I know (edit: three) families of smooth projective connected curves over $\bar{\mathbf{Q}}$ for which the Belyi degree is not hard to bound from above.

  1. The modular curves $X(n)$. They are constructed by compactifying the quotient $Y(n) = \Gamma(n)\backslash \mathbf{H}$. The natural morphism $X(n) \longrightarrow X(1)$ is Belyi of degree $n^2$ (up to a constant factor). This also bounds the Belyi degree of a modular curve given by a congruence subgroup $\Gamma$. In general, Zograf proves that the Belyi degree of a (classical congruence) modular curve is bounded by $128(g+1)$.

  2. The Fermat curves $F(n)$. They are given by the equation $x^n+y^n+z^n =0$ in $\mathbf{P}^2$. The morphism $(x:y:z)\mapsto (x^n:z^n)$ is Belyi of degree $n^2$. It is known that $F(n)$ is not a modular curve for $n$ big enough. So this example is really different than the one above. (Also note that $n^2\leq 10g+10$ by the Plucker formula.)

  3. Wolfart curves are curves $X$ over $\overline{\mathbf{Q}}$ with a Galois Belyi morphism $X\to \mathbf{P}^1$; I took this terminology from a preprint by Pete L. Clark. Such curves are also called Galois Belyi covers or Galois three-point covers in the literature. The Belyi degree of a Wolfart curve is bounded by $84(g-1)$. (In particular, the latter implies that there are only finitely many Wolfart curves of given genus.)

The following family of curves is not so easily dealt with.

  1. For an elliptic curve $E$ over the rational numbers, the Belyi degree can be bounded in the height of the $j$-invariant of $E$ following Belyi's proof of his theorem. This was written down explicitly by Khadjavi and Scharaschkin.

I'm looking for families of curves for which the Belyi degree is ``easy to read off''. That is, a collection (finite or infinite) of smooth projective connected curves $X_i$ over $\bar{\mathbf{Q}}$ for which the Belyi degree can be bounded easily.

Are there any other nice examples?

  • $\begingroup$ Random thought: you could try non-congruence modular curves. $\endgroup$
    – S. Carnahan
    Jun 20, 2011 at 3:37
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    $\begingroup$ Yes. The Fermat curve F(n) is an example of that. But isn't any smooth projective geometrically connected curves X over Q a non-congruence modular curve? (Choose a Belyi morphism X-->P^1 and identify P^1-{0,1,infty} with Y(2). A topological cover of Y(2) is the quotient of the complex upper half plan by a finite index subgroup of Gamma(2)/{1,-1}. ) So this would be a very nice example to ``try'' indeed. $\endgroup$ Jun 20, 2011 at 7:09
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    $\begingroup$ Also modular curves (classical or Shimura) for subgroups of other arithmetic triangle groups; the list, obtained by Takeuchi, is finite but somewhat large (almost 100 cases): Commensurability classes of arithmetic triangle groups, J. Fac. Sci. Univ. Tokyo 24 (1977), 201-212. $\endgroup$ Jun 20, 2011 at 16:52
  • 2
    $\begingroup$ For elliptic curves over Q Khadjavi and Scharaschkin show (Thm. 1a of myweb.lmu.edu/lkhadjavi/belyielliptic.pdf) that the curve $y^2 = x^3 + Ax + B$ has Belyi degree $O(|A|^3 + |B|^2)$. Similarly for a curve with full level-2 torsion and invariant $\lambda$, i.e. $cy^2 = x (x-1) (x-\lambda)$ for some rational $c,\lambda$ with $c \neq 0$ and $\lambda \neq 0, 1$, Belyi's construction gives an explicit Belyi map of degree $O(H(\lambda))$. Here $H$ is the height, $H(m/n) = \max(|m|,|n|)$. This is sharp when $H(\lambda)$ is prime, by a theorem of Beckmann (J.Alg. 125(1989), 236-255). $\endgroup$ Jun 20, 2011 at 23:58

2 Answers 2


Another example, like JSE's, that comes already equipped with a Belyi map but is not as familiar as modular curves and Fermat curves: For any relatively prime integers $m,n$ with $0<m<n$, and any subgroup $G$ of $S_n$, the curve that parametrizes trinomials $x^n + a x^m + b$ up to scaling with Galois group contained in $G$. The Belyi map is the invariant $a^n/b^{n-m}$ of the trinomial, and its degree is $d=[S_n:G]$; it is branched at $0$, $\infty$, and $(-n)^n/(m^m (n-m)^{n-m})$. One may assume $m \leq n/2$ (by symmetry with respect to $x \leftrightarrow 1/x$, $m \leftrightarrow n-m$). Some nontrivial examples with $n=5,7,8$ are given explicitly at http://www.math.harvard.edu/~elkies/trinomial.html; the subsequent paper with N.Bruin on the cases $(m,n) = (1,7)$ and $(1,8)$ with $d = 30$ is

Nils Bruin and Noam D. Elkies, Trinomials $ax^7+bx+c$ and $ax^8+bx+c$ with Galois Groups of Order 168 and $8 \cdot 168$, Lecture Notes in Computer Science 2369 (proceedings of ANTS-5, 2002; C.Fieker and D.R.Kohel, eds.), 172-188.

(These examples all have $G$ transitive, but the construction works for all subgroups $G$.)


I don't think this question is going to have a GREAT answer -- your examples 1 and 2 are HANDED to you as Belyi covers of the line, and I'd think any family that doesn't immediately present itself in this way is unlikely to offer an easy upper bound on Belyi degree.

But that's not an answer, so here's one more -- any Hurwitz curve parametrizing covers of P^1 branched at four points will have a Belyi map (namely, the map to M_{0,4}) whose degree you can read off quite directly.

  • 1
    $\begingroup$ The work of Couveignes and Granboulan, Dessins from a geometric point of view gives some examples, and notes from character theory there is a bound (bottom page 33). math.univ-toulouse.fr/~couveig/publi/CGdes94.pdf They get "families" from dessins that are growing trees (see the last section), and try to compute with them using Puiseux series, though in genus 0 they use other methods. $\endgroup$
    – Junkie
    Jun 20, 2011 at 7:14
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    $\begingroup$ And I should add that, by a handsome theorem of Diaz, Donagi, and Harbater, every curve over Qbar is a Hurwitz curve in this sense (The paper is the one titled "Every curve is a Hurwitz curve.") So what you say about non-congruence modular curves applies to this case too. $\endgroup$
    – JSE
    Jun 20, 2011 at 15:49

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