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I apologize if this is not a research level question (already tried asking https://math.stackexchange.com/questions/1303288/relation-between-parallel-transport-and-jacobi-field-iion stack exchange with no response), but it's not a homework question either: an applied mathematics/medical imaging paper I was reading tried to numerically approximate parallel transport at time step $i$ using Jacobi fields, and when I ran their algorithm and checked the norms of the parallel transport, it's blowing up (literally) exponentially as a function of the time steps $i$ I used to divide $[0,1]$, which tells me it might not be correct.

Here's the the theoretical assumption/question regarding their algorithm:

FIX $t_0>0$.

Let $t>0$ and $W(t_0)\in T_{c(t_0)}M$ be the parallel transport of $W\in T_{c(0)}M$. and $J_{t_0,W(t_0)}$ be the Jacobi field along a geodesic $c$ such that $J_{t_0,W(t_0)}(t_0)=0, J_{t_0,W(t_0)}'(t_0)=W(t_0)$. Let $P_{t_0,t_0+t}$ denote the parallel transport from time $t_0$ to $t_0+t$ for a FIXED $t_0\in \mathbb{R}$

By now, one can prove that:(I agree and I've the proof)

$$\left|\left|\frac{P_{t_0,t_0+t}(W(t_0))-\frac{J_{t_0,W(t_0)}\ \ \ (t_0+t)}{t}}{t}\right| \right|\to0$$ as $t\to 0$.

Now, let $t_0$ VARY, and indeed assume that $t_0=t$. My question is: is the same formula going to be true? That is, do we also have:

$$\left| \left|\frac{P_{t,2t}(W(t))-\frac{J_{t,W(t)}\ \ \ (2t)}{t}}{t}\right| \right|\to0$$ as $t\to 0$? THIS IS MY MAIN QUESTION. The paper seemingly used this. A definite yes or no answer is okay with me for the moment. Proofs very much appreciated.

In the above, $J_{t,W(t)}$ is the unique Jacobi field along $c$ so that $J(t)=0, J'(t)=W(t)$.

In that paper I mentioned, they assume $t=\frac{1}{N}$and $P_{t,2t}(W(t))=P_{0,2t}(W)$ is approximately $\frac{J_{t,W(t)}(2t)}{t}$, and use it to repeatedly calculate Jacobi and parallel transport. But the norms of $P_{t,2t}(W(t))$ is showing huge errors and $P_{t,kt}(W(t))$ is blowing up with $k$.

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The answer is yes. Since the Jacobi field satisfies the second order equation $$ J(s)'' + R(s)J(s)=0,$$ where $R(s)$ is the curvature tensor (details are skipped here); then for $J_{t,W(t)}(s)$ it holds $J(s)=(s-t)W(s) + (s-t)^3 R(s)W(s)/6$ up to higher orders of $(s-t)$ as $s-t$ goes to 0 ($J(t)=0$ is used here). So, if the curvature tensor $R$ of $M$ behaves nicely (say, is bounded) the approximation should work.

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  • $\begingroup$ thanks for your answer, but I'm not sure I understand. I'm doing the "Taylor expansion" of $J_{t,W(t)}$ at t below. $s>t, s=2t$ in particular. $J_{t,W(t)}(s)=J_{t,W(t)}(t)+(s-t)J_{t,W(t)}'(t)+O(s-t)^3=0+(s-t)W(t)+O(s-t)^3=(s-t)W(t)+O(s-t)^3$. But I'm not getting why the last step is: $(s-t)W(s)+O(s-t)^3$? $\endgroup$ – Let's talk math Jun 1 '15 at 12:37
  • $\begingroup$ R is linear operator acting on J; since J=(s-t)W the second derivative of J at s=t is 0, only the third derivative is not: it equals RW. Thus, the Taylor expansion gives you J(s)=(s-t)W + (s-t)^3/6 RW. Is it o'key? $\endgroup$ – valeri Jun 1 '15 at 12:49
  • $\begingroup$ Not sure actually. I do get why in the Taylor expansion there's no $(s-t)^2$ term, because the second derivative of $J_{t,W(t)}$ at $t$ is zero. What I don't get is why you're writing in the final expression $W(s)$ instead of $W(t)$, since it seems you're doing the Taylor expansion around $t$. We don't know that $J_{t,W(t)}'(s)=W(s)$, we only know: $J_{t,W(t)}'(t)=W(t)$, but then can we say $||W(s)-W(t)||=O(t-s)^2 \forall s?$. I guess not. But, if we could, OR, it was indeed $W(s)$, this'd prove the asked formula: by plugging in $s=2t$. But with $W(t)$ instead of your written $W(s)$, not. $\endgroup$ – Let's talk math Jun 1 '15 at 13:06
  • $\begingroup$ If you can give me a reference for the formula you wrote, using $W(s)$ instead of $W(t)$, that'll be good too! $\endgroup$ – Let's talk math Jun 1 '15 at 13:11
  • $\begingroup$ W(s) and W(t) are vectors at different points c(s) and c(t) - they belong to different vector spaces: T_{c(s)} and T_{c(t)}, so W(s)-W(t) has no sense. (To compare them we need parallel transport, but then W(s) equals W(t) being transported). That is why I wrote W(s) in the formulas for J(s) - they have to "live" in the same vector space. Another way - we may introduce Fermi coordinates along c, and then work in usual Euclidean (coordinate) space - then we would have the same formulas and estimates (along c), and euclidean parallel transport - all vectors in the same v.sp. and simpl W(s)=W(t) $\endgroup$ – valeri Jun 1 '15 at 13:13

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