# Prove that $f^*\omega=0$

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

Show that $f^*\omega = 0$

Thanks.

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Do you want $k > m$, rather? Otherwise $\omega$ is already zero. –  Dylan Moreland Jan 3 '12 at 5:08
Can you describe the $k$-forms on $\mathbb R^m$ when $k>m$? –  Mariano Suárez-Alvarez Jan 3 '12 at 5:16
Sorry, it should be $k>m$ –  Jr. Jan 3 '12 at 5:19
I'd say try to answer these two subquestions: Is $f^*\omega$ again a $k$-form? Then do Mariano's question. –  Matt Jan 3 '12 at 5:57
The pullback of a k-form is also a k-form. What do the k-forms on $U$ look like? –  user18063 Jan 3 '12 at 6:25

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

$$\omega=\sum_{i_1<...<i_k}\phi_{i_1,...,i_k}dy_{i_1} \wedge ... \wedge dy_{i_k},$$

therefore, (why?)

$$f^*\omega=\sum_{i_1<...<i_k} \sum_{j_1<...<j_k}(\phi_{i_1,...,i_k} \circ f) \frac{\partial(g_{i_1},...,g_{i_k})}{\partial(x_{j_1},...,x_{j_k})}dx_{j_1} \wedge ... \wedge dx_{j_k}.$$

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

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