Show that $$\omega \wedge v(\left <a_1,a_2,a_3 \right>,\left <b_1,b_2,b_3 \right>) = c_1 dx\wedge dy + c_2 dx\wedge dz + c_3 dy \wedge dz.$$
I wasn't given the form of $\omega$ or $v$. Furthermore, to my knowledge we can only compute the wedge product of a vector right? How come the RHS doesn't evaluate a vector? For instance, shouldn't we get $c_1 dx\wedge dy (vector)$ on the RHS? I am using bachman's book (the book is free on arvix).
So I assume we can let $\omega(a_1,a_2,a_3) = \omega_1a_1 + \omega_2 a_2 + \omega_3 a_3$ and similar definition for $v.$ Should I just group all the $\omega_ia_i, \omega_ib_i, v_ia_i,v_ib_i$ terms together and call that the mixed constants in $a_ib_i$ as $c_i$?
The book can be found here http://arxiv.org/abs/math/0306194 which is made free and the question is found on page 25 in the pdf.
We're given that $\omega,\nu \in \Omega^1( \mathbb{R}^3) $, so we can treat these guys as vectors that take the form ( with basis $dx,dy,dz$ and $a,b,c, s,t,u \in \mathbb{R}$):
$$ \omega = a dx + b dy + c dz \quad \& \quad \nu = s dx + t dy + u dz $$
The exercise says to compute $ \omega \wedge \nu$, so lets do that
\begin{align*} \omega \wedge \nu =& ( a dx + b dy + c dz ) \wedge (s dx + t dy + u dz) \\ =& a t \, dx \wedge dy + b s \, dy \wedge dx + au \, dx \wedge dz + c s \, dz \wedge dx + \ldots \\ = & (at - bs ) dx \wedge dy + (au - cs) dx \wedge dz + (bu-tc) dy \wedge dz \\ =& c_1 dx \wedge dy + c_2 dx \wedge dz + c_3 dy \wedge dz \end{align*}
We used the fact that $dx \wedge dx = 0$.Also! In general the the wedge product acts on differential forms not just vectors but in this case we can identify a 3-vector on the RHS by using the basis $dx\wedge dy, dy \wedge dz, dz \wedge dx$