If $f: U \subset \mathbb{R}^{m} \to \mathbb{R}^{n}$ is a differentiable map and $k > m$, $f^{*}\omega = 0$ for each $k$-form $\omega$.

Question

Let $f: U \subset \mathbb{R}^{m} \to \mathbb{R}^{n}$ be an differentiable map. Suppose that $m < n$ and let $\omega$ be a $k$-form in $\mathbb{R}^{n}$ with $k>m$. Show that $f^{*}\omega = 0$.

My attempt.

By definition

$$\omega(p) = \sum_{i_{1}<\cdots< i_{k}}a_{i_{1}\cdots i_{k}}(p)dx_{i_{1}}\wedge \cdots\wedge dx_{i_{k}},$$

and

$$(f^{*}\omega)(p)(v_{1},...,v_{k}) = \omega(f(p))(df_{p}(v_{1}),...,df_{p}(v_{k}))$$

We know that $f^{*}$ map $k$-forms in $\mathbb{R}^{n}$ in $k$-forms in $\mathbb{R}^{m}$. So, if $f^{*}\omega$ is a $k$-forms in $\mathbb{R}^{m}$ with $k > m$, any combination of $dx_{i_{j}}$ will have at least $dx_{i_{l}}\wedge dx_{i_{l}} = 0$. Then $f^{*}\omega \equiv 0$.

I'm not sure if it is correct. I think that, maybe, I should use the definition of pullback. Moreover, I cannot see the relevance of $m < n$.

Answer 1

Moreover, I cannot see the relevance of $<$

Note $\omega$ is a $k$-form on $\mathbb R^n$: so if $k>n$, then $\omega = 0$ and of course $F^*\omega = F^* 0 = 0$. So if $m\ge n$, the assumption $k>m$ implies $k>n$, so the result is trivial. I guess this is why they ask you for the case $m<n$.

$\text { We know that } f^{*} \text { maps } k \text {-forms in } \mathbb{R}^{n} \text { [to] } k \text {-forms in } \mathbb{R}^{m}$

If you know this, then you are already done with no computation; the only $k$-form on $\mathbb R^m$ for $k>m$ is $0$.