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Dec 20, 2018 at 15:41 answer added Chen Jiang timeline score: 7
Dec 19, 2018 at 9:59 vote accept abx
Dec 18, 2018 at 21:18 history edited Qfwfq CC BY-SA 4.0
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Dec 18, 2018 at 21:09 answer added Jason Starr timeline score: 9
Dec 18, 2018 at 19:54 comment added Sasha @abx: Sure, but when you take the quotient you can kill higher cohomology groups. At list this is what happens for fake quadrics.
Dec 18, 2018 at 19:47 comment added abx @Sasha: $\chi (\mathcal{O})$ is negative for a curve, so for a product of $n$ curves it has the same sign as $(-1)^n$.
Dec 18, 2018 at 19:29 comment added Jason Starr For the pencil of Godeaux surfaces, it looks like the Reid -- Shepherd-Barron -- Tai criterion applies. I will try to write up the details below . . .
Dec 18, 2018 at 19:28 comment added Sasha @abx: Sorry, but I don't understand why is $\chi(\mathcal{O}) < 0$? In the surface case this objection doesn't work, right?
Dec 18, 2018 at 19:17 comment added abx @Sasha: Again this would have $\chi (\mathcal{O})<0$.
Dec 18, 2018 at 19:16 comment added Jason Starr Maybe a general pencil of Godeaux surfaces? . . .
Dec 18, 2018 at 18:34 comment added Sasha @abx: Isn't it possible to cook up an example by taking the quotient of the product of three curves by a finite group action, an analogue of fake $\mathbb{P}^1 \times \mathbb{P}^1$?
Dec 18, 2018 at 18:07 comment added abx @Sasha: forms of all degrees $>0$.
Dec 18, 2018 at 18:05 comment added abx @Jason Starr: no, that doesn't work. $\chi (\mathcal{O})$ is negative for a complete intersection threefold of general type, thus also for any quotient by a free action, while we want $\chi (\mathcal{O})=1$.
Dec 18, 2018 at 17:47 comment added Jason Starr Presumably there are examples that are quotients of general type 3fold complete intersections similar to the Godeaux surface. Do you want examples that are simply connected?
Dec 18, 2018 at 17:42 comment added Sasha Forms of some degree, or forms of all degrees?
Dec 18, 2018 at 17:00 history asked abx CC BY-SA 4.0