Timeline for To what extent does the branch locus determine the covering (Chisini's conjecture)?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Dec 26, 2013 at 14:12 | answer | added | Jason Starr | timeline score: 2 | |
| Dec 23, 2013 at 21:32 | comment | added | Jason Starr | You are right. My covering is not simple: the branch divisor of $h$ shows up multiply in the branch divisor of $f$. | |
| Dec 23, 2013 at 19:06 | comment | added | Serge Lvovski | And yes, I'm interested in positive results, and I am not sure the suggested construction will yield counterexamples. | |
| Dec 23, 2013 at 19:04 | comment | added | Serge Lvovski | @Jason: if the surface $X$ is irreducuble, then the homomorphism $\pi_1(\mathbb P^2\setminus S,p)\to \mathrm{Aut}(f^{-1}(p))$ is automatically surjective. | |
| Dec 23, 2013 at 15:48 | comment | added | Jason Starr | Just to clarify: you are interested in positive results, not in negative results, correct? It seems to me that one can produce negative examples as a composition $f=(g_1\times g_2)\circ h$ of a degree $2$ morphism $h:\mathbb{P}^1\times \mathbb{P}^1\to \mathbb{P}^2$ with a product morphism $g_1\times g_2:C_1\times C_2 \to \mathbb{P}^1\times \mathbb{P}^1$, where $g_i:C_i\to \mathbb{P}^1$ is a simple branched cover. Did you want to add a hypothesis that $\pi_1(\mathbb{P}^2\setminus S, p) \to \text{Aut}(f^{-1}\{p\})$ is surjective? | |
| Dec 23, 2013 at 12:17 | history | asked | Serge Lvovski | CC BY-SA 3.0 |