Derivation of Green's theorem - I have wrong negative sign

Question

I was making my way through Bressoud's 2nd Year Calculus when I got to a derivation of Green's Theorem.

From the book:

If $\omega(\vec{x})$ is a 1-form in $R^2$, $\omega = f \, dx + g \, dy$ and $M$ is a region in the $x,y$ plane with a positive orientation, then $\partial M$ is the closed curve running around $M$ in a counter-clockwise direction and

$d \omega = (\frac{\partial g}{\partial x} - \frac{\partial f}{\partial y}) \, dx\, dy$

It is this derivative I have a problem with deriving.

I would differntiate it as:

$d \omega = \frac{\partial}{\partial x}(f \, dx + g \, dy) \, dx + \frac{\partial}{\partial y}(f \, dx + g \, dy) \, dy$

= ($\frac{\partial f}{\partial x} \, dx + \frac{\partial g}{\partial x} \, dy) \, dx + (\frac{\partial f}{\partial y} \, dx + \frac{\partial g}{\partial y} \, dy) \, dy$

Due to "exterior algebra", $dx \, dx = 0$ and $dy \, dy = 0$, leaving

$d \omega = \frac{\partial g}{\partial x} dy \, dx + \frac{\partial f}{\partial y} dx \, dy$

And, again, due to exterior algebra, $dy \, dx = - dx \, dy$, thus

$d \omega = (\frac{\partial f}{\partial y} - \frac{\partial g}{\partial x}) \, dx \, dy$

But this is wrong by a negative sign. Bressoud has it as:

$d \omega = (\frac{\partial g}{\partial x} - \frac{\partial f}{\partial y}) \, dx \, dy$

Where is my mistake?

Answer 1

You should have: \begin{align} d\omega &=d(f\;dx+g\;dy) \\ &= d(f\;dx) + d(g\;dy) \\ &=\left(\frac{\partial f}{\partial x}\;dx\wedge dx + \frac{\partial f}{\partial y}\;dy\wedge dx\right) + \left(\frac{\partial g}{\partial x}\;dx\wedge dy + \frac{\partial g}{\partial y}\;dy\wedge dy\right) \\ &= \frac{\partial f}{\partial y}\;dy\wedge dx + \frac{\partial g}{\partial x} \;dx\wedge dy \\ &= -\frac{\partial f}{\partial y}\;dx\wedge dy + \frac{\partial g}{\partial x} \;dx\wedge dy \\ &= \left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)\;dx\;dy \end{align}