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One definition of the Killing field is as those vector fields along which the Lie Derivative of the metric vanishes. But for very many calculation purposes the useful way to think of them when dealing with the Riemann-Christoffel connection for a Riemannian metric is that they satisfy the differential equation, $\nabla_\nu X_\nu + \nabla_\nu X_\nu = 0$

I suppose solving the above complicated set of coupled partial differential equations is the only way to find Killing fields given a metric. (That's the only way I have ever done it!) Would be happy to know if there are more elegant ways.

But what would be the slickest way to say prove that the scalar curvature is constant along a Killing field, i.e it will satisfy the equation $X^\mu \nabla_\mu R = 0$ ?

Sometimes it is probably good to think of Killing fields as satisfying a Helmholtz-like equation sourced by the Ricci tensor.

Then again very often one first starts off knowing what symmetries one wants on the Riemannian manifold and hence knows the Lie algebra that the Killing fields on it should satisfy. Knowing this one wants to find a Riemannian metric which has that property. As I had asked in this earlier question

There is also the issue of being able to relate Killing fields to Laplacians on homogeneous spaces and H-connection like earlier raised in this question. Good literature on this has been very hard to find beyond the very terse and cryptic appendix of a paper by Camporesi and Higuchi.

I would like to know what are the different ways of thinking about Killing fields some of which will say help easily understand the connection to Laplacians, some of which make calculations of metric from Killing fields easier and which ones give good proofs of the constancy of scalar curvature along them. It would be enlightening to see cases where the "original" definition in terms of Lie derivative is directly useful without needing to put it in coordinates.

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what would be the slickest way to say prove that the scalar curvature is constant along a Killing Field?

The (local) flow of the Killing field must preserve the metric and hence also the scalar curvature. Is that answer too terse?

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@Michael Not terse but I was looking for a computational proof of this statement (may be I am sounding naive!) I get your intuitive argument but I was looking for something like starting from the differential equation that we know a Killing Field should satisfy can one prove the equation that $X^\mu \nabla _ \mu R=0$ ? – Anirbit Jul 11 '10 at 17:26
Michael's argument is not just intuitive. It's a rigorous proof. If you integrate the Killing field, you get a 1-parameter group of isometries $\phi_t$. So $g_t = g$ for each $t$ near zero, where $g_t = \phi_t^*g$. Therefore, $R\circ\phi_t = R(g_t) = R(g)$. Differentiating with respect to $t$ at $t = 0$ gives the equation you want. But you probably want to prove the equation directly from the definition of a Killing field. This is reasonable and can be extracted from the proof above. – Deane Yang Jul 11 '10 at 18:05

Sometimes it is probably good to think of Killing fields as satisfying a Helmholtz-like equation sourced by the Ricci scalar.

Firstly, the source is not the Ricci scalar, but the Ricci tensor. The equation, up to a minus sign and some constant factors should be $\triangle_g \tau_a \propto R_{ab}\tau^b$.

That expression sometimes is useful when you already know the metric and the Killing vector field, and are interested in a priori esimates in geometric analysis. My Ph.D. thesis contains some related examples. (Though in the Riemannian case on a compact manifold, the Bochner technique is generally more fruitful; the classical paper of Bochner's [in the Killing case] is, however, isomorphic to contracting the elliptic equation against the Killing field itself. )

Anyway, the "Helmholtz like" equation is obtained by taking the trace of the Jacobi equation which is satisfied by any Killing vector field. So you are definitely losing information unless you work with only 2 or 3 manifolds.

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Thanks for the references and pointing out the slip with the Ricci tensor. Corrected that. – Anirbit Jul 11 '10 at 17:20

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