10
$\begingroup$

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.

$\endgroup$
  • 4
    $\begingroup$ Could you clarify what you mean by "affine space"? $\endgroup$ – John Pardon Apr 7 '17 at 2:03
  • 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 Achinger Apr 7 '17 at 9:30
  • 1
    $\begingroup$ What do you mean by homotopy group? The ordinary homotopy groups of the complex points? $\endgroup$ – Sean Tilson Apr 7 '17 at 10:05
  • 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 Arapura Apr 7 '17 at 15:47
  • 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 Grant Apr 7 '17 at 17:09
11
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.