Timeline for Order of Galois action of modular form
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Aug 6, 2019 at 10:01 | answer | added | Henri Cohen | timeline score: 3 | |
| Aug 6, 2019 at 9:46 | comment | added | ililiil | @ François Brunault Thanks!!!! | |
| Aug 6, 2019 at 9:39 | comment | added | François Brunault | @ililiil For an explicit formula for $\sigma(x)$ you can look at Glenn Stevens's book Arithmetic on modular curves, the first sections are available on Google Books. | |
| Aug 6, 2019 at 9:30 | answer | added | François Brunault | timeline score: 5 | |
| Aug 6, 2019 at 8:50 | comment | added | ililiil | @ François Brunault Thanks for kindly reply. I don't know modular curve very well, so maybe I think it's a stupid question... I know that a cusp $x$ can be expressed as element in $P^{1}(\mathbb{Q})$. How do you define $\sigma(x)? | |
| Aug 6, 2019 at 8:31 | comment | added | François Brunault | @ililiil That's right. For any modular form $f$, any cusp $x$ and any $\sigma \in \mathrm{Aut}(\mathbb{C})$, the order of vanishing of $f^\sigma$ at $\sigma(x)$ is equal to that of $f$ at $x$. This follows from the fact that the modular curve $X_0(N)$ is an algebraic curve defined over $\mathbb{Q}$. Cusp forms of weight 2 are differential forms on $X_0(N)$ so are also algebraic. For general weight $k>2$, modular forms are sections of certain line bundles on $X_0(N)$, so are again algebraic. Once you know algebraicity, the proof of this result is formal. | |
| Aug 6, 2019 at 7:46 | history | edited | GH from MO |
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| Aug 6, 2019 at 7:42 | comment | added | ililiil | @François Brunault I don't understand that" It follows that $f$ and $f^{\sigma}$ have different order~". Is it relation between orders of $f$ at conjugate cusps and the order of $f^{\sigma}$ at cusp? | |
| Aug 6, 2019 at 7:31 | comment | added | ililiil | @François Brunault I noticed that my question was weird thanks to your first comment. So I modified my question as below. Let K⊂Q(ξN) be a field... | |
| Aug 6, 2019 at 7:16 | history | edited | ililiil | CC BY-SA 4.0 |
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| Aug 6, 2019 at 7:06 | comment | added | François Brunault | @ililiil If you have a counterexample $f$ with coefficients in some field $K$, you can always multiply $f$ by some constant so that the new field of coefficients contains the cyclotomic field (or even any given number field), this will not change the orders of vanishing you're looking at. | |
| Aug 6, 2019 at 5:24 | comment | added | ililiil | I don't assume that $f$ is a newform. Your examples are so nice.. | |
| Aug 5, 2019 at 21:51 | comment | added | François Brunault | @reuns Here is a reference: arxiv.org/abs/1609.08939v4 (see page 5 for the examples I mention). These newforms $f$ have different orders of vanishing at conjugate cusps. It follows that $f$ and $f^\sigma$ have different order of vanishing at the same cusp. | |
| Aug 5, 2019 at 21:35 | comment | added | reuns | How do you prove your claims ? | |
| Aug 5, 2019 at 17:03 | comment | added | François Brunault | There are newforms $f$ on $\Gamma_0(N)$ such that $f$ and $f^\sigma$ do not always have the same order of vanishing at the cusps. I have examples at level $567,625,891$. But here, do you assume that $f$ is a newform? Do you really assume that the field of coefficients of $f$ contains the $N$-th cyclotomic field? | |
| Aug 5, 2019 at 16:25 | history | edited | ililiil | CC BY-SA 4.0 |
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| Aug 5, 2019 at 15:28 | history | asked | ililiil | CC BY-SA 4.0 |