Consider the manifold $M=\mathbb{R}^3 \setminus B$ where B is the ball with radius 1. Let $f \in C^{ \infty}(M) $ satisfying:
$$f = \frac{C(\theta, \phi)}{r} + O( r^{-2}) $$ Where $(r,\theta,\phi)$ is the usual spherical coordinates on $M$ and $C$ Is smooth function depending only on $\theta$ and $\phi$. Also, suppose derivatives of $f$ are $O(r^{-2})$.
Let $g$ be a metric on $M$ satisfying $$Ric = df^2$$
So the Ricci curvature decays to 0 at infinity. In particular, we have $Ric_{ij} = O(r^{-4})$.
We can use fermi normal coordinates $(\tilde{r}, \tilde{\theta}, \tilde{\phi}) $ satisfying $\tilde{r} = 1 $ on $\partial M$, and $\tilde{\theta} = \theta, \tilde{\phi} = \phi$ on $\partial M$. Then the metric in these coordinates is:
$$g = d\tilde{r}^{2} + \gamma_{\tilde{r}} ( \tilde{\theta}, \tilde{\phi}) $$ Where $\gamma_{\tilde{r_{0}}}$ is the induced metric on $\{ \tilde{r} = \tilde{r_{0}} \}$
1) With these assumptions on $g$, do we know that the coordinates $(\tilde{r}, \tilde{\theta}, \tilde{\phi}) $ do not break down? (Suppose $g$ is complete). Or equivalently do we know that the cut locus is empty? Or equivalently do we know that the distance function $\tilde{r}$ from $\partial M$ (which does exist everywhere on M) is smooth everywhere and satisfies $|\nabla \tilde{r} | = 1$?
2)Suppose those fermi normal coordinates do not break down. In terms of those coordinates, does $f$ enjoy the same asymptotics that it does enjoy in the usual spherical coordinates $(r,\theta,\phi)$? What is the relation between both coordinates? Intuitively, for big r, they should be very close. How do we show that? Finally, will the metric be asymptotically flat? If not, what other assumptions do we need on $g$ to be able to guarantee asymptotic flatness?
Any help is appreciated. Also please send me any references that can help.
