Timeline for Resolution of conical singularities have even-only cohomology?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Nov 15, 2022 at 13:02 | vote | accept | Filip | ||
| Nov 9, 2022 at 14:04 | comment | added | Filip | I am not sure whether your example can be a conical singularity and satisfying $c_1(X)=0$ for the resolution. Anyway, we have found examples satisfying these further conditions as well, so the question is answered (below). | |
| Nov 9, 2022 at 13:32 | history | edited | Filip | CC BY-SA 4.0 |
added 22 characters in body
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| Nov 9, 2022 at 12:52 | comment | added | Piotr Achinger | Trivially this cannot hold (in dimension $>2$) for all resolutions, even if $Y$ is smooth: take any resolution $X'\to Y$, find a smooth curve $C\subseteq X'$ with positive genus contained in the exceptional divisor, and blow it up. The resulting $X$ is smooth and has nonzero $H^3$ by the blowup formula, and the composition $X\to Y$ is a resolution too. | |
| Nov 9, 2022 at 11:58 | comment | added | Filip | Ok, I have written proof of this counterexample in full details now, so that the question does not remain without an official answer. Just to say, you actually need $mH=-K$ instead. | |
| Nov 8, 2022 at 20:31 | answer | added | Filip | timeline score: 2 | |
| Nov 8, 2022 at 19:40 | comment | added | Yosemite Stan | Yes, that should be right. It would be even easier to look at the cone over the quartic threefold because that will live in A^5 | |
| Nov 8, 2022 at 19:17 | comment | added | Filip | This sounds good, thanks! Just to make sure that I got this correct, in the example when $Z\subset \mathbb{C} P^4$ is a cubic threefold, thus $c_1(Z)=2,$ then $X \subset \mathbb{C}^{16}$ can be written as the cone over the image of $Z$ under the morphism $$\mathbb{C} P^4 \rightarrow \mathbb{C} P^{15}, [z_0: z_1 : \dots : z_4]\mapsto [z_0^2:z_0 z_1: \dots :z_4^2].$$ Here I use degree-2 polynomials as $c_1(Z)=2.$ Am I right? | |
| Nov 8, 2022 at 18:00 | comment | added | Yosemite Stan | What about the following. Consider a cubic threefold Z (or some Fano manifold with an embedding given by H such that mH=K for some m>0). Let X be the cone over the |mH|-embedding of X. Then X is conic. It has a resolution given by the total space Y=L of the line bundle O(-mH). The canonical bundle of Y is the pull back of K of X times the relative canonical which is -K (so c_1=0). In the case Z is a cubic threefold, Z has odd cohomology. The line bundle Y is homotopy equivalent to Z. So Y has odd cohomology. | |
| Nov 8, 2022 at 17:22 | comment | added | Filip | That's a fair point -- just added it, thanks! | |
| Nov 8, 2022 at 17:21 | history | edited | Filip | CC BY-SA 4.0 |
edited after comment by @Will Sawin
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| Nov 8, 2022 at 17:17 | comment | added | Will Sawin | If one doesn't demand that $X$ is conic, there will be no reasonable local condition that implies this, as $X$ and $Y$ can have rational cohomology in odd degrees for global reasons, e.g. when $X$ is a smooth affine variety. | |
| Nov 8, 2022 at 17:14 | history | asked | Filip | CC BY-SA 4.0 |