TITLE: Non-vanishing differential form: what does it mean?
QUESTION [9 upvotes]: A $1$-form $\alpha$ over a smooth manifold is non vanishing if for every $p\in M$, $\alpha_p\neq 0$.   
But $\alpha_p$ is a linear map $T_p M\to \mathbb R$ hence $\alpha_p(0)=0$.  So confusion arises and the precise question is:
What does non vanishing mean for differential forms?   
And what does $\alpha\wedge..\wedge\alpha\neq 0$ mean?

REPLY [5 votes]: Non vanishing (at, say, $p$) means that there is a vector $v$ in $T_pM$ such that $\alpha_p(v)\neq 0$. Similarly for the $k$-form, it means that there is a set of $k$ vectors such the form is nonzero if evaluated on these vectors.