Timeline for canonical divisor of a contraction
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 26, 2020 at 19:05 | answer | added | cgodfrey | timeline score: 2 | |
| Jun 26, 2020 at 18:57 | comment | added | cgodfrey | Related to the comment of @Pop: If $\mathrm{codim}(Y \subset X) > 1$ then $Z$ can't be $\mathbb{Q}$-factorial, but $K_Z$ could still be $\mathbb{Q}$-Cartier, for instance if $Z = C(\mathbb{P}^1 \times \mathbb{P}^1)$ (this is a hypersurface so $K_Z$ is Cartier) and $X = \mathrm{Bl}_{C(0 \times \mathbb{P}^1)} Z$ ($X$ is smooth, since the preimage of $C(0 \times \mathbb{P}^1)$ is a smooth Cartier divisor on X, and the exceptional set is a $\mathbb{P}^1$). | |
| Jun 26, 2020 at 13:39 | comment | added | user130022 | By K-negative did you mean that anti canonical is ample ? How to prove that for K-negative contraction with co-dim at least $2$ contracted locus, the canonical divisor is not Q-Cartier ? | |
| Jun 26, 2020 at 9:40 | comment | added | Pop | One has to be a little careful with terminology. It is possible to contract a subvariety of codimension 2 to get $Z$ with $\mathbf Q$-Cartier, even Cartier, canonical divisor. The basic example is the Atiyah flop: here the base of the contraction is an ordinary double point, which is Gorenstein (in particular CM). However, sometimes people use contraction to mean exclusively K-negative contraction: for a $K$-negative contraction (or any $K$-nontrival contraction), if the contracted locus has codimension 2 or more, it's true that the canonical divisor of the base is not $\mathbf Q$-Cartier. | |
| Jun 26, 2020 at 5:28 | comment | added | user130022 | is it true that if $Y$ is of codimension at least $2$ then $Z$ is not Cohen-Macualay ? | |
| Jun 25, 2020 at 17:09 | review | Close votes | |||
| Jun 29, 2020 at 1:47 | |||||
| Jun 25, 2020 at 15:52 | comment | added | Devlin Mallory | The normal thing to do is to compare $K_X$ and the pullback of $K_Z$ (their difference being the discrepancy divisor), but if the codimension of $Y$ is greater than 1, then $K_Z$ fails to be ($\mathbb Q$)-Cartier, so you can't pull back $K_Z$ (at least not in the usual sense). If you're interested in the case where $Z$ is not a divisor, then, it's maybe unclear how one should compare the two canonical divisors. | |
| Jun 25, 2020 at 14:00 | history | asked | user130022 | CC BY-SA 4.0 |