chapter8a.tex

\chapter{Exterior product \label{sec:Exterior-product}}

In this chapter I introduce one of the most useful constructions in
basic linear algebra --- the exterior product, denoted by $\mathbf{a}\wedge\mathbf{b}$,
where $\mathbf{a}$ and $\mathbf{b}$ are vectors from a space $V$.
The basic idea of the exterior product is that we would like to define
an \emph{antisymmetric} and bilinear product of vectors. In other
words, we would like to have the properties $\mathbf{a}\wedge\mathbf{b}=-\mathbf{b}\wedge\mathbf{a}$
and $\mathbf{a}\wedge(\mathbf{b}+\lambda\mathbf{c})=\mathbf{a}\wedge\mathbf{b}+\lambda\mathbf{a}\wedge\mathbf{c}$. 


\section{Motivation\label{sub:Motivation-for-exterior}}

Here I discuss, at some length, the motivation for introducing the
exterior product. The motivation is geometrical and comes from considering
the properties of areas and volumes in the framework of elementary
Euclidean geometry. I will proceed with a formal definition of the
exterior product in Sec.~\ref{sub:Definition-of-the-exterior}. In
order to understand the definition explained there, it is not necessary
to use this geometric motivation because the definition will be purely
algebraic. Nevertheless, I feel that this motivation will be helpful
for some readers.


\subsection{Two-dimen\-sion\-al oriented area\label{sub:Two-dimensional-oriented}}

We work in a two-dimen\-sion\-al Euclidean space, such as that considered
in elementary geometry. We assume that the usual geometrical definition
of the area of a parallelogram is known.

Consider the area $Ar(\mathbf{a},\mathbf{b})$ of a parallelogram
spanned by vectors $\mathbf{a}$ and $\mathbf{b}$. It is known from
elementary geometry that $Ar(\mathbf{a},\mathbf{b})=\left|\mathbf{a}\right|\cdot\left|\mathbf{b}\right|\cdot\sin\alpha$
where $\alpha$ is the angle between the two vectors, which is always
between 0 and $\pi$ (we do not take into account the orientation
of this angle). Thus defined, the area $Ar$ is always non-negative.

Let us investigate $Ar(\mathbf{a},\mathbf{b})$ as a function of the
vectors $\mathbf{a}$ and $\mathbf{b}$. If we stretch the vector
$\mathbf{a}$, say, by factor 2, the area is also increased by factor
2. However, if we multiply $\mathbf{a}$ by the number $-2$, the
area will be multiplied by $2$ rather than by $-2$:\[
Ar(\mathbf{a},2\mathbf{b})=Ar(\mathbf{a},-2\mathbf{b})=2Ar(\mathbf{a},\mathbf{b}).\]
 Similarly, for some vectors $\mathbf{a},\mathbf{b},\mathbf{c}$ such
as shown in Fig.~\ref{fig:The-area-of2}, we have $Ar(\mathbf{a},\mathbf{b}+\mathbf{c})=Ar(\mathbf{a},\mathbf{b})+Ar(\mathbf{a},\mathbf{c})$.
However, if we consider $\mathbf{b}=-\mathbf{c}$ then we obtain \begin{align*}
Ar(\mathbf{a},\mathbf{b}+\mathbf{c}) & =Ar(\mathbf{a},0)=0\\
 & \neq Ar(\mathbf{a},\mathbf{b})+Ar(\mathbf{a},-\mathbf{b})=2Ar(\mathbf{a},\mathbf{b}).\end{align*}


Hence, the area $Ar(\mathbf{a},\mathbf{b})$ is, strictly speaking,
\emph{not} a linear function of the vectors $\mathbf{a}$ and $\mathbf{b}$:
\begin{align*}
Ar(\lambda\mathbf{a},\mathbf{b}) & =\left|\lambda\right|Ar(\mathbf{a},\mathbf{b})\neq\lambda\, Ar(\mathbf{a},\mathbf{b}),\\
Ar(\mathbf{a},\mathbf{b}+\mathbf{c}) & \neq Ar(\mathbf{a},\mathbf{b})+Ar(\mathbf{a},\mathbf{c}).\end{align*}
Nevertheless, as we have seen, the properties of linearity hold in
\emph{some} cases. If we look closely at those cases, we find that
linearly holds precisely when we do not change the orientation of
the vectors. It would be more convenient if the linearity properties
held in all cases. 

The trick is to replace the area function $Ar$ with the \textbf{oriented
area}\index{oriented area} function $A(\mathbf{a},\mathbf{b})$.
Namely, we define the function $A(\mathbf{a},\mathbf{b})$ by \[
A(\mathbf{a},\mathbf{b})=\pm\left|\mathbf{a}\right|\cdot\left|\mathbf{b}\right|\cdot\sin\alpha,\]
where the sign is chosen positive when the angle $\alpha$ is measured
from the vector $\mathbf{a}$ to the vector $\mathbf{b}$ in the counterclockwise
direction, and negative otherwise.


\paragraph{Statement:}

The oriented area $A(\mathbf{a},\mathbf{b})$ of a parallelogram spanned
by the vectors $\mathbf{a}$ and $\mathbf{b}$ in the two-dimen\-sion\-al
Euclidean space is an antisymmetric and bilinear function of the vectors
$\mathbf{a}$ and $\mathbf{b}$:\begin{align*}
A(\mathbf{a},\mathbf{b}) & =-A(\mathbf{b},\mathbf{a}),\\
A(\lambda\mathbf{a},\mathbf{b}) & =\lambda\, A(\mathbf{a},\mathbf{b}),\\
A(\mathbf{a},\mathbf{b}+\mathbf{c}) & =A(\mathbf{a},\mathbf{b})+A(\mathbf{a},\mathbf{c}).\qquad\text{(the sum law)}\end{align*}


%
\begin{figure}
\begin{centering}
\psfrag{0}{0}\psfrag{A}{$A$} \psfrag{B}{$B$} \psfrag{D}{$D$} \psfrag{C}{$C$} \psfrag{E}{$E$} \psfrag{v1}{$\mathbf{b}$} \psfrag{v2}{$\mathbf{a}$} \psfrag{v1lambda}{$\mathbf{b}+\alpha\mathbf{a}$}\includegraphics[width=3in]{./figs/v1v2-vol.eps}
\par\end{centering}

\caption{The area of the parallelogram $0ACB$ spanned by $\mathbf{a}$ and
$\mathbf{b}$ is equal to the area of the parallelogram $0ADE$ spanned
by $\mathbf{a}$ and $\mathbf{b}+\alpha\mathbf{a}$ due to the equality
of areas $ACD$ and $0BE$.\label{fig:The-area-of1}}

\end{figure}



\subparagraph{Proof:}

The first property is a straightforward consequence of the sign rule
in the definition of $A$.

Proving the second property requires considering the cases $\lambda>0$
and $\lambda<0$ separately. If $\lambda>0$ then the orientation
of the pair $\left(\mathbf{a},\mathbf{b}\right)$ remains the same
and then it is clear that the property holds: When we rescale $\mathbf{a}$
by $\lambda$, the parallelogram is stretched and its area increases
by factor $\lambda$. If $\lambda<0$ then the orientation of the
parallelogram is reversed and the oriented area changes sign.

To prove the sum law, we consider  two cases: either $\mathbf{c}$
is parallel to $\mathbf{a}$ or it is not. If $\mathbf{c}$ is parallel
to $\mathbf{a}$, say $\mathbf{c}=\alpha\mathbf{a}$, we use Fig.~\ref{fig:The-area-of1}
to show that $A(\mathbf{a},\mathbf{b}+\lambda\mathbf{a})=A(\mathbf{a},\mathbf{b})$,
which yields the desired statement since $A(\mathbf{a},\lambda\mathbf{a})=0$.
If $\mathbf{c}$ is not parallel to $\mathbf{a}$, we use Fig.~\ref{fig:The-area-of2}
to show that $A(\mathbf{a},\mathbf{b}+\mathbf{c})=A(\mathbf{a},\mathbf{b})+A(\mathbf{a},\mathbf{c})$.
Analogous geometric constructions can be made for different possible
orientations of the vectors $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$.\hfill{}$\blacksquare$

%
\begin{figure}
\begin{centering}
\psfrag{A}{$A$} \psfrag{B}{$B$} \psfrag{D}{$D$} \psfrag{C}{$C$} \psfrag{F}{$F$} \psfrag{E}{$E$}  \psfrag{a}{$\mathbf{a}$} \psfrag{b}{$\mathbf{b}$} \psfrag{c}{$\mathbf{c}$}\psfrag{b+c}{$\mathbf{b}+\mathbf{c}$}\includegraphics[width=3in]{./figs/2darea}
\par\end{centering}

\caption{The area of the parallelogram spanned by $\mathbf{a}$ and $\mathbf{b}$
(equal to the area of $CEFD$) plus the area of the parallelogram
spanned by $\mathbf{a}$ and $\mathbf{c}$ (the area of $ACDB$) equals
the area of the parallelogram spanned by $\mathbf{a}$ and $\mathbf{b}+\mathbf{c}$
(the area of $AEFB$) because of the equality of the areas of $ACE$
and $BDF$.\label{fig:The-area-of2}}

\end{figure}


It is relatively easy to compute the oriented area because of its
algebraic properties. Suppose the vectors $\mathbf{a}$ and $\mathbf{b}$
are given through their components in a standard basis $\left\{ \mathbf{e}_{1},\mathbf{e}_{2}\right\} $,
for instance \[
\mathbf{a}=\alpha_{1}\mathbf{e}_{1}+\alpha_{2}\mathbf{e}_{2},\quad\mathbf{b}=\beta_{1}\mathbf{e}_{1}+\beta_{2}\mathbf{e}_{2}.\]
We assume, of course, that the vectors $\mathbf{e}_{1}$ and $\mathbf{e}_{2}$
are orthogonal to each other and have unit length, as is appropriate
in a Euclidean space. We also assume that the right angle is measured
from $\mathbf{e}_{1}$ to $\mathbf{e}_{2}$ in the counter-clockwise
direction, so that $A(\mathbf{e}_{1},\mathbf{e}_{2})=+1$. Then we
use the Statement and the properties $A(\mathbf{e}_{1},\mathbf{e}_{1})=0$,
$A(\mathbf{e}_{1},\mathbf{e}_{2})=1$, $A(\mathbf{e}_{2},\mathbf{e}_{2})=0$
to compute\begin{align*}
A(\mathbf{a},\mathbf{b}) & =A(\alpha_{1}\mathbf{e}_{1}+\alpha_{2}\mathbf{e}_{2},\beta_{1}\mathbf{e}_{1}+\beta_{2}\mathbf{e}_{2})\\
 & =\alpha_{1}\beta_{2}A(\mathbf{e}_{1},\mathbf{e}_{2})+\alpha_{2}\beta_{1}A(\mathbf{e}_{2},\mathbf{e}_{1})\\
 & =\alpha_{1}\beta_{2}-\alpha_{2}\beta_{1}.\end{align*}


The ordinary (unoriented) area is then obtained as the absolute value
of the oriented area, $Ar(\mathbf{a},\mathbf{b})=\left|A(\mathbf{a},\mathbf{b})\right|$.
It turns out that the oriented area, due to its strict linearity properties,
is a much more convenient and powerful construction than the unoriented
area.


\subsection{Parallelograms in $\mathbb{R}^{3}$ and in $\mathbb{R}^{n}$ \label{sub:Area-of-two-dimensional-parallelograms}}

Let us now work in the Euclidean space $\mathbb{R}^{3}$ with a standard
basis $\left\{ \mathbf{e}_{1},\mathbf{e}_{2},\mathbf{e}_{3}\right\} $.
We can similarly try to characterize the area of a parallelogram spanned
by two vectors $\mathbf{a}$, $\mathbf{b}$. It is, however, not possible
to characterize the orientation of the area simply by a sign. We also
cannot use a geometric construction such as that in Fig.~\ref{fig:The-area-of2};
in fact it is \emph{not true} in three dimensions that the area spanned
by $\mathbf{a}$ and $\mathbf{b}+\mathbf{c}$ is equal to the sum
of $Ar(\mathbf{a},\mathbf{b})$ and $Ar(\mathbf{a},\mathbf{c})$.
Can we still define some kind of {}``oriented area'' that obeys
the sum law?

Let us consider Fig.~\ref{fig:The-area-of2} as a figure showing
the \emph{projection} of the areas of the three parallelograms onto
some coordinate plane, say, the plane of the basis vectors $\left\{ \mathbf{e}_{1},\mathbf{e}_{2}\right\} $.
It is straightforward to see that the projections of the areas obey
the sum law as oriented areas.


\paragraph{Statement:}

Let $\mathbf{a},\mathbf{b}$ be two vectors in $\mathbb{R}^{3}$,
and let $P(\mathbf{a},\mathbf{b})$ be the parallelogram spanned by
these vectors. Denote by $P(\mathbf{a},\mathbf{b})_{\mathbf{e}_{1},\mathbf{e}_{2}}$
the parallelogram within the coordinate plane $\text{Span}\left\{ \mathbf{e}_{1},\mathbf{e}_{2}\right\} $
obtained by projecting $P(\mathbf{a},\mathbf{b})$ onto that coordinate
plane, and similarly for the other two coordinate planes. Denote by
$A(\mathbf{a},\mathbf{b})_{\mathbf{e}_{1},\mathbf{e}_{2}}$ the oriented
area of $P(\mathbf{a},\mathbf{b})_{\mathbf{e}_{1},\mathbf{e}_{2}}$.
Then $A(\mathbf{a},\mathbf{b})_{\mathbf{e}_{1},\mathbf{e}_{2}}$ is
a bilinear, antisymmetric function of $\mathbf{a}$ and $\mathbf{b}$.


\subparagraph{Proof:}

The projection onto the coordinate plane of $\mathbf{e}_{1},\mathbf{e}_{2}$
is a linear transformation. Hence, the vector $\mathbf{a}+\lambda\mathbf{b}$
is projected onto the sum of the projections of $\mathbf{a}$ and
$\lambda\mathbf{b}$. Then we apply the arguments in the proof of
Statement~\ref{sub:Two-dimensional-oriented} to the \emph{projections}
of the vectors; in particular, Figs.~\ref{fig:The-area-of1} and~\ref{fig:The-area-of2}
are interpreted as showing the projections of all vectors onto the
coordinate plane $\mathbf{e}_{1},\mathbf{e}_{2}$. It is then straightforward
to see that all the properties of the oriented area hold for the projected
oriented areas. Details left as exercise.\hfill{}$\blacksquare$

It is therefore convenient to consider the oriented areas of the three
projections --- $A(\mathbf{a},\mathbf{b})_{\mathbf{e}_{1},\mathbf{e}_{2}}$,
$A(\mathbf{a},\mathbf{b})_{\mathbf{e}_{2},\mathbf{e}_{3}}$, $A(\mathbf{a},\mathbf{b})_{\mathbf{e}_{3},\mathbf{e}_{1}}$
--- as three components of a \emph{vector-valued} area $A(\mathbf{a},\mathbf{b})$
of the parallelogram spanned by $\mathbf{a},\mathbf{b}$. Indeed,
it can be shown that these three projected areas coincide with the
three Euclidean components of the vector product $\mathbf{a}\times\mathbf{b}$.
The vector product is the traditional way such areas are represented
in geometry: the vector $\mathbf{a}\times\mathbf{b}$ represents at
once the magnitude of the area and the orientation of the parallelogram.
One computes the unoriented area of a parallelogram as the length
of the vector $\mathbf{a}\times\mathbf{b}$ representing the oriented
area,\[
Ar(\mathbf{a},\mathbf{b})=\left[A(\mathbf{a},\mathbf{b})_{\mathbf{e}_{1},\mathbf{e}_{2}}^{2}+A(\mathbf{a},\mathbf{b})_{\mathbf{e}_{2},\mathbf{e}_{3}}^{2}+A(\mathbf{a},\mathbf{b})_{\mathbf{e}_{3},\mathbf{e}_{1}}^{2}\right]^{\frac{1}{2}}.\]


However, the vector product cannot be generalized to all higher-dimen\-sion\-al
spaces. Luckily, the vector product does not play an essential role
in the construction of the oriented area. 

Instead of working with the vector product, we will generalize the
idea of projecting the parallelogram onto coordinate planes. Consider
a parallelogram spanned by vectors $\mathbf{a},\mathbf{b}$ in an
$n$-dimen\-sion\-al Euclidean space $V$ with the standard basis
$\left\{ \mathbf{e}_{1},...,\mathbf{e}_{n}\right\} $. While in three-dimen\-sion\-al
space we had just three projections (onto the coordinate planes $xy$,
$xz$, $yz$), in an $n$-dimen\-sion\-al space we have $\frac{1}{2}n(n-1)$
coordinate planes, which can be denoted by $\text{Span}\left\{ \mathbf{e}_{i},\mathbf{e}_{j}\right\} $
(with $1\leq i<j\leq n$). We may construct the $\frac{1}{2}n(n-1)$
projections of the parallelogram onto these coordinate planes. Each
of these projections has an oriented area; that area is a bilinear,
antisymmetric number-valued function of the vectors $\mathbf{a},\mathbf{b}$.
(The proof of the Statement above does not use the fact that the space
is \emph{three}-dimen\-sion\-al!) We may then regard these $\frac{1}{2}n(n-1)$
numbers as the components of a vector representing the oriented area
of the parallelogram. It is clear that all these components are needed
in order to describe the actual geometric \emph{orientation} of the
parallelogram in the $n$-dimen\-sion\-al space.

We arrived at the idea that the oriented area of the parallelogram
spanned by $\mathbf{a},\mathbf{b}$ is an antisymmetric, bilinear
function $A(\mathbf{a},\mathbf{b})$ whose value is a vector with
$\frac{1}{2}n(n-1)$ components, i.e.~a vector \emph{in a new space}
--- the {}``space of oriented areas,'' as it were. This space is
$\frac{1}{2}n(n-1)$-dimen\-sion\-al. We will construct this space
explicitly below; it is the space of bivectors, to be denoted by $\wedge^{2}V$. 

We will see that the unoriented area of the parallelogram is computed
as the \emph{length} of the vector $A(\mathbf{a},\mathbf{b})$, i.e.~as
the square root of the sum of squares of the areas of the projections
of the parallelogram onto the coordinate planes. This is a generalization
of the Pythagoras theorem to areas in higher-dimen\-sion\-al spaces.

The analogy between ordinary vectors and vector-val\-ued areas can
be understood visually as follows. A straight line segment in an $n$-dimen\-sion\-al
space is represented by a vector whose $n$ components (in an orthonormal
basis) are the signed lengths of the $n$ projections of the line
segment onto the coordinate axes. (The components are \emph{signed},
or \emph{oriented}, i.e.~taken with a negative sign if the orientation
of the vector is opposite to the orientation of the axis.) The length
of a straight line segment, i.e.~the length of the vector $\mathbf{v}$,
is then computed as $\sqrt{\left\langle \mathbf{v},\mathbf{v}\right\rangle }$.
The scalar product $\left\langle \mathbf{v},\mathbf{v}\right\rangle $
is equal to the sum of squared lengths of the projections because
we are using an orthonormal basis. A parallelogram in space is represented
by a vector $\psi$ whose ${n \choose 2}$ components are the \emph{oriented}
areas of the ${n \choose 2}$ projections of the parallelogram onto
the coordinate planes. (The vector $\psi$ belongs to the space of
oriented areas, not to the original $n$-dimen\-sion\-al space.)
The numerical value of the area of the parallelogram is then computed
as $\sqrt{\left\langle \psi,\psi\right\rangle }$. The scalar product
$\left\langle \psi,\psi\right\rangle $ in the space of oriented areas
is equal to the sum of squared areas of the projections because the
${n \choose 2}$ unit areas in the coordinate planes are an orthonormal
basis (according to the definition of the scalar product in the space
of oriented areas).

The generalization of the Pythagoras theorem holds not only for areas
but also for higher-dimen\-sion\-al volumes. A general proof of
this theorem will be given in Sec.~\ref{proof-of-pythagoras}, using
the exterior product and several other constructions to be developed
below.


\section{Exterior product\label{sub:Definition-of-the-exterior}}

In the previous section I motivated the introduction of the antisymmetric
product by showing its connection to areas and volumes. In this section
I will give the definition and work out the properties of the exterior
product in a purely algebraic manner, without using any geometric
intuition. This will enable us to work with vectors in arbitrary dimensions,
to obtain many useful results, and eventually also to appreciate more
fully the geometric significance of the exterior product. 

As explained in Sec.~\ref{sub:Area-of-two-dimensional-parallelograms},
it is possible to represent the oriented area of a parallelogram by
a vector in some auxiliary space. The oriented area is much more convenient
to work with because it is a \emph{bilinear} function of the vectors
$\mathbf{a}$ and $\mathbf{b}$ (this is explained in detail in Sec.~\ref{sub:Motivation-for-exterior}).
{}``Product'' is another word for {}``bilinear function.'' We
have also seen that the oriented area is an \emph{antisymmetric} function
of the vectors $\mathbf{a}$ and $\mathbf{b}$.

In three dimensions, an oriented area is represented by the cross
product $\mathbf{a}\times\mathbf{b}$, which is indeed an antisymmetric
and bilinear product. So we expect that the oriented area in higher
dimensions can be represented by some kind of new antisymmetric product
of $\mathbf{a}$ and $\mathbf{b}$; let us denote this product (to
be defined below) by $\mathbf{a}\wedge\mathbf{b}$, pronounced {}``a
wedge b.'' The value of $\mathbf{a}\wedge\mathbf{b}$ will be a vector
in a \emph{new} vector space. We will also construct this new space
explicitly.


\subsection{Definition of exterior product}

Like the tensor product space, the space of exterior products can
be defined solely by its algebraic properties. We can consider the
space of \emph{formal} \emph{expressions} like $\mathbf{a}\wedge\mathbf{b}$,
$3\mathbf{a}\wedge\mathbf{b}+2\mathbf{c}\wedge\mathbf{d}$, etc.,
and \emph{require} the properties of an antisymmetric, bilinear product
to hold.

Here is a more formal definition of the exterior product space: We
will construct an antisymmetric product {}``by hand,'' using the
tensor product space.


\paragraph{Definition 1:}

Given a vector space $V$, we define a new vector space $V\wedge V$
called the \textbf{exterior product}\index{exterior product} (or
antisymmetric tensor product, or alternating product, or \textbf{wedge
product}\index{wedge product}) of two copies of $V$. The space $V\wedge V$
is the subspace in $V\otimes V$ consisting of all \textbf{antisymmetric}
tensors, i.e.~tensors of the form\[
\mathbf{v}_{1}\otimes\mathbf{v}_{2}-\mathbf{v}_{2}\otimes\mathbf{v}_{1},\quad\mathbf{v}_{1,2}\in V,\]
and all linear combinations of such tensors. The exterior product
of two vectors $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ is the expression
shown above; it is obviously an antisymmetric and bilinear function
of $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$.

For example, here is one particular element from $V\wedge V$, which
we write in two different ways using the axioms of the tensor product:\begin{align}
\left(\mathbf{u}+\mathbf{v}\right)\otimes\left(\mathbf{v}+\mathbf{w}\right)-\left(\mathbf{v}+\mathbf{w}\right)\otimes\left(\mathbf{u}+\mathbf{v}\right)=\mathbf{u}\otimes\mathbf{v}-\mathbf{v}\otimes\mathbf{u}\nonumber \\
+\mathbf{u}\otimes\mathbf{w}-\mathbf{w}\otimes\mathbf{u}+\mathbf{v}\otimes\mathbf{w}-\mathbf{w}\otimes\mathbf{v}\in V\wedge V.\label{eq:uvw calc 1}\end{align}



\subparagraph{Remark:}

A tensor $\mathbf{v}_{1}\otimes\mathbf{v}_{2}\in V\otimes V$ is not
equal to the tensor $\mathbf{v}_{2}\otimes\mathbf{v}_{1}$ if $\mathbf{v}_{1}\neq\mathbf{v}_{2}$.
This is so because there is no identity among the axioms of the tensor
product that would allow us to exchange the factors $\mathbf{v}_{1}$
and $\mathbf{v}_{2}$ in the expression $\mathbf{v}_{1}\otimes\mathbf{v}_{2}$.


\paragraph{Exercise 1:}

Prove that the {}``exchange map'' $\hat{T}\left(\mathbf{v}_{1}\otimes\mathbf{v}_{2}\right)\equiv\mathbf{v}_{2}\otimes\mathbf{v}_{1}$
is a canonically defined, linear map of $V\otimes V$ into itself.
Show that $\hat{T}$ has only two eigenvalues which are $\pm1$. Give
examples of eigenvectors with eigenvalues $+1$ and $-1$. Show that
the subspace $V\wedge V\subset V\otimes V$ is the eigenspace of the
exchange operator $\hat{T}$ with eigenvalue $-1$

\emph{Hint:} $\hat{T}\hat{T}=\hat{1}_{V\otimes V}$. Consider tensors
of the form $\mathbf{u}\otimes\mathbf{v}\pm\mathbf{v}\otimes\mathbf{u}$
as candidate eigenvectors of $\hat{T}$.\hfill{}$\blacksquare$

It is quite cumbersome to perform calculations in the tensor product
notation as we did in Eq.~(\ref{eq:uvw calc 1}). So let us write
the exterior product as $\mathbf{u}\wedge\mathbf{v}$ instead of $\mathbf{u}\otimes\mathbf{v}-\mathbf{v}\otimes\mathbf{u}$.
It is then straightforward to see that the {}``wedge'' symbol $\wedge$
indeed works like an anti-commutative multiplication, as we intended.
The rules of computation are summarized in the following statement.


\paragraph{Statement 1:}

One may save time and write $\mathbf{u}\otimes\mathbf{v}-\mathbf{v}\otimes\mathbf{u}\equiv\mathbf{u}\wedge\mathbf{v}\in V\wedge V$,
and the result of any calculation will be correct, as long as one
follows the rules:\begin{align}
\mathbf{u}\wedge\mathbf{v} & =-\mathbf{v}\wedge\mathbf{u},\label{eq:uv antisymm}\\
\left(\lambda\mathbf{u}\right)\wedge\mathbf{v} & =\lambda\left(\mathbf{u}\wedge\mathbf{v}\right),\\
\left(\mathbf{u}+\mathbf{v}\right)\wedge\mathbf{x} & =\mathbf{u}\wedge\mathbf{x}+\mathbf{v}\wedge\mathbf{x}.\label{eq:uv distrib}\end{align}
It follows also that $\mathbf{u}\wedge\left(\lambda\mathbf{v}\right)=\lambda\left(\mathbf{u}\wedge\mathbf{v}\right)$
and that $\mathbf{v}\wedge\mathbf{v}=0$. (These identities hold for
any vectors $\mathbf{u},\mathbf{v}\in V$ and any scalars $\lambda\in\mathbb{K}$.)


\subparagraph{Proof:}

These properties are direct consequences of the axioms of the tensor
product when applied to antisymmetric tensors. For example, the calculation~(\ref{eq:uvw calc 1})
now requires a simple expansion of brackets,\[
\left(\mathbf{u}+\mathbf{v}\right)\wedge\left(\mathbf{v}+\mathbf{w}\right)=\mathbf{u}\wedge\mathbf{v}+\mathbf{u}\wedge\mathbf{w}+\mathbf{v}\wedge\mathbf{w}.\]
Here we removed the term $\mathbf{v}\wedge\mathbf{v}$ which vanishes
due to the antisymmetry of $\wedge$. Details left as exercise.\hfill{}$\blacksquare$

Elements of the space $V\wedge V$, such as $\mathbf{a}\wedge\mathbf{b}+\mathbf{c}\wedge\mathbf{d}$,
are sometimes called \textbf{bivectors}\index{bivector}.%
\footnote{It is important to note that a bivector is not necessarily expressible
as a single-term product of two vectors; see the Exercise at the end
of Sec.~\ref{sub:Properties-of-the-ext-powers}.\index{single-term exterior products}%
} We will also want to define the exterior product of more than two
vectors. To define the exterior product of \emph{three} vectors, we
consider the subspace of $V\otimes V\otimes V$ that consists of antisymmetric
tensors of the form\begin{align}
\mathbf{a}\otimes\mathbf{b}\otimes\mathbf{c}-\mathbf{b}\otimes\mathbf{a}\otimes\mathbf{c}+\mathbf{c}\otimes\mathbf{a}\otimes\mathbf{b}-\mathbf{c}\otimes\mathbf{b}\otimes\mathbf{a}\nonumber \\
+\mathbf{b}\otimes\mathbf{c}\otimes\mathbf{a}-\mathbf{a}\otimes\mathbf{c}\otimes\mathbf{b}\label{eq:antisym 3}\end{align}
and linear combinations of such tensors. These tensors are called
\textbf{totally antisymmetric\index{totally antisymmetric}} because
they can be viewed as (tensor-valued) functions of the vectors $\mathbf{a},\mathbf{b},\mathbf{c}$
that change sign under exchange of any two vectors. The expression
in Eq.~(\ref{eq:antisym 3}) will be denoted for brevity by $\mathbf{a}\wedge\mathbf{b}\wedge\mathbf{c}$,
similarly to the exterior product of two vectors, $\mathbf{a}\otimes\mathbf{b}-\mathbf{b}\otimes\mathbf{a}$,
which is denoted for brevity by $\mathbf{a}\wedge\mathbf{b}$. Here
is a general definition.


\paragraph{Definition 2:}

The \textbf{exterior product\index{exterior product} of $k$ copies}
of $V$ (also called the \textbf{$k$-th exterior power} of $V$)
is denoted by $\wedge^{k}V$ and is defined as the subspace of totally
antisymmetric tensors within $V\otimes...\otimes V$. In the concise
notation, this is the space spanned by expressions of the form\[
\mathbf{v}_{1}\wedge\mathbf{v}_{2}\wedge...\wedge\mathbf{v}_{k},\quad\mathbf{v}_{j}\in V,\]
assuming that the properties of the wedge product (linearity and antisymmetry)
hold as given by Statement~1. For instance, \begin{equation}
\mathbf{u}\wedge\mathbf{v}_{1}\wedge...\wedge\mathbf{v}_{k}=\left(-1\right)^{k}\mathbf{v}_{1}\wedge...\wedge\mathbf{v}_{k}\wedge\mathbf{u}\label{eq:uv pull}\end{equation}
({}``pulling a vector through $k$ other vectors changes sign $k$
times'').\hfill{}$\blacksquare$

The previously defined space of bivectors is in this notation $V\wedge V\equiv\wedge^{2}V$.
A natural extension of this notation is $\wedge^{0}V=\mathbb{K}$
and $\wedge^{1}V=V$. I will also use the following {}``wedge product''
notation,\[
\bigwedge_{k=1}^{n}\mathbf{v}_{k}\equiv\mathbf{v}_{1}\wedge\mathbf{v}_{2}\wedge...\wedge\mathbf{v}_{n}.\]


Tensors from the space $\wedge^{n}V$ are also called $n$-\textbf{vectors}\index{$n$-vectors}
or \textbf{antisymmetric tensors}\index{antisymmetric tensor} of
rank $n$.


\paragraph{Question:}

How to compute expressions containing multiple products such as $\mathbf{a}\wedge\mathbf{b}\wedge\mathbf{c}$?


\subparagraph{Answer:}

Apply the rules shown in Statement~1. For example, one can permute
adjacent vectors and change sign,\[
\mathbf{a}\wedge\mathbf{b}\wedge\mathbf{c}=-\mathbf{b}\wedge\mathbf{a}\wedge\mathbf{c}=\mathbf{b}\wedge\mathbf{c}\wedge\mathbf{a},\]
one can expand brackets,\[
\mathbf{a}\wedge(\mathbf{x}+4\mathbf{y})\wedge\mathbf{b}=\mathbf{a}\wedge\mathbf{x}\wedge\mathbf{b}+4\mathbf{a}\wedge\mathbf{y}\wedge\mathbf{b},\]
and so on. If the vectors $\mathbf{a},\mathbf{b},\mathbf{c}$ are
given as linear combinations of some basis vectors $\left\{ \mathbf{e}_{j}\right\} $,
we can thus reduce $\mathbf{a}\wedge\mathbf{b}\wedge\mathbf{c}$ to
a linear combination of exterior products of basis vectors, such as
$\mathbf{e}_{1}\wedge\mathbf{e}_{2}\wedge\mathbf{e}_{3}$, $\mathbf{e}_{1}\wedge\mathbf{e}_{2}\wedge\mathbf{e}_{4}$,
etc.


\paragraph{Question:}

The notation $\mathbf{a}\wedge\mathbf{b}\wedge\mathbf{c}$ suggests
that the exterior product is associative,\[
\mathbf{a}\wedge\mathbf{b}\wedge\mathbf{c}=\left(\mathbf{a}\wedge\mathbf{b}\right)\wedge\mathbf{c}=\mathbf{a}\wedge(\mathbf{b}\wedge\mathbf{c}).\]
How can we make sense of this?


\subparagraph{Answer:}

If we want to be pedantic, we need to define the exterior product
operation $\wedge$ between a single-term bivector $\mathbf{a}\wedge\mathbf{b}$
and a vector $\mathbf{c}$, such that the result is \emph{by} \emph{definition}
the 3-vector $\mathbf{a}\wedge\mathbf{b}\wedge\mathbf{c}$. We then
define the same operation on linear combinations of single-term bivectors,
\[
\left(\mathbf{a}\wedge\mathbf{b}+\mathbf{x}\wedge\mathbf{y}\right)\wedge\mathbf{c}\equiv\mathbf{a}\wedge\mathbf{b}\wedge\mathbf{c}+\mathbf{x}\wedge\mathbf{y}\wedge\mathbf{c}.\]
Thus we have defined the exterior product between $\wedge^{2}V$ and
$V$, the result being a 3-vector from $\wedge^{3}V$. We then need
to verify that the results do not depend on the choice of the vectors
such as $\mathbf{a},\mathbf{b},\mathbf{x},\mathbf{y}$ in the representation
of a bivector: A different representation can be achieved only by
using the properties of the exterior product (i.e.~the axioms of
the tensor product), e.g.~we may replace $\mathbf{a}\wedge\mathbf{b}$
by $-\mathbf{b}\wedge\left(\mathbf{a}+\lambda\mathbf{b}\right)$.
It is easy to verify that any such replacements will not modify the
resulting 3-vector, e.g. \[
\mathbf{a}\wedge\mathbf{b}\wedge\mathbf{c}=-\mathbf{b}\wedge\left(\mathbf{a}+\lambda\mathbf{b}\right)\wedge\mathbf{c},\]
again due to the properties of the exterior product. This consideration
shows that calculations with exterior products are consistent with
our algebraic intuition. We may indeed compute $\mathbf{a}\wedge\mathbf{b}\wedge\mathbf{c}$
as $\left(\mathbf{a}\wedge\mathbf{b}\right)\wedge\mathbf{c}$ or as
$\mathbf{a}\wedge\left(\mathbf{b}\wedge\mathbf{c}\right)$.


\paragraph{Example~1:}

Suppose we work in $\mathbb{R}^{3}$ and have vectors $\mathbf{a}=\left(0,\frac{1}{2},-\frac{1}{2}\right)$,
$\mathbf{b}=\left(2,-2,0\right)$, $\mathbf{c}=\left(-2,5,-3\right)$.
Let us compute various exterior products. Calculations are easier
if we introduce the basis $\left\{ \mathbf{e}_{1},\mathbf{e}_{2},\mathbf{e}_{3}\right\} $
explicitly:\[
\mathbf{a}=\frac{1}{2}\left(\mathbf{e}_{2}-\mathbf{e}_{3}\right),\quad\mathbf{b}=2(\mathbf{e}_{1}-\mathbf{e}_{2}),\quad\mathbf{c}=-2\mathbf{e}_{1}+5\mathbf{e}_{2}-3\mathbf{e}_{3}.\]
We compute the 2-vector $\mathbf{a}\wedge\mathbf{b}$ by using the
properties of the exterior product, such as $\mathbf{x}\wedge\mathbf{x}=0$
and $\mathbf{x}\wedge\mathbf{y}=-\mathbf{y}\wedge\mathbf{x}$, and
simply expanding the brackets as usual in algebra:\begin{align*}
\mathbf{a}\wedge\mathbf{b} & =\frac{1}{2}\left(\mathbf{e}_{2}-\mathbf{e}_{3}\right)\wedge2\left(\mathbf{e}_{1}-\mathbf{e}_{2}\right)\\
 & =\left(\mathbf{e}_{2}-\mathbf{e}_{3}\right)\wedge\left(\mathbf{e}_{1}-\mathbf{e}_{2}\right)\\
 & =\mathbf{e}_{2}\wedge\mathbf{e}_{1}-\mathbf{e}_{3}\wedge\mathbf{e}_{1}-\mathbf{e}_{2}\wedge\mathbf{e}_{2}+\mathbf{e}_{3}\wedge\mathbf{e}_{2}\\
 & =-\mathbf{e}_{1}\wedge\mathbf{e}_{2}+\mathbf{e}_{1}\wedge\mathbf{e}_{3}-\mathbf{e}_{2}\wedge\mathbf{e}_{3}.\end{align*}
The last expression is the result; note that now there is nothing
more to compute or to simplify. The expressions such as $\mathbf{e}_{1}\wedge\mathbf{e}_{2}$
are the basic expressions out of which the space $\mathbb{R}^{3}\wedge\mathbb{R}^{3}$
is built. Below (Sec.~\ref{sub:Properties-of-the-ext-powers}) we
will show formally that the set of these expressions is a basis in
the space $\mathbb{R}^{3}\wedge\mathbb{R}^{3}$.

Let us also compute the 3-vector $\mathbf{a}\wedge\mathbf{b}\wedge\mathbf{c}$,\begin{align*}
 & \mathbf{a}\wedge\mathbf{b}\wedge\mathbf{c}=\left(\mathbf{a}\wedge\mathbf{b}\right)\wedge\mathbf{c}\\
 & =\left(-\mathbf{e}_{1}\wedge\mathbf{e}_{2}+\mathbf{e}_{1}\wedge\mathbf{e}_{3}-\mathbf{e}_{2}\wedge\mathbf{e}_{3}\right)\wedge(-2\mathbf{e}_{1}+5\mathbf{e}_{2}-3\mathbf{e}_{3}).\end{align*}
When we expand the brackets here, terms such as $\mathbf{e}_{1}\wedge\mathbf{e}_{2}\wedge\mathbf{e}_{1}$
will vanish because \[
\mathbf{e}_{1}\wedge\mathbf{e}_{2}\wedge\mathbf{e}_{1}=-\mathbf{e}_{2}\wedge\mathbf{e}_{1}\wedge\mathbf{e}_{1}=0,\]
so only terms containing all different vectors need to be kept, and
we find\begin{align*}
\mathbf{a}\wedge\mathbf{b}\wedge\mathbf{c} & =3\mathbf{e}_{1}\wedge\mathbf{e}_{2}\wedge\mathbf{e}_{3}+5\mathbf{e}_{1}\wedge\mathbf{e}_{3}\wedge\mathbf{e}_{2}+2\mathbf{e}_{2}\wedge\mathbf{e}_{3}\wedge\mathbf{e}_{1}\\
 & =\left(3-5+2\right)\mathbf{e}_{1}\wedge\mathbf{e}_{2}\wedge\mathbf{e}_{3}=0.\end{align*}
We note that all the terms are proportional to the 3-vector $\mathbf{e}_{1}\wedge\mathbf{e}_{2}\wedge\mathbf{e}_{3}$,
so only the coefficient in front of $\mathbf{e}_{1}\wedge\mathbf{e}_{2}\wedge\mathbf{e}_{3}$
was needed; then, by coincidence, that coefficient turned out to be
zero. So the result is the zero 3-vector.\hfill{}$\blacksquare$


\paragraph{Question:}

Our original goal was to introduce a bilinear, antisymmetric product
of vectors in order to obtain a geometric representation of oriented
areas. Instead, $\mathbf{a}\wedge\mathbf{b}$ was defined algebraically,
through tensor products. It is clear that $\mathbf{a}\wedge\mathbf{b}$
is antisymmetric and bilinear, but why does it represent an oriented
area?


\subparagraph{Answer:}

Indeed, it may not be immediately clear why oriented areas should
be elements of $V\wedge V$. We have seen that the oriented area $A(\mathbf{x},\mathbf{y})$
is an antisymmetric and bilinear function of the two vectors $\mathbf{x}$
and $\mathbf{y}$. Right now we have constructed the space $V\wedge V$
simply as the \emph{space of antisymmetric products}. By constructing
that space merely out of the axioms of the antisymmetric product,
we already covered \emph{every} \emph{possible} bilinear antisymmetric
product. This means that \emph{any} antisymmetric and bilinear function
of the two vectors $\mathbf{x}$ and $\mathbf{y}$ is proportional
to $\mathbf{x}\wedge\mathbf{y}$ or, more generally, is a \emph{linear}
\emph{function} of $\mathbf{x}\wedge\mathbf{y}$ (perhaps with values
in a different space). Therefore, the space of oriented areas (that
is, the space of linear combinations of $A(\mathbf{x},\mathbf{y})$
for various $\mathbf{x}$ and $\mathbf{y}$) is in any case mapped
to a subspace of $V\wedge V$. We have also seen that oriented areas
in $N$ dimensions can be represented through ${N \choose 2}$ projections,
which indicates that they are vectors in some ${N \choose 2}$-dimen\-sion\-al
space. We will see below that the space $V\wedge V$ has exactly this
dimension (Theorem~2 in Sec.~\ref{sub:Properties-of-the-ext-powers}).
Therefore, we can expect that the space of oriented areas coincides
with $V\wedge V$. Below we will be working in a space $V$ with a
scalar product, where the notions of area and volume are well defined.
Then we will see (Sec.~\ref{sub:Volumes-of-k-dimensional}) that
tensors from $V\wedge V$ and the higher exterior powers of $V$ indeed
correspond in a natural way to oriented areas, or more generally to
oriented volumes of a certain dimension.


\paragraph{Remark: Origin of the name {}``exterior.''}

The construction of the exterior product\index{exterior product!origin of the name}
is a modern formulation of the ideas dating back to H. Grassmann (1844).
A 2-vector $\mathbf{a}\wedge\mathbf{b}$ is interpreted geometrically
as the oriented area of the parallelogram spanned by the vectors $\mathbf{a}$
and $\mathbf{b}$. Similarly, a 3-vector $\mathbf{a}\wedge\mathbf{b}\wedge\mathbf{c}$
represents the oriented 3-volume of a parallelepiped spanned by $\left\{ \mathbf{a},\mathbf{b},\mathbf{c}\right\} $.
Due to the antisymmetry of the exterior product, we have $(\mathbf{a}\wedge\mathbf{b})\wedge(\mathbf{a}\wedge\mathbf{c})=0$,
$(\mathbf{a}\wedge\mathbf{b}\wedge\mathbf{c})\wedge(\mathbf{b}\wedge\mathbf{d})=0$,
etc. We can interpret this geometrically by saying that the {}``product''
of two volumes is zero if these volumes have a vector in common. This
motivated Grassmann to call his antisymmetric product {}``exterior.''
In his reasoning, the product of two {}``extensive quantities''
(such as lines, areas, or volumes) is nonzero only when each of the
two quantities is geometrically {}``to the exterior'' (outside)
of the other.


\paragraph{Exercise 2:}

Show that in a \emph{two}-dimensional space $V$, any 3-vector such
as $\mathbf{a}\wedge\mathbf{b}\wedge\mathbf{c}$ can be simplified
to the zero 3-vector. Prove the same for $n$-vectors in $N$-dimensional
spaces when $n>N$.\hfill{}$\blacksquare$

One can also consider the exterior powers of the \emph{dual} space
$V^{*}$. Tensors from $\wedge^{n}V^{*}$ are usually (for historical
reasons) called $n$-\textbf{forms}\index{$n$-forms} (rather than
{}``$n$-covectors'').


\paragraph{Question:}

Where is the star here, really? Is the space $\wedge^{n}\left(V^{*}\right)$
different from $\left(\wedge^{n}V\right)^{*}$?


\subparagraph{Answer:}

Good that you asked. These spaces are canonically isomorphic, but
there is a subtle technical issue worth mentioning. Consider an example:
$\mathbf{a}^{*}\wedge\mathbf{b}^{*}\in\wedge^{2}(V^{*})$ can act
upon $\mathbf{u}\wedge\mathbf{v}\in\wedge^{2}V$ by the standard tensor
product rule, namely $\mathbf{a}^{*}\otimes\mathbf{b}^{*}$ acts on
$\mathbf{u}\otimes\mathbf{v}$ as \[
\left(\mathbf{a}^{*}\otimes\mathbf{b}^{*}\right)\left(\mathbf{u}\otimes\mathbf{v}\right)=\mathbf{a}^{*}(\mathbf{u})\,\mathbf{b}^{*}(\mathbf{v}),\]
so by using the definition of $\mathbf{a}^{*}\wedge\mathbf{b}^{*}$
and $\mathbf{u}\wedge\mathbf{v}$ through the tensor product, we find\begin{align*}
\left(\mathbf{a}^{*}\wedge\mathbf{b}^{*}\right)\left(\mathbf{u}\wedge\mathbf{v}\right) & =\left(\mathbf{a}^{*}\otimes\mathbf{b}^{*}-\mathbf{b}^{*}\otimes\mathbf{a}^{*}\right)\left(\mathbf{u}\otimes\mathbf{v}-\mathbf{v}\otimes\mathbf{u}\right)\\
 & =2\mathbf{a}^{*}(\mathbf{u})\,\mathbf{b}^{*}(\mathbf{v})-2\mathbf{b}^{*}(\mathbf{u})\,\mathbf{a}^{*}(\mathbf{v}).\end{align*}
We got a \textbf{combinatorial} \textbf{factor}\index{combinatorial factor}
2, that is, a factor that arises because we have \emph{two} permutations
of the set $\left(\mathbf{a},\mathbf{b}\right)$. With $\wedge^{n}\left(V^{*}\right)$
and $\left(\wedge^{n}V\right)^{*}$ we get a factor $n!$. It is not
always convenient to have this combinatorial factor. For example,
in a finite number field the number $n!$ might be \emph{equal to
zero} for large enough $n$. In these cases we could \emph{redefine}
the action of $\mathbf{a}^{*}\wedge\mathbf{b}^{*}$ on $\mathbf{u}\wedge\mathbf{v}$
as \[
\left(\mathbf{a}^{*}\wedge\mathbf{b}^{*}\right)\left(\mathbf{u}\wedge\mathbf{v}\right)\equiv\mathbf{a}^{*}(\mathbf{u})\,\mathbf{b}^{*}(\mathbf{v})-\mathbf{b}^{*}(\mathbf{u})\,\mathbf{a}^{*}(\mathbf{v}).\]
 If we are not working in a finite number field, we are able to divide
by any integer, so we may keep combinatorial factors in the denominators
of expressions where such factors appear. For example, if $\left\{ \mathbf{e}_{j}\right\} $
is a basis in $V$ and $\omega=\mathbf{e}_{1}\wedge...\wedge\mathbf{e}_{N}$
is the corresponding basis tensor in the one-dimen\-sion\-al space
$\wedge^{N}V$, the dual basis tensor in $\left(\wedge^{N}V\right)^{*}$
could be defined by \[
\omega^{*}=\frac{1}{N!}\mathbf{e}_{1}^{*}\wedge...\wedge\mathbf{e}_{N}^{*},\quad\text{so that}\:\omega^{*}(\omega)=1.\]
The need for such combinatorial factors is a minor technical inconvenience
that does not arise too often. We may give the following definition
that avoids dividing by combinatorial factors (but now we use permutations;
see Appendix~\ref{sub:Properties-of-permutations}).


\paragraph{Definition 3:}

The action of a $k$-form $\mathbf{f}_{1}^{*}\wedge...\wedge\mathbf{f}_{k}^{*}$
on a $k$-vector $\mathbf{v}_{1}\wedge...\wedge\mathbf{v}_{k}$ is
defined by\[
\sum_{\sigma}(-1)^{\left|\sigma\right|}\mathbf{f}_{1}^{*}(\mathbf{v}_{\sigma(1)})...\mathbf{f}_{k}^{*}(\mathbf{v}_{\sigma(k)}),\]
where the summation is performed over all permutations $\sigma$ of
the ordered set $\left(1,...,k\right)$.


\paragraph{Example~2:}

With $k=3$ we have\begin{align*}
 & (\mathbf{p}^{*}\wedge\mathbf{q}^{*}\wedge\mathbf{r}^{*})(\mathbf{a}\wedge\mathbf{b}\wedge\mathbf{c})\\
 & =\mathbf{p}^{*}(\mathbf{a})\mathbf{q}^{*}(\mathbf{b})\mathbf{r}^{*}(\mathbf{c})-\mathbf{p}^{*}(\mathbf{b})\mathbf{q}^{*}(\mathbf{a})\mathbf{r}^{*}(\mathbf{c})\\
 & +\mathbf{p}^{*}(\mathbf{b})\mathbf{q}^{*}(\mathbf{c})\mathbf{r}^{*}(\mathbf{a})-\mathbf{p}^{*}(\mathbf{c})\mathbf{q}^{*}(\mathbf{b})\mathbf{r}^{*}(\mathbf{a})\\
 & +\mathbf{p}^{*}(\mathbf{c})\mathbf{q}^{*}(\mathbf{a})\mathbf{r}^{*}(\mathbf{b})-\mathbf{p}^{*}(\mathbf{c})\mathbf{q}^{*}(\mathbf{b})\mathbf{r}^{*}(\mathbf{a}).\end{align*}



\paragraph{Exercise 3:}

a) Show that $\mathbf{a}\wedge\mathbf{b}\wedge\omega=\omega\wedge\mathbf{a}\wedge\mathbf{b}$
where $\omega$ is any antisymmetric tensor (e.g.~$\omega=\mathbf{x}\wedge\mathbf{y}\wedge\mathbf{z}$).

b) Show that\[
\omega_{1}\wedge\mathbf{a}\wedge\omega_{2}\wedge\mathbf{b}\wedge\omega_{3}=-\omega_{1}\wedge\mathbf{b}\wedge\omega_{2}\wedge\mathbf{a}\wedge\omega_{3},\]
where $\omega_{1}$, $\omega_{2}$, $\omega_{3}$ are arbitrary antisymmetric
tensors and $\mathbf{a},\mathbf{b}$ are vectors. 

c) Due to antisymmetry,  $\mathbf{a}\wedge\mathbf{a}=0$ for any vector
$\mathbf{a}\in V$. Is it also true that $\omega\wedge\omega=0$ for
any bivector $\omega\in\wedge^{2}V$?


\subsection{{*} Symmetric tensor product}


\paragraph{Question:}

At this point it is still unclear why the antisymmetric definition
is at all useful. Perhaps we could define something else, say the
symmetric product, instead of the exterior product? We could try to
define a product, say $\mathbf{a}\odot\mathbf{b}$, with some other
property, such as\[
\mathbf{a}\odot\mathbf{b}=2\mathbf{b}\odot\mathbf{a}.\]



\subparagraph{Answer:}

This does not work because, for example, we would have\[
\mathbf{b}\odot\mathbf{a}=2\mathbf{a}\odot\mathbf{b}=4\mathbf{b}\odot\mathbf{a},\]
so all the {}``$\odot$'' products would have to vanish.

We can define the \emph{symmetric} tensor product, $\otimes_{S}$,
with the property\[
\mathbf{a}\otimes_{S}\mathbf{b}=\mathbf{b}\otimes_{S}\mathbf{a},\]
but it is impossible to define anything else in a similar fashion.%
\footnote{This is a theorem due to Grassmann (1862).%
} 

The antisymmetric tensor product is the eigenspace (within $V\otimes V$)
of the exchange operator $\hat{T}$ with eigenvalue $-1$. That operator
has only eigenvectors with eigenvalues $\pm1$, so the only other
possibility is to consider the eigenspace with eigenvalue $+1$. This
eigenspace is spanned by symmetric tensors of the form $\mathbf{u}\otimes\mathbf{v}+\mathbf{v}\otimes\mathbf{u}$,
and can be considered as the space of symmetric tensor products. We
could write\[
\mathbf{a}\otimes_{S}\mathbf{b}\equiv\mathbf{a}\otimes\mathbf{b}+\mathbf{b}\otimes\mathbf{a}\]
and develop the properties of this product. However, it turns out
that the symmetric tensor product is much less useful for the purposes
of linear algebra than the antisymmetric subspace. This book derives
most of the results of linear algebra using the antisymmetric product
as the main tool!


\section{Properties of spaces $\wedge^{k}V$\label{sec:Properties-of-the-wedgekV}}

As we have seen, tensors from the space $V\otimes V$ are representable
by linear combinations of the form $\mathbf{a}\otimes\mathbf{b}+\mathbf{c}\otimes\mathbf{d}+...$,
but not \emph{uniquely} representable because one can transform one
such linear combination into another by using the axioms of the tensor
product. Similarly, $n$-vectors are not uniquely representable by
linear combinations of exterior products. For example,\[
\mathbf{a}\wedge\mathbf{b}+\mathbf{a}\wedge\mathbf{c}+\mathbf{b}\wedge\mathbf{c}=(\mathbf{a}+\mathbf{b})\wedge(\mathbf{b}+\mathbf{c})\]
 since $\mathbf{b}\wedge\mathbf{b}=0$. In other words, the 2-vector
$\omega\equiv\mathbf{a}\wedge\mathbf{b}+\mathbf{a}\wedge\mathbf{c}+\mathbf{b}\wedge\mathbf{c}$
has an alternative representation containing only a single-term exterior
product, $\omega=\mathbf{r}\wedge\mathbf{s}$ where $\mathbf{r}=\mathbf{a}+\mathbf{b}$
and $\mathbf{s}=\mathbf{b}+\mathbf{c}$.


\paragraph{Exercise:\index{single-term exterior products}}

Show that any 2-vector in a \emph{three}-dimen\-sion\-al space is
representable by a single-term exterior product, i.e.~to a 2-vector
of the form $\mathbf{a}\wedge\mathbf{b}$.

\emph{Hint}: Choose a basis $\left\{ \mathbf{e}_{1},\mathbf{e}_{2},\mathbf{e}_{3}\right\} $
and show that $\alpha\mathbf{e}_{1}\wedge\mathbf{e}_{2}+\beta\mathbf{e}_{1}\wedge\mathbf{e}_{3}+\gamma\mathbf{e}_{2}\wedge\mathbf{e}_{3}$
is equal to a single-term product.\hfill{}$\blacksquare$

What about higher-dimen\-sion\-al spaces? We will show (see the
Exercise at the end of Sec.~\ref{sub:Properties-of-the-ext-powers})
that $n$-vectors cannot be in general reduced to a single-term product.
This is, however, always possible for $(N-1)$-vectors in an $N$-dimen\-sion\-al
space. (You showed this for $N=3$ in the exercise above.)


\paragraph{Statement:}

Any $(N-1)$-vector in an $N$-dimen\-sion\-al space can be written
as a single-term exterior product of the form $\mathbf{a}_{1}\wedge...\wedge\mathbf{a}_{N-1}$.


\subparagraph{Proof:}

We prove this by using induction in $N$. The basis of induction is
$N=2$, where there is nothing to prove. The induction step: Suppose
that the statement is proved for $(N-1)$-vectors in $N$-dimen\-sion\-al
spaces, we need to prove it for $N$-vectors in $(N+1)$-dimen\-sion\-al
spaces. Choose a basis $\left\{ \mathbf{e}_{1},...,\mathbf{e}_{N+1}\right\} $
in the space. Any $N$-vector $\omega$ can be written as a linear
combination of exterior product terms,\begin{align*}
\omega & =\alpha_{1}\mathbf{e}_{2}\wedge...\wedge\mathbf{e}_{N+1}+\alpha_{2}\mathbf{e}_{1}\wedge\mathbf{e}_{3}\wedge...\wedge\mathbf{e}_{N+1}+...\\
 & \quad+\alpha_{N}\mathbf{e}_{1}\wedge...\wedge\mathbf{e}_{N-1}\wedge\mathbf{e}_{N+1}+\alpha_{N+1}\mathbf{e}_{1}\wedge...\wedge\mathbf{e}_{N},\end{align*}
where $\left\{ \alpha_{i}\right\} $ are some constants. 

Note that any tensor $\omega\in\wedge^{N-1}V$ can be written in this
way simply by expressing every vector through the basis and by expanding
the exterior products. The result will be a linear combination of
the form shown above, containing at most $N+1$ single-term exterior
products of the form $\mathbf{e}_{1}\wedge...\wedge\mathbf{e}_{N}$,
$\mathbf{e}_{2}\wedge...\wedge\mathbf{e}_{N+1}$, and so on. We do
not yet know whether these single-term exterior products constitute
a linearly independent set; this will be established in Sec.~\ref{sub:Properties-of-the-ext-powers}.
Presently, we will not need this property.

Now we would like to transform the expression above to a single term.
We move $\mathbf{e}_{N+1}$ outside brackets in the first $N$ terms:\begin{align*}
\omega & =\big(\alpha_{1}\mathbf{e}_{2}\wedge...\wedge\mathbf{e}_{N}+...+\alpha_{N}\mathbf{e}_{1}\wedge...\wedge\mathbf{e}_{N-1}\big)\wedge\mathbf{e}_{N+1}\\
 & \qquad+\alpha_{N+1}\mathbf{e}_{1}\wedge...\wedge\mathbf{e}_{N}\\
 & \equiv\psi\wedge\mathbf{e}_{N+1}+\alpha_{N+1}\mathbf{e}_{1}\wedge...\wedge\mathbf{e}_{N},\end{align*}
where in the last line we have introduced an auxiliary $(N-1)$-vector
$\psi$. If it happens that $\psi=0$, there is nothing left to prove.
Otherwise, at least one of the $\alpha_{i}$ must be nonzero; without
loss of generality, suppose that $\alpha_{N}\neq0$ and rewrite $\omega$
as \[
\omega=\psi\wedge\mathbf{e}_{N+1}+\alpha_{N+1}\mathbf{e}_{1}\wedge...\wedge\mathbf{e}_{N}=\psi\wedge\big(\mathbf{e}_{N+1}+\frac{\alpha_{N+1}}{\alpha_{N}}\mathbf{e}_{N}\big).\]
Now we note that $\psi$ belongs to the space of $\left(N-1\right)$-vectors
over the $N$-dimen\-sion\-al subspace spanned by $\left\{ \mathbf{e}_{1},...,\mathbf{e}_{N}\right\} $.
By the inductive assumption, $\psi$ can be written as a single-term
exterior product, $\psi=\mathbf{a}_{1}\wedge...\wedge\mathbf{a}_{N-1}$,
of some vectors $\left\{ \mathbf{a}_{i}\right\} $. Denoting \[
\mathbf{a}_{N}\equiv\mathbf{e}_{N+1}+\frac{\alpha_{N+1}}{\alpha_{N}}\mathbf{e}_{N},\]
we obtain \[
\omega=\mathbf{a}_{1}\wedge...\wedge\mathbf{a}_{N-1}\wedge\mathbf{a}_{N},\]
i.e. $\omega$ can be represented as a single-term exterior product.\hfill{}$\blacksquare$ 


\subsection{Linear maps between spaces $\wedge^{k}V$\label{sub:Linear-maps-between-spaces}}

Since the spaces $\wedge^{k}V$ are vector spaces, we may consider
linear maps between them. 

A simplest example is a map\[
L_{\mathbf{a}}:\omega\mapsto\mathbf{a}\wedge\omega,\]
mapping $\wedge^{k}V\rightarrow\wedge^{k+1}V$; here the vector $\mathbf{a}$
is \emph{fixed}. It is important to check that $L_{\mathbf{a}}$ is
a \emph{linear} map between these spaces. How do we check this? We
need to check that $L_{\mathbf{a}}$ maps a linear combination of
tensors into linear combinations; this is easy to see,\begin{align*}
L_{\mathbf{a}} & (\omega+\lambda\omega^{\prime})=\mathbf{a}\wedge(\omega+\lambda\omega')\\
 & =\mathbf{a}\wedge\omega+\lambda\mathbf{a}\wedge\omega'=L_{\mathbf{a}}\omega+\lambda L_{\mathbf{a}}\omega'.\end{align*}


Let us now fix a covector $\mathbf{a}^{*}$. A covector is a map $V\rightarrow\mathbb{K}$.
In Lemma~2 of Sec.~\ref{sub:Dimension-of-tensor} we have used covectors
to define linear maps $\mathbf{a}^{*}:V\otimes W\rightarrow W$ according
to Eq.~(\ref{eq:fg rule}), mapping $\mathbf{v}\otimes\mathbf{w}\mapsto\mathbf{a}^{*}\left(\mathbf{v}\right)\mathbf{w}$.
Now we will apply the analogous construction to exterior powers and
construct a map $V\wedge V\rightarrow V$. Let us denote this map
by $\iota_{\mathbf{a}^{*}}$. 

It would be incorrect to define the map $\iota_{\mathbf{a}^{*}}$
by the formula $\iota_{\mathbf{a}^{*}}(\mathbf{v}\wedge\mathbf{w})=\mathbf{a}^{*}\left(\mathbf{v}\right)\mathbf{w}$
because such a definition does not respect the antisymmetry of the
wedge product and thus violates the linearity condition, \[
\iota_{\mathbf{a}^{*}}\left(\mathbf{w}\wedge\mathbf{v}\right)\,{\lyxbuildrel!\above=}\,\iota_{\mathbf{a}^{*}}\left(\left(-1\right)\mathbf{v}\wedge\mathbf{w}\right)=-\iota_{\mathbf{a}^{*}}\left(\mathbf{v}\wedge\mathbf{w}\right)\neq\mathbf{a}^{*}(\mathbf{v})\mathbf{w}.\]
So we need to act with $\mathbf{a}^{*}$ on \emph{each} of the vectors
in a wedge product and make sure that the correct minus sign comes
out. An acceptable formula for the map $\iota_{\mathbf{a}^{*}}:\wedge^{2}V\rightarrow V$
is\[
\iota_{\mathbf{a}^{*}}\left(\mathbf{v}\wedge\mathbf{w}\right)\equiv\mathbf{a}^{*}\left(\mathbf{v}\right)\mathbf{w}-\mathbf{a}^{*}\left(\mathbf{w}\right)\mathbf{v}.\]
(Please check that the linearity condition now holds!) This is how
we will define the map $\iota_{\mathbf{a}^{*}}$ on $\wedge^{2}V$.

Let us now extend $\iota_{\mathbf{a}^{*}}:\wedge^{2}V\rightarrow V$
to a map \[
\iota_{\mathbf{a}^{*}}:\wedge^{k}V\rightarrow\wedge^{k-1}V,\]
defined as follows: \begin{align}
\iota_{\mathbf{a}^{*}}\mathbf{v} & \equiv\mathbf{a}^{*}(\mathbf{v}),\nonumber \\
\iota_{\mathbf{a}^{*}}(\mathbf{v}\wedge\omega) & \equiv\mathbf{a}^{*}(\mathbf{v})\omega-\mathbf{v}\wedge(\iota_{\mathbf{a}^{*}}\omega).\label{eq:inductive}\end{align}
This definition is \emph{inductive}, i.e.~it shows how to define
$\iota_{\mathbf{a}^{*}}$ on $\wedge^{k}V$ if we know how to define
it on $\wedge^{k-1}V$. The action of $\iota_{\mathbf{a}^{*}}$ on
a sum of terms is defined by requiring  linearity, \[
\iota_{\mathbf{a}^{*}}\left(A+\lambda B\right)\equiv\iota_{\mathbf{a}^{*}}\left(A\right)+\lambda\iota_{\mathbf{a}^{*}}\left(B\right),\quad A,B\in\wedge^{k}V.\]


We can convert this inductive definition into a more explicit formula:
if $\omega=\mathbf{v}_{1}\wedge...\wedge\mathbf{v}_{k}\in\wedge^{k}V$
then \begin{align*}
\iota_{\mathbf{a}^{*}} & (\mathbf{v}_{1}\wedge...\wedge\mathbf{v}_{k})\equiv\mathbf{a}^{*}(\mathbf{v}_{1})\mathbf{v}_{2}\wedge...\wedge\mathbf{v}_{k}-\mathbf{a}^{*}(\mathbf{v}_{2})\mathbf{v}_{1}\wedge\mathbf{v}_{3}\wedge...\wedge\mathbf{v}_{k}\\
 & +...+\left(-1\right)^{k-1}\mathbf{a}^{*}(\mathbf{v}_{k})\mathbf{v}_{1}\wedge...\wedge\mathbf{v}_{k-1}.\end{align*}


This map is called the \textbf{interior product}\index{interior product}
or the \textbf{insertion} map\index{insertion map}. This is a useful
operation in  linear algebra. The insertion map $\iota_{\mathbf{a}^{*}}\psi$
{}``inserts'' the covector $\mathbf{a}^{*}$ into the tensor $\psi\in\wedge^{k}V$
by acting with $\mathbf{a}^{*}$ on each of the vectors in the exterior
product that makes up $\psi$.

Let us check formally that the insertion map is linear. 


\paragraph{Statement:}

The map $\iota_{\mathbf{a}^{*}}:\wedge^{k}V\rightarrow\wedge^{k-1}V$
for $1\leq k\leq N$ is a well-defined linear map, according to the
inductive definition.


\subparagraph{Proof:}

First, we need to check that it maps linear combinations into linear
combinations; this is quite easy to see by induction, using the fact
that $\mathbf{a}^{*}:V\rightarrow\mathbb{K}$ is linear. However,
this type of linearity is not sufficient; we also need to check that
the \emph{result} of the map, i.e.~the tensor $\iota_{\mathbf{a}^{*}}(\omega)$,
is defined \emph{independently} \emph{of} \emph{the} \emph{representation}
of $\omega$ through vectors such as $\mathbf{v}_{i}$. The problem
is, there are many such representations, for example some tensor $\omega\in\wedge^{3}V$
might be written using different vectors as \[
\omega=\mathbf{v}_{1}\wedge\mathbf{v}_{2}\wedge\mathbf{v}_{3}=\mathbf{v}_{2}\wedge(\mathbf{v}_{3}-\mathbf{v}_{1})\wedge(\mathbf{v}_{3}+\mathbf{v}_{2})\equiv\tilde{\mathbf{v}}_{1}\wedge\tilde{\mathbf{v}}_{2}\wedge\tilde{\mathbf{v}}_{3}.\]
 We need to verify that any such equivalent representation yields
the same resulting tensor $\iota_{\mathbf{a}^{*}}(\omega)$, despite
the fact that the definition of $\iota_{\mathbf{a}^{*}}$ \emph{appears}
to depend on the choice of the vectors $\mathbf{v}_{i}$. Only then
will it be proved that $\iota_{\mathbf{a}^{*}}$ is a linear map $\wedge^{k}V\rightarrow\wedge^{k-1}V$.

An equivalent representation of a tensor $\omega$ can be obtained
only by using the properties of the exterior product, namely linearity
and antisymmetry. Therefore, we need to verify that $\iota_{\mathbf{a}^{*}}(\omega)$
does not change when we change the representation of $\omega$ in
these two ways: 1) expanding a linear combination,\begin{equation}
(\mathbf{x}+\lambda\mathbf{y})\wedge...\mapsto\mathbf{x}\wedge...+\lambda\mathbf{y}\wedge...;\label{eq:change repr 1}\end{equation}
2) interchanging the order of two vectors in the exterior product
and change the sign,\begin{equation}
\mathbf{x}\wedge\mathbf{y}\wedge...\mapsto-\mathbf{y}\wedge\mathbf{x}\wedge...\label{eq:change repr 2}\end{equation}
It is clear that $\mathbf{a}^{*}(\mathbf{x}+\lambda\mathbf{y})=\mathbf{a}^{*}(\mathbf{x})+\lambda\mathbf{a}^{*}(\mathbf{y})$;
it follows by induction that $\iota_{\mathbf{a}^{*}}\omega$ does
not change under a change of representation of the type~(\ref{eq:change repr 1}).
Now we consider the change of representation of the type~(\ref{eq:change repr 2}).
We have, by definition of $\iota_{\mathbf{a}^{*}}$,\[
\iota_{\mathbf{a}^{*}}(\mathbf{v}_{1}\wedge\mathbf{v}_{2}\wedge\chi)=\mathbf{a}^{*}(\mathbf{v}_{1})\mathbf{v}_{2}\wedge\chi-\mathbf{a}^{*}(\mathbf{v}_{2})\mathbf{v}_{1}\wedge\chi+\mathbf{v}_{1}\wedge\mathbf{v}_{2}\wedge\iota_{\mathbf{a}^{*}}(\chi),\]
where we have denoted by $\chi$ the rest of the exterior product.
It is clear from the above expression that \[
\iota_{\mathbf{a}^{*}}(\mathbf{v}_{1}\wedge\mathbf{v}_{2}\wedge\chi)=-\iota_{\mathbf{a}^{*}}(\mathbf{v}_{2}\wedge\mathbf{v}_{1}\wedge\chi)=\iota_{\mathbf{a}^{*}}(-\mathbf{v}_{2}\wedge\mathbf{v}_{1}\wedge\chi).\]
This proves that $\iota_{\mathbf{a}^{*}}(\omega)$ does not change
under a change of representation of $\omega$ of the type~(\ref{eq:change repr 2}).
This concludes the proof.\hfill{}$\blacksquare$


\paragraph{Remark:}

It is apparent from the proof that the \emph{minus sign} in the inductive
definition~(\ref{eq:inductive}) is crucial for the linearity of
the map $\iota_{\mathbf{a}^{*}}$. Indeed, if we attempt to define
a map by a formula such as\[
\mathbf{v}_{1}\wedge\mathbf{v}_{2}\mapsto\mathbf{a}^{*}(\mathbf{v}_{1})\mathbf{v}_{2}+\mathbf{a}^{*}(\mathbf{v}_{2})\mathbf{v}_{1},\]
the result will \emph{not} be a linear map $\wedge^{2}V\rightarrow V$
despite the appearance of linearity. The correct formula must take
into account the fact that $\mathbf{v}_{1}\wedge\mathbf{v}_{2}=-\mathbf{v}_{2}\wedge\mathbf{v}_{1}$.


\paragraph{Exercise:}

Show by induction in $k$ that\[
L_{\mathbf{x}}\iota_{\mathbf{a}^{*}}\omega+\iota_{\mathbf{a}^{*}}L_{\mathbf{x}}\omega=\mathbf{a}^{*}(\mathbf{x})\omega,\quad\forall\omega\in\wedge^{k}V.\]
In other words, the linear operator $L_{\mathbf{x}}\iota_{\mathbf{a}^{*}}+\iota_{\mathbf{a}^{*}}L_{\mathbf{x}}:\wedge^{k}V\rightarrow\wedge^{k}V$
is simply the multiplication by the number $\mathbf{a}^{*}(\mathbf{x})$.


\paragraph{}


\subsection{Exterior product and linear dependence\label{sub:Properties-of-the-ext-powers}}

The exterior product is useful in many ways. One powerful property
of the exterior product is its close relation to linear independence
of sets of vectors. For example, if $\mathbf{u}=\lambda\mathbf{v}$
then $\mathbf{u}\wedge\mathbf{v}=0$. More generally:


\paragraph{Theorem 1:}

A set $\left\{ \mathbf{v}_{1},...,\mathbf{v}_{k}\right\} $ of vectors
from $V$ is linearly independent if and only if $(\mathbf{v}_{1}\wedge\mathbf{v}_{2}\wedge...\wedge\mathbf{v}_{k})\neq0$,
i.e.~it is a nonzero tensor from $\wedge^{k}V$.


\subparagraph{Proof:}

If $\left\{ \mathbf{v}_{j}\right\} $ is linearly dependent then without
loss of generality we may assume that $\mathbf{v}_{1}$ is a linear
combination of other vectors, $\mathbf{v}_{1}=\sum_{j=2}^{k}\lambda_{j}\mathbf{v}_{j}$.
Then \begin{align*}
\mathbf{v}_{1}\wedge\mathbf{v}_{2}\wedge...\wedge\mathbf{v}_{k} & =\sum_{j=2}^{k}\lambda_{j}\mathbf{v}_{j}\wedge\mathbf{v}_{2}\wedge...\wedge\mathbf{v}_{j}\wedge...\wedge\mathbf{v}_{k}\\
 & =\sum_{j=2}^{k}\left(-1\right)^{j-1}\mathbf{v}_{2}\wedge...\mathbf{v}_{j}\wedge\mathbf{v}_{j}\wedge...\wedge\mathbf{v}_{k}=0.\end{align*}
Conversely, we need to prove that the tensor $\mathbf{v}_{1}\wedge...\wedge\mathbf{v}_{k}\neq0$
if $\left\{ \mathbf{v}_{j}\right\} $ is linearly \emph{in}dependent.
The proof is by induction in $k$. The basis of induction is $k=1$:
if $\left\{ \mathbf{v}_{1}\right\} $ is linearly independent then
clearly $\mathbf{v}_{1}\neq0$. The induction step: Assume that the
statement is proved for $k-1$ and that $\left\{ \mathbf{v}_{1},...,\mathbf{v}_{k}\right\} $
is a linearly independent set. By Exercise~1 in Sec.~\ref{sub:Dual-vector-space}
there exists a covector $\mathbf{f}^{*}\in V^{*}$ such that $\mathbf{f}^{*}\left(\mathbf{v}_{1}\right)=1$
and $\mathbf{f}^{*}\left(\mathbf{v}_{i}\right)=0$ for $2\leq i\leq k$.
Now we apply the interior product map $\iota_{\mathbf{f}^{*}}:\wedge^{k}V\rightarrow\wedge^{k-1}V$
constructed in Sec.~\ref{sub:Linear-maps-between-spaces} to the
tensor $\mathbf{v}_{1}\wedge...\wedge\mathbf{v}_{k}$ and find \[
\iota_{\mathbf{f}^{*}}\left(\mathbf{v}_{1}\wedge...\wedge\mathbf{v}_{k}\right)=\mathbf{v}_{2}\wedge...\wedge\mathbf{v}_{k}.\]
By the induction step, the linear independence of $k-1$ vectors $\left\{ \mathbf{v}_{2},...,\mathbf{v}_{k}\right\} $
entails $\mathbf{v}_{2}\wedge...\wedge\mathbf{v}_{k}\neq0$. The map
$\iota_{\mathbf{f}^{*}}$ is linear and cannot map a zero tensor into
a nonzero tensor, therefore $\mathbf{v}_{1}\wedge...\wedge\mathbf{v}_{k}\neq0$.\hfill{}$\blacksquare$

It is also important to know that any tensor from the highest exterior
power $\wedge^{N}V$ can be represented as just a \emph{single-term}\index{single-term exterior products}
exterior product of $N$ vectors. (Note that the same property for
$\wedge^{N-1}V$ was already established in Sec.~\ref{sec:Properties-of-the-wedgekV}.)


\paragraph{Lemma~1:}

For any tensor $\omega\in\wedge^{N}V$ there exist vectors $\left\{ \mathbf{v}_{1},...,\mathbf{v}_{N}\right\} $
such that $\omega=\mathbf{v}_{1}\wedge...\wedge\mathbf{v}_{N}$.


\subparagraph{Proof:}

If $\omega=0$ then there is nothing to prove, so we assume $\omega\neq0$.
By definition, the tensor $\omega$ has a representation as a sum
of \emph{several} exterior products, say \[
\omega=\mathbf{v}_{1}\wedge...\wedge\mathbf{v}_{N}+\mathbf{v}_{1}^{\prime}\wedge...\wedge\mathbf{v}_{N}^{\prime}+...\]
Let us simplify this expression to just one exterior product. First,
let us omit any zero terms in this expression (for instance, $\mathbf{a}\wedge\mathbf{a}\wedge\mathbf{b}\wedge...=0$).
Then by Theorem~1 the set $\left\{ \mathbf{v}_{1},...,\mathbf{v}_{N}\right\} $
is linearly independent (or else the term $\mathbf{v}_{1}\wedge...\wedge\mathbf{v}_{N}$
would be zero). Hence, $\left\{ \mathbf{v}_{1},...,\mathbf{v}_{N}\right\} $
is a basis in $V$. All other vectors such as $\mathbf{v}_{i}^{\prime}$
can be decomposed as linear combinations of vectors in that basis.
Let us denote $\psi\equiv\mathbf{v}_{1}\wedge...\wedge\mathbf{v}_{N}$.
By expanding the brackets in exterior products such as $\mathbf{v}_{1}^{\prime}\wedge...\wedge\mathbf{v}_{N}^{\prime}$,
we will obtain every time the tensor $\psi$ with different coefficients.
Therefore, the final result of simplification will be that $\omega$
equals  $\psi$ multiplied with some coefficient. This is sufficient
to prove Lemma~1.\hfill{}$\blacksquare$

Now we would like to build a basis in the space $\wedge^{m}V$. For
this we need to determine which sets of tensors from $\wedge^{m}V$
are linearly independent within that space.


\paragraph{Lemma 2:}

If $\left\{ \mathbf{e}_{1},...,\mathbf{e}_{N}\right\} $ is a basis
in $V$ then any tensor $A\in\wedge^{m}V$ can be decomposed as a
linear combination of the tensors $\mathbf{e}_{k_{1}}\wedge\mathbf{e}_{k_{2}}\wedge...\wedge\mathbf{e}_{k_{m}}$
with some indices $k_{j}$, $1\leq j\leq m$.


\subparagraph{Proof:}

The tensor $A$ is a linear combination of expressions of the form
$\mathbf{v}_{1}\wedge...\wedge\mathbf{v}_{m}$, and each vector $\mathbf{v}_{i}\in V$
can be decomposed in the basis $\left\{ \mathbf{e}_{j}\right\} $.
Expanding the brackets around the wedges using the rules~(\ref{eq:uv antisymm})--(\ref{eq:uv distrib}),
we obtain a decomposition of an arbitrary tensor through the basis
tensors. For example, \begin{align*}
\left(\mathbf{e}_{1}+2\mathbf{e}_{2}\right)\wedge\left(\mathbf{e}_{1}-\mathbf{e}_{2}+\mathbf{e}_{3}\right)-2\left(\mathbf{e}_{2}-\mathbf{e}_{3}\right)\wedge\left(\mathbf{e}_{1}-\mathbf{e}_{3}\right)\\
=-\mathbf{e}_{1}\wedge\mathbf{e}_{2}-\mathbf{e}_{1}\wedge\mathbf{e}_{3}+4\mathbf{e}_{2}\wedge\mathbf{e}_{3}\end{align*}
(please verify this yourself!).\hfill{}$\blacksquare$

By Theorem~1, all tensors $\mathbf{e}_{k_{1}}\wedge\mathbf{e}_{k_{2}}\wedge...\wedge\mathbf{e}_{k_{m}}$
constructed out of subsets of vectors from the basis $\left\{ \mathbf{e}_{1},...,\mathbf{e}_{k}\right\} $
are nonzero, and by Lemma~2 any tensor can be decomposed into a linear
combination of these tensors. But are these tensors a basis in the
space $\wedge^{m}V$? Yes:


\paragraph{Lemma 3:}

If $\left\{ \mathbf{v}_{1},...,\mathbf{v}_{n}\right\} $ is a linearly
independent set of vectors (not necessarily a basis in $V$ since
$n\leq N$), then:

\textbf{(1)} The set of ${n \choose 2}$ tensors\[
\left\{ \mathbf{v}_{j}\wedge\mathbf{v}_{k},\:1\leq j<k\leq n\right\} \equiv\left\{ \mathbf{v}_{1}\wedge\mathbf{v}_{2},\mathbf{v}_{1}\wedge\mathbf{v}_{3},...,\mathbf{v}_{n-1}\wedge\mathbf{v}_{n}\right\} \]
is linearly independent in the space $\wedge^{2}V$. 

\textbf{(2)} The set of ${n \choose m}$ tensors\[
\left\{ \mathbf{v}_{k_{1}}\wedge\mathbf{v}_{k_{2}}\wedge...\wedge\mathbf{v}_{k_{m}},\:1\leq k_{1}<k_{2}<...<k_{m}\leq n\right\} \]
 is linearly independent in the space $\wedge^{m}V$ for $2\leq m\leq n$.


\subparagraph{Proof:}

\textbf{(1)} The proof is similar to that of Lemma~3 in Sec.~\ref{sub:Dimension-of-tensor}.
Suppose the set $\left\{ \mathbf{v}_{j}\right\} $ is linearly independent
but the set $\left\{ \mathbf{v}_{j}\wedge\mathbf{v}_{k}\right\} $
is linearly \emph{dependent}, so that there exists a linear combination\[
\sum_{1\leq j<k\leq n}\lambda_{jk}\mathbf{v}_{j}\wedge\mathbf{v}_{k}=0\]
with at least some $\lambda_{jk}\neq0$. Without loss of generality,
$\lambda_{12}\neq0$ (or else we can renumber the vectors $\mathbf{v}_{j}$).
There exists a covector $\mathbf{f}^{*}\in V^{*}$ such that $\mathbf{f}^{*}\left(\mathbf{v}_{1}\right)=1$
and $\mathbf{f}^{*}\left(\mathbf{v}_{i}\right)=0$ for $2\leq i\leq n$.
Apply the interior product with this covector to the above tensor,\[
0=\iota_{\mathbf{f}^{*}}\left[\sum_{1\leq j<k\leq n}\lambda_{jk}\mathbf{v}_{j}\wedge\mathbf{v}_{k}\right]=\sum_{k=2}^{n}\lambda_{1k}\mathbf{v}_{k},\]
therefore by linear independence of $\left\{ \mathbf{v}_{k}\right\} $
all $\lambda_{1k}=0$, contradicting the assumption $\lambda_{12}\neq0$.

\textbf{(2)} The proof of part (1) is straightforwardly generalized
to the space $\wedge^{m}V$, using induction in $m$. We have just
proved the basis of induction, $m=2$. Now the induction step: assume
that the statement is proved for $m-1$ and consider a set $\left\{ \mathbf{v}_{k_{1}}\wedge...\wedge\mathbf{v}_{k_{m}}\right\} $,
of tensors of rank $m$, where $\left\{ \mathbf{v}_{j}\right\} $
is a basis. Suppose that this set is linearly dependent; then there
is a linear combination \[
\omega\equiv\sum_{k_{1},...,k_{m}}\lambda_{k_{1}...k_{m}}\mathbf{v}_{k_{1}}\wedge...\wedge\mathbf{v}_{k_{m}}=0\]
with some nonzero coefficients, e.g.~$\lambda_{12...m}\neq0$. There
exists a covector $\mathbf{f}^{*}$ such that $\mathbf{f}^{*}\left(\mathbf{v}_{1}\right)=1$
and $\mathbf{f}^{*}\left(\mathbf{v}_{i}\right)=0$ for $2\leq i\leq n$.
Apply this covector to the tensor $\omega$ and obtain $\iota_{\mathbf{f}^{*}}\omega=0$,
which yields a vanishing linear combination of tensors $\mathbf{v}_{k_{1}}\wedge...\wedge\mathbf{v}_{k_{m-1}}$
of rank $m-1$ with \emph{some} nonzero coefficients. But this contradicts
the induction assumption, which says that any set of tensors $\mathbf{v}_{k_{1}}\wedge...\wedge\mathbf{v}_{k_{m-1}}$
of rank $m-1$ is linearly independent.\hfill{}$\blacksquare$

Now we are ready to compute the dimension of $\wedge^{m}V$.


\paragraph{Theorem 2:}

The dimension of the space $\wedge^{m}V$ is \[
\dim\wedge^{m}V={N \choose m}=\frac{N!}{m!\left(N-m\right)!},\]
 where $N\equiv\dim V$. For $m>N$ we have $\dim\wedge^{m}V=0$,
i.e.~the spaces $\wedge^{m}V$ for $m>N$ consist solely of the zero
tensor. 


\subparagraph{Proof:}

We will explicitly construct a basis in the space $\wedge^{m}V$.
First choose a basis $\left\{ \mathbf{e}_{1},...,\mathbf{e}_{N}\right\} $
in $V$. By Lemma~3, the set of ${N \choose m}$ tensors\[
\left\{ \mathbf{e}_{k_{1}}\wedge\mathbf{e}_{k_{2}}\wedge...\wedge\mathbf{e}_{k_{m}},\:1\leq k_{1}<k_{2}<...<k_{m}\leq N\right\} \]
 is linearly independent, and by Lemma~2 any tensor $A\in\wedge^{m}V$
is a linear combination of these tensors. Therefore the set $\left\{ \mathbf{e}_{k_{1}}\wedge\mathbf{e}_{k_{2}}\wedge...\wedge\mathbf{e}_{k_{m}}\right\} $
is a basis in $\wedge^{m}V$. By Theorem~\ref{sub:All-bases-have},
the dimension of space is equal to the number of vectors in any basis,
therefore $\dim\wedge^{m}N={N \choose m}$.

For $m>N$, the existence of a nonzero tensor $\mathbf{v}_{1}\wedge...\wedge\mathbf{v}_{m}$
contradicts Theorem~1: The set $\left\{ \mathbf{v}_{1},...,\mathbf{v}_{m}\right\} $
cannot be linearly independent since it has more vectors than the
dimension of the space. Therefore all such tensors are equal to zero
(more pedantically, to the \emph{zero} \emph{tensor}), which is thus
the only element of $\wedge^{m}V$ for every $m>N$.\hfill{}$\blacksquare$


\paragraph{Exercise 1:}

It is given that the set of four vectors $\left\{ \mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d}\right\} $
is linearly independent. Show that the tensor $\omega\equiv\mathbf{a}\wedge\mathbf{b}+\mathbf{c}\wedge\mathbf{d}\in\wedge^{2}V$
\emph{cannot} be equal to a single-term\index{single-term exterior products}
exterior product of the form $\mathbf{x}\wedge\mathbf{y}$.

\emph{Outline of solution}: 

1. Constructive solution. There exists $\mathbf{f}^{*}\in V^{*}$
such that $\mathbf{f}^{*}(\mathbf{a})=1$ and $\mathbf{f}^{*}(\mathbf{b})=0$,
$\mathbf{f}^{*}(\mathbf{c})=0$, $\mathbf{f}^{*}(\mathbf{d})=0$.
Compute $\iota_{\mathbf{f}^{*}}\omega=\mathbf{b}$. If $\omega=\mathbf{x}\wedge\mathbf{y}$,
it will follow that a linear combination of $\mathbf{x}$ and $\mathbf{y}$
is equal to $\mathbf{b}$, i.e.~$\mathbf{b}$ belongs to the two-dimen\-sion\-al
space $\text{Span}\left\{ \mathbf{x},\mathbf{y}\right\} $. Repeat
this argument for the remaining three vectors ($\mathbf{a}$, $\mathbf{c}$,
$\mathbf{d}$) and obtain a contradiction.

2. Non-constructive solution. Compute $\omega\wedge\omega=2\mathbf{a}\wedge\mathbf{b}\wedge\mathbf{c}\wedge\mathbf{d}\neq0$
by linear independence of $\left\{ \mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d}\right\} $.
If we could express $\omega=\mathbf{x}\wedge\mathbf{y}$ then we would
have $\omega\wedge\omega=0$.\hfill{}$\blacksquare$


\paragraph{Remark:}

While $\mathbf{a}\wedge\mathbf{b}$ is interpreted geometrically as
the oriented area of a parallelogram spanned by $\mathbf{a}$ and
$\mathbf{b}$, a general linear combination such as $\mathbf{a}\wedge\mathbf{b}+\mathbf{c}\wedge\mathbf{d}+\mathbf{e}\wedge\mathbf{f}$
does not have this interpretation (unless it can be reduced to a single-term
product $\mathbf{x}\wedge\mathbf{y}$). If not reducible to a single-term
product, $\mathbf{a}\wedge\mathbf{b}+\mathbf{c}\wedge\mathbf{d}$
can be interpreted only as a \emph{formal} linear combination of two
areas.


\paragraph{Exercise 2:}

Suppose that $\psi\in\wedge^{k}V$ and $\mathbf{x}\in V$ are such
that $\mathbf{x}\wedge\psi=0$ while $\mathbf{x}\neq0$. Show that
there exists $\chi\in\wedge^{k-1}V$ such that $\psi=\mathbf{x}\wedge\chi$.
Give an example where $\psi$ and $\chi$ are \emph{not} representable
as a single-term exterior product.

\emph{Outline of solution}: There exists $\mathbf{f}^{*}\in V^{*}$
such that $\mathbf{f}^{*}(\mathbf{x})=1$. Apply $\iota_{\mathbf{f}^{*}}$
to the given equality $\mathbf{x}\wedge\psi=0$:\[
0\,{\lyxbuildrel!\above=}\,\iota_{\mathbf{f}^{*}}(\mathbf{x}\wedge\psi)=\psi-\mathbf{x}\wedge\iota_{\mathbf{f}^{*}}\psi,\]
which means that $\psi=\mathbf{x}\wedge\chi$ with $\chi\equiv\iota_{\mathbf{f}^{*}}\psi$.
An example can be found with $\chi=\mathbf{a}\wedge\mathbf{b}+\mathbf{c}\wedge\mathbf{d}$
as in Exercise 1, and $\mathbf{x}$ such that the set $\{\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d},\mathbf{x}\}$
is linearly independent; then $\psi\equiv\mathbf{x}\wedge\psi$ is
also not reducible to a single-term product.


\subsection{Computing the dual basis\label{sub:Computing-the-dual}}

The exterior product allows us to compute explicitly the dual basis\index{dual basis}
for a given basis.

We begin with some motivation. Suppose $\left\{ \mathbf{v}_{1},...,\mathbf{v}_{N}\right\} $
is a given basis; we would like to compute its dual basis. For instance,
the covector $\mathbf{v}_{1}^{*}$ of the dual basis is the linear
function such that $\mathbf{v}_{1}^{*}(\mathbf{x})$ is equal to the
coefficient at $\mathbf{v}_{1}$ in the decomposition of $\mathbf{x}$
in the basis $\left\{ \mathbf{v}_{j}\right\} $,\[
\mathbf{x}=\sum_{i=1}^{N}x_{i}\mathbf{v}_{i};\quad\mathbf{v}_{1}^{*}(\mathbf{x})=x_{1}.\]
We start from the observation that the tensor $\omega\equiv\mathbf{v}_{1}\wedge...\wedge\mathbf{v}_{N}$
is nonzero since $\left\{ \mathbf{v}_{j}\right\} $ is a basis. The
exterior product $\mathbf{x}\wedge\mathbf{v}_{2}\wedge...\wedge\mathbf{v}_{N}$
is equal to zero if $\mathbf{x}$ is a linear combination only of
$\mathbf{v}_{2}$, ..., $\mathbf{v}_{N}$, with a zero coefficient
$x_{1}$. This suggests that the exterior product of $\mathbf{x}$
with the $(N-1)$-vector $\mathbf{v}_{2}\wedge...\wedge\mathbf{v}_{N}$
is quite similar to the covector $\mathbf{v}_{1}^{*}$ we are looking
for. Indeed, let us compute\[
\mathbf{x}\wedge\mathbf{v}_{2}\wedge...\wedge\mathbf{v}_{N}=x_{1}\mathbf{v}_{1}\wedge\mathbf{v}_{2}\wedge...\wedge\mathbf{v}_{N}=x_{1}\omega.\]
Therefore, exterior multiplication with $\mathbf{v}_{2}\wedge...\wedge\mathbf{v}_{N}$
acts quite similarly to $\mathbf{v}_{1}^{*}$. To make the notation
more concise, let us introduce a special \textbf{complement}\index{Grassmann's complement}
operation%
\footnote{The complement operation was introduced by H. Grassmann (1844).%
} denoted by a star: \[
*\left(\mathbf{v}_{1}\right)\equiv\mathbf{v}_{2}\wedge...\wedge\mathbf{v}_{N}.\]
Then we can write $\mathbf{v}_{1}^{*}(\mathbf{x})\omega=\mathbf{x}\wedge*(\mathbf{v}_{1})$.
This equation can be used for computing $\mathbf{v}_{1}^{*}$: namely,
for any $\mathbf{x}\in V$ the number $\mathbf{v}_{1}^{*}(\mathbf{x})$
is equal to the constant $\lambda$ in the equation $\mathbf{x}\wedge*(\mathbf{v}_{1})=\lambda\omega$.
To make this kind of equation more convenient, let us write\[
\lambda\equiv\mathbf{v}_{1}^{*}(\mathbf{x})=\frac{\mathbf{x}\wedge\mathbf{v}_{2}\wedge...\wedge\mathbf{v}_{N}}{\mathbf{v}_{1}\wedge\mathbf{v}_{2}\wedge...\wedge\mathbf{v}_{N}}=\frac{\mathbf{x}\wedge*(\mathbf{v}_{1})}{\omega},\]
where the {}``division'' of one tensor\index{dividing by tensor}
by another is to be understood as follows: We first compute the tensor
$\mathbf{x}\wedge*(\mathbf{v}_{1})$; this tensor is proportional
to the tensor $\omega$ since both belong to the one-dimen\-sion\-al
space $\wedge^{N}V$, so we can determine the number $\lambda$ such
that $\mathbf{x}\wedge*(\mathbf{v}_{1})=\lambda\omega$; the proportionality
coefficient $\lambda$ is then the result of the division of $\mathbf{x}\wedge*(\mathbf{v}_{1})$
by $\omega$.

For $\mathbf{v}_{2}$ we have\[
\mathbf{v}_{1}\wedge\mathbf{x}\wedge\mathbf{v}_{3}\wedge...\wedge\mathbf{v}_{N}=x_{2}\omega=\mathbf{v}_{2}^{*}(\mathbf{x})\omega.\]
 If we would like to have $x_{2}\omega=\mathbf{x}\wedge*(\mathbf{v}_{2})$,
we need to add an extra minus sign and define\[
*\left(\mathbf{v}_{2}\right)\equiv-\mathbf{v}_{1}\wedge\mathbf{v}_{3}\wedge...\wedge\mathbf{v}_{N}.\]
Then we indeed obtain $\mathbf{v}_{2}^{*}(\mathbf{x})\omega=\mathbf{x}\wedge*(\mathbf{v}_{2})$. 

It is then clear that we can define the tensors $*(\mathbf{v}_{i})$
for $i=1,...,N$ in this way. The tensor $*(\mathbf{v}_{i})$ is obtained
from $\omega$ by removing the vector $\mathbf{v}_{i}$ and by adding
a sign that corresponds to shifting the vector $\mathbf{v}_{i}$ to
the left position in the exterior product. The {}``complement''
map, $*:V\rightarrow\wedge^{N-1}V$, satisfies $\mathbf{v}_{j}\wedge*(\mathbf{v}_{j})=\omega$
for each \emph{basis} vector $\mathbf{v}_{j}$. (Once defined on the
basis vectors, the complement map can be then extended to all vectors
from $V$ by requiring linearity. However, we will apply the complement
operation only to basis vectors right now.)

With these definitions, we may express the dual basis as\[
\mathbf{v}_{i}^{*}(\mathbf{x})\omega=\mathbf{x}\wedge*(\mathbf{v}_{i}),\quad\mathbf{x}\in V,\: i=1,...,N.\]



\paragraph{Remark:}

The notation $*(\mathbf{v}_{i})$ suggests that e.g.~$*(\mathbf{v}_{1})$
is some operation applied to $\mathbf{v}_{1}$ and is a function only
of the vector $\mathbf{v}_{1}$, but this is not so: The {}``complement''
of a vector depends on the entire basis and not merely on the single
vector! Also, the property $\mathbf{v}_{1}\wedge*(\mathbf{v}_{1})=\omega$
is not sufficient to define the tensor $*\mathbf{v}_{1}$. The proper
definition of $*(\mathbf{v}_{i})$ is the tensor obtained from $\omega$
by removing $\mathbf{v}_{i}$ as just explained.


\paragraph{Example:}

In the space $\mathbb{R}^{2}$, let us compute the dual basis to the
basis $\left\{ \mathbf{v}_{1},\mathbf{v}_{2}\right\} $ where $\mathbf{v}_{1}={2 \choose 1}$
and $\mathbf{v}_{2}={-1 \choose 1}$.

Denote by $\mathbf{e}_{1}$ and $\mathbf{e}_{2}$ the standard basis
vectors ${1 \choose 0}$ and ${0 \choose 1}$. We first compute the
2-vector \[
\omega=\mathbf{v}_{1}\wedge\mathbf{v}_{2}=\left(2\mathbf{e}_{1}+\mathbf{e}_{2}\right)\wedge\left(-\mathbf{e}_{1}+\mathbf{e}_{2}\right)=3\mathbf{e}_{1}\wedge\mathbf{e}_{2}.\]
 The {}``complement'' operation for the basis $\left\{ \mathbf{v}_{1},\mathbf{v}_{2}\right\} $
gives $*(\mathbf{v}_{1})=\mathbf{v}_{2}$ and $*(\mathbf{v}_{2})=-\mathbf{v}_{1}$.
We now define the covectors $\mathbf{v}_{1,2}^{*}$ by their action
on arbitrary vector $\mathbf{x}\equiv x_{1}\mathbf{e}_{1}+x_{2}\mathbf{e}_{2}$,\begin{align*}
\mathbf{v}_{1}^{*}(\mathbf{x})\omega & =\mathbf{x}\wedge\mathbf{v}_{2}=\left(x_{1}\mathbf{e}_{1}+x_{2}\mathbf{e}_{2}\right)\wedge\left(-\mathbf{e}_{1}+\mathbf{e}_{2}\right)\\
 & =\left(x_{1}+x_{2}\right)\mathbf{e}_{1}\wedge\mathbf{e}_{2}=\frac{x_{1}+x_{2}}{3}\omega,\\
\mathbf{v}_{2}^{*}(\mathbf{x})\omega & =-\mathbf{x}\wedge\mathbf{v}_{1}=-\left(x_{1}\mathbf{e}_{1}+x_{2}\mathbf{e}_{2}\right)\wedge\left(2\mathbf{e}_{1}+\mathbf{e}_{2}\right)\\
 & =\left(-x_{1}+2x_{2}\right)\mathbf{e}_{1}\wedge\mathbf{e}_{2}=\frac{-x_{1}+2x_{2}}{3}\omega.\end{align*}
Therefore, $\mathbf{v}_{1}^{*}=\frac{1}{3}\mathbf{e}_{1}^{*}+\frac{1}{3}\mathbf{e}_{2}^{*}$
and $\mathbf{v}_{2}^{*}=-\frac{1}{3}\mathbf{e}_{1}^{*}+\frac{2}{3}\mathbf{e}_{2}^{*}$.


\paragraph{Question:}

Can we define the complement operation for all $\mathbf{x}\in V$
by the equation $\mathbf{x}\wedge*(\mathbf{x})=\omega$ where $\omega\in\wedge^{N}V$
is a fixed tensor? Does the complement really depend on the entire
basis? Or perhaps a choice of $\omega$ is sufficient?


\subparagraph{Answer: }

No, yes, no. Firstly, $*(\mathbf{x})$ is not uniquely specified by
that equation alone, since $\mathbf{x}\wedge A=\omega$ defines $A$
only up to tensors of the form $\mathbf{x}\wedge...$; secondly, the
equation $\mathbf{x}\wedge*(\mathbf{x})=\omega$ indicates that $*(\lambda\mathbf{x})=\frac{1}{\lambda}\,*\negmedspace(\mathbf{x})$,
so the complement map would not be linear if defined like that. It
is important to keep in mind that the complement map requires an entire
basis for its definition and depends not only on the choice of a tensor
$\omega$, but also on the choice of all the basis vectors. For example,
in two dimensions we have $*(\mathbf{e}_{1})=\mathbf{e}_{2}$; it
is clear that $*(\mathbf{e}_{1})$ depends on the choice of $\mathbf{e}_{2}$!


\paragraph{Remark:}

The situation is different when the vector space is equipped with
a scalar product (see Sec.~\ref{sub:The-vector-product} below).
In that case, one usually chooses an \emph{orthonormal} basis to define
the complement map; then the complement map is called the \textbf{Hodge\index{Hodge star}
star}. It turns out that the Hodge star is independent of the choice
of the basis as long as the basis is orthonormal with respect to the
given scalar product, and as long as the orientation of the basis
is unchanged (i.e.~as long as the tensor $\omega$ does not change
sign). In other words, the Hodge star operation is invariant under
orthogonal and orientation-preserving transformations of the basis;
these transformations preserve the tensor $\omega$. So the Hodge
star operation depends not quite on the detailed choice of the basis,
but rather on the choice of the scalar product and on the orientation
of the basis (the sign of $\omega$). However, right now we are working
with a general space without a scalar product. In this case, the complement
map depends on the entire basis.


\subsection{Gaussian elimination}


\paragraph{Question:}

How much computational effort is actually needed to compute the exterior
product of $n$ vectors? It looks easy in two or three dimensions,
but in $N$ dimensions the product of $n$ vectors $\left\{ \mathbf{x}_{1},...,\mathbf{x}_{n}\right\} $
gives expressions such as\[
\bigwedge_{i=1}^{n}\mathbf{x}_{n}=\left(x_{11}\mathbf{e}_{1}+...+x_{1N}\mathbf{e}_{N}\right)\wedge...\wedge\left(x_{n1}\mathbf{e}_{1}+...+x_{nN}\mathbf{e}_{N}\right),\]
which will be reduced to an exponentially large number (of order $N^{n}$)
of elementary tensor products when we expand all brackets.


\subparagraph{Answer:}

Of course, expanding all brackets is not the best way to compute long
exterior products. We can instead use a procedure similar to the Gaussian
elimination\index{Gaussian elimination} for computing determinants.
The key observation is that\[
\mathbf{x}_{1}\wedge\mathbf{x}_{2}\wedge...=\mathbf{x}_{1}\wedge\left(\mathbf{x}_{2}-\lambda\mathbf{x}_{1}\right)\wedge...\]
for any number $\lambda$, and that it is easy to compute an exterior
product of the form\[
(\alpha_{1}\mathbf{e}_{1}+\alpha_{2}\mathbf{e}_{2}+\alpha_{3}\mathbf{e}_{3})\wedge(\beta_{2}\mathbf{e}_{2}+\beta_{3}\mathbf{e}_{3})\wedge\mathbf{e}_{3}=\alpha_{1}\beta_{2}\mathbf{e}_{1}\wedge\mathbf{e}_{2}\wedge\mathbf{e}_{3}.\]
It is easy to compute this exterior product because the second vector
($\beta_{2}\mathbf{e}_{2}+\beta_{3}\mathbf{e}_{3}$) does not contain
the basis vector $\mathbf{e}_{1}$ and the third vector does not contain
$\mathbf{e}_{1}$ or $\mathbf{e}_{2}$. So we can simplify the computation
of a long exterior product if we rewrite \begin{align*}
 & \bigwedge_{i=1}^{n}\mathbf{x}_{n}=\mathbf{x}_{1}\wedge\tilde{\mathbf{x}}_{2}\wedge...\wedge\tilde{\mathbf{x}}_{n}\\
 & \equiv\mathbf{x}_{1}\wedge(\mathbf{x}_{2}-\lambda_{11}\mathbf{x}_{1})\wedge...\wedge\left(\mathbf{x}_{n}-\lambda_{n1}\mathbf{x}_{1}-...-\lambda_{n-1,n-1}\mathbf{x}_{n-1}\right),\end{align*}
where the coefficients $\left\{ \lambda_{ij}\,|\,1\leq i\leq n-1,\;1\leq j\leq i\right\} $
are chosen appropriately such that the vector $\tilde{\mathbf{x}}_{2}\equiv\mathbf{x}_{2}-\lambda_{11}\mathbf{x}_{1}$
does not contain the basis vector $\mathbf{e}_{1}$, and generally
the vector \[
\tilde{\mathbf{x}}_{k}\equiv\mathbf{x}_{k}-\lambda_{k1}\mathbf{x}_{1}-...-\lambda_{k-1,k-1}\mathbf{x}_{k-1}\]
 does not contain the basis vectors $\mathbf{e}_{1}$,..., $\mathbf{e}_{k-1}$.
(That is, these basis vectors have been {}``eliminated'' from the
vector $\mathbf{x}_{k}$, hence the name of the method.) Eliminating
$\mathbf{e}_{1}$ from $\mathbf{x}_{2}$ can be done with $\lambda_{11}=\frac{x_{21}}{x_{11}}$,
which is possible provided that $x_{11}\neq0$; if $x_{11}=0$, we
need to renumber the vectors $\left\{ \mathbf{x}_{j}\right\} $. If
none of them contains $\mathbf{e}_{1}$, we skip $\mathbf{e}_{1}$
and proceed with $\mathbf{e}_{2}$ instead. Elimination of other basis
vectors proceeds similarly. After performing this algorithm, we will
either find that some vector $\tilde{\mathbf{x}}_{k}$ is itself zero,
which means that the entire exterior product vanishes, or we will
find the product of vectors of the form\[
\tilde{\mathbf{x}}_{1}\wedge...\wedge\tilde{\mathbf{x}}_{n},\]
 where the vectors $\tilde{\mathbf{x}}_{i}$ are linear combinations
of $\mathbf{e}_{i}$, ..., $\mathbf{e}{}_{N}$ (not containing $\mathbf{e}_{1}$,
..., $\mathbf{e}_{i}$). 

If $n=N$, the product can be evaluated immediately since the last
vector, $\tilde{\mathbf{x}}_{N}$, is proportional to $\mathbf{e}_{N}$,
so\begin{align*}
\tilde{\mathbf{x}}_{1}\wedge...\wedge\tilde{\mathbf{x}}_{n} & =\left(c_{11}\mathbf{e}_{1}+...\right)\wedge...\wedge(c_{nn}\mathbf{e}_{N})\\
 & =c_{11}c_{22}...c_{nn}\mathbf{e}_{1}\wedge...\wedge\mathbf{e}_{N}.\end{align*}
 The computation is somewhat longer if $n<N$, so that \[
\tilde{\mathbf{x}}_{n}=c_{nn}\mathbf{e}_{n}+...+c_{nN}\mathbf{e}_{N}.\]
In that case, we may eliminate, say, $\mathbf{e}_{n}$ from $\tilde{\mathbf{x}}_{1}$,
..., $\tilde{\mathbf{x}}_{n-1}$ by subtracting a multiple of $\tilde{\mathbf{x}}_{n}$
from them, but we cannot simplify the product any more; at that point
we need to expand the last bracket (containing $\tilde{\mathbf{x}}_{n}$)
and write out the terms.


\paragraph{Example 1: }

We will calculate the exterior product\begin{align*}
 & \mathbf{a}\wedge\mathbf{b}\wedge\mathbf{c}\\
 & \equiv(7\mathbf{e}_{1}-8\mathbf{e}_{2}+\mathbf{e}_{3})\wedge(\mathbf{e}_{1}-2\mathbf{e}_{2}-15\mathbf{e}_{3})\wedge(2\mathbf{e}_{1}-5\mathbf{e}_{2}-\mathbf{e}_{3}).\end{align*}
We will eliminate $\mathbf{e}_{1}$ from $\mathbf{a}$ and $\mathbf{c}$
(just to keep the coefficients simpler):\begin{align*}
 & \mathbf{a}\wedge\mathbf{b}\wedge\mathbf{c}=(\mathbf{a}-7\mathbf{b})\wedge\mathbf{b}\wedge(\mathbf{c}-2\mathbf{b})\\
 & =(6\mathbf{e}_{2}+106\mathbf{e}_{3})\wedge\mathbf{b}\wedge(-\mathbf{e}_{2}+9\mathbf{e}_{3})\\
 & \equiv\mathbf{a}_{1}\wedge\mathbf{b}\wedge\mathbf{c}_{1}.\end{align*}
Now we eliminate $\mathbf{e}_{2}$ from $\mathbf{a}_{1}$, and then
the product can be evaluated quickly:\begin{align*}
 & \mathbf{a}\wedge\mathbf{b}\wedge\mathbf{c}=\mathbf{a}_{1}\wedge\mathbf{b}\wedge\mathbf{c}_{1}=(\mathbf{a}_{1}+6\mathbf{c}_{1})\wedge\mathbf{b}\wedge\mathbf{c}_{1}\\
 & =(160\mathbf{e}_{3})\wedge(\mathbf{e}_{1}-2\mathbf{e}_{2}-5\mathbf{e}_{3})\wedge(-\mathbf{e}_{2}+9\mathbf{e}_{3})\\
 & =160\mathbf{e}_{3}\wedge\mathbf{e}_{1}\wedge(-\mathbf{e}_{2})=-160\mathbf{e}_{1}\wedge\mathbf{e}_{2}\wedge\mathbf{e}_{3}.\end{align*}



\paragraph{Example 2:}

Consider\begin{align*}
 & \mathbf{a}\wedge\mathbf{b}\wedge\mathbf{c}\equiv(\mathbf{e}_{1}+2\mathbf{e}_{2}-\mathbf{e}_{3}+\mathbf{e}_{4})\\
 & \quad\wedge(2\mathbf{e}_{1}+\mathbf{e}_{2}-\mathbf{e}_{3}+3\mathbf{e}_{4})\wedge(-\mathbf{e}_{1}-\mathbf{e}_{2}+\mathbf{e}_{4}).\end{align*}
We eliminate $\mathbf{e}_{1}$ and $\mathbf{e}_{2}$:\begin{align*}
 & \mathbf{a}\wedge\mathbf{b}\wedge\mathbf{c}=\mathbf{a}\wedge(\mathbf{b}-2\mathbf{a})\wedge(\mathbf{c}+\mathbf{a})\\
 & =\mathbf{a}\wedge\left(-3\mathbf{e}_{2}+\mathbf{e}_{3}+\mathbf{e}_{4}\right)\wedge\left(\mathbf{e}_{2}-\mathbf{e}_{3}+2\mathbf{e}_{4}\right)\\
 & \equiv\mathbf{a}\wedge\mathbf{b}_{1}\wedge\mathbf{c}_{1}=\mathbf{a}\wedge\mathbf{b}_{1}\wedge(\mathbf{c}_{1}+3\mathbf{b}_{1})\\
 & =\mathbf{a}\wedge\mathbf{b}_{1}\wedge(2\mathbf{e}_{3}+5\mathbf{e}_{4})\equiv\mathbf{a}\wedge\mathbf{b}_{1}\wedge\mathbf{c}_{2}.\end{align*}
We can now eliminate $\mathbf{e}_{3}$ from $\mathbf{a}$ and $\mathbf{b}_{1}$:\begin{align*}
 & \mathbf{a}\wedge\mathbf{b}_{1}\wedge\mathbf{c}_{2}=(\mathbf{a}+\frac{1}{2}\mathbf{c}_{2})\wedge(\mathbf{b}_{1}-\frac{1}{2}\mathbf{c}_{2})\wedge\mathbf{c}_{2}\equiv\mathbf{a}_{2}\wedge\mathbf{b}_{2}\wedge\mathbf{c}_{2}\\
 & =(\mathbf{e}_{1}+2\mathbf{e}_{2}+\frac{7}{2}\mathbf{e}_{4})\wedge(-3\mathbf{e}_{2}-\frac{3}{2}\mathbf{e}_{4})\wedge(2\mathbf{e}_{3}+5\mathbf{e}_{4}).\end{align*}
Now we cannot eliminate any more vectors, so we expand the last bracket
and simplify the result by omitting the products of equal vectors:
\begin{align*}
 & \,\mathbf{a}_{2}\wedge\mathbf{b}_{2}\wedge\mathbf{c}_{2}=\mathbf{a}_{2}\wedge\mathbf{b}_{2}\wedge2\mathbf{e}_{3}+\mathbf{a}_{2}\wedge\mathbf{b}_{2}\wedge5\mathbf{e}_{4}\\
 & =\left(\mathbf{e}_{1}+2\mathbf{e}_{2}\right)\wedge(-\frac{3}{2}\mathbf{e}_{4})\wedge2\mathbf{e}_{3}+\mathbf{e}_{1}\wedge(-3\mathbf{e}_{2})\wedge2\mathbf{e}_{3}\\
 & +\mathbf{e}_{1}\wedge(-3\mathbf{e}_{2})\wedge5\mathbf{e}_{4}\\
 & =3\mathbf{e}_{1}\wedge\mathbf{e}_{3}\wedge\mathbf{e}_{4}+6\mathbf{e}_{2}\wedge\mathbf{e}_{3}\wedge\mathbf{e}_{4}-6\mathbf{e}_{1}\wedge\mathbf{e}_{2}\wedge\mathbf{e}_{3}-15\mathbf{e}_{1}\wedge\mathbf{e}_{2}\wedge\mathbf{e}_{4}.\end{align*}
 


\subsection{Rank of a set of vectors\label{sub:Rank-of-a-set-of-vectors}}

We have defined the rank of a map (Sec.~\ref{sub:Linear-maps-between-different-spaces})
as the dimension of the image of the map, and we have seen that the
rank is equal to the minimum number of tensor product terms needed
to represent the map as a tensor. An analogous concept can be introduced
for sets of vectors.


\paragraph{Definition:}

If $S=\left\{ \mathbf{v}_{1},...,\mathbf{v}_{n}\right\} $ is a set
of vectors (where $n$ is not necessarily smaller than the dimension
$N$ of space), the \textbf{rank} of the set $S$ is the dimension
of the subspace spanned by the vectors $\left\{ \mathbf{v}_{1},...,\mathbf{v}_{n}\right\} $.
Written as a formula,\[
\text{rank}\,(S)=\dim\,\text{Span}\, S.\]


The rank of a set $S$ is equal to the maximum number of vectors in
any linearly independent subset of $S$. For example, consider the
set $\left\{ 0,\mathbf{v},2\mathbf{v},3\mathbf{v}\right\} $ where
$\mathbf{v}\neq0$. The rank of this set is 1 since these four vectors
span a one-dimen\-sion\-al subspace,\[
\text{Span}\left\{ 0,\mathbf{v},2\mathbf{v},3\mathbf{v}\right\} =\text{Span}\left\{ \mathbf{v}\right\} .\]
Any subset of $S$ having two or more vectors is linearly dependent.

We will now show how to use the exterior product for computing the
rank of a given (finite) set $S=\left\{ \mathbf{v}_{1},...,\mathbf{v}_{n}\right\} $. 

According to Theorem~1 in Sec.~\ref{sub:Properties-of-the-ext-powers},
the set $S$ is linearly independent if and only if $\mathbf{v}_{1}\wedge...\wedge\mathbf{v}_{n}\neq0$.
So we first compute the tensor $\mathbf{v}_{1}\wedge...\wedge\mathbf{v}_{n}$.
If this tensor is nonzero then the set $S$ is linearly independent,
and the rank of $S$ is equal to $n$. If, on the other hand, $\mathbf{v}_{1}\wedge...\wedge\mathbf{v}_{n}=0$,
the rank is less than $n$. We can determine the rank of $S$ by the
following procedure. First, we assume that all $\mathbf{v}_{j}\neq0$
(any zero vectors can be omitted without changing the rank of $S$).
Then we compute $\mathbf{v}_{1}\wedge\mathbf{v}_{2}$; if the result
is zero, we may omit $\mathbf{v}_{2}$ since $\mathbf{v}_{2}$ is
proportional to $\mathbf{v}_{1}$ and try $\mathbf{v}_{1}\wedge\mathbf{v}_{3}$.
If $\mathbf{v}_{1}\wedge\mathbf{v}_{2}\neq0$, we try $\mathbf{v}_{1}\wedge\mathbf{v}_{2}\wedge\mathbf{v}_{3}$,
and so on. The procedure can be formulated using induction in the
obvious way. Eventually we will arrive at a subset $\{\mathbf{v}_{i_{1}},...,\mathbf{v}_{i_{k}}\}\subset S$
such that $\mathbf{v}_{i_{1}}\wedge...\wedge...\mathbf{v}_{i_{k}}\neq0$
but $\mathbf{v}_{i_{1}}\wedge...\wedge...\mathbf{v}_{i_{k}}\wedge\mathbf{v}_{j}=0$
for any other $\mathbf{v}_{j}$. Thus, there are no linearly independent
subsets of $S$ having $k+1$ or more vectors. Then the rank of $S$
is equal to $k$. 

The subset $\{\mathbf{v}_{i_{1}},...,\mathbf{v}_{i_{k}}\}$ is built
by a procedure that depends on the order in which the vectors $\mathbf{v}_{j}$
are selected. However, the next statement says that the resulting
subspace spanned by $\{\mathbf{v}_{i_{1}},...,\mathbf{v}_{i_{k}}\}$
is the same regardless of the order of vectors $\mathbf{v}_{j}$.
Hence, the subset $\{\mathbf{v}_{i_{1}},...,\mathbf{v}_{i_{k}}\}$
yields a basis in $\text{Span}\, S$. 


\paragraph{Statement: }

Suppose a set $S$ of vectors has rank $k$ and contains \emph{two}
different linearly independent subsets, say $S_{1}=\left\{ \mathbf{v}_{1},...,\mathbf{v}_{k}\right\} $
and $S_{2}=\left\{ \mathbf{u}_{1},...,\mathbf{u}_{k}\right\} $, both
having $k$ vectors (but no linearly independent subsets having $k+1$
or more vectors). Then the tensors $\mathbf{v}_{1}\wedge...\wedge\mathbf{v}_{k}$
and $\mathbf{u}_{1}\wedge...\wedge\mathbf{u}_{k}$ are proportional
to each other (as tensors from $\wedge^{k}V$).


\subparagraph{Proof:}

The tensors $\mathbf{v}_{1}\wedge...\wedge\mathbf{v}_{k}$ and $\mathbf{u}_{1}\wedge...\wedge\mathbf{u}_{k}$
are both nonzero by Theorem~1 in Sec.~\ref{sub:Properties-of-the-ext-powers}.
We will now show that it is possible to replace $\mathbf{v}_{1}$
by one of the vectors from the set $S_{2}$, say $\mathbf{u}_{l}$,
such that the new tensor $\mathbf{u}_{l}\wedge\mathbf{v}_{2}\wedge...\wedge\mathbf{v}_{k}$
is nonzero and proportional to the original tensor $\mathbf{v}_{1}\wedge...\wedge\mathbf{v}_{k}$.
It will follow that this procedure can be repeated for every other
vector $\mathbf{v}_{i}$, until we replace all $\mathbf{v}_{i}$'s
by some $\mathbf{u}_{i}$'s and thus prove that the tensors $\mathbf{v}_{1}\wedge...\wedge\mathbf{v}_{k}$
and $\mathbf{u}_{1}\wedge...\wedge\mathbf{u}_{k}$ are proportional
to each other.

It remains to prove that the vector $\mathbf{v}_{1}$ can be replaced.
We need to find a suitable vector $\mathbf{u}_{l}$. Let $\mathbf{u}_{l}$
be one of the vectors from $S_{2}$, and let us check whether $\mathbf{v}_{1}$
could be replaced by $\mathbf{u}_{l}$. We first note that $\mathbf{v}_{1}\wedge...\wedge\mathbf{v}_{k}\wedge\mathbf{u}_{l}=0$
since there are no linearly independent subsets of $S$ having $k+1$
vectors. Hence the set $\left\{ \mathbf{v}_{1},...,\mathbf{v}_{k},\mathbf{u}_{l}\right\} $
is linearly \emph{dependent}. It follows (since the set $\left\{ \mathbf{v}_{i}\,|\, i=1,...,k\right\} $
was linearly independent before we added $\mathbf{u}_{l}$ to it)
that $\mathbf{u}_{l}$ can be expressed as a linear combination of
the $\mathbf{v}_{i}$'s with some coefficients $\alpha_{i}$:\[
\mathbf{u}_{l}=\alpha_{1}\mathbf{v}_{1}+...+\alpha_{k}\mathbf{v}_{k}.\]
If $\alpha_{1}\neq0$ then we will have\[
\mathbf{u}_{l}\wedge\mathbf{v}_{2}\wedge...\wedge\mathbf{v}_{k}=\alpha_{1}\mathbf{v}_{1}\wedge\mathbf{v}_{2}\wedge...\wedge\mathbf{v}_{k}.\]
The new tensor is nonzero and proportional to the old tensor, so we
can replace $\mathbf{v}_{1}$ by $\mathbf{u}_{l}$. 

However, it could also happen that $\alpha_{1}=0$. In that case we
need to choose a different vector $\mathbf{u}_{l'}\in S_{2}$ such
that the corresponding coefficient $\alpha_{1}$ is nonzero. It remains
to prove that such a choice is possible. If this were impossible then
all $\mathbf{u}_{i}$'s would have been expressible as linear combinations
of $\mathbf{v}_{i}$'s with zero coefficients at the vector $\mathbf{v}_{1}$.
In that case, the exterior product $\mathbf{u}_{1}\wedge...\wedge\mathbf{u}_{k}$
would be equal to a linear combination of exterior products of vectors
$\mathbf{v}_{i}$ with $i=2,...,k$. These exterior products contain
$k$ vectors among which only $\left(k-1\right)$ vectors are different.
Such exterior products are all equal to zero. However, this contradicts
the assumption $\mathbf{u}_{1}\wedge...\wedge\mathbf{u}_{k}\neq0$.
Therefore, at least one vector $\mathbf{u}_{l}$ exists such that
$\alpha_{1}\neq0$, and the required replacement is always possible.\hfill{}$\blacksquare$


\paragraph{Remark:}

It follows from the above Statement that the subspace spanned by $S$
can be uniquely characterized by a nonzero tensor such as $\mathbf{v}_{1}\wedge...\wedge\mathbf{v}_{k}$
in which the constituents --- the vectors $\mathbf{v}_{1}$,..., $\mathbf{v}_{k}$
--- form a basis in the subspace $\text{Span}\, S$. It does not matter
which linearly independent subset we choose for this purpose. We also
have a computational procedure for determining the subspace $\text{Span}\, S$
together with its dimension. Thus, we find that a $k$-dimen\-sion\-al
subspace is adequately specified by selecting a nonzero tensor $\omega\in\wedge^{k}V$
of the form $\omega=\mathbf{v}_{1}\wedge...\wedge\mathbf{v}_{k}$.
For a given subspace, this tensor $\omega$ is unique up to a nonzero
constant factor. Of course, the decomposition of $\omega$ into an
exterior product of vectors $\left\{ \mathbf{v}_{i}\,|\, i=1,...,k\right\} $
is not unique, but any such decomposition yields a set $\left\{ \mathbf{v}_{i}\,|\, i=1,...,k\right\} $
spanning the same subspace. 


\paragraph{Exercise 1:}

Let $\left\{ \mathbf{v}_{1},...,\mathbf{v}_{n}\right\} $ be a linearly
independent set of vectors, $\omega\equiv\mathbf{v}_{1}\wedge...\wedge\mathbf{v}_{n}\neq0$,
and $\mathbf{x}$ be a given vector such that $\omega\wedge\mathbf{x}=0$.
Show that $\mathbf{x}$ belongs to the subspace $\text{Span}\left\{ \mathbf{v}_{1},...,\mathbf{v}_{n}\right\} $. 


\paragraph{Exercise 2:}

Given a nonzero covector $\mathbf{f}^{*}$ and a vector $\mathbf{n}$
such that $\mathbf{f}^{*}(\mathbf{n})\neq0$, show that the operator
$\hat{P}$ defined by\[
\hat{P}\mathbf{x}=\mathbf{x}-\mathbf{n}\frac{\mathbf{f}^{*}(\mathbf{x})}{\mathbf{f}^{*}(\mathbf{n})}\]
 is a projector\index{projector} onto the subspace $\mathbf{f}^{*\perp}$,
i.e.~that $\mathbf{f}^{*}(\hat{P}\mathbf{x})=0$ for all $\mathbf{x}\in V$.
Show that\[
(\hat{P}\mathbf{x})\wedge\mathbf{n}=\mathbf{x}\wedge\mathbf{n},\quad\forall\mathbf{x}\in V.\]



\subsection{Exterior product in index notation\label{sub:Exterior-product-in-index}}

Here I show how to perform calculations with the exterior product
using the index notation\index{exterior product!in index notation}
(see Sec.~\ref{sub:Index-notation}), although I will not use this
later because the index-free notation is more suitable for the purposes
of this book. 

Let us choose a basis $\left\{ \mathbf{e}_{j}\right\} $ in $V$;
then the dual basis $\left\{ \mathbf{e}_{j}^{*}\right\} $ in $V$
and the basis $\left\{ \mathbf{e}_{k_{1}}\wedge...\wedge\mathbf{e}_{k_{m}}\right\} $
in $\wedge^{m}V$ are fixed. By definition, the exterior product of
two vectors $\mathbf{u}$ and $\mathbf{v}$ is \[
A\equiv\mathbf{u}\wedge\mathbf{v}=\mathbf{u}\otimes\mathbf{v}-\mathbf{v}\otimes\mathbf{u},\]
 therefore it is written in the index notation as $A^{ij}=u^{i}v^{j}-u^{j}v^{i}$.
Note that the matrix $A^{ij}$ is antisymmetric: $A^{ij}=-A^{ji}$.

Another example: The 3-vector $\mathbf{u}\wedge\mathbf{v}\wedge\mathbf{w}$
can be expanded in the basis as\[
\mathbf{u}\wedge\mathbf{v}\wedge\mathbf{w}=\sum_{i,j,k=1}^{N}B^{ijk}\mathbf{e}_{i}\wedge\mathbf{e}_{j}\wedge\mathbf{e}_{k}.\]
What is the relation between the components $u^{i}$, $v^{i}$, $w^{i}$
of the vectors and the components $B^{ijk}$? A direct calculation
yields\begin{equation}
B^{ijk}=u^{i}v^{j}w^{k}-u^{i}v^{k}w^{j}+u^{k}v^{i}w^{j}-u^{k}w^{j}v^{i}+u^{j}w^{k}v^{i}-u^{j}w^{i}w^{k}.\label{eq:Bijk formula}\end{equation}
In other words, every permutation of the set $\left(i,j,k\right)$
of indices enters with the sign corresponding to the parity of that
permutation. 


\paragraph{Remark:}

Readers familiar with the standard definition of the matrix determinant
will recognize a formula quite similar to the determinant of a $3\times3$
matrix. The connection between determinants and exterior products
will be fully elucidated in Chapter~\ref{sec:Determinants-and-all}.


\paragraph{Remark:}

The {}``three-dimen\-sion\-al array'' $B^{ijk}$ is antisymmetric
with respect to \emph{any} pair of indices: \[
B^{ijk}=-B^{jik}=-B^{ikj}=...\]
Such arrays are called \textbf{totally antisymmetric\index{totally antisymmetric}}.\hfill{}$\blacksquare$

The formula~(\ref{eq:Bijk formula}) for the components $B^{ijk}$
of $\mathbf{u}\wedge\mathbf{v}\wedge\mathbf{w}$ is not particularly
convenient and cannot be easily generalized. We will now rewrite Eq.~(\ref{eq:Bijk formula})
in a different form that will be more suitable for expressing exterior
products of arbitrary tensors.

Let us first consider the exterior product of three vectors as a map
$\hat{E}:V\otimes V\otimes V\rightarrow\wedge^{3}V$. This map is
linear and can be represented, in the index notation, in the following
way:\[
u^{i}v^{j}w^{k}\mapsto\left(\mathbf{u}\wedge\mathbf{v}\wedge\mathbf{w}\right)^{ijk}=\sum_{l,m,n}E_{lmn}^{ijk}u^{l}v^{m}w^{n},\]
where the array $E_{lmn}^{ijk}$ is the component representation of
the map $E$. Comparing with the formula~(\ref{eq:Bijk formula}),
we find that $E_{lmn}^{ijk}$ can be expressed through the Kronecker
$\delta$-symbol as\[
E_{lmn}^{ijk}=\delta_{l}^{i}\delta_{m}^{j}\delta_{n}^{k}-\delta_{l}^{i}\delta_{m}^{k}\delta_{n}^{j}+\delta_{l}^{k}\delta_{m}^{i}\delta_{n}^{j}-\delta_{l}^{k}\delta_{m}^{j}\delta_{n}^{i}+\delta_{l}^{j}\delta_{m}^{k}\delta_{n}^{i}-\delta_{l}^{j}\delta_{m}^{i}\delta_{n}^{k}.\]
 It is now clear that the exterior product of two vectors can be also
written as\[
(\mathbf{u}\wedge\mathbf{v})^{ij}=\sum_{l,m}E_{lm}^{ij}u^{l}v^{m},\]
where\[
E_{lm}^{ij}=\delta_{l}^{i}\delta_{m}^{j}-\delta_{l}^{j}\delta_{m}^{i}.\]
By analogy, the map $\hat{E}:V\otimes...\otimes V\rightarrow\wedge^{n}V$
(for $2\leq n\leq N$) can be represented in the index notation by
the array of components $E_{j_{1}...j_{n}}^{i_{1}...i_{n}}$. This
array is totally antisymmetric with respect to all the indices $\left\{ i_{s}\right\} $
and separately with respect to all $\left\{ j_{s}\right\} $. Using
this array, the exterior product of two general antisymmetric tensors,
say $\phi\in\wedge^{m}V$ and $\psi\in\wedge^{n}V$, such that $m+n\leq N$,
can be represented in the index notation by\[
(\phi\wedge\psi)^{i_{1}...i_{m+n}}=\frac{1}{m!n!}\sum_{(j_{s},k_{s})}E_{j_{1}...j_{m}k_{1}...k_{n}}^{i_{1}...i_{m+n}}\phi^{j_{1}...j_{m}}\psi^{k_{1}...k_{n}}.\]
The combinatorial factor $m!n!$ is needed to compensate for the $m!$
equal terms arising from the summation over $\left(j_{1},...,j_{m}\right)$
due to the fact that $\phi^{j_{1}...j_{m}}$ is totally antisymmetric,
and similarly for the $n!$ equal terms arising from the summation
over $\left(k_{1},...,k_{m}\right)$.

It is useful to have a general formula for the array $E_{j_{1}...j_{n}}^{i_{1}...i_{n}}$.
One way to define it is\[
E_{j_{1}...j_{n}}^{i_{1}...i_{n}}=\begin{cases}
\left(-1\right)^{\left|\sigma\right|} & \text{ if }\left(i_{1},...,i_{n}\right)\text{ is a permutation }\sigma\text{ of }\left(j_{1},...,j_{n}\right);\\
0 & \text{ otherwise}.\end{cases}\]
We will now show how one can express $E_{j_{1}...j_{n}}^{i_{1}...i_{n}}$
through the Levi-Civita symbol $\varepsilon$.

The \textbf{Levi-Civita symbol}\index{Levi-Civita symbol} is defined
as a totally antisymmetric array with $N$ indices, whose values are
$0$ or $\pm1$ according to the formula \[
\varepsilon^{i_{1}...i_{N}}=\begin{cases}
\left(-1\right)^{\left|\sigma\right|} & \text{ if }\left(i_{1},...,i_{N}\right)\text{ is a permutation }\sigma\text{ of }\left(1,...,N\right);\\
0 & \text{otherwise.}\end{cases}\]
Comparing this with the definition of $E_{j_{1}...j_{n}}^{i_{1}...i_{n}}$,
we notice that\[
\varepsilon^{i_{1}...i_{N}}=E_{1...N}^{i_{1}...i_{N}}.\]
Depending on convenience, we may write $\varepsilon$ with upper or
lower indices since $\varepsilon$ is just an array of numbers in
this calculation. 

In order to express $E_{j_{1}...j_{n}}^{i_{1}...i_{n}}$ through $\varepsilon^{i_{1}...i_{N}}$,
we obviously need to use at least two copies of $\varepsilon$ ---
one with upper and one with lower indices. Let us therefore consider
the expression\begin{equation}
\tilde{E}_{j_{1}...j_{n}}^{i_{1}...i_{n}}\equiv\sum_{k_{1},...,k_{N-n}}\varepsilon^{i_{1}...i_{n}k_{1}...k_{N-n}}\varepsilon_{j_{1}...j_{n}k_{1}...k_{N-n}},\label{eq:E tilda def}\end{equation}
where the summation is performed \emph{only} over the $N-n$ indices
$\left\{ k_{s}\right\} $. This expression has $2n$ free indices
$i_{1}$, ..., $i_{n}$ and $j_{1}$, ..., $j_{n}$, and is totally
antisymmetric in these free indices (since $\varepsilon$ is totally
antisymmetric in all indices). 


\paragraph{Statement:}

The exterior product operator $E_{j_{1}...j_{n}}^{i_{1}...i_{n}}$
is expressed through the Levi-Civita symbol as\begin{equation}
E_{j_{1}...j_{n}}^{i_{1}...i_{n}}=\frac{1}{\left(N-n\right)!}\tilde{E}_{j_{1}...j_{n}}^{i_{1}...i_{n}},\label{eq:E def}\end{equation}
where $\tilde{E}$ is defined by Eq.~(\ref{eq:E tilda def}).


\subparagraph{Proof:}

Let us compare the values of $E_{j_{1}...j_{n}}^{i_{1}...i_{n}}$
and $\tilde{E}_{j_{1}...j_{n}}^{i_{1}...i_{n}}$, where the indices
$\left\{ i_{s}\right\} $ and $\left\{ j_{s}\right\} $ have some
fixed values. There are two cases: either the set $\left(i_{1},...,i_{n}\right)$
is a permutation of the set $\left(j_{1},...,j_{n}\right)$; in that
case we may denote this permutation by $\sigma$; or $\left(i_{1},...,i_{n}\right)$
is not a permutation of $\left(j_{1},...,j_{n}\right)$. 

Considering the case when a permutation $\sigma$ brings $\left(j_{1},...,j_{n}\right)$
into $\left(i_{1},...,i_{n}\right)$, we find that the symbols $\varepsilon$
in Eq.~(\ref{eq:E tilda def}) will be nonzero only if the indices
$\left(k_{1},...,k_{N-n}\right)$ are a permutation of the complement
of the set $\left(i_{1},...,i_{n}\right)$. There are $\left(N-n\right)!$
such permutations, each contributing the same value to the sum in
Eq.~(\ref{eq:E tilda def}). Hence, we may write%
\footnote{In the equation below, I have put the warning {}``no sums'' for
clarity: A summation over all repeated indices is often \emph{implicitly}
assumed in the index notation.%
} the sum as \[
\tilde{E}_{j_{1}...j_{n}}^{i_{1}...i_{n}}=\left(N-n\right)!\,\varepsilon^{i_{1}...i_{n}k_{1}...k_{N-n}}\varepsilon_{j_{1}...j_{n}k_{1}...k_{N-n}}\text{ (no sums!)},\]
where the indices $\left\{ k_{s}\right\} $ are chosen such that the
values of $\varepsilon$ are nonzero. Since \[
\sigma\left(j_{1},...,j_{n}\right)=\left(i_{1},...,i_{n}\right),\]
we may permute the first $n$ indices in $\varepsilon_{j_{1}...j_{n}k_{1}...k_{N-n}}$\begin{align*}
\tilde{E}_{j_{1}...j_{n}}^{i_{1}...i_{n}} & =\left(N-n\right)!(-1)^{\left|\sigma\right|}\varepsilon^{i_{1}...i_{n}k_{1}...k_{N-n}}\varepsilon_{i_{1}...i_{n}k_{1}...k_{N-n}}\text{ (no sums!)}\\
 & =\left(N-n\right)!(-1)^{\left|\sigma\right|}.\end{align*}
(In the last line, we replaced the squared $\varepsilon$ by $1$.)
Thus, the required formula for $\tilde{E}$ is valid in the first
case.

In the case when $\sigma$ does not exist, we note that\[
\tilde{E}_{j_{1}...j_{n}}^{i_{1}...i_{n}}=0,\]
because in that case one of the $\varepsilon$'s in Eq.~(\ref{eq:E tilda def})
will have at least some indices equal and thus will be zero. Therefore
$\tilde{E}$ and $E$ are equal to zero for the same sets of indices.\hfill{}$\blacksquare$

Note that the formula for the top exterior power ($n=N$) is simple
and involves no summations and no combinatorial factors:\[
E_{j_{1}...j_{N}}^{i_{1}...i_{N}}=\varepsilon^{i_{1}...i_{N}}\varepsilon_{j_{1}...j_{N}}.\]



\paragraph{Exercise:}

The operator $\hat{E}:V\otimes V\otimes V\rightarrow\wedge^{3}V$
can be considered within the subspace $\wedge^{3}V\subset V\otimes V\otimes V$,
which yields an operator $\hat{E}:\wedge^{3}V\rightarrow\wedge^{3}V$.
Show that in this subspace,\[
\hat{E}=3!\,\hat{1}_{\wedge^{3}V}.\]
Generalize to $\wedge^{n}V$ in the natural way. 

\emph{Hint}: Act with $\hat{E}$ on $\mathbf{a}\wedge\mathbf{b}\wedge\mathbf{c}$.


\paragraph{Remark:}

As a rule, a summation of the Levi-Civita symbol $\varepsilon$ with
any antisymmetric tensor (e.g.~another $\varepsilon$) gives rise
to a combinatorial factor $n!$ when the summation goes over $n$
indices.


\subsection{{*} Exterior algebra (Grassmann algebra)}

The formalism of exterior algebra is used e.g.~in physical theories
of quantum fermionic fields and supersymmetry.


\paragraph{Definition:}

An \textbf{algebra}\index{algebra} is a vector space with a distributive
multiplication. In other words, ${\cal A}$ is an algebra if it is
a vector space over a field $\mathbb{K}$ and if for any $a,b\in{\cal A}$
their product $ab\in{\cal A}$ is defined, such that $a\left(b+c\right)=ab+ac$
and $\left(a+b\right)c=ac+bc$ and $\lambda\left(ab\right)=\left(\lambda a\right)b=a\left(\lambda b\right)$
for $\lambda\in\mathbb{K}$. An algebra is called \textbf{commutative}
if $ab=ba$ for all $a,b$. 

The properties of the multiplication in an algebra can be summarized
by saying that for any fixed element $a\in{\cal A}$, the transformations
$x\mapsto ax$ and $x\mapsto xa$ are linear maps of the algebra into
itself.


\paragraph{Examples of algebras:}
\begin{enumerate}
\item All $N\times N$ matrices with coefficients from $\mathbb{K}$ are
a $N^{2}$-dimen\-sion\-al algebra. The multiplication is defined
by the usual matrix multiplication formula. This algebra is not commutative
because not all matrices commute. 
\item The field $\mathbb{K}$ is a one-dimen\-sion\-al algebra over itself.
(Not a very exciting example.) This algebra is commutative.
\end{enumerate}

\paragraph{Statement:}

If $\omega\in\wedge^{m}V$ then we can define the map $L_{\omega}:\wedge^{k}V\rightarrow\wedge^{k+m}V$
by the formula\[
L_{\omega}\left(\mathbf{v}_{1}\wedge...\wedge\mathbf{v}_{k}\right)\equiv\omega\wedge\mathbf{v}_{1}\wedge...\wedge\mathbf{v}_{k}.\]
For elements of $\wedge^{0}V\equiv\mathbb{K}$, we define $L_{\lambda}\omega\equiv\lambda\omega$
and also $L_{\omega}\lambda\equiv\lambda\omega$ for any $\omega\in\wedge^{k}V$,
$\lambda\in\mathbb{K}$. Then the map $L_{\omega}$ is linear for
any $\omega\in\wedge^{m}V$, $0\leq m\leq N$.


\subparagraph{Proof: }

Left as exercise.\hfill{}$\blacksquare$


\paragraph{Definition: }

The \textbf{exterior algebra}\index{exterior algebra} (also called
the \textbf{Grassmann algebra}\index{Grassmann algebra}) based on
a vector space $V$ is the space $\wedge V$ defined as the direct
sum, \[
\wedge V\equiv\mathbb{K}\oplus V\oplus\wedge^{2}V\oplus...\oplus\wedge^{N}V,\]
with the multiplication defined by the map $L$, which is extended
to the whole of $\wedge V$ by linearity.

For example, if $\mathbf{u},\mathbf{v}\in V$ then $1+\mathbf{u}\in\wedge V$,\[
A\equiv3-\mathbf{v}+\mathbf{u}-2\mathbf{v}\wedge\mathbf{u}\in\wedge V,\]
and\[
L_{1+\mathbf{u}}A=\left(1+\mathbf{u}\right)\wedge\left(3-\mathbf{v}+\mathbf{u}-2\mathbf{v}\wedge\mathbf{u}\right)=3-\mathbf{v}+4\mathbf{u}-\mathbf{v}\wedge\mathbf{u}.\]
Note that we still write the symbol $\wedge$ to denote multiplication
in $\wedge V$ although now it is not necessarily anticommutative;
for instance, $1\wedge x=x\wedge1=x$ for any $x$ in this algebra.


\paragraph{Remark: }

The summation in expressions such as $1+\mathbf{u}$ above is \emph{formal}
in the usual sense: $1+\mathbf{u}$ is not a new vector or a new tensor,
but an element of a \emph{new} \emph{space}. The exterior algebra
is thus the space of formal linear combinations of numbers, vectors,
2-vectors, etc., all the way to $N$-vectors.\hfill{}$\blacksquare$

Since $\wedge V$ is a direct sum of $\wedge^{0}V$, $\wedge^{1}V$,
etc., the elements of $\wedge V$ are sums of scalars, vectors, bivectors,
etc., i.e.~of objects having a definite {}``grade'' --- scalars
being {}``of grade'' 0, vectors of grade 1, and generally $k$-vectors
being of grade $k$. It is easy to see that $k$-vectors and $l$-vectors
either commute or anticommute, for instance\begin{align*}
\left(\mathbf{a}\wedge\mathbf{b}\right)\wedge\mathbf{c} & =\mathbf{c}\wedge\left(\mathbf{a}\wedge\mathbf{b}\right),\\
\left(\mathbf{a}\wedge\mathbf{b}\wedge\mathbf{c}\right)\wedge1 & =1\wedge\left(\mathbf{a}\wedge\mathbf{b}\wedge\mathbf{c}\right),\\
\left(\mathbf{a}\wedge\mathbf{b}\wedge\mathbf{c}\right)\wedge\mathbf{d} & =-\mathbf{d}\wedge\left(\mathbf{a}\wedge\mathbf{b}\wedge\mathbf{c}\right).\end{align*}
The general law of commutation and anticommutation can be written
as \[
\omega_{k}\wedge\omega_{l}=\left(-1\right)^{kl}\omega_{l}\wedge\omega_{k},\]
where $\omega_{k}\in\wedge^{k}V$ and $\omega_{l}\in\wedge^{l}V$.
However, it is important to note that sums of elements having different
grades, such as $1+\mathbf{a}$, are elements of $\wedge V$ that
do \emph{not} have a definite grade, because they do not belong to
any single subspace $\wedge^{k}V\subset\wedge V$. Elements that do
not have a definite grade can of course still be multiplied within
$\wedge V$, but they \emph{neither} commute \emph{nor} anticommute,
for example:\begin{align*}
\left(1+\mathbf{a}\right)\wedge\left(1+\mathbf{b}\right) & =1+\mathbf{a}+\mathbf{b}+\mathbf{a}\wedge\mathbf{b},\\
\left(1+\mathbf{b}\right)\wedge\left(1+\mathbf{a}\right) & =1+\mathbf{a}+\mathbf{b}-\mathbf{a}\wedge\mathbf{b}.\end{align*}
So $\wedge V$ is a \emph{noncommutative} (but associative) algebra.
Nevertheless, the fact that elements of $\wedge V$ having a pure
grade either commute or anticommute is important, so this kind of
algebra is called a \textbf{graded algebra}\index{graded algebra}.


\paragraph{Exercise 1:}

Compute the dimension of the algebra $\wedge V$ as a vector space,
if $\dim V=N$.


\subparagraph{Answer: }

$\dim\left(\wedge V\right)=\sum_{i=0}^{N}{N \choose i}=2^{N}$.


\paragraph{Exercise 2:}

Suppose that an element $x\in\wedge V$ is a sum of elements of \emph{pure
even} grade, e.g.~$x=1+\mathbf{a}\wedge\mathbf{b}$. Show that $x$
commutes with any other element of $\wedge V$. 


\paragraph{Exercise 3:}

Compute $\exp\left(\mathbf{a}\right)$ and $\exp\left(\mathbf{a}\wedge\mathbf{b}+\mathbf{c}\wedge\mathbf{d}\right)$
by writing the Taylor series using the multiplication within the algebra
$\wedge V$.

\emph{Hint}: Simplify the expression $\exp(x)=1+x+\frac{1}{2}x\wedge x+...$
for the particular $x$ as given.


\subparagraph{Answer: }

$\exp\left(\mathbf{a}\right)=1+\mathbf{a}$; \[
\exp\left(\mathbf{a}\wedge\mathbf{b}+\mathbf{c}\wedge\mathbf{d}\right)=1+\mathbf{a}\wedge\mathbf{b}+\mathbf{c}\wedge\mathbf{d}+\mathbf{a}\wedge\mathbf{b}\wedge\mathbf{c}\wedge\mathbf{d}.\]