3
$\begingroup$

Let $X\rightarrow B$ be a family of Kaehler manifolds with possibly singular fibers. Let $G$ be the monodromy group on $H^n(X_b,\mathbb{Z})$, where $n=\dim X_b$ with the smooth fiber $X_b$ over some $b \in B$. We say that an element $g \in G$ is neat if none of the eigenvalue of $g$ is a non-trivial root of unity. We say that $G$ is neat if all elements in $G$ are neat.

Here is my question: Is it true that if each local monodromy is neat then $G$ is neat? In other words, is it true that $G$ is neat if it is generated by neat elements?

I think my question is false if one removes the geometric picture that the group $G$ is the monodromy group of a family of Kahelr manifolds. We may use the fact that $G \subset Sp(k,\mathbb{Z})$ if $n$ is odd, for example.

Edit As Jason pointed out below, I need to assume that the base space $B$ is simply connected and the global monodromy group is generated by the local monodromies.

$\endgroup$
4
  • $\begingroup$ You ask two different questions: (1) "Is it true that if each local monodromy is neat then $G$ is neat?", (2) "... is it true that $G$ is neat if it is generated by neat elements?" Typically these are different questions: the global monodromy group is not necessarily generated by local monodromies. There are geometric hypotheses that insure that the global monodromy group is generated by local monodromies, e.g., for the family of hyperplane sections of a fixed projective manifold obtained from a Lefschetz pencil of hypereplane sections. Did you want to impose such a hypothesis? $\endgroup$ Commented Dec 26, 2013 at 15:20
  • $\begingroup$ You are right. I am interested in a more specific example, where $B$ is simply connected and the full monodromy group is generated by the local monodromies. $\endgroup$
    – Ariel
    Commented Dec 26, 2013 at 21:20
  • $\begingroup$ If the base is simply connected, is not the monodromy group also trivial? $\endgroup$ Commented Dec 27, 2013 at 15:15
  • $\begingroup$ Not necessarily. The mirror quintic gives an example of a family over $\mathbb{P}^1$ whose monodromy group is non-trivial. $\endgroup$ Commented Feb 25, 2014 at 8:12

3 Answers 3

3
$\begingroup$

No. Consider the family of elliptic curves over $\mathbb P^1$, $y^2=x(x-1)(x+1-t^2)$. This is a pullback of the Legendre family $y^2=x(x-1)(x-\lambda)$ by the cover $\mathbb P^1 \to \mathbb P^1$ with equation $\lambda=t^2-1$.

The Legendre family has local monodromy a $2 \times 2$ unipotent Jordan block at $0$ and $1$, and minus that block at $\infty$, so its monodromy is not neat. Since this cover is ramified of degree $2$ at $\infty$, the monodromy becomes unipotent again, hence the local monodromy of the cover is everywhere neat.

But I claim the global monodromy of the cover is the same as the Legendre family, and hence is not neat. To see this, we will use $t=0$ or $\lambda=-1$ as our base points. Choose any loop in the fundamental group of the base of the Legendre family, and lift it to the cover. You face no problem except that the loop might go from one sheet to the next. But because the two sheets meet at the base point $t=0$, one can always return to the base point and close the loop. So the map on fundamental groups is surjective, hence the global monodromy is the same, and isn't neat.

$\endgroup$
1
$\begingroup$

I believe this is trivially false: there are families over a compact Kähler base having no singular fibers yet with nontrivial, finite monodromy group. The simplest example I know of is an isotrivial family where $B$ is an elliptic curve and the fibers $X_b$ are hyperelliptic curves.

Edit. Although the OP does not specify this in the statement, I suspect the OP is interested in cases where the local monodromies generate the global monodromy group. Obviously if the global monodromy group is finite, then every local monodromy is a finite order element in $\textbf{GL}_n(\mathbb{Z})$. The only such element that is "neat" is the identity element. If every local monodromy is the identity, and if the local monodromies generate the full monodromy group, then also the global monodromy group is trivial.

$\endgroup$
3
  • $\begingroup$ Thank you for the answer pointing out the mistake in my question. I do not understand why you assume that the global monodromy group is finite. $\endgroup$
    – Ariel
    Commented Dec 26, 2013 at 21:26
  • $\begingroup$ @Ariel: If the global monodromy group is finite and nontrivial, then there are non-neat elements in the global monodromy group. $\endgroup$ Commented Dec 27, 2013 at 2:44
  • $\begingroup$ OK. You are talking about a special case. I don't think my question is trivial if the monodromy group is not finite (this is the general case) and generated by the local monodromies. $\endgroup$
    – Ariel
    Commented Dec 27, 2013 at 5:22
1
$\begingroup$

Even the modified version (with the base simply connected) is false. It is possible to construct an elliptic surface over $CP^1$ with any prescribed local and global monodromy (see e.g. http://www.math.colostate.edu/~miranda/BTES-Miranda.pdf, Proposition VI.3.4). Choose local monodromy unipotent (e.g. of type $I_1$ in Kodaira classification), in such a way that these unipotent elements generate the whole $SL(2,Z)$. The local monodromy is torsion-free, but there are torsion elements in the global monodromy group.

$\endgroup$
1
  • $\begingroup$ I don't think this works. $\endgroup$ Commented Apr 15, 2014 at 19:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.