I am currently reading Carmo's book about differential forms. I tried excercise 6 from chapter 1, but I am a bit confused. Here is the excercise:
Let $f: U \subset \mathbb{R}^m \to \mathbb{R}^n$ be a diff. map. Assume that $m<n$ and let $\omega$ be a $k$-form in $\mathbb{R}^n$, with $k>m$. Show that $f^*\omega=0$.
My idea was that we have more different $df_{i_k}$ (we should have $n$ many, right?) and then $k<n$ so there is some $i_k$ in the definition of $f^*\omega$ where $df_{i_k} \wedge df_{i_k}$ so this is $0$.
In the exercise, they said that $k>m$ not $k>n$ so my argument would be false. In the book he always uses the notation $f: U \subset \mathbb{R}^n \to \mathbb{R}^m$. Is this a mistake in the exercise or is there a different way that I do not see?
I'm glad if someone helps me. Thanks a lot :)
To make this post officially closed here the summary.
In general
for every k-form in $\mathbb{R}^m$ there are $\binom{m}{k}$ different basis vectors. This implies there are no non-trivial k-forms in $\mathbb{R}^m$ if $k > m$.
For this exercise
Since $f^*\omega$ is a k-form in $\mathbb{R}^m$, but $k>m$ we can only have the k-form $f^*\omega = 0$