Is there some kind of classification of (connected) smooth complex varieties such that every homotopy group of the manifold of complex points is torsion-free? Any reference on this topic will be most welcome.
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4$\begingroup$ Could you clarify what you mean by "affine space"? $\endgroup$– John PardonCommented Apr 7, 2017 at 2:03
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5$\begingroup$ There are many interesting $K(\pi, 1)$ examples, e.g. certain complements of hyperplane arrangements. I don't know examples with nontrivial torsion-free higher homotopy groups. $\endgroup$– Piotr AchingerCommented Apr 7, 2017 at 9:30
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1$\begingroup$ What do you mean by homotopy group? The ordinary homotopy groups of the complex points? $\endgroup$– Sean TilsonCommented Apr 7, 2017 at 10:05
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3$\begingroup$ A smooth complex variety has the homotopy type of a finite CW complex. So I doubt you'll find any examples with all homotopy groups torsion free. $\endgroup$– Donu ArapuraCommented Apr 7, 2017 at 15:47
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2$\begingroup$ Further to Donu Arapura's comment, perhaps David Chataur's answer to mathoverflow.net/questions/207448/… can be used to show that the only examples are $K(\pi,1)$'s as mentioned by Piotr Achinger. $\endgroup$– Mark GrantCommented Apr 7, 2017 at 17:09
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2 Answers
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Following Mark Grant's comment, referencing David Chataur's answer here: McGibbon and Neisendorfer proved that a finite-dimensional 1-connected space with any nonzero reduced homology has infinitely many homotopy groups with torsion. In particular, this applies to the universal cover $\tilde{X}$ of the space $X$ we care about. So $\tilde{X}$ must be acyclic (and is simply-connected), and is thus contractible by Hurewicz's theorem. Therefore $X$ is aspherical.
answered Apr 7, 2017 at 23:50
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The answer is likely to be a bit messy, since every projective curve is a $K(\pi, 1),$ and so are cartesian products and (iterated) fibrations of such (classifying which of these are algebraic is, presumably, still open, even for surface bundles over surfaces). I assume, in my ignorance, that complements of hyperplane arrangements are not of this form, which makes me think that the question is hopeless.
answered Apr 8, 2017 at 1:22