Suppose I have a projective plane curve $C = V(F)$ defined over $\mathbb{C}$, where $F$ is some homogeneous polynomial in three variables. For the sake of simplicity, let's assume that $C$ is irreducible, and that $C$ contains a rational point $P$.

If $C$ is smooth of degree 2 (so genus $0$), then we can obtain a universal conformal covering map $\mathbb P^1(\mathbb C) \rightarrow C$ by identifying each point in $\mathbb P^1$ with a line $\ell$ through $P$ in the usual way, and outputting the other intersection point of $\ell$ with $C$.

If $C$ is smooth of degree 3 (so genus $1$), then we can obtain a universal conformal covering map $\mathbb C \rightarrow C$ by applying a change of variables to convert $C$ to Weierstrass form $$ F(x,y,z) = y^2 z - x^3 + g_2 xz^2 - g_3z^3, $$ and then using the Weierstrass $\wp$-function for the associated lattice: $$ z \longmapsto [\wp(z) : \wp'(z) : 1], $$ where we send the poles of $\wp$ to the limiting points at infinity.

If $C$ is singular of geometric genus $\le 1$, then we can resolve the singularities via a normalization in order to obtain a smooth curve, followed by parameterizing as above.

Consider now the case where $C$ has genus $g\ge 2$. We then obtain a Riemann surface of genus $g$ punctured at the singular points. By the uniformization theorem, there is a conformal map $\varphi_C:\mathbb H\rightarrow C$, where $\mathbb H$ is the upper half-plane, corresponding to the universal covering of $C$. However, I am not aware of any explicit formula for this map, even for a specific instance of $C$.

Is there any known way to explicitly parameterize these higher genus curves by a conformal map?

  • $\begingroup$ naive comment: These short sources may interest you. In the opposite direction, Serre, in A course in arithmetic, p.83, gives Eisenstein series for the coefficients of a plane elliptic curve in terms of its lattice, which curve is uniformized by the corresponding P-function. Partially analogous, Shafarevich, in Basic algebraic geometry, 1974 ed., p.393 ff., gives Poincare series for canonical embedding of a curve in terms of the discrete subgroup of Aut(D) uniformizing it. But how to make that group explicit, describe the embedded curve explicitly, or go from curve to uniformization? $\endgroup$
    – roy smith
    Commented Apr 26, 2022 at 17:45

1 Answer 1


The answer is no, except some very special cases. There is a comprehensive book dedicated to this: H. P. de Sain-Gervais, Uniformisation des surfaces de Riemann, ENS Ed., 2010. There is an English translation.

Of course, to make it precise, one has define "explicit". Why uniformization of genus 1 curves is "explicit" from your point of view? Elliptic functions are transcendental (defined by rather complicated series). What makes them "explicit" is probably the fact that we know a lot about them. But uniformisation of higher genus curves is more complicated by orders of magnitude.


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