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\begin{center}\vspace{1in}{\bf MATH 209, MANIFOLDS II, WINTER 2014}\\
\bigskip
{\bf Homework Assignment 3: Differential Forms}\\

\end{center}


\noindent{\bf 1.}  [HAND IN] 
The following exercise relates  $\divv, \grad$ and $\curl$
of vector calculus to  the exterior 
derivative  on $M=\R^3$.     Equip $\R^3$ with the standard inner
product $\left<\cdot,\cdot\right>$. Let $\CX$ be the space of
smooth vector fields on $\R^3$ and $\Omega^0$ be the space of smooth
functions thereon.  Define isomorphisms

\begin{itemize}
\item $\Psi_1\colon \CX\to\Omega^1(\R^3)$ by 
 $\Psi_1(v)=\left<v,\cdot\right>$.  

\item $\Psi_2\colon \CX\to\Omega^2(\R^3)$ by $\Psi_2(v)=i_v(dx\wedge
dy\wedge dz)$.  

\item $\Psi_3\colon \Omega^0 \to\Omega^3(\R^3)$ 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 

\bigskip
 \noindent{\bf 2.}   Rewrite Maxwell's equations in terms of exterior derivatives.
 The electromagnetic field is a two form $F= \Sigma F_{\mu \nu } dx^{\mu}  \wedge dx^{\nu} $,
 where the indices $\mu, \nu$ run through $0,1,2,34$ and 
 the coordinates of space-time are $x^0, x^1, x^2, x^3$ with $x^0 = ct$ being the time coordinate.
 Write $i, j$ for any of the spatial indices $1,2,3$.  
 The components $F_{0i}$ correspond to the electric field and $F_{ij}$ to the magnetic field.
 The charge- current density is a one-form $J = \Sigma J_{\mu}  dx^{\mu}$
 with $J_0$ corresponding to charge density $\rho$.
 
 You will have to learn about the Hodge star operator for Minkowski space.  
 
 
   
 \bigskip

\noindent{\bf 3.} 
 Prove that for a vector field $v$ on $M$ and
$\alpha\in\Omega^k(M),  
 \beta\in\Omega^l(M)$,
we have $i_v(\alpha\wedge\beta)=(i_v\alpha)\wedge\beta+
(-1)^k \alpha\wedge(i_v\beta)$.
 \bigskip

\noindent{\bf 4.} 
A smooth map $F\colon M\to N$  induces the algebra homomorphism
  $F^*\colon \Omega^*(N)\to \Omega^*(M)$. Prove
that $dF^*=F^*d$.

\bigskip

\bigskip



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