Questions about hyperbolic structures on a sphere with cone point singularities

Question

1. How exactly do we put hyperbolic structures on a sphere with cone point singularities. Should I consider that sphere with cone points as an extended complex plane with punctures endowed with a suitable metric of curvature -1 ?

2. I am having trouble with the following questions when I was reading a paper : the paper mentioned the facts I am looking for a proof without proof : Could somebody give me the proofs ?

a) Let us take a sphere S with 5 cone points ( you can assume that the cone points are zero such that we get a sphere with 5 cusps ) . Let F denote the family of all simple closed non-trivial geodesics $ \delta $ such that each connected component of $ S \backslash \delta $ contains all least 2 cone points. Now choose $ \gamma $ such that $\gamma $ minimizes the length among all $ \delta $ 's in F. Cut S open along $\gamma $ and call the two pieces $ S_1 $ and $ S_2 $ .Choose any two points p and q on $ \gamma $. Prove that : there exists another geodesic path c joining p and q in $S_2$ such that c is not a subpath of $\gamma $.

b) Call $\gamma $ and $ \gamma' $ the two subpaths of $ \gamma $ separated by p and q.Prove that : ( or explain ) Either the concatenation of c with $\gamma $ or the concatenation of c with $\gamma' $ is is a simple closed curve whose simple closed geodesic representatives lie in F. [ my question is about "lying in F" part ].

c) Why do the above two facts imply that $ l(c) \geq $ minimum of l($\gamma$) ,l($\gamma'$) ?

I know these questions might be a bit technical, but I must apprecaiate your answer ,Thanks !

Answer 1

Concerning your first question, if you're used to Riemann surfaces then one way to understand hyperbolic metrics with cone singularities on the sphere is through a version of the uniformization theorem: given a closed Riemann surface $S$ with marked points $p_1, \cdots, p_n$ and angles $\theta_1,\cdots,\theta_n\in (0,2\pi)$ satisfying the obvious condition coming from the Gauss-Bonnet relation, there is a unique metric on $S$ compatible with the Riemann surface structure, with cone singularities of angle $\theta_i$ at each $p_i$.

This is a special case of Theorem A in: Troyanov, Marc, Prescribing curvature on compact surfaces with conical singularities. Trans. Amer. Math. Soc. 324 (1991), no. 2, 793–821.

In other terms, once you have chosen the $n$ angles in $(0,2\pi)$ with and $\sum_i (2\pi-\theta_i)>4\pi)$, hyperbolic metrics on the sphere with cone points of angles equal to the $\theta_i$ are in one-to-one correspondence with configurations of $n$ distinct points on $CP^1$.

Answer 2

Here are some pointers in the right direction. For 1), I would suggest that you read Peter Scott's classic article on the eight Thurston geometries. The article can be found on his webpage and it includes a highly readable introduction to hyperbolic geometry in the plane, as well as a discussion of orbifolds.

Next: Your family $F$ is the family of all simple closed geodesics in $S$. Next you chose a shortest element of $F$. (Note that this really is a choice - there may be a tie for shortest.) Now you choose two points $p,q$ in $\gamma$. Let $X$ be a component of $S - \gamma$. Note that (the closure of) $X$ is homeomorphic to either a twice-punctured disk or a three-times punctured disk. So draw this in the plane. Now prove that there is a simple arc (not necessarily a geodesic!) connecting $p$ to $q$ that is not properly isotopic into the boundary. (Say, using facts about the classification of surfaces, or using what you know about the fundamental group.) Now use what you know about geodesics to "pull the arc tight" to get the desired geodesic arc $c$.

Now, if $X$ is a twice-punctured disk then both of the concatenations are parallel into a puncture. So hopefully you took $X$ to be the three-time punctured disk. Again use the classification of surfaces to show that one of the two concatenations is non-trivial (while the other is parallel to a puncture.

Finally, pull that concatentation (call it $\omega$) tight to get an element of $F$. So $\ell(\omega) \geq \ell(\gamma)$ because $\gamma$ was shortest. Now, if $\ell(c)$ was less than the lengths of both $\gamma'$ and $\gamma''$ then $\omega$ would have length less than $\gamma$, a contradiction. Pulling tight strictly decreases length in this situation, because $c$ meets $\gamma$ at a definite angle.