I would like to pose a question regarding the distance between a geodesic $\gamma(t)$ and a perturbed geodesic $\gamma_{\epsilon}(t)$ on a Riemannian manifold. Specifically, is the distance controlled by a bound on the associated Jacobi field?

Following @Deane and @Otis (With heartfelt thanks), I'd like to attempt this question from a different perspective.. It may not be as strict, but I will do my best to gradually refine it. Starting from the definition (Actually, I want to point out that when $t$ is sufficiently large, the distance between the geodesic still has an upper bound) :

Let $ M $ be a Riemannian manifold and there exists a smooth map $\Gamma: [0,\epsilon] \times [0,t] \rightarrow M$, such that \begin{align*} \Gamma(0,t) = \gamma(t), \quad \Gamma(\epsilon,t) = \gamma_\epsilon(t), \end{align*}

Let $\sigma(s)$ denote a unit speed minimizing geodesic from $\gamma(0)$ to $\gamma_\epsilon(0)$, and $\epsilon$ be the length of $\sigma$.

Suppose we have a bound on the Jacobi field $\partial_s\Gamma(s,t)$ for each $s \in (0,\epsilon)$ of the form \begin{equation} |\partial_s\Gamma| \leq C. \end{equation} Note: This is essential; see @Deane's answer for details.

Try to demonstrate that (It also holds when $t$ is sufficiently large.)

\begin{equation} \frac{1}{2} d^2 (\gamma_\epsilon(t),\gamma(t))\leq C \epsilon. \end{equation}

Attempts (Update)


\begin{equation*} D(\epsilon) = \frac{1}{2}\int_0^\epsilon d^{{2}}(\gamma_s(t),\gamma(t)) ds. \end{equation*}

For each $ s \in (0, \epsilon) $, the Jacobi field $ \partial_s \Gamma(s, t) $ satisfies the bound \begin{equation*} \left|\partial_s \Gamma\right| \leqslant C, \end{equation*} where $C>0$ is a constant.

\begin{align*} (D'(\epsilon)) & = \frac{1}{2} d^2 (\gamma(t),\gamma_\epsilon(t))= \frac{1}{2} \int_0^\epsilon \langle \nabla d^{2}(\gamma(t),\gamma_s(t)), \partial_s\Gamma(s,t) \rangle ds\\ & \leq \int_0^\epsilon \langle d(\gamma(t),\gamma_s(t)) , \vert \partial_s\Gamma(s,t) \vert \rangle ds \\ & \leq C \int_0^\epsilon d(\gamma(t),\gamma_s(t)) ds. \end{align*}

According to the inequality $\frac{1}{2} d^2 (\gamma(t),\gamma_\epsilon(t)) \leq C \int_0^\epsilon d(\gamma(t),\gamma_s(t)) ds$, we can obtain the following bound \begin{equation*}d(\gamma(t),\gamma_\epsilon(t)) \leq C\epsilon. \end{equation*}

  • $\begingroup$ By the way, are there any similar results on dg? This outcome feels somewhat intuitive and perhaps shouldn't require proof, given that the Jacobi field equations describe the distance between neighboring geodesics. $\endgroup$
    – lumw
    Aug 4 at 15:40
  • 2
    $\begingroup$ You should define the exact relationship between $\gamma_\epsilon$ and $J$. But I think this method answers your question: mathoverflow.net/a/446677/1540 . $\endgroup$ Aug 4 at 15:51
  • $\begingroup$ @OtisChodosh Thank you, I'll have a look at that. In my context, I define the Jacobi field $J$ as the derivative of the geodesic $\gamma(s,t)$ with respect to $s$. Is this an application of the comparison theorem? $\endgroup$
    – lumw
    Aug 4 at 16:13
  • $\begingroup$ Indeed, it appears similar. In analogous problems, the distance between two geodesics is calculated using curvature, which aligns more with what I am trying to demonstrate. This indirectly suggests that it can also be shown using a bound on the Jacobi field. $\endgroup$
    – lumw
    Aug 4 at 16:18
  • 2
    $\begingroup$ You still have not defined $\gamma_\epsilon(t)$ sufficiently well to answer the question. Is $\gamma_\epsilon(t)$ ANY $1$-parameter family of geodesics with $\frac{d}{d\epsilon}\gamma_\epsilon|_{\epsilon=0}=J$? If so, the bound you want is clearly false (you cannot bound a function if you just know its derivative at a point!). If $\gamma_\epsilon$ is constructed from $J$ like in my answer to the other question (parallel transport $J$ along $\exp(J(0))$) then the answer in my other question precisely gives your answer (just don't use the upper bound for $J$). $\endgroup$ Aug 4 at 16:40

1 Answer 1


You should formulate your question in more detail and rigor. Let me try to do this here and elaborate on Otis's comments.

Let $M$ be a Riemannian manifold and $\gamma: [0,1] \rightarrow M$ be a constant speed geodesic. You say that $\gamma_\epsilon: [0,1] \rightarrow M$ is a perturbed geodesic but do not say what that means. I'll assume that this means there is a smooth map $$ \Gamma: [0,\epsilon] \times [0,1] \rightarrow M $$ such that for any $t\in [0,1]$, \begin{align*} \Gamma(0,t) &= \gamma(t)\\ \Gamma(\epsilon,t) &= \gamma_\epsilon(t) \end{align*} and for each $s \in [0,\epsilon]$, the curve $$t \mapsto \Gamma(s,t)$$ is a constant speed geodesic.

Your goal is to estimate the distance between $\gamma(t)$ and $\gamma_\epsilon$ and propose to use the observation that $$ d(\gamma(t),\gamma_\epsilon(t)) \le \ell(\Gamma(\cdot,t)) = \int_{0}^{\epsilon} |\partial_s\Gamma(s,t)|\,ds. $$ By the Cauchy-Schwarz inequality, the following bound also holds: $$ (d(\gamma(t),\gamma_\epsilon(t)))^2 \le \int_{0}^{\epsilon} |\partial_s\Gamma(s,t)|^2\,ds. $$

Along each geodesic $\Gamma(s,\cdot)$, the vector field $\partial_s\Gamma(s,\cdot)$ is a Jacobi field. You denote $$ J = \partial_s\Gamma(0,\cdot). $$ It is clear that a bound on $|J(t)|=|\partial_s\Gamma(0,t)|$ does not imply any bound on $|\partial_s\Gamma(s,t)|$ for $0 < s \le \epsilon$. As Otis observes, this implies that your assumptions are not enough to obtain a bound on $d(\gamma(t),\gamma_\epsilon(t))$.

If, on the other hand, if you have a bound on the Jacobi field $\partial_s\Gamma(s,\cdot)$ for each $s \in (0,\epsilon)$ of the form $$ |\partial_s\Gamma| \le C, $$ then you obtain trivially that $$ d(\gamma(t),\gamma_\epsilon(t)) \le C\epsilon. $$

If you assume a sectional curvature bound, then, as Otis observes, you still need to assume more. One possible assumption consists of (sufficiently small) bounds on $d(\gamma(0),\gamma_\epsilon(0))$ and $d(\gamma(1),\gamma_\epsilon(1))$. Standard comparison theorems then imply the bounds you want. There are other assumptions you could make that imply similar bounds.

Last, I want to say again that if you want to ask a question on math.stackexchange.com (where I think this question really belongs) or MathOverflow, you should write it as carefully as I have done above. Often, by doing only that, you can see what's going on more clearly.

  • $\begingroup$ Thank you for your help. I appreciate the detailed and rigorous explanation of the issue; I've learned a great deal from your answer. As a beginner with limited scientific training, I recognize that I have much to improve, and I will strive to do so. $\endgroup$
    – lumw
    Aug 5 at 15:21

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