Comparing fundamental groups of a complex orbifolds and their resolutions.

Let $X$ be a complex manifold with quotient singularities, and let $\tilde X$ be its resolution (that exists, for example, by Hironaka). Then I am pretty sure that $\pi_1(X)\cong \pi_1(\tilde X)$.

• Question. Is there a reference for such a statement? At least in dimension 3?

The reason why this should be true is that it should be possible to find such a resolution of $X$ that the preimage of each point in $\tilde X$ is simply connected ($\mathbb CP^n/G$ is simply-connected for finite $G$). Then everything follows from a standard topological lemma. I guess this statement should be true as well if $X$ is a complex analytic variety with arbitrary Kawamata log terminal singularities (because Fano manifolds are simply connected). I would be grateful for a reference for any kind of such statement.

ADDED. As Francesco says in his answer, this statement is classical for surfaces. I would like to have a reference what would cover at least the case of $3$-folds with non-isolated quotient singularities (say with abelian stabilisers).

UPD. The question is now 100% settled by the reference provided by the answer of Benoit.

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A reference is Theorem 7.8 of the article by Kollar: "Shafarevich maps and plurigenera of algebraic varieties", Invent. Math. 113. This proves the equality of fundamental groups for quotient singualrities in all dimensions and also for algebraic fundamental groups in the case of klt singularities.

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Many thanks!!! –  Dmitri Jan 1 '11 at 11:55
Would you mind to give me your name, for example by sending an email to dmitri.panov@kcl.ac.uk ? I will cite this result in an article and would like to thank you. –  Dmitri Jan 3 '11 at 2:10

Since it is not clear if there is a reference for this statement, I'll give a proof here (we just worked it out with Sergey Galkin). This proof is complete for the case when the stabilizers of all points are abelian, otherwise it works modulo the following lemma (which should of course be true).

• Lemma. Let $G\in GL(n,\mathbb C)$ be a finite group. Then any resolution of $\mathbb C^n/G$ is simply connected.

To prove the case when $G$ is abelian we notice that in this case $\mathbb C^n/G$ is toric and it has a toric resolution, which is simply connected.

Recall a statement from Griffiths and Harris (Chapter 4, page 494)

• Proposition. Let $M$ and $N$ be birational complex analytic varieties. Then $\pi_1(M)\cong \pi_1(N)$.

The proof is very short, around 10 lines. Now let as use these two statements in order to prove the promised result.

• Lemma. Let $X$ be a complex orbifold, then for any resolution $\hat X$ of $X$ we have $\pi_1(\hat X)\cong \pi_1(X)$.

Proof. The idea of the proof is simple. We will cover $X$ by a collection of contractible neighbourhoods $U_i$, such that all their mutual intersections are also contractible. In this case the homothopy type of $X$ equals the homothopy type of the nerve of the covering. In particular, $\pi_1(X)$ equals the fundamental group of the nerve. In addition we will chose $U_i$ so that for each resolution of $X$ all preimages of $U_i$ together with their intersections are simply connected. Then again we can calculate $\pi_1(\hat X)$ as the fundamental group of the nerve (surely, this will not work for higher homothopy groups, since the preimages of $U_i$ will not be contractible any more).

Choice of $U_i$. Since $X$ is an orbifold, each point $x\in X$ has a neighbourhood isomorphic to $\mathbb B^n/G_x$ where $\mathbb B^n$ is a unite complex ball in $\mathbb C^n$ and $G_x$ is the finite stabiliser of $x$ that is acting linearly on $\mathbb C^n$ (and hence on $\mathbb B^n$). Now we chose a finite cover of $X$ by such neighbourhoods.

Now by the above lemma $U_i$ has a simply connected resolution. Hence each of its resolution is simply connected by the above proposition.

End of proof.

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The proof of simple connectedness of a resolution of quotient singularities is in this paper of mine http://arxiv.org/abs/math/9903175 Theorem 4.1 (published in Asian J. Math. 4, 2000, no. 3, 553-563), but I guess it was known to Bogomolov long ago (and maybe even published in some of his papers)

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I did not know about Kollar's proof - it would save me much bother. Thanks to the anonymous commenter in the first answer. –  Misha Verbitsky Jan 7 '11 at 12:34
Misha, spasibo! –  Dmitri Jan 7 '11 at 13:09

Dear Dmitri, the result you hope is also true (ie when X has only klt singularities, its fundamental group is isomorphic to the one of any of its desingularization). This theorem is due to Takayama (Local simple connectedness of resolution of log-terminal singularities, International journal of Math 2003).

Bests, Benoît

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Benoit, thanks a lot! Nice to know that this result is proved by now. –  Dmitri Apr 29 '11 at 17:43

Your statement is surely true in the following two cases:

1. $\dim X=2$. In this case something stronger is true; in fact, if $X$ is a surface with only rational singularities and $\tilde{X}$ is a resolution of singularities, then $\pi_1(\tilde{X})=\pi_1(X)$. This because the resolution of a rational $2$-dimensional singularity is alwais given by a bunch of rational curves, in particular the exceptional locus is simply connected, so you can conclude by Seifert-Van Kampen theorem. Thi argument is used for instance in the paper of R. Barlow

"A symply connected surface of general type with $p_g=0$", Invent. Math.79.

1. $\dim X$ is arbitrary and all singularities of $X$ are isolated cyclic quotient singularities. In fact, by a result of Fujiki, an isolated cyclic quotient singularity can always be resolved by a normal crossing divisor whose components are smooth rational varieties, so you can apply again Seifert-Van Kampen theorem.

This argument is used in the paper of Bauer, Catanese, Pignatelli, Grunewald

Quotients of products of curves, new surfaces with $p_g=0$ and their fundamental groups

in order to compute the fundamental group of a quotient of the form $(C_1 \times C_2 \times \cdots \times C_n)/G$, where the $C_i$ are smooth curves (see in particular Remark 2.4).

I do not know whether the result is true in full generality. In fact, given a finite group $G$ acting on a simply connected manifold $V$, by a result of Armstrong (Proc. Amer. Math. Soc. 84) one has

$\pi_1(V/G)=G/E$,

where $E$ is the subgroup of elements having fixed points.

In your case $V=\mathbb{C}^n$, and if $\mathbb{C}^n/G$ is simply connected, i.e. if $G=E$, then you can apply Seifert-Van Kampen again. Therefore it seems to me that your question is related to the following one:

Does any cyclic group $\langle g \rangle$ acting on $\mathbb{C}^n$ have a fixed point?

If $g$ is a polynomial automorphism then the answer is yes, but in general this is an open problem for $n \geq 3$ (for $n=2$ it is true since any finite group acting on $\mathbb{C}^2$ is linearizable by a result of Suzuki).

You can find more on this topic (and several related references) in the paper by Kraft and Schwarz "Finite automorphisms of affine $n$-spaces".

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Francesco, thanks a lot for your detailed answer! I'll have a look on the result of Fujiki. I do believe that the statement is always true, for example because the result of Fujiki should hold not only for isolated cyclic quotient singularities. (Though, if BCPG only have a reference to this result, maybe it is the best one written down... :( ). I have not quite got what you mean when you say that in my case $V=\mathbb C^n$ ... (maybe I have not formulated my question properly, but my question is about Any complex manifold with quotient singularities). –  Dmitri Dec 29 '10 at 13:22
Dimitri, I just meant that locally analitically a quotient singularity is of the form $\mathbb{C}^n/G$, where $G \subset \textrm{Aut}(\mathbb{C}^n)$. So if for every singular point you know the local $G$-action and you are able to verify that $G=E$, then it seems to me that you can apply the usual Seifert-Van Kampen argument to conclude... –  Francesco Polizzi Dec 29 '10 at 13:53
Francesco, sorry, I try to understand what you say. I just have not got what is conclusion of the second part of the answer... (starting form the result of Armstrong) –  Dmitri Dec 29 '10 at 14:15
dimitri, you are right, my argument using armstrong result only proves that if G=E then there exists a neighborhood of the singularity which is simply connected, but it says nothing about the resolution... –  Francesco Polizzi Dec 29 '10 at 21:17