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Consider $\mathbb{C}^2 = (r,\theta,\rho,\varphi)$ with the following cone metric in polar coordinates

\begin{equation*} ds^2= (dr^2 + \alpha^2 r^2 d\theta^2) + (d\rho^2 + \beta^2\rho^2 d\varphi^2) \end{equation*}

It is apparently well known that the regular part $(\mathbb{C}^*)^2$ is geodesic convex i.e for any two points $p,q \ in (\mathbb{C}^*)^2$, the minimal geodesic between them doesnt go through the "bad" sets $\{r=0\}$ or $\{\rho=0\}$. I know one line of argument proving this which exploits the fact that the above metric is toric (invariant under translations in $\theta$ and $\varphi$), and is similar in spirit to the case of orbifolds. I was wondering if there was any direct proof, for instance by explicitly showing that going around the singularity might be less expensive than going through it. In fact I am not even able to show this in the complex one dimensional case (i.e the standard two-dimensional cone).

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It is true only if $\alpha,\beta<1$.

Your space is isometric to a product of two cones. Therefore it is enough to prove that geodesic can not pass through the tip of two-dimensional cone.

Assume contrary. Then geodesic cuts cone into two angles with measure at least $\pi$ each. This implies that $\alpha\ge 1$.

By the way, your space is also Alexandrov space and the regular locus in Alexandrov space is convex. (This is my result and for polyhedral spaces the same was proved earlier by Milka).

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  • $\begingroup$ Yes, I should have said $\alpha,\beta < 1$. Could you please elaborate on your answer? Why does the geodesic cut the cone into two angles with measure greater than $\pi$? $\endgroup$ Feb 4 '14 at 18:26
  • $\begingroup$ If one of the angles is smaller than $\pi$ then you can make shortcut in the angle. $\endgroup$ Feb 4 '14 at 18:49

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