How does a differential form looks like on $\mathbb{R}^n \times [0,1]$?

Solution 1:

This is mainly in issue of notation, I think: Firstly $\mathbb{R}^n\times[0,1]\subset \mathbb{R}^{n+1}$. Then a $k$-form $\alpha$ on $\mathbb{R}^n\times[0,1]$ can written w.r.t. to a basis of the space of $k$-vectors, hence $$\alpha = \sum_{\ell} \alpha_{\ell} dx_{\ell_1}\wedge\ldots\wedge dx_{\ell_k}.$$ Since we usually have linearity, we only take a look at one of the summands and try to write down what could happen. Since we like to emphasise the variable connected to $[0,1]$ we call it $t$ instead of $x_{n+1}$. Then in $dx_{\ell_1}\wedge\ldots\wedge dx_{\ell_k}$ this $dx_{n+1}=dt$ can appear or it may not appear. To reflect both cases, we write down two summands: $$fdx_{i_1}\wedge\ldots\wedge dx_{i_k}+gdx_{n+1}\wedge dx_{j_1}\wedge\ldots\wedge dx_{j_{k−1}}.$$ Note that here $i_1,\ldots,i_k, j_1,\ldots,j_{k-1}\neq n+1$. Since we set $x_{n+1}=t$ we replace $dx_{n+1}=dt$ and then you have your desired form.

Also $dt\wedge dx_i$ is thereby defined as $dx_{n+1}\wedge dx_i$. Hope this helps.