Timeline for Pushforward of a log canonical pair
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15 events
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| Jul 31, 2014 at 12:47 | comment | added | Karl Schwede | I don't think people typically define $\omega_X = \mu_* \O_{X'}(K_{X'})$. That is not a sheaf which corresponds to a divisor, it does not coincide with the $\text{Ext}$ definition. It does satisfy various Kodaira-type vanishing theorems though. I would NOT use that definition as that of the canonical sheaf (it is really a early version of the multiplier ideal). Hopefully this helps. | |
| Jul 31, 2014 at 6:09 | comment | added | Li Yutong | @KarlSchwede Sorry, I realized that in the definition of $\omega_X$, it is written to be $\omega_X: = \mu_* \mathcal{O}_{X'}(K_X)$ for a resolution $\mu: X' \to X$, then does this consistent with the usual definition using $Ext$? (Because my version of the book "Positivity of algebraic geometry" is a draft version, I am not sure if this is the correct definition... ) | |
| Jul 31, 2014 at 6:02 | vote | accept | Li Yutong | ||
| Jul 31, 2014 at 6:00 | comment | added | Li Yutong | Moreover, I had a stupid question: if $D$ is a Weil divisor on $X$, then $\mathcal{O}(D)$ is defined to be the subsheaf of $K(X)$ by $\mathcal{O}(D)(U)= \{f \mid (f)|_U+D|_U \geq 0, f\in K(X)\}$ (where $(f)$ evaluated with respect to the structure sheaf of $X$). Is that correct? | |
| Jul 31, 2014 at 5:56 | comment | added | Li Yutong | @KarlSchwede Thank you so much!! However, I am a little worry about the consistency on the definition of $\omega_X$. One the one hand, it is defined using $Ext$; on the other, as in the Example 4.3.12 in "Positivity of algebraic geometry", they define ""Grauert-Reimenschneider canonical sheaf" of $X$, by choosing any resolution $\mu: X' \to X$, and define $\omega_X = \mu_* \mathcal{O}_{X'}(K_{X'})$. If one use this definition, then it certainly satisfies $\phi_*(\omega_X) = \omega_Y$. This means the two definitions are not the same? | |
| Jul 31, 2014 at 5:06 | comment | added | Karl Schwede | Of course, there are other choice of $K_Y$ too (although each one comes from a choice of some $K_X$). It is worth emphasizing that just because $\phi_* K_X = K_Y$ does not mean that $\phi_* \omega_X = \omega_Y$ (this latter condition is basically rational singularities). | |
| Jul 31, 2014 at 5:06 | comment | added | Karl Schwede | $\phi_* K_X = K_Y$ has nothing to do with Grauert-Riemenschneider vanishing I think. The edit to the question gives one way to see it. Here is another. Both $X$ and $Y$ are normal and they coincide on an open set $U$ whose complement has codimension 2 on $Y$. $K_X$ is any divisor such that $O_X(K_X) \cong \omega_X$ and $K_Y$ is any divisor such that $O_Y(K_Y) \cong \omega_Y$. If I have a divisor $K_X$, this gives me a divisor $K_U = K_X|_U$. But since the complement of $U$ has codimension 2 on $Y$, $K_U$ uniquely determines a choice of $K_Y$ such that $\phi_* K_X = K_Y$. | |
| Jul 31, 2014 at 4:14 | history | edited | Chen Jiang | CC BY-SA 3.0 |
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| Jul 30, 2014 at 15:30 | comment | added | dhy | I usually take $\phi_*(K_X)=(K_Y)$ as part of the G-R vanishing theorem statement. Actually though, I think G-R usually has stronger conditions than the ones you assume... | |
| Jul 30, 2014 at 14:09 | comment | added | Li Yutong | @dhy Which form of the theorem do you mean, for the G-R vanishing theorem I check in "Positivity of algebraic geometry" (THM 4.3.9), it only states that $R^i\phi_*(K_X)=0$ for $i>0$. | |
| Jul 30, 2014 at 8:07 | comment | added | dhy | I think the key words here are "Grauert-Reimenschneider vanishing theorem." | |
| Jul 30, 2014 at 7:29 | comment | added | Li Yutong | Why $\phi_*(K_X)=K_Y$? Could you point out some reference? | |
| Jul 30, 2014 at 4:40 | review | Low quality posts | |||
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| Jul 30, 2014 at 4:25 | review | First posts | |||
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| Jul 30, 2014 at 4:21 | history | answered | Chen Jiang | CC BY-SA 3.0 |