I'm in the following situation:
Let $\Sigma^{n-1}$ be a hyper-surface of a riemannian manifold $M^n$ and $\nu:\Sigma\to T\Sigma$ is a vector field normal to $\Sigma$ with $|\nu(p)|=r>0, \forall p\in \Sigma$. Let $\alpha:(-\epsilon,\epsilon)\to \Sigma$ with $\alpha(0)=p$ and $\alpha'(0)=u\in T_p\Sigma$.
Is it true that $$\left.\frac{d}{dt}\right|_0\exp_{\alpha(t)}[\nu(\alpha(t))]=P_{\gamma}(u),$$ where $\gamma(s)=\exp_p(s\nu(p))$ is the geodesic going out from $p$ with vector $\nu(p)$ and $P_{\gamma}(u)$ is the parallel transport of $u$ along $\gamma$ from $\gamma(0)=p$ to $\gamma(1)$?
If so, how do I compute this?
EDIT: It seems it is false: take $\alpha(t)=(\cos t, \sin t, 0)$ on the unit sphere $\Sigma=\mathbb{S}^2$ into $M=\mathbb{R}^3$ and $\nu(p)$ the outward normal vector, so $\nu(p)=p$. Then $\left.\frac{d}{dt}\right|_0\exp_{\alpha(t)}[\nu(\alpha(t))]=\left.\frac{d}{dt}\right|_02\alpha(t)=2u$, which is not the parallel transport of $u$ referred above.
So, what would be the correct answer?