1
$\begingroup$

I would like to know:

What does the fundamental group of a quasi-projective algebraic variety look like?

I remember that I have seen somewhere that for a connected, finite-type CW-complex $X$, its fundamental group is the direct product of a free group of finite rank and a finite abelian group.

Is this statement true? Why?

$\endgroup$
  • $\begingroup$ The statement you wrote is not true. Pick a torus as a counterexample (the fundamental group is $\mathbb{Z}\oplus\mathbb{Z}$). In fact every finitely presented group is the fundamental group of some connected finite CW of dimension at most 2. $\endgroup$ – Denis Nardin Apr 24 at 11:43
  • 2
    $\begingroup$ $\mathbb{Z}\oplus\mathbb{Z}$ is a free abelian group, not a free group. $\endgroup$ – Denis Nardin Apr 24 at 11:46
  • 1
    $\begingroup$ I do mean a complex torus (which topologically is just a torus). $\endgroup$ – Denis Nardin Apr 24 at 11:50
  • 7
    $\begingroup$ There are indeed restrictions on the fundamental group of smooth complex algebraic varieties, besides being finitely presentable. The first known one is described in a paper of J. Morgan, The topology of Algebraic Varieties. If you drop "smooth" then any finitely presented group can appear. Other restrictions are known but no complete characterization. What is the precise question? $\endgroup$ – Louis-Clément LEFÈVRE Apr 24 at 12:16
  • 2
2
$\begingroup$

Much of the OP's questions were addressed in the comments. Here is a summary with some additional references and remarks.

In Hatcher's book Algebraic Topology, Corollary 1.28 states (which follows from van Kampen's Theorem): For every group $G$ there is a 2-dimensional cell complex $X_G$ with $\pi_1(X_G)\cong G$. In particular, if $G$ is finitely presentable, then there is a finite CW complex with $G$ as its fundamental group.

So your statement "a connected, finite-type CW-complex $X$, [has] it's fundamental group the direct product of a free group of finite rank and a finite abelian group" is about as false as possible.

You also asked about the fundamental group of quasi-projective varieties.

The first class to think about is smooth projective varieties (addressed by J. Morgan here in 1978, who attributes this question to Serre). Since these varieties are Kähler, the restrictions on Kähler groups is perhaps the first thing to think about. There is much known and you can read about it at this MO question: Fundamental groups of compact Kähler manifolds. For example, free groups can't be Kähler, but free abelian groups of even rank are. Another worthwhile resource is: Fundamental Groups of Smooth Projective Varieties by Donu Arapura. For example, all finite groups are the fundamental group of a smooth projective variety (a result of Serre), but no group that is not finitely presentable is. I am only highlighting some of the simple examples in this very useful exposition.

Moving on, if one allows singularities, then Simpson proved in 2010 that for every finitely presentable group $G,$ there exists an irreducible projective variety with that fundamental group (Theorem 12.1). He asked whether this can be done with restrictions on the singularities. Using a result of Goldman and Millson, Kapovich and Kollár show in 2011 that for every finitely-presented group $G$ there is a complex, projective surface $S_G$ with simple normal crossing singularities only such that $\pi_1(S_G)\cong G$ (Theorem 2). However, they did not prove $S_G$ could be taken to be irreducible. So a year later (2012), Kapovich shows: for any $G$ that is finitely-presented, there exists a 2-dimensional irreducible complex-projective variety $W$ with the fundamental group $G$, so that the only singularities of $W$ are normal crossings and Whitney umbrellas (Theorem 1.2).

As explained in answers to this MO post, Donu Arapura has some papers (for example this one) showing certain groups cannot be the fundamental group of a normal projective variety. And here we find a generalization of a result of Millson and Kapovich (by Dimca, Papadima, & Suciu): Among right-angled Artin groups, there are infinitely many mutually non-isomorphic groups which are not isomorphic to fundamental groups of smooth, quasiprojective complex varieties (Corollary 11.8).

Some open problems are given here: Links of Complex Analytic Singularities by J. Kollár.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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