Timeline for The intersection multiplicity of the canonical divisor of a surface with a fibre of a map to a curve
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 2, 2013 at 1:37 | comment | added | Donu Arapura | $F$ and $G$ are algebraically equivalent by definition, but you can also use the argument you mention to show that $F\cdot D=G\cdot D$. | |
| Jan 1, 2013 at 23:56 | vote | accept | Matt Grimes | ||
| Jan 1, 2013 at 23:47 | comment | added | Matt Grimes | Thank you for the response. To see that $F$ and $G$ are algebraically equivalent, do we just observe that they are fibres over two points, which are topologically equivalent, and so the Chern classes of $F$ and $G$ agree? | |
| Jan 1, 2013 at 22:38 | comment | added | Donu Arapura | The general fibre, say $G$, and $F$ are algebraically equivalent. Therefore $F\cdot D= G\cdot D$ for any divisor $D$, in particular for $D=K$. Since $G\cong \mathbb{P}^1$, we can use the adjunction formula $G\cdot K=G^2+G\cdot K=2(0)-2$. | |
| Jan 1, 2013 at 21:27 | answer | added | Jonathan | timeline score: -1 | |
| Jan 1, 2013 at 21:19 | history | asked | Matt Grimes | CC BY-SA 3.0 |