I mean, is it possible to have a connection on the 2-sphere with vanishing curvature but not vanishing torsion? In a more general sense, it is know that every Riemannian manifold has a Levi-Civita connection,is this true for the Weitzenböck connection?

  • $\begingroup$ I suggest you to ask this question on MathOverflow.net $\endgroup$ – Yuri Vyatkin Feb 22 '13 at 5:57

It seems to me that a vector bundle on a simply connected manifold that has a flat connection must be trivial. (Parallel transport a frame at a base point to all other points.) But $S^2$ is far from parallelizable.

  • $\begingroup$ Thank you. Your comment makes things clear. Is it correct then to say that there are topological obstructions for the existence of Weitzenböck connection? If I understand correctly, if one assume $S^{2}$ has a flat connection then one can choose a frame and parallel transport it all over the sphere. The condition of being flat implies the parallel transport is path independent. Moreover,by glueing geodesics one can form a non vanishing vector field on $S^{2}$. The result given by the problem of combing the sphere give us a contradiction. $\endgroup$ – yess Mar 25 '14 at 1:27
  • $\begingroup$ Well, parallel transport is path-independent because of both flatness and simple-connectivity. I would not mention the word geodesic — it's normally associated to an affine connection. Right, $\chi(S^2)=2$ tells us the tangent bundle is not trivial. $\endgroup$ – Ted Shifrin Mar 25 '14 at 1:37

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