Timeline for Given a positive-definite integral unimodular Gram matrix, how to find a basis of the associated lattice (over $\mathbf Q$)?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 17, 2014 at 9:15 | vote | accept | Oblomov | ||
| Nov 20, 2014 at 20:47 | answer | added | few_reps | timeline score: 4 | |
| Sep 15, 2014 at 9:36 | history | edited | Oblomov | CC BY-SA 3.0 |
added 1 character in body
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| Sep 8, 2014 at 15:01 | comment | added | Oblomov | In general, p/q=pq/q^2 and use Lagrange's theorem to decompose a the (positive) numerator as a sum of at most 4 squares. | |
| Sep 8, 2014 at 14:59 | comment | added | Oblomov | 1/7=7/49 and 7=2²+1²+1²+1², so 1/7=(2/7)²+(1/7)^2+(1/7)^2+(1/7)^2. | |
| Sep 8, 2014 at 14:53 | comment | added | Chua KS | Yes. Now I understand what you mean : Gram Schmidt is essentially completing squares but how to write 1/7 as a sum of squares of rationals ? | |
| Sep 8, 2014 at 13:55 | comment | added | Oblomov | Yes, that's more or less implicit in my "note n°2". | |
| Sep 8, 2014 at 13:37 | comment | added | Chua KS | If you do "completing squares" on the quadratic form $f(x)=x^TGx$, you can rewrite it as $x^TGx=\sum D_j(x_j+a_{j,j+1}x_{j+1}+...+a_{j,n}x_n)^2$ where all the coefficients are rationals. This is equivalent to the expression $G=M_1^tDM_1$ where $D$ is diagonal and $M_1$ is triangular. So $M=\sqrt{D}M_1$ but this is not rational because of the square root. | |
| Sep 8, 2014 at 9:25 | history | asked | Oblomov | CC BY-SA 3.0 |