How to show $d(\exp_p(tv),\exp_p(tw))=\vert t \vert\cdot\Vert v-w \Vert+O(t^2)$?

I'm self-studying Peter Petersen's Riemannian geometry book. This is Exercise 5.(14) in the book:

Exercise 5.14) $M$ is a $n$-dimensional Riemannian manifold, $p\in M$, and $\exp_p:B(0,\varepsilon)\subset T_pM\rightarrow B(p,\varepsilon)$ is diffeomorphism. Then, whenever $tv,tw\in B(0,\varepsilon)$, we have the following result.

$$d(\exp_p(tv),\exp_p(tw))=\vert t \vert\cdot\Vert v-w \Vert+O(t^2)$$

So far, I've been trying to use the normal coordinate. In the normal coordinate, one can express $g_{ij}(x^1,\cdots,x^n)=\delta_{ij}+\frac{1}{3}R_{ikjl}x^kx^l+O(\Vert x \Vert^3)$ where $R_{ikjl}=\langle R(\partial_i,\partial_k)\partial_j,\partial_l \rangle$. And, choose the minimizing geodesic $\gamma(s)=(x^1(s),\cdots,x^n(s))$ between $\gamma(0)=tv$ and $\gamma(1)=tw$.

Now, since the geodesic distance $d(tv,tw)=\int _0 ^1 \Vert \dot{\gamma}(s)\Vert ds$, I tried to directly compute $\Vert \dot{\gamma}(s)\Vert$. It is $$\sqrt{g_{ij}\,\dot{x}^i(s)\,\dot{x}^j(s)}=\sqrt{(\dot{x}^i)^2+\frac{1}{3}R_{ikjl}x^kx^l\dot{x}^i\dot{x}^j+\cdots}$$ But I don't know how to proceed after this. It seems like some analysis trick is needed? Or do I need totally different approach?

Solution 1:

You may want to check out:

https://mathoverflow.net/questions/215573/square-of-the-distance-function-on-a-riemannian-manifold

In particular appendix A of:

$(*)~~$ https://www3.nd.edu/~lnicolae/RandCrVal_p2.pdf

where they show that in normal coordinates, the 3rd jet in the expansion of the distance squared function on the diagonal has no order 3 terms.

Here are a few details, in expanding distance squared upto order 2.

Let $M\times M \stackrel{\eta}{\to}\mathbb{R}$ be the distance squared, $\eta(p,q) := d^2(p,q)$.

1. Think of $T_{(p,p)}(M\times M)$ as $T_pM\times T_pM$. Since $\eta\equiv 0$ on the diagonal, we have $d_{(p,p)}\eta (x,x) = 0$ for any $x\in T_pM$. Since $\eta(p,q) = \eta(q,p)$, we have $d_{(p,p)}\eta (x,-x) = 0$ for any $x\in T_pM$. Hence $$d_{(p,p)}\eta\equiv 0$$ since $\{ (x,x): x\in T_pM\}$ and $\{(x,-x): x\in T_pM\}$ span $T_{(p,p)}(M\times M)$.

2. Since $d\eta = 0$ on the diagonal, we have a symmetric bilinear form, $d^2\eta$, on $T_{(p,p)}(M\times M)$: for $U, V\in T_{(p,p)}(M\times M)$, we set $d^2\eta(U,V) :=\tilde U\tilde V\eta|_p$, where $\tilde U, \tilde V$ are extensions of $U,V$ to local vector fields. For $(x,x)\in T_{(p,p)}(M\times M)$, we have $$d^2\eta ((x,x), \cdot ) \equiv 0$$ since $(\tilde x, \tilde x)$ flows tangent to the diagonal where $\eta$ is identically zero.

3. By part 2. we get $$d^2\eta((x,y), (x,y)) = d^2\eta ((x-y, 0), (x-y, 0))$$ for $x,y\in T_pM$. And lastly for $z\in T_pM$ we compute: $$d^2\eta((z,0), (z,0)) = \frac{d^2}{dt^2}\eta(\exp_p(tz), p) = \frac{d^2}{dt^2} t^2\|z\|^2 = 2\|z\|^2.$$

Now Taylor expansion gives (let $p + tv$ be a curve through $p$ with initial velocity $v$):

$\eta( p +tv, p + tw) = $

$= \eta(p,p) + t d_{(p,p)}\eta(v,w) + \frac{t^2}{2}d^2\eta((v,w), (v,w)) + O(t^3) =$

$= t^2\|v-w\|^2 + O(t^3)$.

The expansion to this point has no need of normal coordinates. The appendix A of $(*)$ uses normal coordinates to show the $O(t^3)$ term vanishes. (to help with the computation $(*)$ uses that for $y$ fixed, $S(x) = d(x,y)$ satisfies the Eikonal equation: $\|dS\| = 1$. Hence $\eta = S^2$ satisfies $\|d\eta\|^2 = 4\eta$.)

Now your formula follows from:

$d^2(\exp_p(tv), \exp_p(tw)) = t^2\|v-w\|^2 + O(t^4)$