2
$\begingroup$

I'd like some help to prove the following: $f\colon U \subset \mathbb{R}^m \to \mathbb{R}^n$, differentiable; $m ; $\omega$ is a $k$-form in $\mathbb{R}^n$, $k>m$.

Show that $f^*\omega = 0$

Thanks.

  • 2
    Do you want $k > m$, rather? Otherwise $\omega$ is already zero.2012-01-03
  • 2
    Can you describe the $k$-forms on $\mathbb R^m$ when $k>m$?2012-01-03
  • 0
    Sorry, it should be $k>m$2012-01-03
  • 4
    I'd say try to answer these two subquestions: Is $f^*\omega$ again a $k$-form? Then do Mariano's question.2012-01-03
  • 0
    The pullback of a k-form is also a k-form. What do the k-forms on $U$ look like?2012-01-03

1 Answers 1