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I am searching for a non-Killing vector field $\zeta \in\frak X\rm (M)$ where $(M,g)$ is a Riemannian manifold such that
$$\frac12 \frak L_\zeta \rm g+Ric=\lambda g$$ $$ \frak L_\zeta \rm Ric=\lambda \frak L_\zeta \rm g$$ The first equation is called soliton equation and $\zeta$ ia called a potential vector field. It is clear that if a Killing vector field is a solution of the first equation, then $(M,g)$ is Einstein with factor $\lambda$ and the second equation holds. I want to consider a non-Killing vector field. I checked 8 examples with non-constant ricci curvature. In each case I got a system of difficult PDEs so I considered many special cases. All solutions are Killing vector fields.
My question
Is there an example with a non-Killing vector field satisfying these equations?
Or any hind to a non-existence result.
Thanks in advance.

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  • $\begingroup$ I made some edits thanks to Holonomia $\endgroup$
    – Semsem
    Commented Jan 9, 2016 at 19:52

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I don't know about general dimensions, but, for surfaces (i.e., when $M$ has dimension $2$), it is not hard to show, using the Cartan structure equations, that there is no solution to this system for which $\zeta$ is not a Killing field. It is an overdetermined system, and the compatibility conditions rapidly eliminate any possibility of there being a non-trivial solution in this case.

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  • $\begingroup$ Thank you for your answer. Examples lead to the same conclusion for 4-dimensional cases. But I don't know how to proceed. $\endgroup$
    – Semsem
    Commented Jan 9, 2016 at 15:05
  • $\begingroup$ to Sameh Shenawy: if a Killing vector field is a solution your metric is Einstein, isn't it?. So a Killing vector field of a non Einstein manifold is not a solution. Perhaps I am misunderstanding something. $\endgroup$
    – Holonomia
    Commented Jan 9, 2016 at 17:54
  • $\begingroup$ @Holonomia I think it should be as follows. The first part is: If $\zeta$ is Killing, then $M$ is Einstein with factor $\lambda$. Its equivalent is: If $M$ is not Einstein with factor $\lambda$, then $\zeta$ is not Killing. $\endgroup$
    – Semsem
    Commented Jan 9, 2016 at 19:29
  • $\begingroup$ @Sameh Shenawy So your claim: "It is clear that a Killing vector field is a solution" is true just for Einstein manifolds and not in general for a Riemannian manifold as you wrote in your question. $\endgroup$
    – Holonomia
    Commented Jan 9, 2016 at 19:42
  • $\begingroup$ @Holonomia yes you are right and i will edit my question now thanks to you $\endgroup$
    – Semsem
    Commented Jan 9, 2016 at 19:43

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