Differential forms in $ \mathbb{R}^n \times \mathbb{R} $

Question

In page 34 of "Differential forms in Algebraic Topology," the author states that every forms on $ \mathbb{R}^n \times \mathbb{R} $ is uniquely a linear combination of the following two types of forms:

$$(I) (\pi^* \phi)f(x,t), $$ $$(II) (\pi^* \phi)f(x,t)dt, $$

(The notation of wedge product is omitted) Where $ \pi$ is a projection from $ \mathbb{R}^n \to \mathbb{R}$ and $ \phi$ is a form on the base $ \mathbb{R}^n$

My question is, can't you represent the two types of forms the following way?

$$(I) (\prod_I dx^I)f(x,t), $$ $$(II) (\prod_I dx^I)f(x,t)dt, $$

Where $(\prod_I dx^I)$ is some combination $ dx_{i_1},dx_{i_2} ... dx_{i_n}$?

I was just wondering because I thought writing the two the second way seems more natural.

Answer 1

No. The first type has no $dt$ in it and the second type has $dt$ in it. In particular, if you evaluate any form of the first type on $k$ vectors, one of which is $(0,1)\in\Bbb R^n\times\Bbb R$, you always get $0$.

EDIT: Sorry I didn't read your original question carefully enough. Any form of the first variant is indeed a linear combination of forms of the second variant, as the $dx^I$ give a basis for the $k$-forms (as a module over the ring of smooth functions on $\Bbb R^n$).

Answer 2

For $(I)$, note that since $\pi$ projects $(x,t)\in \mathbb R^n\times \mathbb R$ to $x\in \mathbb R^n,\ \pi^*$ sends forms $\phi\in \Omega^k(\mathbb R^n)$ to forms $\pi^*\phi\in \Omega^k(\mathbb R^n\times \mathbb R)$, and as pointed out in the comments, "nothing happens".

For example, suppose $\phi(x,y)=f(x,y)dx\wedge dy.$ Then,

$\pi:\mathbb R^2\times \mathbb R\to\mathbb R^2:(x,y,t)\mapsto (x,y)\Rightarrow\pi^*\phi(x,y,t)=f\circ \pi(x,y,t) \ d(x\circ \pi)\wedge\ d(y\circ \pi)=f(x,y)dx\wedge\ dy$.

And this is why we need to include terms of the form $(II)$, because with $(I)$ alone we do not get the full vector space (for each $k\le n+1$) of forms on $\mathbb R^n\times \mathbb R.$