# Prove that if $η$ is exact, then $η∧β$ is also exact.

Prove that if $η$ is exact, then $η∧β$ is also exact. Please give a clear way to solve?

-
Plkease tell what in the world you're talking about...?! –  DonAntonio Mar 18 '13 at 18:12
@DonAntonio From the tags, I believe $\eta$ and $\beta$ are exact differential forms on some differentiable manifold. –  A.P. Mar 18 '13 at 18:19
Oh, A.P.:I'm almost certain that's what the OP meant, yet college/university students must be encouraged to stop writing down in a sloppy way and to define whatever they want to talk about... –  DonAntonio Mar 18 '13 at 18:34
@DonAntonio I couldn't agree more, although I fear the OP is somewhat struggling with the language... –  A.P. Mar 18 '13 at 18:37
Perhaps you're right, A.P., yet even in that case I'd demmand at least some effort to tell what the very question is about. Thanks. –  DonAntonio Mar 18 '13 at 18:40

The following hint works if you suppose $\beta$ to be closed. I'm not sure (though I doubt) that $\eta \wedge \beta$ is in general exact if $\eta$ is exact.
Try to write out what it means for $\eta$ to be exact, i.e. that $\exists \omega$ such that $\eta=d\omega$. Then recall the formula: $$d(\alpha \wedge \gamma)=d\alpha \wedge \gamma + (-1)^k \alpha \wedge d\gamma$$ where $\alpha$ and $\gamma$ are differential forms and $k$ is the degree of $\alpha$. Now apply it to $\omega\wedge\beta$: \begin{align*} d(\omega\wedge\beta)&=d\omega\wedge\beta+(-1)^k\omega\wedge d\beta\\ &=\eta\wedge\beta+(-1)^k\omega\wedge 0\\ &=\eta\wedge\beta \end{align*} since $\beta$ closed means that $d\beta=0$. Therefore $\eta\wedge\beta$ is exact.
I just wanted to add that the hypothesis that $\beta$ is closed is necessary. In $\mathbb{R}^2$, use $\eta = dx$ and $\beta = y$. Then $\eta \wedge \beta = ydx$ is not even closed, must less exact. If the OP finds this too trivial, in $\mathbb{R}^3$, try $\eta = dx$ and $\beta = ydz$. Then $\eta\wedge \beta$ is again, not closed, so can't be exact. –  Jason DeVito Mar 18 '13 at 19:36
I will apply his formula for $η∧β$. And then I cannot understand how will I do? @A.P. –  Mathlover4 Mar 18 '13 at 21:20
@Mrckh Sorry, I miswrote. Apply it to $\omega \wedge \beta$. –  A.P. Mar 18 '13 at 21:29