TITLE: Killing vector fields are affine
QUESTION [3 upvotes]: Let $(M, g)$ be a Riemannian manifold and let $X$ be a smooth vector field on $M$. We say that $X$ is affine if $L_X \nabla = 0$, where $\nabla$ is the Riemannian connection on $M$. How do we prove that Killing vector fields are affine?

I know that if $X$ is a Killing vector field, then $L_X g = 0$, which is equivalent to saying that for all other smooth vector fields $Y$ and $Z$, $$g(\nabla_Y X, Z) + g(Y, \nabla_Z X) = 0. $$ We can also say that the flow of $X$ preserves $g$, that is, $(\phi_X^t)^*g = g$, where $\phi_X^t$ is the flow of $X$ at time $t$. However, I don't know how to relate this to the Lie derivative of the connection.
I also read about a formula relating the coefficients of $L_X \nabla$ with the curvature tensor (in local coordinates). However, I would like to avoid using it.

REPLY [2 votes]: In principle, there is a simple anwser: As you noted, the local flows of a Killing filed all are isometries, so they preserve the metric. Since the Levi-Civita connection is naturally derived from the metric, it is also preserved by the local flows of a Killing field. (You can make this explict either via the Koszul-formula mentioned in the comments. Alternatively, you can observe that pulling back the Levi-Civita connection along an isometry, you obtain a torsion free connection that is compatible with the metric. Then the result follows from uniqueness of the Levi-Civita connection.) The simplest definition of an affine vector field actually is that all local flows prefer the connection. If you want to make to phrase this in terms of a Lie derivative, you first have to say what exactly you mean by the Lie derivative of a linear connection (since this is not a tensor field). I guess, the easiest way to get to this is to differentiate the property that the connection is preserved by the local flows ...
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