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Is it true that for a smooth $m$-manifold $M$ and for $\bigwedge^m M := \bigwedge^m(T^{dual}M)$ (which I understand to be its 'top bundle')

$M$ is compact and oriented/orientable if and only if $\bigwedge^m M$ is trivial?

Well, for any smooth $m$-manifold $M$, I think $(\bigwedge^m M)_p \cong \bigwedge^m(T^{dual}_pM \cong \mathbb R^m) \cong \mathbb R^{\binom{m}{m}} = \mathbb R$, so I guess $\bigwedge^m M \cong M \times \mathbb R$, so I think specifically we have $\bigwedge^m M$ not only a trivial bundle (defined: isomorphic to $M \times \mathbb R^n$, for some $n$) but like, a trivial line bundle (defined: isomorphic to $M \times \mathbb R$). I don't quite see what compact or oriented/orientable has to do with this.

Also: I think $M$ is indeed manifold without boundary, so not sure if Stokes' theorem is relevant. I think what might be relevant is that

  1. A top smooth form on a compact smooth $m$-manifold (smooth $m$-form on compact smooth $m$-manifold) which is never zero is never exact.

  2. Smooth $m$-manifold $M$ is orientable if and only if $M$ has a smooth top form that is never zero.

  3. Top deRham cohomology group of a compact orientable manifold is 1-dimensional

  4. The relationship between de rham cohomology and exterior algebra, which I've forgotten. But I think it's to do with that forms are sections of wedge bundles

  5. Maybe something to do with Poincaré duality and that compact de rham cohomology = regular de rham cohomology for compact manifolds or something

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If $M$ is a manifold of dimension $n$, then $\Lambda^n(T^*M)$ is a rank $\binom{n}{n}=1$ vector bundle over $M$, that is, a line bundle. But one does not have, in general, $\Lambda^n(T^*M) \simeq M \times \mathbb{R}$. This would imply that it is a trivial line bundle. On a given manifold, there could exist many non-trivial line bundles. For example, the Mobius band can be viewed as a non-trivial line bundle over the cirlce $\mathbb{S}^1$.

The answer to your question will depend on the definition of orientability you are using. One can say that $M$ is orientable if it has an orientation atlas, that is an atlas where change of charts have positive jacobian.

What is true, is that, with this definition of orientability $M$ is orientable if and only if $\Lambda^n(T^*M)$ is a trivial line bundle. The compactness has nothing to do with this.

The fact that $\Lambda^n (T^*M)$ is a trivial line bundle is equivalent to the fact that it has a non-vanishing section. Thus, $M$ is orientable if and only if there exists a nowhere-vanishing differential $n$-form, that is a volume form.

Moreover, Stokes theorem is a theorem about integration on manifolds, which is defined related to an orientation... So it seems irrelevant to apply Stokes theorem to a manifold to see if it is orientable (maybe I misunderstood what you were talking about while considering Stokes theorem).

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  • $\begingroup$ Thanks DIdier_! I'll analyse this later. $\endgroup$
    – BCLC
    Dec 10 '20 at 5:30
  • $\begingroup$ which part exactly here is wrong please 'I think $(\bigwedge^m M)_p \cong \bigwedge^m(T^{dual}_pM \cong \mathbb R^m) \cong \mathbb R^{\binom{m}{m}} = \mathbb R$, so I guess $\bigwedge^m M \cong M \times \mathbb R$' ? $\endgroup$
    – BCLC
    Dec 12 '20 at 2:30
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    $\begingroup$ The last part is false ("so I guess"). A line bundle is not in general a direct product $M\times \mathbb{R}$. $\endgroup$
    – Didier
    Dec 12 '20 at 12:25
  • $\begingroup$ Ok I seem to have forgotten my bundles. thanks DIdier_! $\endgroup$
    – BCLC
    Dec 13 '20 at 3:41

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