Let $M \subset \mathbb R^d$ be k dimensional submanifold and $\Phi:U\to M \subset V$ continuously differentiable while $U \subset \mathbb R^n, V \subset \mathbb R^d $ open.
Why is $\Phi^*(\omega)=0$ for every l-Form $\omega$ on V with $l>k$?
I tried to prove this, but failed. Any help or linked prove is much appreciated.
This boils down to the linear algebra observation that a $l$-multilinear alternating map $T \colon V \times V \times \dots \times V \rightarrow \mathbb{R}$ on a $k$-dimensional vector space $V$ must be zero. The reason is that a list $(v_1,\dots,v_l)$ of $l$ vectors from $V$ must be linearly dependent so if
$$ v_i = \sum_{j \neq i} a_j v_j $$
then
$$ T(v_1,\dots,v_l) = \sum_{j \neq i} a_{j} T(v_1,\dots,v_j,\dots,v_l) = 0$$
because in each summand the vector $v_j$ appears twice (once in the $j$th place and once in the $i$th place).
Applying this to $T = \Phi^{*}(\omega)|_p$ (which is by definition an $l$-multilinear alternating map on a $k$-dimensional vector space $T_pM$) we see that $\Phi^{*}(\omega)|_p = 0$ for all $p \in M$ so $\Phi^{*}(\omega) = 0$.