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\newcommand{\myTitle}{HW 4, due Friday, March 1}
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\begin{enumerate}
\item {\bf Computations.}
  \begin{enumerate}
    \item  
      \begin{align*}
	f\colon\R^2&\to\R^4\\
            (x,y)&\mapsto(x^2,2x+y, y^4,xy)
      \end{align*}
      \begin{enumerate}
	\item Let $\omega = y_1 dy_1\wedge dy_2 +(y_1y_3) dy_3\wedge
      dy_4\in\Omega^2\R^4$. 
     Compute $f^*\omega$ and express your answer in terms of the standard basis
     for $\Omega^2\R^2$.
   \item Consider the vector field $v=y\frac{\partial}{\partial x}+
     x\frac{\partial}{\partial y}$ on 
     $T\R^2$.  Compute $f_{*,(1,1)}v$, i.e., $df_{(1,1)}(v)$, in terms of the standard
     basis for $T\R^4$.
      \end{enumerate}
    \item Consider the polar coordinates map
      \begin{align*}
	f\colon I:=(0,1)\times(0,2\pi)&\to\R^2\\
	(r,\theta)&\mapsto(r\,\cos\theta,r\,\sin\theta)
      \end{align*}
      and the ``volume form'', $\omega:=dx\wedge dy\in\Omega^2\R^2$.  Compute
      $f^*\omega\in\Omega^2I$.
  \end{enumerate}
  {\sc solution:}

\item Suppose that $\omega$ is a $1$-form on $\proj^1$.  Let $(U_x,\phi_x)$ and
  $(U_y,\phi_y)$ denote the two standard charts for $\proj^1$.  So $\phi_x\colon
  U_x\stackrel{\sim}{\to}\R^1$ and similarly for $\phi_y$.  Say $f(a)\,da$ is
  $\omega$ in $(U_x,\phi_x)$ coordinates and $g(b)\,db$ is $\omega$ in $(U_y,\phi_y)$ coordinates.  On the overlap, $U_x\cap U_y$ this gives
  representations for $\omega$, so it make sense to compare them.
  \begin{enumerate}
    \item What is
  $f(a)\,da$ in terms of $g(b)\,db$?  In other words, compute the pullback
  $(\phi_y\circ\phi_x^{-1})^*(g(b)\,db)$.
\item  In light of your answer to part (a), construct a nonzero (globally
  defined) $1$-form on $\proj^1$.
  \end{enumerate}
  {\sc solution:}

\item Consider the $2$-sphere with its usual embedding in space, $\iota\colon
  S^2\to\R^3$.  Let $\omega=x\,dx+y\,dy+z\,dz\in T^*\R^3$.  What is
  $\iota^*\omega\in T^* S^2$?  
  \begin{enumerate}
    \item Compute the pullback with respect to the
  charts for $S^2$ given in the last homework assignment.    
    \item Explain why your answer to (a) makes sense (i.e., could have been
      surmised without calculation).
\end{enumerate}
  {\sc solution:}

\item {\bf Another characterization of tangent space.}  Let $M$ be an
  $n$-dimensional manifold.  For each $p\in M$, let $\xi_p$ be the $\R$-algebra
  of germs of functions at $p$.  Let $\mathfrak{m}_p\subset\xi_p$ denote the
  ideal of germs vanishing at $p$. (Recall that the value of $f\in\xi_p$ at
  $p$ is well-defined; so in particular, the notion of a germ being zero at $p$ is
  well-defined). The purpose of this exercise is to show
  \[
  \left(\mathfrak{m}_p/\mathfrak{m}^2_p\right)^*\approx T_pM,
  \]
  where $\mathfrak{m}_p^2$ is the square of the ideal $\mathfrak{m}_p$.
  \begin{enumerate}
    \item Think of $T_pM$ as the space of derivations of germs and define 
      \begin{align*}
	\alpha\colon (\mathfrak{m}_p/\mathfrak{m}_p^2)^*&\to T_pM\\
	\phi&\mapsto \alpha(\phi)
      \end{align*}
      where
      \begin{align*}
	\alpha(\phi)\colon\xi_p&\to\R\\
	f&\mapsto\phi(f-f(p)).
      \end{align*}
      Linearity of both $\alpha$ and $\alpha(\phi)$ is straightforward (check it on your
      own).  Prove that $\alpha(\phi)$ is a derivation.  Hint:
      \[
       fg-f(p)g(p)=(f-f(p))(g-g(p))+f(p)(g-g(p))+g(p)(f-f(p)).
      \]
    \item Now define
      \begin{align*}
	\beta\colon T_pM&\to(\mathfrak{m}_p/\mathfrak{m}_p^2)^*\\
	v&\mapsto \beta(v)
      \end{align*}
      where
      \begin{align*}
	\beta(v)\colon \mathfrak{m}_p/\mathfrak{m}_p^2&\to\R\\
	f&\to v(f).
      \end{align*}
      \begin{enumerate}
	\item Show that $\beta(v)$ is well-defined.
	\item Show that $\alpha$ and $\beta$ are inverses.
      \end{enumerate}
  \end{enumerate}
  {\sc solution:}

\item Let $\R^{(\omega)}:=\bigoplus_{i=1}^\infty\R$, the collection of all sequences
  of real numbers with only a finite number of nonzero terms.  Let
  $\R^{\omega}:=\prod_{i=1}^\infty\R$, the collection of all sequences
  of real numbers. 
\begin{enumerate}
\item Show that $\R^{(\omega)}$ and $\R^{\omega}$ are a categorical
  coproduct and product, respectively, in the category of vector
  spaces.  For instance, first consider $\R^{(\omega)}$.  For
  $i=1,2\dots$, there are canonical injections $\ell_i\colon
  \R\to\R^{(\omega)}$ sending $x\in\R$ to the sequence whose $i$-th
  term is $x$ and whose other terms are zeroes.  Suppose $X$ is a
  real vector space and you are given (linear) mappings
  $f_i\colon\R\to X$
  for each $i$.  Show there is a unique mapping $g$ so
  that the following diagram commutes for each $i$:
\begin{center}
  \begin{tikzpicture}[node distance=2cm, auto, >=latex']
  \node (X) {$X$};
  \node (Rw) [right of=X] {$\R^{(\omega)}$};
  \node (R) [below of=Rw] {$\R$};
  \draw[densely dotted,->,thick] (Rw) to node [swap] {$\exists!g$} (X);
  \draw[->] (R) to node {$f_i$} (X);
  \draw[->] (R) to node [swap] {$\ell_i$} (Rw);
\end{tikzpicture}
\end{center}
The mapping $g$ is usually denoted $\oplus_i f_i\colon\R^{(\omega)}\to X$.
  To show that $\R^\omega$ is the product, you need to show the
  ``dual'' result, turning all the arrows around.  There
  are canonical projections $\pi_i\colon\R^\omega\to\R$ sending a
  sequence to its $i$-th term.  Show that
  given mappings $f_i\colon X\to\R$ for each $i$, there exists a
  unique mapping $g$ so that the following diagram commutes for each $i$:
\begin{center}
  \begin{tikzpicture}[node distance=2cm, auto, >=latex']
  \node (X) {$X$};
  \node (Rw) [right of=X] {$\R^{\omega}$};
  \node (R) [below of=Rw] {$\R$};
  \draw[densely dotted,->,thick] (X) to node {$\exists!g$} (Rw);
  \draw[->] (X) to node [swap] {$f_i$} (R);
  \draw[->] (Rw) to node {$\pi_i$} (R);
\end{tikzpicture}
\end{center}
\item Show that $(\R^{(\omega)})^*\approx\R^\omega$.  So here is an
  example of a vector space $V$ for which $V^*$ is not isomorphic to
  its dual.  [It is impossible to have a linear isomorphism between
  $\R^{(\omega)}$ and $\R^\omega$ since only one has countable
  dimension.]  Recall that $V^*\approx V$ whenever $V$ is
  finite-dimensional.  
\end{enumerate}
  {\sc solution:}

\end{enumerate}
\end{document}
