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Analysis and Algebra on Differentiable Manifolds, first edition, chapter 7.3.1, defines orientation on a vector space and orientable manifolds. There is a part of the definition that I do not understand:

Let $V$ be a real vector space of dimension $n$. An orientation of $V$ is a choice of component of $\Lambda^n V-\{0\}$. [...] Let $O$ be the "0-section" of the exterior n-bundle $\Lambda^n M^*$; that is,

\begin{equation} O = \bigcup_{p \in M} \{0 \in \Lambda^n T^*_p M\} \end{equation}

Then since $\Lambda^n T^*_p M - \{0\}$ has exactly two components, it follows that $\Lambda^n T^*M - O$ has at most two components.

Why do they say that there are two components ? Why not just one ? On a 3-manifold, for example, $dx_1 \wedge dx_2 \wedge dx_3$ is a component, what would the other one be, $- dx_1 \wedge dx_2 \wedge dx_3$ ?

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up vote 5 down vote accepted

On a vector space of dimension $n$, $\Lambda^n V$ has dimension $1$, so it is isomorphic to $\mathbb{R}$. If you remove the point $0$ from $\mathbb{R}$, how many components does it have?

Also, I'm not sure what you mean by "$dx_1\wedge dx_2\wedge dx_3 $ is a component". In the example of $\mathbb{R} -\{0\}$, an analogous statement is "$1$ is a component."

Perhaps what you meant to say is that $dx_1\wedge dx_2 \wedge dx_3$ lies in one component while, presumably, $-dx_1 \wedge dx_2 \wedge dx_3$ lies in another. If $M$ is orientable, then this is true, but if $M$ is nonorientable, this isn't true.

The picture in the case where $M$ is nonorientable (dropped down several dimensions for visualization purposes) is analogous to that of a Moebius band. Projecting a Moebius band onto its core circle $S^1$ gives it the structure of an $\mathbb{R}$ bundle over $S^1$. Removing the core $S^1$ (i.e., removing the $0$ section) locally breaks the Moebius band into two components, but by going all the way around it, you see that it's really just a single component.

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Thank you for the explanation, I did not understand component in the topological sense, but mistakenly in the sense of components of a vector. – vkubicki Feb 18 '13 at 8:55

If $V$ is $n$-dimensional, then $\Lambda^nV$ is $1$-dimensional, so removing the origin disconnects it into two components.

In your $3$-manifold example, $dx_1\wedge dx_2\wedge dx_3$ only makes sense on a local neighbourhood $U$ diffeomorphic to an open set in $\mathbb{R}^n$. Such things are always orientable, so $\Lambda^n T^*M-O$ will have two components. The element $dx_1\wedge dx_2\wedge dx_3$ is not strictly a component itself, but it determines one, i.e. the component containing it. Then it is indeed true that $-dx_1\wedge dx_2\wedge dx_3$ lies in the other component.

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Thank you for your explanation ! – vkubicki Feb 18 '13 at 8:56

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