# Decomposition of pullback metric

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.

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.

• Thanks for your answer @WillieWong. Can you explain a little bit? What's means "extended by the product structure"? I need to consider the pushforward or the pullback of $V$? And why that implies the existence of a smooth family of metrics in $\Sigma$? I'll appreciate. – Irddo Jul 8 '16 at 4:06
• "Extended by the product structure" is the same as $[\dot{\gamma}_p,V] = 0$. Basically $T_{(p,t)}(\Sigma \times [0,\epsilon))$ can be canonically identified with $T_p\Sigma \times T_t[0,\epsilon)$ using that we have a Cartesian product. Then the extension of $V\in T_p\Sigma$ is the vector field $(V,0)\in T_p\Sigma \times T_t[0,\epsilon)$. – Willie Wong Jul 8 '16 at 13:23
• You need to consider the pushforward. $V$ is defined along $\Sigma \times [0,\epsilon)$ you push it forward to $M$ by $\phi$. The function $t$ is defined on $M$ in the $\epsilon$ neighborhood of $M$ by essentially some version of the tubular neighborhood theorem (that $\phi$ is in fact a $C^1$ diffeomorphism onto its image). – Willie Wong Jul 8 '16 at 13:26
• For each $t$ the diffeo $\phi$ induces a metric on $\Sigma$ by the pullback along $\phi(\cdot,t): \Sigma \to M$. These give you the $\mathrm{d}\sigma_t^2$. In fact given any immersion $\Psi: \Sigma \times (a,b) \to M$ where $\Sigma$ is arbitrary and $M$ is a Riemannian manifold, for each $t\in (a,b)$ you get a pullback metric $g_t$ on $\Sigma$ defined by $\Psi(\cdot,t)^* g$ and the family is smooth. The only part of your claim not immediately obvious from the definitions is the fact that the pullback metric on $\Sigma\times (a,b)$ splits. – Willie Wong Jul 8 '16 at 13:28
• I leave that as an exercise. But a hint: $V(t)|_{t = 0} = 0$ since $V$ is tangent to $\Sigma$ initially. So it suffices to show $$\nabla_{\dot{\gamma}_p} (V(t)) = 0$$ (I abuse notation by identifying $V$ and its pushforward). This equation follows by pushing some definitions around. – Willie Wong Jul 8 '16 at 20:22