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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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    You have (by definition of $d$) $d(y\, dx) = dy\wedge dx + y\;d^2x = dy \wedge dx$.2012-07-13

1 Answers 1

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$dƒ$ 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 $d(y \cdot dx)$ ?2012-07-13
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    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 $y\, dx$.2012-07-13
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    In general $fd\alpha$ is the same as $f \wedge d\alpha$2012-07-13