Manifolds with geodesics which minimize length globally

Question

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.

score 7 · Accepted Answer · 2015-11-23T02:16:18.590

7

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.

edited Nov 23 '15 at 02:16

answered Nov 15 '15 at 12:03

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). – Asaf Shachar Nov 15 '15 at 18:40
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 https://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 $(2) \Rightarrow (3)$ – Asaf Shachar Nov 22 '15 at 16:04
So the question whether $(2)$ is equivalent to $(1),(3)$ remains open for now. – Asaf Shachar Nov 22 '15 at 16:06
@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? – Nov 23 '15 at 02:17
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

Asaf Shachar

Nov 23 '15 at 09:38

Yes, that's the result that I am looking for. Thanks for pointing that out anyway. I'll take a look at Jack lee's book. @AsafShachar – Nov 23 '15 at 09:48

@JohnMa I have surface in 3D which can be paramertrized as $(u,v)\mapsto (u,v, f(u, v))$ where $f$ is smooth, and the domain of the parametrization is the unit disk. This animation shows that multiple geodesics connects the blue and the black point. Is it because of incompleteness that the points are not joined by a unique geodesic? – MeMyselfI Aug 02 '18 at 20:03

score 6 · Answer 2 · answered Nov 15 '15 at 18:25

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

Manifolds with geodesics which minimize length globally

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