Cross posted from my question:


This is a formula/theorem (written below) that I found mentioned in a medical imaging paper, without any proof or any further detail. I'd appreciate if anyone could please provide either the detail or the sketch to prove it.

Let $c=\exp_p(tV)$ be a geodesic on a Riemannian manifold $M$. Let $W(t)$ be a parallel vector field $c$ with $W(0)=W$. Consider the Jacobi field $J(t)=(D\exp_p)_{tV} ( {tW} )$.

Then: $\lim_{t\to 0}|\frac{W(t)-\frac{J(t)}{t}}{t}|=0.$

For constant curvature spaceforms, when I take normal Jacobi fields the formula holds good. But I'm not sure how to prove it for general manifolds.

P.S. in the paper that I mentioned, the formula is mentioned somewhat heuristically, but this is my interpretation of their words.

EDIT I: I can also see that if we take compare the norms of $W(t)=W$ and norms of $J(t)$, which is $|J(t)|=t|W|+O(t^3)$ (See Do Carmo, P. 115), then clearly,

$lim_{t\to 0}\frac{|W(t|)-\frac{|J(t)|}{|t|}}{|t|}=0.$

But still it doesn't prove the result, but shows that the claim could be correct.

EDIT II:I think I've got an idea, it's working so far, but not fully yet. I considered the function $f(t):=\langle tW(t)-J(t),tW(t)-J(t)\rangle =||tW(t)-J(t)||^2$ and showed so far that $f(0)=f'(0)=f''(0)=0$. So if I do some careful work on the fourth derivative $f''''(0)$ now, hopefully I'll have it!


1 Answer 1


(1) Consider Chapter 2 exercise 2 (57p.) in the book "Riemannian Geometry" by Do Carmo: So $$ \bigg(W-\frac{J}{t}\bigg)' = \lim_t \frac{ P^{-1}_{0,t} \bigg(W-\frac{J}{t} \bigg) + \bigg(W-\frac{J}{t}\bigg)(0) }{t} = \lim_t \frac{W-\frac{J}{t} }{t} $$ where $P_{0,t} : T_{p} M \rightarrow T_{c(t)}M$ is a parallel transport.

Here we used : $P_{0,t}^{-1} $ goes to $id$ if $t$ goes to $0$. And $$ J(0)=0,\ J'(0)=W(0) \Rightarrow J = tW(t) + o(t^1) $$

(2) And $$ \bigg(W-\frac{J}{t}\bigg)' = \frac{J't-J}{t^2} $$

Let $X:=J't-J$ so that $$ X(0)=0,\ X'(0)= (J't-J)' = J'' t =-tR(c',Y)c'=0$$ $$ X''(0)=J'''t+J''=-R(c',Y)c'=0$$

Hence $$ X(t)= \sum_{i=3}^n \frac{t^i}{i!} E_i(t)+ o(t^n) $$ for some parallel vector fields $E_i$ s.t. $E_i(0) =\nabla_{c'}^{(i)} X(0)$. Hence $\lim_t \frac{X}{t^2} =0$.

  • 1
    $\begingroup$ I had to change the question a bit without changing anything significantly: I put absolute value sign in the limit I wanted to prove to be $0$. It otherwise could not make sense to take limits of $X(t)$ as $t\to 0$, since you're comparing quantities belonging to different tangent spaces, which might make sense only after we define metric on the tangent bundle $TM$, which we should probably avoid for this question. Also, for the same reason, you cannot write $lim_{t \to 0}\frac{J't-J}{t^2}$ and not apply l'Hospital like you did twice. So, in my EDIT II, I worked with absolute values. $\endgroup$ Apr 28, 2015 at 13:00
  • $\begingroup$ I see. I correct it $\endgroup$
    – HK Lee
    Apr 28, 2015 at 13:34
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    $\begingroup$ Sorry HK Lee, still not clear. How do you write in (1) of your answer: $$\lim_t \frac{ P^{-1}_{0,t} \bigg(W-\frac{J}{t} \bigg) + \bigg(W-\frac{J}{t}\bigg)(0) }{t} = \lim_t \frac{W-\frac{J}{t} }{t} $$ ? By $W$ here, did you mean $W(0)$ or $W(t)$? It seems that no $W$ term will remain in the numerator, since there are two $W(0)=W$ apprearing in the numerator, but they cancel out. In any case, you're again taking limits of vectors belonging to different tangent spaces here, which should not be done. Also, in the numerator right after $=$ sign in (1), it should be - sign, not +. $\endgroup$ Apr 28, 2015 at 23:08
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    $\begingroup$ Continued comment: In (2) of your answer, I do see, by Taylor expansion, that $X(t)= O(t^3)$ from the facts that you've proven the zeroth, first and second derivative at $0$ vanishes. Why should that mean something stronger, i.e. $X(t)=o(t^3)$, which is equivalent to saying that $ \bigg(W(t)-\frac{J(t)}{t}\bigg)'=o(t)$? Finally, suppose, you could prove that, $ \bigg(W(t)-\frac{J(t)}{t}\bigg)'=o(t)$ holds, then how does it \bold{rigorously} prove that $\lim_t \frac{|W-\frac{J}{t} |}{|t|}=0$. I think it's safe to work with absolute value sign. $\endgroup$ Apr 28, 2015 at 23:23
  • $\begingroup$ I add some in my answer. $\endgroup$
    – HK Lee
    Apr 29, 2015 at 0:29

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