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\begin{center}\vspace{1in}{\bf MATH 209, MANIFOLDS II, WINTER 2011}\\
\bigskip
{\bf Homework Assignment I: One-forms and integration}\\
\end{center}
\bigskip

\noindent{\bf 1.} Let $f\colon \R\to\R$ be given by $y=x^3$. 
Calculate $f_*(\p/\p x)$ and $f^*\,dy$. Is $f_*(\p/\p x)$ smooth?



\bigskip

\noindent{\bf 2.} Let $f$ be a smooth function on a manifold $M$.
Show that $df=f^*dt$, where $t$ is the natural coordinate on
$\R$. 

\noindent {\bf Hints.}  The  solution depends on how you choose
to define $df$.   I suggest you use the definition  $df (v) = L_v f$, so at a point $p$
$df_p :  T_p M \to \R$ is the  linear functional given by   $df(p) (v_p) = v_p[f], v_p \in T_p M$.  
How are you going to represent the $v_p$'s?  One choice is as an equivalence
class of curves through $p$.
\bigskip

\noindent{\bf 3.}  Continuing with the notation of the previous problem show that
$$
df=\sum_{i=1}^n \frac{\p f}{\p x_i}dx_i .
$$
whenever $(x_1,\ldots,x_n)$ be a system of local
coordinates on $M$.

{\bf Key remark.} The
one-forms $dx_1,\ldots,dx_n$  form the  basis of $T_x^* M$
dual to the coordinate  basis $\p/\p x_1,\ldots \p/\p x_n$ for $T_x M$.  
Alternatively, define     $dx_i$ by $dx_i(v)=L_v x_i$, where
$x_i$ is viewed  as a real-valued smooth  function defined on an open subset of
$M$. 
\bigskip
 
 

\bigskip


\noindent{\bf 4.} A one-form  $\alpha$   on a smooth connected
manifold $M$ is called  {\it exact} if and only if  there
exists a function $f$ such that $\alpha=df$. Prove that

\begin{itemize}

\item
$\alpha$ is exact $\Longleftrightarrow$ $\int_\gamma\alpha=0$ for any smooth closed 
                                        curve $\gamma$.

\end{itemize}

\bigskip

\noindent {\bf Hint for }  $\Leftarrow$. The goal is to construct
$f$ such that $\alpha=df$.  Fix a reference point $x_0\in M$. For
$x\in M$, set $f(x)=\int_\eta \alpha$, where $\eta$ is any  curve
connecting $x_0$ to $x$. You will need to show that  $f$ is well
defined,  i.e.,   independent of the choice of curve $\eta$\ connecting $x_0$ and $x$.
Then prove that $df=\alpha$. 

\bigskip


\noindent{\bf 5.} 
Let $\alpha=(x\,dy - ydx)/(x^2 + y^2)$.    

a) Show that $d \alpha = 0$.

b)  Show that $\int_c \alpha = 2 \pi$ where $c$ is the unit circle.

c) Show that, when rewritten in polar coordinates, $(r, \theta)$ we have that
$\alpha = d \theta$.   

d) Explain why (c) does not contradict problem {\bf 4}.





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