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Nov 7, 2021 at 15:36 vote accept Garrett Credi
Nov 7, 2021 at 12:06 answer added François Brunault timeline score: 7
Nov 7, 2021 at 10:03 comment added François Brunault Applying the Aktkin-Lehner operator may introduce denominators: the Fourier coefficients of $f|W_2$ are essentially those of $f|\sigma$ with $\sigma=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$, and the cusp $\sigma\infty=0$ is not integral. At a conceptual level, using Katz's algebraic modular forms, the matrix $\sigma$ acts on the level structures $\mu_2 \times \mathbb{Z}/2 \to E[2]$, and there is an isomorphism of group schemes between $\mu_2=\mathrm{Spec}(\mathbb{Z}[x]/(x^2-1))$ and $\mathbb{Z}/2=\mathrm{Spec}(\mathbb{Z}^2)$ only over $\mathbb{Z}[1/2]$, which explains the denominators.
Nov 6, 2021 at 23:57 history edited YCor CC BY-SA 4.0
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Nov 6, 2021 at 22:05 history asked Garrett Credi CC BY-SA 4.0