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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 $dx^1_p \wedge ... \wedge dx^n_p(X_{1p},...,X_{np}) \neq 0 \forall p$?

roi_saumon
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    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. – Aphelli Jan 11 '19 at 19:29
  • @Mindlack Sorry, I don't get your point, could you expand a bit? – roi_saumon Jan 11 '19 at 19:34
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    How would you define $dx^1 \wedge dx^2$ on the 2-sphere for instance? – Aphelli Jan 11 '19 at 19:36
  • I chose some charts $(U,(x^1,x^2))$ around each point $p$ no? I don't understand – roi_saumon Jan 11 '19 at 19:47
  • You need to define it on the whole sphere, not on some charts. @roi_saumon – Angina Seng Jan 11 '19 at 19:51
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    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. – Aphelli Jan 11 '19 at 19:53
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    Oh,thanks, I think I get it. The object that we create will be be nowhere vanishing but not smooth so not a form. – roi_saumon Jan 11 '19 at 20:34
  • But then in $\mathbb{R}^3$, we would have no problem of gluing it if we take the standard atlas ${\mathbb{R}^3, Id_{\mathbb{R}^3}}$ so $dx \wedge dy \wedge dz$ would be a non-vanishing form (so an orientation form) here right? – roi_saumon Jan 11 '19 at 20:37
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    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 '19 at 23:37
  • https://math.stackexchange.com/questions/2025086/partition-of-unity-and-volume-form-on-a-manifold – user135520 May 30 '19 at 23:36
  • https://math.stackexchange.com/questions/252068/non-vanishing-k-form-on-a-k-manifold-in-mathbbrn-implies-orientability – user135520 May 30 '19 at 23:42
  • The easiest way I know how to construct an oriented volume form on an oriented manifold is to use a Riemannian metric. – Deane Aug 04 '23 at 23:15

1 Answers1

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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.

Luiz Felipe
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