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I am interested in complete Riemannian manifolds whose geodesics minimize length globally. Such manifolds must be non-compact (otherwise there is always a self-intersecting geodesic)

However, I suspect this property is much more restrictive.

Question: Assume $(M,g)$ is complete and has this property.

Must $M$ be simply connected? The exponential map has to be a diffeomorphism on all $T_pM$? Does the sectional curvature has to be non-positive? Are there unique geodesics between any two points?

Update: From John Ma's answer, it turns out that the $exp_p$ is a diffeomorphism, and in particular there are unique geodesics between any two points.

I would still like to know if anything intelligent can be said about the curvature though.

My guess is that it does not has to be non-positive everywhere, but only 'mostly everywhere' in some sense. (i.e I can imagine a surface with small regions of positive curvature which doesn't violate our condition )


I am looking in general for necessary and sufficient conditions (topological\curvature constraints) for this property to hold.

One sufficient conditions is provided by Hadamard's Theorem.

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2 Answers 2

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Partial answer

Let $M$ be a complete Riemannian manifold. Then every two points can be joined by a length minimizing geodesic ($\gamma$ is called length minimizing between $p$, $q$ if for any piece-wise smooth curves $\eta$ joining $p$ and $q$ we have $L(\eta)\ge L(\gamma)$). The following are equivalent:

  • All geodesics in $M$ are length-minimizing.

  • All points in $M$ are joined by a unique geodesic.

  • The exponential map $\exp_p : T_pM \to M$ is a diffeomorphism for all $p \in M$.

$(1)\Rightarrow (2)$: If not, let $p, q\in M$ and $\gamma_1, \gamma_2 : [0,1] \to M$ be two geodesics joining $p, q$. Then both $\gamma_1, \gamma_2$ cannot be length minimizing when $t>1$.

$(2)\Rightarrow (1)$ is obvious, since length minimizing curve must be a geodesic and every two points can be joined by at least one geodesic as $M$ is complete.

$(2)\Rightarrow (3)$: By completeness, $\exp_p$ is surjective. If it is not injective, then there are two geodesics which starts at $p$ and interest at some points. Indeed it has to be a diffeomorphism. If not, then there are two points that are conjugate to each other. Thus there are (yet another) two points $p, q$ so that they are joined by a geodesic $\gamma$ which is not length minimizing. By completeness, $p$ and $q$ are also joined by another geodesic $\eta$, thus it contradicts $(2)$.

$(3)\Rightarrow (2)$ is also obvious.

So there are huge topological constraint given by $(3)$. Not completely sure about the curvature constraint, except the one you mentioned.

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  • $\begingroup$ Thanks. $(1) \Rightarrow (2)$ was not entirely obvious to me. Is what I have written in my answer similar to what you were thinking about? Also, you could have proved $(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (1)$ circularly, since $(3) \Rightarrow (1)$ by a well known claim. (you can see my answer for details). $\endgroup$ Commented Nov 15, 2015 at 18:40
  • $\begingroup$ Actually, I thought more about your argument $(2) \Rightarrow (3)$, and something bothers me. Existence of a nontrivial Jacoby field does not imply existence of a family of geodesics joining $p$ and $q$. (See en.wikipedia.org/wiki/Conjugate_points). However, it's true that no geodesic is minimizing past its first conjugate point. Hence, if $exp_p$ has a singular point, we obtain a geodesic which does not minimize length at all times. Thus I think this argument shows that $(1)+(2) \Rightarrow (3)$. As far as I can see it does not prove $(2) \Rightarrow (1)$ or $(2) \Rightarrow (3)$ $\endgroup$ Commented Nov 22, 2015 at 16:04
  • $\begingroup$ So the question whether $(2)$ is equivalent to $(1),(3)$ remains open for now. $\endgroup$ Commented Nov 22, 2015 at 16:06
  • $\begingroup$ @AsafShachar : Please take a look at the edit. Thanks for your link about conjugate points. However I am a bit skeptical about the result stated in that page (I avoided it anyway). Do you have a proof to that result? $\endgroup$
    – user99914
    Commented Nov 23, 2015 at 2:17
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    $\begingroup$ 1) I agree with your edit. Very nice! 2) I guess you meant for a proof that two conjugate points are not necessarily joined by more than one geodesic? I do not have an example for this, and I am in fact also interested in finding one... 3) There is some discussion about this, though no concrete example in Lee's book "Riemannian manifolds An introduciton to curvature" - page 188, where he also proves geodesics are not length minimizers after they pass conjugate points $\endgroup$ Commented Nov 23, 2015 at 9:38
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I am just elaborating on some points related to John's answer:

$(1)\Rightarrow (2)$: By completeness, every two points are joined by a geodesic. Assume there exists $p,q \in M, \gamma_1, \gamma_2 : [0,L] \to M$ two geodesics (parametrized by arclength) joining $p,q$. Then both $\gamma_1,\gamma_2$ cannot be length minimizing when $t>L$:

Assume $\gamma_1$ is minimizing between $p=\gamma_1(0),\gamma_1(L+\epsilon)$. Define a path $\beta=\gamma_1|_{[L,L+\epsilon]} *\gamma_2|_{[0,L]}$.

$L(\beta)=L+\epsilon=L(\gamma_1|_{[0,L+\epsilon]})=d(p,\gamma_1(L+\epsilon))$

So $\beta$ is length minimizing between $p,\gamma_1(L+\epsilon)$ so it's a geodesic*, and in particular smooth. But this implies $\dot \gamma_1(L)=\dot \gamma_2(L)$, so by uniqueness $\gamma_1=\gamma_2$.

$(2) \Rightarrow (1)$: Assume there is a geodesic between $p,q$ which is not minimizing. Hopf-Rinow theorem implies there is a minimizing geodesic between $p,q$, so there is more than one which is a contradiction.

Note: $(3) \Rightarrow (1)$ is also immediate, since we know geodesics minimize distance from a given point $p$ as long as they stay in a normal ball around $p$. (See Do-carmo's Riemannian geometry, proposition 3.6). If $exp$ is a diffeomorphism, then all $M$ can considered as a normal ball.


*See Do-carmo's Riemannian geometry, proposition 3.9

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