Timeline for Is $\mathbb{C}^2$ homeomorphic to $\mathbb{C}^2 - (0,0)$ with the Zariski topology?
Current License: CC BY-SA 3.0
13 events
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| Mar 4, 2017 at 1:28 | answer | added | Moishe Kohan | timeline score: 15 | |
| Oct 21, 2011 at 21:53 | comment | added | Gro-Tsen |
I believe the question is equivalent to this: consider the simplicial complexes whose vertices are the (irreducible) curves in $\mathbb{C}^2$ (resp. $\mathbb{C}^2\setminus\{(0,0)\}$) with a face connecting certain curves when they have a nonempty intersection — are these two structures isomorphic? It may not be simpler that way, but it puts the emphasis on curves, and also suggests looking at it from the model-theoretic point of view: the back-and-forth argument can be phrased by saying in terms of an Ehrenfeucht-Fraïssé game between these structures. (And I'm running out of space.)
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| Oct 21, 2011 at 18:39 | comment | added | Bruno Martelli | Parallel lines in $\mathbb C^2$ also form a compact family in the Zarisky topology ... | |
| Oct 21, 2011 at 17:49 | comment | added | Greg Kuperberg | @Sasha The answer that I posted is similar to remark, but you have to do more work because you are not automatically granted the parameterization or even its topology. | |
| Oct 21, 2011 at 17:44 | comment | added | Sasha | In ${\mathbb C}^2 \setminus (0,0)$ there is a compact family of nonintersecting curves (lines through the origin). I would guess that there is no such family in ${\mathbb C}^2$. | |
| Oct 21, 2011 at 17:17 | comment | added | Tyler Lawson | @Greg Ah! Yes, of course. | |
| Oct 21, 2011 at 17:12 | comment | added | Greg Kuperberg | @Tyler I thought so or hoped so at first, but i's not that simple because there are lots of algebraic automorphisms of the affine plane, for instance $(x,y) \mapsto (x,y+p(x))$. | |
| Oct 21, 2011 at 17:11 | answer | added | Greg Kuperberg | timeline score: 6 | |
| Oct 21, 2011 at 17:08 | history | edited | Tyler Lawson |
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| Oct 21, 2011 at 17:06 | comment | added | Tyler Lawson | Can you cook up some way to distinguish the generic points of lines from those of general curves by generic intersection numbers under Bezout's theorem? Then you could say that before removing the point, "either intersecting trivially or being identical" is an equivalence relation on lines (cutting them into equivalence classes of parallel lines), whereas after removing the origin it is not (since $(y=0) \simeq (x=0) \simeq (x=1)$). | |
| Oct 21, 2011 at 16:48 | comment | added | S. Carnahan♦ | As an easier example, $\mathbb{C}^2$ is not homeomorphic to $\mathbb{CP}^2$, because any two 1-dimensional closed sets have nonempty intersection in the latter. You might be able to devise a more complicated intersection problem that obstructs homeomorphism in the question under consideration. | |
| Oct 21, 2011 at 16:04 | comment | added | Jack Huizenga | Actually $\mathbb{C}^2 - (0,0)$ is a perfectly fine object in algebraic geometry--it's just a quasiprojective variety. The non-natural bit is that merely continuous maps really play no role in algebraic geometry. Not sure on the answer to your question though. | |
| Oct 21, 2011 at 15:31 | history | asked | Steven Gubkin | CC BY-SA 3.0 |