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Feb 19, 2021 at 1:22 comment added user35593 Proposition 1.2.8 of research-collection.ethz.ch/bitstream/handle/20.500.11850/… seems related.
Feb 17, 2021 at 11:03 comment added Uzu Lim I think there is a misunderstanding. My question is on whether the error term of the 1st order Taylor expansion of $\exp_x$ can be realized as $(t^2/2) Q_x(w)$ for some appropriate $w \in T_x M$ with $\|w\| \le 1$. I don't think that this follows directly from Taylor's theorem? The best shot seems to be applying Taylor's theorem coordinate-wise to $t \mapsto \exp_x(tv)$ but that won't imply the desired result directly. Also, $J_x: T_x M \rightarrow \mathbb{R}^n$ is a required notation if we regard the tangent space as not a priori embedded in $\mathbb{R}^n$.
Feb 17, 2021 at 3:15 comment added Ryan Budney Your question looks essentially like the definition of the Taylor expansion of 1st order. So, provided I'm not misunderstanding your question, the answer is yes. Moreover, J_x(v)=v for any reasonable definition of the tangent space.
Feb 17, 2021 at 1:39 comment added Uzu Lim I think I have a resolution to the question asked, by applying Taylor's theorem coordinate-wise to the exponential map $t \mapsto \exp_x(tv)$ and bounding the error term crudely. This seems to yield that the error term is bounded by $Bt^2$ where $B = \frac n2 \sup_{w \in TM} \| \nabla_w w \| $ and $n$ is the ambient dimension. I'll clarify after I get some sleep.
Feb 17, 2021 at 1:28 history edited Uzu Lim CC BY-SA 4.0
typo deleted
Feb 17, 2021 at 1:27 comment added Uzu Lim ah, someone added a typo while editing. I'm going to delete that $>$.
Feb 17, 2021 at 1:15 comment added Ryan Budney Your notation does not make sense to me. What does the inequality mean between a point of $M$ and the function on the right, which appears to be vector-valued?
Feb 16, 2021 at 21:27 history edited Francesco Polizzi CC BY-SA 4.0
added 9 characters in body
Feb 16, 2021 at 19:13 history asked Uzu Lim CC BY-SA 4.0