Timeline for Is there a nice choice-free argument to count the number of sublattices?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Feb 2, 2016 at 8:24 | comment | added | Tom De Medts | @QiaochuYuan Sorry for the typo, I meant matrices in $\mathrm{Mat}_2(\mathbb{Z})$ of determinant $\pm n$. | |
| Feb 1, 2016 at 18:23 | answer | added | Qiaochu Yuan | timeline score: 2 | |
| Feb 1, 2016 at 18:22 | comment | added | Qiaochu Yuan | @Tom: matrices in $GL_2(\mathbb{Z})$ can only have determinant $\pm 1$. | |
| Feb 1, 2016 at 16:14 | comment | added | Simon Rose | That's what I mean, yeah. Maybe my description isn't very clear. | |
| Feb 1, 2016 at 15:29 | comment | added | Tom De Medts | I'm not sure whether I agree that your first method involves a choice. You are counting the number of subgroups of $\mathbb{Z}^2$ of index $n$, and you observe that this corresponds to the number of $\mathrm{SL}_2^\pm\mathbb{Z}$-equivalence classes of matrices in $\mathrm{GL}_2\mathbb{Z}$ of determinant $\pm n$. (I don't see the choice yet.) Then you compute that this number is $\sigma_1(n)$, by coming up with a "normal form" for each of these equivalence classes. Is it this last step that you interpret as a "choice"? | |
| Feb 1, 2016 at 10:55 | history | edited | YCor |
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| Feb 1, 2016 at 10:52 | history | asked | Simon Rose | CC BY-SA 3.0 |