Timeline for On the automorphism group of binary quadratic forms
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Oct 22, 2019 at 21:27 | vote | accept | Stanley Yao Xiao | ||
| Oct 27, 2017 at 21:06 | answer | added | Luc Guyot | timeline score: 3 | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jul 28, 2016 at 6:20 | comment | added | individ | There is another approach. But there is quite a long formula and You formula not like. The meaning is simple to find solutions for the equation $Z=\pm1$ math.stackexchange.com/questions/1513733/… You can use another idea. To reduce the equation to another form. $ax^2+bxy+cy^2=(p^2-ks^2)(t^2-qn^2)=j$ | |
| Jul 28, 2016 at 4:39 | comment | added | individ | The formula I've shown, but it is still not satisfied. It is not clear why? For quadratic forms $ax^2+bxy+cy^2=j$ It is necessary that the ratios between them were connected through the square. Or so $t^2=ja$ Or so $t^2=j(a+bk+ck^2)$ . You can pick up a different configuration, but the task still comes down to this. | |
| Jul 27, 2016 at 20:57 | vote | accept | Stanley Yao Xiao | ||
| Jul 27, 2016 at 20:58 | |||||
| Jul 27, 2016 at 19:23 | comment | added | Will Jagy | From page 144, Theorem 103, in Dickson 1929, the number of ambiguous classes is at least as large as the number of genera. This is in section 83, where he is just discussing forms with even middle coefficient. So, if there is just one genus, as in $x^2 - p y^2$ when prime $p \equiv 1 \pmod 4,$ the only ambiguous class is that of the principal form. For $x^2 - q y^2$ when $q \equiv 3 \pmod 4,$ there are just two ambiguous classes, principal and $-x^2 + q y^2.$ | |
| Jul 27, 2016 at 18:42 | answer | added | Will Jagy | timeline score: 2 | |
| Jul 27, 2016 at 18:22 | answer | added | Will Jagy | timeline score: 1 | |
| Jul 27, 2016 at 18:20 | comment | added | Stanley Yao Xiao | @WillJagy I am interested in doing asymptotic estimates, not algorithms in checking specific cases, so I will need some sort of theorem that can count over a long range of discriminants which forms $f$ has non-empty $G_f^-$. For example, it would be nice if the diagonal forms above are pairwise distinct for different divisors, but it seems to be not the case... | |
| Jul 27, 2016 at 18:18 | comment | added | Will Jagy | the second part I can do: both your forms are "diagonal" and so "ambiguous." They are $GL_2 \mathbb Z$ equivalent if and only if they are $SL_2 \mathbb Z.$ For any specific forms, this can be checked with the Lagrange cycles of reduced forms, or more recent Zagier cycles of reduced forms. | |
| Jul 27, 2016 at 18:17 | history | edited | Stanley Yao Xiao | CC BY-SA 3.0 |
added 44 characters in body
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| Jul 27, 2016 at 18:01 | history | asked | Stanley Yao Xiao | CC BY-SA 3.0 |