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The question is inspired by Carnahan's comment in the following MO question

What is the intuition for $\mathbb{Q}^{ab}$ having cohomological dimension $1$?

Precisely, Borel proved (not Grothendieck, I think, who proved the arithmetic version) that, when S is a punctured disk, and H is a variation of polarized Hodge structures on S, then the monodromy operator is quasi-unipotent. Are there any counter-examples when S is a complex annulus? If yes, any examples of geometric origin? Namely f: X ---> S is a holomorphic family of projective manifolds ... .

To get started, what is the universal covering (as a complex manifold) of an annulus a < |z| < b ?

Thank you.

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By complex manifold, do you mean Riemann surface? In which case… – Yemon Choi Jun 30 '13 at 23:23
I get confused: the universal covering of the punctured disk, which is a special case of an annulus, is the hyperbolic disk rather than the complex plane, right? So, the link you showed me doesn't seem quite right... – shenghao Jun 30 '13 at 23:34
Yes: map the disk to the left half-plane and then exponentiate – Yemon Choi Jun 30 '13 at 23:36
Shenghao, the universal cover of the annulus is the still the upper half plane. As far as I can tell, the argument [Schmid, Variations of Hodge ..., p 230] still applies, but I'm too lazy to check carefully. (As an aside, geometric examples would extend all the way to the punctured disk.) – Donu Arapura Jun 30 '13 at 23:49
Dear Shenghao, You can move any three points in $\mathbb P^1(\mathbb C)$ to any other three points. So change coords. on $\mathbb P^1(\mathbb C)$ so that $0,$ $1$, are moved into some small disk, and $\infty$ is moved outside a larger disk containing it, and consider the annulus between these two disks. Consider the family $y^2 = x(x-1)(x- \lambda)$, which has singular points at $0$, $1$, and $\infty$, but now in the new coords., so that its singular points corresponding to $0$ and $1$ are inside the small disk. The monodromy matrices around $0$ and $1$ are ... – Emerton Jul 2 '13 at 6:36

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