Timeline for When a Killing vector field on Riemannian manifold $(M,g)$ is gradient?
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6 events
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| Sep 2, 2017 at 19:23 | comment | added | Willie Wong | @C.F.G. that is elementary calculus. A smooth function on a compact manifold must attain its maximum at some point, and at that point its gradient vanishes and cannot have length one relative to any Riemannian metric. // What does Perelman's theorem have to do with this? | |
| Sep 1, 2017 at 21:20 | comment | added | Willie Wong | In fact no compact manifolds (w/o boundary) admit a unit vector field that is a gradient, so we can rule all of them out without even touching the Killing assumption. // The gradient condition implies that the killing field is hypersurface orthogonal, so the manifold must split as $(a,b) \times \Sigma$ with the product metric. This is an if and only if. | |
| Sep 1, 2017 at 21:10 | comment | added | Igor Khavkine | Benoît, of course you're right about the terminology (fixed it). Sorry for the momentary vocabulary lapse! I should also make the point that my comments were meant to be considered locally, where there is no difference. | |
| Sep 1, 2017 at 21:03 | history | edited | Igor Khavkine | CC BY-SA 3.0 |
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| Sep 1, 2017 at 20:36 | comment | added | Benoît Kloeckner | I am not familiar with your notation, but what you call exactness rather looks like closedness. Examples asked should be even more constrained than what you implied (e.g. a flat torus has no killing field which is a gradient). | |
| Sep 1, 2017 at 15:07 | history | answered | Igor Khavkine | CC BY-SA 3.0 |