
\section{An Introduction to Differential Forms}
We will end this course with a short introduction to differential forms. Our goal is to develop a unified framework for the main concepts (line integrals, flux integrals) and theorems (Green's, Divergence, Stokes's) that we have covered. It turns out that all of those can be viewed as special cases of ``Generalized Stokes's Theorem", which also extends to arbitrary dimensions.

We will work from the bottom up, using coordinate systems and concrete examples at every step. This will allow us to ground the abstract theory in our earlier work in 2 and 3 dimensions. This also means that we define some of the main concepts (wedge product, exterior derivative) in terms of how they are computed. The same concepts could also be defined in a coordinate-free way (in terms of their functional properties), but that would require building up more theory and then doing additional work to derive the coordinate representation. There is not enough time for it in this course, but we encourage the interested student to consider further study in differential geometry. 

\subsection{Manifolds: Definitions and Examples} We will provide a rigorous definition of a manifold a little bit later; for the purpose of this introduction, it suffices to think of manifolds as a common term for objects like curves in $\rr^n$, surfaces in $\rr^3$, hypersurfaces in 
$\rr^n$, and so on. It also includes 0-dimensional objects (points) and $n$-dimensional domains in $\rr^n$. 

A manifold might or might not have a boundary. For example, the boundary of a ball $$\{(x,y,z)\in\rr^3:\ 
x^2+y^2+z^2\leq a^2\}$$ is the sphere $$\{(x,y,z)\in\rr^3:\ 
x^2+y^2+z^2= a^2\}.$$Stokes's Theorem involves a surface $\calS$ and a curve $C$ on its boundary. 
The boundary of a curve $C$ from $P$ to $Q$ consists of its endpoints $P,Q$. Generally, if our manifold
is $k$-dimensional, its boundary (if non-empty) has dimension $k-1$. An example of a manifold with no boundary is a closed surface or curve.

Let's first look at a few types of integrals on manifolds that we have seen.

\begin{itemize}
\item 
If $f:D\to\rr$ is a function defined on a domain $D\subset\rr^n$, we could fix a point $P\in D$ and 
consider $f(P)$ (the value of $f$ at $P$) as the ``integral" of $f$ on the zero-dimensional set $P$. In fact,
using definitions from analysis courses such as Math 320, $f(P)$ is the integral of $f$ with respect to the ``delta-measure" at $P$---an idea we encountered in connection with the Divergence Theorem and Stokes's Theorem for surfaces.

\medskip

\item In single-variable calculus, we consider integrals $\int_a^b f(x) dx$ of functions $f:[a,b]\to\rr$, where $[a,b]\subset \rr$ is an interval.
Moreover, we have the Fundamental Theorem of Calculus: if $f=F'$ then
$$
\int_a^b f(x)dx = F(b)-F(a).
$$
This gives a relation between the integral of a function $F$ on the 0-dimensional set of endpoints $\{a, b\}$ and the integral of another function $f$, related to $F$ through differentiation, on a manifold of dimension 1. (The set of endpoints $\{a,b\}$ is ``oriented" in the sense that we take the {\em difference} $F(b)-F(a)$, so that it matters which term has the minus sign.)

\medskip

\item If ${\bf F}$ is a vector field on a domain $D\subset \rr^n$, and if $C$ is an oriented curve in $D$ connecting the point $P$ to  the point $Q$, we can consider the integral
$$
\int_C {\bf F}\cdot d{\bf r}.
$$
Moreover, if ${\bf F}=\nabla f$ is conservative, then
$$
\int_C {\bf F}\cdot d{\bf r} = f(Q)-f(P),
$$
a relation similar to the Fundamental Theorem of Calculus.

\smallskip

\item \textit{Green's Theorem}: If ${\bf F}$ is a vector field and if $D$ is a 
2-dimensional domain bounded by a closed curve $C$, then
$$
\iint_D {\bf curl\,F} \cdot {\bf k} \, dA = \int_{C}{\bf F} \cdot  d{\bf r},
$$
a relation between a one-dimensional line integral of a vector field and a two-dimensional integral of a function
involving the derivatives of the component functions of that vector field. The Fundamental Theorem of Calculus was featured prominently in the proof.

\smallskip

\item \textit{The Divergence Theorem}: If ${\bf F}$ is a vector field on $\mathbb{R}^3$ and if $D$ is a 
domain bounded by a closed surface $\calS$, then
$$
\iiint_D {\bf div\,F} \, dV = \iint_{\calS}{\bf F} \cdot  d{\bf S}.
$$
Again, this relates an integral of a vector field on the boundary of a domain (this time, 2-dimensional) to the integral
of a function derived from it on that domain (3-dimensional).

\end{itemize}

The justification for introducing differential forms is to establish a common framework for all of these. A {\em differential form} is
going to be ``something we can integrate on a manifold".  Differential forms will have a certain dimensionality that has to be matched to the dimension of the manifold where we want to integrate them. For example, functions on domains or vector fields on curves can be rewritten as differential forms satisfying these dimensionality conditions.

We will work towards \textit{Generalized Stokes's Theorem}---a relation between 
integrals 
$$
\int_{\partial \mathcal{M}} \omega = \int_{\mathcal{M}} d\omega,
$$
where $\omega$ is a differential form, $d\omega$ is another differential form derived from it, $\mathcal{M}$ 
is a manifold, and $ \partial \mathcal{M}$ is the boundary of $\mathcal{M}$. Without thinking (for now)
about what a differential form is or what it means to integrate it, you can check that all of the above theorems
have this general form.

\subsection{Differential $k$-forms} We introduce the basic language used to define differential forms and operations on them. While the general theory works in $\rr^n$ 
for any $n \geq 1$, the case $n=4$ is the lowest dimension in which you will see new phenomena that we did not cover
previously. In this section, we will usually work out 2-dimensional and 3-dimensional cases of our general formulas in detail, to ensure that our abstract general theory matches the theorems we already know in $\mathbb{R}^2$ and $\mathbb{R}^3$. Additionally, we will often refer to examples in dimension $n = 4$, to make sure we go beyond our previous work.

We start with $k$-forms with constant coefficients. In general, we will use ${\bf x}=(x_1,x_2,\dots,x_n)$ to denote vectors in $\rr^n$, but we will also continue to use the coordinates $(x,y)$ in $\rr^2$ and $(x,y,z)$ in $\rr^3$ as needed.

\begin{itemize}
\item \textit{$1$-forms} with constant coefficients are linear mappings $\rr^n\to\rr$ (in functional analysis, such mappings are called {\it linear functionals}), written as $\omega=a_1dx_1+\dots+a_ndx_n$,
acting on vectors via
$$
dx_i({\bf v})=v_i
$$
and then extended via linearity so that $(a_1dx_1+\dots+a_ndx_n)({\bf v})=\sum a_i v_i$.
For example, if $\omega=3dx_1-5dx_4 $ in $\rr^4$, and if ${\bf v}=\langle 1,2,3,4\rangle$, then 
$$\omega({\bf v}) =3\cdot 1 -5\cdot 4 =3-20=-17.$$

\medskip

\item \textit{$2$-forms} with constant coefficients are functions $\phi : \rr^n\times\rr^n \to \rr$ (so that $\phi$ acts on two vectors simultaneously) that are \textit{bilinear}:
$$
\phi(a{\bf v}+ b {\bf w}, {\bf u}) = a \phi({\bf v}, {\bf u})  + b \phi( {\bf w}, {\bf u}) ,
$$
$$
\phi({\bf u}, a{\bf v}+ b {\bf w}) = a \phi({\bf u}, {\bf v})  + b \phi( {\bf u}, {\bf v}) ,
$$
and \textit{antisymmetric}:
$$
\phi({\bf v}, {\bf u}) = - \phi({\bf u}, {\bf v}) .
$$
The elementary $2$-forms are 
$$
dx_i\wedge dx_j (  {\bf u}, {\bf v})=
    \left| \begin{matrix}
u_i& v_i \\ u_j& v_j\\
\end{matrix} \right|
=   \left| \begin{matrix}
u_i& u_j \\ v_i& v_j\\
\end{matrix} \right|.
$$
By the basic properties of determinants, we have
\begin{equation}\label{51eq-antisymm1}
dx_i\wedge dx_j =- dx_j\wedge dx_i ,\ \ dx_i\wedge dx_i=0.
\end{equation}
(A determinant is 0 if two or its rows or columns are identical, and changes sign if two of its rows or columns are interchanged.)

\begin{example}
If ${\bf v}=\langle 1,2,3,4\rangle$ and ${\bf w}=\langle -5,6,-7,8\rangle$, then
$$
(2dx_1 \wedge dx_4 - dx_2\wedge dx_3)({\bf v}, {\bf w}) 
= 2    \left| \begin{matrix}
1&-5 \\ 4 & 8\\
\end{matrix} \right|
-    \left| \begin{matrix}
2 & 6 \\ 3 & -7\\
\end{matrix} \right|=24.
$$
\end{example}

\begin{example}
Let $\phi=dx_2\wedge dx_3  +dx_3\wedge dx_1 + dx_1\wedge dx_2$ in $\rr^3$,
then
$$
\phi({\bf u}, {\bf v})= 
 \left| \begin{matrix}
u_2& u_3 \\ v_2& v_3\\
\end{matrix} \right|
+  \left| \begin{matrix}
u_3& u_1 \\ v_3& v_1\\
\end{matrix} \right|
+  \left| \begin{matrix}
u_1& u_2 \\ v_1& v_2\\
\end{matrix} \right|.
$$
\end{example}

The formal similarity to the cross product of two vectors is 100\% intended; in fact, we will use differential forms to generalize the concept of the cross-product to higher dimensions.

\medskip

\item \textit{$k$-forms} with constant coefficients are defined as mappings
$$
\phi :\underbrace{\mathbb{R}^n \times \cdots \times \mathbb{R}^n}_{k \textrm{ times}} \rightarrow \mathbb{R},
$$
generalizing the above to $k$ vector-valued variables.  We continue to require \textit{multilinearity},
\begin{align*}
& \phi (\mathbf{v}_1,\dots , \mathbf{v}_{i-1}, a\mathbf{v}_{i} + b\mathbf{v}'_i,\mathbf{v}_{i+1}, \dots, \mathbf{v}_k) 
\\
&\ \ \ = a \phi(\mathbf{v}_1,\dots,\mathbf{v}_{i},\dots, \mathbf{v}_k) + b \phi (\mathbf{v}_1,\dots,\mathbf{v}'_{i},\dots, \mathbf{v}_k)
\end{align*}
and \textit{antisymmetry},
$$
\phi(\mathbf{v}_1,...,\mathbf{v}_{i},...,\mathbf{v}_{i'},...,\mathbf{v}_k) = - \phi(\mathbf{v}_1,...,\mathbf{v}_{i'},...,\mathbf{v}_i,...,\mathbf{v}_k).
$$
In words: multilinearity requires $\phi$ to be linear in each variable; antisymmetry requires $\phi$ to change signs if any two variables are exchanged (e.g. first vector is switched with the last; 2nd vector switched with 8th vector). We use $\Lambda_k(\rr^n)$ to denote the linear space of all $k$-forms with constant coefficients on $\rr^n$.

\smallskip

\textit{Elementary $k$ forms} are the basic building blocks for $k$-forms. They are:
\begin{eqnarray*}
d x_{i_1} \wedge \cdots \wedge d x_{i_k} (\mathbf{v}_1, ...,\mathbf{v}_k) & : = & \begin{vmatrix}
v_{1 i_1} & v_{2 i_1} & \cdots & v_{k i_1} \\
v_{1 i_2} & v_{2 i_2} & \cdots & v_{k i_2} \\
\vdots & \vdots & \ddots & \vdots \\
v_{1 i_k} & v_{2 i_k} & \cdots & v_{k i_k}
\end{vmatrix}
\end{eqnarray*}
and any $k$-form can be written as a linear combination of such forms. 
Another way to write the above formula is
$$
d x_{i_1} \wedge \cdots \wedge d x_{i_k} (\mathbf{v}_1, ...,\mathbf{v}_k) = {\rm det} (d x_{i_s} (\mathbf{v}_t ) )_{1 \leq s,t \leq k},
$$
if we remember that the $1$-form  $d x_i$ were defined as $dx_i (\mathbf{x}) = x_i$. 
We give an example below to illustrate this.

\begin{example}
Let ${\bf u}=\langle 1,2,3,4\rangle$,
${\bf v}=\langle 5,6,7,8\rangle$, and ${\bf w}=\langle 1,-1,1,-1\rangle$, then
$$
dx_1 \wedge dx_3 \wedge dx_4({\bf u}, {\bf v}, {\bf w}) 
= \left| \begin{matrix}
1&5& 1 \\ 3&7&1\\ 4&8&-1  \\ 
\end{matrix} \right|
$$
\end{example}



As in (\ref{51eq-antisymm1}), an elementary $k$-form is zero if two of the indices $i_j$ are equal, and changes sign if two of the expressions $d x_{i_j}$ and $d x_{i_{j'}}$ are interchanged. For example, $dx\wedge dy\wedge dx=0$ and 
$dx\wedge dy\wedge dz=-dz\wedge dy\wedge dx$.


\begin{example}\label{51-lin-indep}
The only nonzero elementary $n$-form in $\rr^n$, up to $\pm$ signs, is $dx_1\wedge dx_2\wedge\dots\wedge dx_n$. If ${\bf v}_1,
\dots,{\bf v}_n$ are vectors in $\rr^n$, then
$$
{\bf v}_1,\dots,{\bf v}_n \hbox{ are linearly independent }\ \Leftrightarrow\  dx_1\wedge\dots\wedge dx_n({\bf v}_1,\dots,{\bf v}_n)\neq 0.
$$
(See the determinant formula above.)

\end{example}




\end{itemize}


A general \textbf{differential $k$-form} is a linear combination of the elementary $k$-forms defined above, but we allow variable coefficients which are functions of $\mathbf{x} = (x_1,...,x_n)$. 
For example,
$$
x^2dx\wedge dy,\ \ ydx-zdy,\ \ x_2\cos(x_3) dx_1\wedge dx_3\wedge dx_4
$$
are differential forms.
Unless we say explicitly otherwise, we will assume that all coefficient functions are $C^\infty$ on their domain $D$, meaning that all partial derivatives of any order exist and are continuous on $D$. Many of the results below are valid under weaker assumptions, for example Proposition \ref{51prop:exteriorcalculus} below remains true if the coefficient functions are $C^2$ instead of $C^\infty$. The point of the $C^\infty$ assumption is to allow us to focus on the concepts without getting distracted by counting derivatives.


When $k = 0$, a differential $0$-form is simply a $C^\infty$ function $D \to \mathbb{R}$.
For $k\geq 1$, a general differential $k$-form on a domain $D \subset \mathbb{R}^n$ can be written as
$$
\Phi (\mathbf{x}) = \sum\limits_{1 \leq i_1 < i_2 < \cdots < i_k \leq n} a_{i_1\cdots i_k} (\mathbf{x}) d x_{i_1} \wedge \cdots \wedge d x_{i_k},
$$
where the coefficients $a_{i_1...i_k}$ are $C^\infty$ functions on $D$. Notice that we only use the indices $1 \leq i_1 < i_2 < \cdots < i_k \leq n$ in the summation. This is because, first, $d x_{i_1} \wedge \cdots \wedge d x_{i_k}=0$ if any two of the indices are equal. Second, for any remaining non-zero terms, we can rearrange the $dx_i$'s so that the indices come in increasing order, at the cost of possibly multiplying that term by $-1$. For instance,
$$
x_1^2 dx_3\wedge dx_4\wedge dx_1 = - x_1^2 dx_1\wedge dx_4\wedge dx_3 = x_1^2 dx_1\wedge dx_3\wedge dx_4,
$$
where at each step we swapped two of the $dx_i$'s and changed the sign of the form. 



Similarly to the coefficients, we will also allow the \textit{arguments} of differential forms to depend on ${\bf x}$, so that differential $k$-form may act on a $k$-tuple of vector fields (rather than just $k$-tuples of fixed vectors).  For example,
if ${\bf F}=e^x \ii -e^y \jj + e^{2z} \kk$ and ${\bf G}= x\ii +x^2\jj +x^3\kk$, then
$$
y dx \wedge dz ({\bf F}, {\bf G}) 
=y  \left| \begin{matrix}
e^x & e^{2z} \\ x & x^3  \\ 
\end{matrix} \right|
=y(x^3 e^x - xe^{2z}).
$$





\subsection{Exterior (wedge) product} 
If $\phi \in \Lambda_k (\mathbb{R}^n)$ and $\psi \in \Lambda_l (\mathbb{R}^n)$ (so that $\phi$ is a $k$-form and $\psi$ is an $l$-form on $\mathbb{R}^n$), we define their wedge product $\phi \wedge \psi \in \Lambda_{k+l} (\mathbb{R}^n)$ as follows:

\begin{itemize}
\item if $\phi$ and $\psi$ are elementary forms, just concatenate them and write a wedge in between,
\item use linearity to extend it to more general forms.

\end{itemize}

For example, if $\phi=dx_1\wedge dx_2$ and $\psi=dx_3\wedge dx_4\wedge dx_5$, then 
$$\phi \wedge \psi= dx_1\wedge dx_2\wedge dx_3\wedge dx_4\wedge dx_5.$$
A couple of (slightly) more complicated examples are below.

\begin{example}
Suppose $\phi = a_1 d x_1 + a_2 d x_2 + a_3 d x_3$ and $\psi = b_1 d x_1 + b_2 d x_2$.  Then, 
\begin{eqnarray*}
\phi \wedge \psi & = & \cancelto{0}{a_1 b_1 d x_1 \wedge d x_1} +  a_2 b_1 d x_2 \wedge d x_1 + a_3 b_1 d x_3 \wedge dx_1 \\
\quad & & \quad + a_1 b_1 d x_1 \wedge d x_2 + \cancelto{0}{a_2 b_2 dx_2 \wedge dx_2} + a_3 b_2 d x_3 \wedge d x_2 \\[2ex]
\quad & = & (a_1 b_2 - a_2 b_1) d x_1 \wedge dx_2 + a_3 b_1 d x_3 \wedge d x_1 + a_3 b_2 d x_3 \wedge d x_2.
\end{eqnarray*}
\end{example}

\begin{example}
Let $\phi = a_1 dx_1 + a_2 dx_2 + a_3dx_3$ and
$$
\psi = b_1 (dx_2 \wedge dx_3) + b_2 (dx_3 \wedge dx_1) + b_3 (dx_1 \wedge dx_2).
$$
Then, 
$$
\phi \wedge \psi = (a_1b_1 + a_2b_2 + a_3b_3) (dx_1 \wedge dx_2 \wedge dx_3).
$$
(Exercise: verify this!)
\end{example}


Here are a couple of examples with variable coefficients. 

\begin{example}
We work in $\rr^3$, with coordinates $\mathbf{r} = (x,y,z)$. Then
$\Phi  = 3xz dx \wedge dy + y^2 dx \wedge dz$ is a differential $2$-form in $3$ variables, and  $\Psi : = e^x dx + e^z dz$ is a differential $1$-form in $3$ variables.  Their wedge product is
\begin{eqnarray*}
\Phi \wedge \Psi & = & (3xz dx \wedge dy + y^2 dx \wedge dz) \wedge (e^x dx + e^z dz) \\
\quad & = & \cancelto{0}{3xz e^x dx \wedge dy \wedge dx} + \cancelto{0}{y^2 e^x dx \wedge dz \wedge dx} \\
\quad & \quad & \quad +{3xze^z dx \wedge dy \wedge dz} + \cancelto{0}{y^2 e^z dx \wedge dz \wedge dz} \\[2ex]
\quad & = & 3xz e^z dx \wedge dy \wedge dz.
\end{eqnarray*}
Three terms in this calculation cancel to zero because $dx \wedge dx = dz \wedge dz = 0$.  Since $\Phi$ was a $2$-form and $\Psi$ was a $1$-form, the resulting wedge product $(\Phi \wedge \Psi)$ is a $3$-form.
\end{example}

\begin{example}
Let $\Phi  = xy dx + ydz$ and $\Psi= x dx + z^2 dy$ be differential $1$-forms in $3$ variables. Then, their wedge product is the 2-form
\begin{eqnarray*}
\Phi \wedge \Psi & = & (xy dx + ydz) \wedge (x dx + z^2 dy) \\
\quad & = &  xy dz \wedge dx + xyz^2 dx \wedge dy+ y z^2 dz \wedge dy ,
\end{eqnarray*}
where again we used that $dx \wedge dx = 0$.
\end{example}







%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


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




\item
{\it  Let ${\bf u}= (1, 2, -1, 3)$, ${\bf v}=(3, 0, 1, 0)$, ${\bf w}=(2, 2, 1, -1)$. Evaluate:}

(a) 
$dx_1\wedge dx_3 \wedge dx_4({\bf u}, {\bf v}, {\bf w})$,

\begin{align*}
dx_1\wedge dx_2 \wedge dx_4({\bf u}, {\bf v}, {\bf w}) &=
\begin{vmatrix}
1 & 3 & 2\\
-1 & 1 & 1\\
3 & 0 & -1
\end{vmatrix}
\\
&= 3
\begin{vmatrix}
3 & 2\\
1 & 1
\end{vmatrix}
-
\begin{vmatrix}
1 & 3\\
-1 & 1
\end{vmatrix}
\\
&= 3(1) - 1 (4) \\
&= -1
\end{align*}
where we expanded the $3\times 3$ determinant along the third row.



(b) 
$2 dx_1\wedge dx_2({\bf u}, {\bf v})
- dx_3 \wedge dx_4({\bf u}, {\bf w})$.

\begin{align*}
2 dx_1\wedge dx_2({\bf u}, {\bf v}) - dx_3 \wedge dx_4({\bf u}, {\bf w}) &=
2 \begin{vmatrix}
1 & 3\\
2 & 0
\end{vmatrix}
-
\begin{vmatrix}
-1 & 1\\
3 & -1
\end{vmatrix}
\\
&= 2 (-6) - (-2)\\
&= -12+2=10
\end{align*}


 
\item
{\it
Calculate:}

(a) $\Phi({\bf F}, {\bf G})$, if $\Phi= y dx\wedge dy - z^2  dx\wedge dz$,
${\bf F}=\cos x \ii + \sin x \jj $, and ${\bf G}=xy \jj + yz\kk$.

\begin{align*}
\Phi({\bf F}, {\bf G}) &= y dx\wedge dy ({\bf F}, {\bf G}) - z^2  dx\wedge dz ({\bf F}, {\bf G})\\
&= y
\begin{vmatrix}
\cos x & 0\\
\sin x & xy
\end{vmatrix}
-z^2
\begin{vmatrix}
\cos x & 0\\
0 & yz
\end{vmatrix}
\\
&= x y^2\cos x - yz^3 \cos x.
\end{align*}


(b) $\Psi({\bf F}, {\bf G}, {\bf H})$, if $\Psi=  dx\wedge dy\wedge dz$, ${\bf F}$ and $\bf G$ are
as in (a), and ${\bf H}=2y\ii -x\kk$.


\begin{align*}
\Psi({\bf F}, {\bf G}, {\bf H}) =  dx\wedge dy\wedge dz ({\bf F}, {\bf G}, {\bf H})
&=
\begin{vmatrix}
\cos x & 0 & 2y\\
\sin x & xy & 0\\
0 & yz & -x
\end{vmatrix}
\\
&= \cos x
\begin{vmatrix}
xy & 0 \\
yz & -x
\end{vmatrix}
+2y
\begin{vmatrix}
\sin x & xy \\
0 & yz
\end{vmatrix}
\\
&= -x^2y \cos x  + 2y^2 z \sin x.
\end{align*}











\item
{\it  Calculate and simplify the wedge product $\phi \wedge \psi$. Write the $dx_i$'s in the answers in increasing subscript order.}

(a) $\phi=2 dx_1\wedge dx_3 - 5 dx_3 \wedge dx_4,$
$\psi=  - dx_1 \wedge dx_2 + 4 dx_2 \wedge dx_4$

\begin{align*}
\phi \wedge \psi &= (2 dx_1\wedge dx_3- 5 dx_3 \wedge dx_4
) \wedge (-  dx_1 \wedge dx_2 + 4 dx_2 \wedge dx_4)\\
&= 5 dx_3\wedge dx_4 \wedge dx_1\wedge dx_2
+8 dx_1\wedge dx_3 \wedge dx_2\wedge dx_4\\
&= -5 dx_1\wedge dx_4 \wedge dx_3\wedge dx_2
-8 dx_1\wedge dx_2 \wedge dx_3\wedge dx_4
\\
&= 5 dx_1\wedge dx_2 \wedge dx_3\wedge dx_4
-8 dx_1\wedge dx_2 \wedge dx_3\wedge dx_4
\\
&= -3 dx_1\wedge dx_2 \wedge dx_3\wedge dx_4.
\end{align*}
The other terms vanish because they each have two appearances of some $dx_i$.


(b) $\phi= dx_1\wedge dx_3\wedge dx_4,$
$\psi= 7 dx_2 -  dx_3$


\begin{align*}
\phi \wedge \psi &= (dx_1\wedge dx_3\wedge dx_4) \wedge (7 dx_2 -  dx_3)\\
&= 7 dx_1\wedge dx_3\wedge dx_4 \wedge dx_2\\
&= - 7 dx_1\wedge dx_2\wedge dx_4 \wedge dx_3\\
&= -7 dx_1 \wedge dx_2\wedge dx_3\wedge dx_4.
\end{align*}







\end{enumerate}


\section*{Practice Problems}
\begin{enumerate}[1.]
\item
{Let ${\bf u}= (1,-1,0,5)$, ${\bf v}=(3,2,6,-1)$, ${\bf w}=(5,-2,3,1)$. Evaluate
$dx_1\wedge dx_3 \wedge dx_4({\bf u}, {\bf v}, {\bf w})$.}

\bigskip

\item
{Let ${\bf u}= (2,1,4,-1)$, ${\bf v}=(1,3,5,0)$, ${\bf w}=(2,7,-2,3)$. Evaluate:

\smallskip

(a) 
$dx_1\wedge dx_2 \wedge dx_4({\bf u}, {\bf v}, {\bf w})$,

\smallskip

(b) 
$dx_2\wedge dx_3({\bf u}, {\bf v})
-2 dx_1 \wedge dx_4({\bf u}, {\bf w})$.
}

\bigskip

\item
{Calculate and simplify the wedge product $\phi\wedge \psi$. Write the $dx_i$'s in the answers in increasing subscript order.
\smallskip

(a) $\phi=dx_1\wedge dx_3\wedge dx_4,$
$\psi= a_1 dx_1+ a_2 dx_2 + a_3 dx_3$.

\smallskip

(b) $\phi= dx_2\wedge dx_3 +  dx_1\wedge dx_4,$
$\psi= 4 dx_1\wedge dx_2 - dx_1
\wedge dx_3 + 5 dx_2\wedge dx_3$.
}

\bigskip


\item
{Calculate and simplify the wedge product $\phi \wedge \psi$. Write the $dx_i$'s in the answers in increasing subscript order.

\smallskip

(a) $\phi=3 dx_2\wedge dx_3 - 7 dx_2 \wedge dx_4,$
$\psi= 2 dx_1+ 3 dx_2 - dx_4$.

\smallskip

(b) $\phi= dx_2\wedge dx_3\wedge dx_4,$
$\psi= 3 dx_1 - 5 dx_2$

\smallskip

(c) $\phi=2 dx_1\wedge dx_3$, $\psi= - 2 dx_1 \wedge dx_4 + 3 dx_2 \wedge dx_4$.

}

\bigskip




\item
{Compute the exterior product 
$\Phi_1 \wedge \Phi_2$, if
$\Phi_1 = x_1  \, dx_1 \wedge dx_2 \wedge dx_4$ and
$\Phi_2 = x_4 dx_1 + x_2 dx_3 + x_3 dx_4$ in $\rr^4$.
Give your answer in the form $f(x_1,x_2,x_3,x_4) 
dx_1\wedge dx_2 \wedge dx_3 \wedge dx_4$.}

\bigskip




\end{enumerate}



