I am currently studying minimal surface theory from Robert Osserman's text Survey of Minimal Surfaces. I am hung up on the following definiton:
Let $\vec{v}=(v_{1},v_{2},\ldots, v_{n})$ and $\vec{w}=(w_{1},w_{2},\ldots, w_{n})$ be two vectors in $\mathbf{R}^{2}$. Then we denote the exterior product by
$\vec{v}\wedge \vec{w}\:;\:\vec{v}\wedge\vec{w}\in \mathbf{R}^{N},\:N=\left(\begin{array}{cc}
n\\
2
\end{array}\right)$.
What exactly is the definitions of the exterior product? Is it in any way related to the cross product of vectors? Thanks in advance for any help!
Best Answer
The exterior algebra can be defined by a universal property that any alternating, $k$-multilinear map $V^k \to W$ factors through a linear map from $\wedge^k V \to W$. But that doesn't really tell you what it is.
Start with a vector space $V$, and install a skew-commutative multiplication $\wedge$ that distributes over addition. So $\wedge^2V$ is generated by expressions of the form $v\wedge w$, with relations like $w\wedge v = -v \wedge w$, $(u+v)\wedge w = u \wedge w + v\wedge w$, and $(\alpha v)\wedge w = \alpha (v \wedge w) = v\wedge(\alpha w)$.
This is still a vector space. What is its dimension? If $v_1, \dots, v_n$ is a basis for $V$, then $\left\{v_i \wedge v_j \mid i,j=1,\dots,n\right\}$ span $\wedge^2 V$. But this is too many vectors to form a basis. Notice $v_1 \wedge v_1 = 0$ by the skew-commutativity, and $v_1 \wedge v_2 = -v_2 \wedge v_1$. So the number of linearly independent $v_i \wedge v_j$ is $(n^2 - n)/2$, one for each distinct pair of $i$ and $j$. And, $$ \frac{n^2-n}{2} = \frac{n(n-1)}{2} = \binom{n}{2} $$
In general, $\wedge^k V$ is generated by expressions of the form $v_1 \wedge v_2 \wedge \dots \wedge v_k$, where each $v_i$ is in $V$. The skew-commutativity relations are \begin{align*} v_1 \wedge \dots \wedge v_{i-1} \wedge v_j \wedge v_{i+1} \wedge\dots v_{j-1} \wedge v_i \wedge v_{j+1} \wedge \dots \wedge v_n &= - v_1 \wedge v_2 \wedge \dots \wedge v_n \end{align*} for any $i$ and $j$. (Basically, swapping any pair of vectors in the wedge product costs a minus sign.) If $v_1,\dots v_n$ are a basis, then $\wedge^k V$ has as a basis the expression $v_{i_1} \wedge v_{i_2} \wedge \dots \wedge v_{i_k}$, where $1 \leq i_1 < i_2 < \dots < i_k \leq n$ form an increasing $k$-tuple from $\{1,2,\dots,n\}$. Thus $$ \dim \wedge^k V = \binom{\dim V}{k} $$
In the case that $n=3$ and $k=2$, we see that $\wedge^2 \mathbb{R}^3$ has a basis $e_1 \wedge e_2$, $e_2 \wedge e_3$, and $e_3 \wedge e_1$. So $\wedge^2 \mathbb{R}^3$ is isomorphic to $\mathbb{R}^3$. The isomorphism $e_1 \wedge e_2 \mapsto e_3$, $e_2 \wedge e_3 \mapsto e_1$, $e_3 \wedge e_1 \mapsto e_2$ is known as the Hodge star, and carries the wedge product to the cross product. This isomorphism, as well as the fact that the curl of a vector field in $\mathbb{R}^3$ is a vector field, is a coincidence due to the fact that $\binom{3}{2} = 3$.