Prove that $f^*\omega=0$

Question

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.

Answer 1

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.