Timeline for Adjunction formula on pair
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Jul 2, 2017 at 13:14 | vote | accept | Kevin | ||
| Jan 17, 2017 at 4:19 | comment | added | Karl Schwede | Yup, for a map of normal varieties that is a correct definition. | |
| Jan 16, 2017 at 3:11 | comment | added | user21574 | You mean f is Galois , i.e, $f$ as etale morphism is finite and the function field extension is Galois? | |
| Jan 16, 2017 at 3:04 | history | edited | user21574 | CC BY-SA 3.0 |
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| Jan 16, 2017 at 0:51 | comment | added | Karl Schwede | yes, and that is the Galois case. The ramification divisor is $0 + \infty$ on $X$. For a more complicated example, imagine the ramification divisor on a curve $X$ is $2P + Q$ for $P$ and $Q$ both mapping to the same point $A$ on $Y$ (say in char 0, tame ramification). Then there is no multiple of $A$ that pulls back to $2P + Q$ since $f^* A = 3P + 2Q$. If the map is Galois however, then if $f * A = \sum e_i P_i$ , it follows that all $e_i$ are the same (since the Galois group moves all the $P_i$ to each other). Hopefully that makes sense. | |
| Jan 15, 2017 at 21:34 | history | edited | user21574 | CC BY-SA 3.0 |
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| Jan 15, 2017 at 18:24 | comment | added | user21574 | Take for example $X=\mathbb P^1\to \mathbb P^1=Y$ with ramified at two point $p_1$ and $p_2$ then $K_X=f^*(K_Y+\frac{1}{2}p_1+\frac{1}{2}p_2)$ | |
| Jan 15, 2017 at 18:20 | comment | added | Karl Schwede | Whoops, in the second formula, that's not the ramification divisor and a divisor doesn't exist in that way generally, for that you'd want $K_Z - R = f^* K_X$ (this exists canonically for $f$ a separable finite map between normal varieties). Here $R$ is the ramification divisor. If $f$ is Galois, then such a $B$ exists (a branch divisor), but in most other cases it doesn't components of $B$ will pull back to components of $R$ with different multiplicity. | |
| Jan 15, 2017 at 17:49 | comment | added | Karl Schwede | The pair being log canonical doesn't have anything to do with the first formula. In the second formula, the divisor B is usually the ramification divisor. | |
| Jan 15, 2017 at 16:34 | history | edited | user21574 | CC BY-SA 3.0 |
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| Jan 15, 2017 at 16:25 | history | edited | user21574 | CC BY-SA 3.0 |
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| Jan 15, 2017 at 16:17 | history | answered | user21574 | CC BY-SA 3.0 |