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Given two forms α and β, let α be closed and β be exact, how do you prove that αβ is exact? I can see that αβ is closed, but that is only sufficient fact from the product being exact. Any suggestions?

jfcjohn
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    Hint: if $\alpha$ is a $k$-form, $\beta$ is an $l$-form, and you're given that $\alpha = d \gamma,$ $\beta = d\tau$ for $\gamma$ a $k-1$ form and $ \tau$ an $l-1$ form, there are only so many things you can do to construct a $k+l-1$ form from the ones that you have. So write one down, apply $d$ to it, and hope you get $\alpha \wedge \beta$. If you don't, adjust as necessary. – vociferous_rutabaga Oct 01 '15 at 02:44
  • But I am not given α is exact, so I can't assume α=dγ for a γ of order (k-1) right? Since I only know α is closed. – jfcjohn Oct 01 '15 at 05:10
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    Sorry, I misread your question! However the general principal still applies. Updated hint: if $\alpha$ is a $k$-form, $\beta$ is an $l$-form, and you're given that $\alpha = d \gamma $ for $\gamma$ a $k-1$ form, then there are only so many things you can do to construct a $k+l-1$ form from the ones that you have. Write one down, apply $d$ to it, and hope you get $\alpha \wedge \beta$. If you don't, adjust as necessary. – vociferous_rutabaga Oct 01 '15 at 11:37

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