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Let $p$ be a point in a Riemannian manifold, and take a normal neighborhood nearby $p$. If a field $X$ on a Riemannian manifold satisfies the condition

$$\mathcal L_X\partial r=[X,\partial r]=0$$

where $\partial r$ is the radial coordinate (which is also $\nabla r$, where $r$ is the distance from $p$) then $X$ is a Jacobi field along all radial geodesics from $p$ (in the normal neighborhood).

(I'm talking about the situation of this post.)

Now the question is: under which conditions does the converse hold? It should be true that the Jacobi equation has more solutions than the given condition (cfr. Petersen, Riemannian geometry). Is this really true? Does a dimension argument work for something like Lie derivative which is not tensorial (Petersen says "the first equation is first order, the Jacobi equation is second order, so the latter has more solutions")?

And, if this is true, is there any known condition which implies the equation with the Lie derivative under the hypotesis "X is Jacobi"?

Sorry for the vague kind of question: the point is that I would like to have the condition in order to prove something, but I can't assume it in general.

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There are a few different ways to see this. Here is one.

Take a hypersurface $S$ contained in the $d$-sphere around $p$ and parametrize it with coordinates $x_1,\ldots,x_{n-1}$. Then extend this parametrization to an open neighborhood $U$ by adding the coordinate $x_n:=r$, letting the coordinates $x_1,\ldots,x_{n-1}$ be constant on radial lines. (As in the post, $r$ is the distance from $p$). The equation $[X,\partial r]=0$, where $X=X^i\frac{\partial}{\partial x^i}$ is a vector field, simply means that the coefficients $X^1,\ldots X^n$ are constant along radial lines (this is only true using this kind of coordinates). Hence, any vector field along the inclusion $S\subset M$ extends uniquely to a vector field on $U$ which commutes with $\partial r$.

On the other hand, a Jacobi field $J$ along a geodesic is determined by two initial conditions - $J(0)$ and $\dot{J}(0)=\frac{D}{dt}J(0).$ So, given any two vector fields, $\tilde{X}$ and $\tilde{Y}$ along the inclusion $S\subset M$, there is a unique vector field $X$ on $U$ satisfying$$X|_S=\tilde{X},\quad\frac{D}{dr}X|_S=\tilde{Y},$$which is a Jacobi field along all the radial lines.

As a matter of fact, the right way to express everything written in this answer is by a single sentence that appears in the question - "a second order equation has more solutions than a first order one".

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