# exterior product of forms is exact.

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.

• Do you mean that $\alpha\wedge\beta$ is exact? – Olivier Bégassat Nov 9 '14 at 1:45
• That's it, just fixed :) thank you – Vinicius Rodrigues Nov 9 '14 at 1:46
• Hint: if $\beta=db$, then look at $\pm\alpha\wedge b$... – Olivier Bégassat Nov 9 '14 at 1:46
• 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 :) – Vinicius Rodrigues Nov 9 '14 at 1:49
• ok : ) ${}{}{}$ – 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.