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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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    @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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