TITLE: Zero Sectional Curvature implies exp is a local isometry
QUESTION [7 upvotes]: 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)

REPLY [8 votes]: 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 $t$.