I'm working on an exercise where I have to calculate the wedge product of differential forms with vectors and I feel like I'm missing something very basic.
The question is asking to calculate $(\phi_1\wedge dx_2+\phi_2\wedge\phi_3)(u,v)$. (All of these are given of course.)
So far I've developed $(\phi_1\wedge dx_2+\phi_2\wedge\phi_3)$ using the properties of wedge product that we saw in class, but what does it mean to calculate it for $u$ and $v$? What exactly am I calculating here?
(For reference $u=(1,0,2,-1)$, $v=(3,2,1,0)$ and I developed $(\phi_1\wedge dx_2+\phi_2\wedge\phi_3)=dx_2\wedge dx_1+3dx_4\wedge dx_1 +9dx_4\wedge dx_3$.)
I'm assuming, since you have a $dx_1,\dots$ basis, that $u=(1,0,2,-1)$ is written in the corresponding $\partial_{x_1}$ basis. So $u=1\partial_{x_1} + 2\partial_{x_3} -1\partial_{x_4}$, and similarly for $v$.
Do you know the following rule for how a 2-form acts on 2 vectors? $$(\alpha\wedge\beta)(X,Y) = \alpha(X)\beta(Y)-\beta(X)\alpha(Y)$$ i.e. $\alpha\wedge\beta = \alpha\otimes\beta - \beta\otimes\alpha$, if that makes any sense to you (I don't know how exactly your class chose to define the wedge product).
Now you can use $dx_1(\partial_{x_1})= 1$ and $dx_1(\partial_{x_j})= 0$ if $j\neq 1$, and use the linearity of everything.