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Is every orientable $I$-bundle over an orientable surface $F$ trivial, $I \times F$? Is this also the case for vector all bundles?

Similarly, is every orientable $I$-bundle over an nonorientable surface equal to the twisted bundle $I \tilde{\times} F$?

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    $\begingroup$ I-bundles are the unit disc bundles of (normed) real line bundles, because the inclusion $O(1) \to \text{Diff}(I)$ is a homotopy equivalence. Real line bundles are trivial if and only if they are orientable. $\endgroup$
    – mme
    Commented Feb 19, 2019 at 15:55
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    $\begingroup$ I misread. Your question is about the total space. In the case of non-orientable base, instead the above comment should conclude that the total space is orientable iff $w_1(\lambda) = w_1(F)$, as desired. $\endgroup$
    – mme
    Commented Feb 19, 2019 at 17:54
  • $\begingroup$ Can you answer using simpler terms? Lets say I have a $n$-torus $S$. Is every I-bundle homeomorphic to $I \times S$ or are there different interval (line) bundles? $\endgroup$
    – Jake B.
    Commented Feb 21, 2019 at 19:53
  • $\begingroup$ What I said is that the classification of real line bundles is the same as the classification of I-bundles (where "classification of I-bundles" means up to fiber-preserving diffeomorphisms). Real line bundles are determined up to isomorphism by their first Stiefel-Whitney class, $w_1(\lambda) \in H^1(M;\Bbb Z/2)$. (you can find this in many sources, eg Hatcher's notes on vector bundles.) The n-torus thus supports $2^n$ real line bundles, and in particular, some non-trivial ones. The discussion on orientability of the total space follows from additivity: $w_1(L \otimes L') = w_1(L) + w_1(L')$. $\endgroup$
    – mme
    Commented Feb 21, 2019 at 20:01
  • $\begingroup$ if if helps $w_1$ can be thought of as the homomorphism $\pi_1 M \to \pm 1$ which records whether or not the bundle reverses orientation or not as you traverse a loop. $\endgroup$
    – mme
    Commented Feb 21, 2019 at 20:07

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