Timeline for Rational congruence of binomial coefficient matrices
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 18, 2017 at 11:36 | history | edited | darij grinberg | CC BY-SA 3.0 |
added 53 characters in body
|
| Jun 3, 2010 at 22:51 | comment | added | darij grinberg | Yes - I meant I have no idea how to do this with our method. | |
| Jun 3, 2010 at 22:23 | comment | added | Wadim Zudilin | Darji, Skip gives in a comment above the links to his original contribution (Representation Theory 12 (2008) 435-446; arxiv.org/abs/math/0702731]) where he proves the equivalence in any char $\ne2$. | |
| Jun 3, 2010 at 11:18 | comment | added | darij grinberg | Skip: now that comes as a surprise. My proof requires char to be $0$ indeed (or at least $4n$, I believe), and at the moment I have no idea what to do against this. By the way, are the forms still equivalent over $\mathbb Z\diagup p^k\mathbb Z$ for $p>2$ prime? | |
| Jun 2, 2010 at 23:33 | comment | added | Wadim Zudilin | @Skip: I guess you are Skip G., the author of the theorem in the OP. Without Darji's solution the problem can be lost, so I would suggest that you agree on a note (at least for the arXiv); there is no work to do but to take the already existing texts. @Darji: Binomial coefficients are too interdisciplinary, I wonder whether we had an MO question "Why binomials are so important in maths". :-) | |
| Jun 2, 2010 at 15:17 | comment | added | Skip | That a great argument, Darij. One additional comment... Tim posed the problem over the rational numbers, but the theorem works over every field of characteristic $\ne 2$. (If char = 2, then you see immediately that it is wrong.) I guess your proof requires char $=0$ or $>n$. | |
| Jun 2, 2010 at 14:35 | comment | added | darij grinberg | Oh, nice, I forgot that finite differences are used in analysis all the time - but still I count them to combinatorial algebra. Well, probably finite differences belong to all of mathematics. By the way, the problem also illustrates very nicely that the $K$-vector space of all polynomials in two variables $X$ and $Y$ of degree $\leq n-1$ in each of these variables is the tensor product of the $K$-vector space of all polynomials in one variable $X$ of degree $\leq n-1$ with itself. (Together with the fact that bilinear maps on a vector space are linear maps on its tensor product with itself.) | |
| Jun 2, 2010 at 12:08 | comment | added | Wadim Zudilin | Oh yes, analysis as well. (I forgot to add linear algebra.) A very similar trick of using the difference operator can be found in Pade approximation business. I learned this from Marc Huttner (the paper is [M. Hata and M. Huttner, Pad\'e approximation to the logarithmic derivative of the Gauss hypergeometric function, Analytic number theory, eds. C. Jia and K. Matsumoto (Kluwer 2002), pp. 157--172]) and used myself in wain.mi.ras.ru/PS/nupi-SbM2005.pdf (proof of Lemma 1). | |
| Jun 2, 2010 at 0:36 | comment | added | Wadim Zudilin | These are really a solution (and the problem itself) belonging to my taste. It can serve as a very good problem for textbooks in analysis/combinatorics. Congratulations, Darji! | |
| Jun 1, 2010 at 21:24 | vote | accept | Timothy Chow | ||
| Jun 1, 2010 at 20:58 | history | edited | darij grinberg | CC BY-SA 2.5 |
deleted 8 characters in body; edited body
|
| Jun 1, 2010 at 20:18 | comment | added | fherzig | That's a nice solution. So one can describe an explicit change of basis between the two quadratic forms. A couple of typos: it should say $Q_{2i+1}$ when you rename near the top. Also $2(2(n-1)+1) = 4n-2$ (!) and this is less than $4n$. You really need $< 4n$, since otherwise the alternating sum you wrote wouldn't vanish. | |
| Jun 1, 2010 at 17:28 | history | edited | darij grinberg | CC BY-SA 2.5 |
added 61 characters in body; added 300 characters in body
|
| Jun 1, 2010 at 14:35 | history | edited | darij grinberg | CC BY-SA 2.5 |
edited body
|
| Jun 1, 2010 at 11:51 | comment | added | Wadim Zudilin | Darji, this is indeed a wonderful proof, and much more than 50% (my post was just a reformulation of the original problem). It should also work for the second case. Thanks a lot for giving me rest of the problem! :) | |
| Jun 1, 2010 at 10:54 | history | answered | darij grinberg | CC BY-SA 2.5 |