
\section{Exterior Derivative}





\subsection{Defining the exterior derivative}\label{subsec-defexterior}

We now define an operation that will generalize several of the differential operators (such as the gradient, curl, and divergence) we used previously.


\begin{definition}\label{51def:exteriorderivative}
The \textbf{exterior derivative} of a differential $0$-form $f : D \subset \mathbb{R}^n \rightarrow \mathbb{R}$ is the differential $1$-form
$$
d f (\mathbf{x}) : = \sum\limits_{i=1}^{n} \frac{\partial f}{\partial x_i} dx_i.
$$
Formally, this is the same as the differential of a function $f$ that we used in earlier parts of this course, but now we interpret it as a differential form.


For a differential $k$-form $\Phi (\mathbf{x}) =  a (\mathbf{x}) d x_{i_1} \wedge \cdots \wedge d x_{i_k}$,
its exterior derivative $d \Phi$ is the differential $(k+1)$-form given by
$$
d \Phi (\mathbf{x}) : =
da (\mathbf{x}) \wedge d x_{i_1} \wedge \cdots \wedge d x_{i_k},
$$
with $da (\mathbf{x})$ defined as above. We then extend it to general $k$-forms by linearity.
\end{definition}

\noindent
For $1$-forms, this means $d(fdx_i)= df \wedge dx_i$, with $df$ being the differential of the function $f$. This should get us a 2-form. For 2-forms, we have $d(f dx_i\wedge dx_j)= df \wedge dx_i\wedge dx_j$ (a 3-form), and so on. We then use linearity to extend it to the general case. Here is a concrete example.

\begin{example}
We have
$$
d (xy) = y dx + x dy \quad \quad \quad d (x^2 + y^2) = 2x dx + 2y dy,
$$
and
\begin{eqnarray*}
d\Big(xydx + (x^2 + y^2)dy\Big) 
& = & d(xy)  \wedge dx + d(x^2+y^2) \wedge dy \\[1ex]
\quad & = & (y dx + x dy) \wedge dx + (2x dx + 2y dy) \wedge dy \\[1ex]
\quad & = & x dy \wedge dx + 2x dx \wedge dy \\[1ex]
\quad & = & x dx \wedge dy,
\end{eqnarray*}
where at the last step we used that $dy \wedge dx = - (dx \wedge dy)$. Note that if we take another exterior derivative, we have
$$
d(x dx \wedge dy) = dx \wedge dx \wedge dy = 0.
$$
\end{example}



The exterior derivative generalizes several of the definitions we have used before, if vector fields are replaced by differential forms with the same component functions.  This will be important in matching our previous fundamental theorems (i.e. Green's, Divergence, Stokes) to the appropriate special cases in {Generalized Stokes's Theorem}.


\begin{enumerate}[(1)]
\item \underline{\textit{Gradient}}: If $f$ is a $0$-form on $\mathbb{R}^n$, then 
$$df = \sum_{i=1}^n \frac{\partial f}{\partial x_i} dx_i$$
 is a differential $1$-form with coefficients $\frac{\partial f}{\partial x_i}$. Note that these coefficients match the corresponding components of the gradient vector field $\nabla f$. If we apply the differential 1-form $df$ to a unit vector ${\bf u}$, we get
 $$
 df({\bf u})=  \sum_{i=1}^n \frac{\partial f}{\partial x_i} u_i =D_{\bf u}f,
 $$
 the directional derivative of $f$ in the direction of ${\bf u}$.
 
 
\medskip

\item \underline{\textit{Curl in $\mathbb{R}^3$}}: Let $\mathbf{F}=F_1\ii+F_2\jj+F_3\kk$ be a smooth vector field on $\mathbb{R}^3$. Can we match $\mathbf{F}$ to a differential form on $\rr^3$ with the same coefficients $F_1,F_2,F_3$?
One natural choice is 
$$
\Phi = F_1 dx + F_2 dy + F_3 dz,
$$
but a second choice that might be a little bit less obvious is to use the 2-form
$$
F_1 dy \wedge dz + F_2 dz \wedge dx + F_3 dx \wedge dy.
$$
We illustrate the usefulness of both types of matching as follows.
Let $\mathbf{G} = \curl \mathbf{F}$.  In components, we have $\mathbf{G}=G_1\ii + G_2\jj + G_3\jj$. 
Then, with $\Phi$ defined as above, we have 
$$
d \Phi = G_1 dy \wedge dz + G_2 dz \wedge dx + G_3 dx \wedge dy.
$$
We show this calculation in the Worked Problems Section below. Therefore, curl is a special case of exterior derivative if we use the first kind of matching for $\mathbf{F}$ and the second kind for $\curl \mathbf{F}$.

\medskip

\item \underline{\textit{Divergence in $\mathbb{R}^3$}}: Suppose that $\mathbf{G}$ is a smooth vector field on $\mathbb{R}^3$. Let
$$
\Psi = G_1 dy \wedge dz + G_2 dz \wedge dx + G_3 dx \wedge dy.
$$
Then
$$
d \Psi = (\divg \mathbf{G} )(dx \wedge dy \wedge dz).
$$
We will show the calculation in the Worked Problems Section below.  (Exercise: how would you modify this example to make it work in $\rr^2$?)
\end{enumerate}

\subsection{Closed and exact differential $k$-forms}\label{subsec-closedexact}

We are ready for our first common generalization of two different results about vector fields.  

\begin{definition}
A differential $k$-form $\Phi$ is called \textbf{closed} if $d \Phi = 0$ and \textbf{exact} if there is some $(k-1)$-form $\Psi$ such that $\Phi = d \Psi$.
\end{definition}



\begin{proposition}\label{51prop:exteriorcalculus}
Let $\Phi$ be a differential $k$-form on some domain $D \subset \mathbb{R}^n$, and $a$ and $b$ are real numbers.  Then
\begin{equation}\label{51-ddeq}
d^2 (\Phi) : = d (d \Phi) = 0,
\end{equation}
where $0$ denotes the zero differential form.  Put another way, every exact form is closed.

\end{proposition}

\begin{proof}[Outline of proof]
The proof uses the equality of the mixed partials. By linearity, it is enough to prove (\ref{51-ddeq}) when $\Phi=a (\mathbf{x}) d x_{i_1} \wedge \cdots \wedge d x_{i_k}$. Suppose for the moment that $\Phi$ is a 0-form, so that $\Phi=a({\bf x})$
is a smooth function on $\mathbb{R}^n$. Then
$$
d\Phi = d a = \sum\limits_{i=1}^n \frac{\partial a}{\partial x_i} dx_i.
$$
By direct computation (you should check this!), 
$$
d(d\Phi)=\sum_{i<j} \left(\frac{\partial^2 a}{\partial x_i \partial x_j} - \frac{\partial^2 a}{\partial x_j \partial x_i}\right) dx_i\wedge dx_j,
$$
with coefficients identically zero by the equality of mixed partials.  Hence $d(d\Phi) = 0$. The general case is similar, but with more ``trailing" $dx_i$'s that do not change the calculation. We leave it as an exercise for you to write out the details.
\end{proof}


For the converse, we have the following result, which is known as \textit{Poincar{\'e}'s Lemma}.

\begin{theorem}\label{51thm:poincarelemma}
Let $\Phi$ be a closed differential $k$-form on a \textbf{starlike} domain $D \subset \mathbb{R}^k$ (see below). Then $\Phi$ is exact on $D$.
\end{theorem}

The domain $D$ is starlike if there exists a point $P\in D$ such that every other point in $D$ can be joined to $P$ by a line segment that lies entirely in $D$. There may be more than one such point---for example, if $D$ is convex, we may take $P$ to be \underline{any} point in $D$. However, starlike domains do not have to be convex.

\begin{figure}[H]
    \centering
    \begin{subfigure}[b]{.5\textwidth}
        \centering
        \includegraphics[width=.5\textwidth]{figures/starlike1.png}
        \caption{A starlike domain}
    \end{subfigure}
    ~
    \begin{subfigure}[b]{.5\textwidth}
        \centering
        \includegraphics[width=.5\textwidth]{figures/starlike2.png}
        \caption{A simply connected, non-starlike domain}
    \end{subfigure}
    \caption{Starlike is a stronger condition than being simply connected.}
\end{figure}



We omit the proof of the theorem. Instead, we discuss how the above results apply to vector fields in 3 dimensions.


\begin{enumerate}[(1)]
\item \underline{\textit{Conservative fields}}: Let $f$ be a smooth function on $\mathbb{R}^3$, and let $\mathbf{F}=\nabla f$.
Define the 1-form
$$
\Phi = d f = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy+ \frac{\partial f}{\partial z} dz
=F_1dx+F_2dy+F_3dz.
$$
Proposition \ref{51prop:exteriorcalculus} tells us that $d\Phi=d(df)=0$.
But we have seen that 
$$d\Phi  = G_1 dy \wedge dx + G_2 dz \wedge dx + G_3 dx \wedge dy,
$$
where ${\bf G}=\curl{\bf F}$. Hence, $d \Phi = 0$ means that $\curl \mathbf{F} = {\bf 0}$.


Proposition \ref{51prop:exteriorcalculus} thus includes, as a special case, the fact that conservative vector fields have zero curl.
Poincar\'e's Lemma says that, on a starlike domain $D$, the converse is true: if ${\bf F}$ has zero curl, it must in fact be conservative.


\medskip

\item \underline{\textit{Divergence and Curl}}: Let $\mathbf{F}$ be a smooth vector field in $\mathbb{R}^3$, and let $\mathbf{G} = \curl \mathbf{F}$. We have seen that the differential forms
$$
\Phi = F_1 dx + F_2 dy + F_3 dz, \textrm{ and } \Psi : = G_1 dy \wedge dx + G_2 dz \wedge dx + G_3 dx \wedge dy.
$$
satisfy $d \Phi = \Psi$.  Proposition \ref{51prop:exteriorcalculus} then says that
$$
d \Psi = d(d\Phi) = 0.
$$
But, we have calculated that $d \Psi =( \divg \mathbf{G}) \,dx \wedge dy \wedge dz$! So, in this case, Proposition \ref{51prop:exteriorcalculus} shows that the divergence of curl is zero. For starlike domains $D$, Poincar{\'e}'s Lemma provides a converse: if a smooth vector field $\mathbf{G}$ satisfies $\divg \mathbf{G}=0$ on $D$, then $\mathbf{G} = \curl \mathbf{F}$ for some smooth vector field $\mathbf{F}$ on $D$.

\end{enumerate}


There are important special cases when the starlike assumption
in Poincar{\'e}'s Lemma can be weakened and the conclusion still holds. As an example, we already know that if $n = 2$ or $3$ and if the domain $D \subset \mathbb{R}^n$ is simply connected, then 
$\curl \mathbf{F} = 0$ on $D$ implies that $\mathbf{F}$ is conservative on $D$.  The property of being simply connected is strictly weaker than the property of being starlike, in the sense that all starlike domains are simply connected, but the converse is not true.  Starlike domains are a general class of domains for which Poincar{\'e}'s Lemma holds for \textit{all differential forms}, regardless of order and ambient dimension.



\section*{Worked Problems}
\begin{enumerate}[1.]

\item
{\it Calculate the exterior derivative of the differential forms. Write the $dx_i$'s in the answers in increasing subscript order.}

(a) $\Phi= x_1^2e^{x_3} dx_1\wedge dx_2+ (x_2+x_3)  dx_3\wedge dx_4$



\begin{align*}
d\Phi &= d\left(x_1^2e^{x_3} dx_1\wedge dx_2+ (x_2+x_3)  dx_3\wedge dx_4\right)\\
&= d(x_1^2e^{x_3}) \wedge dx_1\wedge dx_2 + d(x_2+x_3) \wedge dx_3\wedge dx_4\\
&= (2x_1e^{x_3}dx_1 +x_1^2e^{x_3} dx_3) \wedge dx_1\wedge dx_2 
%&\ \ \ \ \ 
+ (dx_2 +dx_3) \wedge dx_3\wedge dx_4\\
&=x_1^2e^{x_3} dx_3 \wedge dx_1\wedge dx_2 +  dx_2  \wedge dx_3\wedge dx_4
\\
&= x_1^2e^{x_3} dx_1 \wedge dx_2\wedge dx_3
+dx_2 \wedge dx_3 \wedge dx_4.
\end{align*}

(b) $\Phi = x_1x_2^2\cos(x_3) dx_1 \wedge dx_2 \wedge dx_3 
 + x_2^2\sin(x_3)  dx_1 \wedge dx_3 \wedge dx_4$. 

 
\begin{align*}
d\Phi &= d(x_1x_2^2\cos(x_3)) \wedge dx_1 \wedge dx_2 \wedge dx_3 + d(x_2^2\sin(x_3) ) \wedge dx_1 \wedge dx_3 \wedge dx_4\\
&=2x_2\sin(x_3) dx_2 \wedge dx_1 \wedge dx_3 \wedge dx_4\\
&= - 2x_2\sin(x_3) dx_1 \wedge dx_2 \wedge dx_3 \wedge dx_4.
\end{align*}




\item{\it Let $\mathbf{F} = F_1 \mathbf{i} + F_2 \mathbf{j} + F_3 \mathbf{k}$ be a smooth vector field, and define the differential $1$-form $\Phi = F_1 dx + F_2 dy + F_3 dz$. Calculate $d \Phi$ and identify its associated component functions $G_1, G_2$ and $G_3$ as coefficients of $dy \wedge dz, dz \wedge dx$ and $dx \wedge dy$.}

\medskip
This is the calculation for Special Case (2) of Subsection \ref{subsec-defexterior}.  Recall that, for each $i = 1,2,3$, we have
$$
d F_i = \frac{\partial F_i}{\partial x}dx + \frac{\partial F_i}{\partial y}dy + \frac{\partial F_i}{\partial z}dz,
$$
and
$$
d \Phi = d F_1 \wedge dx + d F_2 \wedge dy + d F_3 \wedge dz.
$$
Using linearity of the wedge product,  the above becomes
\begin{eqnarray*}
d \Phi & = & \cancelto{0}{\frac{\partial F_1}{\partial x} dx \wedge dx} + \frac{\partial F_1}{\partial y} dy \wedge dx + \frac{\partial F_1}{\partial z} dz \wedge dx \\
\quad & \quad & + \frac{\partial F_2}{\partial x} dx \wedge dy + \cancelto{0}{\frac{\partial F_2}{\partial y} dy \wedge dy} + \frac{\partial F_2}{\partial z} dz \wedge dy \\
\quad & \quad & + \frac{\partial F_3}{\partial z} dx \wedge dz + \frac{\partial F_3}{\partial y} dy \wedge dz + \cancelto{0}{\frac{\partial F_3}{\partial z} d z \wedge dz}.
\end{eqnarray*}
We then re-write the above to match the basis $dy \wedge dz$, $dz \wedge dx$ and $dx \wedge dy$, keeping in mind that $dz \wedge dy=-dy \wedge dz$ and similarly for other variables:
\begin{equation*}%\label{51eq:curlcalc}
d \Phi = \underbrace{\bigg(\frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z} \bigg)}_{G_1} dy \wedge dz + \underbrace{\bigg(\frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x} \bigg)}_{G_2} dz \wedge dx + \underbrace{\bigg(\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \bigg)}_{G_3} dx \wedge dy.
\end{equation*}
But the coefficients $G_1,G_2,G_3$ are the same as the components of $\curl  \mathbf{F}$! Letting
$\mathbf{G} = G_1 \mathbf{i} + G_2 \mathbf{j} + G_3 \mathbf{k}=\curl \mathbf{F}$, this identifies $\mathbf{G}$ (using our second kind of matching) with the exterior derivative $d\Phi$.


\bigskip

\item{\it Suppose that
$$
\Theta = H_1 dy \wedge dz + H_2 dz \wedge dx + H_3 dx \wedge dy,
$$
is a differential $2$-form with smooth component functions $H_i$.  Calculate the differential $3$-form $d \Theta$. We use this calculation in Special Case (3) of Subsection \ref{subsec-defexterior}. }
 
\medskip
 
We know that
$$
d \Theta : = d H_1 \wedge dy \wedge dz + d H_2 \wedge dz \wedge dx + d H_3 \wedge dx \wedge dy.
$$
Now, we have $d H_i : = \frac{\partial H_i}{\partial x} dx + \frac{\partial H_i}{\partial y} dy + \frac{\partial H_i}{\partial z} dz$,  with terms involving each of the elementary forms $dx$, $dy$ and $dz$.  But since $dx \wedge dx = dy \wedge dy = dz \wedge dz = 0$,  the only nonzero terms in $d\Theta$ are
$$
d \Theta : = \frac{\partial H_1}{\partial x} dx \wedge dy \wedge dx + \frac{\partial H_2}{\partial y} dy \wedge dz \wedge dx + \frac{\partial H_3}{\partial z} dz \wedge dx \wedge dy.
$$
To simplify this, we notice that
$$
dy \wedge dz \wedge dx = - dy \wedge dx \wedge dz = + dx \wedge dy \wedge dz,
$$
and similarly
$$
dz \wedge dx \wedge dy = + dx \wedge dy \wedge d z.
$$
Therefore
$$
d \Theta = \bigg(\frac{\partial H_1}{\partial x} + \frac{\partial H_2}{\partial y} + \frac{\partial H_3}{\partial z} \bigg) dx \wedge dy \wedge dz.
$$
So, if we define the vector field $\mathbf{H} = \langle H_1, H_2, H_3 \rangle$, then
$$
d \Theta = (\divg \mathbf{H}) \, dx \wedge dy \wedge dz.
$$
\end{enumerate}


\section*{Practice Problems}
\begin{enumerate}[1.]


\item 
{Let $\Psi=x_1^2\, dx_1\wedge dx_3 -x_3 \,dx_2 \wedge dx_4$ 
in $\rr^4$. 

\smallskip

(a) Compute $\Psi \wedge \Phi$, if
$\Phi= x_3^5  \, dx_2 \wedge dx_4$.
Give your answer in the form $f(x_1,x_2,x_3,x_4) 
dx_1\wedge dx_2 \wedge dx_3 \wedge dx_4$.

\smallskip

(b) Find $d\Psi$. Simplify your answer.}

\bigskip


\item
{Calculate the exterior derivative of the differential forms:

\smallskip

(a) $\Phi=xydy + x^2z^2dz$


\smallskip

(b) $\Phi= (x_1+x_2^2) dx_1\wedge dx_2+ x_2\cos(x_3) dx_2\wedge dx_4$
}

\bigskip


\item
{ Calculate the exterior derivative of the differential forms:

\smallskip

(a) $\Phi= 5x_2 \, dx_1\wedge dx_3+ (3x_3 - x_1)  \, dx_3\wedge dx_4$

\smallskip

(b) $\Psi = x_1 x_2^2  \,dx_1 \wedge dx_3 \wedge dx_4 
 + x_3^2 x_4  \, dx_2 \wedge dx_3 \wedge dx_4$. 
 
\smallskip
 
(c) $\Phi = e^{x_1-x_3}  dx_1 \wedge dx_2 \wedge dx_4 
 + e^{4x_2-5x_3}  dx_1 \wedge dx_3 \wedge dx_4$. 
 }

\bigskip


\item
{Find the exterior derivative of the 3-form
$x_1x_2^2 x_3^4 \, dx_1\wedge dx_3 \wedge dx_4$ in $\rr^4$.
Give your answer in the form $g(x_1,x_2,x_3,x_4)\,  
dx_1\wedge dx_2 \wedge dx_3 \wedge dx_4$.}


\bigskip

\item {
 Let $\Psi=xy dx + z^2 dy$ and $\Phi= (2y+z)dx\wedge dy + (xy+z^2) dy \wedge dz$.
 
\smallskip

(a)  Find $\Psi \wedge \Phi$. Simplify your answer.

\smallskip

(b) Find $d\Phi$ and $d\Psi$.}



\end{enumerate}




