The following comes from the proof in differentiable manifolds that $d^2\omega=0$.

Let $f$ belong to the set of $0$-forms. From definition I have that

$\displaystyle df = \frac{\partial f}{\partial x^j}dx^j$

Then from the definition for $d$ of a one-form, apparently we have that

$\displaystyle d(df)=d\Big(\frac{\partial f}{\partial x^j}\Big) \wedge dx^j$.

I cannot see how this can been derived.

I tried $\displaystyle d(df)=d\Big(\frac{\partial f}{\partial x^j} dx^j \Big)$. You then use some kind of product rule?

The definition I have been given is if

$\displaystyle \omega = \frac{1}{k!} \omega_{i_1 ... i_k} dx^{i_1}\wedge...\wedge dx^{i_k}$ then its derivative is

$\displaystyle d\omega = \frac{1}{k!} d\omega_{i_1\cdots i_k} \wedge dx^{i_1}\wedge\cdots\wedge dx^{i_k}$

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    $\begingroup$ Its basically how the exterior derivative is defined. $\endgroup$ – BigM Aug 26 '14 at 20:12

I'll write out an answer for a form of three variables. This might yield some intuition. Thing to notice is that $dx^i\wedge dx^i=0$, and that $dx^i\wedge dx^j=-dx^j\wedge dx^i$, and $f_{xy}=f_{yx}$.

Thus, because $dx^i\wedge dx^i=0$

$$d^2f=f_{xy}dy\wedge dx+f_{xz}dz\wedge dx+f_{yx}dx\wedge dy+f_{yz}dz\wedge dy+f_{zx}dx\wedge dz+f_{zy}dy\wedge dz $$

Do you see how these terms cancel out to give $0$?


$$ \begin{split} d(df)&=d(\frac{\partial f}{\partial x^j})\wedge dx^j\\ &=\frac{\partial^2 f}{\partial x^jx^k}dx^k\wedge dx^j\\ &=0 \end{split} $$ since $$ dx^k\wedge dx^j=-dx^j\wedge dx^k $$ and $$ \frac{\partial^2 f}{\partial x^jx^k}=\frac{\partial^2 f}{\partial x^kx^j} $$


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