Timeline for Trigonometry / Euclidean Geometry for natural numbers?
Current License: CC BY-SA 4.0
31 events
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| Oct 7, 2020 at 9:29 | answer | added | mathoverflowUser | timeline score: 0 | |
| Feb 7, 2020 at 9:26 | answer | added | user6671 | timeline score: 2 | |
| Oct 13, 2019 at 16:16 | answer | added | user6671 | timeline score: 4 | |
| Oct 12, 2019 at 6:33 | vote | accept | CommunityBot | ||
| Oct 12, 2019 at 5:50 | answer | added | user142929 | timeline score: 4 | |
| Oct 11, 2019 at 20:18 | answer | added | Rodrigo | timeline score: 8 | |
| Oct 11, 2019 at 2:23 | comment | added | Tim Campion | Ah, I see, my mistake. The cognitive dissonance is high, but the Nash embedding theorem concerns a weaker type of "isometry" which only makes sense for manifolds. | |
| Oct 11, 2019 at 2:14 | comment | added | user6671 | @TimCampion with the law of cosines | |
| Oct 11, 2019 at 2:13 | comment | added | Tim Campion | @orgesleka That's helpful to see where you're coming from, but it doesn't address my question. Just because the same word "flat" is used doesn't mean that the space is "flat" like a flat manifold. For example, spherical space and hyperbolic space both embed isometrically into Euclidean space and yet violate the equation $\alpha + \beta + \gamma = \pi$ for all triangles. And I still don't understand how you're computing angles. | |
| Oct 11, 2019 at 2:07 | comment | added | user6671 | @TimCampion I think this might adress your question? mathoverflow.net/a/12409/6671 | |
| Oct 11, 2019 at 0:20 | comment | added | Tim Campion | Concerning Edit 2, where does this come from? I don't understand how you're computing angles. Moreover the law $\alpha + \beta + \gamma = \pi$ only holds in flat space. Even if this metric space is isometrically embeddable in Euclidean space, it need not be flat. For instance, every manifold (regardless of its curvature) embeds isometrically in Euclidean space by Nash. | |
| Oct 10, 2019 at 16:03 | history | edited | user6671 |
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| Oct 10, 2019 at 13:46 | history | edited | user6671 | CC BY-SA 4.0 |
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| Oct 10, 2019 at 13:34 | history | edited | user6671 | CC BY-SA 4.0 |
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| Oct 10, 2019 at 11:38 | answer | added | Anixx | timeline score: 4 | |
| Oct 10, 2019 at 11:29 | history | edited | user6671 | CC BY-SA 4.0 |
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| Oct 9, 2019 at 18:57 | history | edited | user6671 | CC BY-SA 4.0 |
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| Oct 9, 2019 at 16:51 | history | edited | user6671 | CC BY-SA 4.0 |
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| Oct 9, 2019 at 16:38 | history | edited | user6671 | CC BY-SA 4.0 |
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| Oct 9, 2019 at 14:42 | comment | added | user6671 | @GreginGre: Thanks for your comment. I don't think so. See the related MSE question. There are metrics $d$ on the natural numbers which do not embedd in euclidean space hence the corresponding matrix is not semidefinite. For example $a_{ij} = d_{Cos}(i,j)$ satisfies your requirement, but there are counterexamples where the correspoinding matrix is not positive semidefinite. | |
| Oct 9, 2019 at 14:34 | comment | added | GreginGre | Maybe more genrally, you could examine the following question: Assume that $0\leq a_{ij}\leq 1$ , $a_{ij}=a_{ji}$, $a_{ii}=0$, for all $0\leq i,j\leq n$, and assume that $a_{ij}\leq a_{0i}+a_{0j}$ for all $0\leq i,j\leq n$. Is the symmetric matrix $M=(a_{0i}^2+a_{0j}^2-a_{ij}^2)_{1\leq i,j\leq n}$ positive semidefinite ? | |
| Oct 9, 2019 at 10:08 | comment | added | user6671 | @GreginGre: Not yet. But I have done some computer experiments for different (random) and not random sets, and the matrix was always positive semidefinite. Thanks for your suggestion. I will try this out. | |
| Oct 9, 2019 at 10:04 | comment | added | GreginGre | Ok, i thought so, but wanted to be sure. The easiest case seems to take $x_0=1$ and $x_1,x_2,x_3$ pairwise coprime. Have you tried to write down the $4\times 4$ matrix and see if it is positive semidefinite ? | |
| Oct 9, 2019 at 9:54 | history | edited | user6671 | CC BY-SA 4.0 |
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| Oct 9, 2019 at 9:54 | comment | added | user6671 | With "natural numbers", I mean $\mathbb{N}$ without $0$. Sorry for not being clear on this point. | |
| Oct 9, 2019 at 9:53 | comment | added | GreginGre | Ok. How do you define your metric if $a=0$ ? | |
| Oct 9, 2019 at 9:44 | comment | added | user6671 | @GreginGre: Yes it is a metric: mathoverflow.net/a/343061/6671 | |
| Oct 9, 2019 at 9:42 | comment | added | GreginGre | Do you have a proof that your $d$ is indeed a metric ? | |
| Oct 9, 2019 at 8:15 | history | edited | user6671 | CC BY-SA 4.0 |
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| Oct 9, 2019 at 8:00 | history | edited | user6671 |
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| Oct 9, 2019 at 7:52 | history | asked | user6671 | CC BY-SA 4.0 |