On page 34 where the author proves
The maps $\pi: \Bbb R ^n \times \Bbb R^1 \rightarrow \Bbb R$, $s: \Bbb R \rightarrow \Bbb R^n \times \Bbb R^1$ ($x \mapsto (x,0))$, induces isomoprhism. $$H^*(\Bbb R^n \times \Bbb R^1) \rightarrow H^*(\Bbb R^1)$$ in the de Rham cohomology.
he stated that
Every form on $\Bbb R^n \times \Bbb R$ is uniquely a linear combination of the following two types.
- $(\pi^* \phi ) f(x,t)$
- $(\pi^* \phi) f(x,t) dt. $
where $\phi$ is a form on the base $\Bbb R^n$.
I do not understand what is meant here. May someone elaborate? Is it what I think below?
We know a $k$-form is written uniquely as a sum of $$ \sum_{|I|=k-1} f_Idx_I dt + \sum_{|I|=k} f_I dx_I $$ such that $I$ is an increasing subset of $1$ to $n$. It seems that $\phi = dx_I$ otherwise we would not have uniqueness. Is this right?
Yes, you're right, but I would recommend being a bit more explicit and not abusing notation. If you have a $k$-form $\omega$, then you can write $$\omega = \sum_{|I|=k-1}f_I(x,t)dx_I\,dt + \sum_{|J|=k} g_J(x,t)dx_J,$$ where you restrict, as you said, to increasing multi-indices. (In general, $\phi$ could be a combination $\sum h_I(x)dx_I$, but we can absorb the functions into the $f_I(x,t)$, and so on.)