Let be $(M^2,g)$ a closed riemannian manifold and $c:[0,L]\to M$ a simple closed geodesic on $M$. For each $s\in [0,L]$, let be $n(s)$ a unit normal vector field along to $c(s)$ and $\beta(s)$ the cut distance of $c(s)$ along to the geodesic $\gamma_s(t)=\exp_{c(s)}(tn(s))$.

Consider the application $$f(s,t)=\exp_{c(s)}(tn(s)), \ \ 0\le s\le L, \ \ 0\le t\le \beta(s).$$

How can I show that $f$ is a coordinate system to the disk $N\subset M$ bounded by the geodesic $c$ with and inner normal $n$? I.e., why the boundary of $M$, $\partial M$, is just the geodesic $c$?

Because, a priori I have no information about the application $s\mapsto \beta(s).$

I tried to prove that

$$\frac{\partial f}{\partial s}\left( \exp_{c(s)}(\beta(s)n(s)) \right)\equiv q\in M, \ \ \forall \ s\in [0,L],$$

but I don't know if $s \mapsto \beta(s)$ is smooth and how to explain $\beta'(s)$, for example.

Any help? Thanks.

ERRATA: $(M^2,g)$ is a topological sphere.


1 Answer 1


This is not true. The map may be onto, but it's for sure not a diffeomorphism up to $L$. And regarding smoothness, the question is not whether $\beta$ is smooth, the question is whether the directional derivative of $f$ with respect to $t$ is at $(s, \beta(s))$

The simplest case is, quite obviously, a round sphere in Euclidean space. In that case any geodesic is a great circle, and you map $[0,a]\times [0,b]$ onto a hemisphere. One side of the square will be mapped to a point, this cannot be a diffeomorphism, not even locally.

If your question is "why is the image of $f$ a closed disc?" then this is, in general, also not true, only if you choose the normal $n$ in such a way that deforming the geodesic into that direction allows you to contract it to a single point. On a sphere this is true for both directions, but in general this need not be true. So the question could be "let $\gamma$ be a losed geodesic bounding a disc $D$ in $M$, then why is $f$ onto, when choosing $n$ in such a way that it points into $D$".

With these clarifications it might make sense (but is not true) to try to prove that

$$\exp_{c(s)}(\beta(s)n(s)) = q$$ for some $q$, but not (as you wrote) that this is true for the derivative of this expression with respect to $s$.

In general the image of $\exp_{c(s)}(\beta(s)n(s)) = q$ will be difficult to analyze (there is an interesting paper of Philip Hartmann from the 50ies of last century iIrc about the distance set ot a curve in a plane which does exactly this in detail). The reason is that the cut locus may contain conjugate point, in which case $f$ fails to be a local diffeomophism (like in the example of the sphere, there every point is conjugate point).

It can be shown that the derivative of $f$ has full rank in the preimage of the cut locus minus focal points, this is done, e.g. in that paper of Hartmann.

Having said all this you can easily show that $f$ is onto by assuming there is some point $a$ in $D$ which is not in the image of $f$. This is in contradiction to the completenss of $M$ which implies there is minimizing geodesic from $a$ to $\gamma$.

Edit in response to a comment: if $a\in D$ is not in the image of $f$ there is (at least one) length minimizing geodesic $c_a$ from $a$ to $\gamma$.

If this is not clear consider, for every $s$ in the domain of definition of $\gamma$, a length minimizing geodesic $c_{as}$ from $a$ to $c(s)$ . Each of these geodesics can be written as $t\mapsto \exp_a tv(s)$ for some unit length tangent $v(s)\in T_a M$. Now minimize the lengths among theses curves (to see this consider a sequence $v(s_n)$ such that the lengths of the corresponding geodesics tends towwards a minimum. Since the vectors $v(s_n)$ have length $1$ a subsequence will converge to some $v(s)$ by compactness. It then follows from standard arguments about the exponential map (continuity) that $exp_a tv(s)$ is a geodesic from $a$ to $\gamma$ with minimal lengths among all such geodesics) It is then not difficult to see that this length minimizing geodesic has to meet $\gamma$ orthogonally. By uniqueness of geodesics with given initial direction this geodesic has to be of the form $t\mapsto f(s,t)$, and for symmetry reasons it has to be minimizing from $\gamma $ to $a$, contradicting the defining assumptions about $a$.

  • $\begingroup$ Thanks so such for your answer @Thomas. I have a really sorry, cause I forget to say in the that question $(M^2,g)$ is a topological sphere. So, can I tell me more about "only if you choose the normal nn in such a way that deforming the geodesic into that direction allows you to contract it to a single point."? I will add the hypothesis, sorry again. $\endgroup$
    – Irddo
    Apr 9, 2016 at 6:06
  • $\begingroup$ @Irddo In that case you can look at either side of the geodesic, but the rest of my answer still applies (with the same reasoning). i) $f$ won't be a diffeomorphism up to the boundary, ii) the image of the cut locus won't be a single point iii) $f$ will be onto by the same reasoning. $\endgroup$
    – Thomas
    Apr 9, 2016 at 6:11
  • $\begingroup$ Then I can't conclude that the image of $f$ a closed disc bounded by the geodesic $c$? I'm not sure that I understand your point. $\endgroup$
    – Irddo
    Apr 9, 2016 at 6:17
  • $\begingroup$ @Irddo Sure you can, $f$ is just not a diffeomorphism up to the boundary, nor is it one to one. I'm not sure what exactly you are trying to show. The (simple) reasoning from my answer starts out from the assumption that the geodesic bounds a topological disc and then shows it's onto. If you need to show that $\gamma$ bounds a disc, additional reasiong is necessary, but this is a topological question. If you want to answer that using your $f$ then you have to figure out what the image for the values $(s\beta(s))$ is (and as mentioned it's usually not a point nor is it a smooh curve in gener $\endgroup$
    – Thomas
    Apr 9, 2016 at 9:02
  • $\begingroup$ @Irddo (and it's not as straightforward as you seemingly thought it would be) $\endgroup$
    – Thomas
    Apr 9, 2016 at 9:19

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