2
$\begingroup$

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).
$\endgroup$
5
  • $\begingroup$ 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. $\endgroup$ May 16, 2016 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. $\endgroup$ May 16, 2016 at 22:39
  • 1
    $\begingroup$ 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. $\endgroup$ May 16, 2016 at 22:39
  • $\begingroup$ Noted, I'll ask a separate question. $\endgroup$
    – Michael
    May 16, 2016 at 22:39
  • $\begingroup$ Actually you just answered the second question, thank you. $\endgroup$
    – Michael
    May 16, 2016 at 22:50

1 Answer 1

8
$\begingroup$

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 $w_2$.

$\endgroup$
3
  • 1
    $\begingroup$ 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. $\endgroup$ May 16, 2016 at 22:00
  • 1
    $\begingroup$ 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." $\endgroup$ May 16, 2016 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. $\endgroup$ May 17, 2016 at 2:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.