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Dec 15, 2020 at 5:30 comment added Aoi Koshigaya Excuse me. Hodge theory says that $H^m(X, \mathcal O_X) = \bigoplus_{p+q=m} H^q(X, \Omega_X^p)$. In Debarre's book, he shows that $H^0(X, (\Omega_X^p)^{\otimes k}) = 0$, why does this condition imply that $H^q(X, \Omega_X^p) = 0$.
Mar 30, 2019 at 21:29 comment added user74900 @TimPerutz but do you know an example of a connected smooth projective complex variety that is etale simply connected but is not classically simply connected? I have heard that is an open problem.
Mar 15, 2014 at 21:01 vote accept Mohammad Farajzadeh-Tehrani
Jan 31, 2013 at 23:10 comment added Tim Perutz Mohammad: being algebraically simply connected means that there are no non-trivial connected etale covers. Such covers are finite, while in topology one makes no finiteness condition on the covers. See en.wikipedia.org/wiki/Étale_fundamental_group
Jan 31, 2013 at 20:30 comment added Mohammad Farajzadeh-Tehrani @Tim: My question was about the classical fundamental group. What is the definition algebraic one? @Vesselin: I still believe we can say some thing if the singularities are terminal or something nice.
Jan 31, 2013 at 20:29 comment added Vesselin Dimitrov Indeed, I intended to write $y^2 = x^3 + x^2$ (for a nodal rational curve). Thanks for the correction!
Jan 31, 2013 at 20:24 comment added Jason Starr @Vesselin: The curve $y^2-x^3$ is cuspidal, not nodal, although your point is correct. In my comment, the variety should be unibranch.
Jan 31, 2013 at 19:39 comment added Jason Starr @Tim. The topological fundamental group of any rationally chain connected projective variety is a priori finite by Chow variety / Hilbert scheme methods: a degree of a certain Chow variety of curves gives an upper bound on the size of the fundamental group. Combined with the arguments in Debarre's article (which are originally due to Koll\'ar), this implies simple connectedness of the fundamental group of a smooth, projective rationally connected variety.
Jan 31, 2013 at 19:36 comment added Vesselin Dimitrov By the way, the nodal curve ${y^2 = x^3}$ is not simply connected, so the result does not extend to singular rationally connected varieties.
Jan 31, 2013 at 19:35 comment added Vesselin Dimitrov Yes indeed; the full proof of the classical simple connectedness of a complex rationally connected smooth projective variety can be found in Chapter 4 of Debarre's book.
Jan 31, 2013 at 19:30 comment added Tim Perutz Do I understand right that your paragraph beginning "Alternatively" shows that over the complex numbers the algebraic fundamental group is trivial, but leaves open the possibility that the classical fundamental group is non-trivial and without finite-index subgroups? Is the classical fundamental group of a rationally connected, smooth, complex projective variety known by other means to be trivial?
Jan 31, 2013 at 19:13 history edited Vesselin Dimitrov CC BY-SA 3.0
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Jan 31, 2013 at 19:04 history edited Vesselin Dimitrov CC BY-SA 3.0
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Jan 31, 2013 at 18:56 history answered Vesselin Dimitrov CC BY-SA 3.0