Timeline for The degree of the cube root of the $j$-invariant
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| May 7, 2018 at 15:17 | comment | added | Jeremy Rouse | By the way, there's a "reason" that for any CM point with discriminant coprime to $3$ one can find an ${\rm SL}_{2}(\mathbb{Z})$ orbit of $\omega$ that gives a solution to $j = x^{3}$ of degree $h(D)$. This is because the equation $j = x^{3}$ is the equation of the modular curve corresponding to the normalizer of a non-split Cartan subgroup modulo $3$. The normalizer of the split Cartan subgroup is contained in this, and an elliptic curve with CM by $\mathbb{Z}[\omega]$ will have mod $3$ image contained in one of these two if the discriminant is coprime to $3$. | |
| May 6, 2018 at 12:40 | vote | accept | Shimrod | ||
| May 6, 2018 at 11:57 | answer | added | Luca Ghidelli | timeline score: 3 | |
| May 6, 2018 at 10:34 | comment | added | Shimrod | @Luca Ghidelli: The discriminant $D$ is always negative. | |
| May 6, 2018 at 10:32 | history | edited | Shimrod | CC BY-SA 4.0 |
added 7 characters in body
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| May 6, 2018 at 0:06 | comment | added | Luca Ghidelli | D is positive or negative or it depends? | |
| May 5, 2018 at 12:42 | comment | added | Shimrod | @Laurent Moret-Bailly: The $j$-invariant is real on the imaginary axis, for example we have $j(i)=1728=12^3$. It turns out that the function $j$ has a holomorphic cube root. The function $\gamma_2$ is defined to be that cube root which takes real values on the imaginary axis. Thus we have for instance $\gamma_2(i)=12$. | |
| May 5, 2018 at 11:38 | comment | added | Laurent Moret-Bailly | How can a cube root be real on the imaginary axis? | |
| May 4, 2018 at 23:39 | history | asked | Shimrod | CC BY-SA 4.0 |