# Expressing a differential form in terms of a scalar function

We can express every k-form in the form $\omega(x) = \sum_Id_Idx_I$ where $I$ is k-tuple and $d_I$ is just some scalar function of x. That's entirely understandable for me. But while reading Munkres "Analysis on manifolds" I stumbled upon something quite confusing. In the discussion of the pullback function $\alpha^*$, there is the following part: The form $\alpha_*(dy_I)$ is a k-form defined on an open set of $R^k$, so it has the form $h dx_1\wedge dx_2 \wedge ... \wedge dx_k$ for some scalar function $h$.

I completely don't see where that comes from. Can somebody clear this up for me?

Here are 2 pages of the book, puttin the problem in context: http://img31.imageshack.us/img31/5096/outfp.png

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The key thing here is that it's specifically a $k$-form on $\Bbb{R}^k$. Since $dx^i \wedge dx^i=0$ for all $i$, this means there's only one possible term in the sum: you need to use all the different coordinates or you won't have enough.
Also, any other presentation of $dx^1 \wedge dx^2 \wedge \cdots \wedge dx^k$ can be permuted to go back to the $1,2, \dots , 3$ order due to the algebra of the wedge product. It suffices to consider just scalar function multiples of the cannonical volume form $dx^1 \wedge dx^2 \wedge \cdots \wedge dx^k$ for this reason. –  James S. Cook Dec 1 '12 at 15:41