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I'm having trouble understanding how orientability of vector bundles work. The book I'm reading, Spivak's A comprehensive introduction to differential geometry, is not very clear on this.

Edit: Definition If $V$ is a real vector space, and $(b_1 \ldots b_n), (c_1 \ldots c_n)$ are two ordered bases with $b_j=m^i_j c_i$, we say that they are equally oriented if $\det(m^i_j)>0$. An orientation of $V$ is an equivalence class of ordered bases.

Let us take a rank $n$ vector bundle $E(B, \mathbb{R}^n, \pi)$ ($E$=total space, $B$=base space, $\mathbb{R}^n$=typical fibre, $\pi$=projection). We say that $E$ is orientable if we can find a trivializing atlas $\mathcal{A}=\{(U_\alpha, t_\alpha)\}_{\alpha}$ such that $t_\alpha \circ t_{\beta}^{-1}$ is orientation-preserving on every fibre of $U_{\alpha}\cap U_{\beta}\times \mathbb{R}^n$. Ok. What I don't get very well is how this allows us to consistently choose an orientation on each fibre of $E$. I guess we need to import the oriented structure of $U \times \mathbb{R}^n$ into $E$ someway, but I can't seem to focus the details very well.

Can you explain this to me? Alternatively, you can give me a good reference too. Thank you.


Bonus question (secondary)

Spivak's book adopts a different definition, that is: a family $\{\mu_p\}_{p \in B}$ of orientations for $\pi^{-1}(p)$ is an orientation of $E$ if the following compatibility condition is satisfied:

If $t \colon \pi^{-1}(U)\to U \times \mathbb{R}^n$ is an equivalence (=vector bundle isomorphism) and the fibres of $U \times \mathbb{R}^n$ are given the standard orientation, then $t$ is either orientation preserving or orientation reversing on all fibres.

I cannot understand the link between this definition and the one above.

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    Minor point: How exactly does Spivak (do you) define an orientation? Equivalence class of bases or an element of of the top exterior power of the dual?2011-07-09
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    @Theo: Equivalence class of bases. If $V$ is a real vector space, and $(b_1 \ldots b_n), (c_1 \ldots c_n)$ are two ordered bases with $b_j=m^i_j c_i$, we say that they are *equally oriented* if $\det(m^i_j)>0$.2011-07-09
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    Okay, so fix an orientation on $\mathbb{R}^n$ once and for all, and fix an oriented trivializing atlas $\mathcal{A}$. Given a point $p \in B$ in the base, and any chart $t_\alpha: U_{\alpha} \times \mathbb{R}^n$, you get an isomorphism $\pi^{-1}(p) \to \mathbb{R}^n$. Call a basis in $\pi^{-1}(p)$ positively oriented if this isomorphism sends its class of $\pi^{-1}(p)$ to the one chosen in $\mathbb{R}^n$. This is well-defined by assumption on orientation of $\mathcal{A}$. I think this should answer both of your questions, no?2011-07-09
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    @Theo: The first one is answered, ok! It was easy, it is just a little confusing to the beginner like me (this is true of all differential geometry, I guess). As for the second question, it remains a point to be clarified. Suppose we have an oriented trivializing atlas, which gives us an orientation on every fibre. Let $t\colon \pi^{-1}(U) \to U \times \mathbb{R}^n$ be a vector bundle isomorphism. *Why is $t$ orientation-preserving or orientation-reversing on all fibres*? I'll think about it a bit.2011-07-09
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    Must be something like the following. Every vector bundle automorphism of a trivial bundle $X \times \mathbb{R}^n$ preserves or reverses orientation on all fibres, by a matter of continuity of determinant function. Now take our vector bundle isomorphism $t \times \pi^{-1}(U) \to U \times \mathbb{R}^n$ and write $t=(t \circ t_{\alpha}^{-1})\circ t_{\alpha}$, where $t_{\alpha}$ is one of the trivializations in $\mathcal{A}$. Since $t_\alpha$ takes every chosen base of single fibers $\pi^{-1}(p)$ in the canonical base of $\{p\}\times \mathbb{R}^n$, (continues...)2011-07-09
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    (...) and since $t \circ t_{\alpha}$ is a vector bundle automorphism of a trivial bundle, $t$ is always orientation-preserving or orientation-reversing in the whole of $\pi^{-1}(U \cap U_{\alpha})$. By the condition on $t_{\alpha}\circ t_{\beta}^{-1}$ $t$ is always orientation-preserving or orientation-reversing in the whole of $\pi^{-1}(U)$. I know this is not precise but I don't think this annoying little question deserves more effort than this. @Theo: Thank you very much for your help! Ciao!2011-07-09
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    You can look at all these orientation stuff from two angles - the point is that you want to give each fiber an orientation so that they are compatible as you move continuously on the manifold. One way to implement this is to say that locally, the orientation is fixed, and I want the orientation to be compatible as I go from one chart to another. As the manifold is covered by charts, we are good. Another way to implement this is to a priori give an orientation to each fiber. Then you want to say that the orientation is sort of continuous, i.e. locally consistent. That's your second definition.2011-07-09
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    @Soarer: Yes, now I get the idea. Thank you. Those things look terrible at first sight but in the end they are quite natural.2011-07-09

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