# Differential forms on $M\times\mathbb{R}$

Let $\alpha$ be a differential form of degree $p+1$ on $M\times\mathbb{R}$, where $M$ is an arbitrary smooth manifold, and $p$ a non negative integer.

Can $\alpha$ be always written as $\beta+\gamma\wedge dt$, where $\beta$ and $\gamma$ are differential forms on $M\times\mathbb{R}$ of degree $p+1$ and $p$ respectively, such that $\frac{\partial}{\partial t}\lrcorner\beta=0$ and $\frac{\partial}{\partial t}\lrcorner\gamma=0$? and, in the affirmative case, are $\beta$ and $\gamma$ uniquely determined?

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You might want to take a look into Bott and Tu's book. – Rasmus Apr 11 '11 at 17:21
You can easily see that this holds locally (pick appropriate local coordinates and an appropriate local basis of the space of all 2-forms on $M \times \mathbb R$). Now use this to prove the global statement. Hint: $\mathbb R$ admits an atlas with only one chart. – Alexander Thumm Apr 11 '11 at 18:06
Take $M = \mathbb{R}$. Is it possible to write $xt\,dx \wedge dt$ in the form you want? – Santiago Canez Apr 11 '11 at 19:39
@Rasmus: Bott & Tu is big reference, but I think there is not an answer to my silly question.@Alexander Thumm: Thanks for your comment, then Should I prove the global statement using partitions of unity. @Santiago Canez: yes it is possible if you look at my original post that I have restored. – Giuseppe Apr 11 '11 at 20:51
Good, this is now correct. My intention was simply to point out that the answer is no if you want $\beta$ and $\gamma$ to be forms on $M$ alone :) – Santiago Canez Apr 11 '11 at 22:30

yes and yes, for $\alpha$ of arbitrary degree. I'll rather write $\alpha=\beta+dt\wedge\gamma$ to save a sign (i.e. I possibly changed the sign of $\gamma$). Then $\gamma=\frac{\partial}{\partial t}\lrcorner \alpha$ and $\beta=\alpha-dt\wedge\gamma$. (you can easily see that these equations must be true if your conditions are satisfied, and also that they give $\beta$ and $\gamma$ satisfying your conditions)