Exterior Product of Two Vectors 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!
 A: 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 $\binom{3}{2} = 3$.
