Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$\def\d{\mathrm{d}}\def\R{\mathbb{R}}$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.

share|improve this question
1  
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
1  
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

1 Answer 1

up vote 3 down vote accepted

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 $$ \rho \wedge \rho = \sum_{i = 1}^n \sum_{j = 1}^{i-1} (\rho_i \rho_j - \rho_j \rho_i) \mathrm{d}x_i \wedge \mathrm{d}x_j + \sum_{i = 1}^n \rho_i \rho_i \mathrm{d}x_i \wedge \mathrm{d}x_i = 0 $$

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.