Equivalent definition of orientability

Question

I came across the following definition of orientability.

Let $O$ be the "$0$-section" of the exterior n-bundle $\Lambda^nM^*$ with $M$ being a connected differentiable manifold of dimension $n$. It is said that $M$ is orientable if $\Lambda^nM^* \smallsetminus O$ has two components; an orientation is a choice of one of the two components of $\Lambda^nM^* \smallsetminus O$. It is said that $M$ is non-orientable if $\Lambda^nM^* \smallsetminus O$ is connected.

Now, this seems like a really good definition. But I have a problem understanding how it is equivalent to saying that an orientable manifold $M$ admits a nowhere-vanishing smooth $n$-form. In particular, if the definitions are to be equivalent, the following statement must be true.

$$ \Gamma(\Lambda^nM^* \smallsetminus O) = \varnothing \Leftrightarrow \Lambda^nM^* \smallsetminus O \text{ is connected}$$

I don't see why that should be the case. For example, I could take a Möbius band (which can be seen as the $2$-dimensional cotangent bundle over the $1$-dimensional Möbius circle) and start cutting through it without touching the base circle (say a constant $2 cm$. off of it). I would go around twice before completing the section. So I do have a smooth non-vanishing $1$-form here, although I believe it is supposed to be non-orientable by the given definition.

Another example could be an exterior $n$-bundle which has the topology of a torus (does that exist?) which clearly admits a smooth section of $\Lambda^nM^* \smallsetminus O$.

Answer 1

First, to address your example: there is no such thing as a "Möbius circle". The central circle of a Möbius band is just an ordinary circle: it is diffeomorphic to $S^1$, and in particular it is orientable (and the Möbius band is not its cotangent bundle!). In any case, it is unclear how you are claiming that what you have described would give a nonvanishing section of the Möbius bundle--it would seem to have two different values in each fiber (since you go around the whole circle twice before coming back to where you started).

Now let's look at your general question. One direction of the equivalence is pretty simple: if there is a nowhere vanishing $n$-form $\omega$ on $M$, define a subset $U\subseteq\Lambda^n M^*\setminus O$ by saying $\alpha\in U$ iff $\alpha/\omega(\pi(\alpha))>0$, where $\pi:\Lambda^nM^*\to M$ is the projection. This definition makes sense since $\alpha$ and $\omega(\pi(\alpha))$ are both nonzero elements of a $1$-dimensional vector space (the fiber of $\Lambda^nM^*$ over $\pi(\alpha)$), so they differ by a nonzero scalar factor. The set $U$ is open, and its complement is also open since its complement is just the set of $\alpha$ such that $\alpha/\omega(\pi(\alpha))<0$. So $U$ is a nontrivial clopen subset of $\Lambda^n M^*\setminus O$, so $\Lambda^n M^*\setminus O$ is disconnected. (In fact, $U$ and its complement are exactly the two connected components of $\Lambda^n M^*\setminus O$.)

The converse is more complicated, since you have to actually somehow construct a nonvanishing $n$-form. Here's a sketch of how it works. Suppose $\Lambda^n M^*\setminus O$ is disconnected and let $U$ be one of its components. You can show that $\pi(U)$ is all of $M$ (using the fact that $M$ is connected), and the intersection of $U$ with each fiber is closed under positive linear combinations (basically, each fiber of $\Lambda^n M^*$ looks like a copy of $\mathbb{R}$, and in each fiber one of the two components of $\Lambda^n M^*\setminus O$ must contain all the positive numbers and the other must contain all the negative numbers). In any coordinate patch on $M$, you can construct a nonvanishing $n$-form which takes values in $U$ anywhere. Using a partition of unity, you can add together these local $U$-valued $n$-forms to get a global $n$-form. The global $n$-form will still be $U$-valued everywhere, and in particular will be nonvanishing.