+x^2+y^2 \right)^2}. 176 | \end{align*} 177 | Notice in particular that this is a function of the standard Euclidean metric. 178 | \end{example} 179 | 180 | % subsection riemannian_metrics (end) 181 | 182 | \pagebreak 183 | 184 | \subsection{Geometry with the Riemannian metric} % (fold) % formerly Riemannian geometry 185 | \label{sub:geometry_with_the_riemannian_metric} 186 | 187 | Let $g$ be a Riemannian metric on $U\subset\R^2$. 188 | 189 | \begin{definition} 190 | A path $\gamma:[0,1]\to\R^2$ is \emph{piecewise smooth} if it is continuous on $[0,1]$ and smooth except at finitely many points $0=t_00$ so that $B_r(\pp\p)\in U$, there is $c>0$ such that if $\gamma:[0,1] \to B_r(\pp\p)$ is a path, then $L_g(\gamma) \geq c\left\vert \gamma(0)-\gamma(1) \right\vert$ (Euclidean distance). 242 | \end{lemma} 243 | 244 | \begin{proof} 245 | [Proof of lemma] Consider the general form of $g_\pp$: 246 | \begin{equation*} 247 | g_\pp = \mat{E & F \\ F & G}. 248 | \end{equation*} 249 | This is symmetric and positive definite, so it has strictly positive eigenvalues $\lambda_1(\pp), \lambda_2(\pp)$ and eigenvectors $\vv_1(\pp), \vv_2(\pp)$ which form an orthonormal basis of $\R^2$. 250 | 251 | If $\vv=a\vv_1(\pp) + b\vv_2(\pp)$, then 252 | \begin{align*} 253 | g_\pp(\vv,\vv) 254 | &= a^2 \lambda_1(p) + b^2 \lambda_2(\pp) \\ 255 | &\geq \min(\lambda_1,\lambda_2) \left.(a^2+b^2)\right. \\ 256 | &= \min(\lambda_1,\lambda_2) \, \vv\cdot\vv. 257 | \tag{$*$} 258 | \end{align*} 259 | Now $\lambda_1,\lambda_2$ are continuous functions of $\pp$ and $\closure{B_r(\pp\p)}$ is compact, so there exists some $\qq_1 \in \closure{B_r(\pp\p)}$ with $\lambda_1(\qq_1) \leq \lambda_1(\rr)$ for all $\rr\in B_r(\pp)$. Similarly, there is some $\qq_2$ with $\lambda_2(\qq_2) \leq \lambda_2(\rr)$ for all $\rr\in B_r(\pp)$. 260 | 261 | So take $\lambda=\min(\lambda_1(\qq_1),\lambda_2(\qq_2))$, then $g_\pp(\vv,\vv)≥\lambda \vv\cdot\vv$ for all $\pp\in B_r(\pp)$ (from $(*)$). Then 262 | \begin{align*} 263 | L_g(\gamma) 264 | &= \int_0^1 \sqrt{g_{\gamma(t)}(\gamma\p(t),\gamma\p(t))} \dif{t} \\ 265 | &\geq \int_0^1 \sqrt{\lambda \gamma\p(t) \cdot \gamma\p(t)} \dif{t} \\ 266 | &= \sqrt{\lambda}\, L_\text{Euclidean}(\gamma) \\ 267 | &\geq \sqrt{\lambda}\,\left\vert \gamma(0)-\gamma(1) \right\vert. 268 | \end{align*} 269 | So we take $c=\sqrt{\lambda}$. 270 | \end{proof} 271 | 272 | \begin{proof} [Proof of proposition] 273 | \mbox{} 274 | \begin{enumerate} 275 | \item For any $\gamma$, we have $L_\gamma(\gamma) \geq 0$, so 276 | \begin{equation*} 277 | d(\pp,\qq) = \inf_\gamma {L_g(\gamma)} \geq 0. 278 | \end{equation*} 279 | Pick $r>0$ with $\closure{B_r(\pp)} \subset U$, and choose $c>0$ as in the lemma. 280 | 281 | Spose $\qq \neq \pp$. If $\qq\in\closure{B_r(\pp)}$, $\gamma(0)=\pp$, $\gamma(1)=\qq$, then the lemma tells us that 282 | \begin{equation*} 283 | L_g(\gamma) 284 | \geq c \left\vert \gamma(0)-\gamma(1) \right\vert 285 | = c\,d_\text{Euclidean}(\pp,\qq) 286 | \end{equation*} 287 | and so we have 288 | \begin{equation*} 289 | d(\pp,\qq) = \int_\gamma {L_g(\gamma)} \geq c\,d_\text{Euclidean}(\pp,\qq) > 0, 290 | \end{equation*} 291 | since $\pp \neq \qq$. 292 | 293 | If $\qq\not\in \closure{B_r(\pp)}$, then by the intermediate value theorem, if $\gamma(0)=\pp$, $\gamma(1)=\qq$, then there exists some $t\in(0,1)$ with $\left\vert \pp-\gamma(t) \right\vert=\rr$. Then 294 | \begin{equation*} 295 | L_g(\gamma) \geq L_g(\gamma|_{[0,t]}) \geq c\left\vert \pp-\gamma(t) \right\vert = cr > 0, 296 | \end{equation*} 297 | so again $\inf L_g(\gamma) \geq cr > 0$. 298 | 299 | \item There's a bijection between 300 | \begin{equation*} 301 | \left\{ \text{paths from $\pp$ to $\qq$} \right\} \longleftrightarrow \left\{ \text{paths from $\qq$ to $\pp$} \right\} 302 | \end{equation*} 303 | taking $\gamma(t) \mapsto \gamma(1-t) = \gamma^{-1}(t)$; that is, just traversing the paths in the opposite directions. So we have 304 | \begin{equation*} 305 | L_g(\gamma) = L_g(\gamma^{-1}) \implies d(\pp,\qq) = d(\qq,\pp). 306 | \end{equation*} 307 | 308 | \pagebreak 309 | 310 | \item We want to show that $d(\pp,\qq)+d(\qq,\rr) \geq d(\pp,\rr)$. Pick: 311 | \begin{itemize} 312 | \shortskip 313 | \item $\gamma_1$ with $\gamma_1(0)=\pp$, $\gamma_1(1)=\qq$, and $L_g(\gamma_1) \leq d(\pp,\qq)+\epsilon$ ($\epsilon>0$), and; 314 | \item $\gamma_2$ with $\gamma_2(0)=\qq$, $\gamma_2(1)=\rr$, and $L_g(\gamma_2) \leq d(\qq,\rr)+\epsilon$. 315 | \end{itemize} 316 | % \missingfigure{Geo 10/1} 317 | Define $\gamma$ by 318 | \begin{equation*} 319 | \gamma(t) = 320 | \begin{cases} 321 | \gamma_1(2t) & \text{if } t \leq 1/2, \\ 322 | \gamma_2(2t-1) & \text{if } t>1/2. 323 | \end{cases} 324 | \end{equation*} 325 | Then $\gamma$ is piecewise smooth and 326 | \begin{equation*} 327 | L_g(\gamma) 328 | = L_g(\gamma_1) + L_g(\gamma_2) 329 | = d(\pp,\qq) + d(\qq,\rr) + 2\epsilon 330 | \geq d(\pp,\rr), 331 | \end{equation*} 332 | and letting $\epsilon\to 0$ gives 333 | \begin{equation*} 334 | d(\pp,\qq) + d(\qq,\rr) \geq d(\pp,\rr). \qedhere 335 | \end{equation*} 336 | \end{enumerate} 337 | \end{proof} 338 | 339 | So now this definitely defines a metric. Now we move to consider angles. If $\gamma_1,\gamma_2:[0,1]\to U$, with $\gamma_i(t_i) = \pp$, then let $\vv_i = \gamma\p_i(t_i)$. 340 | 341 | \begin{center} 342 | \begin{tikzpicture}[scale=2.5, rotate=22] 343 | \draw plot [smooth] coordinates {(-0.8, -0.512) (-0.7, -0.343) (-0.6, -0.216) (-0.5, -0.125) 344 | (-0.4, -0.064) (-0.3, -0.027) (-0.2, -0.008) (-0.1, -0.001) (0,0) 345 | (0.1, 0.001) (0.2, 0.008) (0.3, 0.027) (0.4, 0.064) (0.5, 0.125) 346 | (0.6, 0.216) (0.7, 0.343) (0.8, 0.512)}; % 347 | 348 | \draw (0.8,0.512) node [above right] {$\gamma_1$}; 349 | 350 | \begin{scope} [rotate=-72] 351 | \draw plot [smooth] coordinates {(-0.8, -0.512) (-0.7, -0.343) (-0.6, -0.216) (-0.5, -0.125) 352 | (-0.4, -0.064) (-0.3, -0.027) (-0.2, -0.008) (-0.1, -0.001) (0,0) 353 | (0.1, 0.001) (0.2, 0.008) (0.3, 0.027) (0.4, 0.064) (0.5, 0.125) 354 | (0.6, 0.216) (0.7, 0.343) (0.8, 0.512)}; % 355 | 356 | \draw (0.8,0.512) node [right] {$\gamma_0$}; 357 | \end{scope} 358 | 359 | \draw (0,0) node {$\bullet$}; 360 | \draw (0.05,0) node [above=3pt] {$\pp$}; 361 | 362 | \bigarrow [thick] (0,0) -- (0.5,0); 363 | \bigarrow [thick] (0,0) -- ++ (-72:0.5); 364 | 365 | \draw (0.2,0) arc (0:-72:0.2); 366 | \draw (0.16,-0.14) node [right] {$\theta$}; 367 | \end{tikzpicture} 368 | \end{center} 369 | 370 | % \missingfigure{Geo 10/2} 371 | 372 | The angle $\theta$ between $\gamma_1$ and $\gamma_2$ at $\pp$ is defined to be 373 | \begin{equation*} 374 | \cos\theta = \f{g_\pp(\vv_1,\vv_2)}{\left\vert \vv_1 \right\vert_g \left\vert \vv_2 \right\vert_g}, \text{ where } \left\vert \vv_i \right\vert_g = \sqrt{g_\pp(\vv_i,\vv_i)}. 375 | \end{equation*} 376 | If $g$ is induced by $\sigma:U\to\R^3$, then $\theta$ is the Euclidean angle between $\Gamma_1=\sigma\circ\gamma_1$ and $\Gamma_2=\sigma\circ\gamma_2$ at $\sigma(\pp)$. 377 | 378 | % subsection geometry_with_the_riemannian_metric (end) 379 | 380 | \pagebreak 381 | 382 | \subsection{Isometries} % (fold) 383 | \label{sub:riemannian_isometries} 384 | 385 | Let $U_i\subset\R^2$. Suppose $\varphi:U_1\to U_2$ is bijective, and that $\varphi,\varphi^{-1}$ are smooth. 386 | 387 | If $g_2$ is an Riemannian metric on $U_2$, then there is an induced metric $g_2\p$ on $U_1$, given by 388 | \begin{equation*} 389 | g\p_{2\pp}(\va,\vb) = g_{2\,\varphi(\pp)} (\ddif{\varphi}{\pp}(\va), \ddif{\varphi}{\pp}(\vb)). 390 | \end{equation*} 391 | In terms of matrices, we have 392 | \begin{equation*} 393 | g_2\p = \dif{\varphi^\Trans} \mat{E_2 & F_2 \\ F_2 & G_2} \dif{\varphi}, 394 | \end{equation*} 395 | where 396 | \begin{equation*} 397 | g_2 = \mat{E_2 & F_2 \\ F_2 & G_2}. 398 | \end{equation*} 399 | 400 | \begin{definition} 401 | If $\varphi$ is as above and $g_i$ is an Riemannian metric on $U_i$, we say that $\varphi$ is a \emph{Riemannian isometry} if $g_1 = g_2\p$; that is, 402 | \begin{equation*} 403 | g_1(\va,\vb) = g_2(\dif{\varphi(\va)}, \dif{\varphi(\vb)}). 404 | \end{equation*} 405 | \end{definition} 406 | 407 | \begin{proposition} 408 | If $\varphi:U_1 \to U_2$ is a Riemannian isometry, then 409 | \begin{enumerate} 410 | \shortskip 411 | \item $L_{g_1}(\gamma) = L_{g_2}(\varphi \circ \gamma)$. 412 | \item $d_1(\pp,\qq) = d_2(\varphi(\pp),\varphi(\qq))$, where $d_i$ is the metric induced by $g_i$. 413 | \item The angle between $\gamma_1$ and $\gamma_2$ at $\pp$ is the angle between $\varphi\circ \gamma_1$ and $\varphi\circ\gamma_2$ at $\varphi(\pp)$. 414 | \item If $A\subset U_1$, then $\Area_1(A) = \Area_2(\varphi(A))$. 415 | \end{enumerate} 416 | \end{proposition} 417 | 418 | \begin{proof} 419 | \mbox{} 420 | \begin{enumerate} 421 | \item First we have 422 | \begin{equation*} 423 | g_2((\varphi \circ \gamma)', (\varphi \circ \gamma)') 424 | = g_2(\dif{\varphi(\gamma\p)}, \dif{\varphi(\gamma\p)}) 425 | = g_1(\gamma\p,\gamma\p). 426 | \end{equation*} 427 | Thus we have 428 | \begin{equation*} 429 | L_{g_2}(\varphi \circ \gamma) 430 | = \int_0^1 \sqrt{g_2((\varphi \circ \gamma)', (\varphi\circ\gamma)')} \dif{t} 431 | = \int_0^1 \sqrt{g_1(\gamma\p, \gamma\p)} \dif{t} 432 | = L_{g_1}(\gamma). 433 | \end{equation*} 434 | \item There's a bijection 435 | \begin{equation*} 436 | \left\{\text{paths from $p$ to $q$ in $U_1$}\right\} \longleftrightarrow \left\{\text{paths from $\varphi(\pp)$ to $\varphi(\qq)$ in $U_2$}\right\} 437 | \end{equation*} 438 | taking $\gamma \mapsto \varphi\circ\gamma$. Since $L_{g_1}(\gamma) = L_{g_2}(\varphi \circ \gamma)$, the infinima are the same. 439 | \item Similar to (i). 440 | \item In matrix form, 441 | \begin{equation*} 442 | g_1 = \dif{\varphi^\Trans} g_2 \dif{\varphi} \implies \det g_1 = (\det \dif{\varphi})^2 \det g_2. 443 | \end{equation*} 444 | With this in hand, we have 445 | \begin{align*} 446 | \Area_1(A) 447 | &= \iint_A \sqrt{\det g_1} \dif{A} \\ 448 | &= \iint_A \left\vert \det \dif{\varphi} \right\vert \sqrt{\det g_2} \dif{A} \\ 449 | &= \iint_{\varphi(A)} \sqrt{\det g_2} \dif{A} \\ 450 | &= \Area_2(\varphi(A)). \qedhere 451 | \end{align*} 452 | \end{enumerate} 453 | \end{proof} 454 | 455 | This naturally leads us to consider conformal maps. \lecturemarker{11}{25 Feb} % Hopefully these are familiar from \emph{Complex Analysis}, although we'll look at them slightly differently. 456 | 457 | \begin{definition} 458 | If $g_1,g_2$ are Riemannian metrics on $U\subset \R^2$, then we say that $g_1$ and $g_2$ are \emph{conformal} if 459 | \begin{equation*} 460 | g_{1\pp} = \lambda(\pp)\,g_{2\pp}, 461 | \end{equation*} 462 | where $\lambda:U\to\R^+$. 463 | \end{definition} 464 | 465 | Notice that if $g_1,g_2$ are conformal, then 466 | \begin{equation*} 467 | \f{g_{1\pp}(\va,\vb)}{\left\vert \va \right\vert_{g_{1\pp}} \left\vert \vb \right\vert_{g_{1\pp}}} 468 | = \f{\lambda(\pp)\,g_{2\pp}(\va,\vb)}{ \sqrt{\lambda(\pp)}\, \left\vert \va \right\vert_{g_{2\pp}} \sqrt{\lambda(\pp)}\,\left\vert \vb \right\vert_{g_{2\pp}}} 469 | = \f{g_{2\pp}(\va,\vb)}{\left\vert \va \right\vert_{g_{2\pp}} \left\vert \vb \right\vert_{g_{2\pp}}}, 470 | \end{equation*} 471 | so the angle between $\va, \vb$ is the same under $g_1$ and $g_2$. Conformal maps preserve angles. 472 | 473 | \begin{example} 474 | Consider the Euclidean metric $g^E$, and the spherical metric $g^S$. These are conformal: 475 | \begin{equation*} 476 | g^E = \dif{x^2} + \dif{y^2} \qqand 477 | g^S = \f{4\left.( \dif{x^2} + \dif{y^2})\right.}{\left.( 1+x^2+y^2 )\right.^2}. 478 | \end{equation*} 479 | \end{example} 480 | 481 | There's more than one definition. If $g_i$ is a metric on $U_i$, and $\phi:U_1\to U_2$, then we say that $\phi$ is conformal if $g_2\p$ defined by 482 | \begin{equation*} 483 | g_2\p(\va,\vb) = g_2(d\phi(\va), d\phi(\vb)) 484 | \end{equation*} 485 | is conformal to $g_1$. 486 | 487 | % \begin{example} 488 | % Consider the stereographic projection $\pi:S^{\,2}\backslash\{\NN\} \to \C$. This defines a conformal map. 489 | % \end{example} 490 | 491 | \begin{proposition} 492 | If $f:U_1\to U_2$ is holomorphic with $f\p(w)\neq 0$ for all $w\in U_1$, then it is conformal to the Euclidean metric $g^E$ or the spherical metric $g^S$. 493 | \end{proposition} 494 | 495 | \begin{proof} 496 | Let $z=x+iy$. Then $\zbar=x-iy$, and we have 497 | \begin{equation*} 498 | \dif{x^2} + \dif{y^2} = \dif{z} \dif{\zbar}. 499 | \end{equation*} 500 | If $z=f(w)$, then 501 | \begin{equation*} 502 | dz = \pd{f}{w} \dif{w} = f\p(w) \dif{w}, 503 | \end{equation*} 504 | % since the second term is zero by the Cauchy-Riemann equations. 505 | Similarly, we have 506 | \begin{equation*} 507 | \dif{\zbar} = f\p(\wbar) \dif{\wbar} = \overline{f\p(w)} \dif{\wbar}. 508 | \end{equation*} 509 | Combining these two results, we have 510 | \begin{equation*} 511 | \dif{z} \dif{\zbar} = \left.|f\p(w)|\right.^2 \dif{w} \dif{\wbar}, 512 | \end{equation*} 513 | and so $\dif{z} \dif{\zbar}$ is conformal to $\dif w \dif{\wbar} = g^E$. 514 | \end{proof} 515 | 516 | \begin{corollary} 517 | Mobius transformations are conformal with respect to $g^E$. 518 | \end{corollary} 519 | 520 | \vspace{-6pt} 521 | 522 | \emph{Converse.} If $f:U_1\to U_2$ is conformal, then either 523 | \begin{enumerate} 524 | \shortskip 525 | \item $f$ is orientation preserving, and hence holomorphic, or; 526 | \item $f$ is orientiation reversing, and $f(z)=g(\zbar)$, where $g$ is holomorphic. 527 | \end{enumerate} 528 | 529 | \emph{Idea of proof.} Since $\pi$ is conformal, and isometries are conformal, we see that $\pi\circ A \circ\pi^{-1}$ is a conformal map $\C_\infty \to \C_\infty$. Every orientation preserving conformal map of $\C_\infty$ is a Mobius map, which is proved properly in \emph{Complex Analysis} or \emph{Complex Methods}. 530 | 531 | % subsection riemannian_isometries (end) -------------------------------------------------------------------------------- /ib-met-top-spaces/mettop-chap-02.tex: -------------------------------------------------------------------------------- 1 | %!TEX root = met-top-spaces.tex 2 | \stepcounter{lecture} 3 | \setcounter{lecture}{2} 4 | \sektion{Topological spaces} 5 | 6 | \subsection{Introduction} % (fold) 7 | \label{sub:top_introduction} 8 | 9 | We've already discussed some of the properties of open subsets of metric spaces. We can abstract these for a definition of a \emph{topological space}. 10 | 11 | \begin{definition} 12 | A \emph{topological space} $(X,\tau)$ consists of a set $X$ and a set (called the \emph{topology}) $\tau$ of subsets of $X$ (hence $\tau \subset \powerset(X)$, the power set of $X$). By definition, we call the elements of $\tau$ the \emph{open sets}, satisfying the three properties 13 | \begin{enumerate} 14 | \shortskip 15 | \item $X,\emptyset \in \tau$; 16 | \item If $U_i \in \tau$ for all $i\in I$, then $\bigcup_{i\in I} U_i \in \tau$; 17 | \item If $U_1, U_2 \in \tau$, then $U_1 \cap U_2 \in \tau$ (or similarly for finite intersections). 18 | \end{enumerate} 19 | In this sense, a metric space $(X,\rho)$ gives rise to a topology, which we call the \emph{metric topology}. 20 | 21 | Two metrics $\rho_1$ and $\rho_2$ on a set $X$ are called (topologically) \emph{equivalent} if the associated topologies are the same. 22 | \end{definition} 23 | 24 | \begin{exercise} 25 | Show that Lipschitz equivalence implies equivalence. 26 | \end{exercise} 27 | 28 | \begin{example} 29 | The discrete metrix on a set $X$ gives rise to the \emph{discrete topology}, in which \emph{all} subsets are open; that is, $\tau = \powerset(X)$. 30 | \end{example} 31 | 32 | \begin{examples} 33 | [Non-metric topologies] \label{eg:non-metric-topologies} \mbox{} 34 | \begin{enumerate} 35 | \shortskip 36 | \item Let $X$ be a set with at least two elements, and take $\tau=\{\emptyset, X\}$. This is the \emph{indiscrete topology}. 37 | \item Let $X$ be any (infinite) set, and take 38 | \begin{equation*} 39 | \tau = \left\{\emptyset\right\} \cup \left\{Y \cup \text{$X$ such that $X\backslash Y$ is finite}\right\}. 40 | \end{equation*} 41 | Then $(X,\tau)$ is a topological space, and $\tau$ is the \emph{cofinite topology}. 42 | 43 | If $X$ is $\R$ or $\C$, then this is called the \emph{Zariski topology}, and open sets are ``complements of zeroes of polynomials''. This, and similar Zariski topologies on $\Rn$ and $\Cn$, are very important in Part~II \emph{Algebraic Geometry}. 44 | \item Let $X$ be any (uncountable) set, such as $\R$ or $\C$, and take 45 | \begin{equation*} 46 | \tau = \left\{\emptyset\right\} \cup \left\{Y \cup \text{$X$ such that $X\backslash Y$ is countable}\right\}. 47 | \end{equation*} 48 | This is the \emph{co-countable topology}. 49 | \item Finally, we consider some finite topologies. Take $X=\{a,b\}$. Then there are four distinct topologies: 50 | \begin{itemize} 51 | \shortskip 52 | \item Discrete (a metric topology); 53 | \item Indiscrete; 54 | \item $\{\emptyset, \{a\}, \{a,b\}\}$; 55 | \item $\{\emptyset, \{b\}, \{a,b\}\}$. 56 | \end{itemize} 57 | \end{enumerate} 58 | \end{examples} 59 | 60 | \pagebreak 61 | 62 | \begin{remark} 63 | A subset $Y\subset X$ of a topological space $(X,\tau)$ is called \emph{closed} if $X\backslash Y$ is open, just as we defined in metric spaces. 64 | 65 | We can describe a topology on a set $X$ by specifying the \emph{closed sets} in $X$; these will satisfy 66 | \begin{enumerate} 67 | \shortskip 68 | \item Both $\emptyset$ and $X$ are closed; 69 | \item If $F_i$, for $i\in I$, are closed, then so too is $\bigcap_{i\in I} F_i$. 70 | \end{enumerate} 71 | This description is sometimes more natural, such as with examples~(ii) and (iii) above. 72 | \end{remark} 73 | 74 | \begin{example} 75 | The \emph{half-open interval topology} $\tau$ on $\R$ consists of the arbitrary unions of half-open intervals $[a,b)$, for $a}[dd]^{\exists\,!\fbar} \\ 368 | \\ 369 | & & Z 370 | } 371 | \end{equation*} 372 | \end{remark} 373 | 374 | \begin{examples} 375 | \mbox{} 376 | \begin{enumerate} 377 | \item Define $\sim$ on $\R$ by $x\sim y$ if and only if $x-y\iZ$. Then the map 378 | \begin{equation*} 379 | \fullfunction{\phi}{R/\sim}{\bb{T} = \{z\iC: \left\vert z \right\vert=1\}}{[x]}{e^{2\pi ix}} 380 | \end{equation*} 381 | is both well-defined and a homeomorphism. (See Examples Sheet~1, Question~15.) 382 | 383 | \item Define the two-dimensional torus $T^{\,2}$ to be $\R^2/\sim$, where $(x_1,y_1) \sim (x_2,y_2)$ if and only if $x_1-x_2 \iZ$ and $y_1-y_2\iZ$. The topology comes from a metric on $T^{\,2}$ (examples sheet~1, question~18) and so is well-defined. (Note that we sometimes denote $T^{\,2}$ as $\R^2/\Z^2$). 384 | 385 | \emph{Remark.} In general, we can get some rather nasty (non-Hausdorff, for example) topologies for an arbitrary equivalence relation. 386 | 387 | \item \emph{Special case.} If $A\subset X$, then we can define $\sim$ on $X$ by $x \sim y$ if and only if $x=y$ or $x,y\in A$. The quotient space is sometimes written as $X/A$, in which we scrunch $A$ down to a point. (Note the conflict with example (ii).) Usually we take $A$ to be closed. 388 | 389 | For example, if $D$ is the closed unit disc in $\C$, then the boundary is the unit circle $C$, and $D/C$ is homeomorphic to $S^{\,2}$, the unit sphere. (See examples sheet~2, question~13.) 390 | \end{enumerate} 391 | \end{examples} 392 | 393 | \begin{lemma} 394 | Suppose $(X,\tau)$ is Hausdorff and $A \subset X$ is closed. Suppose further that for any $x\not\in A$, there are open sets $U,V$ with $U\cap V=\emptyset$, $U \supseteq A$ and $V \ni x$. Then $X/A$ is Hausdorff. 395 | \end{lemma} 396 | 397 | \begin{proof} 398 | Given two points $\xbar \neq \ybar$ in $X\backslash A$, we have two possibilities: 399 | \begin{enumerate} 400 | \item Neither $\xbar$ nor $\ybar$ correspond to $A$. In this case, there are unique $x,y\in X$ corresponding to $\xbar$ and $\ybar$. Thus there are 401 | \begin{itemize} 402 | \shortskip 403 | \item [] $U_x\supset A$, $V_x\ni x$ such that $U_x \cap V_x=\emptyset$; 404 | \item [] $U_y\supset A$, $V_y\ni y$ such that $U_y \cap V_y=\emptyset$. 405 | \end{itemize} 406 | Since $X$ is Hausdorff, without loss of generality we may assume that $V_x \cap V_y = \emptyset$. Thus the correpsonding open sets $q(V_x)$ and $q(V_y)$ in $X/A$ separate $\xbar$ and $\ybar$. 407 | 408 | \item We have $\xbar=q(x)$, where $x\in X\backslash A$, and $\ybar$ corresponding to $A$. Then there exist open sets $U\supset A$ and $V \ni x$ such that $U\cap V = \emptyset$. The corresponding open sets $q(U)$, $q(V)$ in $X/A$ separate $\xbar$ and $\ybar$. \qedhere 409 | \end{enumerate} 410 | \end{proof} 411 | 412 | % subsection quotient_spaces (end) 413 | 414 | \subsection{Product topologies} % (fold) 415 | \label{sub:product_topologies} 416 | 417 | \begin{definition} 418 | Given topological space $(X,\tau)$ and $(Y,\sigma)$, we define the \emph{product topology} $\tau\times \sigma$ on $X\times Y$ as follows: $W\subseteq X\times Y$ is open if and only if, for all $x,y\in X$, there exist open sets $U\subseteq X$ and $V\subseteq Y$ such that $x\in U$, $y\in V$ and $U\times V \subseteq W$. 419 | 420 | In other words, $X\times Y$ has a base of open sets of the form $U\times V$, where $U$ is open in $X$ and $V$ is open in $Y$. 421 | \end{definition} 422 | 423 | Notice that $(U_1\times V_1) \cap (U_2 \times V_2) = (U_1 \cap U_2) \times (V_1 \cap V_2)$, and so this does indeed form the base of a topology. We can extend our definition to a product of countably many spaces: 424 | 425 | \begin{definition} 426 | If $(X_i,\tau_i)_{i=1}^n$ are topological spaces, then the product topology on $\prod_{i=1}^n X_i$ is defined by having a base of open sets of the form $\prod_{i=1}^n U_i$, where $U_i$ is open in $X_i$, for $i=1,\ldots,n$. 427 | \end{definition} 428 | 429 | Again, this forms a topology. 430 | 431 | \begin{example} 432 | Consider $\R$ with the usual topology. The product topology on $\R \times \R = \R^2$ is just the usual metric topology. Basic open sets in the product topology include the open rectangles $I_1 \times I_2$ (a product of open intervals $I_1$ and $I_2$ in $\R$), and these form a base for the usual topology on $\R^2$. 433 | \end{example} 434 | 435 | \pagebreak 436 | 437 | \begin{lemma} 438 | Given topological spaces $(X_1,\tau_1)$ and $(X_2,\tau_2)$, the projection maps $\pi_i: (X_1 \times X_2, \tau_1 \times \tau_2) \to (X_i,\tau_i)$ are continuous. Moreover, given a topological space $(Y,\tau)$ with continuous map $f_i:Y\to X_i$, there is a unique factorisation, and $f$ (in the diagram) is continuous. 439 | \begin{equation*} 440 | \xymatrix{ 441 | & & & & X_1 \\ 442 | Y \ar@{-->}[rr]^{\exists\,!f} \ar@/^1.5pc/[rrrru]^{f_1} \ar@/_1.5pc/[rrrrd]_{f_2} & & 443 | {X_1 \times X_2} \ar[rru]^{\pi_1} \ar[rrd]_{\pi_2} \\ 444 | & & & & X_2 445 | } 446 | \end{equation*} 447 | \end{lemma} 448 | 449 | \begin{proof} 450 | The first part is easy: we use $\pi_1^{-1}(U)=U\times X_2$ and $\pi_2^{-1}(V)=X_1\times V$. 451 | 452 | For the second part, define $f(y)=(f_1(y),f_2(y)) \in X_1\times X_2$. The fact that $f$ is unique, making the diagram commute, is obvious. For any basic open set $U\times V \subseteq X_1 \times X_2$, where $U$ is open in $X_1$ and $V$ is open in $X_2$, we have $f^{-1}(U \times V) = f_1^{-1}(U) \cap f_2^{-1}$ open in $Y$. 453 | 454 | Further, since any open set $W$ in $X_1 \times X_2$ is the union of basic open sets, $f^{-1}(W)$ is the union of their inverse, and so $f^{-1}(W)$ is open in $W$. Thus $f$ is continuous. 455 | \end{proof} 456 | 457 | % subsection product_topologies (end) -------------------------------------------------------------------------------- /ib-geometry/geo-chap-06.tex: -------------------------------------------------------------------------------- 1 | %!TEX root = geometry.tex 2 | \stepcounter{lecture} 3 | \setcounter{lecture}{6} 4 | \sektion{Geodesics} 5 | 6 | \newcommand{\tgamma}{\tilde{\gamma}} 7 | 8 | Let $g$ be a Riemannian metric on some open set $U\subset \R^2$, say 9 | \begin{equation*} 10 | g = E \dif{x^2} + 2F \dif{x} \dif{y} + G \dif{y^2}. 11 | \end{equation*} 12 | The basis problem we want to answer is: given $p,q\in U$, how can we find the shortest path with respect to $g$ from $p$ to $q$, supposing it exists? These shortest paths are the analogues of lines in hyperbolic space. 13 | 14 | \subsection{Energy functionals} % (fold) 15 | \label{sub:energy_functionals} 16 | 17 | Let $\gamma:[0,1] \to U$ be a smooth path. As we've seen before the length is 18 | \begin{equation*} 19 | L_g(\gamma) = \int_0^1 \left.|\gamma\p(t)|\right._g \dif{t}. 20 | \end{equation*} 21 | This is invariant under reparameterisation: 22 | \begin{equation*} 23 | L_g(\gamma\circ f) = L_g(\gamma), 24 | \end{equation*} 25 | where $f:[0,1] \to [0,1]$ is monotone and continuous. 26 | 27 | Now we introduce a new function, which is not invariant under reparameterisation. This might seem like a bad thing, but actually it makes our lives a lot easier. 28 | 29 | \begin{definition} 30 | The \emph{energy} of a smooth path $\gamma:[0,1] \to U$ is 31 | \begin{equation*} 32 | E_g(\gamma) = \int_0^1 \left.|\gamma\p(t)|\right._g^2 \dif{t}. 33 | \end{equation*} 34 | \end{definition} 35 | 36 | Recall the Cauchy-Schwarz inequality, which we've met in many different contexts: 37 | \begin{equation*} 38 | \textstyle \int ab \leq \sqrt{\int a^2} \, \cdot \sqrt{\int b^2}, 39 | \end{equation*} 40 | with equality if and only if $a=\lambda b$ or $b=\lambda a$, for some $\lambda$. Then 41 | \begin{equation*} 42 | L_g(\gamma) 43 | = \int_0^1 \left.|\gamma\p(t)|\right._g \dif{t} 44 | \leq \sqrt{\int_0^1 \left.|\gamma\p(t)|\right._g^2 \dif{t}} \, \cdot \sqrt{\int_0^1 \dif{t}} 45 | = \sqrt{E_g(\gamma)}, 46 | \end{equation*} 47 | with equality if and only if $\left.|\gamma_p(t)|\right._g = \lambda \cdot 1$ (a constant function); that is, if $\gamma$ has constant speed. Now, every $\gamma$ with $\gamma\p(t) \neq 0$ for all $t$ has a constant speed reparametrisation; consider 48 | \begin{equation*} 49 | \tgamma = \gamma \circ f, \qquad 50 | f:[0,1] \to [0,1], \qquad 51 | f(s) = F^{-1}(s), 52 | \end{equation*} 53 | where $F:[0,1] \to [0,1]$ is given by 54 | \begin{equation*} 55 | F(s) = \f{\int_0^s \left.|\gamma_p(t)|\right._g \dif{t}}{\int_0^1 \left.|\gamma_p(t)|\right._g \dif{t}}. 56 | \end{equation*} 57 | 58 | \begin{definition} 59 | If $p,q \in U$, then the set of paths from $p$ to $q$ is given by $\Omega_{p,q}$; formally, 60 | \begin{equation*} 61 | \Omega_{p,q} = \left\{\text{$\gamma:[0,1]\to U$ smooth}: \gamma(0)=p, \gamma(1)=q\right\}. 62 | \end{equation*} 63 | \end{definition} 64 | 65 | \begin{proposition} 66 | The following two conditions are equivalent: 67 | \begin{enumerate} 68 | \shortskip 69 | \item $E(\gamma_0) \leq E(\gamma)$ for all $\gamma\in\Omega_{p,q}$; 70 | \item $L(\gamma_0) \leq L(\gamma)$ for all $\gamma\in \Omega_{p,q}$, and $\gamma$ has constant speed. 71 | \end{enumerate} 72 | \end{proposition} 73 | 74 | \begin{proof} 75 | (i) $\implies$ (ii). If $\tgamma_0$ has constant then, then 76 | \begin{equation*} 77 | E(\tgamma_0) 78 | = \left[ L(\tgamma_0) \right]^2 79 | = \left[ L(\gamma_0) \right]^2 80 | \leq E(\gamma_0), 81 | \end{equation*} 82 | with equality if and only if $\gamma_0$ has constant speed; that is, $\gamma_0 = \tgamma_0$. 83 | 84 | (ii) $\implies$ (i). We have 85 | \begin{equation*} 86 | E(\gamma_0) 87 | = \left[ L(\gamma_0) \right]^2 88 | \leq \left[ L(\tgamma_0) \right]^2 89 | \leq E(\gamma), 90 | \end{equation*} 91 | which is what we require. 92 | \end{proof} 93 | 94 | \vspace{3pt} 95 | 96 | \begin{note} 97 | If $\gamma\p(a) = 0$ for some $a\in[0,1]$, then can always find $F$ such that $E(\gamma \circ f) < E(\gamma)$. 98 | \end{note} 99 | 100 | % subsection energy_functionals (end) 101 | 102 | \subsection{Calculus of variations} % (fold) 103 | \label{sub:calculus_of_variations} 104 | 105 | Given $H=H(x,y,z,w)$, suppose $\gamma\in\Omega_{p,q}$ minimises 106 | \begin{equation*} 107 | \Phi(y) = \int_0^1 H(\gamma_1(t), \gamma_2(t), \gamma\p_1(t), \gamma\p_2(t)) \dif{t} 108 | \end{equation*} 109 | if, for example, 110 | \begin{equation*} 111 | H(x,y,z,w) = E(x,y)\,z^2 + 2F(x,y) \, zw + G(x,y)\,w^2. 112 | \end{equation*} 113 | Then $\Phi(\gamma) = E_g(\gamma)$. 114 | 115 | For any $\delta:[0,1] \to \R^2$ with $\delta(0)=\delta(1)=0$, if 116 | \begin{equation*} 117 | (\gamma+\epsilon\delta)(t) = \gamma(t) + \epsilon\,\delta, 118 | \end{equation*} 119 | then $\gamma+\epsilon\delta \in \Omega_{p,q}$, when $\epsilon\ll1$. Thus $\epsilon=0$ minimises $\Phi(\gamma+\epsilon\delta)$: 120 | \begin{align} 121 | 0 122 | &= \dod{}{\epsilon}\Phi(\gamma+\epsilon\delta) \notag \\ 123 | &= \int_0^1 \dod{}{\epsilon}\left[ (H(\gamma_1+\epsilon\delta_1, \gamma_2+\epsilon\delta_2, \gamma_1\p + \epsilon\delta_1\p, \gamma_2\p+\epsilon\delta_2\p) \right] \dif{t} \notag \\ 124 | &= \int_0^1 \left[ H_x \delta_1 + H_y \delta_2 + H_z \delta_1\p + H_w \delta_2\p \right] \dif{t}. \label{eq:calc-variations} 125 | \end{align} 126 | Now we have 127 | \begin{equation*} 128 | \int_0^1 H_z \delta_1\p \dif{t} 129 | = \left[ H_z \delta_1 \right]_0^1 - \int_0^1 \od{}{t}(H_z) \,\delta_1 \dif{t} 130 | = -\int_0^1 \od{}{t}(H_z)\,\delta_1 \dif{t}, 131 | \end{equation*} 132 | as $\delta$ is a closed curve. 133 | 134 | Returning to~\eqref{eq:calc-variations}, we see that 135 | \begin{equation*} 136 | \int_0^1 \left[ \left( H_x - \dod{H_z}{t} \right)\delta_1 + \left( H_y - \dod{H_w}{t} \right)\delta_2 \right] \dif{t} 137 | \equiv 0, 138 | \end{equation*} 139 | for any $\delta_1,\delta_2$ with $\delta_i(0)=\delta_i(1) = 0$, $i=1,2$. 140 | 141 | This gives us the \emph{Euler-Lagrange equations}: 142 | \begin{equation*} 143 | H_x=\od{H_z}{t} \qqand 144 | H_y = \od{H_w}{t}. 145 | \end{equation*} 146 | 147 | % subsection calculus_of_variations (end) 148 | 149 | \subsection{Geodesic equations} % (fold) 150 | \label{sub:geodesic_equations} 151 | 152 | \newcommand{\dgamma}{\dot{\gamma}} 153 | 154 | In our case, we have 155 | \begin{equation*} 156 | H(x,y,z,w) = E(x,y) \, z^2 + 2F(x,y)\,zw + G(x,y)\,w^2. 157 | \end{equation*} 158 | Simple differentiation gives us 159 | \begin{equation*} 160 | H_x = E_x\,z^2 + 2F_x\,zw + G_x\,w^2 \qqand 161 | H_z = 2Ez + 2Fw. 162 | \end{equation*} 163 | Now we write $E(x,y) = E(\gamma_1(t), \gamma_2(t))$, and similar for $F$ and $G$. Letting a dot denote differentiation with respect to $t$, and substituting into the Euler-Lagrange equations, we obtain the \emph{geodesic equations} 164 | \begin{align*} 165 | E_x \dot{\gamma}_1^2 + 2F_x \dgamma_1 \dgamma_2 + G_x \dgamma_2^2 &= \dod{}{t}(2E\dgamma_1 + 2F\dgamma_2), \\ 166 | E_y \dgamma_1^2 + 2 F_y \dgamma_1 \dgamma_2 + G_y \dgamma_2^2 &= \dod{}{t}(2F\dgamma_1 + 2G\dgamma_2). 167 | \end{align*} 168 | This is a (nasty!) system of second-order differential equations. 169 | 170 | \begin{definition} 171 | A path $\gamma:[a,b]\to U$ is a \emph{geodesic} if it satisfies the geodesic equations, or if is a critical points for the energy functional. 172 | 173 | A shortest length, constant speed path is a geodesic. 174 | \end{definition} 175 | 176 | \begin{theorem} 177 | Given $\pp\in U$ and $\xx\iR^2$, there is a unique geodesic $\gamma:(-\epsilon,\epsilon)\to U$ with $\gamma(0)=\pp$, $\gamma\p(0)=\xx$. 178 | \end{theorem} 179 | 180 | \begin{proof} 181 | This is an immediate consequence of the existence and uniqueness of solutions for ordinary differential equations. 182 | \end{proof} 183 | 184 | % subsection geodesic_equations (end) 185 | 186 | \pagebreak 187 | 188 | \subsection{Exponential map} % (fold) 189 | \label{sub:exponential_map} 190 | 191 | \lecturemarker{14}{6 Mar} 192 | \newcommand{\gvt}{\gamma_{\vv_\theta}} 193 | \newcommand{\dGamma}{\dot{\Gamma}} 194 | 195 | For $\pp\in U$, $\vv\iR^2$, there's a unique geodesic $\gamma_\vv:(-\epsilon,\epsilon) \to U$ with $\gamma(0)=\pp$, $\gamma\p(0)=\vv$. So why can't we extend this over all of $\R$? There are lots of reasons. Consider, for example, $U=\R^2\backslash\{0\}$. There is not geodesic linking the points $-x$ and $x$, $x\iR$, since we cannot go through the point $0$. 196 | 197 | \begin{lemma} 198 | $\gamma_\vv(\lambda t) = \gamma_{\lambda\vv}(t)$, $\lambda\iR$. 199 | \end{lemma} 200 | 201 | \begin{proof} 202 | If $\gamma(t)$ satisfies the geodesic equations, so does $\gamma(\lambda t)$ (Both sides get multiplied by $\lambda^2$.) So $\overline{\gamma}=\gamma(\lambda t)$ is a geodesic with $\overline{\gamma}(0) = \gamma(0) = \pp$, $\overline{\gamma}\p(0) = \lambda\,\overline{\gamma}\p(0) = \lambda\vv$, and this gives us $\overline{\gamma} = \gamma_{\lambda\vv}$. 203 | \end{proof} 204 | 205 | \begin{definition} 206 | The \emph{exponential map} $\exp_\pp:B_\epsilon(0) \to U$ is given by 207 | \begin{equation*} 208 | \exp_\pp(\vv) = \gamma_\vv(1) 209 | \end{equation*} 210 | \end{definition} 211 | 212 | Note $\exp_\pp(\lambda \vv) = \gamma_{\lambda\vv}(1) = \gamma_\vv(\lambda)$, so $\exp_\pp(\vv)$ is defined for $|\vv|$ small. 213 | 214 | \begin{proposition} 215 | $\eval[0]{\dif{\,\exp_\pp}}_0 = I$. 216 | \end{proposition} 217 | 218 | \begin{proof} 219 | Working from the definition, we have: 220 | \begin{align*} 221 | \eval[0]{\dif{\,\exp_\pp}}_0 222 | &= \lim_{\epsilon\to 0} \f{\exp_p(\epsilon\ww)-\exp_p(0)}{\epsilon} \\ 223 | &= \lim_{\epsilon\to 0} \f{\gamma_{\epsilon\ww}(1) - \gamma_0(1)}{\epsilon} \\ 224 | &= \lim_{\epsilon\to 0} \f{\gamma_\ww(\epsilon) - \gamma_\ww(0)}{\epsilon} \\ 225 | &= \gamma\p_\ww(0) = \ww. \qedhere 226 | \end{align*} 227 | \end{proof} 228 | 229 | \begin{corollary} 230 | There are open sets $V_1 \subset \R^2$, $V_2\subset U$ with $0\in V_1, \pp\in V_2$, such that $\exp_p:V_1 \to V_2$ is a diffeomorphism; that is, differentiable, bijective and the inverse is differentiable. 231 | \end{corollary} 232 | 233 | \begin{proof} 234 | This follows from the inverse function theorem, since $I$ is invertible. Equivalently, we cause $\exp_p$ to define a new set of coords on $V_2$. 235 | \end{proof} 236 | 237 | % subsection exponential_map (end) 238 | 239 | \subsection{Geodesic polar coordinates} % (fold) 240 | \label{sub:geodesic_polar_coordinates} 241 | 242 | Pick $\vv_1,\vv_2$ orthogonal with respect to the Riemannian metric $g_\pp$. Then we define 243 | \begin{equation*} 244 | \vv_\theta = \vv_1 \cos\theta + \vv_2 \sin\theta. 245 | \end{equation*} 246 | This allows us to define the map 247 | \begin{equation*} 248 | \fullfunction{T}{[0,\epsilon)}{[0,2\pi) \times U}{(r,\theta)}{\exp_p(r\vv_\theta)}. 249 | \end{equation*} 250 | These define a set of \emph{geodesic polar coordinates}. 251 | 252 | Let $\overline{g} = \overline{E} \dif{r^2} + 2\overline{F} \dif{r}\dif{\theta} + \overline{G} \dif{\theta^2}$ be the metric induced from $g$ using $T$; that is, 253 | \begin{equation*} 254 | \overline{E} = g(T_r,T_r), 255 | \qquad \overline{F} = g(T_r,T_\theta), 256 | \qquad \overline{G} = (T_\theta,T_\theta). 257 | \end{equation*} 258 | 259 | \begin{example} 260 | Consider the Euclidean metric $g = g^E = \dif{x^2} + \dif{y^2}$, with exponential map $T(r,\theta) = \exp_0(r\vv_\theta) = r\vv_\theta$. Then 261 | \begin{align*} 262 | x = r\cos\theta, & \qquad \dif{x} = \dif{r} - r\sin\theta \dif{\theta}, \\ 263 | y = r\sin\theta, & \qquad \dif{y} = \dif{r} + r\cos\theta \dif{\theta}. 264 | \end{align*} 265 | Thus we have 266 | \begin{equation*} 267 | \overline{g} = \dif{x^2} + \dif{y^2} = \dif{r^2} + r^2 \dif{\theta^2}. 268 | \end{equation*} 269 | \end{example} 270 | 271 | We consider a similar problem on the third examples sheet: 272 | 273 | \begin{example} 274 | We consider the metric on the disc: 275 | \begin{equation*} 276 | g = g^D = \f{4\left( \dif{x^2}+\dif{y^2} \right)}{\left( 1-x^2-y^2 \right)^2}. 277 | \end{equation*} 278 | Then our map is given by $T(r,\theta) = 2\tanh^{-1} r \vv_\theta$, and the induced metric is 279 | \begin{equation*} 280 | \overline{g} = \dif{r^2} + \sinh^2 r \dif{\theta}^2. 281 | \end{equation*} 282 | \end{example} 283 | 284 | There's a common pattern in both of these examples: 285 | 286 | \vspace{3pt} 287 | 288 | \begin{proposition} 289 | Under the notation established thus far, we have $\overline{E}=1$, $\overline{F}=0$, $\overline{G} = r^2 \tilde{G}(r,\theta)$ where $\lim_{r\to 0} \tilde{G}(r,\theta) = 1$. 290 | \end{proposition} 291 | 292 | \begin{proof} 293 | First we need a lemma: 294 | 295 | \vspace{8pt} 296 | 297 | \begin{lemma} 298 | Geodesics have constant speed: 299 | \begin{equation*} 300 | \left.|\gamma\p_\vv(t)|\right._g = \left.|\gamma\p_\vv(0)|\right._g = \left.|\vv|\right._{g_0}. 301 | \end{equation*} 302 | \end{lemma} 303 | 304 | \vspace{-3pt} 305 | 306 | Proof of the lemma is on the examples sheet. Now we can prove the proposition, tackling each function in turn: 307 | \begin{enumerate} 308 | \item From our previous work, we have 309 | \begin{equation*} 310 | T_r = \pd{}{r}(\gamma_{\vv_\theta}(r)) = \gamma\p_{\vv_\theta}(r). 311 | \end{equation*} 312 | Using our lemma, we thus have 313 | \begin{equation*} 314 | \overline{E} 315 | = g(T_r,T_r) 316 | = g(\gamma\p_{\vv_\theta}, \gamma\p_{\vv_\theta}) 317 | = g(\gamma\p_{\vv_\theta}(0), \gamma\p_{\vv_\theta}(0)) 318 | = g_0(\vv_\theta, \vv_\theta) 319 | = 1. 320 | \end{equation*} 321 | 322 | \item First consider the energy functional 323 | \begin{equation*} 324 | E_g^{[0,r]}(\gamma_{\vv_\theta}) 325 | = \int_0^r g(\gamma\p_{\vv_\theta}, \gamma\p_{\vv_\theta}) \dif{t} 326 | = \int_0^r 1 \dif{t} 327 | = r. 328 | \end{equation*} 329 | Thus we have 330 | \begin{equation*} 331 | 0 332 | = \pd{}{\theta}\left[ E_g^{[0,r]}(\gamma_{\vv_\theta}) \right] 333 | = \pd{}{\epsilon}\left[ E_g^{[0,r]}(\gvt + \epsilon\delta) \right], 334 | \end{equation*} 335 | where $\delta(t) = \pd{}{\theta}(\gvt(t)) = T_\theta(t,\theta)$. 336 | 337 | From our derivativation of the geodesic equations, we know 338 | \begin{align} 339 | &\dpd{}{\epsilon}\left[ E_g^{[0,r]}(\gvt + \epsilon\delta) \right] \notag \\ 340 | &\qquad= \underbrace{\left[ H_z \delta_1 \right]_0^r + \left[ H_w \delta_2 \right]_0^r}_{\eqref{eq:calc-variations}} + \int_0^r \left( H_x - \dod{H_z}{t} \right) \delta_1 + \left( H_y - \dod{H_w}{t} \right) \delta_2 \dif{t}. 341 | \label{eq:induced-metric-geodesic} 342 | \end{align} 343 | The integral cancels to zero, since $\gvt$ is a geodesic. Hence,~\eqref{eq:calc-variations} and~\eqref{eq:induced-metric-geodesic} give 344 | \begin{equation*} 345 | 2 \left[ \left.(E\dgamma_1 + F\dgamma_2)\right.\delta_1 + \left.(F\dgamma_1 + G\dgamma_2)\right. \delta_2 \right] 346 | = 2g(\gvt\p, \delta) = 0. 347 | \end{equation*} 348 | But $2g(T_r,T_\theta) = 2g(\gvt\p,\delta)$, so $\overline{F}=0$. 349 | 350 | \item First we consider 351 | \begin{equation*} 352 | T_\theta(0,\theta) = \pd{}{\theta}(\gvt(0)) = \pd{\pp}{\theta} = \vec{0} 353 | \end{equation*} 354 | Then we have 355 | \begin{align*} 356 | \dpd{T}{r}(T_\theta)(0,\theta) = T_{\theta r}(0,\theta) = T_{r\theta}(0,\theta) 357 | &= \dpd{}{\theta}T_r(0,\theta) \\ 358 | &= \dpd{}{\theta}(\gvt\p(0)) \\ 359 | &= \dpd{}{\theta}(\vv_\theta) \\ 360 | &= -\vv_1 \sin\theta + \vv_2 \cos\theta = \vv_\theta^\perp 361 | \end{align*} 362 | Unpacking this, we deduce that 363 | \begin{equation*} 364 | T_\theta(r,\theta) = r\vv(r,\theta), 365 | \end{equation*} 366 | where $\lim_{r\to 0} \vv(r,\theta) = \vv_\theta^\perp$. Now 367 | \begin{equation*} 368 | \overline{G} = g(T_\theta, T_\theta) = r^2 g(\vv,\vv) = r^2 \tilde{G}(r,\theta), 369 | \end{equation*} 370 | where 371 | \begin{equation*} 372 | \lim_{r\to 0} \tilde{G}(r,\theta) = \lim_{r \to 0} g(\vv,\vv) = \eval[0]{g}_0(\vv_\theta^\perp, \vv_\theta^\perp) = 1. \qedhere 373 | \end{equation*} 374 | \end{enumerate} 375 | \end{proof} 376 | 377 | % subsection geodesic_polar_coordinates (end) 378 | 379 | \subsection{Local Gauss-Bonnet} % (fold) 380 | \label{sub:local_gauss_bonnet} 381 | 382 | First we set up an open set $U\subset \R^2$ with geodesic polar coordinates $g=\dif{r^2} + G\dif{\theta^2}$, as discussed in the previous section. Let $\triangle OBC$ be a geodesic triangle. 383 | 384 | \begin{center} 385 | \begin{tikzpicture}[scale=2.2] 386 | \draw (0,0) node {$\bullet$} node [below left] {$O$} 387 | -- ++ (0:2) node {$\bullet$} node [below right] {$B$} 388 | -- ++ (120:2) node {$\bullet$} node [above=2pt] {$C$} -- cycle; 389 | 390 | \draw [thick] (0,0) 391 | arc (140:90:1) 392 | arc (270:307.5:1.4) node {$\bullet$} 393 | arc (307.5:340:1.4); 394 | 395 | \draw [thick] (0,0) 396 | arc (140:90:1) 397 | arc (270:307.5:1.4) ++ (0,-0.05) node [right=4pt] {$P_\theta$}; 398 | 399 | \draw [dashed] (0,0) -- ++ (50:0.8); 400 | \draw (0.4,0) arc (0:50:0.4); 401 | \draw (0.35,0.15) node [right=2pt] {$\alpha$}; 402 | 403 | \draw (1.6,0) arc (180:120:0.4); 404 | \draw (2,0) ++ (120:1.6) arc (300:240:0.4); 405 | 406 | \draw (1,1.4) node [below=4pt] {$\gamma$}; 407 | \draw (1.76,0.22) node [left=5pt] {$\beta$}; 408 | 409 | \draw [dashed] (0,0) arc (140:90:1) arc (270:307.5:1.4) -- ++ (37.5:0.8); 410 | \draw (0,0) arc (140:90:1) arc (270:307.5:1.4) -- ++ (37.5:0.3) arc (37.5:120:0.3); 411 | 412 | \draw (1.7,0.92) node [above] {$\phi(\theta)$}; 413 | \end{tikzpicture} 414 | \end{center} 415 | 416 | Geodesic triangles are formed by the arcs of three geodesics on a curved surface; straight lines are used above only for illustrative purposes. Here we have 417 | \begin{align*} 418 | OB &= \left\{\theta=0, r\in[0,b]\right\}, \\ 419 | OC &= \left\{\theta=\alpha, r\in[0,c]\right\}, \\ 420 | BC &= \left\{\Gamma(\theta) = (f(\theta,\theta)), \theta\in[0,1]\right\}. 421 | \end{align*} 422 | Now let $P_\theta=\Gamma(\theta)$. Then $\phi(\theta)$ is the angle between $OP_\theta$ and $BC$. Note that $\phi(0) = \pi-\beta$ and $\phi(\alpha)=\gamma$. 423 | 424 | The length of this curve $BP_\theta$, given by 425 | \begin{equation*} 426 | s(\theta) = \int_0^\theta |\Gamma\p(u)| \dif{u}. 427 | \end{equation*} 428 | We then define 429 | \begin{equation*} 430 | h(\theta) := \od{s}{\theta} = \left.|\Gamma\p(\theta)|\right._g 431 | \qquad 432 | \text{which gives us} 433 | \qquad 434 | \od{f}{s} = \od{f}{\theta} \od{\theta}{s} = \f{f\p(\theta)}{h(\theta)}. 435 | \end{equation*} 436 | 437 | \begin{lemma} 438 | $\disp \dod{}{s}\left( \f{f\p}{h} \right) = \f{G_r}{2h}.$ 439 | \end{lemma} 440 | 441 | \begin{proof} 442 | If we parametrise by arc length $\Gamma$, then it satisfies the geodesic equations. Letting a dot denote differentiation with respect to $s$: 443 | \begin{equation*} 444 | \od{}{s}\left[ 2E\,\dGamma_1 + 2F\,\dGamma_2 \right] = E_r\,\dGamma_1^2 + F_r\,\dGamma_1 \, \dGamma_2 + G_r\,\dGamma_2^2. 445 | \end{equation*} 446 | Most terms didsppaear, leaving 447 | \begin{equation*} 448 | \od{}{s}\left( 2\dod{f}{s} \right) = G_r \left( \dod{\theta}{s} \right)^2. 449 | \end{equation*} 450 | Finally, this gives us 451 | \begin{equation*} 452 | 2\,\od{}{s}\left( \f{f\p}{h} \right) = \f{G_r}{h^2}, 453 | \end{equation*} 454 | which is easily rearranged to give the result. 455 | \end{proof} 456 | 457 | \begin{lemma} 458 | $\od{\phi}{\theta} \equiv \phi\p = (-\sqrt{G}\,)r.$ 459 | \end{lemma} 460 | 461 | \begin{proof} 462 | In the diagram above, $OP_\theta$ is a ray of constant $\theta$, parameterised by $\rho(u) = (u,\theta)$ and $\rho\p(u) = (1,0)$. Now $\Gamma\p=(f\p,1)$, and then 463 | \begin{equation*} 464 | \cos \phi 465 | = \f{g(\Gamma\p, \rho)}{|\Gamma\p|_g |\rho\p|_g} 466 | = \f{f\p}{h \cdot 1} 467 | = \f{f\p}{h}. \tag{$*$} 468 | \end{equation*} 469 | Now we also consider 470 | \begin{equation*} 471 | \sin^2\phi 472 | = 1-\cos^2\phi 473 | = 1 - \left( \f{f\p}{h} \right)^2 474 | = 1-\f{(f\p)^2}{(f\p)^2 + G} 475 | = \f{G}{(f\p)^2 + G} 476 | = \f{G}{h^2}. \tag{$**$} 477 | \end{equation*} 478 | We thus have $\sin \phi = \sqrt{G}/h$. Then we differentiate $(*)$: 479 | \begin{equation*} 480 | -\phi\p \sin\phi = \left( \f{f\p}{h} \right)\p = \f{G_r}{2h}, 481 | \end{equation*} 482 | by the previous lemma. Then 483 | \begin{equation*} 484 | \phi\p 485 | = -\f{G_r}{2h \sin\theta} 486 | = -\f{G_r}{2h} \left( \f{h}{\sqrt{G}} \right) 487 | = -\f{G_r}{2\sqrt{G}} 488 | = -(\sqrt{G}\,) r. \qedhere 489 | \end{equation*} 490 | \end{proof} 491 | 492 | This leads us to one of the main theorems of this chapter: 493 | 494 | \begin{theorem} 495 | [Local Gauss-Bonnet theorem] For a geodesic triangle as described previously, 496 | \begin{equation*} 497 | \defect(OBC) 498 | = \alpha+\beta+\gamma-\pi 499 | = \iint_{OBC} \f{-(\sqrt{G}\,)_{rr}}{\sqrt{G}} \dif{A_g}. 500 | \end{equation*} 501 | \end{theorem} 502 | 503 | \begin{proof} 504 | We have $\dif{A_g} = \sqrt{\deg g} = \sqrt{G}$\,, so 505 | \begin{align*} 506 | \iint_{OBC} \f{(-\sqrt{G}\,)_{rr}}{\sqrt{G}} \dif{A_g} 507 | &= \int_0^\alpha \int_0^{f(\theta)} (-\sqrt{G}\,)_{rr} \dif{r} \dif{\theta} \\ 508 | &= \int_0^\alpha \left[ -(\sqrt{G}\,)_r \right]_0^{f(\theta)} \dif{\theta} \\ 509 | &= \int_0^\alpha \left[ (\sqrt{G}\,)_r \right]_{r=0} - \left[ (\sqrt{G}\,)_r \right]_{r=f(\theta)} \dif{\theta}. 510 | \intertext{Note $G=r^2 \tilde{G}$, with $\tilde{G} \to 1$ as $r\to 0$. Thus $\sqrt{G} = r \sqrt{\tilde{G}}$\,, and so $(\sqrt{G}\,)_r = \sqrt{\tilde{G}} + r(\sqrt{\tilde{G}}\,)_r \to 1$ as $r\to 0$. Then} 511 | &= \int_0^\alpha (1+\phi\p) \dif{\theta} \\ 512 | &= \left[ \theta+\phi(\theta) \right]_0^\alpha \\ 513 | &= \alpha+\gamma-\left( \pi-\beta \right) \\ 514 | &= \alpha+\gamma+\beta-\pi. \qedhere 515 | \end{align*} 516 | \end{proof} 517 | 518 | \begin{corollary} 519 | For $\triangle ABC \subset U$ with geodesic sides, we have 520 | \begin{equation*} 521 | \defect(ABC) = \iint_{\triangle ABC} \f{-(\sqrt{G}\,)_{rr}}{\sqrt{G}} \dif{A_g}. 522 | \end{equation*} 523 | \end{corollary} 524 | 525 | \begin{proof}[Sketch proof] 526 | Introduce a point $O$ as follows: 527 | 528 | \begin{minipage}{0.85\textwidth} 529 | \centering 530 | \begin{tikzpicture}[scale=2] 531 | 532 | \coordinate (A) at (0,0); 533 | \coordinate (B) at (2,0); 534 | \coordinate (C) at ++(120:1.2); 535 | \coordinate (O) at ++(230:2); 536 | 537 | \draw (A) -- (B) -- (C) -- (A) -- (O) -- (B) -- (C) -- (O); 538 | 539 | \draw (0.15,0) arc (0:120:0.15); 540 | \draw (0,0) ++ (120:0.2) arc (120:230:0.2); 541 | \draw (0,0) ++ (230:0.25) arc (230:360:0.25); 542 | 543 | \draw (B) ++ (-0.6,0) arc (180:158.217:0.6); 544 | \draw (B) ++ (-0.7,0) arc (180:205:0.7); 545 | 546 | \draw (B) ++ (169.1085:0.8) node {$\beta_2$}; 547 | \draw (B) ++ (192.5:0.85) node {$\beta_1$}; 548 | 549 | \draw (A) ++ (230:1.6) arc (60:85.06:0.4); 550 | \draw (A) ++ (230:1.5) arc (60:35:0.5); 551 | 552 | \draw (C) ++ (-21.79:0.4) arc (-21.79:-60:0.4); 553 | \draw (C) ++ (-60:0.5) arc (-60:-104.92:0.5); 554 | 555 | \draw (A) ++ (60:0.3) node {$x_3$}; 556 | \draw (A) ++ (175:0.35) node {$x_1$}; 557 | \draw (A) ++ (295:0.4) node {$x_2$}; 558 | 559 | \draw (O) ++ (37.5:0.75) node {$\alpha_2$}; 560 | \draw (O) ++ (62.525:0.65) node {$\alpha_1$}; 561 | 562 | \draw (C) ++ (-40.895:0.55) node {$\gamma_2$}; 563 | \draw (C) ++ (-82.475:0.65) node {$\gamma_1$}; 564 | 565 | \draw (B) node [right] {$B$}; 566 | \draw (A) ++ (0.4,0.05) node [above] {$A$}; 567 | \draw (O) node [below left] {$O$}; 568 | \draw (C) node [above=2pt] {$C$}; 569 | \end{tikzpicture} 570 | \end{minipage} 571 | \begin{minipage}{0.13\textwidth} 572 | \begin{align*} 573 | \triangle_1 &= \triangle OAC \\ 574 | \triangle_2 &= \triangle BAO \\ 575 | \triangle_3 &= \triangle CAB 576 | \end{align*} 577 | \end{minipage} 578 | 579 | Let $\psi=\triangle_1 \cup \triangle_2 \cup \triangle_3$. Then 580 | \begin{align*} 581 | \defect(\triangle_1) + \defect(\triangle_2) + \defect(\triangle_3) 582 | &= \gamma_1 + \gamma_2 + \alpha_1 \alpha_2 + \beta_1 + \beta_2 + x_1 + x_2 + x_3 - 3\pi \\ 583 | &= \gamma_1 + \gamma_2 + \alpha_1 \alpha_2 + \beta_1 + \beta_2 - \pi 584 | = \defect(\triangle_4). 585 | \end{align*} 586 | We can apply local Gauss-Bonnet to $\triangle_1, \triangle_2, \triangle_4$: 587 | \begin{align*} 588 | \defect(\triangle_3) 589 | &= \defect(\triangle_4)-\defect(\triangle_1)-\defect(\triangle_2) \\ 590 | &= \iint_{\triangle_4} (-1)\dif{A_g} - \iint_{\triangle_1} (-1) \dif{A_g} - \iint_{\triangle_2} (-1)\dif{A_g} \\ 591 | &= \iint_{\triangle_3} \f{-(\sqrt{G}\,)_{rr}}{\sqrt{G}} \dif{A_g}, 592 | \end{align*} 593 | as required. This is only a sketch proof because we need to consider different configurations of points and triangles, but the other cases are very similar. 594 | \end{proof} 595 | 596 | \begin{corollary} 597 | \begin{equation*} 598 | \lim_{A,B,C\to p} \f{\defect(ABC)}{\Area(ABC)} = \eval[-1]{-\f{(\sqrt{G}\,)_{rr}}{\sqrt{G}}}_p. 599 | \end{equation*} 600 | \end{corollary} 601 | 602 | \begin{definition} 603 | If $g$ is a Riemannian metric on $U$, then the \emph{Gauss curvature} at $p$ is given by 604 | \begin{equation*} 605 | K_p(g) := \lim_{A,B,C\to p} \f{\defect(ABC)}{\Area(ABC)}. 606 | \end{equation*} 607 | \end{definition} 608 | 609 | Now, isometries preserve angles and areas, so if $\phi:(U_1,g_1) \to (U_2,g_2)$, then $K_p(g_1) = K_{\phi(p)}(g_2)$. The corollary shows that the limit in the definition exists (which is the hard part to prove), and is given by 610 | \begin{equation*} 611 | -(\sqrt{G}\,)_{rr}/\sqrt{G} \qquad \text{if} \qquad g=\dif{r^2} + G\dif{\theta^2}. 612 | \end{equation*} 613 | 614 | \begin{example} 615 | Consider $g=g^D$, the hyperbolic metric on $D$. In geodesic polar coordinates, this is equivalent to 616 | \begin{equation*} 617 | \overline{g} = \dif{r^2} + \sinh^2 r \dif{\theta} \qqand 618 | \sqrt{G} = \sinh r, 619 | \end{equation*} 620 | and the curvature is given by 621 | \begin{equation*} 622 | K = -\f{(\sqrt{G}\,)_{rr}}{\sqrt{G}} = -\f{\sinh r}{\sinh 1} \equiv -1. 623 | \end{equation*} 624 | \end{example} 625 | 626 | \begin{corollary} 627 | If $ABC$ is a triangle in $\HH$, then $\defect(ABC) = -\Area(ABC)$. 628 | \end{corollary} 629 | 630 | % subsection local_gauss_bonnet (end) -------------------------------------------------------------------------------- /ib-met-top-spaces/mettop-chap-01.tex: -------------------------------------------------------------------------------- 1 | %!TEX root = met-top-spaces.tex 2 | \stepcounter{lecture} 3 | \setcounter{lecture}{1} 4 | \sektion{Metric spaces} 5 | 6 | \subsection{Introduction} % (fold) 7 | \label{sub:introduction} 8 | 9 | We start by considering Euclidean space $\Rn$, equipped with the standard Euclidean inner product: given vectors $\xx,\yy\iRn$ with coordinates $x_i,y_j$, respectively, we define 10 | \begin{equation*} 11 | \left( \xx,\yy \right) := \sum_{i=1}^n x_i \, y_i, 12 | \end{equation*} 13 | which is also referred to as the dot product $\xx \cdot \yy$. 14 | 15 | From this, we can define the \emph{Euclidean norm} on $\Rn$: 16 | \begin{equation*} 17 | \norm{\xx} := \left( \xx,\xx \right)^{1/2}, 18 | \end{equation*} 19 | which represents the length of the vector $\xx$. 20 | 21 | This allows us to define a \emph{distance function}: 22 | \begin{equation*} 23 | d_2(\xx,\yy) 24 | := \norm{\xx-\yy} 25 | = \left( \sum_{i=1}^n \left( x_i-y_i \right)^2 \right)^{1/2}. 26 | \end{equation*} 27 | This is an example of a \emph{metric}. 28 | 29 | \begin{definition} 30 | A \emph{metric space} $(X,d)$ consists of a set $X$ and a function, called the \emph{metric}, $d:X\times X\to\R$ such that for all $P,Q,R \in X$: % 31 | \begin{enumerate} 32 | \shortskip 33 | \item $d(P,Q)\geq 0$ with equality if and only if $P=Q$; 34 | \item $d(P,Q)=d(Q,P)$; 35 | \item $d(P,Q) + d(Q,R) \geq d(P,R)$. 36 | \end{enumerate} 37 | \end{definition} 38 | 39 | Condition~(iii) is called the \emph{triangle inequality}. This comes from a simple result in Euclidean space. Any triangle with vertices $P$, $Q$ and $R$ satisfies the following property: 40 | 41 | \begin{itemize} 42 | \shortskip 43 | \item [] the sum of the lengths of two sides of the triangle will be at least the length of the third side. 44 | \end{itemize} 45 | 46 | In other words, travelling along straight line segments from $P$ to $Q$, and from $Q$ to $R$, the length of the journey is at least that of travelling directly from $P$ to $R$. 47 | 48 | \begin{proposition} 49 | The Euclidean distance function $d_2$ on $\Rn$ is a metric (called the \emph{Euclidean metric}). % 50 | \end{proposition} 51 | 52 | \begin{proof} 53 | Conditions (i) and (ii) are obvious in thse case, so we only need to prove (iii). For (iii), we use the Cauchy-Schwarz inequality: % 54 | \begin{equation*} 55 | \left( \sum_{i=1}^n x_i \, y_i \right)^2 \leq \left( \sum_{i=1}^n x_i^2 \right) \left( \sum_{j=1}^n y_j^2 \right), % 56 | \end{equation*} 57 | or in the inner product notation, 58 | \begin{equation*} 59 | \left( \xx,\yy \right)^2 \leq \norm{\xx} \norm{\yy}. % 60 | \end{equation*} 61 | To prove the triangle inequality, we take $P=\bf{0}\iRn$, and $Q$ to have position vector $\xx$ with respect to $P$, and $R$ to have position vector $\yy$ with respect to $Q=\bf{0}$, so $R$ has position vector $\xx+\yy$ with respect to $P$. % 62 | 63 | Cauchy-Schwarz then gives 64 | \begin{align*} 65 | \norm{\xx+\yy}^2 66 | &= \left( \xx+\yy, \xx+\yy \right) \\ 67 | &= \norm{\xx}^2 + 2\left( \xx,\yy \right) + \norm{\yy}^2 \\ % 68 | &\leq \norm{\xx}^2 + 2 \norm{\xx} \norm{\yy} + \norm{\yy}^2 \\ % 69 | &= \left( \norm{\xx} + \norm{\yy} \right)^2. 70 | \end{align*} 71 | Taking square roots gives 72 | \begin{equation*} 73 | d(P,R) = \norm{\xx+\yy} \leq \norm{\xx} + \norm{\yy} = d(P,Q) + d(Q,R). \qedhere % 74 | \end{equation*} 75 | \end{proof} 76 | 77 | For completeness, we now state and prove Cauchy-Schwarz: 78 | 79 | \begin{lemma} 80 | [Cauchy-Schwarz inequality] For all $\xx,\yy \iRn$, we have % 81 | \label{lem:cauchy-schwarz} 82 | \begin{equation*} 83 | \left( \xx,\yy \right)^2 \leq \norm{\xx}^2 \norm{\yy}^2. % 84 | \end{equation*} 85 | \end{lemma} 86 | 87 | \begin{proof} 88 | The quadratic polynomial in the real variable $\lambda$ given by % 89 | \begin{equation*} 90 | \left( \lambda\xx+\yy, \lambda\xx+\yy \right) = \norm{\xx}^2 \lambda^2 + 2\left( \xx,\yy \right)\lambda + \norm{\yy}^2 % 91 | \end{equation*} 92 | is positive semi-definite (that is, non-negative for all $\lambda$). A quadratic polynomial $a\lambda^2 + b\lambda+c$ is positive semi-definite if and only if $a\geq 0$ and $b^2\leq ac$. This gives us the desired inequality. % 93 | \end{proof} 94 | 95 | \begin{remarks} 96 | \mbox{} 97 | \begin{enumerate} 98 | \item In the Euclidean case, we have equality in the triangle inequality if and only if $Q$ lies on the straight line segment $PR$. % (Examples Sheet 1, Question 2). % 99 | \item This argument just given also proves Cauchy Schwarz for integrals: if $f$ and $g$ are continuous functions on $[0,1]$, then % 100 | \begin{equation*} 101 | \int \left( \lambda f+g \right)^2 \Longrightarrow \left( \int_0^1 fg \right)^2 \leq \int_0^1 f^2 \int_0^1 g^2 % 102 | \end{equation*} 103 | as in the previous argument. 104 | \end{enumerate} 105 | \end{remarks} 106 | 107 | \begin{examples} 108 | \mbox{} 109 | \begin{enumerate} 110 | \item For $X=\Rn$, the functions 111 | \begin{equation*} 112 | d_1(\xx,\yy) := \sum_{i=1}^n \left\vert x_i-y_i \right\vert 113 | \qquad \text{or} \qquad 114 | d_\infty(\xx,\yy) := \max_i \left\vert x_i-y_i \right\vert, 115 | \end{equation*} 116 | are both metrics. % 117 | 118 | \item For any set $X$, and $\xx,\yy \in X$, we can define the \emph{discrete metric} 119 | \begin{equation*} 120 | d_\text{disc}(\xx,\yy) := 121 | \begin{cases} 122 | 1 & \text{if } \xx\neq \yy, \\ % 123 | 0 & \text{if } \xx = \yy. 124 | \end{cases} 125 | \end{equation*} 126 | 127 | \pagebreak 128 | 129 | \item If we take $X=C[0,1]$ to be the set of continuous functions on $[0,1]$, then we can define metrics $d_1,d_2,d_\infty$ on $X$: % 130 | \begin{align*} 131 | d_1(f,g) &:= \int_0^1 \left\vert f-g \right\vert\dif{x}, \\ 132 | d_2(f,g) &:= \left( \int_0^1 \left( f-g \right)^2 \dif{x} \right)^{1/2}, \\ 133 | d_\infty(f,g) &:= \sup_{x\in[0,1]} \vert f(x)-g(x) \vert. 134 | \end{align*} 135 | For $d_2$, the triangle inequality follows from Cauchy-Schwarz for integrals and the same argument as in the proof of lemma~\ref{lem:cauchy-schwarz}. % 136 | 137 | \item \emph{British Rail metric}. Consider $\Rn$ with the Euclidean metric $d$ (in the case $n=2$) and let $O$ denote the origin $\bf{0}$. Define a new metric $\rho$ on $\Rn$ by % 138 | \begin{equation*} 139 | \rho(P,Q) := 140 | \begin{cases} 141 | d(P,\bf{0}) + d(\bf{0},Q) & \text{if } P\neq Q, \\ % 142 | 0 & \text{if } P=Q, 143 | \end{cases} 144 | \end{equation*} 145 | that is, all the journeys from $P$ to $Q\neq P$ must go via $\bf{0}$. (This is called the British Rail metric because ``All rail journeys have to go via London''.) 146 | \end{enumerate} 147 | \end{examples} 148 | 149 | Some metrics satisfy a stronger triangle inequality. 150 | 151 | \begin{definition} 152 | A metric space $(X,d)$ is \emph{ultra-metric} if $d$ satisfies a stronger condition (iii)$'$: for all $P,Q,R \in X$, % 153 | \begin{equation*} 154 | d(P,R)\leq \max \left\{ d(P,Q),d(Q,R) \right\} 155 | \end{equation*} 156 | \end{definition} 157 | 158 | \begin{example} 159 | Take $X=\Z$ and $p$ a prime number. The \emph{$p$-adic metric} is then defined as % 160 | \begin{equation*} 161 | d(m,n) := 162 | \begin{cases} 163 | 0 & \text{if } m=n, \\ % 164 | 1/p^r & \text{if } m\neq n, \text{where } r=\max\{s\iN \text{ with } p^s \divides (m-n)\}. % 165 | \end{cases} 166 | \end{equation*} 167 | We claim that this is an ultrametric. For proof, suppose $d(m,n)=1/p^{r_1}$ and $d(n,q) = 1/p^{r_2}$. Then % 168 | \begin{align*} 169 | \begin{rcases} 170 | p^{r_1} \divides (m-n) \\ % 171 | p^{r_2} \divides (n-q) 172 | \end{rcases} 173 | \implies p^{\min\{r_1,r_2\}} \divides (m-q). 174 | \end{align*} 175 | So for some $r\geq \min\{r_1,r_2\}$, we have 176 | \begin{equation*} 177 | d(m,q) 178 | = \f{1}{p^r} 179 | \leq \f{1}{p^{\min\{r_1,r_2\}}} 180 | = \max \left\{ \f{1}{p^{r_1}}, \f{1}{p^{r_2}} \right\} 181 | = \max\left\{d_p(m,n), d_p(n,q) \right\}, 182 | \end{equation*} 183 | as desired. 184 | % \begin{align*} 185 | % d(m,q) 186 | % &= \f{1}{p^r} \\ 187 | % &\leq \f{1}{p^{\min\{r_1,r_2\}}} \\ 188 | % &= \max\left\{\f{1}{p^{r_1}},\f{1}{p^{r_2}} \right\}. 189 | % \end{align*} 190 | % (Wlog, $r_1\leq r_2$, so $\min\{r_1,r_2\}=r_1$). 191 | 192 | % But $1/p^{r_1}\geq 1/p^{r_2}$ implies $\max\{1/p^{r_1}, 1/p^{r_2}\} = 1/p^{r_1}$. 193 | 194 | This can be extended to a $p$-adic metric on $\Q$. For any $x,y\iQ$ with $x\neq y$, we can write $x-y = p^r m/n$, $r\iZ$, with $m,n$ coprime to $p$. Then we define $d(x,y) = 1/p^r$ as before. Minor modifications to the prior proof will show that this yields $(\Q,d_p)$ as an ultra-metric space. 195 | \end{example} 196 | 197 | % \begin{example} 198 | % The sequence $a=1+p+\cdots+p^{n-1}$ is a convergent sequence in $(\Q,d_p)$ with limit $1/(1-p)=a$. (Since $p^n \divides (a-a_n)$ for all $n$.) That is, $d_p(a_n,a) \to 0$ as $n\to\infty$. % 199 | % \end{example} 200 | 201 | \begin{definition} 202 | We say that two metrics $\rho_1$ and $\rho_2$ on a set $X$ are \emph{Lipschitz equivalent} if there are some $0< \lambda_1 \leq \lambda_2 \iR$ such that 203 | \begin{equation*} 204 | \lambda_1 \rho_1 \leq \rho_2 \leq \lambda_2 \rho_1. 205 | \end{equation*} 206 | % $\exists\,0<\lambda_1\leq \lambda_2\iR$ such that $\lambda_1\rho_1\leq \rho_2\leq \lambda_2\rho_1$ (an equivalence relation). % 207 | \end{definition} 208 | 209 | \begin{remark} 210 | For metrics $d_1, d_2$ and $d_\infty$ on $\Rn$, we can show that 211 | \begin{equation*} 212 | d_1 \geq d_2 \geq d_\infty \geq d_2/\sqrt{n} \geq d_1/n, 213 | \end{equation*} 214 | and so these are Lipschitz equivalent. 215 | \end{remark} 216 | 217 | Of course, not all metrics are Lipschitz equivalent. Consider the following counterexample: 218 | 219 | \begin{proposition} 220 | On $C[0,1]$, the metric $d_1$ and $d_\infty$ are not Lipschitz equivalent. 221 | \label{prop:not-lip-equiv} 222 | \end{proposition} 223 | 224 | \begin{proof} 225 | For $n \geq 2$, let $f_n\in C[0,1]$ be given by 226 | 227 | \begin{center} 228 | \begin{minipage}{0.45\textwidth} 229 | \centering 230 | \begin{tikzpicture}[scale=0.8] 231 | 232 | \bigarrow (0,0) -- (5,0); 233 | \bigarrow (0,0) -- (0,5); 234 | 235 | \draw [very thick] (0,0) -- (0.8,4) -- (1.5,0) -- (4.5,0); 236 | 237 | \draw (-0.15,4) node [left] {$\sqrt{n}$} -- (0.15,4); 238 | \draw (0.8,-0.15) node [below] {$\f{1}{n}$} -- (0.8,0.15); 239 | \draw (1.5,-0.15) node [below] {$\f{2}{n}$} -- (1.5,0); 240 | 241 | \draw (4.5,0.15) -- (4.5,-0.15) node [below] {$1$}; 242 | \draw (0,0) -- (0,-0.15) node [below] {$0$}; 243 | 244 | \end{tikzpicture} 245 | \end{minipage} 246 | \begin{minipage}{0.53\textwidth} 247 | \begin{equation*} 248 | f_n(x) = 249 | \begin{cases} 250 | x/\sqrt{n} & \text{if } 0\leq x<1/n, \\ 251 | 2\sqrt{n} - x/\sqrt{n} & \text{if } 1/n\leq x<2/n, \\ 252 | 0 & \text{if } 2/n \leq x \leq 1. 253 | \end{cases} 254 | \end{equation*} 255 | \end{minipage} 256 | \end{center} 257 | 258 | Then $d_1(f_n,0)$ is the area of the triangle, while $d_\infty(f_n,0)$ is $\sqrt{n}$. Thus we have 259 | \begin{equation*} 260 | \lim_{n \to \infty} d_1(f_n,0) = 0 261 | \qqand 262 | \lim_{n \to \infty} d_\infty(f_n,0) = \infty, 263 | \end{equation*} 264 | and so these two metrics cannot possibly be Lipschitz equivalent. 265 | \end{proof} 266 | 267 | \begin{exercise} 268 | Continuing the example above, show that $d_2(f_n,0)=\sqrt{2/3}\,$ for all $n$, and so $d_2$ is not Lipschitz equivalent to $d_1$ or $d_\infty$ on $C[0,1]$. 269 | \end{exercise} 270 | 271 | % subsection introduction (end) 272 | 273 | \pagebreak 274 | 275 | \subsection{Open balls and open sets} % (fold) 276 | \label{sub:open_balls_and_open_sets} 277 | 278 | \begin{definition} 279 | Let $(X,d)$ be a metric space, $P\in X$ and $\delta>0$. The \emph{open ball of radius $\delta$ about $p$} is given by 280 | \begin{equation*} 281 | B_d(P,\delta) := \left\{Q\in X : d(P,Q)<\delta\right\}, 282 | \end{equation*} 283 | often also denoted by $B(P,\delta)$ or $B_\delta(P)$. 284 | \end{definition} 285 | 286 | \begin{examples} 287 | \mbox{} 288 | \begin{enumerate} 289 | \shortskip 290 | \item In $(\R,d_1)$, open balls are open intervals of the form $(P-\delta,P+\delta)$. 291 | \item In $\R^2$, we obtain different open balls depending on our choice of metric: 292 | \begin{itemize} 293 | \shortskip 294 | \item In $(\R^2,d_1)$, we obtain tilted squares or ``diamonds''. 295 | \item In $(\R^2,d_2)$, we obtain open discs of radius $\delta$. 296 | \item In $(\R^2,d_\infty)$, open balls are squares. 297 | \end{itemize} 298 | \mbox{} 299 | \begin{center} 300 | \begin{minipage}{0.3\textwidth} 301 | \centering 302 | \begin{tikzpicture}[scale=1.2] 303 | \draw (0,-1) -- (1,0) -- (0,1) -- (-1,0) -- cycle; 304 | \draw (0,0) node {$\bullet$}; 305 | 306 | \draw [<->] (-1.5,1) -- (-1.5,0); 307 | \draw (-1.5,0.5) node [right] {$\delta$}; 308 | \end{tikzpicture} 309 | 310 | $(\R_2,d_1)$ 311 | \end{minipage} 312 | \begin{minipage}{0.3\textwidth} 313 | \centering 314 | \begin{tikzpicture}[scale=1.2] 315 | \draw (0,0) circle (1); 316 | \draw (0,0) node {$\bullet$}; 317 | 318 | \draw [<->] (0,0.95) -- (0,0.1); 319 | \draw (0,0.5) node [right] {$\delta$}; 320 | \end{tikzpicture} 321 | 322 | $(\R_2,d_2)$ 323 | \end{minipage} 324 | \begin{minipage}{0.3\textwidth} 325 | \centering 326 | \begin{tikzpicture}[scale=1.2] 327 | \draw (-1,-1) rectangle (1,1); 328 | \draw (0,0) node {$\bullet$}; 329 | 330 | \draw [<->] (0,0.95) -- (0,0.1); 331 | \draw (0,0.5) node [right] {$\delta$}; 332 | \end{tikzpicture} 333 | 334 | $(\R_2,d_\infty)$ 335 | \end{minipage} 336 | \end{center} 337 | \mbox{} 338 | 339 | \item In $(C[0,1],d_\infty)$, the open ball of radius $\delta$ is the area swept out by translating the image of the function up and down by a distance $\delta$. 340 | 341 | \mbox{} 342 | 343 | \begin{center} 344 | \begin{tikzpicture}[scale=2,xscale=3, domain=0:1, yscale=1.2] 345 | \draw [->] (0,0) -- (1.2,0); 346 | \draw [->] (0,0) -- (0,1.8); 347 | 348 | \foreach \s in {0,1} { 349 | \draw (\s,0) -- (\s,-0.05) node [below] {$\s$}; 350 | } 351 | 352 | \draw [thick] plot [smooth, id=main] function {0.05*(2*x-1)*(8*x-1)*(8*x-7)*(x-1)*(x-2)+1.1}; 353 | \draw [dashed] plot [smooth, id=lower] function {0.05*(2*x-1)*(8*x-1)*(8*x-7)*(x-1)*(x-2)+0.8}; 354 | \draw [dashed] plot [smooth, id=upper] function {0.05*(2*x-1)*(8*x-1)*(8*x-7)*(x-1)*(x-2)+1.4}; 355 | 356 | \draw [<->] (1.05,1.1) -- (1.05,1.4); 357 | \draw (1.05,1.25) node [right] {$\delta$}; 358 | \end{tikzpicture} 359 | \end{center} 360 | \mbox{} 361 | 362 | \item For any set $X$, in $(X,d_\text{disc})$, we have $B(P,\f{1}{2}) = \{P\}$ for all $P\in X$. 363 | \end{enumerate} 364 | \end{examples} 365 | 366 | \begin{definition} 367 | A subset $U\subset X$ of a metric space $(X,d)$ is called an \emph{open subset} if, for all $P \in U$, there is some $\delta>0$ such that the open ball $B(P,\delta)$ is contained in $U$. (Note that the empty set $\emptyset$ is open, as is the whole space $X$.) 368 | 369 | Under this definition, an open subset is a union of (usually infinitely many) open balls. 370 | 371 | As the opposite to this definition, a subset $F\subset X$ is a \emph{closed subset} if $X\backslash F$ is open. 372 | \end{definition} 373 | 374 | \begin{example} 375 | Analogously to open balls, we can define \emph{closed balls}: 376 | \begin{equation*} 377 | \closure{B}(P,\delta) := \left\{Q\in X: d(P,Q) \leq \delta\right\}, 378 | \end{equation*} 379 | which is the union of the open ball $B(P,\delta)$ and its boundary. The name is appropriate: these are indeed closed. 380 | 381 | Consider: if $Q\not\in\closure{B}(P,\delta)$, then $d(P,Q)>\delta$, and we can find $\delta\p d(P,Q) - \delta\p 386 | > \delta. 387 | \end{equation*} 388 | Thus $B(Q,\delta\p) \subset X\backslash\closure{B}$. Then $\closure{B}(P,\delta)$ is closed since the complment is open. 389 | \end{example} 390 | 391 | \begin{lemma} 392 | \mbox{} 393 | \begin{enumerate} 394 | \shortskip 395 | \item Both $X$ and $\emptyset$ are open subsets of $(X,d)$. 396 | \item If $\{U_i:i\in I\}$ are open subsets of $(X,d)$, then so is $\bigcup_{i\in I} U_i$. 397 | \item If $U_1,U_2$ are open subsets, then so is $U_1 \cap U_2$. 398 | \end{enumerate} 399 | \end{lemma} 400 | 401 | \begin{proof} 402 | Both (i) and (ii) are easy, and left as exercises. 403 | 404 | For (iii): if $P \in U_1 \cap U_2$, then there are open balls $B(P,\delta_1)\subset U_1$ and $B(P,\delta_2) \subset U_2$. Thus, if $\delta=\min\{\delta_1,\delta_2\}$, then $B(P,\delta) \subset U_1 \cap U_2$. % 405 | \end{proof} 406 | 407 | \begin{definition} 408 | If $P$ is a point in $(X,d)$, then an \emph{open neighbourhood $N$ of $P$} is an open subset $N\ni P$, such as the open balls centred on $P$. 409 | \end{definition} 410 | 411 | \begin{example} 412 | Under the British Rail metric $\rho$ on $\R^2$, what are the open neighbourhoods of a point $P$? Recall that we have a point $0\in X$, % 413 | \begin{equation*} 414 | \rho(P,Q) = 415 | \begin{cases} 416 | d(P,0) + d(0,Q) & \text{if } P\neq Q, \\ % 417 | 0 & \text{if } P=Q, 418 | \end{cases} 419 | \end{equation*} 420 | where $d$ is the Euclidean metric. 421 | 422 | Hence, if $P\neq 0$, $\delta0$ ($U$ contains a Euclidean disc). % 425 | \end{example} 426 | 427 | % subsection open_balls_and_open_sets (end) 428 | 429 | \pagebreak 430 | 431 | \subsection{Limits and continuity} % (fold) 432 | \label{sub:limits_and_continuity} 433 | 434 | \begin{definition} 435 | Suppose $x_1,x_2,\ldots$ is a sequence of points in a metric space $(X,d)$. We say that $x_n$ \emph{converges to a limit} $x$ (denoted $x_n \to x$) if $d(x_n,x) \to 0$ as $n\to\infty$. 436 | 437 | Equivalently, for any $\epsilon>0$, there is some $N(\epsilon)$ such that $x_n \in B(x,\epsilon)$ for all $n\geq N$. Both of these definitions should be familiar from \emph{Analysis}. % 438 | \end{definition} 439 | 440 | \begin{examples} 441 | \begin{enumerate} 442 | \item $1+p+p^2+\cdots+p^{n-1} \to \df{1}{1-p}$ in $(\Q,d_p)$. 443 | \item Consider the sequences of functions $f_n$ defined in the proof of proposition~\ref{prop:not-lip-equiv}. We have different limits, depending on our choice of metric. Clearly in $(X,d_1)$, we have $f_n \to 0$, but this is not the case in $(X,d_2)$ or $(X,d_\infty)$. 444 | \end{enumerate} 445 | \end{examples} 446 | 447 | \begin{proposition} 448 | We have $x_n \to x$ in $(X,d)$ if and only if, for any open neighbourhood $U\ni x$, there is some $N$ such that $x_n \in U$ for all $n\geq N$. 449 | \end{proposition} 450 | 451 | \begin{proof} 452 | The ``if'' direction is clear by taking $U=B(x,\epsilon)$, for arbitrary $\epsilon$. For the converse, given an open set $U\ni x$, there is some $\epsilon>0$ such that $B(x,\epsilon) \subset U$. Hence there is some $N$ such that $x_n \in B(x,\epsilon) \subset U$ for all $n\geq N$. 453 | \end{proof} 454 | 455 | This allows us to rephrase $x_n \to x$ in terms of open subsets in $(X,d)$. 456 | 457 | \begin{example} 458 | Take $X=C[0,1]$, $d=d_1$, and the function $g_n$ given by: % 459 | 460 | \begin{center} 461 | \begin{minipage}{0.45\textwidth} 462 | \centering 463 | \begin{tikzpicture}[scale=0.8] 464 | 465 | \bigarrow (0,0) -- (5,0); 466 | \bigarrow (0,0) -- (0,5); 467 | 468 | \draw [very thick] (0,4) -- (1.5,0) -- (4.5,0); 469 | 470 | \draw (-0.15,4) node [left] {$1$} -- (0.15,4); 471 | \draw (1.5,-0.15) node [below] {$\f{1}{n}$} -- (1.5,0); 472 | 473 | \draw (4.5,0.15) -- (4.5,-0.15) node [below] {$1$}; 474 | \draw (0,0) -- (0,-0.15) node [below] {$0$}; 475 | 476 | \end{tikzpicture} 477 | \end{minipage} 478 | \begin{minipage}{0.53\textwidth} 479 | \begin{equation*} 480 | g_n(x) = 481 | \begin{cases} 482 | 1-nx & \text{if } 0\leq x<1/n, \\ 483 | 0 & \text{if } 1/n\leq x\leq 1. 484 | \end{cases} 485 | \end{equation*} 486 | \end{minipage} 487 | \end{center} 488 | 489 | Then $g_n(0)=1$ for all $n$, but $d_1(g_n,0) \to 0$; that is, $g_n \to 0$ in $(X,d_1)$ as $n\to\infty$. 490 | \end{example} 491 | 492 | \begin{definition} 493 | We say that a function $f:(X,\rho_1) \to (Y,\rho_2)$ is 494 | \begin{itemize} 495 | \shortskip 496 | \item \emph{continuous} at $x\in X$ if, given $\epsilon > 0$, there exists $\delta>0$ such that $\rho_1(x,x\p) < \delta$ implies $\rho_2(f(x), f(x\p))<\epsilon$ for all $x\p\in X$. 497 | \item \emph{uniformly continuous on $X$} if, given $\epsilon > 0$, there exists $\delta>0$ such that $\rho_1(x,x\p) < \delta$ implies $\rho_2(f(x), f(x\p))<\epsilon$ for all $x,x\p\in X$. 498 | \end{itemize} 499 | \end{definition} 500 | 501 | \vspace{3pt} 502 | 503 | \begin{remark} 504 | We may rephrase this: $f: (X,\rho_1) \to (Y,\rho_2)$ is continuous at $x\in X$ if, given $\epsilon>0$, there exists $\delta>0$ such that $f(B(x,\delta)) \subset B(f(x),\epsilon)$; that is, $B(x,\delta) \subset f^{-1}(B(f(x),\epsilon))$. 505 | \end{remark} 506 | 507 | \pagebreak 508 | 509 | \begin{lemma} 510 | If $f:(X,\rho_1) \to (Y,\rho_2)$ is continuous and $x_n\to x$ in $(X,\rho_1)$, then $f(x_n) \to f(x)$ in $(Y,\rho_2)$. 511 | \end{lemma} 512 | 513 | \begin{proof} 514 | Given $\epsilon>0$, there exists $\delta>0$ such that $\rho_1(x,x\p)<\delta$ implies $\rho_2(f(x),f(x\p)) < \epsilon$. As $x_n\to x$, we know there exists $N$ such that for all $n\geq N$, $\rho_1(x_n,x)<\delta$. Hence, for $n\geq N$, $\rho_2(f(x_n),f(x))<\epsilon$, and so $f(x_n) \to f(x)$. 515 | \end{proof} 516 | 517 | \begin{example} 518 | Consider the identity map $\id: (C[0,1],d_\infty) \to (C[0,1],d_1)$. Since $d_\infty(f,g) < \epsilon$ is equivalent to $\sup_{x\in [0,1]} \left\vert f(x)-g(x) \right\vert< \epsilon>$, which implies $d_1(f,g)<\epsilon$, we see that $\id$ is continuous. 519 | 520 | However, we can use the functions $f_n$ in the proof of proposition~\ref{prop:not-lip-equiv} to see that the identity in the opposite direction, $\id:(C[0,1], d_1) \to (C[0,1], d_\infty)$, is not continuous. We note that $d_1(f_n,0) \to0$ but $d_\infty(f_n,0) \to \infty$ as $n\to\infty$. 521 | \end{example} 522 | 523 | Now we wish to express continuity of maps purely in terms of open sets. 524 | 525 | \begin{proposition} 526 | \mbox{} 527 | \begin{enumerate} 528 | \shortskip 529 | \item A map $f:(X,\rho_1) \to (Y,\rho_2)$ of a metric space is continuous if and only if, for any open subset $U\subset Y$, the pre-image $f^{-1}(U)$ is open in $(X,\rho_1)$. 530 | \item The map $f$ is continuous if and only if, for every closed subset $F\subset Y$, the pre-image $f^{-1}(F)$ is closed in $(X,\rho_1)$. 531 | \end{enumerate} 532 | \end{proposition} 533 | 534 | \begin{proof} 535 | \mbox{} 536 | \begin{enumerate} 537 | \item ($\Leftarrow$) Take $U=B_{\rho_2}(f(x),\epsilon)$. If $f^{-1}(U)$ is open, then there exists $\delta>0$ such that $B_{\rho_1}(x,\delta)\subset f^{-1}(U)$; that is, $\rho_1(x',x)<\delta$ implies $\rho_2(f(x'),f(x))<\epsilon$. % 538 | 539 | ($\Rightarrow$) If $U\subset Y$ is open, then consider a point $x\in f^{-1}(U)$. Since $U$ is open, we can choose a open ball $B_{\rho_2}(f(x),\epsilon)\subset U$. Since $f$ is continuous at $x$, there exists $\delta>0$ such that $B_{\rho_1}(x,\delta) \subset f^{-1}(B_{\rho_2}(f(x),\epsilon)) \subset f^{-1}(U)$. Since this is true for all $x\in f^{-1}(U)$, we have $f^{-1}(U)$ open. % 540 | 541 | \item Note that $f^{-1}(Y\backslash F)=X\backslash f^{-1}(F)$. So if $F$ is closed, then $Y\backslash F$ is open and hence $f^{-1}(Y\backslash F)$ is open. Then $X\backslash f^{-1}(F)$ is open and $f^{-1}(F)$ is closed. % 542 | 543 | Conversely, if $U$ is open in $Y$, then $Y\backslash U$ is closed in $Y$. Then $f^{-1}(Y\backslash U)=X-f^{-1}(U)$ is closed in $X$, and hence $f^{-1}(U)$ is open in $X$; that is, $f$ is continuous. \qedhere 544 | \end{enumerate} 545 | \end{proof} 546 | 547 | % subsection limits_and_continuity (end) 548 | 549 | \pagebreak 550 | 551 | \subsection{Completeness} % (fold) 552 | \label{sub:completeness} 553 | 554 | \begin{definition} 555 | A metric space $(X,\rho)$ is called \emph{complete} if, for any sequence $x_1,x_2,\ldots \in X$ such that, given $\epsilon>0$, there exists $N$ such that for all $m,n\geq N$, $\rho(x_m,x_n)<\epsilon$, we have $x_n\to x$ for some limit point $x$. That is, every Cauchy sequence in $X$ converges in $X$. 556 | \end{definition} 557 | 558 | Recall that $(\R,d_1)$ is complete; this is sometimes referred to as \emph{Cauchy's principle of convergence}. However, neither $(\Q,d_1)$ nor $((0,1) \subset \R, d_\text{Eucl})$ are complete. 559 | 560 | \begin{example} 561 | Let $X=C[0,1]$ and $\rho=d_1$. This is not complete, for consider $f_n$ as shown: 562 | 563 | \begin{center} 564 | \begin{minipage}{0.45\textwidth} 565 | \centering 566 | \begin{tikzpicture}[scale=0.8] 567 | 568 | \bigarrow (0,0) -- (5,0); 569 | \bigarrow (0,0) -- (0,5); 570 | 571 | \draw [very thick] (0,4) -- (1.5,4) -- (3,0) -- (4.5,0); 572 | 573 | \draw (-0.15,4) node [left] {$1$} -- (0.15,4); 574 | \draw (1.5,-0.15) node [below] {$\f{1}{2}$} -- (1.5,0.15); 575 | \draw (3,-0.15) node [below] {$\f{1}{2}+\f{1}{n}$} -- (3,0.15); 576 | 577 | \draw (4.5,0.15) -- (4.5,-0.15) node [below] {$1$}; 578 | \draw (0,0) -- (0,-0.15) node [below] {$0$}; 579 | 580 | \end{tikzpicture} 581 | \end{minipage} 582 | \begin{minipage}{0.53\textwidth} 583 | \begin{equation*} 584 | f_n(x) = 585 | \begin{dcases} 586 | 1 & \text{if } 0\leq x<\tf{1}{2}, \\ 587 | 1-nx & \text{if } \tf{1}{2} \leq x <\tf{1}{2}+\tf{1}{n}, \\ 588 | 0 & \text{if } \tf{1}{2}+\tf{1}{n} \leq x \leq 1. 589 | \end{dcases} 590 | \end{equation*} 591 | \end{minipage} 592 | \end{center} 593 | 594 | Then for $m,n \geq N$, we have $\rho(f_m,f_n) \leq 1/N$. 595 | 596 | Now, if $f_n \to f \in C[0,1]$, then $\int_0^1 \left\vert f_n-f \right\vert \to f$. But 597 | \begin{align*} 598 | \int_0^1 \left\vert f_n - f \right\vert 599 | &\geq \int_0^{1/2} \left( \left\vert f-1 \right\vert - \left\vert f_n-1 \right\vert \right) + \int_{1/2}^1 \left( \left\vert f \right\vert-\left\vert f_n \right\vert \right) \\ 600 | &\to \int_0^{1/2} \left\vert f-1 \right\vert + \int_{1/2}^1 \left\vert f \right\vert \geq 0. 601 | \end{align*} 602 | Thus we must have 603 | \begin{equation*} 604 | \int_0^{1/2} \left\vert f-1 \right\vert = 0 605 | \qqand 606 | \int_{1/2}^1 \left\vert f \right\vert = 0, 607 | \end{equation*} 608 | and our limit is given by 609 | \begin{equation*} 610 | f(x) = 611 | \begin{cases} 612 | 1 & x\leq 1/2, \\ 613 | 0 & x>1/2, 614 | \end{cases} 615 | \end{equation*} 616 | but this contradicts $f \in C[0,1]$. Hence this space is not complete. 617 | \end{example} 618 | 619 | % subsection completeness (end) --------------------------------------------------------------------------------