# differential form

one form $\alpha$ over a smooth manifold is non vanishing means for every $p\in M$, $\alpha_p\neq 0$.

But $\alpha_p$ is linear map $T_M\to \mathbb R$, hence $\alpha_p(0)=0$. So confusion arises and precise question is:

What does mean by non vanishing one form. What does mean by $\alpha\wedge..\wedge\alpha\neq 0$

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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 $\alpha_p(v) \neq 0$. – Paul Siegel May 28 '12 at 11:54
@PaulSiegel thnks for the comment.. i got it now. – Junu May 28 '12 at 12:39

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.