Completely Geodesic

I'm having trouble showing the following implication:

Let $M$ be a Riemannian manifold, let $L\subset M$ be a submanifold such that the following holds:

• If $\gamma: I \to M$ is a geodesic s.t. $\gamma(0) \in L$, $\dot{\gamma}(0) \in T_{\gamma(0)}L$, then there is an $\epsilon >0$ such that for $|t| < \epsilon$ we have $\gamma(t) \in L$.

Then for any smooth curve $\eta: I \to L$ we have

$$P_\eta (s,t) T_{\eta(t)}L = T_{\eta(s)}L$$

where $P_\eta (s,t)$ denotes parallel transport along $\eta$.

I don't really have an approach to this one. The differential equations satisfied by a geodesic looks sort of like the ones satisfied by the parallel transport of a vector, but I didn't get anywhere with this.

Any pointers would be greatly appreciated. Thanks in advance!

S.L.

-
BTW, the usual English terminology is "totally geodesic", not "completely geodesic". – Willie Wong Jan 24 '11 at 13:44

Let $II$ denote the second fundamental form of $L$. That is, given $\nabla$ the Levi-Civita connection of $M$ and its Riemannian metric, given vector fields $v,w$ tangent to $L$, define $II(v,w)$ to be the normal projection of $\nabla_vw$ (normal projection meaning the projection to the orthogonal complement of $TL$).
$II$ is tensorial and symmetric.
1. The parallel transport condition you want to show is equivalent to $II = 0$
2. The geodesic condition you are given tells you that $II(v,v) = 0$ for all $v$. Now consider $II(v+w,v+w)$.