Timeline for Rational curves on varieties of general type
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| May 17, 2016 at 12:49 | history | edited | Francesco Polizzi | CC BY-SA 3.0 |
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| Aug 10, 2012 at 5:34 | comment | added | sanokun | I have not read Bogomolov's paper, but assuming the condition on the chern numbers on surfaces of general type, can you find the finitely many rational curves explicitly? Is there any explicit description? | |
| Jan 27, 2011 at 19:27 | comment | added | Tong | @inkspot: What you mean is that there ARE only finitely many smooth rational curve on surfaces of general type? | |
| Jan 19, 2011 at 20:02 | comment | added | inkspot | For smooth rational $C$ on a minimal non-ruled complex surface $X$ the Bogomolov-Miyaoka-Yau inequality $c_1^2\le 3c_2$ for the logarithmic cotangent bundle $\Omega^1_X(\log C)$ gives $K_X^2-3c_2(X)+2\le C^2$, so there are only finitely many such $C$ on $X$. But this says nothing about their (in)dependence in the Neron-Severi group. | |
| Jan 18, 2011 at 21:42 | vote | accept | Tong | ||
| Jan 18, 2011 at 19:45 | comment | added | Francesco Polizzi | I think you miscalculated. The genus formula actually yields $KC=-2-C^2 \geq 0$, so $C$ is not a component of $|K|$ in general. | |
| Jan 18, 2011 at 19:42 | comment | added | J.C. Ottem | I don't get this question. If $C^2<0$ then by the genus formula $C.K<-2$ and so C is a fixed component in $|K|$? Surely, there must be finitely many such curves (if any). What does this have to do with the Picard number? | |
| Jan 18, 2011 at 19:41 | comment | added | Francesco Polizzi | ...So Bogomolov's theorem is really a non-trivial result. At any rate, if hyperbolicity conjecture is true for surfaces of general type, then one always has only finitely many rational curves (i.e. the numerical condition on $c_1^2$ and $c_2$ would be superflous). | |
| Jan 18, 2011 at 19:39 | comment | added | Francesco Polizzi | If $C$ is a smooth rational curve on a minimal surface of general type, then $C^2 \leq -2$ by the genus formula. The rank of NS(S) is always finite, but in order to prove finiteness you should also prove that the rational curves are independent in the Neron Severi. This is easy if $C^2$b is bounded from below; but if you allows $C^2$ to vary, than a priori you could have countably many smooth rational curves... | |
| Jan 18, 2011 at 19:25 | comment | added | Tong | I think it is. It is the Picard number. | |
| Jan 18, 2011 at 19:07 | comment | added | Tong | Great answer! Thanks! Well, if only assuming that any of these smooth rational curve has negetive intersection number without the condition on $c_1$ and $c_2$, can I say that the number is finite using the finiteness of the rank of Neron Severi group? | |
| Jan 18, 2011 at 19:01 | vote | accept | Tong | ||
| Jan 18, 2011 at 21:42 | |||||
| Jan 18, 2011 at 19:01 | vote | accept | Tong | ||
| Jan 18, 2011 at 19:01 | |||||
| Jan 18, 2011 at 18:37 | history | edited | Francesco Polizzi | CC BY-SA 2.5 |
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| Jan 18, 2011 at 18:30 | history | answered | Francesco Polizzi | CC BY-SA 2.5 |