When is a vector field on a surface a Lie bracket - MathOverflow

When is a vector field on a surface a Lie bracket

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On a Riemannian 2 manifold, when is a vector field (possibly defined only on the manifold minus a finite number of points) the Lie bracket of two mutually orthogonal vector fields?

flag	asked Sep 14 2011 at 11:31 marc 289●1●3

	I'm not sure why you are allowing deleting a finite number of points. How is this different from a vector field defined on a (possibly noncompact or incomplete) Riemannian manifold. Are you assuming that the Riemannian metric on the original manifold is complete? Are you requiring something about the 2 unknown vector fields at the 'missing points'? Also, why use the word 'mutually'? It doesn't appear to do anything. – Robert Bryant Sep 15 2011 at 15:10
	I want the metric to be extendable to the entire compact surface. However the vector field may not be with some zeros. i assume that the metric is complete. Bracket products of of orthomornmal vector fields are pull backs of the Levi-Cevita connection on the unit circle bundle. The questions asks if there is any global constraint on which vector fields can be such pulls backs. – marc Oct 8 2011 at 11:19
	@marc: So you want the two vector fields to be orthonormal, not just orthogonal? – Robert Bryant Nov 4 2011 at 12:14

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