# how prove $\rho\wedge d\rho=0$ and how to show if $d(f\rho)=0$ for $f$ on $\Bbb R^{n}$ then $\rho\wedge d\rho=0$

$\def\d{\mathrm{d}}\def\R{\mathbb{R}}$Let $ρ$ be a $1$-form on $\R^{n}$ then

Firstly how to prove $\rho\wedge \d\rho=0$

Secondly, how to show that If $\d(f\rho)=0$ for some nowhere vanishing smooth function $f$ on $\R^{n}$ then $\rho\wedge \d\rho=0$

Please give an idea and solution to prove both of these two cases? Thank you.

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Do you know any rules for expanding $\d(f\rho)$? If so, what happens when you take the wedge product with $\rho$? – Chris Taylor Mar 18 '13 at 15:18
@ChrisTaylor I know how to take wedge product. But I dont know (or understand) how to expand $d(f\rho)$? Please explain. Thank you. – B11b Mar 18 '13 at 15:36
Let $\rho = x \mathrm{d}y + \mathrm{d}z$, $\mathrm{d}\rho = \mathrm{d}x \wedge \mathrm{d}y$. So $\mathrm{d}\rho \wedge \rho = \mathrm{d}x \wedge \mathrm{d}y \wedge \mathrm{d}z \neq 0$. – Willie Wong Mar 18 '13 at 15:44
@WillieWong is this the way to expand d(f\rho)? I learn this topic new. I have solved many problems. But ı cannot solve 3questions of these. This is one of them. So can you please solve this problem expicitly? I want to understand. Thank you willie – B11b Mar 18 '13 at 15:54

I will keep referring to the Leibniz-rule for exterior derivatives. That is axiom 3 here.

1. The expression $\rho \wedge \mathrm{d}\rho = 0$ is only true for all one-forms $\rho$ in less than or equal to two dimensions. In dimension $\geq 3$, let $x,y,z$ be the first three of the coordinate functions, we can consider the one form $$\rho = x \wedge \mathrm{d}y + \mathrm{d}z$$ in local coordinates. (Recall that for a scalar function $x$ and a differential form $\omega$, the notations $x\omega = x\wedge \omega$.) Its exterior derivative is, using the Leibniz-rule for exterior derivatives $$\mathrm{d}\rho = \mathrm{d}x \wedge \mathrm{d}y + (-1)^0 x\wedge \mathrm{d}(\mathrm{d}y) + \mathrm{d}(\mathrm{d}z) = \mathrm{d}x \wedge \mathrm{d}y$$ So computing we have $$\rho \wedge \mathrm{d}\rho = x \mathrm{d}y \wedge (\mathrm{d}x \wedge \mathrm{d}y) + \mathrm{d}z \wedge (\mathrm{d}x \wedge \mathrm{d}y) = \mathrm{d}x \wedge \mathrm{d}y \wedge \mathrm{d}z \neq 0$$

On the other hand, in dimension 2, we have that $\mathrm{d}\rho$ would be a two-form, and $\rho \wedge \mathrm{d}\rho$ would be a three form. And we know that for $k$-forms on a manifold of dimension $n$, whenever $k > n$ the form must vanish (by the pigeonhole principle, if you will).

2. Now, supposing that we have a one-form $\rho$ and a nowhere-vanishing function $f$ such that $\mathrm{d}(f\rho) = \mathrm{d}(f\wedge \rho) = 0$. Using the Leibniz rule we have $$\mathrm{d}(f\wedge \rho) = \mathrm{d}f \wedge \rho + (-1)^0 f \wedge \mathrm{d}\rho = 0$$ Now take wedge product with $\rho$, we have $$0 = \rho \wedge 0 = \rho \wedge \mathrm{d}(f\wedge \rho) = \rho \wedge \mathrm{d}f \wedge \rho + \rho \wedge f \wedge \mathrm{d}\rho$$ Using the property of the wedge product we can re-organize $$0 = (-1)^1 \mathrm{df} \wedge \rho \wedge \rho + (-1)^0 f \wedge \rho \wedge \mathrm{d}\rho$$ Now for one forms we know that $\rho \wedge \rho = 0$, which implies that the first term vanishes. So we conclude that $$0 = f (\rho \wedge \mathrm{d}\rho)$$ Since $f$ is nowhere vanishing, we can divide by $f$, and this gives $$0 = \rho \wedge \mathrm{d}\rho$$

In case you don't know that $\rho \wedge \rho = 0$: Let us express in coordinates $(x_1, \ldots, x_n)$. We can write $\rho = \sum_{i = 1}^n \rho_i \mathrm{d}x_i$. Then in coordinates $$\rho \wedge \rho = \sum_{i = 1}^n \sum_{j = 1}^n \rho_i \rho_j \mathrm{d}x_i \wedge \mathrm{d}x_j$$ Using that $\mathrm{d}x_i \wedge \mathrm{d}x_j = - \mathrm{d}x_j \wedge \mathrm{d}x_i$ we have $$\rho \wedge \rho = \sum_{i = 1}^n \sum_{j = 1}^{i-1} (\rho_i \rho_j - \rho_j \rho_i) \mathrm{d}x_i \wedge \mathrm{d}x_j + \sum_{i = 1}^n \rho_i \rho_i \mathrm{d}x_i \wedge \mathrm{d}x_i = 0$$

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Really thank you so much for your help!!:) @williewong – B11b Mar 18 '13 at 17:45