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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?

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    $\begingroup$ You're confusing "non-zero linear map" with "non-vanishing linear map" (there are no non-vanishing linear maps). As you say, $\alpha_p$ is non-vanishing means $\alpha_p$ is nonzero for every $p$, or in other words $\alpha_p: T_p M \to \mathbb{R}$ is not the zero linear map. This just means that there exists $v \in T_p M$ such that $\alpha_p(v) \neq 0$. $\endgroup$ – Paul Siegel May 28 '12 at 11:54
  • $\begingroup$ @PaulSiegel thnks for the comment.. i got it now. $\endgroup$ – Junu May 28 '12 at 12:39
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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.

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