I came across this idea at the very beginning of a book and I don't quite seem to grasp it. It states given $\omega \in \bigwedge ^2 (V) $ you can define a linear map $ \omega^\#: V \to V^* $ by $\omega(v, \cdot)$ .
My first question is just a verification. For example, if $\dim(V) = 2$ then $\omega = a e^1 \wedge e^2$. If we take a vector $v = v_1 e_1 + v_2 e_2$ would $\omega (v, \cdot) = v_1 e^2 - v_2 e^1 $ ?
While this seems all well and good, it continues on and states that it wishes to show that $\omega = dx \wedge dy$ is non-degenerate. It does this by saying to considered $$\omega ^\# \left ( \frac{\partial }{\partial x} \right ) = \iota \left ( \frac{\partial }{\partial x} \right ) (dx \wedge dy) = dy $$
First, I am not certain where the iota came from or if it was a typo.
Assuming $\frac{\partial}{\partial x} = \frac{\partial}{\partial x} e_1$, then from my first question, it would follow that $\left ( \frac{\partial }{\partial x} \right ) (dx \wedge dy) = \iota \frac{\partial }{\partial x}dy$. Why is this simply $dy$ or is my approach simply wrong?
Thanks again for the help.
The symbol $\iota $ is a common notation for the insertion of a vector fields. It maps $k$-forms to $k-1$-forms having the eating the vector, that is \begin{equation} \iota_v(\omega)(u_1, \ldots, u_{k-1})=\omega(v,u_1,\ldots,u_{k-1}) \end{equation} Different conventions are around and sometimes the vector is fed into the last argument. So in your case \begin{equation} \iota_{\partial/\partial x}\ \omega=\omega(\partial/\partial x, \cdot)=(\mathrm{d}x\wedge\mathrm{d}y)(\partial/\partial x,\cdot)=\mathrm{d}y \end{equation}