Expressing a differential form in terms of a scalar function

100 Views Asked by At

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 $ \mathbb{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:

enter image description here

1

There are 1 best solutions below

2
On BEST ANSWER

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.