We know that on a Riemannian manifold, exponential map is locally a diffeomorphism and moreover, it's locally radial isometry: Let $p\in M$, then in a neighborhood in $T_pM$ where exponential map is well defined, pick a vector $v\in T_pM$, $q=\exp_p(v)$, $w\in T_v(T_pM)\sim T_pM$,(i.e., we identify these two vector spaces), then we have: $$ \left<(d \exp_p)_v(v),(d \exp_p)_v(w)\right>_q=\left<v,w\right>_p $$ Regarding the above equality, I believe that radial isometry does not imply isometry, am I right? In fact, I'm not certain about my understanding about the radial isometry, so if you could clarify the idea of radial isometry for me, that would also be great.
Final question, if that is correct, then could you please give me an explicit example for this? i.e. I want to say that there exists some $u\in T_v(T_pM)$ such that: $$ \left<(d \exp_p)_v(u),(d \exp_p)_v(w)\right>_q=\left<u,w\right>_p $$ does not hold.
Thanks in advance.