\documentclass{amsart}
%\usepackage{kbordermatrix}

\newtheorem*{defn}{Defintion}
\newtheorem*{prop}{Proposition}
\newtheorem*{thm}{Theorem}
\newtheorem*{cor}{Corollary}
%\theoremstyle{definition}
\newtheorem*{defin}{Definition}
\newtheorem*{example}{Example}
\newtheorem*{remark}{Remark}

\newcommand{\F}{\mathbb F}
\renewcommand{\L}{\mathcal L}
\newcommand{\bilin}[1]{\langle#1\rangle}

\title{Lecture 19: Exterior Products II}

\begin{document}
\maketitle

We ended last time looking at a basis for the exterior powers. Today we will finish with that and describe the symmetric power. First we want a generalization of the Reisz representation theorem.

\begin{prop}
If $f\colon U\otimes V\to W$ is bilinear, then we have canonical maps $U\to\L(V,W)$ and $V\to L(U,W)$ given by
\[
\vec{u}\mapsto f(\vec{u},-)\text{ and }\vec{v}\mapsto f(-,\vec{v}).
\]
\end{prop}

Now if $W$ is $1$-dimensional, then $\L(V,W)\cong V^\ast$. Additionally, if for all $0\neq \vec{u}\in U$, we have a $\vec{v}\in V$ such that $f(\vec{u},\vec{v})\neq 0$, then our map is actually an injection $U\to V^\ast$. Similarly, if for every $0\neq\vec{v}\in V$ we have a $\vec{u}\in U$ such that $f(\vec{u},\vec{v})\neq 0$, then $V\to U^\ast$ is an injection. Thus if both conditions are satisfied, and if everything in sight is finite dimensional, then our bilinear function $f$ establishes isomorphisms $U\to V^\ast$ and $V\to U^\ast$. This is the case with exterior products, as we shall soon see.

If $f$ establishes an isomorphism $U\to V\ast$ and $V\to U^\ast$, then we say that $f$ is a \underline{perfect pairing}.

Let's assume for now that $\Lambda^m(V)$ is one dimensional. Our earlier argument shows that it is at most $1$-dimensional. In fact, if we know that this is $1$-dimensional, then we can prove that the listed elements are all linearly independent, as we'll soon see.

\begin{prop}
The canonical map 
\[
\Lambda^i(V)\otimes\Lambda^{n-i}(V)\to\Lambda^n(V)
\]
given by $\vec{v}\otimes\vec{w}\mapsto\vec{v}\wedge\vec{w}$, is a perfect paring.
\end{prop}

Before continuing, we spend a little time describing the map. Let $I_i$ denote the subspace of $V^{\otimes i}$ spanned by vectors of the form 
$\vec{v}_1\otimes\dots\vec{v}_i\otimes\dots\vec{v}_i\otimes\dots\vec{v}_n$. We'll use the canonical identification $V^{\otimes i}\otimes V^{\otimes (n-i)}$ with $V^{\otimes n}$ and henceforth blur the distinction between the two. This gives us subspaces $I_i\otimes V^{\otimes (n-i)}$ and $V^{\otimes i}\otimes I_{n-i}$ of $V^{\otimes n}$. It is immediate that
\[
\Lambda^{i}(V)\otimes\Lambda^{n-i}(V)\cong V^{\otimes n}/(I_i\otimes V^{\otimes(n-i)}+V^{\otimes i}\otimes I_{n-i}),
\]
and the map realizing this is the tensor product of the two projection maps 
\[
V^{\otimes i}\to\Lambda^i(V)\text{ and }V^{\otimes(n-i)}\to\Lambda^{n-i}(V).
\]

Since $(I_i\otimes V^{\otimes(n-i)}+V^{\otimes i}\otimes I_{n-i})$ is visibly a subspace of $I_n$, we conclude from the defining property of the quotient that there is a map
\[
\Lambda^i(V)\otimes\Lambda^{n-i}(V)\to\Lambda^n(V).
\]
This is our map above.

We can spell out a little more what this means and why we would expect such a thing. The object $\Lambda^i(V)$ represents ``alternating, multilinear maps from $V^i$'':
\[
\L(\Lambda^i(V),W)=\{f\colon V^{\times i}\to W | f\text{ is alternating, multilinear}\}.
\]
Tensor product has a similar property. Thus $\L(\Lambda^i(V)\otimes\Lambda^{n-i}(V),W)$ is the set of multilinear maps $V^{\times i}\times V^{\times (n-i)}\cong V^{\times n}$ to $W$ that are alternating in the first $i$ variables and alternating in the last $(n-i)$-variables. On the other hand, if we are given a function $V^{\times n}\to W$ that is multilinear and alternating in all variables, then in particular, it is multilinear and alternating in the first $i$ and last $(n-i)$ variables. Thus we see that for all $W$,
\[
\L(\Lambda^n(V),W)\subset\L(\Lambda^i(V)\otimes\Lambda^{n-i}(V),W).
\]
We actually know more. This is ``natural'' in $W$. If you give a map $W\to W'$, then composing with this map gives a natural square with inclusions. That alone tells us for purely formal reasons that we have a linear map
\[
\Lambda^i(V)\otimes\Lambda^{n-i}(V)\to\Lambda^n(V),
\]
precomposition by which is realizes the inclusions (and in fact, the ``inclusions'' part guarantees that this map is a surjection). Where does it come from? If we let $W=\Lambda^{n}(V)$, then we have the identity map in 
\[
\L\big(\Lambda^{n}(V),\Lambda^n(V)\big)\subset\L\big(\Lambda^{i}(V)\otimes\Lambda^{n-i}(V),\Lambda^n(V)\big).
\]
The image of the identity map is again our map.

Now back to the proof.

\begin{proof}
We must show that for each $\vec{v}\in\Lambda^i(V)$, there is a $\vec{v}'\in\Lambda^{n-i}(V)$ such that $\vec{v}\wedge\vec{v}'\neq 0$. Interchanging the roles of $i$ and $n-i$ will then show the other direction too. We start with our spanning set, and make a few minidefinitions.

A small bit of notation. Let $I\subset [n]=\{1,\dots,n\}$ be a set of cardinality $i$, and let $I^c$ denote the complement. For a subset $J$ of $[n]$, let
\[
\vec{v}_J=\bigwedge_{j\in J}\vec{v}_{j}.
\]
Then the following is an easy exercise in alternatingness: If $|I|+|J|=n$, then
\[
\vec{v}_I\wedge\vec{v}_J=\begin{cases}
0 & J\neq I^c \\
\pm \vec{v}_M & J=I^c.
\end{cases}
\]

Now since $\vec{v}_I$ spans where $I$ ranges over the subsets of cardinality $i$, we know that any $\vec{v}$ can be written as a linear combination 
\[
\vec{v}=a_{I_1}\vec{v}_{I_1}+\dots+a_{I_m}\vec{v}_{I_m},
\]
where without loss of generality, we may assume $a_{I_1}\neq 0$. Then by the previous result, if we let $\vec{v}'=\vec{v}_{I_1^c}$, then
\[
\vec{v}\wedge\vec{v}'=\pm a_{I_1}\vec{v}_{[n]}\neq 0,
\]
as required.
\end{proof}

As an easy corollary, we conclude that our spanning sets are also linearly independent.
\begin{prop}
If we assume $\dim\Lambda^m(V)=1$, then
\[
\{\vec{v}_{i_1}\wedge\dots\wedge\vec{v}_{i_n} | i_1<\dots<i_n\}
\]
is a basis for $\Lambda^n(V)$.
\end{prop}
\begin{proof}

Now consider a linear dependence relation
\[
a_{I_1}\vec{v}_{I_1}+\dots+a_{I_k}\vec{v}_{I_k}=0.
\]
If we wedge this with $\vec{v}_{I_1^c}$, then the above observation shows us that all terms except the first drop out. Now since $
\vec{v}_M$ spans $\Lambda^m(V)$ and since this is a $1$-dimensional space, we know $\vec{v}_M\neq\vec{0}$. Thus our dependence relation becomes
\[
a_{I_1}\vec{v}_M=0,
\]
which means $a_{I_1}=0$. Induction on $k$ gives the result.
\end{proof}

As a consequence, we see that wedge product map $\Lambda^n(V)\otimes\Lambda^{m-n}(V)\to\Lambda^{m}(V)$ establishes an isomorphism between $\Lambda^m(V)$ and $\Lambda^{n-m}(V)$ as described above. In fact, we have an even better result: the basis dual to $\{\vec{v}_I\}$ is the basis $\{\vec{v}_{I^c}\}$. Thus we recover a very classical result:
\[
\binom{m}{n}=\dim\Lambda^n(V)=\dim\Lambda^{m-n}(V)=\binom{m}{m-n}.
\]

To complete the proof in the exterior case, we have to produce a single non-zero, alternating, multilinear function on $m$-tuples of vectors in $V$. This will show that $\Lambda^m(V)\neq 0$, since if it were zero, we'd know that all such functions are zero (as they factor through the zero space). $V$ is an $m$-dimensional space, so our choice of basis identifies this with $\F^m$. Now we will be done if we can find a single non-zero alternating multilinear function on $m$-tuples on column vectors. Of course, an $m$-tuple of $m$-dimensional column vectors is the same thing as an $m\times m$-matrix, and this gives us a great example.

\begin{prop}
The determinant, viewed as a function from $m$-tuples of $m$-dimensional column vectors, is alternating, multilinear, and non-zero.
\end{prop}

\begin{cor}
The vector space $\Lambda^m(V)$ is one dimensional.
\end{cor}

\end{document} 