Question. Does there exist a smooth complex projective variety with infinite and perfect fundamental group?
A group $G$ is perfect if its Abelianisation $G/[G,\, G]$ is the trivial group.
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2$\begingroup$ There's an extensive theory of Kähler groups, i.e. groups that can occur as the fundamental group of a compact Kähler manifold (which includes all smooth complex projective varieties). Searching for "Kähler groups" will find many papers. I don't remember ever seeing anything about infinite perfect groups, but I am certainly not an expert in this area. $\endgroup$– Neil StricklandMay 11, 2021 at 11:21
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1$\begingroup$ Thompson groups $T$ and $V$ don't work arxiv.org/abs/math/0506254 $\endgroup$– laconMay 11, 2021 at 11:33
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$\begingroup$ Can $\mathrm{SL}_n(\mathbb{Z})$ for $n\geq 3$ occur? $\endgroup$– laconMay 11, 2021 at 11:36
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2$\begingroup$ @lacon $SL_n(\mathbb{Z})$ won't occur by work of Simpson [Higgs Bundles and local systems]. $\endgroup$– Donu ArapuraMay 11, 2021 at 12:23
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2$\begingroup$ There are many infinite perfect 3-manifold groups but these don't occur. See work of Friedl, Kotschick, Suciu... $\endgroup$– Danny RubermanMay 11, 2021 at 15:37
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2 Answers
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The simplest example I know of such a group is given by the presentation $$ \langle a, b, c| a^2, b^3, c^7, abc\rangle. $$ This group $G$ is obviously perfect and (less obviously) is isomorphic to the fundamental group of a smooth complex-projective variety. The reason for the latter is that $G$ acts holomorphically, properly and cocompactly on the unit disk and contains a finite index torsion-free subgroup.
With this in mind, one uses the following lemma a form of which one can find in the book "Kahler groups":
Lemma. Suppose that $G$ is a group containing a torsion-free normal subgroup $H$ of finite index. Suppose that $G$ is acting holomorphically, properly on a simply-connected complex manifold $X$ such that $X/H$ is biholomorphic to a smooth complex-projective variety. Then $G$ is isomorphic to the fundamental group of a smooth complex projective variety.
Proof. There exists a simply-connected projective variety $Z$ with a free holomorphic action of $G/H$. Now, form the fiber product $$ Y=(X\times Z)/G $$ where the action of $G$ on $X$ is as above and $G$ acts on $Z$ via the action of $G/H$. Then $\pi_1(Y)\cong G$ and $Y$ is biholomorphic to a smooth projective variety, since it is the quotient of the projective variety $W=(X/H) \times Z$ by a free holomorphic action of a finite group, namely $G/H$.
answered May 11, 2021 at 15:35
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1$\begingroup$ Looks good! (For those who don't see the last part of the argument. If $G$ is the above group and $H\subset G$ a finite index normal subgroup, we can choose a smooth projective var. $Y$ with $\pi_1(Y)=G/H$. Then $X=(D\times Y)/G$ is a smooth projective variety with fund gp $G$.) $\endgroup$ May 11, 2021 at 16:07
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1$\begingroup$ Perhaps it's worth saying that explicitly that this is the "(2,3,7)-triangle group". It's a famous group! Do I understand correctly that your answer works for any hyperbolic triangle group? (Which will of course be perfect if $gcd(p,q,r)=1$.) $\endgroup$– HJRWMay 11, 2021 at 16:56
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1$\begingroup$ In your proof, you probably want to assume that $Z$ is simply-connected. $\endgroup$ May 11, 2021 at 17:44
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1$\begingroup$ @MoisheKohan: I think more often than note people mean exactly this group. Indeed, to quote wikipedia (en.wikipedia.org/wiki/(2,3,7)_triangle_group), 'the "(2,3,7) triangle group" most often refers, not to the full triangle group $\Delta(2,3,7)$ ... but rather to the ordinary triangle group $D(2,3,7)$ of orientation-preserving maps (the rotation group), which is index 2.' $\endgroup$– HJRWMay 11, 2021 at 19:04
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1$\begingroup$ By the way, I once wrote a paper about (generalisations of) triangle groups with Jason Manning and Alex Lubotzky. I suggested we use the term "Von Dyck group", and Lubotzky told me that, in all his decades in group theory, he had never heard the term before. So I don't think it's really "standard" these days (though I grant that wikipedia says it is). We went with the term "ordinary triangle group", and usually omitted the word "ordinary". $\endgroup$– HJRWMay 11, 2021 at 19:07
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A number of such examples are known. For example, if $X$ is a fake projective plane, then its fundamental group would be perfect and infinite (it's a lattice in $PU(2,1)$).
Postscript As abx points out, these fundamental groups have nontrivial finite abelianized fundamental groups, so they don't quite work. However, Moishe Kohan's answer gives a very simple example.
answered May 11, 2021 at 12:19
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4$\begingroup$ I don't think that the group is perfect. Its abelianization is $H_1(X,\Bbb{Z})$, which is never zero according to Prasad-Yeung (see e.g. this paper, 7.3, Corollary 1. $\endgroup$– abxMay 11, 2021 at 12:35
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$\begingroup$ @abx OK, thanks. I'll edit when I get a chance, and include some bonafide examples. $\endgroup$ May 11, 2021 at 12:41
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$\begingroup$ The issue with such lattices (as well as lattices in higher-dimensional Lie groups) is that most known theorems guarantee vanishing of the 1st Betti numbers, but allow for finite nontrivial abelianization. $\endgroup$ May 11, 2021 at 15:37