Distance function on complete Riemannian manifold.

Question

Let $\left(M, g\right)$ be a complete Riemannian manifold. Let us fix a point $p \in M$ and consider the distance function $$ r(x) := \operatorname{dist}(x, p). $$ I would like to characterize the point where $r$ is not smooth. Is it true that they must be critical points of the function $\exp_p$? And is it true that a critical point of $\exp_p$ is a point where $r$ is not smooth?

Any help will be really appreciated. Thanks!

Answer 1

Let $C_p$ be the cut locus of $p$. $C_p$ contains two type of points: (i) points $q$ such that there exist at least two minimizing geodesics from $p$ to $q$; (ii) points $q$ that are conjugate to $p$. It can be proven that $C_p$ is closed and also a null set (i.e. if $(h,U)$ is a chart, then $h (U \cap C_p)$ is a null set with respect to the usual Lebesgue measure).

There is an alternative description of $C_p$. Let $I_p$ be the connected component of $\{ v \in T_p M \mid \Bbb d _v \exp_p \text{ is an isomorphism} \}$ containing $0 \in T_p M$. Let $c_p = \partial I_p$, the boundary of $I_p$. Then $C_p = \exp _p (c_p)$.

Finally, to answer your question: it is well known that $d(p, \cdot)$ is smooth on $M \setminus (C_p \cup \{p\})$ (that $d(p, \cdot)$ cannot be smooth in $p$ is obvious: choose normal coordinates around $p$ and think of what happens to the Euclidean norm around $0$). The square $d(p, \cdot)^2$ is smooth on $M \setminus C_p$.

For further details on the cut locus, see for instance sub-chapter III.2 of "Riemannian Geometry" by Isaac Chavel (I have a feeling, though, that you will not enjoy his style). A clearer exposition can be found in sub-chapter VIII.7 of vol. 2 of "Foundations of Differential Geometry" by Kobayashi & Nomizu (conjugate points are introduced in VIII.3).

A brief exposition of these concepts, together with a proof of the differentiability of $d(p, \cdot)$ can be found in theorem 3.1 of the lecture notes of a course given by Zuoqin Wang at the University of Science and Technology of China.

Answer 2

Even if the curvature is $\leq 0$, hence the exponential is a diffeomorphism, one can find a point where $r$ is not smooth. Intuitively, if $r$ is smooth its gradient would give you the shortest way to go to the origin. Consider the shortest loop based at $p$ of length $l$. On this loop consider the point $q$ on this geodesic at the distnace $l/2$ of $p$ ; if the exponential map computed at the point $q$ is a diffeo (for instance if the manifold has $K\leq 0$), in the neigbourhoof of $q$ the gradient of $r$ would give you two opposite direction.

For example, consider a surface with curvature $0$ outside a small neigbourhood of some point with huge negative curvature, (even a flat torus if the genus is 0), and such that our geodesic loop remains in the $K=0$ region.

Passing to the univeral cover at the point $q$ via the exponentail map at this point, one see that the distance to $p$ can be computed in a chart : is the minium of the distance to two points say $P, P'$ such that the origin (the lift of $q$) is the middle. Using Pythagore formula the distance in this euclidan open set will be $r(x,y)= Min (\sqrt {(x-l/2)^2+y^2} ; \sqrt {(x +l/2)^2+y^2)}= Min (r_1, r_2)$ (around $(0,0)$. But $\partial _x r_1 (0,0)=-1$, and $\partial _x r_2 (0,0)=+1$, and $f$ is not smooth.

For your second question, on the 2-sphre with constant curvature, if $p$ is the south pole and $q$ the north pole, $q$ is a crtical value of $\exp _p$ but around $q$ the distance to $p$ is smooth.