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 $dx^1_p \wedge ... \wedge dx^n_p(X_{1p},...,X_{np}) \neq 0 \forall p$?
Y will separe the questions.
1)Isn't $dx_1\wedge...\wedge dx_n$ always a nowhere vanishing?
No, $dx_1\wedge...\wedge dx_n$ is only locally defined.
2)Indeed, for me we should have $dx_{1p}\wedge\dots\wedge dx_{np}\neq0, \ \forall p$ since $\{dx_{1p},...,dx_{np}\}$ is a basis of $T^∗pM$, no?
No, because the same reason of the last question.
3)If $X1,\dots,Xn$ is a vector field such that $X_{1p},\dots,X_{np}$ are linearly independent for all $p$, does this imply that $dx_{1p}\wedge\dots\wedge dx_{np}(X_{1p},...,X_{np})\neq0, \ \forall p$?
Yes, if $p$ is in the domain of the chart.