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Let $(M, g)$ be a Riemannian manifold, not necessarily complete. Let $x$ be a point in $M$, and let $r>0$ be such that the exponential map $\operatorname{exp}_x$ is defined on an open ball $B=B(0,r)\subseteq T_xM$. That is, all geodesics from $x$ exist to distance $r$.

If $y$ is a point in $M$ whose Riemannian distance from $x$ is less than $r$, does there necessarily exist a minimizing geodesic from $x$ to $y$?

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  • $\begingroup$ Maybe you want $M$ to be connected? $\endgroup$ – Neal Dec 12 '18 at 17:36
  • $\begingroup$ Sure -- if $y$ is at finite Riemannian distance from $x$ then it is in the same connected component as $x$. $\endgroup$ – macbeth Dec 12 '18 at 18:18
  • $\begingroup$ Can't you just apply the usual proof of Hopf-Rinow to deduce the result: Everything you need is that all arc-length geodesics starting at $x$ are defined up to time $d(x,y)$ which is guaranteed by your assumption. $\endgroup$ – Sebastian Dec 13 '18 at 9:26
  • $\begingroup$ Sort of! I am looking now at the proof of Hopf-Rinow, (2)=>(3), in Petersen, and indeed it's pretty similar to the argument I came up with in the CW answer below. The proof in Lee is a little different. $\endgroup$ – macbeth Dec 13 '18 at 12:30
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This is indeed the case. The basic reason is that within the domain of the exponential map, there are no issues of completeness (the minimizing geodesic needs to stay within this set.) As such, minimizing curves are indeed geodesics. I don't know of an elegant proof for this, but it's possible to just brute force the issue.

For a brute force approach, we want to find a complete manifold $\bar Q$ whose Riemannian metric coincides with that of $M$ on an open ball of $x$. From there, we can use the theory for complete manifolds to finish the argument.

Since $y$ is in the image of the exponential map, we have that $y = \exp_x(s\cdot \mathbf{v})$ for $\|\mathbf{v}\|=1$ and $s<r$. Pick $ q$ with $s< q<r$ and consider the set $U_q=\exp_x(B(0,q))$. This is an compact subset of $M$ with Lipschitz boundary.

As such, we can find an open domain $Q$ so that $$\exp_x(B(0,s)) \subset Q \subset U_q$$ and $Q$ has smooth boundary. Now, each of the $j$ boundary components $\partial Q_j$ are sub-manifolds, so we can attach cylinders $(\partial Q_j) \times [0,\infty)$ to $Q$ so that the resulting space $\bar Q$ is complete. We can do this attachment smoothly so that $\bar Q$ is a complete Riemannian manifold.

By our construction, there is a natural inclusion $ i:\exp_x(B(0,s)) \to \bar Q$. Via the triangle inequality, the minimal curve from $i(x)$ to $i(y)$ is completely contained in the image of $i$ and the Riemmanian metric of $M$ is the same as the Riemannian metric of $\bar Q$ on the set $\exp_x(B(0,s))$.

Therefore, if $\gamma$ is the minimizing curve from $x$ to $y$, then $i(\gamma)$ is the minimizing curve from $i(x)$ to $i(y)$ (and vice-versa). Since $\bar Q$ is complete and length minimizing curves in complete manifolds are geodesics, the curve $i(\gamma)$ is a geodesic. As such, the minimizing curve from $x$ to $y$ is also a geodesic.

Edit to address a question:

Macbeth asked how we know that $y$ is in the image of the exponential map, so I'm including an edit to address this question. This is only a sketch because it's really hard to give a clear proof without drawing a picture. As before, this a brute force argument that doesn't do the geometry justice. Intuitively, geodesics travel away from a point as fast as possible, so there is no way for a curve of the same length to be a further distance away.

Suppose that there were a point $y$ with $d(x,y)<r$ and $y$ was not in the image of the exponential map. Then there we can construct a sequence of points $y_i$ so that all of the points $y_i$ are not in the image of the exponential map and so that $d(x,y_i)$ decreases to the infimum of distances for which the exponential map is not surjective. We can extract a convergent subsequence and consider the limit point $y_\infty$. Since all points $x'$ with $d(x,x') < d(x,y_\infty)$ can be joined to $x$ via a geodesic, the path minimizing the distance from $x$ to $y_\infty$ is composed of at most two geodesic segments (there's something left to prove here). By applying the triangle inequality and taking a sequence of points on the second segment of the piecewise geodesic curve, this actually shows that there is a unique geodesic from $x$ to $y_\infty$.

To get a genuine contradiction, consider $m$ large so that $y_m$ close enough to $y_\infty$ so that we can connect $y_m$ to $y_\infty$ via a single geodesic. Then there is a two geodesic path from $x$ to $y_m$, but not a single geodesic path from $x$ to $y_m$, by assumption. To get a contradiction, we will find a sequence of piecewise geodesic paths which converge to a geodesic from $y_m$ to $x$.

To start this procedure, find a geodesic connecting $y_m$ to a point $x_1$ on the geodesic connecting $x$ to $y_\infty$. The geodesic connecting $y_m$ to $x_1$ followed by the geodesic connecting $x_1$ to $x$ is shorter than our original segments via the triangle inequality. Then, find a point $y'_1$ of distance $d(x,y_\infty)$ from $x$ on the geodesic segment from $y_m$ to $x_1$. The geodesic connecting $y_m$ to $y'_1$ followed by the geodesic connecting $y'_1$ to $x$ is shorter than the previous geodesic segments. Repeating this to obtain points $x_i$ and $y'_i$, we find a sequence of such paths. I definitely recommend drawing a picture here, because it's hard to follow without one.

If one does this procedure in a reasonable way, then the sequence of paths will converge to a geodesic. A priori, it's only piecewise geodesic, but the direction on either side of $y'_\infty$ will be equal if you do this reasonably. As such, there is a geodesic from $x$ to $y_m$ whose length is less than $r$, which is a contradiction.

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    $\begingroup$ Thanks for the answer, the approach seems clever. I'm missing something simple, maybe -- why is $y$ in the image of the exponential map of $x$? $\endgroup$ – macbeth Dec 12 '18 at 21:10
  • $\begingroup$ Ah. Good question. You can do it using the triangle inequality, but let me edit my answer to address this. $\endgroup$ – Gabe K Dec 12 '18 at 21:41
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    $\begingroup$ Thanks for the details! A question -- when you extract a convergent subsequence of the sequence $\{y_n\}$, I imagine you are implicitly using that they are all contained in a common compact set. What is that set? The naive choice, to take the closed Riemannian ball $B^g_x(r)$, seems to me like it might fail, because in an incomplete manifold such a ball need not be compact. $\endgroup$ – macbeth Dec 12 '18 at 22:52
  • $\begingroup$ The idea I had in mind was to that it's a compact sequence of unit tangent vectors from $x$. To make that precise, you have to consider all the $y_n$ which can be joined by two segments, then use the triangle inequality on the second segment to get the right unit tangent vector. Since the distances from the geodesic ball are going to zero, one can take a limit of the directions to get a point in the boundary of the geodesic ball. $\endgroup$ – Gabe K Dec 12 '18 at 23:26
  • $\begingroup$ To be honest, I don't think it's necessary to prove that $y_\infty$ is the image of a geodesic. It's possible to just repeat the second half of the argument while having the $y'_i$ just inside geodesic ball where the exponential map is surjective, instead of on the boundary. Having them on the boundary just makes life a bit easier I think. $\endgroup$ – Gabe K Dec 12 '18 at 23:28
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I can answer the question under the additional assumption that $exp_x$ is a diffeomorphism on $B(0, r) \subseteq T_x M$, that is, if its image is a geodesic ball. All you need to do then is to invoke the following standard

Lemma. In the notation as above, let $U = exp_x (B(0, r)) \subseteq M$, and let $y \in U$. Then (up to reparametrization) the radial geodesic from $x$ to $y$ is minimizing (in fact, it is the unique minimizing curve from $x$ to $y$ in $M$, but we do not need that).

Reference: Proposition 6.10, John M. Lee, Riemannian Manifolds: An Introduction to Curvature, Springer (1997).

Having this said, we need only to show that $d(x,y) < r$ implies $y \in U$. Assume the converse: $y \notin U$. Let $\gamma \colon [a, b] \to M$ be a piecewise-smooth curve from $x$ to $y$ of length less than $r$. Let $c = \sup\{t \in [a, b] \, \colon \; \gamma(t) \in U\}$. On can easily see (from the lemma) that $d(x, \gamma(c)) \geqslant r$, which leads to a contradiction.

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  • $\begingroup$ Thank you, I am aware of this fact, but I am interested in the general case. $\endgroup$ – macbeth Dec 12 '18 at 18:17
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I am writing my reformulation of Gabe's argument here (community wiki), in case it is useful to others!

Let $\Sigma$ be the set of real numbers $t\in[0,r)$ such that for all $y$ with $d(x,y)\leq t$, there is a minimizing geodesic from $x$ to $y$. Clearly $\Sigma$ contains $0$ and is connected. We will show it is closed and open, which implies $\Sigma=[0,r)$.

Lemma. If $t\in\overline{\Sigma}$, then for any point $y$ with $d(x,y)\geq t$, there exists a vector $v\in T_xM$ with $|v|=t$, $d(x,\exp_x(v))=t$, and $d(\exp_x(v),y)=d(x,y)-t$.

Proof. Let $(t_n)$ be a sequence in $\Sigma$ tending to $t$. For each $n$ there exists a point $y_n$ such that $d(x,y_n)=t_n$ and $d(x,y)-t_n\leq d(y_n,y)<d(x,y)-t_n+\tfrac{1}{n}$. Since $t_n\in\Sigma$, there exists a minimizing geodesic from $x$ to $y_n$, hence a vector $v_n\in T_xM$ such that $|v_n|=t_n$ and $\exp_x(v_n)=y_n$. Passing to a subsequence, we may assume $(v_n)$ converges to some $v\in T_xM$ with $|v|=t$. This $v$ has the desired property.

Claim 1: $\Sigma$ is closed. (This works in any Riemannian manifold.)

Proof. Suppose that $[0,t)\subseteq\Sigma$. Let $y$ be a point with $d(x,y)=t$. By the Lemma, there exists a vector $v\in T_xM$ with $|v|=t$, $d(x,\exp_x(v))=t$, and $d(\exp_x(v),y)=0$. Thus $\exp_x(v)=y$. So the exponential map in direction $v$ is a minimizing geodesic from $x$ to $y$. We conclude $t\in\Sigma$.

Claim 2: $\Sigma$ is open.

Proof. Suppose that $t\in\Sigma$. Then the closed Riemannian ball $B:=\{\overline{y}:d(x,\overline{y})\leq t\}$ is equal to the image of the closed ball $\overline{B(0,t)}\subseteq T_xM$ under $\operatorname{exp}_x$, and therefore is compact. (This is where we use the hypothesis that $\operatorname{exp}_x$ is defined throughout $B(0,r)\subseteq T_xM$.) There therefore exists $\epsilon>0$, such that for all $\overline{y}\in B$, there are geodesic co-ordinates about $\overline{y}$ to radius $\epsilon$.

Let $y$ be a point with $t<d(x,y)<t+\epsilon$. By the Lemma, there exists a vector $v\in T_xM$ with $|v|=t$, such that, letting $\overline{y}:=\exp_x(v)$, we have $d(x,\overline{y})=t$, and $d(\overline{y},y)=d(x,y)-t<\epsilon$.

So, using the geodesic co-ordinates about $\overline{y}$, we can construct a minimizing geodesic from $\overline{y}$ to $y$. Since $d(x,y)=d(x,\overline{y})+d(\overline{y},y)$, and both these distances are achieved by minimizing geodesics, the distance from $x$ to $y$ is also achieved, necessarily by a minimizing geodesic. We conclude that $[t,t+\epsilon)\subseteq \Sigma$.

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Let $\overline{M}$ be the completion of $M$ as a metric space. For any $x, y \in M$, there exists a length-minimizing curve $c \subset \overline{M}$ from $x$ to $y$. Since $c \cap M$ is locally length-minimizing, it must consist of geodesic segments.

Assume $B = \exp_x B(0,r) \subset M$ and $d(x,y) < r$. If $y \notin B$, then the geodesic segment of $c$ that starts at $x$ must hit the boundary of $B$ and therefore has length at least $r$, which contradicts the length-minimizing property of $c$. It follows that $y \in B$ and $c$ is a length-minimizing geodesic in $B$ that connects $x$ to $y$.

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    $\begingroup$ I did not see this answer at the time it was posted, sorry. This is an interesting approach, and seems to rely on a fact which I had not previously known was true: that the completion of a Riemannian manifold is a geodesic space. Can you suggest a reference for this fact? It is surprising to me, since the standard criterion for a length space to be a geodesic space is properness, however the completion of a Riemannian manifold is not always proper. $\endgroup$ – macbeth Jan 17 at 19:44
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    $\begingroup$ I think you found the gap in my "proof". The argument works at stated, only if $\overline{M}$ is locally compact and that does not have to be the case for the metric completion of a Riemannian manifold. $\endgroup$ – Deane Yang Jan 18 at 2:39
  • $\begingroup$ @macbeth, is there any chance you know an example of a Riemannian manifold whose metric completion is not a geodesic space? $\endgroup$ – Deane Yang Jan 18 at 4:34
  • $\begingroup$ No, but I would like to .... You should ask this question! $\endgroup$ – macbeth Jan 18 at 14:47
  • $\begingroup$ OK. I will. I want to know, too! $\endgroup$ – Deane Yang Jan 18 at 14:58

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