Timeline for Integral quaternary forms and theta functions
Current License: CC BY-SA 3.0
20 events
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| Feb 22, 2016 at 21:05 | history | edited | few_reps | CC BY-SA 3.0 |
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| Feb 22, 2016 at 14:58 | history | edited | few_reps | CC BY-SA 3.0 |
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| Feb 22, 2016 at 2:16 | history | edited | few_reps | CC BY-SA 3.0 |
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| Feb 22, 2016 at 1:54 | comment | added | Y. Zhao | @few_reps: Thanks for your calculation! In the $d=37^2$ case, the involution fixes the first and the third lattice, while it swaps the second and the fourth one. | |
| Feb 22, 2016 at 1:19 | history | edited | few_reps | CC BY-SA 3.0 |
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| Feb 22, 2016 at 1:04 | history | edited | few_reps | CC BY-SA 3.0 |
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| Feb 21, 2016 at 23:45 | comment | added | few_reps | @WillJagy But the class number is often even (I'm performing heuristics right now) ... | |
| Feb 21, 2016 at 23:44 | comment | added | few_reps | @WillJagy Yes, it is an involution since $(L^\sharp)^\sharp=L$ and $(pL)^\sharp=\frac 1p L^\sharp$. | |
| Feb 21, 2016 at 23:31 | comment | added | Will Jagy | few, I think that performing this operation twice should always return you to the original form, with discriminant $p^2.$ If so, we are guaranteed a fixed form (as the OP wants) when the class number is odd, to add to the two trivial infinite families in my answer. | |
| Feb 21, 2016 at 21:03 | comment | added | few_reps | @WillJagy All right ! | |
| Feb 21, 2016 at 21:02 | comment | added | Will Jagy | few_reps, give me a few minutes, I may have something. Also i think you answered this in good faith and should leave this here, it is good information. | |
| Feb 21, 2016 at 21:00 | comment | added | few_reps | @WillJagy Would you make your comment in an answer ? | |
| Feb 21, 2016 at 20:59 | comment | added | few_reps | @zy_ ok, so I guess I should delete this answer ... | |
| Feb 21, 2016 at 20:29 | comment | added | Y. Zhao | @WillJagy Thank you so much. I checked all primes $p<41$, and found no counterexamples. | |
| Feb 21, 2016 at 20:26 | comment | added | Will Jagy | @zy_ the identical binaries can be the principal form $x^2 + xy + \left( \frac{p+1}{4} \right) y^2$ | |
| Feb 21, 2016 at 20:16 | comment | added | Will Jagy | @zy_, you did not word this well. If all you want is one $A$ of the discriminant, whenever $p \equiv 3 \pmod 4$ there will always be one, because there is a quaternary made from the sum of two identical binaries. | |
| Feb 21, 2016 at 20:13 | comment | added | few_reps | hmm sorry, I misunderstood the question ... I'm gonna modify the algorithm to search for a counter-example to your very question ... | |
| Feb 21, 2016 at 20:09 | comment | added | Y. Zhao | Eh..... Why do you think the answer is "no" for determinant $37^2$? At least two matrices in your calculation support my idea. | |
| Feb 21, 2016 at 19:54 | history | edited | few_reps | CC BY-SA 3.0 |
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| Feb 21, 2016 at 19:46 | history | answered | few_reps | CC BY-SA 3.0 |