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As the title. Geometrically, is there a projective complex manifold(or more generally an projective algebraic variety) accepting only infinite nontrivial cover(which may not be projective)? Thanks.

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There is a terminological confusion in this question: A complex manifold need not be an algebraic variety and a complex algebraic variety need not be a complex manifold. Thus, this question can be interpreted in at least 3 different ways. It is a well known open problem if there are smooth complex projective varieties with nontrivial fundamental group and no nontrivial finite covers. Toledo proved that fundamental group need not be residually finite, but his examples have many finite covers. – Misha Apr 4 '13 at 16:07
For the question exactly as stated in the body of the question the answer is yes since any finitely presented group is the fundamental group of a compact complex manifold. – Mohan Ramachandran Apr 4 '13 at 16:29
@Misha Sorry, I have made the question precise. I only concern projective case. – stjc Apr 7 '13 at 11:50
up vote 10 down vote accepted

There are several classes of spaces for which this question can be asked, here are the answers:

  1. Compact complex-projective manifolds (also frequently called manifolds admitting flat complex-projective structures): These are n-manifolds admitting an atlas where transition maps are elements of $PGL(n+1, C)$. For such manifolds, either the fundamental group is trivial or it admits a nontrivial projective-linear representation (holonomy of the structure). It follows from Malcev's theorem, that the group always contains a finite-index subgroup (different from the original group). Thus, the answer to OP's question in this setting in NO.

  2. Complex-projective varieties. It was observed by Serre (and many others) that every finitely-presented group $\pi$ appears as fundamental group of such a variety. One does not need Simpson's paper for this (he was answering a much harder question). Here is the construction. Take finite simplicial complex with the given fundamental group. Embed it as a subcomplex of a standard affine k-simplex. Replace each face of this subcomplex with its complex-projective span. Take the union of these subspaces of $CP^k$. That's your variety. Then, as Ricky noted, the answer to OP's question is YES.

  3. Smooth complex-projective varieties. Then it becomes a well-known open problem. It is also open for compact Kahler manifolds.

  4. Compact complex manifolds. It was proven by Taubes in 1992 that every finitely-presented group is the fundamental group of such a 3-dimensional manifold. Thus, the answer again is, YES.

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I rewrite the answer taking into account anon's comment below.

The are two questions here, the one in the title and the one in the body of the question. The difference is that in the title the word variety is used, while in the body the word manifold is used (so to answer the question in the title we can use singular varieties).

Let me start with the question of the title. The answer is yes. Any finitely presented group is the topological fundamental group of a complex algebraic variety. This is Theorem 12.1 in the paper "local systems on proper algebraic $V$-manifolds" by Carlos Simpson. Now consider the so called Higman's group $G$ defined by $$ G := \langle a,b,c,d \;| \; aba^{−1} = b^2 , \; bcb^{−1} = c^2 , \; cdc^{−1} = d^2 , \; dad^{−1} = a^2 \rangle. $$

One can prove that $G$ is not trivial (it is infinite) but the only finite index subgroup of $G$ is $G$ itself (see for example Serre's "Trees", Proposition 6 in Section 1.4 of Chapter I). In particular, the profinite completion of $G$ is trivial. Then any complex manifold with $G$ as fundamental group gives an answer to your question, since the algebraic fundamental group is the profinite completion of the topological one.

The question for manifolds is of course harder, and I think it is open.

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I am confused about the question. a (smooth) Variety is different from a complex manifold. Are these examples of $G$ also the topological fundamental groups of varieties or only of complex manifolds? – Venkataramana Apr 4 '13 at 12:21
I don't understand your comment. The (topological) fundamental group depends only on the structure of topological space. A complex manifold is of course a topological space (with extra structure), so it has a topological fundamental group. – Ricky Apr 4 '13 at 12:25
You have mentioned this result that every is the fundamental group of a complex manifold. But is it true that it is the fundamental group of some algebraic variety? For example, if we take smooth projective varieties, their fundamental groups are restricted (for example, the abelianisation of $\pi _1$ must have even rank as an abelian group). – Venkataramana Apr 4 '13 at 13:00
Yes, you're right. I've added a reference. – Ricky Apr 4 '13 at 13:04
An open problem is if such variety could be made smooth. – Misha Apr 4 '13 at 13:39

This should be a comment on Ricky's answer. I think you have to appeal to a paper of Carlos Simpson (PAMQ 2011) to know that there is an algebraic variety with $G$ as its fundamental group. Added: by variety I mean irreducible variety.

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Yes, of course, the manifold must be algebraic! – Ricky Apr 4 '13 at 12:37

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