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\begin{document}



\title{ Math 3220-3 Homework 5}
\author{Due April 22 before class }
\maketitle
\begin{enumerate}
\item Recall the parametrized torus $T$ of Homework 4 
 obtained by fixing $0<b<a$ and defining $\Phi:\R^2\to\R^3$ with image $T$ by $\Phi(\phi,\theta)=(x(\phi,\theta),y(\phi,\theta),z(\phi,\theta)) $ where
\begin{eqnarray}
x & = & (a+b\ \cos\phi)\cos\theta \nonumber  \\
y & = & (a+b\ \cos\phi)\sin\theta   \nonumber  \\
z & = & b\ \sin\phi  \nonumber
\end{eqnarray}
  By periodicity, $T = \Phi(D)$ where $D=I_1\times I_2$ for any two intervals $I_1,I_2$ of length $2\pi$.
  
  For this problem and the next you'll need to use the definition of singular cubes in $U$(as maps of the standard cube to $U$), singular cubical chains, their boundaries.  See the posted lecture notes for week 12, starting page 46.   
\vskip10pt
  
 
\begin{enumerate}
\item Define a singular  square $\sigma:[0,1]^2\to \R^3$ with image in $T$ (briefly, a singular square in $T$)  by
$$
\sigma (s,t) = \Phi( 2 \pi s, 2 \pi t)
$$

Check that $\partial \sigma = 0$.
 
 
  \item For any $2$-form $\alpha\in A^2(\R^3)$ define $\int_T\alpha$ to be $\int_D\Phi^*\alpha$ for any $D=I_1\times I_2$ as above, for example. $D=[0,2\pi]\times [0,2\pi]$.  Prove that
  $$
  \int_\sigma \alpha = \int_T \alpha 
  $$ 
  
     \item Show, without any calculation, that $\int_T dy\wedge dz = 0.$
     \item More generally, apply Theorem 10.39 in Rudin to show if $\alpha \in A^2(\R^3)$ is any \emph{closed} form, that is, $d\alpha = 0$, then  $\int_T\alpha = 0 $.
\end{enumerate}
\vskip10pt
\item (See Rudin, Chapter 10,Exercise 22)   Let  $\zeta\in A^2(\R^3\setminus 0)$ be defined by
$$
\zeta = \frac{1}{r^3} \Big{(} x\  dy\wedge dz -y\  dx\wedge dz + z\  dx\wedge dy\Big{)}
$$
where $r = r(x,y,z) = (x^2 + y^2 + z^2 )^\frac{1}{2}$ is distance from the origin.  Let  $D = [0,\pi]\times [0,2\pi]$ and $\Sigma:D\to\R^3$ the parametrized surface
$$
\Sigma(u,v) = (x,y,z) = (\sin u \cos v, \sin u \sin v, \cos u)
$$
 \begin{enumerate}
  \item Prove that $d\zeta = 0$, that is,  $\zeta$ is \emph{closed}.
  \item Show that $\Sigma(D) = S^2 = \{ x^2 + y^2 + z^2 = 1\}$ the unit sphere in $\R^3$ centered at $0$.  Show that $\Sigma$ maps $(0,\pi)\times [0,2\pi)$ bijectively onto $S^2\setminus\{N,S\}$. where $N,S$ are the north and south poles $(0,0,\pm 1)$ and collapses $0\times [0,2\pi]$ to $N$ and $\pi\times [0.2\pi ]$ to $S$.   Sketch.
  \item Check that $\Sigma^* \zeta = \sin u du\wedge dv$,  the area form of $S^2$ in spherical coordinates, so  
   $\int_\Sigma \zeta = 4\pi\ne 0$.
  
   \item  Prove that $\zeta$ is not exact.   Follow the  steps  of Problem 1(a),(b).  First let $\sigma:[0,1]\times [0,1]\to S^2 \subset \R^3\setminus \{0\}$ be the singular square defined by the map  $\sigma(s,t) = \Sigma(\pi s, 2\pi t)$.
   Show that $\partial \sigma = \phi_N - \phi_S$, where $\phi_N,\phi_S$ are the maps $[0,1]\to S^2\subset \R^3\setminus \{0\}$ that send $[0,1]$ to $N$, $S$.
    
    \item Then show that for any  $\alpha\in A^2(\R^3\setminus\{0\})$, $\int_\sigma \alpha = \int_\Sigma \alpha$.  Then $\forall\eta\in A^1(\R^3\setminus\{0\})$, 
    $$
    \int_\Sigma d\eta = \int_\sigma d\eta = \int_{\partial \sigma} \eta = \int_0^1 \phi_N^*\eta - \int_0^1 \phi_S^*\eta = 0
 $$
 since $\phi_N^*\eta = 0 =  \phi_S^*\eta$ because $\phi_N,\phi_S$  are constant maps.  Thus $\zeta$ is not exact.
 
  
 \item But $\zeta$ is exact in $\R^3\setminus Z$, where $Z=$ the $z$-axis. In fact, let
 $$
 \lambda =( -\frac{z}{r}) \Big{(}  \frac{x\ dy - y \ dx}{x^2 + y^2}\Big{)} \in A^1(\R^3\setminus Z)
 $$
 Show, by direct calculation, that $d\lambda = \zeta$. 
 
 {\sl Motivation}: On $(0,\pi)\times[0,2 \pi] $, $\Sigma^*\zeta = \sin u du\wedge dv = d(-\cos u dv)$ and
 $$ 
 \cos u = \frac{z}{r}, \ \ \  dv = \frac{x\ dy - y \ dx}{x^2 + y^2}.
 $$
 
 
 
  
 \item Show that, if $T$ is the torus of Problem 1, $\int_T\zeta = 0$
 
    \end{enumerate}
\vskip10pt

\item Two problems relevant to Lebesgue measure and integration:
\begin{enumerate}
  \item Let $E = \Q\cap[0,1]$ and let $I_1,\dots,I_k$ ve a \emph{finite} collection of open intervals in  $\R$ covering $E$, that is, $E\subset \cup_{i=1}^k I_i$.  Prove that $\sum_{i=1}^k \ell(I_i) \ge 1$, where $\ell(I_i)$ is the length of $I_i$.
    \item (Rudin, Chapter 11, problem 5) Let
    \begin{eqnarray}
g(x) & = & \begin{cases}
    0  & 0\le x \le \frac{1}{2}, \nonumber  \\
     1 & \frac{1}{2} <x\le 1,
\end{cases} \\
f_{2i}(x) & = & g(x) \ \ \ \ \ \ 0\le x \le 1  \nonumber   \\
f_{2i-1}(x) & = &  g(1-x)\ \ \ \ \  0\le x\le 1   \nonumber 
\end{eqnarray}
Show that
$$
\liminf_{n\to\infty}f_n(x) =0\ \ \text{ for }  \ 0\le x\le 1\ \ \text{ but }\  \ 
\int_0^1 f_n(x) dx = \frac{1}{2}
$$
  \end{enumerate}
  \end{enumerate}

 \end{document}

