Relation between Jacobi field and Killing field

Question

Let $M$ be a manifold, $X$ a vector field on $M$. Question: If for every point $p\in M$, there exist a neighborhood $U$, such that for any radial geodesic $\gamma$ in $U$, $X(\gamma)$ is a Jacobi field, then, must $X$ be a Killing field?

Answer 1

No - Jacobi fields only preserve geodesics, not all geometry. It's easiest to see the distinction in the flat case - any continuous family of linear transformations of $\mathbb R^n$ will preserve geodesics and thus produce such an $X$, but most linear transformations are not isometries.

For an explicit counterexample, consider the radial vector field $X = x \partial_x + y \partial_y$ on the Euclidean plane $\mathbb R^2$, which is clearly not Killing. An arbitrary geodesic has the form $\gamma(t) = (a+vt,b+wt)$, along which we have $$X(\gamma(t)) = (a+vt)\partial_x + (b+wt)\partial_y.$$ Since the Christoffel symbols are vanishing, the Jacobi equation is simply $(X \circ \gamma)''(t) = 0$, which is clearly satisfied.

What you can conclude is that $X$ is an "infinitesimal affine transformation", i.e. its flow pulls back the Levi-Civita connection. See e.g. Chapter 13 of Lang's Fundamentals of Differential Geometry for some discussion of the relationship - in particular you can conclude $X$ is Killing if you assume $M$ is compact, by a theorem of Yano.

Edit: addressing the comments, here's a (cumbersome) proof that for a torsion-free connection, the variation $\gamma(t,s) = F^X_s \gamma_0(t)$ is geodesic whenever $\gamma_0$ is geodesic and $X$ is Jacobi along every geodesic. This fact about $X$ implies the "global Jacobi equation" $\nabla^2_{v,v} X = R(v,X)v$ for any vector $v.$ I use the notation $\dot \gamma$ for the vector field $D\gamma(\partial_t)$, which is the velocity vector of $\gamma_s$ when $s$ is fixed. Note that $X$ and $\dot \gamma$ commute.

$\def\n{\nabla} \def\dg{\dot \gamma}$Applying the definition of the curvature and using $\n_X \dg = \n_\dg X$ we get $$\n_X \n_{\dg} \dg = R(X, \dg) \dg + \n_\dg \n_\dg X.$$ Using the global equation $\n^2_{\dg,\dg}X = \n_{\dot \gamma} (\n_{\dot \gamma} X) - \nabla_{\nabla_{\dot \gamma} \dot \gamma} X = R(\dot \gamma,X)\dot \gamma$ we can write this as

$$\n_X(\n_{\dot \gamma} \dot \gamma) = \n_{\n_{\dot \gamma} \dot \gamma} X;$$

i.e. $A = \n_{\dot \gamma} \dot \gamma$ and $X$ commute. Since $A$ is zero along $\gamma_0$, we therefore have $$F^A_t F^X_s p = F^X_s F^A_tp = F^X_s p$$ for all $p$ on $\gamma_0$, and thus $A$ is zero on all points that can be reached by flowing along $X$ from $\gamma_0$, as desired.