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It is the second term that is always zero (because $d^2a = 0$), no matter what $k$ is. The first term is zero because $da$ has odd degree. In general $$ x \wedge y = (-1)^{|x||y|} y \wedge x$$ where $|x|,|y|$ are the degrees of $x,y$. In particular, if $x$ has odd degree then $x \wedge x = 0$.


The cotangent bundle of $M \times F$, as a vector bundle, is the direct sum of the cotangent bundles of $M$ and $F$ (more precisely, of their pullbacks along the natural projections), so you're asking how to describe the exterior powers of a direct sum $V \oplus W$ of two vector bundles. The answer is that the exterior algebra is a graded tensor product ...


$\bigwedge^r T^*(M\times F)=\sum_{s+t=r}\bigwedge^s T^*M\otimes\bigwedge^t T^*F$


The tuple ${x^1,...,x^k}$ has clearly determined order and elements. The tuple ${}{x^{i_1},...,x^{i_p}}$ means that every element of the tuple can be any coordinate, the indices are numbered however to distinguish them, so that each indexed element can take on values independent of other indexed elements. If $\omega$ is an $n$-form on $\mathbb{R}^n$ (eg. it ...

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