# A manifold admits a nowhere vanishing volume form if and only if it is orientable?

on this wikipedia article, it is said :

A manifold admits a nowhere vanishing volume form if and only if it is orientable

I don't really understand why. Isn't $$dx^1 \wedge ... \wedge dx^n$$ always a nowhere vanishing? Indeed, for me we should have $$dx^1_p \wedge ... \wedge dx^n_p \neq 0 \forall p$$ since $$\{dx^1_p,...,dx^n_p\}$$ is a basis of $$T_p^*M$$, no?

Also another question :
If $$X_1,...,X_n$$ is a vector field such that $$X_{1p},...,X_{np}$$ are linearly independent for all $$p$$, does this imply that $$dx^1_p \wedge ... \wedge dx^n_p(X_{1p},...,X_{np}) \neq 0 \forall p$$?

• Note that $dx^1 \wedge \ldots \wedge dx^n$ is defined locally and depends on the choice of coordinates. And you cannot define coordinates globally except in trivial cases. – Mindlack Jan 11 at 19:29
• @Mindlack Sorry, I don't get your point, could you expand a bit? – roi_saumon Jan 11 at 19:34
• How would you define $dx^1 \wedge dx^2$ on the 2-sphere for instance? – Mindlack Jan 11 at 19:36
• Yes, this is the point: your coordinates are local, they are only defined in charts. And the corresponding $dx^1 \wedge dx^2$ at each $p$ will depend of which chart you took. So you will not have one well-defined $2$-form at each point, or if you do, there is no trivial way to make sure it is smooth, or even continuous. If you wish, orientability corresponds to being able to « glue together » enough nowhere-vanishing forms defined in different charts. – Mindlack Jan 11 at 19:53
• What you say about $\mathbb R^3$ is true. However, it is quite rare for an $n$-manifold to possess an atlas having only a single coordinate chart. In fact that holds if and only if the manifold is diffeomorphic to an open subset of $\mathbb R^n$. In particular, it is never true for compact manifolds. – Lee Mosher Jan 11 at 23:37