# geodesic convexity of regular sets for metrics with cone singularities

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).

It is true only if $\alpha,\beta<1$.
Assume contrary. Then geodesic cuts cone into two angles with measure at least $\pi$ each. This implies that $\alpha\ge 1$.
• 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 $\pi$? – Poincare-Lelong Feb 4 '14 at 18:26
• If one of the angles is smaller than $\pi$ then you can make shortcut in the angle. – Anton Petrunin Feb 4 '14 at 18:49