Zero Sectional Curvature implies exp is a local isometry

Question

Im studying DoCarmo's book Riemannian Geometry, the first problem of the chapter 5 (Jacobi Fields) states that

If $(M,g)$ is a riemannian manifold with sectional curvature identically zero, show that for every $p \in M$, $exp_p: B_\varepsilon(0) \subseteq T_pM \rightarrow B_\varepsilon(p)$ is an isometry.

I do not figure out how to relate Jacobi Fields with this question in particular, I appreciate any hint.

(PD this is not a homework, Im studying Riemannian geometry for my own just for research purposes)

Answer 1

What I am gonna do is to show that in normal coordinates the metric is the euclidean one

Given a vector $v$ of $T_{q}M,$ for some $q$ in a normal neighborhood of $p,$ we are going to compute $g(v,v).$ To this end, we decompose $v$ orthognally as $v=a\frac{\partial}{\partial r}+u,$ where the second part, $u,$ is tangente to the geodesic sphere (we know the decomposicition is orthogonal thanks to Gauss' lemma). The norm of the radial part is just $a^{2}.$ I will now show how to compute the tangential part.

Let $\gamma(s)$ be the unique geodesic joining $p$ with $q$ (Is unique beacause we are in a normal neighborhood). Notice that in normal coordinates the componentes of a Jacobi field $J$ along a geodesic $\gamma(s)$ such that $J(0)=0$ and $J^{\prime}(0)=V,$ are expresed as $J^{i}(s)=sV^{i}.$ Hence, if we call $d$ the distant from $p$ to $q,$ $u$ is the value at $q$ of the Jacobi field whose components in normal coordinates are: $J^{i}=\frac{s}{d}u^{i}.$

Fuerthermore, letting $n$ be a normal vector to $\gamma^{\prime}(0)$ and $n(s)$ its pararllell transport along $\gamma,$ we will have $J(s)=sn(s),$ because our manifold has curvature zero.

Taking covariant derivatives, $D J (0)=n(0).$ But $n(s)$ is pararlell; that is, in particular, its lenght is constant.

$$\vert u \vert ^{2}=\vert J(d) \vert ^{2}=d^{2}\vert n(d) \vert ^{2}=d^{2}\vert D J(0) \vert ^{2},$$ But also, form the expresion in normal coordinates we see: $$ (D J(0))^{i}=\frac{1}{d}u^{i}.$$

Now well, the metric coincides with the euclidean at $p$ (we are in normal coordinates given by $\text{exp}_{p}$). Therefore: $$\vert u \vert_{eucl}= d \vert DJ(0) \vert,$$ Where the subscript means we are taking the eculidean norm of the components.

Substituting, we reach: $$\vert u \vert ^{2}=d^{2}\vert DJ(0) \vert ^{2}=d^{2}\frac{1}{d^{2}}\vert u \vert_{eucl} ^{2}=\vert u \vert_{eucl}^{2},$$ An we are done

Answer 2

Here's another possible way.

Let $v\in B_\epsilon(0)$ and $w\in T_v(T_pM)\cong T_pM$. Consider the variation of geodesics given by \begin{equation*} \gamma_s(t)=\exp_p\bigl(t(v+sw)\bigr) \end{equation*} It follows that \begin{equation*} J(t)=\frac{\partial\gamma_s}{\partial s}\biggr|_{s=0}=(d\exp_p)(tw) \end{equation*} is a Jacobi field along $\gamma_0$. Since sectional curvature is identically zero, $J$ satisfies the following PDE \begin{equation*} \frac{\partial^2J}{\partial t^2}=0 \end{equation*} Let $\{e_i\}$ be an orthnormal basis for $T_pM$ and extend it to a parallel frame along $\gamma_0$ such that $e_i(0)=e_i$. In terms of the frame, $J(t)=a^i(t)e_i(t)$ and $w=w^ie_i$. Solving the above equation, we have $J(t)=(w^it)e_i(t)$, so that \begin{equation*} w^ie_i(1)=(d\exp_p)(w) \end{equation*} Following the same argument, if $u\in T_v(T_pM)\cong T_pM$ and $u=u^ie_i$, then \begin{equation*} u^ie_i(1)=(d\exp_p)(u) \end{equation*} Therefore, \begin{align*} \langle(d\exp_p)(w),(d\exp_p)(u)\rangle &=w^iu^j\langle e_i(1),e_j(1)\rangle\\ &=w^iu^j\langle e_i,e_j\rangle\\ &=\langle w,u\rangle, \end{align*} proving that $\exp_p\colon B_\epsilon(0)\to B_\epsilon(p)$ is an isometry. Note that we used the fact that $\langle e_i(t),e_j(t)\rangle$ is indepdent of $t$.