Application of Gauss Lemma In the proof of the following theorem:




First I don't understand why $L(\gamma) = R$. How do we differentiate the exponential map here? $\gamma'(t) = \frac{d}{dt} \exp_p (tX) = d(\exp_p)_{tX} X$. I think it should be evaluated at 0, so we get the identity and then $|\gamma'(t)| = |X|$, but I don't know why you have to evaluate this at 0.
My second question is why the Gauss Lemma implies:
$|\gamma'(t)|^2 = |\rho'(t) d(\exp(v(t))|^2 + |\rho(t)d(\exp(v'(t))|$.
We must have that the cross term vanishes, but I don't see how this follows from the Gauss Lemma. $v'(t)$ is tangent to some geodesic sphere, so if the $\rho'(t) d(\exp(v(t))$ would somehow represent a geodesic emanating from the center of the geodesic ball, the Gauss Lemma would give this, but I don't see why this term would correspond to such a geodesic.
 A: If $\gamma$ is a  geodesic, by definition ${D\over dt}\gamma'(t) = 0$, where ${D \over dt} = \nabla_{\gamma'}$, with $\nabla$ the Levi-Civita connection. Hence, 
$$ {d\over dt} \langle \gamma'(t),\gamma'(t)\rangle = \langle  {D\over dt} \gamma'(t),\gamma'(t)\rangle + \langle   \gamma'(t),{D\over dt}\gamma'(t)\rangle = 0.$$
Hence $|\gamma'(t)|^2 = |\gamma'(0) |^2 = |X|^2 = R^2.$ 
For your second question about the claim, where $\gamma$ is no longer assumed to be a geodesic: 


*

*By construction, $t\mapsto v(t)$ is a curve on $S^{n-1} \subset T_pM$. 

*Therefore, viewed as a vector, $v(t)$ is perpendicular to $v'(t)$ (i.e., the radial vector is normal to the sphere where it crosses/points). 

*View $v(t)$ and $v'(t)$ as (fixed) vectors 'starting' at the point $\xi(t)$ in $T_pM$  [i.e., by translation.].  

*Then, by the Gauss lemma, $d\exp_{\xi(t)}v(t)$ is perpendicular to $d\exp_{\xi(t)}v'(t)$, as vectors in $T_{\gamma(t)}M$. 


[I think there is a misprint in the above:the  $TM $ should be $T_pM$.]
