Timeline for Non-projective smooth complete threefolds with a pair of points intersecting every surface
Current License: CC BY-SA 3.0
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Nov 15, 2013 at 22:43 | comment | added | jacob | @SándorKovács: Since I'm assuming the surface I pick does not go through P, so pulling it back is easy as X\C is isomorphic to Y\P. Am i missing something? | |
Nov 14, 2013 at 19:44 | comment | added | Sándor Kovács | @jacob: although, as Olivier pointed out, indeed I wasn't claiming that the result is a scheme, I am not sure that your argument in your first comment works. If the surface you pick through the image point $P$ is not $\mathbb Q$-Cartier, then what is its pull-back? In other words it appears that what you are proving is that the target cannot be $\mathbb Q$-factorial. | |
Nov 14, 2013 at 10:43 | comment | added | jacob | Ah, Ok. I was just confused because I thought they meant singular schemes, rather than algebraic spaces. No, I don't really know what to expect either. I'm intrigued by the strategy proposed, though. Is it known which curves can be contracted on a threefold? If one can find a projective X with a bunch of curves $C_1,...,C_n$ all of which can be contracted, and such that their sum intersects each surface positively, then one will have made a threefold violating $NC_{n}$. Is such a thing feasible to carry out? | |
Nov 14, 2013 at 9:30 | comment | added | Olivier Benoist | @jacob : Jason's and Sándor's examples are indeed not schemes, but algebraic spaces (equivalently, if working over $\mathbb{C}$, compact Moishezon spaces). As for your question in the "smooth proper scheme" setting, toric varieties are not going to provide counterexamples either, as every pair of points have a common affine neighbourhood. I have not been able to make up my mind what the answer to your question should be: do you have a guess ? | |
Nov 14, 2013 at 8:58 | comment | added | jacob | Thaanks for your answer! I'm a little confused though: Can you just blow down curves and still get a scheme? your example can't work too well, because say X is a projective threefold with picard group Z and I blow down some curve C to get a scheme Y, where the image of C is some point P. Now take any surface on Y avoiding P (you can certainly do this, just take an affine open around P, take a surface there and then close up). Pulling this back to X gives you a contradiction. I realize of course I'm just doing your argument backwards! But doesn't this show that you can't contract such curves? | |
Nov 12, 2013 at 17:21 | history | edited | Sándor Kovács | CC BY-SA 3.0 |
added 598 characters in body
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Nov 12, 2013 at 8:00 | history | answered | Sándor Kovács | CC BY-SA 3.0 |