# Confused about why $dd=0$ in de rham cohomology?

In Bredon's proof that $dd=0$, he lets $\omega=fdx_1\wedge\cdots\wedge dx_p$, and calculates $$dd\omega=\sum_{j=1}^n\sum_{i=1}^n\frac{\partial^2 f}{\partial x_i\partial x_j}dx_j\wedge dx_i\wedge dx_1\wedge\cdots\wedge dx_p.$$

He says to rearrange the double sum so that it ranges over $j<i$, and then everything cancels. But what happens to the terms where $i=j$?

I thought the wedge product of two $p$-forms is zero only when $p$ is odd?

• They disappear. The wedge product of two copies of the same thing is zero. – Qiaochu Yuan Jul 15 '14 at 3:22
• @QiaochuYuan Corollary 1.4 on the preceding page, says that if $\omega\in A^p(V)$ is a $p$-form, and $p$ is odd, then $\omega\wedge\omega=0$. Is there something different happening when $p$ is even in this case? – Glen Glen Glen Jul 15 '14 at 3:26
• Sorry, I should have been more specific by "same thing." I mean $dx_i \wedge dx_i = 0$ (here $p = 1$). – Qiaochu Yuan Jul 15 '14 at 3:37
• Oooh, the $dx_i$ are all $1$-forms, thanks. – Glen Glen Glen Jul 15 '14 at 3:39
• It is a general fact of tensor arithmetic, when a symmetric pair of indices is contracted over an antisymmetric pair they annihilate in puff of zero. $\sum_{i,j}S_{ij}A^{ij}=0$. The partial derivatives are symmetric whereas $dx^i \wedge dx^j=-dx^j \wedge dx^i$ expresses the antisymmetry. Of course, you are working out the details in this case you consider here, but this pattern is worth noting for future reference. – James S. Cook Jul 15 '14 at 6:49

If $i=j$ the $dx_i\wedge dx_j\wedge \ldots = 0$ vanish.