\documentclass[11
pt]{amsart}
%\usepackage{amssymb,amscd}\textheight=23cm


\usepackage{graphicx}
\usepackage{amsmath}
\usepackage{epstopdf}
\usepackage{amsthm}
\usepackage{times}

\usepackage{geometry}                % See geometry.pdf to learn the layout options. There are lots.
\geometry{letterpaper}                   % ... or a4paper or a5paper or ... 
%\geometry{landscape}                % Activate for for rotated page geometry
%\usepackage[parfill]{parskip}    % Activate to begin paragraphs with an empty line rather than an indent
\usepackage{graphicx}
%\usepackage{amssymb}
%\usepackage{epstopdf}
\DeclareGraphicsRule{.tif}{png}{.png}{`convert #1 `dirname #1`/`basename #1 .tif`.png}

\textwidth=15cm
%\titlepage
\topmargin=0pt
\headheight=0pt
\headsep=0pt
%\footskip=0pt
%\oddsidemargin=0pt
%\marginparsep=0pt
%\marginparwidth=0pt
\pagestyle{empty}

\newcommand{\R}{{\mathbb R}}
\newcommand{\C}{{\mathbb C}}
\newcommand{\Z}{{\mathbb Z}}
\newcommand{\Q}{{\mathbb Q}}
\newcommand{\N}{{\mathbb N}}

\newcommand{\T}{{\mathcal T}}
\newcommand{\B}{{\mathcal B}}
\newcommand{\X}{{\mathcal X}}

\newcommand{\vx}{{\mathbf x}}
\newcommand{\vy}{{\mathbf y}}
\newcommand{\vz}{{\mathbf z}}

\begin{document}
\parindent=0pt


{{\noindent Math 5510} \hfill {Homework 6--  Due  Wednesday 11/22/17}
\hfill {\ \ \ \ }\hfill}



\vskip.2cm

 \hrule



\bigskip

\begin{enumerate}

\item (5pts)  \emph{ An atlas for $S^2$: }   Let $S^2 =\{(x,y,z)\in\R^3 \ | \  x^2 + y^2 + z^2 = 1\}$ be the unit sphere in $\R^3$.  Let $N = (0,0,1)$ be the north pole, and $S = (0,0,-1)$ be the south pole.  Cover $S^2$ by two open sets $U_N = S^2\setminus \{N\}$ and $U_S = S^2\setminus \{S\}$.  Let $\phi_N:U_N\to \R^2$ and $\phi_S:U_S\to\R^2$ be stereographic projection from $N,S$ respectively. 
This means: if $(x,y,z)\in S^2\subset \R^3$, $\phi_N(x,y,z)\in  \{z=0\} = \R^2$ is the point  where the line from $N$ to $(x,y,x)$ intersects the plane $\{z=0\}$, and $\phi_S(x,y,z)$ the point where the line from $S$ to $(x,y,z)$ intersects the same plane $\{z=0\}$.  See the picture of a cross-section, say $x=0$. 

%\begin{figure}[htbp] %  figure placement: here, top, bottom, or page
   %\centering
    %\includegraphics[width=3.2in]{Stereo.png}
  % \caption{}
  % \label{fig-threedimcomp}
%\end{figure}

\emph{The problem is to show that the two charts $\phi_N:U_N\to \R^2$ and $\phi_S:U_S\to\R^2$ form an atlas that make $S^2$ into a smooth surface.} Follow the following steps:

\begin{enumerate}
  \item Find formulas for $\phi_N, \phi_S$.  These will show in particular that they are continuous.
  \item Find formulas for $\phi_N^{-1},\phi_S^{-1}$. These will show that $\phi_N,\phi_S$ are homeomorphisms.
  \item Show that the domain of $\phi_S\circ\phi_N^{-1}$ is $\R^2\setminus\{0\}$ and find a formula for this map.  This will show that the transition function $\phi_{SN}$ is smooth.
  (The formula  is  $\phi_{SN}(x,y) = (x,y)/(x^2 + y^2)$.  This map is called \emph{inversion in the unit circle}.  In polar coordinates is given by $(r,\theta)\to (1/r,\theta)$, in complex notation by $z\to1/\overline{z}$  Thus it fixes the unit circle, interchanging inside and outside.)
  \vskip.1cm
  \emph{The following two problems extra credit}:
  \vskip.1cm
  
  \item If $f:S^2\to\R$ is a smooth function, then so is  $f\circ\phi_N^{-1}:\R^2\to\R$.  Question: which functions $g:\R^2\to\R$ are obtained in this way?  Prove that the answer is the following: a smooth function $g:\R^2\to \R$ is $f\circ\phi_N^{-1}$ for some smooth function $f:S^2\to\R$ if and only if the function $g(\frac{x}{x^2 + y^2},\frac{y}{x^2 + y^2})$, defined for $0<x^2+y^2<1$, extends over the origin  to a smooth function defined on $x^2+y^2<1$.
  \item Which of the following functions $g:\R^2\to\R$ satisfy the above criterion:$g(x,y) = 1$, $g(x,y) = x$, $g(x,y) = x^2 + y^2$, $g(x,y) = 1/(x^2 + y^2)$, $g(x,y) = 1/\sqrt{x^2 + y^2}$.
 \end{enumerate}


\vskip.3cm

\item (5pts)  \emph{An atlas for the torus}:  Let $T = \R^2/\Z^2$ be the torus.  This means that $T$ is the quotient space of $\R^2$ by the equivalence relation $(x,y)\sim(x',y')\Longleftrightarrow x-x'\in \Z$ and $y-y'\in\Z$. Thus $T$ is the set of equivalence classes of $\R^2$ under $\sim$.  Let $[(x,y)]$ denote the equivalence class of $(x,y)$, that is, $[(x,y)] =\{(x+m,y+n) \  | \  m,n\in\Z\}$.  We have a surjective map $q:\R^2\to T$ defined by $q(x,y)=[(x,y)]$ and $T$ has the quotient topology.
\begin{enumerate}
\item Let $T'=[0,1]\times [0,1]/\cong$, where $\cong$ is the equivalence relation generated by $(x,0)
\cong (x,1)$ for all $x\in [0,1]$ and $(0,y)\cong (1,y)$ for all $y\in[0,1]$.  Let $<(x,y)>$ denote the $\cong$ - equivalence class of $(x,y)$.  Prove that the inclusion $[0,1]\times [0,1]\to \R^2$ takes $\cong$ equivalent points to $\sim$-equivalent points, therefore it gives a map $
f:T'\to T$. Prove that this map is a homeomorphism.  In particular, $T$ is compact.

\item Define four open subsets of $\R^2$:
\begin{enumerate}
  \item $V_{00} = \{ 0<x<1,0<y<1\}$
  \item $V_{10} = \{ 1/2 <x<3/2,0<y<1\}$
  \item $V_{01} = \{ 0<x<1,1/2<y<3/2\}$
  \item $V_{11} = \{ 1/2<x<3/2,1/2<y<3/2\}$
\end{enumerate}
and let $U_{ij} = q(V_{ij})\subset T$, $i,j \in\{0,1\}\times\{0,1\}$. Prove that $q:V_{ij}\to U_{ij}$ is a homeomorphism in all four cases, and that the four sets $U_{ij}$ cover $T$. 

%\begin{figure}[htbp] %  figure placement: here, top, bottom, or page
   %\centering
    %\includegraphics[width=4.9 in]{TransTorus.png}
  % \caption{}
  % \label{fig-threedimcomp}
%\end{figure}


\item Let $\phi_{ij} = (q|_{V_{ij}})^{-1}$.  Prove that $\{U_{ij},\phi_{ij} \}_{ i,j\in \{0,1\}}$ is an atlas making $T$ a smooth surface by showing that the transition functions $\phi_{ij}\circ\phi_{i'j'}^{-1}$ are given by a translation on each connected component of their domain, therefore smooth.  (The intersections $U_{ij}\cap U_{i'j'}$ are not connected).  It would help to prove that $\phi_{ij}(U_{ij}\cap U_{i'j'}) = V_{ij}\cap [V_{i'j'}]$, where, for $A\subset\R^2$,   $[A]=\{[(x,y)] \  | \  (x,y)\in A\} =\{(x+m,y+n)\  | \   (x,y)\in A, m,n\in\Z\}$. See figure for $U_{00}\cap U_{10}$.




\item \emph{Extra Credit}: We know that the map that assigns to a  function $f:T\to\R$ the function $f\circ q:\R^2\to \R$ gives a bijective correspondence between continuous functions on $T$ and doubly-periodic functions on $\R^2$.   Prove that the same holds with \lq\lq continuous" replaced by \lq\lq smooth".

\end{enumerate}



    \end{enumerate}  
   
 

                

 \end{document}
