Timeline for Structures that turn out to exhibit a symmetry even though their definition doesn't
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 19, 2013 at 7:56 | comment | added | Wolfgang | @NoamD.Elkies Granted. That reminds me of the relation between $\zeta(1-s)$ and $\zeta(s)$, cast as $\Xi(1-s)=\Xi(s)$ with appropriate $\Xi$. | |
| Dec 19, 2013 at 7:54 | comment | added | Wolfgang | @Matt: yes, that is exactly the point, and I guess that is also why Terry Tao's mention of Quadratic Reciprocity got so many "great comment" votes... Now if we started a thread about this kind of "simplicity", that one would be endless (not in a mathematical sense). | |
| Dec 18, 2013 at 20:25 | comment | added | Noam D. Elkies | @Wolfgang asks a fair question. To add to Matt Young's answer, we can define $s'(b,c) = s(b,c) + 1/8 - b/12c - 1/24bc$, and then the reciprocity formula says that $s'(b,c)$ is antisymmetric: $s'(b,c) = -s'(c,b)$. | |
| Dec 18, 2013 at 20:11 | comment | added | Matt Young | I think the point is that each of $s(b,c)$ and $s(c,b)$ is complicated, but once added together, one obtains an extremely simple formula. It's the simplicity of the right hand side rather than the symmetry. | |
| Dec 18, 2013 at 18:01 | comment | added | Wolfgang | But I don't understand what is so special about this, at least in terms of symmetry: for about any function $s(\cdot,\cdot)$, including the Legendre symbol, $s(b,c)+s(c,b)$ or $s(b,c)s(c,b)$ is symmetric in $b$ and $c$. Where is the surprise? | |
| S Dec 18, 2013 at 16:38 | history | answered | Noam D. Elkies | CC BY-SA 3.0 | |
| S Dec 18, 2013 at 16:38 | history | made wiki | Post Made Community Wiki by Noam D. Elkies |