Wedge product and determinants I am attempting to self-study differential forms this summer, but I ran into this definition for the wedge product in my book, and it doesn't make any sense to me. Note that $\Bbb R^3_p$ is the tangent space of $p$ in $\Bbb R^3$, and $(\Bbb R^3_p)^{\ast}$ is the space of linear functionals from $\Bbb R^3_p \to \Bbb R$.
Is the determinant being referred to here the same as the determinant used in linear algebra? If so, isn't it in most cases a multilinear function? I'm sorry if this is unclear, I'm just very lost...

 A: Yes, it's the same determinant.  The notation
$$(\phi_1 \wedge \phi_2)(v_1, v_2) = \det(\phi_i(v_j))$$
means that the $2$-form $\phi_1 \wedge \phi_2$ (which is, as defined, a function of two vector variables) has as its value on the pair $(v_1, v_2)$ the number
$$\det\begin{pmatrix} \phi_1(v_1) & \phi_1(v_2) \\ \phi_2(v_2) & \phi_2(v_2) \end{pmatrix}.$$
The notation "$\phi_i(v_j)$" is just shorthand for "the matrix with this number as its $(i,j)$ entry."  That's an alternating bilinear function of its rows (and thus of the $\phi_i$'s) as well as of its columns (and thus of the $v_j$'s).  So this formula defines an alternating bilinear function of two variables $\phi_1 \wedge \phi_2$, hence a 2-form.  In addition, this construction has the property that, as a function of $\phi_1$ and $\phi_2$, it is bilinear and alternating.  This is equivalent to the distributivity property:
$$(\phi_1 + \psi) \wedge \phi_2 = \phi_1 \wedge \phi_2 + \psi \wedge \phi_2$$
as well as the anticommutativity property:
$$\phi_1 \wedge \phi_2 = -\phi_2 \wedge \phi_1.$$
A: If you write it out you get, 
$$(\varphi_1 \wedge \varphi_2)(v_1,v_2)=
\begin{vmatrix}
\varphi_1(v_1)&\varphi_1(v_2)\\
\varphi_2(v_1)&\varphi_2(v_2)\\
\end{vmatrix} =\varphi_1(v_1)\varphi_2(v_2)-\varphi_1(v_2)\varphi_2(v_1)$$
then you can check that this is a multilinear functional.
