Timeline for Continuity in terms of lines
Current License: CC BY-SA 2.5
17 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
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| Dec 17, 2012 at 6:34 | history | edited | Rodrigo A. Pérez |
edited tags
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| Nov 22, 2010 at 3:11 | comment | added | Gerald Edgar | This sort of map is called a collineation en.wikipedia.org/wiki/Collineation | |
| Nov 21, 2010 at 21:16 | comment | added | Kevin Buzzard | @Willie: Qing Liu's answer (which just appeared at my end) says that conversely this trick of taking field automorphisms is basically "the only trick there is" other than using affine translations. | |
| Nov 21, 2010 at 21:14 | comment | added | Kevin Buzzard | @Willie: I'm not entirely sure what you're asking. Think about it this way. Say there was a crazy discontinuous field automorphism $\sigma$ from the reals to itself. It's not hard to convince yourself (unless I made a slip) that applying $\sigma$ coordinatewise to $\mathbb{R}^n$ would give the desired map. Unfortunately no such $\sigma$ exists. However the complexes have plenty of discontinuous automorphisms, because as an abstract field the complexes are just the rationals, adjoin (size of complexes) many independent transcendentals, and then take the alg closure. | |
| Nov 21, 2010 at 21:09 | vote | accept | trutheality | ||
| Nov 21, 2010 at 21:01 | answer | added | Qing Liu | timeline score: 22 | |
| Nov 21, 2010 at 20:36 | comment | added | Willie Wong | @Kevin: is there a more detailed description written down somewhere in the case of $\mathbb{C}$ which you just touched on? Even that is not obvious to me. | |
| Nov 21, 2010 at 20:27 | comment | added | Kevin Buzzard | For what it's worth, if we place all $\mathbf{R}$s by $\mathbf{C}$s, and consider maps from $\mathbf{C}^2$ to itself sending every affine line (something of the form $a+b\lambda$ for $a,b\in\mathbf{C}^2$ and $\lambda$ varying in $\mathbf{C}$) to an affine line, then (assuming some weak form of the axiom of choice) one can choose a discontinuous field automorphism of $\mathbf{C}$ and apply it coordinatewise to get a discontinuous map sending affine lines to affine lines. Unfortunately this automorphism must move the reals to somewhere else so can't be used in this form for the problem at hand. | |
| Nov 21, 2010 at 20:26 | comment | added | trutheality | I think that the standard terminology is that lines go off to infinity, while [0,1] and (0,1) would be segments. | |
| Nov 21, 2010 at 20:23 | comment | added | Kevin Buzzard | I'm sure it's standard---but what is a "line"? Is [0,1] or (0,1) a line in $\mathbf{R}$, or does a line go off to infinity in both directions? | |
| Nov 21, 2010 at 20:19 | history | edited | trutheality | CC BY-SA 2.5 |
added 3 characters in body
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| Nov 21, 2010 at 20:19 | comment | added | Andy Putman | Ah, I see. That is an interesting question! | |
| Nov 21, 2010 at 20:17 | comment | added | Willie Wong | Take the second meaning, and restrict to $n > 1$, please. The original question at math.SE is about $\mathbb{R}^2$. | |
| Nov 21, 2010 at 20:15 | history | edited | Willie Wong | CC BY-SA 2.5 |
Added trivial restriction.
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| Nov 21, 2010 at 20:15 | comment | added | Andy Putman | What do you mean by "every line is mapped to a line"? Do you mean that the restriction of $f$ to any line is linear and injective, or do you just mean that the image of every line is a line? If you mean the latter, then it is easy to construct counterexamples even for $n=1$. | |
| Nov 21, 2010 at 20:11 | history | asked | trutheality | CC BY-SA 2.5 |