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Let $\omega = P\ dx + Q\ dy$ be a 1-form on $\mathbb{R}^2$. Also, define a 0-form $I\omega({\bf x}) = I\omega(x, y)$ by

$$ I\omega({\bf x}) = \int_0^1 P(t {\bf x}) x + Q(t {\bf x}) y\ dt.$$

I would like to show that $\omega = d(I\omega)$, provided that $d\omega = 0$ (here, $d$ is the exterior derivative).


Here is my approach: $I\omega$ is a 0-form, therefore we get

$$d(I\omega) = D_1(I\omega) dx + D_2(I\omega) dy = \frac{\partial (I\omega)}{\partial x} dx + \frac{\partial (I\omega)}{\partial y} dy.$$

Since this is supposed to equal $\omega = P\ dx + Q\ dy$, it seems that it requires $P = D_1(I\omega)$ and $Q = D_2(I\omega)$. Also, since $d\omega = 0$ implies $D_1Q = D_2P$, we get

$$ \begin{eqnarray} D_1(I\omega)({\bf x}) & = & \int_0^1 \frac{\partial}{\partial x} (P(t {\bf x}) x + Q(t {\bf x})) y\ dt \\ & = & \int_0^1 D_1P(t{\bf x}) t x + P(t{\bf x}) + D_1Q(t{\bf x}) t y\ dt \\ & = & \int_0^1 D_1P(t{\bf x}) t x + P(t{\bf x}) + D_2P(t{\bf x}) t y\ dt \end{eqnarray} $$

Now, this is where I got stuck, because I fail to see how this expression should be equal to $P$?

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1 Answer 1

up vote 2 down vote accepted

$(x,y)$ is fixed; define $g(t):=P(tx,ty)$. Then $$\frac d{dt}g(t)=xD_1P(tx,ty)+yD_2P(tx,ty),$$ so $$D_1(I\omega)(x,y)=\int_0^1tg'(t)+g(t)dt=[tg(t)]_0^1-\int_0^1g(t)dt+\int_0^1g(t)dt=g(1).$$

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Davide, are you sure this is right? I feel you might missed a factor. –  Bombyx mori Sep 17 '12 at 21:11
    
I think it's right, but I'm not sure. Which factor do you think is missing? –  Davide Giraudo Sep 18 '12 at 8:02
    
Very nice, thanks. –  koletenbert Sep 19 '12 at 20:33
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