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Jun 9, 2011 at 22:58 vote accept aglearner
Jun 9, 2011 at 19:24 comment added Piotr Achinger Thank you for the reference, it's very interesting. The author leaves open the question whether there exists an algebraically simply connected algebraic variety which is not topologically simply connected. Such an example would be much more surprising!
Jun 9, 2011 at 17:36 comment added Francesco Polizzi For higher-dimensional projective varieties the result is false. In fact, D. Toledo has constructed examples with non-residually finite fundamental group, see archive.numdam.org/article/PMIHES_1993__77__103_0.pdf
Jun 9, 2011 at 17:31 comment added Piotr Achinger So the statement is that for Riemann surfaces, the topological group injects into the algebraic fundamental group (the profinite completion of the former). If it is OK to ask questions in comments, I would like to know if this holds more generally for smooth complex algebraic varieties.
Jun 9, 2011 at 17:28 history edited aglearner CC BY-SA 3.0
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Jun 9, 2011 at 17:05 comment added aglearner Charlie, thanks for the comment! Unfortunately it is not simple enough for me to understand... You don't suggest to use residual finiteness of linear groups? Also, how do you prove what you wrote? You are more than welcome to develop your comment into a full answer!
Jun 9, 2011 at 17:05 comment added user6976 Certainly, the surface groups are lattices in $SL_2({\mathbb C})$. But that is harder to prove than residual finiteness (I think).
Jun 9, 2011 at 16:58 answer added user6976 timeline score: 11
Jun 9, 2011 at 16:54 comment added Charlie Frohman It is a finitely generated subgroup of invertible matrices with coefficients in an algebraic extension of the rationals?
Jun 9, 2011 at 16:37 history asked aglearner CC BY-SA 3.0