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Let $M$ be a Riemannian manifold and $\nabla$ its Levi-Civita connection. Let $c : [0,1] \to M$ be a geodesic, and let $J : [0,1] \to TM$ be a Jacobi field along it. (If it matters, $J(t) \perp \dot c (t)$ for all $t$.)

Is the curve $t \mapsto \exp_{c(t)} J(t)$ a geodesic?

I am assuming that the exponential is defined (this happens, for instance, if $\max _{t \in [0,1]} \|J\| < \max _{t \in [0,1]} \operatorname{injrad} c(t)$; or if $M$ is supposed complete).

One major difficulty comes from the presence of $t$ inside $c(t)$ in the lower argument of $\exp$: I have no idea about how to deal with the derivative with respect to it in that position. I have thought about using Fermi coordinates adapted to $c$, but so far this looks like a dead end.

In general, if $\gamma (t) = \exp _{c(t)} J(t)$, checking that $\nabla _{\dot \gamma} \dot \gamma = 0$ seems very uncomfortable, and the naive direct approaches get stuck very quickly. How to proceed?

Ѕᴀᴀᴅ
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Alex M.
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  • In a bit of a rush so hopefully this is correct: Without mentioning the exponential map to begin with, for any point $p\in M$ and $v\in T_p M$ there is a unique geodesic $\gamma_v$ such that $\gamma_v(0)=p$ and $\gamma_v'(0)=v$. Then by definition of the exponential map $\exp_{c(t)}J(t)=\gamma_{J(t)}(1)=:\tilde c(t).$ Since $M$ geodesically complete, between $\tilde c(t)$ and $\tilde c(t')$ there must be a unique geodesic $\gamma_{\tilde c'(t)}$. Maybe from here one can relate this geodesic to $\exp_{c(t)}J(t)$? – manifoldcurious Mar 21 '25 at 11:29
  • @Saad: When I added those tags, I did so in order to create them because there are plenty of questions about Jacobi fields and about the Riemannian exponential map that are too generically tagged and would benefit from having these new and specific tags. Removing them without asking me first was quite brutal. – Alex M. Mar 21 '25 at 11:39
  • @manifoldcurious: Your last question asks whether one can relate the geodesic between $\tilde c (t)$ and $\tilde c (t')$ to the point $\tilde c (t)$. This does not sound very helpful. – Alex M. Mar 21 '25 at 11:46
  • How about the following. Fix two points $p,q\in M$ sufficiently close to be connected by a geodesic $c_p^q$ such that $c_p^q(1)=q$. Let $J_p^q:[0,1]\to TM$ be the corresponding Jacobi field. In your notation $J_p^{c(t)}(1)=J(t)$. At $t=1$ you get the vector $J_p^q(1)\perp c'(1).$ Then define $\phi_p:M\to M$ such that $\phi_p(q):=\exp_q J_p^q(1)$. It maps $c$ to a curve $\phi_p\circ c$. By the Gauss Lemma this curve must be a geodesic since $\phi_p$ must be an isometry for $q$ sufficiently close to $p$. But $\phi_p(c(t)) = \exp_{c(t)}J_p^{c(t)}(1). $ – manifoldcurious Mar 21 '25 at 16:25
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    @manifoldcurious: Notice that you do not use anywhere the fact that $J$ is Jacobi. You also constrain $\phi_p$ only by specifying its action on $q$, which is too little to be useful. (From Moishe Cohan's link given below, it seems that my question has a negative answer.) – Alex M. Mar 22 '25 at 10:30

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Not in general. For instance, suppose that $0$ and $1$ are conjugate for the geodesic $c$. Let $J$ be the corresponding Jacobi field along $c$ vanishing at $0$ and $1$. Then your construction yields a curve $$ b: [0,1]\to M, b(t)=\exp_{c(t)}J(t) $$ such that $p=b(0)=c(0), q=b(1)=c(1)$. If $b$ were geodesic, we would obtain two distinct (even up to a reparameterization) geodesics connecting $p$ to $q$. But such geodesics need not exist, see for instance this question.

Moishe Kohan
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