This question consists of two parts. The first part is simply linear algebra whereas the second requires some geometry.

1) Let $E$ and $F$ be two finite dimensional vector spaces and suppose $\textrm{dim}(F)=1$. Let $\omega\in \Lambda^k(E\oplus F)$ and $\theta\in F\setminus\{0\}$. How to show there exists unique $\tau\in \Lambda^k E$ and $\eta\in \Lambda^{k-1}E$ such that: $$\omega=\tau+\theta\wedge \eta?\quad\quad\quad\quad\quad (I)$$

2) Let $\pi_A:A\longrightarrow M$ and $\pi_B:B\longrightarrow N$ be two vector bundles. We can build a new vector bundle $E$ setting: $$E:=\bigsqcup_{(p, q)\in M\times N} A_p\oplus B_q\longrightarrow M\times N,$$ where $A_p$ and $B_q$ are the fibers over $p$ and $q$, respectively. Let us suppose $B$ is a line bundle, that is, its rank is $1$. Consider the space of smooth sections $$\Omega^k(E):=\Gamma(\Lambda^k E^*).$$ Can we decompose elements $\omega\in \Omega^k(E)$ in a form like $(I)$?

Indeed, for me, $B$ will be the trivial line bundle so you can add this hypothesis if you wish.


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    $\begingroup$ For $1$: pick a basis of $E\oplus F$ which includes $\theta$. From that, construct as usual a basis of $\Lambda^k(E\oplus F)$, and write in that basis your form $\omega$ and group terms according to whether they contain or not $\theta$. $\endgroup$ – Mariano Suárez-Álvarez Feb 9 '17 at 16:52
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    $\begingroup$ For $2$, and assuming $B$ is trivial, then just do the same thing in every fiber, of course. To get the result globally you have to be a bit careful: pick a nonzero section of $B$, and use it in every fiber, and show that the decomposition in $1$ is unique, so as to be able to glue. $\endgroup$ – Mariano Suárez-Álvarez Feb 9 '17 at 16:56

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