Timeline for homology class of an algebraic curve versus homology class of a genus zero holomorphic curve
Current License: CC BY-SA 3.0
12 events
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| May 24, 2013 at 20:09 | comment | added | Sándor Kovács | For smooth projective curves being birational is equivalent to being isomorphic. Take the normalization of $C$. If $C$ is rational, then its normalization is $\mathbb CP^1$. | |
| May 24, 2013 at 18:42 | comment | added | user15512 | Thank you. By definition, $C$ is rational if and only if there exists a birational transformation $f$ from $\mathbb{C}P^1$ to $C$. ``$f$ is birational'' is equivalent to the fact that $f$ is an isomorphism between an open set in $\mathbb{C}P^1$ and an open set in $C$. Why is $f$ defined on the entire $\mathbb{C}P^1$ [as opposed to a smaller Zariski-open subset]? | |
| May 24, 2013 at 5:34 | comment | added | Sándor Kovács | The answer to what you are asking is trivial: the map that makes $C$ rational is a holomorphic map from $\mathbb CP^1$ to $X$ that satisfies what you ask for. The non-trivial question is whether this holds for any $C$, that is, including non-rational ones. In the case when $X$ is Fano, then this actually happens, but that's a highly non-trivial fact. It follows from Bend & Break, but even that is non-trivial. | |
| May 23, 2013 at 20:03 | vote | accept | user15512 | ||
| May 23, 2013 at 20:03 | comment | added | user15512 | Thank you for the explanations. Assume that $C$ is a rational curve in $X$ [i.e. birational to $\mathbb{C}P^1$]. Why is it possible to find holomorphic curves $u_1,\ldots,u_k:\mathbb{C}P^1\longrightarrow X$ such that $$[C]=(u_1)_*[\mathbb{C}P^1]+\ldots+(u_k)_*[\mathbb{C}P^1]?$$ [in $H_2(X;\mathbb{Q}$] | |
| Jun 15, 2011 at 11:40 | comment | added | diverietti | Of course! Anyway, it seemed to me that the fact that generic projective hypersurfaces of high degree are "algebraically hyperbolic" (ie they do not contain any subvariety not of general type) was sufficient for your aims. | |
| Jun 15, 2011 at 10:31 | history | edited | Sándor Kovács | CC BY-SA 3.0 |
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| Jun 15, 2011 at 10:27 | comment | added | Sándor Kovács | Hi Simone, you are absolutely correct and I actually knew that. I don't know why I put it that way. Laziness, probably. Anyway, to make the point one does not need to know the Kobayashi conjecture, just that there exist Brody hyperbolic spaces and that is certainly true. | |
| Jun 15, 2011 at 10:25 | history | edited | Sándor Kovács | CC BY-SA 3.0 |
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| Jun 15, 2011 at 8:49 | comment | added | diverietti | Hi Sándor, in fact it is not known whether a general projective hypersurface of high degree is Brody hyperbolic or not, as soon as its dimension is at least four. On the other hand it is conjecturally true (Kobayashi, '70). What is known is that they do not contain any subvariety not of general type. | |
| Jun 15, 2011 at 6:45 | history | edited | Sándor Kovács | CC BY-SA 3.0 |
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| Jun 1, 2011 at 6:07 | history | answered | Sándor Kovács | CC BY-SA 3.0 |