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

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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?

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