Let $(M^n,g)$ be a Riemannian manifold that admit a unit Killing vector field $X$. i.e., $\mathscr{L}_Xg=0$. Is it possible that there exist a smooth function $f$ on $M$ such that $X=\mathrm{grad}f$?
A good reference will be very appreciated.
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2 Answers
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It is possible for this to happen (for example, the unit translation vector fields in $\mathbb{R}^n$), but it does not necessarily hold (for example, there are unit Killing vector fields on the stanard Riemannian $3$-sphere, but, obviously, none of them could be gradients, since they never vanish and the $3$-sphere is compact).
I'm not sure what fact or theory you want a reference for.
answered Sep 1, 2017 at 11:02
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$\begingroup$ @C.F.G: I'm sorry, but I don't understand your question. Necessary for what? $\endgroup$ Commented Sep 2, 2017 at 9:28
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$\begingroup$ for Killing vector field! if this vector field is not unit then your examples are correct? $\endgroup$– C.F.GCommented Sep 2, 2017 at 13:23
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1$\begingroup$ @C.F.G: I really don't understand you. My examples are correct examples of what they claim: The first is a gradient vector field that is also a unit Killing field, the second is a unit Killing Field that is not a gradient vector field (even locally). In your comment/question, you didn't say what you meant by 'it', so I couldn't answer your question. Are you asking whether a gradient Killing field is necessarily a unit vector field? (Answer: no.) Are you asking whether a Killing field is necessarily a unit vector field? (Answer: no.) Please ask your question in a complete self-contained way. $\endgroup$ Commented Sep 2, 2017 at 15:17
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The Killing condition $\nabla_a X^\flat_b + \nabla_b X^\flat_a = 0$, together with the gradient condition (really, the closedness condition for a $1$-form) $\nabla_a X^\flat_b - \nabla_b X^\flat_a = 0$, imply that $X$ is covariantly constant, $\nabla_a X^b = 0$. Note that I'm using the relation $X^\flat(-) = \langle X, - \rangle$ to define the $1$-form $X^\flat$.
The existence of covariantly constant vectors implies various restrictive conditions on the geometry of the Riemannian manifold $(M^n, g)$. If there's any particular restriction that you are interested in, you may want to refine your question.
answered Sep 1, 2017 at 15:07
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1$\begingroup$ 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). $\endgroup$ Commented Sep 1, 2017 at 20:36
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1$\begingroup$ 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. $\endgroup$ Commented Sep 1, 2017 at 21:10
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1$\begingroup$ 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. $\endgroup$ Commented Sep 1, 2017 at 21:20
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2$\begingroup$ @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? $\endgroup$ Commented Sep 2, 2017 at 19:23