Timeline for How to embed genus 4 surface inside $\mathbb{C}P^2\# \mathbb{C}P^2$ representing nontrivial homology class
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| S May 18, 2014 at 22:42 | history | suggested | Marco Golla |
Removed the complex-geometry tag, added the 4-manifolds tag instead.
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| May 18, 2014 at 22:20 | vote | accept | Agustín Moreno | ||
| May 18, 2014 at 22:13 | review | Suggested edits | |||
| S May 18, 2014 at 22:42 | |||||
| May 18, 2014 at 22:11 | answer | added | Marco Golla | timeline score: 6 | |
| May 18, 2014 at 21:25 | comment | added | Igor Rivin | @MarcoGolla please go ahead (and if you do, please do elaborate on your second comment/answer). | |
| May 18, 2014 at 21:00 | comment | added | Marco Golla | @IgorRivin: I was sure that was what you meant. Should one of us go ahead and post an answer? | |
| May 18, 2014 at 20:04 | comment | added | Igor Rivin | @MarcoGolla that's precisely what I had in mind (adding handles), hence the question :) | |
| May 18, 2014 at 19:01 | comment | added | Marco Golla | You can actually also take a cubic and a quartic, and take their embedded connected sum. This would represent the class $(3,4)$ and, by the degree-genus formula, have genus $1+3$. | |
| May 18, 2014 at 18:31 | comment | added | Marco Golla | Then you can just take any homologically nontrivial surface of genus $g<2$ and add trivial handles. For example, you can take a complex line in $\mathbb{CP}^2$ and add two handles contained in a small 4-ball (if you want to be more formal, you're taking a connected sum (of pairs) with a genus-2 surface in $S^4$). The problem is not with increasing the genus, but rather with decreasing it. | |
| May 18, 2014 at 17:57 | history | edited | Agustín Moreno | CC BY-SA 3.0 |
edited body
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| May 18, 2014 at 17:42 | comment | added | Agustín Moreno | Just a topological surface. I tried the algebraic curve thing because I thought I could handle homology that way, but maybe there is some other approach. | |
| May 18, 2014 at 16:50 | comment | added | Igor Rivin | You just want a topological surface, or an algebraic curve? | |
| May 18, 2014 at 15:31 | comment | added | Agustín Moreno | I think I managed to convince myself of the fact that a degree $d$ curve represents $d[H]$: I know that if $D=Z(s)$ for a degree $d$ homogeneous polynomial $s$, then the homology class $[Z(s)]$ is the Poincare dual of the first Chern class of the twisted bundle $\mathcal{O}(D)= \mathcal{O}(d)$, that is, we have $$c_1(\mathcal{O}(D))=c_1(\mathcal{O}(d))=d$$ and hence by applying Poincare duality again I get $[Z(s)]=d[H]$. Am I right? | |
| May 18, 2014 at 15:28 | review | First posts | |||
| May 18, 2014 at 15:51 | |||||
| May 18, 2014 at 15:12 | history | asked | Agustín Moreno | CC BY-SA 3.0 |