# Existence of Fermi coordinates on a Riemannian manifold

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?

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I hope $\nabla$ here is gradient in a coordinate system, but NOT Levi-Civita connection (otherwise it has no sense). – Anton Petrunin Mar 19 '10 at 2:46
Anton: Absolutely. I'm implicitly working in a coordinate system, but I thought my post would be clear without me explicitly stating that. But yes: In my preferred coordinate system, g and its derivatives are bounded. – Tom LaGatta Mar 19 '10 at 7:01

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.

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1) I'm content with the thickness of the neighborhood shrinking. That's great to hear. 2) I'm working in a probabilistic context where the background coordinate system plays a crucial role. – Tom LaGatta Mar 19 '10 at 22:15

You need yet

• a lower bound for injectivity radius, otherwise small torus would be a counterexample.
• make this bound to be uniform on the manifold.
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Anton: Could you expand on this? By the way, my manifold is topologically equivalent to the plane, so there are no topological obstructions as with the torus. What do you mean by "make this bound to be uniform on the manifold?" – Tom LaGatta Mar 19 '10 at 7:05
Your condition is equivalent to bound on curvature at one point. It is outside of arbitrary nbhd of $p$ you space can be arbitrary bad --- you need a global condition, a bound on curvature in a fixed nbhd of $p$ will do. The topological condition will not help either, you may construct a metric on the plane which looks like cylinder in a nbhd of a point and has arbitrary small curvature everywhere. – Anton Petrunin Mar 19 '10 at 15:28

It is not enough to be within the injective radius. Let $(M,g)$ be an $(n+1)$-dimensional manifold and $\gamma\colon (-a,a) \to M$ a geodesic. One way to define Fermi coordinates

$\phi \colon (-a,a)\times D^n_{\epsilon} \to M$

is by

$\phi(t,x_1,\dots,x_n)$ = exp$\strut_{\gamma(t)}(x_1v_1(t)+\cdots + x_nv_n(t)$,

where $D^n_{\epsilon}$ is the open disc with radius $\epsilon$ and $v_i(s)$ is the $i^{\textrm{th}}$ vector of an othonormal frame of the orthogonal complement of $\gamma'(t) \in T_{\gamma(t)} M$, which is given by parallel transporting $v_i(0)$ back and forth to all of $\gamma$.

If we let $\delta>0$ be so small that any geodesic triangle with all sides less than $\delta$ cannot have two right angles (this exists locally, and may have to be strictly smaller than the injective radius) then if $2a$ and $\epsilon$ is smaller than this $\delta$ then $\phi$ will be injective and in fact a diffeomorphisms onto its image. To prove injectivity assume it is not then you get a geodesic triangle with two right angles - contradiction.

To the concrete question: The answer is NO. Knowing bounds at the point $p$ is never enough to get even an injective radius depending on these bounds, and certainly not this $\delta$.

HOWEVER: If you know concrete bounds on all of $M$ or within a concrete radius of $p$ then this is possible. I.e. you $f(h)$ does not exist, but if you assume the bounds in a radius depending also on $h$ it does.

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Precisely, in general, not knowing anything about the manifold (M,g), knowing any bound of any k-jet of the metric at one point is not enough to get any control away from the point, since you can always modify (M,g) arbitrarily close to p in any way you want while maintaining the value of any k-jet. If, on the other hand, you have some a priori knowledge of (M,g) [say, certain partial differential relation on g], then you may have a chance. Though usually existence is difficult to prove if you are just given data at a point. – Willie Wong Mar 19 '10 at 12:16

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.)

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