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?

  • 2
    $\begingroup$ They disappear. The wedge product of two copies of the same thing is zero. $\endgroup$ – Qiaochu Yuan Jul 15 '14 at 3:22
  • $\begingroup$ @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? $\endgroup$ – Glen Glen Glen Jul 15 '14 at 3:26
  • $\begingroup$ Sorry, I should have been more specific by "same thing." I mean $dx_i \wedge dx_i = 0$ (here $p = 1$). $\endgroup$ – Qiaochu Yuan Jul 15 '14 at 3:37
  • $\begingroup$ Oooh, the $dx_i$ are all $1$-forms, thanks. $\endgroup$ – Glen Glen Glen Jul 15 '14 at 3:39
  • $\begingroup$ 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. $\endgroup$ – James S. Cook Jul 15 '14 at 6:49

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

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.