Timeline for "Pythagoras number" for integral matrices
Current License: CC BY-SA 3.0
21 events
| when toggle format | what | by | license | comment | |
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| Oct 15, 2015 at 21:17 | comment | added | few_reps | @WKC That was also clear ... | |
| Oct 15, 2015 at 21:16 | comment | added | WKC | @few_reps OK, thanks. In that case I would like to emphasize that the $g(n)$ in all the results I mentioned depends only on $n$. | |
| Oct 15, 2015 at 19:04 | comment | added | few_reps | @WKC Just as I said in my previous comment, there's no claim of independence on $M$ in the remark I made. This should be clear since the quantity $n_0$ explicitely depends on $M$. | |
| Oct 15, 2015 at 18:29 | comment | added | WKC | @few_reps We would like to have a bound on $g(n)$ (or the $n_0$ in your answer) independent of $M$. But since your argument involves the quantity $d$ and I do not see how one can get rid of the dependence of $M$ at the end. | |
| Oct 15, 2015 at 17:59 | comment | added | few_reps | @WKC Yes, it grows quite quickly with de determinant of $M$. Icaza's bounds are a better answer, as well of course as Kim and Oh's. | |
| Oct 15, 2015 at 17:12 | comment | added | WKC | Does your $d$ depend on $M$ (and so does the final $n_0$)? It seems to me that $d$ grows with the discriminant of $M$. | |
| Oct 14, 2015 at 11:40 | vote | accept | Hans | ||
| Oct 14, 2015 at 8:25 | comment | added | few_reps | The thing about $\mathbf Z[\frac 12]$ follows from the fact (due to Kneser, I guess) that "any unimodular $\mathbf Z[\frac 12]$-lattice on $\mathrm{I}_n\otimes\mathbf Q$ is isometric to $\mathrm{I}_n\otimes\mathbf Z[\frac 12]$" ... and that can be proved by using the Strong Approximation Theorem ... I should say : Cassel's book is (really) easier to read if you are not familiar with the material it contains ... and the Strong Approximation Theorem appears in it. | |
| Oct 14, 2015 at 7:54 | comment | added | Hans | if not in general, perhaps if we further assume that $M$ is invertible in $\mathbb{Z}_{(2)}$? | |
| Oct 14, 2015 at 7:49 | comment | added | Hans | thank you for that detailed answer! the thing you say about $\mathbb{Z}[\frac{1}{2}]$, is it somehow hidden in O'Meara's book? when I went through it, I couldn't find it, or have I just overlooked it? Can one say something similar about $\mathbb{Z}_{(2)}$? | |
| Oct 13, 2015 at 18:48 | history | edited | few_reps | CC BY-SA 3.0 |
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| Oct 13, 2015 at 18:41 | history | edited | few_reps | CC BY-SA 3.0 |
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| Oct 13, 2015 at 18:01 | history | edited | few_reps | CC BY-SA 3.0 |
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| Oct 13, 2015 at 13:57 | comment | added | few_reps | It's just a teaching day. | |
| Oct 13, 2015 at 12:04 | comment | added | Gerry Myerson | @Denis, lack of time, not space – Galois, not Fermat. | |
| Oct 13, 2015 at 11:45 | comment | added | Igor Rivin | @DenisSerre He did promise to write up the proof tomorrow, so I don't think you should be snarky yet... | |
| Oct 13, 2015 at 11:21 | comment | added | Denis Serre | Do you mean that the margin is too narrow for your beautiful prof ? | |
| Oct 13, 2015 at 10:18 | comment | added | few_reps | Everything is in O'Meara's Introduction to quadratic forms, but you might as well be content with Cassel's Rational quadratic forms. | |
| Oct 13, 2015 at 10:04 | comment | added | Hans | Thanks, that sounds great. Can you give me some references for the two statements that you use? | |
| Oct 13, 2015 at 10:00 | history | edited | few_reps | CC BY-SA 3.0 |
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| Oct 13, 2015 at 9:43 | history | answered | few_reps | CC BY-SA 3.0 |