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For an embedded Riemannian manifold $M \subseteq \mathbb{R}^m$ and a point $x \in M$, there is a series expansion (page 8 of Monera's paper): $$\exp_x(t v) = x + t J_x(v) + \frac{t^2}{2!} Q_x(v) + \cdots$$ where $v \in T_x M$ satisfies $\|v\|=1$, $t \ge 0$, $u \mapsto x + J_x(u)$ is the (linear) embedding of the tangent space $T_x M$, and $Q_x(v) = (\nabla_v v)^\perp$, the component of $\nabla_v v$ that is normal to $T_x M$.

Question. Is the following or a variant of it true? For some constant $C$, we have that whenever $0 < t < C$, there is a vector $w > \in T_x M$ with $\|w \|\le 1$ such that: $$\exp_x(tv) = x + t J_x(v) + \frac{t^2}{2!} Q_x(w)$$ A higher order (more than quadratic) analogue is also sought, if it exists.

My end goal of this analysis is to obtain a bound on the local deviation of an embedded manifold from being linear. Namely, upon accepting the claimed result, we have that $$\| \exp_x(tv) - (x + t J_x(v)) \| \le \frac{A}{2}\cdot t^2 \text{ where } A := \sup_{w \in TM, \|w \| \le 1 } (\nabla_w w)^\perp $$

And honestly, all this feels like they are likely to be some exercise problems to a Riemannian geometry course, but I can't find a reference for this type of result. Pointing towards a classical reference for this would be highly welcome as well.

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  • $\begingroup$ 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? $\endgroup$
    – Ryan Budney
    Commented Feb 17, 2021 at 1:15
  • $\begingroup$ ah, someone added a typo while editing. I'm going to delete that $>$. $\endgroup$
    – Uzu Lim
    Commented Feb 17, 2021 at 1:27
  • $\begingroup$ 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. $\endgroup$
    – Uzu Lim
    Commented Feb 17, 2021 at 1:39
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    $\begingroup$ 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. $\endgroup$
    – Ryan Budney
    Commented Feb 17, 2021 at 3:15
  • $\begingroup$ 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 $\mathbb{R}^n$. $\endgroup$
    – Uzu Lim
    Commented Feb 17, 2021 at 11:03

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