Since I'm lazy, I'm shamelessly referring to the following question:
Given a Riemannian manifold $M$, let $\gamma: (a,b) \to M$ be a geodesic and $E$ a parallel vector field along $\gamma$. Define $\phi: (a,b) \to M$ by $t \mapsto \exp_{\gamma(t)}(E(t))$.
I completely agree, as Michael Pauley answered, that $\phi'(t)$ is nothing but $J_t(1)$, where $J_t$ is the Jacobi field with $J_t(0)=\gamma'(t), \nabla_{\frac{\partial{x}}{\partial{u}}} {J_t}(0)=0$. But what is $\nabla_{\phi'(t)} {\phi'(t)}?$
Notice that, we can think of $\phi(t)=J_t(1)$ as a "parallel translate" of the geodesic $\gamma(t)$. Indeed in the Euclidean spaces or their quotient spaces, the answer is $\nabla_{\phi'(t)} {\phi'(t)}=0$, and the "parallel translate" is a geodesic, which is a staright line in Euclidean space. But what can we say if we work with general manifolds?
I've two questions:
1) Are there milder conditions on the manifold, than being Euclidean, that can guarantee that $\phi(t)$is a geodesic?
2) Are there computations that could relate $\nabla_{\phi'(t)} {\phi'(t)}$ to $\gamma, J, R$, $R$ being the curvature tensor, which I'm assuming will arise in the expression.
You may ignore the rest.
What I tried to do so far, without much success:
On P. 174, Lemma 10.1 of of John Lee's Riemannian manifolds and curvature book (available for free download online), I tried to put $V=S$ and used the symmetry lemma: $D_s{T}=D_t{S}$ to obtain:
$D_s^2 (T) - D_t(D_s(S)) = R(S,T)S$. Note that $D_s(S)(s=1,t)=\nabla_{\phi'(t)} {\phi'(t)}$. But that's how far I'm only getting!
Thanks for your answers/inputs!
