Let $M$ be a connected orientable open 4-manifold (noncompact, without boundary).

  1. Is it possible for $M$ to be non-parallelizable ? If yes, what example of such $M$ is there ?

[EDIT : The answer to this is yes : see the answer of Danny Ruberman]

  1. Suppose now that $M$ admits a lorentzian metric and a spin structure [EDIT]. Is it then possible for $M$ to be non-parallelizable ? (This wikipedia page seems to say that this is not possible, but I don't understand the argument).
  • $\begingroup$ Generally it's considered bad form to change your question after it has been answered. It would be more appropriate to check Ruberman's answer as correct, and then post a follow-up (separate) question. $\endgroup$ – Ryan Budney May 16 '16 at 22:36
  • $\begingroup$ A classification of $SO(n)$ bundles with $n\le 4$ over a complex of dimension $\le 4$ is given by Dold-Whitney in maths.ed.ac.uk/~aar/papers/doldwhit.pdf, see Theorem 1. The classification is in terms of certain charactersitic classes which all vanish for open orientable spin manifolds. $\endgroup$ – Igor Belegradek May 16 '16 at 22:39
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    $\begingroup$ To answer your 2nd question, perhaps the Wikipedia description of spin structures and characteristic classes would answer it? If your manifold is spin the tangent bundle trivializes over the 2-skeleton. It automatically trivializes over the 3-skeleton as $\pi_2 SO_4$ is trivial. Since its non-compact it admits a cell structure with no $4$-cells so you are done. $\endgroup$ – Ryan Budney May 16 '16 at 22:39
  • $\begingroup$ Noted, I'll ask a separate question. $\endgroup$ – Michael May 16 '16 at 22:39
  • $\begingroup$ Actually you just answered the second question, thank you. $\endgroup$ – Michael May 16 '16 at 22:50

Yes to your first question; the Stiefel-Whitney classes obstruct parallelizability, even for open manifolds. So for instance a non-orientable manifold (eg a Mobius band cross R^2) is not parallelizable. An oriented example would be $CP^2$ minus a point, which has nonzero $w_2$.

I think you may have interpreted that wikipedia page incorrectly. It says that if your manifold is of the form $M^3 \times R$, with $M$ orientable, then it's parallelizable. (This follows from the fact that oriented 3-manifolds are parallelizable; you can find this in eg Milnor-Stasheff, Characteristic Classes.) But I don't think that every Lorentzian manifold is of that form; I imagine that there are some with non-trivial $w_2$.

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    $\begingroup$ I suppose taking a non-orientable 3-manifold and building the orientable (total space) $S^1$-bundle over it would give you a non-parallelizable lorentzian manifold. You could then puncture it once to make it open. $\endgroup$ – Ryan Budney May 16 '16 at 22:00
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    $\begingroup$ It seems the wikipedia article assumes the manifold is spin, in which case there is no $w_2$. It says: "Usually, one also requires that a world manifold admits a spinor structure in order to describe Dirac fermion fields in gravitation theory. There is the additional topological obstruction to the existence of this structure. In particular, a noncompact world manifold must be parallelizable." $\endgroup$ – Igor Belegradek May 16 '16 at 22:02
  • $\begingroup$ @Ryan: Perhaps a more natural construction is to take a compact non-orientable surface (with a Riemannian metric) $\times R^2$ where the latter has the Minkowski metric. This version is (presumably--I don't know much about Lorentz metrics) complete. $\endgroup$ – Danny Ruberman May 17 '16 at 2:04

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