Question about notation in differential forms. If you have
$\Omega = dx\wedge dy$
what does
$\Omega(x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}, \frac{\partial}{\partial x})$
mean exactly?
I know that $\frac{\partial}{\partial x}$ and so forth are the basis. So then you have $d(x+y)\wedge d(1)$. That is not right, is it?
How do you actually evaluate this?
 A: This goes back to the definitions:


*

*a two form is a skew symmetric bilinear mapping $T_pM \times T_pM \to \mathbb R$, 

*$dx$ is a linear mapping so that $dx\left( \frac{\partial}{\partial x}\right) = 1$, $dx\left( \frac{\partial}{\partial y}\right) = 0$ (similar for $dy$), and 

*the wedge product of two one forms $\alpha, \beta$ is defined by 
$$ \alpha \wedge \beta (X, Y) = \alpha(X)\beta(Y) - \alpha(Y)\beta(X).$$
So we have 
$$\begin{split}
dx\wedge dy \left( x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}, \frac{\partial}{\partial x}\right) &\overset{1.}{=}dx\wedge dy \left( x\frac{\partial}{\partial x}, \frac{\partial}{\partial x}\right) + dx\wedge dy \left( y\frac{\partial}{\partial y}, \frac{\partial}{\partial x}\right) \\
&\overset{3.}{=} dx\wedge dy \left(y\frac{\partial}{\partial y}, \frac{\partial}{\partial x}\right) \\
&\overset{1.}{=} y \ dx\wedge dy \left(\frac{\partial}{\partial y}, \frac{\partial}{\partial x}\right) \\
&\overset{2., 3.}{=}-y
\end{split}$$
