Skip to main content
12 events
when toggle format what by license comment
May 16, 2021 at 0:12 comment added k.j. Thank you very much! I understand that!
May 15, 2021 at 9:02 comment added François Brunault You may need to adjust the parameter $d$ and the precision of the $q$-expansion (number of rows). For $N=11$, the modular curves $X_1(N)$ and $X_0(N)$ are elliptic curves so there are other methods. The morphism $\pi$ is just $11.a3 \to 11.a2$ in lmfdb.org/EllipticCurve/Q/11/a It is the quotient of $X_1(11)$ by the $5$-torsion subgroup (which is the orbit of $\infty$ under the diamond operators). In this case you can get equations for $\pi$ using Vélu's formulas, it is implemented in Pari/GP with the function ellisogeny.
May 15, 2021 at 8:59 comment added François Brunault With a computer algebra system like Pari/GP or Sage, it is not difficult. You set up a matrix whose columns are indexed by your functions $F^i X$ and $F^i G^j$ (create a separate vector which remembers how you order the columns). Each column gives the coefficients of the $q$-expansion of the corresponding function (taking into account that there may be negative exponents, so you should consider the minimal valuation of all these functions, at which every column should start). Then compute the (right) kernel of the matrix.
May 15, 2021 at 8:26 comment added k.j. Thank you. Would you explain it for explicit $X_1(N) \to X_0(N)$, for example $N = 11$? I tried it, but because there are so many terms I can't find the relation well...
May 15, 2021 at 7:35 vote accept k.j.
May 15, 2021 at 7:30 comment added François Brunault @k.j. You're welcome. I should have explained it: just compute the $q$-expansions of all the functions $F^i X$ and $F^i G^j$, and then find a non-trivial linear relation between them by solving the linear system over $\mathbb{Q}$.
May 15, 2021 at 7:03 comment added k.j. Thank you very much! I've read the paper and computed the q expansions of some $F$, $G$, $X = \Sigma_\gamma F |_\gamma$ and similar for $Y$. But how can I find the relation as in (*)?
May 10, 2021 at 20:43 history edited François Brunault CC BY-SA 4.0
Slight additions and clarifications
May 10, 2021 at 15:57 history edited François Brunault CC BY-SA 4.0
Expanded the answer and corrected typos
May 8, 2021 at 20:05 history edited François Brunault CC BY-SA 4.0
added 371 characters in body
May 8, 2021 at 19:45 history edited François Brunault CC BY-SA 4.0
Added explanation on how to certify the relation
May 8, 2021 at 19:31 history answered François Brunault CC BY-SA 4.0