6

Let $ M $ be a complete Riemannian manifold and let $ d: M \rightarrow R $ be the distance function from a given point $ 0 \in M $. I want to prove that $ d $ is a smooth function on $ M -(C(p)\cup {p}) $ where $ C(p) $ is the cut locus of M with respect to $ p $.

I have made the following arguments:

1) if $ r \leq dist(p, C(p)) $ then $ exp_p: B_r(0) \rightarrow B_r(p) $ is a diffeomeorphism (since it can be proved that $ exp_p $ is injective on $ B_r(0) $ and it has no critical points)

2) if $ r \leq dist(p, C(p)) $ then $ d= |exp_p^{-1}| $ on $ B_r(p) $. Therefore $ d $ is smooth on $ B_r(p) - {p} $.

Now i'm not able to extend this argument from $ B_r(p) - {p} $ to $ M -(C(p)\cup {p}) $. The principal problem is that we cannot give an inverse of $ exp_p $ on $ M -(C(p)\cup {p}) $...

Thanks

Karthik C
  • 2,581
  • If $q \in M - C(p)$ there's a unique minimizing geodesic from $p$ to $q$ -- that is to say, $\exp_p$ really is "invertible" on $M - C(p)$ (in the same way it's invertible on $B_r(p)$, anyway). – mollyerin Jul 10 '13 at 18:29
  • A smooth function can be a homeomorphism without being a diffeomorphism, for example $f(x)=x^3$, so the comment needs more explanation – David Epstein Feb 10 '14 at 12:35

1 Answers1

8

This question is several months old, so those who were interested probably know the answer by now. The crucial notion is that of a conjugate point, whose definition is usually given using Jacobi fields, but an equivalent definition is that $q$ is conjugate to $p$ if $\exp_p(t)=q$ and the derivative of $\exp_p$ at $t$ is not an injective linear map. It is a standard theorem about conjugate points that a point conjugate to $p$ is in the cut locus of $p$. In other words, there is a sequence of points $(q_i)$ converging to $q$, such that, for each $i$, there are at least two distinct shortest geodesics from $p$ to $q_i$. (It is not necessarily the case that there are two distinct shortest geodesics from $p$ to $q$.)

If $q$ is not in the cut locus of $p$, then $q$ is not a conjugate point. Therefore the derivative of $\exp_p$ is an isomorphism at $t$, and therefore an isomorphism nearby by continuity of the derivative. The inverse function theorem then shows that $\exp_p$ has an inverse near $q$.

David Epstein
  • 285