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Let $(M, g)$ be a Riemannian manifold and $\gamma :[0, \delta] \to M$ a $C^1$-curve, $\gamma(0) = y$ and $\dot\gamma(0) \neq 0$. Then we have

$$\lim_{t\to 0} \frac{d_M(y, \gamma(t))}{\int_0^t |\dot \gamma|} = 1.$$

The proof suggested below uses the Taylor expansion of metric in a normal coordinate.

I post this question and answer here since

  • it helps complete a proof I wrote here, and

  • this result has some independent interest (I think), and

  • Would someone please check if the answer is correct, I am bit unfamiliar with the big $O$ argument.

On the other hand, it would be great if someone points me to a reference where this is proved.

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1 Answers1

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Let $\delta >0$ be small so that $\gamma$ lies completely in the normal coordinate centered at $y$. In this normal coordinate, we write

$$\gamma(0) = 0, \ \ \ g_{ij}(x) = \delta_{ij} + O(|x|^2).$$

Write $v = \dot\gamma(0)$. Then we have

$$ \gamma (t) = tv + O(t^2), \ \ \ \dot\gamma = v+ O(t)$$

$$\begin{split} \int_0^t \|\dot\gamma (s)\|ds &= \int_0^t \| v + O(s)\| ds \\ &= \int_0^t \sqrt{\|v\|^2 + O(s)} ds \\ &= \int_0^t \left( \|v\| + O(s)\right) ds \\ &= t\| v\| + O(t^2) \\ &= \|tv\| + O(t^2)\\ &= |\gamma (t) - O(t^2)| + O(t^2) \\ &= |\gamma(t)| + O(t^2). \end{split}$$

Now in normal coordinate, $|\gamma (t)| = d_M(y, \gamma(t)).$ Thus

$$\frac{\int_0^t \|\dot\gamma\|}{d_M(y, \gamma(t))} = 1 + \frac{O(t^2)}{t} \to 1$$

as $t\to 0$.