Timeline for Rational points on a sphere in $\mathbb{R}^d$
Current License: CC BY-SA 3.0
32 events
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| May 5, 2017 at 13:00 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
"on" vs. "in."
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| May 5, 2017 at 11:24 | history | edited | Qfwfq | CC BY-SA 3.0 |
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| May 5, 2017 at 11:01 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Image link broken; now fixed.
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| Jan 26, 2017 at 7:50 | comment | added | user62562 | If you replace the sphere by a different algebraic set, for example the unit cylinder, do you still have such a nice theory? | |
| Aug 21, 2014 at 1:27 | comment | added | P Vanchinathan | If the radius $r$ is an integer express it as a sum of squares $r= \sum_{i=1}^k x_i^2 $ with $1\leq k \leq 4$. If the ambient dimension is $d$ then we can find ${d\choose k}$ integer points from the above expression and they should be affinely independent as required in Misha's comment above. | |
| Aug 21, 2014 at 1:26 | history | edited | Ricardo Andrade | CC BY-SA 3.0 |
removed deprecated tag 'geometry'
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| Sep 8, 2013 at 0:30 | vote | accept | Joseph O'Rourke | ||
| Sep 8, 2013 at 0:30 | vote | accept | Joseph O'Rourke | ||
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| Sep 8, 2013 at 0:29 | vote | accept | Joseph O'Rourke | ||
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| Jul 17, 2013 at 22:37 | comment | added | stan wagon | Yes, I can confirm that Klee/Wagon took an overly complicated approach. The projection idea works well, and I found today that I needed to know this fact about the sphere for another project! | |
| Apr 9, 2013 at 19:56 | answer | added | D.Kleinbock | timeline score: 12 | |
| Mar 27, 2013 at 17:56 | answer | added | Tim Browning | timeline score: 11 | |
| Mar 25, 2013 at 9:16 | comment | added | François Brunault | Are the spheres of Problem 10.8 centered at the origin? If so then unless I'm missing something, a necessary and sufficient condition is that $\rho^2$ should be a sum of $d$ squares. Indeed if there is a rational point then applying $O(n,\mathbf{Q})$ which is dense in $O(n,\mathbf{R})$ you get a dense set of rational points. | |
| Mar 23, 2013 at 17:16 | answer | added | Stefan Kohl♦ | timeline score: 33 | |
| Mar 23, 2013 at 13:19 | comment | added | Joseph O'Rourke | Thank you, Misha, for explaining this connection to Moebius transformations so clearly. | |
| Mar 23, 2013 at 12:04 | comment | added | Misha | Joseph: One more thing. A sphere contains dense set of rational points if and only if it contains a finite set of rational points whose affine span is the entire affine space. Proving this is an exercise in using rational Moebius transformations, generalizing the argument with stereographic projection which is a restriction of such a Moebius transformation. I guess, Klee and Wagon did not know about this group (which is PO(n,1; Q)). | |
| Mar 23, 2013 at 11:45 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
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| Mar 23, 2013 at 11:05 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
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| Mar 23, 2013 at 1:43 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
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| Mar 23, 2013 at 1:28 | comment | added | Misha | Joseph: Of course the answer depends on the radius. In order for a rational point to exist the radius has to be in a quadratic extension of Q. | |
| Mar 23, 2013 at 0:36 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
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| Mar 22, 2013 at 18:12 | answer | added | Daniel m3 | timeline score: 12 | |
| Mar 22, 2013 at 18:03 | comment | added | Joseph O'Rourke | Thanks, everyone, for the answers. The connection to stereographic projections is beautiful! I found a paper that addresses in some form my second question, "Rational Points on the Unit Sphere," by Eric Schmutz, PDF here: math.drexel.edu/~eschmutz/PAPERS/cejm.pdf . (He also points out (p.2) that the inverse of the stereographic projection maps rational points to rational points.) | |
| Mar 22, 2013 at 12:06 | comment | added | R.P. | I don't understand why Peter Michor's comment got two upvotes. If a quadric hypersurface $X \subset \mathbf{A}^{n+1}$ defined over $\mathbf{Q}$ has a point $P \in X(\mathbf{Q})$, then projecting away from $P$ gives a birational map $X \stackrel{\sim}{\dashrightarrow} \mathbf{A}^n$ that is defined over $\mathbf{Q}$. Restricting this birational map gives an isomorphism between open subsets of $X$ and $\mathbf{A}^n$ that is defined over $\mathbf{Q}$. In particular, the rational point son $X$ are dense in the real locus of $X$ iff the same holds for $\mathbf{A}^n$, which is trivially the case. | |
| Mar 22, 2013 at 11:52 | comment | added | Tom Goodwillie | The inverse of stereographic projection does map rational points to rational points. | |
| Mar 22, 2013 at 11:31 | history | edited | Joseph O'Rourke |
Added number theory, reflecting responses so far.
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| Mar 22, 2013 at 8:31 | comment | added | François Brunault | Does this answer your first question? mathoverflow.net/questions/90070/… | |
| Mar 22, 2013 at 8:08 | answer | added | Peter Michor | timeline score: 7 | |
| Mar 22, 2013 at 2:14 | comment | added | Jérémy Blanc | Yes. And the formula of the stereographic projection is not more complicated in high dimension, so this also answers to question 2. | |
| Mar 22, 2013 at 1:48 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
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| Mar 22, 2013 at 1:39 | comment | added | Tom Goodwillie | Stereographic projection makes rational points on the sphere correspond to rational points in $\mathbb R^{d-1}$. | |
| Mar 22, 2013 at 1:34 | history | asked | Joseph O'Rourke | CC BY-SA 3.0 |