2
$\begingroup$

What is the geography of Kähler manifolds with negative sectional curvature? More precisely, can any hyperbolic group be realized as the fundamental group of a Kähler manifold with negative sectional curvature?

$\endgroup$
4
  • 6
    $\begingroup$ Kähler manifolds have even first Betti number, so any hyperbolic group with odd first Betti number is a counterexample. For instance, $\mathbb{Z}$ or, less trivially, the free group $F_3$. Other counterexamples can probably be deduced from work of Gromov, Schoen and Delzant, which I think restricts how Kähler groups can act on cube complexes. (And lots of hyperbolic groups are the fundamental groups of npc cube complexes.) In general, the problem of characterizing which groups are Kähler is very hard. $\endgroup$
    – HJRW
    Jun 29, 2016 at 15:22
  • 2
    $\begingroup$ The OP did not consider compact Kähler manifolds. $\endgroup$
    – abx
    Jun 29, 2016 at 19:47
  • 1
    $\begingroup$ @abx, apologies, I should have clarified that I was talking about compact Kähler manifolds. If the manifold is not compact then negative sectional curvature does not imply hyperbolicity, so the hypothesis does seem to be implicit in the question, if not explicit. It would help if the OP could clarify. $\endgroup$
    – HJRW
    Jun 30, 2016 at 8:40
  • $\begingroup$ Without compactness (or, at least, completeness) assumption, every countable group can be realized as the fundamental group of a Kahler manifold of negative sectional curvature. $\endgroup$ Jan 26, 2023 at 20:22

1 Answer 1

3
$\begingroup$

$\DeclareMathOperator\PO{PO}\DeclareMathOperator\PU{PU}$This is more a comment than an answer, but if one does not impose compactness of the manifold (but still asks for the Kähler manifold to be complete, with pinched negative curvature or even more to be complete locally symmetric of rank 1) then one sees that certain hyperbolic groups occurs in this way whereas they cannot occur as fundamental groups of any closed Kähler manifold.

More precisely take a cocompact torsion free lattice $\Gamma$ in the group $\PO(n,1)$ with $n\ge 3$. Then $\Gamma$ is not isomorphic to the fundamental group of a closed Kähler manifold by an old theorem of Carlson and Toledo (see their paper Harmonic mappings of Kähler manifolds to locally symmetric spaces). Yet by embedding $\PO(n,1)$ into $\PU(n,1)$ and letting $\Gamma$ act on complex hyperbolic space of (complex) dimension $n$, one sees that $\Gamma$ is the fundamental group of a negatively curved Kähler manifold.

So without any further assumption on the negatively curved Kähler manifold one really gets more examples of hyperbolic groups than by simply looking at fundamental groups of closed Kähler manifolds and asking them to be Gromov hyperbolic.

By the way, even studying convex cocompact subgroups of $\PU(n,1)$ is a difficult topic and there are few examples (see Granier - Groupes discrets en géométrie hyperbolique — Aspects effectifs for an interesting one).

$\endgroup$

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.