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I'd like some help to prove the following: $f\colon U \subset \mathbb{R}^m \to \mathbb{R}^n$, differentiable; $mm$.

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

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Definition is enough.

You can see that the pull-back of a $k$-form (by a differentiable map) is a $k$-form again.

Let $x_1,...,x_m$ be a local coordinate at $x \in U$ and $y_1,...,y_n$ be at $f(x),$ related by $y_i=g_i(x_1,...,x_m)$ for $1 \leq i \leq n.$ Then, locally $\omega$ can be expressed as

$$\omega=\sum_{i_1<...

therefore, (why?)

$$f^*\omega=\sum_{i_1<...

While any $k$-form on an $m$-dimensional manifold, where $k>m$ is zero.