Consider a metric space $(X,d)$ with a distinguished selection of geodesics, i.e. a geodesic bicombing $\sigma:X\times X\times [0,1]\rightarrow X$. We call a geodesic bicombing conical if it satisfies the following "convexity-type condition" $$ d(\sigma_{xy}(t), \sigma_{x^\prime y^\prime}(t))\leq (1-t)d(x,x^\prime)+t d(y,y^\prime) $$ for all $x,x',y,y'\in X$ and $t\in [0,1]$.

If $(X,d)$ is a complete, separable, contractible Gromov ($\delta$) hyperbolic intrinsic metric space, for some $\delta>0$, then does $(X,d)$ admit a conical geodesic bicombing?

  • $\begingroup$ This condition implies contractibility of $X$, are you sure that you really mean it? $\endgroup$ Sep 18 at 16:28
  • $\begingroup$ I think you should write "Gromov-hyperbolic" rather than $\delta$-hyperbolic. $\delta$ is a parameter, which does not play any role in your question. $\endgroup$
    – YCor
    Sep 18 at 16:29
  • $\begingroup$ @MoisheKohan Ah, in the case I had in mind, my space is a contracible ANR which is additionally $\delta$-hyperbolic. Otherwise, yes, I would not have a shot. $\endgroup$ Sep 18 at 16:47
  • $\begingroup$ Ok, there is still an easy counter-example: take the unit sphere with the standard (Riemannian) metric and remove a point from it. Maybe you are making further assumptions to rule out such examples. Note that metric completeness is not enough (remove a small open disk from the unit sphere). $\endgroup$ Sep 18 at 16:54
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    $\begingroup$ As I explained, the added assumptions are insufficient. Maybe you should explain in detail the actual example you are interested in. $\endgroup$ Sep 18 at 17:33

1 Answer 1


None of your extra assumptions will suffice. An easy example is obtained as follows. Start with the unit 2-sphere $S^2$ with the standard Riemannian metric $ds^2$ induced from the embedding in ${\mathbb R}^3$. Remove from $S^2$ a small open metric disk (of radius $<\pi/2$, cetered at a point $p\in S^2$). Equip the resulting space $X$ with the length structure coming from $ds^2$. The corresponding metric $d$ on $X$ will be complete, etc, since $X$ is compact. However, it is clear that $(X,d)$ fails the required convexity property: Take the point $x=-p$. Then every $y\in (X,d)$ is connected to $x$ by a unique geodesic. But the convexity property fails for such geodesics (just look at $y, y'$ which belong to the boundary circle of $X$).


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