# Complex varieties with non-torsion homotopy groups

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.

• Could you clarify what you mean by "affine space"? – John Pardon Apr 7 '17 at 2:03
• 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. – Piotr Achinger Apr 7 '17 at 9:30
• What do you mean by homotopy group? The ordinary homotopy groups of the complex points? – Sean Tilson Apr 7 '17 at 10:05
• 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. – Donu Arapura Apr 7 '17 at 15:47
• 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. – Mark Grant Apr 7 '17 at 17:09

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.
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.