I'd like some help to prove the following: $f\colon U \subset \mathbb{R}^m \to \mathbb{R}^n$, differentiable; $m
Show that $f^*\omega = 0$
Thanks.
I'd like some help to prove the following: $f\colon U \subset \mathbb{R}^m \to \mathbb{R}^n$, differentiable; $m
Show that $f^*\omega = 0$
Thanks.
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.