How to differentiate a differential form?

Please explain me the idea of differentiating differential forms (tensors). Example: compute d(xdy + ydx) The answer is known, we should have 0. What's the rule?

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Hint: a differential form is a bit like a product. How do you differentiate a product? –  Asal Beag Dubh Mar 18 '13 at 18:02

There is a formula of computing exterior derivative of any differential form (which is assumed to be smooth). In your case, if $\sigma$ is a 1-form, and $$\sigma = \sum_{j=1}^n f_j \mathrm{d}x^j.$$ Then the exterior derivative of $\omega$ is: $$\mathrm{d}{\sigma} =\sum_{j=1}^n \sum_{i=1}^n \frac{\partial f_j}{\partial x^i} \mathrm{d}x^i \wedge \mathrm{d}x^j .$$

Following this formula the exterior derivative of the 1-form you gave is: \begin{aligned} \mathrm{d}(x\mathrm{d}y + y\mathrm{d}x)&= \mathrm{d}(x\mathrm{d}y) + \mathrm{d}(y\mathrm{d}x) \\ &= \left(\frac{\partial x}{\partial x}\mathrm{d}x\wedge\mathrm{d}y + \frac{\partial x}{\partial y}\mathrm{d}y\wedge\mathrm{d}y\right) + \left(\frac{\partial y}{\partial x}\mathrm{d}x\wedge\mathrm{d}x + \frac{\partial y}{\partial y}\mathrm{d}y\wedge\mathrm{d}x\right) \\ &= (\mathrm{d}x\wedge\mathrm{d}y+0) + (0+ \mathrm{d}y\wedge\mathrm{d}x)=0. \end{aligned} Notice we used $\mathrm{d}x^i\wedge\mathrm{d}x^i = 0$ and $\mathrm{d}x^i\wedge\mathrm{d}x^j= -\mathrm{d}x^j\wedge\mathrm{d}x^i$.

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at the second addend of first bracket should be dy^dy if I understood you in a right way –  maggot092 Mar 18 '13 at 18:21
@maggot092 Yes, thanks for correcting the typo. –  Shuhao Cao Mar 18 '13 at 19:31

Visually, the differential of a differential form

$$\color{red}{f} \, \color{blue}{\text{d}x}$$

is determined by taking the differential the function part $f$ (in red) in place:

$$\text{d}(\color{red}{f} \, \color{blue}{\text{d}x}) = \color{red}{\text{d}f} \, \color{blue}{\text{d}x}$$

if you write wedge products out expictly, you need to insert a wedge:

$$\text{d}(\color{red}{f} \, \color{blue}{\text{d}x}) = \color{red}{\text{d}f} \wedge \color{blue}{\text{d}x}$$

The same is true with higher forms:

$$\text{d}(\color{red}{f} \, \color{blue}{\text{d}x\,\text{d}y}) = \color{red}{\text{d}f} \, \color{blue}{\text{d}x \, \text{d}y}$$

And, of course, the differential distributes across sums as usual.

Because the double differential $\text{d}(\text{d}f) = 0$ for all functions $f$ (and $\text{d}(\text{d}\omega)$ is also true for all differential forms $\omega$), $\text{d}f$ behaves very much like a constant; e.g. it factors out of the derivative. The only trick is that it's not commutative, so you have to make sure it factors out on the correct side: to the right, not to the left.

This all can be viewed as a special case of the product rule

$$\text{d}(\omega \theta) = (\text{d} \omega)\, \theta + (-1)^{\deg \omega} \omega \, (\text{d}\theta)$$

where $\deg \omega$ is the 'degree' of the form $\omega$: it is $n$ if $\omega$ is an $n$-form (and so $0$ if $\omega$ is a scalar).

Alternatively, written with wedge products:

$$\text{d}(\omega \wedge \theta) = (\text{d} \omega) \wedge \theta + (-1)^{\deg \omega} \omega \wedge (\text{d}\theta)$$

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You have:

$d(xdy + ydx)=dx\wedge dy+ x\wedge d^2 y+ dy\wedge dx+y\wedge d^2 x=dx\wedge dy +dy\wedge dx=0$.

For the general formula see (for example): Calculus on manifolds (Spivak)