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Let $x$, $y$ be 0-forms, thus $dx$, $dy$ are 1-forms. Since 1-forms compose an algebra over 0-forms ring, expressions like

$$y dx$$

make perfect sense. Now I ask what is

$$d(y dx)$$

I suggest it to be $y d(dx) = 0$, since $d$ is linear, however I feel it is likely to be wrong. Is there any other meaningful product except $\land$ between forms that would generalize the situation above?

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4  
You have (by definition of $d$) $d(y\, dx) = dy\wedge dx + y\;d^2x = dy \wedge dx$. –  martini Jul 13 '12 at 10:05

1 Answer 1

up vote 5 down vote accepted

$dƒ$ is the differential of $ƒ$ for smooth functions $ƒ$.

$d(dƒ) = 0$ for any smooth function $ƒ$.

$d(α∧β) = dα∧β + (−1)^p(α∧dβ) $ where $\alpha$ is a p-form. Take $\alpha=y$ and $\beta=dx$ in this last formula.

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The same question as to Chris, why did you substitute $\cdot$ by $\land$ in $d(y \cdot dx)$ ? –  Yrogirg Jul 13 '12 at 10:38
2  
There is no substitution. When you write $y\cdot dx$ (where $y$ is a 0-form) you really mean $y\wedge dx$. It just happens that 0-forms commute, so it is often written as $y\cdot dx$ or just $y\, dx$. –  Chris Taylor Jul 13 '12 at 10:45
    
In general $fd\alpha$ is the same as $f \wedge d\alpha$ –  PAD Jul 13 '12 at 10:48
    
thanks, that helped. –  Yrogirg Jul 13 '12 at 10:49

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