Let $(M^3,g)$ be a complete riemannian manifold and $\Sigma ^2\subset M^3$ a embedded minimal compact surface. Consider the immersion $\phi: \Sigma \times [0,\varepsilon)\to M$ given by


$$\phi(p,t)=\exp\_p(tN(p)),$$


when $N$ is a unit normal vector field along to $\Sigma$.


I would like to show that if we take the pullback metric $\phi^\*g$ in $\Sigma\times [0,\varepsilon)$, then $\phi^\*g=d\sigma\_t^2+dt^2$, where $d\sigma\_t^2$ is a smooth family of metrics in $\Sigma$.


I saw already argument like that in many papers, but I failed in prove it.


Anyone has a little help? Thanks so much.



Decomposition of pullback metric
This has nothing to do with minimality of $\Sigma$. 


By definition of the exponential map, the mapping $\gamma\_p(t) = \phi(p,t)$, as a map $[0,\epsilon)\to M$ is a geodesic ray, with unit speed, initial position $p\in \Sigma$, and initial velocity $N\_p$. 


Let $V$ be a vector in $T\_p \Sigma$ extended to $\Sigma \times [0,\epsilon)$ along $\gamma\_p(t)$ by the product structure. Equivalently, you have that $V$ is Lie-transported along $\gamma\_p(t)$, or that $[\dot{\gamma\_p}, V] = 0$. (Vector field commutators are independent of metric!)


The pushforward of $V$ is a vector field that satisfies $V(t) = 0$. To get your splitting it suffices to show that $\langle V, \dot{\gamma\_p}\rangle\_g = 0$. But this follow because at time $t = 0$ the equality is true, and 
$$ \nabla\_{\dot{\gamma\_p}} \langle V, \dot{\gamma\_p}\rangle\_g = \langle \nabla\_V \dot{\gamma\_p}, \dot{\gamma\_p}\rangle\_g = 0 $$
by construction and geodesy. 


