# Understanding orientability of vector bundles

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? – t.b. Jul 9 '11 at 17:43
@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 $\det(m^i_j)>0$. – Giuseppe Negro Jul 9 '11 at 17:47
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? – t.b. Jul 9 '11 at 17:57
@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. – Giuseppe Negro Jul 9 '11 at 18:16
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...) – Giuseppe Negro Jul 9 '11 at 18:53