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.

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



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. 


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. 

