The following interesting question came up in a discussion I was having with Alex Wright.

Suppose given a branched cover C -> P^1 with four branch points. It's not hard to see that the field of definition of C has transcendence degree at most 1 over $\mathbf{\bar{Q}}$.

Which leads one to ask:

Give an example of a field K of transcendence degree 1 over $\mathbf{\bar{Q}}$, and a geometrically connected curve C/K, such that C does not admit a branched covering to P^1 with four branch points.

Update: Ben Weiland suggests that taking C to be the universal curve over the function field of a a compact curve in $M_g$ might give such an example. Indeed, one might ask for lower bounds on the number of singular fibers (or, thinking of a curve in $\bar{M}_g$, on the intersection with various boundary components.)

Re-update: Both Ben and Jason present examples of compact families of 4-branched covers. So I think the question still stands open as written. Based on the discussion in the answers and comments, I would think that a promising candidate would be provided by a compact Shimura curve parametrizing genus-2 curves whose Jacobians have multiplication by an indefinite quaternion algebra. Can this be a component of a Hurwitz curve in M_2? (Warning: as Dan Peterson points out these are not actually compact in M_2, only in A_2.)

Some remarks:

$\bullet$ If you replace "four branch points" with "three branch points" and "transcendence degree 1" with "transcendence degree 0," the nonexistence of such an example is Belyi's theorem.

$\bullet$ There is no obstruction coming from the choice of K; a theorem of Diaz, Donagi, and Harbater guarantees that for any field K of transcendence degree 1 over $\mathbf{\bar{Q}}$, there exists a geometrically connected curve C/K which admits a 4-branched cover to P^1.

$\bullet$ There is a somewhat subtle issue which doesn't arise in the Belyi case: a 4-branched cover C/K->P^1/K yields a map from Spec K to $\mathcal{M}_{{0,4}}$

("forget the cover, remember the branch points") which, after passage to the generic point gives you a choice of inclusion of the function field $\mathbf{\bar{Q}}(\mathcal{M}_{0,4})$ into K.

One can either consider K as an abstract field, or as a fixed extension of the rational function field $\mathbf{\bar{Q}}(\mathcal{M}_{0,4})$, which places a stronger condition on C. E.G. if you ask that this inclusion be an isomorphism, you are requiring that C be a 4-branched cover which is "determined uniquely by its branch points," like y^2 = x(x-1)(x-t).

  • $\begingroup$ Nice question, Jordan. You are assuming that the base points are rational over $K$, I imagine. $\endgroup$ – Angelo Aug 2 '11 at 17:19

Edit: this doesn't work.

Original: I think that a compact curve in $M_g$ provides an example. In such a family, the length of the shortest geodesic is bounded. Whereas, in a family produced by ramifying over four points in $P^1$, as two ramification points collide, there is a loop whose image downstairs stays near that pair of points, looping about them several times, and thus has arbitrarily small length. There's probably a variant of this argument with Mumford-Tate groups.


Jason objects that branch points downstairs can collide without ramification points upstairs colliding. This is true, but easy to patch. There are four branch points and some pairs of them might collide without the ramification points colliding, but if all pairs collide without the ramification points colliding, then they must permute different sets of sheets and the curve is not (geometrically) connected.

Jordan objects that even if the ramification points collide, the curve may remain smooth. In particular, if the curve is a $d$-fold cover of $P^1$ and all the monodromy is a power of a fixed $d$-cycle, if two points labeled by $a$ and $b$ so that all three of $a$, $b$, and $a+b$ are relatively prime to $d$, then the family is smooth. If this is true for all pairs of collisions, this gives a complete family of curves that map to $P^1$ with only four branch points, contradicting my claim. In particular, $d=5$ and $1,1,1,2$ is a complete family of genus $4$ curves.

Such examples don't exist for smaller $d$. Thus a complete curve in $M_3$ (which exist, right?) is a good candidate for not having a map to $P^1$ with $4$ branch points. Collisions of branch points with other forms of monodromy are harder for me to understand, but they look more likely to result in degeneration.

  • $\begingroup$ Excellent point! I am modifying the question to take this into account. $\endgroup$ – JSE Aug 2 '11 at 19:00
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    $\begingroup$ What if the "ramification points" don't collide, just the "branch points", e.g., something like $(y^2-x)(y^2-x-t)$? More generally, what does the number 4 have to do with this answer? It seems to me that the same argument would imply that the generic member cannot map to $\mathbb{P}^1$ with any number of branch points, which is absurd. $\endgroup$ – Jason Starr Aug 2 '11 at 19:59
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    $\begingroup$ Dear Ben, If we define ramification points to be points in the cover where the morphism to $\mathbb{P}^1$ is non-\'etale, and if we define branch points to be the image points in $\mathbb{P}^1$, then I completely agree that for any complete family in $\overline{M}_{0,n}$, the branch points must collide. However, this does <I>not</I> imply that ramification points collide. My example is a one-parameter family of branched covers of the $x$-line, the parameter is $t$, where branch points collide but ramification points do not. continued $\endgroup$ – Jason Starr Aug 3 '11 at 1:38
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    $\begingroup$ OK, I'll repeat myself in your language. If we fix ramification data, so that we have a family over $M_{0,4}$ and the extension to $\overline{M}_{0,4}$ is smooth, so that no pair of branch point collisions involves ramification collisions, then it is not (geometrically) connected. If branch points collide without ramification points colliding, then they are permuting different sets of sheets. If the curve is (geometrically) connected, some pair of branch points must be permuting the same sheets. $\endgroup$ – Ben Wieland Aug 3 '11 at 3:19
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    $\begingroup$ Huh, I am now confused too as to whether I believe Ben's original statement. I am going to bed but will comment again tomorrow morning. $\endgroup$ – JSE Aug 3 '11 at 5:10

Dear Jordan,

If the curve admits a 4-to-1 map to $\mathbb{P}^1$, then it has a rational point after a solvable extension. For $g$ large, there are many known genus $g$ curves that have no solvable points.

MR2057289 (2005f:14044) Pál, Ambrus(3-MTRL-R) Solvable points on projective algebraic curves. (English summary) Canad. J. Math. 56 (2004), no. 3, 612–637. 14G05 (11G20 14H25)

Edited: I misread the question! Of course Jordan is not asking about 4-to-1 maps, he is asking about covers branched over 4 points. Sorry for the confusion.

  • $\begingroup$ But by the way, I like the question about solvable points, too, and I hasten to point out that over a global field (Q or F_p(t)) we have NO IDEA whether there are genus-g curves without solvable points. Even the genus 1 case is very hard! (It turns out there are always solvable points in that case -- Mirela Ciperiani proved this in her Ph.D. thesis.) Pal, if I remember right, always works over a "big" base field like Q(T), which gives you much more Galois-theoretic wiggle room. $\endgroup$ – JSE Aug 2 '11 at 17:34
  • $\begingroup$ Yes, Pal works over $\mathbb{Q}(t)$, not over $\overline{\mathbb{Q}}(t)$. There are also examples over $\overline{\mathbb{Q}}(s,t)$. I think I can also see how to produce curves over $\overline{\mathbb{Q}}(s,t)$ which admit no morphism to $\mathbb{P}^1$ branched over four (rational) points. But of course that is also not the question you asked! $\endgroup$ – Jason Starr Aug 2 '11 at 19:05
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    $\begingroup$ Çiperiani, Mirela and Wiles, Andrew, Solvable points on genus one curves. Duke Math. J. 142 (2008), no. 3, 381–464. $$ $$ This important article shows that any genus one curve $C$ over ${\bf Q}$ with a rational point over every $p$-adic completion of ${\bf Q}$ and semistable Jacobian has a point defined over a solvable extension of ${\bf Q}$. $$ $$ Reviewed by Henri Darmon $\endgroup$ – Chandan Singh Dalawat Aug 3 '11 at 2:51

Breaking protocol and answering my own question: this is based on Ben Wieland's answer and also owes much to what Dawei Chen explained to me over e-mail.

The question: is it possible for a family of 4-branched covers of P^1 to have no singular fibers? If not, then, as Ben pointed out, any such family gives an example of the sort I requested.

First of all, let's observe that indeed the branch points can come together while the cover stays smooth, and be a little more precise about when this happens. You can describe your genus g degree-d cover by giving four elements (g_1, g_2, g_3, g_4) of S_d which generate a transitive subgroup and which sastisfy g_1 g_2 g_3 g_4 = 1.

What happens when two branch points come together? I'll think of this as pinching off a loop which separates punctures 1 and 2 from punctures 3 and 4. Then the base becomes a union of two P^1s, meeting at a node. Our branched cover now specializes to the union of a 3-branched cover of each P^1, the monodromy being

g_1, g_2, (g_1 g_2)^{-1}

(g_3 g_4)^{-1}, g_3, g_4


When is the cover smooth? When one of these 3-branched covers has genus g, which forces the other one to be a union of rational tails. For each g_i write ind(g_i) for d minus the number of orbits of g_i (so that e.g. ind(g_i) = 1 precisely when g_i is a transposition.) Then the condition for smoothness is precisely that

(S) "ind(g_1 g_2) = ind(g_3 g_4) must be equal either to ind(g_1) + ind(g_2) or to ind(g_3) + ind(g_4)."

Now this can clearly happen, as in the examples already discussed. But can it happen for every degeneration, as is necessary for the family to miss the boundary entirely? This is equivalent to asking that (S) is true after you hit (g_1, g_2, g_3, g_4) with an arbitrary element of the braid group. That seems unlikely.

One thing you could do is let c_i be the conjugacy class of g_i and consider the Hurwitz space parametrizing four-branched covers of P^1 with monodromy type c_1, c_2, c_3, c_4. To show this space meets the boundary, you only need to show that there exists a choice of g_1, g_2, g_3, g_4 in those four conjugacy classes (or any permutation) which makes (S) false. I would guess you could prove by character theory or something that such a choice always exists?

But the family we're contemplating is a connected component of the Hurwitz space. And what the connected components look like (equivalently -- what the orbits of the braid group on the set of 4-tuples drawn from (c_1,c_2,c_3,c_40 look like) is actually incredibly hard to get at. From this point of view it seems easy to show that the Hurwitz space hits the boundary, but hard to show that it doesn't have some crazy component which misses the boundary.

  • $\begingroup$ So are you saying that you believe Ben's argument above is rigorous? $\endgroup$ – Jason Starr Aug 4 '11 at 21:49
  • $\begingroup$ No, I'm saying that it's very believable that compact curves in M_g provide examples, but it might be hard to prove. $\endgroup$ – JSE Aug 4 '11 at 22:33
  • $\begingroup$ I apologize, but I don't see anything in the argument above that could not be generalized to covers of $\mathbb{P}^1$ branched over $n$ points, where $n$ is any integer. We know that there are complete curves in $M_g$ parameterizing covers of $\mathbb{P}^1$ branched over $n$ points. Why is $n=4$ special? By the way, although I completely agree that connected components of Hurwitz spaces are complicated, there is quite a bit known about them, for instance, cf. Kluitman's article in the Braids volume (reporting on Deligne's results), Richard Hamilton's thesis, work of Gabai and Kazez, ... $\endgroup$ – Jason Starr Aug 5 '11 at 12:26
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    $\begingroup$ Start with an elliptic curve which is a Belyi-branched cover of $\mathbb{P}^1$, say $f:E\to \mathbb{P}^1$. Let $g:E'\to E$ be some isogeny of elliptic curves of even degree $d$. For every point $x$ in $E$, let $D_x$ be the preimage divisor in $E'$. Let $u:B \to E$ be the parameter space for a point $x=u(b)$ together with an invertible sheaf $L_b$ on $E'$ such that $L_b^2 = O_{E'}(D_x)$. Associated to $D_x$ and $L_b$, there is a degree 2 branched cover $h_b:C_b \to E'$ branched over only $D_x$. So the composition $f\circ g \circ h_b:C_b\to \mathbb{P}^1$ is branched over only $4$ points. $\endgroup$ – Jason Starr Aug 5 '11 at 13:02
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    $\begingroup$ You are correct that the abstract curve $C_b$ is not changing! But we can run the same argument with higher genus curves. Just let $f:E\to \mathbb{P}^1$ be a genus $g>1$ Belyi-curve. Let $g:E'\to E$ be an 'etale map of even degree. Let $B$ parameterize divisors $D_x$ together with a square root of $\OO_{E'}(D_x)$, etc. This will produce a complete family where the moduli of the curves $C_b$ actually does change. $\endgroup$ – Jason Starr Aug 5 '11 at 15:49

Without pretending to understand Ben Wieland's argument, let me take the Shimura curve suggestion and run with it. In particular, modulo a few details I think the following should be true:

The Shimura curve $X^{218}$ defines a complete curve in $M_2$

My first reaction upon seeing Jordan's Shimura curve suggestion was "No way". The reason is that a priori Shimura curves $X^D$ don't parametrize curves of genus 2, but abelian surfaces which admit the action of a maximal order in the unique $\mathbf{Q}$-quaternion algebra of discriminant $D$ (and the fixed choice of a polarization compatible with the action of the choice of maximal order). Indeed, infinitely many of these abelian surfaces are products of elliptic curves with complex multiplication, and these of course do not need to necessarily be Jacobians.

Thankfully we have at our disposal the work of Hayashida and Nishi who asked the question: For what pairs of isogenous elliptic curves $E,E'$ is there a genus 2 curve inside $E\times E'$?

In particular they show the following:

If $E$ does not have CM, $E\times E$ is not a Jacobian

If $E,E'$ have CM by a maximal order in $\mathbf{Q}(\sqrt{-m})$ then $E\times E'$ contains a genus 2 curve if and only if $m \ne 1,3,7,15$

by reducing the above question to one about a certain real valued 4-variable quadratic form over the integers. A priori this doesn't answer the question, because we have to concern ourselves with nonmaximal orders in $\mathbf{Q}(\sqrt{-m})$. However, it seems (at least in the cursory read of their paper I've been able to give this afternoon) that their methods can be generalized to nonmaximal orders with a little bit of care. If the apparent qualms people have about Ben Wieland's argument can be resolved, I'll do exactly that in a few days.

But assuming that all works out, life is great! In particular we can come up with a Shimura curve $X^D$ which avoids those particular products of CM elliptic curves if and only if there is a prime $p|D$ such that $\left(\dfrac{-m}{p}\right) = 1$ for $m= 1,3,7,15$. We can compute that the minimal such prime is $109$, which is $1 \bmod 12$, $4 \bmod 5$ and $4\bmod 7$.

We recall momentarily that $X^D$ is a complete curve if and only if the unique rational quaternion algebra of discriminant $D$ is a division algebra if and only if $D$ is the squarefree product of an even number of primes. For convenience we take $D = 2p$.

Therefore (again assuming the bit about nonmaximal orders) we have found $X^{218}$ to be a curve lying in the image of the Torelli map in $\mathcal{A}_2$, and thus in $M_2$.

Comments are very welcome!

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    $\begingroup$ Something is wrong here. M_2 is affine, so it can not contain a complete curve. $\endgroup$ – Dan Petersen Aug 5 '11 at 22:19
  • $\begingroup$ My comment is that I forgot to think about the decomposable locus, so thanks for working this out! Re the qualms with Ben's argument, surely it can't be fully right because both Ben and Jason gave examples of compact families of 4-branched covers. $\endgroup$ – JSE Aug 5 '11 at 22:29
  • $\begingroup$ Oh, and in this respect (my forgetting to think about the decomposable locus) Dan's point is also germane. The question of whether universal genus-2 curves over Shimura curves are 4-branched covers remains active, though! $\endgroup$ – JSE Aug 5 '11 at 22:31

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