Question about notation in differential forms.

Question

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?

Answer 1

This goes back to the definitions:

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

2. $dx$ 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

3. 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

$$\begin{split} dx\wedge dy \left( x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}, \frac{\partial}{\partial x}\right) &\overset{1.}{=}dx\wedge dy \left( x\frac{\partial}{\partial x}, \frac{\partial}{\partial x}\right) + dx\wedge dy \left( y\frac{\partial}{\partial y}, \frac{\partial}{\partial x}\right) \\ &\overset{3.}{=} dx\wedge dy \left(y\frac{\partial}{\partial y}, \frac{\partial}{\partial x}\right) \\ &\overset{1.}{=} y \ dx\wedge dy \left(\frac{\partial}{\partial y}, \frac{\partial}{\partial x}\right) \\ &\overset{2., 3.}{=}-y \end{split}$$