TITLE: Existence of Fermi coordinates on a Riemannian manifold
QUESTION [7 upvotes]: Let $(M,g)$ be a Riemannian manifold, $p$ a point on the manifold and $v \in T_p M$.  Let $\gamma$ be the geodesic starting at $p$ in the direction $v$.  There exists a time $t_f$ such that there exists a Fermi coordinate system adapted to $\gamma$ up to time $t_f$.  
My question: Does there exist a lower bound for $t_f$ in terms of the 2-jet of $g$ at $p$?
That is, I have solid estimates on $g$ up to its second derivatives at $p$:  $$\|g\| + \|\nabla g\| + \|\nabla^2 g\| \le h \qquad (1)$$ for some $h$.  I would like to show that there exists $f(h)$ such that $$t_f \ge f(h)$$ for all Riemannian metrics satisfying (1) at $p$.
Edit (Mar 19):  Taking the helpful advice of Anton, Deane, TK and Willie into account, I've reworded the question:
Let $U = B(0,r)$ be the closed Euclidean ball of radius $r$ in $\mathbb R^n$.  Write $$\lambda = \inf_{x \in U} \inf_{\|v\|=1} \langle v,v \rangle_{g(x)}$$ as the minimum eigenvalue of the metric in $U$, and suppose that $$\frac{1}{\lambda} + \|g\|_{C^2(U)} \le h.$$  This is a more refined version of (1) above.  Since $$\ddot \gamma^k = -\Gamma^k_{ij} \dot \gamma^i \dot \gamma^j,$$ our estimate gives a control on the acceleration of a geodesic $\gamma$ in $U$, so there exists a minimum self-intersection time $t_i$ depending on $h$ and $r$ (i.e., if $t, t' \le t_i$ then $\gamma(t) \ne \gamma(t')$).  
Does this imply the existence of a uniform lower bound on $t_f$ (depending only on $r$ and $h$)?  If so, can we relax the control on the second derivative of $g$?
More succinctly:  Are existence and non-self-intersection of a geodesic the only obstructions to the existence of Fermi coordinates?

REPLY [3 votes]: Let me start with definitions:  Given a submanifold $N$ in a Riemannian manifold $(M,g)$, one can consider exp as a map from the normal bundle of $N$ to $M$.  If $N$ is compact (possibly with boundary), then for small values of $t$, exp is a diffeomorphism from the normal disk bundle of radius $t$ to a neighborhood of $N$.  (I believe that this diffeomorphism is more or less what is meant by "Fermi coordinates" or "Gaussian coordinates."  Note that when $N$ is a point, these are just normal coordinates.)  I'll define the injectivity radius of $N$ to be the sup over all such $t$ that give you a diffeomorphism.  This is reasonable terminology, since I believe that (just as in the case when $N$ is a point) injectivity of exp restricted on the normal disk bundle of radius $t$ implies that exp is a diffeo (i.e. that there are no "focal points" or "conjugate points").
Proposition:  Suppose that the sectional curvature of $(M,g)$ is bounded above by $K$, and that the injectivity radius of each point in $M$ is bigger than $R_0$.  Then for any minimizing geodesic $\gamma$, the injectivity radius of $\gamma$ is at least as big as $R_1=\min(R_0,\pi/2\sqrt{K})$.
Proof:  Non-injectivity of exp means that you have two geodesics shorter than $R_1$ that are normal to $\gamma$ and intersect each other.  This gives you a geodesic triangle in $M$.  Consider a comparison triangle with the corresponding lengths in the sphere of curvature $K$.  Since two of the sides are shorter than $\pi/2\sqrt{K}$, the comparison triangle must have an acute angle on the edge corresponding to $\gamma$.  But this violates Toponogov's Theorem since our original triangle has right angles on the $\gamma$ edge.
The proof above is pretty much applicable to your situation.  You can prove a lower bound on the injectivity radii of points that are not too close to the boundary of your chart.  Similarly, if the endpoints of your geodesic are far away enough from the boundary, then your geodesic is minimizing.  Finally, the bounds on your metric certainly give you an upper bound on curvature.  (Moreover, you do not really need the full-strength version of Toponogov.)