Suppose I have a Riemann surface $C$ given by an explicit equation or equations as a complex plane curve or space curve. $C$ has a(n unramified) covering space $C'$ associated to each subgroup of finite index in $\pi_1(C)$. General principles make $C'$ first a Riemann surface and then an algebraic curve. But is there an explicit method for writing down an equation for $C'$ starting from the equation(s) for $C$ and, say, generators for the subgroup?
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asked Mar 24, 2016 at 7:45
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$\begingroup$ For curves of genus 1, this is called "transformations of elliptic functions". Equations can be made explicit. $\endgroup$– Alexandre EremenkoCommented Mar 24, 2016 at 11:58
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I can propose an algorithm for doing this for ``Belyi curves'', the curves whose coefficients are algebraic numbers. According to Belyi's theorem, such a curve of genus $g$ is defined by a combinatorial triangulation of an abstract genus $g$ surface. And this triangulation can be explicitly computed (in principle). Then if you have a covering, it is defined by another triangulation which can be easily constructed from the first one. And passing from a triangulation back to the equation is a procedure which requires only solving algebraic equations.
We obtain that a finite covering of a Belyi curve is a Belyi curve again, and coefficients can be (in principle) computed. Of course this is a theoretical possibility, the explicit algorithm can be hard.
Belyi curves are dense in the space of all curves of given genus. When you slightly vary your curve, you may slightly vary the covering. This shows that in principle there are formulas which express the coefficients of the covering curve in terms of the coefficients of the original one in the general case.
answered Mar 24, 2016 at 12:07
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$\begingroup$ Thank you Alexandre! I have a question though. After a projection or two, my $C$ comes to you explicitly as a branched cover of the Riemann sphere, and thus also $C'$. What's the theoretical advantage of first approximating, then finding a model of $C$ and then $C'$ branched over $0,1 \infty$, thus throwing away the original branching data? $\endgroup$ Commented Mar 24, 2016 at 19:55
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1$\begingroup$ @David Feldman: it is a minor technical advantage. Correspondence between curves ramified over $0,1,\infty$ and triangulations is essentially $1-1$. In recovering some other curve from a triangulation some additional parameters will mess around. $\endgroup$ Commented Mar 24, 2016 at 20:03