Timeline for When is the ratio of Jacobi theta functions algebraic?
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
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| Nov 6, 2014 at 14:28 | comment | added | Noam D. Elkies | Good point; I should have described the dependence on $n$ a bit more carefully: The degree is within a constant factor of the class number of the order ${\bf Z}[\sqrt{-n}\,]$. This does increase with $\;n$, even in a family such as $n = m^2 n_0$. | |
| Nov 6, 2014 at 7:19 | comment | added | joro | @NoamD.Elkies Thanks, but $n$ need not be squarefree and might be square leading to infinitely many isomorphic number fields. | |
| Nov 6, 2014 at 1:16 | comment | added | Noam D. Elkies | On the contrary, it's known that the degree tends to $\infty$ as $n \rightarrow \infty$, because it's within a bounded factor of the class number of ${\bf Q}(\sqrt{-n}\,)$. | |
| Nov 5, 2014 at 17:08 | comment | added | joro | @NoamD.Elkies is the degree of $f(n)^4$ bounded (possibly conjecturally) infinitely often? | |
| Nov 4, 2014 at 16:54 | comment | added | Noam D. Elkies | They're all algebraic, by the theory of complex multiplication. For example, for $\ n=14$ (the first number not in your list) I find that $2f(n)^4$ is a root of $$ x^8 - 8x^7 - 1964x^6 + 11896x^5 - 27034x^4 + 28680x^3 - 14764x^2 + 3976x + 1. $$ Values at "Heegner numbers" are easiest to recognize because the degree is small. | |
| Nov 4, 2014 at 16:46 | comment | added | joro | @FredrikJohansson Thank you :-). Why not 1-to-1? | |
| Nov 4, 2014 at 16:32 | comment | added | Fredrik Johansson | $f(n)$ is essentially the "lambda function" (mathworld.wolfram.com/EllipticLambdaFunction.html) or the "elliptic modulus", and I think the indices correspond (maybe not 1-to-1) to "elliptic integral singular values" (mathworld.wolfram.com/EllipticIntegralSingularValue.html) | |
| Nov 4, 2014 at 15:24 | history | asked | joro | CC BY-SA 3.0 |