Timeline for Duke and Schulze-Pillot condition for equidistribution
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Oct 5, 2022 at 14:03 | comment | added | GH from MO | @MathqA I think you are right. In my response, I secretly assumed that $n$ was coprime to $\det(a_{ij})$. In that case, I believe that modulo $4\det(a_{ij})$ is sufficient. See Hilfssatz 13 in Siegel: Über die analytische Theorie der quadratischen Formen, Ann. of Math. 36 (1935), 527-606. Note that when applying this theorem, $b=0$ for $p>2$, and $b=1$ for $p=2$ (because $p\mid S$ implies $p\nmid T$ by our coprimality assumption). | |
| Oct 5, 2022 at 8:38 | comment | added | MathqA | I needed to come back to this topic and a question about your answer just hit me. Could you please justify the statement "The latter is equivalent to: $n$ is primitively represented by $Q$ modulo $4det(a_{ij})$. All the references I have consulted seem to need a congruence with $8det(a_{ij})^2$. What I am understanding wrong? | |
| Jun 8, 2022 at 5:41 | comment | added | MathqA | That is exactly what I was looking for. I needed some help to follow Duke and Schulze-Pillot paper. Thank you! | |
| Jun 8, 2022 at 5:39 | vote | accept | MathqA | ||
| Jun 7, 2022 at 0:18 | history | edited | GH from MO | CC BY-SA 4.0 |
deleted 29 characters in body
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| Jun 7, 2022 at 0:09 | history | answered | GH from MO | CC BY-SA 4.0 |