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In his book "The geometry of geodesics" H. Busemann defines the notion of a G-space to be a space which satisfies the following axioms:

  1. The space is metric
  2. The space is finitely compact, i.e., a bounded infinite set has at least one accumulation point
  3. [metric convexity] For every $x\neq z$ there exists a third point $y$ different from $x$ and $z$ such that $d(x,y)+d(y,z)=d(x,z)$
  4. [local prolongation] To every point $p$ there corresponds $\rho_p>0$ such that for every two point $x,y\in S(p,\rho_p)$ there exists a point $z$ such that $d(x,y)+d(y,z)=d(x,z)$
  5. [uniqueness of prolongation] If $d(x,y)+d(y,z_1)=d(x,z_1)$ and $d(x,y)+d(y,z_2)=d(x,z_2)$ and $d(y,z_1)=d(y,z_2)$ then $z_1=z_2$.

Busemann conjectured that every $G$-space is a topological manifold. My question is does every topological/smooth/Riemannian manifold is also a $G$-space?

As for connected complete Riemannian manifold, I figured out that 1 holds since by the metric. 3 holds since every two points can be joined by a minimal geodesic, and then we can pick $y$ to be a point on it. 4 holds since it is a manifold and locally it is homeomorphic to some Euclidean space. Unfortunately, even in this case, I couldn't figure out 5 and 2.

Dror Atariah
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  • A couple of pedanticisms:

    1. Does Busemann state his conjecture only for connected $G$-spaces? By the usual definition, which specifies a dimension, the $G$-space that is a disconnected union of $S^1$ and $S^2$ is a not a topological manifold.

    2. The question, more precisely, is whether every manifold can be given a $G$-space metric. (The starting metric might be bad: Any cone point with angle greater than $2\pi$ spoils unique extension of geodesics through that point, or since deleting isolated points requires local reparameterization to a complete metric.

     – Tracy Hall Aug 04 '10 at 16:02
  • @Tracy: I'm not sure what do you mean by the "usual definition" of dimensionality. In his book, Busemann uses the Menger-Urysohn definition.

    As for the example you gave, it is not a G-space, since the 3rd axiom fails. If you have one point on $x\in S^1$ and the other on $z\in S^2$, then you cannot find a $y$ such that the equality holds. However, Busemann doesn't address the issue of connectedness as far I I could see in the book.

    Finally, you're saying that there can be a Riem. manifold which is not a G-space using the induced metric structure?

     – Dror Atariah Aug 18 '10 at 08:41
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    What a confusing term. A $G$-space should be a space equipped with an action of $G$... – Qiaochu Yuan Apr 14 '14 at 05:09
  • @TracyHall, Busemann proves from 1,2,3 that every two points in a G-space have a geodesic connecting them, and therefore that the space is connected. –  Jan 19 '17 at 20:38
  • @QiaochuYuan, "G-space" sounds confusing now, but it probably was not when Busemann started using the term in 1955. –  Jan 19 '17 at 20:51

2 Answers2

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On a *complete*smooth Riemannian manifold,

  1. Any bounded (with respect to the distance function induced by the Riemannian metric) closed set in a manifold is compact.

  2. This is telling you that there is a minimal geodesic joining $x$ to $y$ that, when extended, is also a minimal geodesic joining $x$ to $z_1$. And there is another minimal geodesic joining $x$ to $y$ that when extended is a minimal geodesic joining $x$ to $z_2$. But if there are two distinct geodesics joining $x$ to $y$, neither is minimal beyond $y$. So the two geodesics have to be the same and therefore $z_1 = z_2$.

CORRECTION: "complete" added to assumption above.

For a smooth manifold, you need to construct a distance function to get a G-space. One way to do this is to construct a complete Riemannian metric. I'm not certain that this can be done, but offhand if you take a locally finite covering by open sets diffeomorphic to the Euclidean ball, use the standard Euclidean metric on each ball (where each ball has radius $1$), and use a partition of unity subordinate to this cover to glue together these metrics, it seems to me that the resulting metric is complete.

For a topological manifold, I don't know.

Deane Yang
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  • So, as expected, a Riemannian manifold is also a G-space. What about topological or smooth manifolds? – Dror Atariah Aug 04 '10 at 06:09
  • @Yang: regarding your first point. In order to claim that closed and bounded sets are compact, you have to have a complete Riemannian manifold. And then, indeed, an infinite bounded closed set is finitely compact, since compactness implies limit point compactness and the set is bounded. However, how one proves that infinite bounded but open sets are finitely compact? This has to be addressed in order to show that the 2nd axiom in the definition holds. – Dror Atariah Aug 04 '10 at 08:58
  • I agree that completeness is a requirement for the distance function induced by a Riemannian metric to define a G-space. It's obviously not so otherwise. – Deane Yang Aug 04 '10 at 11:58
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    Deane -- the metric you constructed by a partition of unity might not be complete. (For a counterexample, just take a bounded open interval in R, cover it with finitely many unit intervals, and use the standard metric in each open set. When you glue them together with a POU, you get back the standard metric.) But it's true that every smooth manifold admits a complete Riemannian metric. One way to see this is because every smooth manifold admits a proper (hence closed) embedding into some Euclidean space; the metric induced by such an embedding is complete. – Jack Lee Aug 04 '10 at 14:40
  • Jack, thanks! that was pretty dumb of me. At the risk of being dumb twice, what if I mapped each ball onto $R^n$ and pulled back the standard metric on $R^n$? – Deane Yang Aug 04 '10 at 19:04
  • Here's another way maybe? Start with a locally finite cover by balls. Take a partition of unity subordinate to the cover. You get a new covering by balls using the support of each partition function thickened slightly. Use the metric that makes each of these balls radius 1 or infinite. – Deane Yang Aug 04 '10 at 19:09
  • Hmmm... Your first suggestion (mapping each ball onto $R^n$) certainly sounds plausible, and the second one might work too; but at the moment I don't see how to prove either one or to find a counterexample. Let me think about it a little more. – Jack Lee Aug 04 '10 at 21:46
  • I still don't understand how to prove the 2nd axiom. In the case of closed and bounded the issue is clear. What about the open and bounded case? – Dror Atariah Aug 18 '10 at 08:52
  • The closure of any open bounded set is closed and bounded, i.e. compact. – Deane Yang Aug 18 '10 at 14:17
  • But can't it happen then that the accumulation point is in the boundary of the set and not in the open set itself? Am I missing something? – Dror Atariah Aug 18 '10 at 15:11
  • Of course, it can. There is no reason to expect the accumulation point to lie in the open set itself. That isn't true even for the real line. – Deane Yang Aug 18 '10 at 16:53
  • So, in this case we still have to show why an open bounded set has an accumulation point in order to complete the proof that a Riemannian manifold is a G-space. – Dror Atariah Aug 19 '10 at 08:13
  • So the real line is not a G-space? Could you give me an example of something that is? – Deane Yang Aug 19 '10 at 13:28
  • @Yang: As the real line is a 1-manifold, it is also, as proved by Busemann himself, a G-space. Moreover, for the real line, infinite, bounded sets have accumulation points.

    My current problem with your explanation is how do you prove that an open infinite bounded subset of a Riemannian manifold has accumulation point. This has to be addressed in order to complete the proof that a Riemannian manifold is also a G-space.

     – Dror Atariah Aug 19 '10 at 14:00
  • "how do you prove that an open infinite bounded subset of a Riemannian manifold has accumulation point": The same proof that works for the real line works for a Riemannian manifold. But notice that even for the real line, the accumulation point can lie on the boundary and therefore outside the open set itself. – Deane Yang Aug 19 '10 at 14:28
  • I'm am fairly certain that in Busemann's definition he is not assuming that the accumulation point must lie in the open set itself. – Deane Yang Aug 19 '10 at 14:55
  • If this is the case, then I'm not sure that I understand how to show that the second axiom holds for a Riemannian manifold. Shouldn't it be showed that every infinite, bounded set has at least one accumulation point (in the set). On the real line, for an infinite bounded set may have accumulation points on the boundary, but there will also be accumulation points in the set. Can there be an infinite subset of $R$ for which all accumulation points are boundary points? – Dror Atariah Aug 19 '10 at 15:00
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    Thanks to Google Books, I can actually find the relevant passages. I would agree that the text is confusing. He does state explicitly that he wants a definition that applies to Euclidean space. It appears to me that his term "finitely compact" is essentially what we today call "locally compact". In particular, he simply wants to assume that any bounded sequence in a G-space has at least one accumulation point in the space (but not necessarily in the sequence itself). – Deane Yang Aug 19 '10 at 19:42
  • So if we agree upon this, it means, that one can always refer to infinite bounded closed sets, and use compactness. – Dror Atariah Aug 20 '10 at 14:20
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If I am not wrong, I believe that every topological manifold admits a complete metric. With this metric, it is possible to give a Path Metric Space structure (in the sense of Gromov, see the book Metric Structures for Riemannian and non-Riemannian spaces). I guess that this structure allows to make the same arguments as for Riemannian manifolds for topological manifolds.

Anyway, I recomend this survey about Busemann conjecture which also discusses a stronger conjecture (the Bing-Borsuk conjecture).

rpotrie
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