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This problem is from John Lee's Introduction to Riemannian Manifold Problem 6-7(a)

Let $(M, g)$ be a connected Riemannian Manifold. Show that for every $p \in M$ and $v, w \in T_p M$, we have $$ \lim _{t \rightarrow 0} \frac{d_g\left(\exp _p t v, \exp _p t w\right)}{t}=|v-w|_g . $$ [Hint: Use the Taylor series for $g$ in Riemannian normal coordinates on a convex geodesic ball centered at $p$.]

Follow the hint. Consider normal coordinates at $p$. Then we can write $g_{ij} = \delta_{ij} + O(|x|^2)$. The key step is to show that $d_g\left(\exp _p t v, \exp _p t w\right) \approx t|v-w|_g$ as $t$ is sufficiently small. This makes sense as around $p$ with normal coordinate $g$ looks like Euclidean space. However, I would like to make it rigorous.

More background:

  • $d_g(p, q)$ denotes the infimum length among all curves connecting $p$ and $q$.
  • With normal coordinates, all directional derivative of $g_{ij}$ is zero at the center point. In this case, it is $p$.

What I have tried so far:

Let $\epsilon > 0$ such that $B_\epsilon(p)$ is a geodesic ball and a uniformly $\delta$-normal. Suppose $\gamma_t(s)$, ($s \in [0, t]$) be the minimum curve between $\exp_ptv$ and $\exp_ptw$. Therefore, $$ \begin{align} d_g\left(\exp _p t v, \exp _p t w\right) &= \int_0^t <\gamma_t'(s), \gamma_t'(s)>^{1/2}_gds \\ & = \int_0^t <\gamma_t'(s), \gamma_t'(s)>^{1/2} ds + O(t^2) \end{align} $$ where $<\cdot,\cdot>$ denote the standard Euclidean inner product. Using the intermediate value theorem, we have $$ d_g\left(\exp _p t v, \exp _p t w\right) = t <\gamma_t'(s_0), \gamma_t'(s_0)>^{1/2} + O(t^2). $$ Thus $$ \lim _{t \rightarrow 0} \frac{d_g\left(\exp _p t v, \exp _p t w\right)}{t}= <\gamma_0'(0), \gamma_0'(0)>^{1/2} = |\gamma_0'(0)|_g $$ Don't know how to justify $|\gamma_0'(0)|_g = |v-w|_g$

Yan Zhu
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    What have you tried so far? – Deane Dec 31 '23 at 02:38
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    @Deane I edited my original post. – Yan Zhu Dec 31 '23 at 03:25
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    You don't say anything about the parameterization used for $\gamma_t$, and I'm not sure why you use the interval $[0,t]$ for the parameterization domain. This makes things more difficult because both the endpoints of integration and the integrand depend on $t$. I suggest using a constant speed parameterization $\gamma_t: [0,1]\rightarrow M$ with $\gamma_t(0) = \exp_ptv$ and $\gamma_t(1)=\exp_ptw$. Also, I suggest using the energy integral $$\int_0^1 \langle\gamma'_t,\gamma_t'\rangle,ds$$ instead of the length integral. – Deane Dec 31 '23 at 03:39

1 Answers1

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I figured it out.

Let $q = \exp_pvt$ and let $u(t) = \exp_q^{-1}(\exp_p(wt))$. Thus $\exp_p(wt) = \exp_qu(t)$, or $wt = \exp_p^{-1}(\exp_qu(t))$. Take derivative of $t$ and evaluate at $t= 0$: $$ \begin{align} w &= d\exp_p^{-1}|_o \left.\frac{d}{dt}\right|_{t=0} \exp_qu(t) \\ & = \left.\frac{d}{dt}\right|_{t=0} \exp_qu(t) \end{align} $$ Denote $\exp_qu(t) = \gamma(t, q)$. Then $$ \left. \frac{d}{dt}\right|_{t=0} \gamma(t, q) = D_1|_{t=0}\gamma(t, q) + D_2|_{t=0} \gamma(t, q) \frac{dq}{dt} = u'(0) + v $$ Thus $u'(0) = w-v$. Note that $$ \lim _{t \rightarrow 0} \frac{d_g\left(\exp _p t v, \exp _p t w\right)}{t} = \lim _{t \rightarrow 0} \frac{|u(t)|_q}{t} = \lim _{t \rightarrow 0} \frac{|u(t)|_p + O(t^2)}{t} = |u'(0)|_p = |w-v|_p $$

Yan Zhu
  • 723