Consider a 3-sheeted Riemann surface without a Z_3 symmetry. The first and second sheets are sewn together at an interval (u1,v1), and the second and third are sewn at (u2,v2). This is a Riemann sphere. What is the map that uniformizes this surface? In other word, I am looking for a map w(z) that maps the n-sheeted surface to the Riemann sphere, and has trivial monodromy around all loops that circle singularities u1,v1,u2,v2.
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$\begingroup$ It seems to me that you are studying some Renyi entropy literature and that has lead you to this. Can you share some of your motivation for doing this weird kind of branching? (this isn't how the branching would be if you were say calculating entropies for two intervals) So in your description there is no branching across the (u_2,v_2) on sheet-1 and (u_1,v_1) on sheet-3? That looks weird to me... $\endgroup$– user6818Commented Mar 14, 2014 at 22:11
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$\begingroup$ My understanding of the Renyi entropy literature is that the sphere uniformization exists only for the n-sheeted branching across a single interval. With more than 1-interval the required branched Riemann surface doesn't uniformize. [...although as I said above the required branched surface is different from what you are describing...] $\endgroup$– user6818Commented Mar 14, 2014 at 22:13
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$\begingroup$ This question has come up in a calculation closely related to Renyi entropies in CFTs. Indeed, I am interested in a 3-sheeted surface with no branching across the (u_2,v_2) on sheet-1 and (u_1,v_1) on sheet-3. This is topologically a Riemann sphere. So I believe there has to exist a meromorphic map that uniformizes it. $\endgroup$– Nima LashkariCommented Mar 14, 2014 at 22:52
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$\begingroup$ And how do you see that this is the sphere? Can you explain the calculation that you were doing which lead to this? $\endgroup$– user6818Commented Mar 14, 2014 at 23:13
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$\begingroup$ The second sheet is connected with a cylinder to the first sheet, and a separate cylinder to the third one. Since each sheet is a Riemann sphere, it is topologically equivalent to three spheres connected with two cylinders. Therefore, the whole surface is a Riemann sphere. This came up in an effort to generalize Replica trick to a new class of entropy functions different from Renyi entropies. $\endgroup$– Nima LashkariCommented Mar 14, 2014 at 23:25
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2 Answers
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Just few remarks to Alex's answer. If all $u_1,u_2,v_1,v_2$ are distinct, the function mapping the Riemann sphere on your surface is a rational function of degree $3$. And conversly, every rational function of degree $3$, with simple critical points and distinct critical values, maps the sphere on such Rimemann surface. Up to conformal automorphisms in the domain and in the image, it depends on one parameter, for example the $4$-th cricital point, if we normalize so that three critical points are $0,1,\infty$ and three critical values are also $0,1,\infty$. Then the $4$-th critical value is a quadratic function of this $4$-th critical point, so for each critical value you have two choices for your normalized rational function. All computation can be done by hand, and was actually done in the paper L. Goldberg, Catalan numbers and branched coverings by the Riemann sphere. Adv. Math. 85 (1991), no. 2, 129–144.
answered Mar 15, 2014 at 14:05
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If I understood your question correctly, here is what Maple says. Up to Moebius transformation, you can assume that your four critical values are $0$, $1$, $\infty$, and something. Then, this map $$z\mapsto-\frac{z^2(z-a)}{2az-3z+2-a}$$ has critical points $0\mapsto0$, $1\mapsto1$, $\infty\mapsto\infty$, and $\dfrac{a(a-2)}{2a-3}\mapsto\dfrac{a^3(a-2)}{(2a-3)^3}$. Thus, given the fourth critical value, you get four maps, which should differ by the monodromy, but the rest should be an easy computation.
answered Mar 15, 2014 at 5:02