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\begin{center}\vspace{1in}{\bf MATH 209, MANIFOLDS II, WINTER 2011}\\
\bigskip
{\bf Homework Assignment IV: Differential Forms}\\

\end{center}
\bigskip

\noindent Throughout this assignment we assume that $M$ a smooth
manifold of dimension $n$.

\bigskip

\noindent{\bf 1.} Prove that for a vector field $v$ on $M$ and
$\alpha\in\Omega^k(M)$ and 
$\beta\in\Omega^l(M)$,
we have $i_v(\alpha\wedge\beta)=(i_v\alpha)\wedge\beta+
(-1)^k \alpha\wedge(i_v\beta)$.

\bigskip

\noindent Remark. This is a consequence of a similar identity in linear
algebra: $i_v(\alpha\wedge\beta)=(i_v\alpha)\wedge\beta+
(-1)^k \alpha\wedge(i_v\beta)$, where $v\in V$ and $\alpha\in\bigwedge^k V^*$ and
$\beta\in\bigwedge^l V^*$.


\bigskip

\noindent{\bf 2.} Prove that every $k$-form $\omega$ on $M$ can be written
as a (locally finite) sum of products of compactly supported one-forms.
More precisely, there exist compactly supported one-forms $\alpha_i^j$
with $i$ ranging within some countable set and $j=1,\ldots,k$ such that
the sets $F_i=\cup_j\supp(\alpha_i^j)$ form a locally finite cover of $M$
and
$$
\omega=\sum_i \alpha_i^1\wedge\cdots\wedge\alpha_i^k.
$$

\bigskip

\noindent Hint. Here is one possible approach. Using a partition of
unity associated with a locally finite cover by coordinate charts
reduce the question to the case where $\omega$ is supported in a
coordinate chart $U$ with coordinates $(x_1,\ldots,x_n)$.  Then, use a
cut-off function equal to one on $\supp(\omega)$ and vanishing outside
of $U$ to extend each $dx_l$ to a smooth one form $\beta_l$ on $M$
equal to $dx_l$ near $\supp \omega$ and vanishing outside of $U$.

\bigskip

\noindent{\bf 3.} 
Let $F\colon M\to N$ be a smooth map. As we have seen, it induces the
pull-back map of algebras $F^*\colon \Omega^*(N)\to \Omega^*(M)$. Prove
that $dF^*=F^*d$.

\bigskip

\noindent Hint. Observe that, utilizing the properties of $d$ and $F$ and 
Problem 2, it is enough to prove this for functions and one-forms.

\bigskip

\noindent{\bf 4.} Problem 12-6 (page 320) from Chapter 12 of the textbook.

\bigskip

\noindent{\bf 5.} The goal of this problem is to show that $\grad$,
$\curl$, and $\divv$ are just particular cases of the de Rham
differential on $M=\R^3$.  Let us equip $\R^3$ with the standard inner
product $\left<\cdot,\cdot\right>$ and denote by $\CX$ the space of
smooth vector fields on $\R^3$. Define:

\begin{itemize}
\item $\Psi_1\colon \CX\to\Omega^1(\R^3)$ by 
 $\Psi_1(v)=\left<v,\cdot\right>$ or, more explicitly, 
$\Psi_1(a\p_x+b\p_y+c\p_z)=a\,dx+b\,dy+c\,dz$.

\item $\Psi_2\colon \CX\to\Omega^2(\R^3)$ by $\Psi_1(v)=i_v(dx\wedge
dy\wedge dz)$. (Work out an explicit expression for $\Psi_2$.)

\item $\Psi_3\colon C^{\infty}(\R^3)\to\Omega^3(\R^3)$ by 
$\Psi_3(f)=f\,dx\wedge dy\wedge dz$.
\end{itemize}

Prove that the following diagram commutes (up to signs):
$$\begin{CD}
C^{\infty}(\R^3) @>\grad>> \CX @>\curl>> \CX @>\divv>> C^{\infty}(\R^3) \\
 @VV{\mathit id}V @VV\Psi_1 V @VV\Psi_2 V @VV\Psi_3 V \\
C^{\infty}(\R^3) @>d>>\Omega^1(\R^3)@>d>>\Omega^2(\R^3)@>d>>\Omega^3(\R^3)
\end{CD}$$

\bigskip 

\noindent
Remark. The first and the last square of the diagram make sense (how?)
and commute for $\R^n$, but the middle one does not. Note also that in
$\R^3$ the identities $\curl\circ \grad=0$ and $\divv\circ \curl=0$
together are equivalent to $d^2=0$.

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