Orbifold Beauville-Siu

The Beauville-Siu theorem states that for a compact Kahler manifold the following two statements are equivalent:

1. $M$ admits a surjective holomorphic map with connected fibers to a compact Riemann surface of genus at least $2.$

2. $M$ admits a surjective homomorphism to the fundamental group of a compact Riemann surface of genus at least $2.$

If one thinks of fibrations with exceptional fibers ("Seifert fibrations") it is natural to ask whether the theorem holds where "compact Riemann surface" is replaced with "compact hyperbolic orbifold", and "fundamental group" is replaced by "orbifold fundamental group".

I assume that should not be too difficult, but have trouble finding references.

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Yes, this sounds right. Beauville in his original paper "Annulation pour $H^1$...", actually does work with orbifolds, but only implicitly. So you extract it out from there. An alternative would be to deduce from Corlette-Simpson, [Compositio 2008]. They do use orifolds explicitly. –  Donu Arapura Apr 22 '11 at 17:09
Thanks! I will check out the two references... –  Igor Rivin Apr 22 '11 at 17:50

1 Answer

Actually, I'm not sure the references I gave above are going to help that much. So let me just do it here, since it's useful to have a record somewhere. The key is to have an equivariant form the the standard Castelnuovo-de Franchis (CdF). This follows by looking at the usual proof.

Theorem. Let $M$ be a compact Kahler manifold on which a finite group $G$ acts holomorphically. Suppose that $\omega_1,\ldots \omega_n$ is a collection of linearly independent holomorphic $1$-forms such that $n>1$, $\omega_i\wedge \omega_j=0$, their span $V$ is $G$-stable. Then there exists a $G$-equivariant holomorphic map, with connected fibres, $f:M\to C$ to a curve such that $\omega_i$ pulls back from $C$.

Proof. The statement without $G$ can be found in many places, such as Catanese, Inventiones 104 (1991). Writing $\omega_i = g_i(z)dz$ (locally), we get a map $f':M\dashrightarrow \mathbb{P}^{n-1}$ given by $f'(z) = [g_1(z),\ldots g_n(z)]$. The usual arguments show that this is defined everywhere, the image is a curve $C'$, and that forms pullback from it. Stein factor to get $f:M\to C$. It is clear that $G$ acts on $C$ in such a way that $f$ is equivariant.

Now suppose that $M$ is compact Kahler, and that there is a surjection $h:\pi_1(M)\to \Gamma$ onto the orbifold fundamental group of a compact hyperbolic orbifold (i.e suppose that $\Gamma$ embeds into $SL_2(\mathbb{R})$ as a cocompact properly discontinuous subgroup).

Then I claim that $h$ can be realized as by a holomorphic map of $M$ to an orbifold.

To see this, observe that we can find a torsion free normal subgroup $\Gamma_1\subset \Gamma$ of finite index. Let $G=\Gamma/\Gamma_1$. Let $M_1$ be the $G$-cover corresponding to $h^{-1}(\Gamma_1)$. Let $W\subset H^1(\Gamma_1,\mathbb{C})$ be a $G$-invariant Lagrangian subspace. If either $V=(h^*W)^{10}$ or $V=\overline{(h^*W)^{01}}$ has dimension at least $2$, we get to appeal to CdF to get an equivariant map $M_1$ to a Riemann surface. There is an exceptional case, which can be handled as in Catanese page 269. I won't reproduce it here.

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Thanks! Truly you are wise in the ways of science! –  Igor Rivin Apr 23 '11 at 18:23
I'm not sure about that, but thank you. –  Donu Arapura Apr 23 '11 at 22:23