I don't know what to do to prove the following statement:

Let $U \subset \mathbb R^n$ be an open set and let $\alpha$ be a $k$-form on $U$ and $\beta$ be an $l$-form on $U$. Suppose both $\alpha, \beta$ are closed forms and that $\beta$ is exact. Then $\alpha \wedge \beta$ is exact.

Every assistance would be very much appreciated.

  • $\begingroup$ Do you mean that $\alpha\wedge\beta$ is exact? $\endgroup$ – Olivier Bégassat Nov 9 '14 at 1:45
  • $\begingroup$ That's it, just fixed :) thank you $\endgroup$ – Vinicius Rodrigues Nov 9 '14 at 1:46
  • 1
    $\begingroup$ Hint: if $\beta=db$, then look at $\pm\alpha\wedge b$... $\endgroup$ – Olivier Bégassat Nov 9 '14 at 1:46
  • $\begingroup$ Oh, now I see exactly what I must do. If you wish, post this message as an answer so I can mark it as the answer :) $\endgroup$ – Vinicius Rodrigues Nov 9 '14 at 1:49
  • $\begingroup$ ok : ) ${}{}{}$ $\endgroup$ – Olivier Bégassat Nov 9 '14 at 1:55

If $b$ satisfies $db=\beta$, and $\alpha$ is a closed $k$-form, then $$d\omega=\alpha\wedge\beta$$ where $\omega=(-1)^k\alpha\wedge b$. The proof works for forms on any manifold.


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.