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Is the tangent bundle of $S^2 \times S^1$ trivial or not?

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    $\begingroup$ It is trivial: it is isomorphic to $pr_1^*T_{S^2}\oplus pr_2^*T_{S^1}$, that is, to the pull back of $T_{S^2}\oplus \varepsilon $, which is trivial. $\endgroup$
    – abx
    Dec 6, 2015 at 12:10
  • $\begingroup$ @abx: Bravo! I suppose the ambiguity of the word"trivial" in your statement was deliberate, although I find it ingenious to show how trivial the proof of triviality of the vector bundle can be made :-) $\endgroup$ Dec 7, 2015 at 13:57
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    $\begingroup$ @Georges Elencwajg: Actually I missed the pun, I used "trivial" in the mathematical sense. $\endgroup$
    – abx
    Dec 7, 2015 at 19:17

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It is trivial.

More generally, by a result of Kervaire a product of any number $n \geq 2$ of spheres is a parallelizable manifold if one of them has odd dimension.

For a short proof see

E. B. Staples, A short and elementary proof that a product of spheres is parallelizable if one of them is odd, Proc. Amer. Math. Soc. 18 (1967), 570--571.

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    $\begingroup$ Actually every orientable $3$-manifold is parallelizable. $\endgroup$
    – ThiKu
    Dec 6, 2015 at 12:25
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    $\begingroup$ Right, $S^2 \times S^1$ is parallelizable for several reasons. So there are several possible generalizations. $\endgroup$ Dec 6, 2015 at 12:29