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](http://en.wikipedia.org/wiki/Fermi_coordinates) 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?



Existence of Fermi coordinates on a Riemannian manifold
I would answer this question this way:


First, without any further assumptions about the Riemannian metric on $M$, you don't even have a lower bound on the time $t\_f$ for which the geodesic *exists* and does not intersect itself. The geodesic may fail to exist either simply because the metric is incomplete.


Second, if for some reason you know that the geodesic exists for time $t\_f$ and does not hit any boundary or singular set of $M$, then you can definitely find Fermi co-ordinates on a neighborhood of the geodesic, *but* you have no control over the "thickness" of the neighborhood along the geodesic, so it may approach zero.


Third, the answers above hold regardless of whether you know anything about the metric at a single point $p$. That is far too little information, and you can make the metric do anything you want at a point arbitrarily close to the point $p$. As others have indicated, you must assume some information on an neighborhood of $p$, and the conclusion holds only for that neighborhood.


ADDED: Your question is analogous to the following one: Suppose I have a real-valued function $f$ on an interval, and for some (large) value of $k$, I know the $k$-th order Taylor polynomial of $f$ at a point $p$ in the interval. Is there a bound on the first derivative of $f$ on an open interval containing $p$, where the bound and the size of the interval depend only on the coefficients of the Taylor polynomial?


It is easy to show that the answer is no, and essentially the same proof can be used to show that the answer to your question is also no.


RESPONSE TO REVISED VERSION:
1) Yes, if your manifold is open, and you start with a geodesic segment, then there exist Fermi co-ordinates in a neighborhood of the geodesic. But you have no control over the thickness of the neighborhood around the geodesic on which the Fermi co-ordinates exist. It will vary and may approach zero as you approach one of the endpoints of the geodesic segment.


2) I'm a little baffled by why you would want to do everything relative to a background flat metric. Is this metric a natural part of what you need this for? This makes the question a bit contrived for me.


