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**.



G-spaces and manifolds
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. 


