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.

  • $\begingroup$ 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. $\endgroup$
    – Tracy Hall
    Aug 4, 2010 at 16:02
  • $\begingroup$ @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? $\endgroup$ Aug 18, 2010 at 8:41
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    $\begingroup$ What a confusing term. A $G$-space should be a space equipped with an action of $G$... $\endgroup$ Apr 14, 2014 at 5:09
  • $\begingroup$ @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. $\endgroup$
    – user44143
    Jan 19, 2017 at 20:38
  • $\begingroup$ @QiaochuYuan, "G-space" sounds confusing now, but it probably was not when Busemann started using the term in 1955. $\endgroup$
    – user44143
    Jan 19, 2017 at 20:51

2 Answers 2


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.

  • $\begingroup$ So, as expected, a Riemannian manifold is also a G-space. What about topological or smooth manifolds? $\endgroup$ Aug 4, 2010 at 6:09
  • $\begingroup$ @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. $\endgroup$ Aug 4, 2010 at 8:58
  • $\begingroup$ 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. $\endgroup$
    – Deane Yang
    Aug 4, 2010 at 11:58
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    $\begingroup$ 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. $\endgroup$
    – Jack Lee
    Aug 4, 2010 at 14:40
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    $\begingroup$ 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). $\endgroup$
    – Deane Yang
    Aug 19, 2010 at 19:42

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).

  • $\begingroup$ Sounds right to me! But I haven't worked out the details. $\endgroup$
    – Deane Yang
    Aug 4, 2010 at 13:58

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