Timeline for Rational congruence of binomial coefficient matrices
Current License: CC BY-SA 2.5
15 events
| when toggle format | what | by | license | comment | |
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| Jun 3, 2010 at 5:54 | comment | added | Wadim Zudilin | Skip, thanks for the reference. As far as I understand, the elementary derivation of the rational equivalence implies some other results in the paper. | |
| Jun 2, 2010 at 15:14 | comment | added | Skip | The machinery proof is in "Orthogonal representations of twisted forms of $SL_2$", Representation Theory, vol. 12 (2008), 435-446. A preprint version is at arxiv.org/abs/math/0702731 Published version at dx.doi.org/10.1090/S1088-4165-08-00335-X | |
| Jun 1, 2010 at 21:24 | vote | accept | Timothy Chow | ||
| Jun 1, 2010 at 16:38 | answer | added | fherzig | timeline score: 3 | |
| Jun 1, 2010 at 10:54 | answer | added | darij grinberg | timeline score: 10 | |
| Jun 1, 2010 at 1:03 | answer | added | Wadim Zudilin | timeline score: 5 | |
| May 31, 2010 at 17:39 | answer | added | Will Jagy | timeline score: 2 | |
| May 31, 2010 at 4:34 | comment | added | Will Jagy | Thank you, Timothy. I was way off. So for n=8 we have A=diag(1,28) and then B=diag(8,56). | |
| May 31, 2010 at 3:23 | comment | added | Timothy Chow | @Will Jagy: For n = 2, A = B = diag(1). For n = 4, A = diag(1) and B = diag(4). For n = 6, A = diag(1, 15) and B = diag(6, 10). In general if 4|n then the dimension is n/4 and if n = 2 mod 4 then the dimension is (n+2)/4. | |
| May 31, 2010 at 2:33 | comment | added | Pete L. Clark | Indeed, it is necessary and sufficient that the discriminants are equal, the signatures are equal, and for all primes $p$ the Hasse-Witt invariants at $p$ are equal. For two particular quadratic forms $A$ and $B$, there are certainly algorithms to compute all these invariants. For an infinite family as in this case, more work and/or cleverness may be required. | |
| May 31, 2010 at 2:27 | history | edited | Pete L. Clark | CC BY-SA 2.5 |
added 12 characters in body
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| May 31, 2010 at 2:27 | comment | added | fherzig | It's not sufficient: the question asks about two rational quadratic forms to be isometric. The square class of the determinant is the discriminant of the quad. form, but there are other invariants. Ex.: diag(3,3) not isometric to diag(1,1). The main theorem on rational quadratic forms is Hasse-Minkowski (see wikipedia). | |
| May 31, 2010 at 2:20 | comment | added | Peter Shor | Is anything known about the rational congruence of matrices? Clearly, for two diagonal matrices to be rationally congruent, their determinants must be equal up to multiplication by a rational square. Is it possible that this is sufficient? | |
| May 31, 2010 at 0:53 | comment | added | Wadim Zudilin | Sounds nice: the corresponding diagonal quadratic forms are rationally equivalent. Reminds me a little of mathoverflow.net/questions/22399. Is there a link to the "high-powered machinery" proof? | |
| May 31, 2010 at 0:28 | history | asked | Timothy Chow | CC BY-SA 2.5 |