Prove that for any differential form $w$ $d(d(w))=0$
I managed to prove it for $w$ being a $0$-form:
$$
d(d(w))=d\left(\frac{\partial f}{\partial x_i}dx_i\right)=\sum_{i=1}^n\sum_{j=1}^n \frac{\partial f}{\partial x_i\partial x_j}dx_i\wedge dx_j=\sum_{i<j}\left(\frac{\partial f}{\partial x_i\partial x_j}dx_i\wedge dx_j + \frac{\partial f}{\partial x_j\partial x_i}dx_j\wedge dx_i\right)=0
$$
But I don't understand how to generalize this fact to any non-zero differential form. Could someone give me a clue?
Best Answer
Show it for forms that look like $f \mathrm{d}x^{i_1}\wedge\cdots \wedge \mathrm{d}x^{i_r}$. This is pretty straight-forward. Then, locally, every form is a linear combination of terms that have this form.
This is because : for an exact $1$-form $\mathrm{d}f$, $\mathrm{d}\mathrm{d}f = 0$ by definition. Then, for a local $r$-form, $\mathrm{d}\left(f \mathrm{d}x^{i_1}\wedge\cdots \wedge \mathrm{d}x^{i_r} \right) = \mathrm{d}f \wedge\mathrm{d}x^{i_1}\wedge\cdots \wedge \mathrm{d}x^{i_r}$, thus $\mathrm{d}^2 \left( f \mathrm{d}x^{i_1}\wedge\cdots \wedge \mathrm{d}x^{i_r}\right) = \mathrm{d}\left(\mathrm{d}f \wedge\mathrm{d}x^{i_1}\wedge\cdots \wedge \mathrm{d}x^{i_r}\right) = 0$ (remind thatif $\alpha$ and $\beta$ are differential form and $\alpha$ is of degree $p$, $\mathrm{d}\left( \alpha\wedge \beta\right) = \mathrm{d}\alpha \wedge \beta + (-1)^p \alpha \wedge \mathrm{d}\beta$)