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Lefschetz fixed point theorem (for example) implies that a compact manifold with a nowhere vanishing vector field must have Euler characteristic zero. Is there a way to draw stronger algebraic conclusion from the existence of a basis of vector fields / a trivial tangent bundle?
I would like to apply something like that to show the tangent bundle of the Klein bundle is nontrivial (which I guessing to be the case from trying to draw some), but I don't know any relevant theorem.
A smooth manifold whose tangent bundle is trivializable is called parallelizable or frameable. On such a manifold, not only does the Euler characteristic vanish, but all characteristic classes vanish. Hence the nonvanishing of any characteristic class prevents a manifold from being parallelizable.
In particular, the Klein bottle is non-orientable, so its first Stiefel-Whitney class $w_1$, which measures non-orientability, doesn't vanish. More directly you can use the argument anomaly gives in the comments: a smooth manifold is orientable iff the top exterior power of its tangent bundle is trivializable.
There are more precise things you can say if you know that there are, say, $k$ linearly independent vector fields; this is a good exercise in characteristic classes.
