Suppose I have a finite set of points in $\mathbb{P}^1$ (over the complex numbers), and suppose that at each point, I am given a [Edit: quasi-unipotent] conjugacy class in $Sp(2g,\mathbb{Z})$ for $g$ a fixed positive integer. Then near each point, I have an analytic neighborhood where I can construct a family of complex tori whose $H^1$ varies according to the monodromy. Furthermore, if there are representatives of the conjugacy classes whose product is $1 \in Sp(2g, \mathbb{Z})$, then I can at least glue these local families into a $C^\infty$ family of $2g$-dimensional tori over the punctured $\mathbb{P}^1$.

First question: If we have representatives whose product is identity, does there exist a family of abelian varieties over the punctured line whose $H^1$ has the specified monodromy at the points? I think we can do this by choosing a variation of Hodge structure on the corresponding local system of rank $2g$ groups, and taking a quotient, but it's all a bit cloudy to me.

Second question: Given a particular puncture and its assigned conjugacy class in $Sp(2g,\mathbb{Z})$, does there exist a family of genus $g$ curves over a small neighborhood of the puncture whose $H^1$ has monodromy in the specified conjugacy class? (I suppose I should just ask this about the punctured affine line.) In the $C^\infty$ world, this can be done with a mapping torus, but I don't know how it works holomorphically.

Third question: If the answers to the first two questions are "yes", can we make our family of abelian varieties out of Jacobians of genus $g$ curves? Up to some finite cover problem, I think this is basically asking if there exists a family of genus $g$ curves whose $H^1$ has the specified monodromy.

From David Brown's answer here, it seems too much to ask for an explicit minimal model curve over the punctures when $g > 2$, but I'm just hoping for existence away from those points.

  • 4
    $\begingroup$ Maybe I'm missing something, but don't you need each conjugacy class in $Sp_{2g}(Z)$ to be quasi-unipotent (Grothendieck, SGA 7)? So e.g. the matrix $(5,3;3,2)$ for $g=1$ cannot be realized locally. $\endgroup$
    – inkspot
    Feb 28, 2011 at 9:25
  • $\begingroup$ From a moduli-theoretic point of view, you are asking for a map from $\mathfrak P^1$ to the moduli of abelian varieties which meets the boundary at your specified points; I'm not sure exactly how much, and in exactly what way, specifying the monodromy further restricts the intersection with the boundary. I wonder though if this could help? $\endgroup$
    – Emerton
    Feb 28, 2011 at 18:03
  • $\begingroup$ Perhaps a more flexible question to ask is when such a local system is motivic. I wonder if the geometric Langlands correspondence can be interpreted as the assertion that it always is? $\endgroup$ Feb 28, 2011 at 22:04
  • $\begingroup$ I guess the previous comment is silly. Even over function fields, I presume being automorphic is weaker than being motivic. $\endgroup$ Feb 28, 2011 at 22:18
  • 3
    $\begingroup$ On the other hand, to continue my silly thread of comments, a naive function field analogue of the Fontaine-Mazur conjecture would say that quasi-unipotence of the local monodromy is sufficient for the representation to be motivic. Is it feasible to extract this out of geometric Langlands? $\endgroup$ Feb 28, 2011 at 23:34

4 Answers 4


There are several restrictions. First, the existence of potential semistable reduction (Grothendieck-Mumford) implies that your representative $\sigma$ at every puncture must be quasi-unipotent of level 2, i.e, there exists a positive integer $N$ such that $(\sigma^N-1)^2=0$. Second, the Zariski closure $G\subset Sp_{2g,Q}$ of the global monodromy group $\Gamma \subset Sp(2g,Z)$ must satisfy the following properties (Deligne).

  1. Its identity component $G^0$ is semisimple.

  2. All absolutely simple quotients of $G^0$ (over the field $C$ of complex numbers) are classical algebraic groups (A_r,B_r,C_r,D_r) and ``their" natural nontrivial irreducible subrepresentations in $C^{2g}$ are (fundamental) minuscule, i.e., the corresponding set of weights is an orbit of the Weyl group.

  • $\begingroup$ Thank you. This is very informative. It sounds like without an explicit construction we don't have much in the way of existence results for general $g$. Is that a reasonable summary? $\endgroup$
    – S. Carnahan
    Feb 28, 2011 at 15:13
  • $\begingroup$ You are welcome. Well, what I know about that is mostly related to the $Q$-algebraic group $G$ and its natural representation in $Q^{2g}$. Michael Fried and/or Helmut Voelklein may know more about the existence results for the construction you are looking for. $\endgroup$ Feb 28, 2011 at 17:23

This is sort of an antianswer (sorry). I think that most representations $$\rho:\pi_1(\mathbb{P}^1-\{p_0,\ldots p_n\})\to Sp(2g,\mathbb{Z})$$ will not occur as monodromy representations of families of abelian varieties. There are local constraints as inkspot points out: the local monodromies $\rho(\gamma_i)$ should be quasiunipotent and for $g=1$, and perhaps $2$, much more is known (Kodaira, Namikawa). There is also a global constraint, since by Deligne $\rho\otimes \mathbb{Q}$ must be semisimple. But I suspect that this is the tip of the iceberg. Determining exactly which representations occur is probably very hard, but interesting.

  • $\begingroup$ Antianswers are good, too. Thank you. $\endgroup$
    – S. Carnahan
    Feb 28, 2011 at 15:14

A special case of this situation: suppose one gives a sequence of simple closed curves $(\gamma_1,\dots, \gamma_N)$ on an oriented $C^\infty$ surface, such that the product of Dehn twists $\tau_{\gamma_N}\circ \dots \circ \tau_{\gamma_1}$ is isotopic to the identity. One can then build a closed 4-manifold $X$ and a topological Lefschetz fibration $f\colon X\to \mathbb{CP}^1$ whose monodromy, for a chosen basis of vanishing paths, is given by the specified Dehn twists. Moreover, $X$ carries essentially canonical symplectic forms (Gompf).

A variant of question 3 (which shortcuts issues concerning the Torelli group) asks: when one can take $X$ to be a Kaehler surface and $f$ to be an algebraic map?

Donaldson proved that every symplectic 4-manifold with rational symplectic class admits (after blowing up enough points) a topological Lefschetz fibration. From this point of view, the question is about the disparity between symplectic and Kaehler structures on 4-manifolds - and this disparity is very wide.

But here's a positive result about the genus 2 case. The genus 2 mapping class group maps onto the braid group on 6 strings, and if the $\gamma_i$ are non-separating and the twists $\tau_{\gamma_i}$ act transitively on the strings, then Siebert and Tian show that $(X,f)$ can be made algebraic.

  • $\begingroup$ This is a direction that had not crossed my mind. Very interesting! $\endgroup$
    – S. Carnahan
    Mar 1, 2011 at 2:15

I agree with Donu. Indeed, I think even the much weaker question of whether a mod-p representation of the fundamental group of the base on Sp(2g,Z/pZ) occurs as a monodromy representation might typically have a negative answer. Given such a representation rho, you get a fibration X_rho -> P^1, whose fibers are isomorphic to the moduli space of abelian g-folds with full p-level structure; this will be general type for p large. Any abelian g-fold A/C(t) with monodromy rho corresponds to a section from P^1 back to X_rho, and I don't see why there would be such a section in general.

Oh yeah, and; the answer to your second question is yes, I think. When the monodromy is of the form


0 I

with M of full rank; you can construct an abelian variety over C((t)) with totally multiplicative reduction which has any desired monodromy, as in Mumford's paper "Degenerating abelian varieties...."

If M has smaller rank maybe you can just use a product of a constant a.v. with a totally multiplicative one of dimension rank(M)?

  • $\begingroup$ Sorry for not replying. I'm afraid I am still trying to digest what the level structure argument says in the case of a genus 1 family. At any rate, for the local construction you just added, can you choose those abelian varieties to be Jacobians? $\endgroup$
    – S. Carnahan
    Mar 1, 2011 at 7:50
  • $\begingroup$ Not sure whether you can achieve any possible local monodromy with Jacobians. $\endgroup$
    – JSE
    Mar 2, 2011 at 2:33

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