3
$\begingroup$

Let $f:X\to Y$ be a holomorphic map of holomorphic manifolds. You can assume that $dimY=1$. Let $\tilde X$ and $\tilde Y$ be universal covers of $X$ and $Y$ with group of holomorphic automorphisms $Aut(\tilde X)$ and $Aut(\tilde Y)$. Do we get a homomorphism $Aut(\tilde X)\to Aut(\tilde Y)$ in general?

Improve this question
$\endgroup$
3
  • $\begingroup$ Automorphism group in which category? You may be asking about the induced map on fundamental groups. $\endgroup$
    – Fernando Muro
    Commented Sep 27, 2013 at 13:41
  • $\begingroup$ I think I clarrified this in my question. Automorphism in the category of holomorphic manifolds. So $Aut(\tilde X)$ is the group of holomorphic automorphisms of $\tilde X$. Of course $\pi_{1}(X)\subseteq Aut(\tilde X)$ but I am interested in the full group of automorphisms of $\tilde X$ not only those which cover the coverings map $\tilde X\to X$ (i.e. $\pi_{1}(X)$) $\endgroup$
    – Darius Math
    Commented Sep 27, 2013 at 13:50
  • $\begingroup$ The question is definitely unclear, it admits some trivial answers which, of course, is not what you're looking for (e.g. you always have the trivial homomorphism). $\endgroup$
    – Fernando Muro
    Commented Sep 27, 2013 at 22:02

1 Answer 1

Reset to default
2
$\begingroup$

The answer is negative. For instance, let $X$ be a compact quotiont of the unit ball in ${\mathbb C}^2$ by a discrete torsion-free subgroup of $PU(2,1)$, and $Y$ be a hyperbolic Riemann surface. There are many examples where there exists a nonconstant holomorphic map $f: X\to Y$ (say, if $b_1(X)\ne 0$ then there is always one, for some $Y$), but, clearly, there are no nontrivial homomorphisms $PU(2,1)\to PU(1,1)$.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thanks a lot! In this example $Y$ is non-compact. Can it be that if we assume $Y$ to be a compact curve, there exists a homomorphism? $\endgroup$
    – Darius Math
    Commented Sep 27, 2013 at 18:29
  • 1
    $\begingroup$ @DariusMath: In my examples $Y$ is compact. $\endgroup$
    – Misha
    Commented Sep 27, 2013 at 19:45
  • $\begingroup$ Ah, OK. I thought by hyperbolic Riemann surface you mean a Riemann surface with a number of points removed. $\endgroup$
    – Darius Math
    Commented Sep 27, 2013 at 20:01
  • 1
    $\begingroup$ The standard definition of a hyperbolic Riemann surface is the one whose universal cover is biholomorphic to the unit disk. For instance, among compact Riemann surfaces this is equivalent to genus $>1$. $\endgroup$
    – Misha
    Commented Sep 27, 2013 at 20:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .