# $M$ is compact and oriented/orientable if and only if top bundle trivial?

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

## 1 Answer

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).

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