## The graph of congruences between modular forms

Let $S$ be the (countable) set of holomorphic cuspidal new eigenforms of weight $\geq 2$. Any $f\in S$ has a level $N_f$ and a canonically normalized Fourier expansion $f(z)=\sum_{n=1}^{\infty}a_f(n)e^{2\pi i nz}$ with $a_f(1)=1$ and $a_f(n) \in \overline{\mathbf{Z}}$.

Form a graph $\mathcal{G}$ as follows: Take the vertex set of $\mathcal{G}$ to be the set $S$, and join $f$ and $g$ by an edge if the algebraic integers $(a_f(n)-a_g(n))_{n \nmid N_f N_g}$ generate a nontrivial ideal in $\overline{\mathbf{Z}}$; in other words, $f$ and $g$ are joined if their Fourier coefficients are related by a nontrivial congruence.

Is $\mathcal{G}$ connected?

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It would be interesting to ask the same question but requiring the ideal to be nontrivial in $\overline{\mathbb{Z}}[1/N]$. – Dror Speiser Jan 10 2012 at 8:55
BTW, to add to Hansen's answer, it might be worth noting that, if $p$ divides $N$, your definition of $F$ and $G$ being connected by an edge coming from a prime above $p$ is equivalent to asking that the mod $p$ (or some prime above $p$) Galois representations $\overline{\rho}_G$ and $\overline{\rho}_F)$ have isomorphic semi-simplifications. Thus Hansen's remark that all forms are connected to a form of level prime to $N$ follows from a theorem of Ash-Stevens. You might want to ask Ash about it - or even better, ask his student(?) David Hansen. – Lavender Honey Jan 12 2012 at 2:32
Wanax, you are correct to guess that I'm working with Ash. But the answer below was written by The Hamburglar, not by me. – David Hansen Jan 12 2012 at 19:41

Suppose that $f$ has level $N$, and suppose that $N$ is divisible by $p$. Then it is well known that $f$ is congruent modulo (some prime above) $p$ to a form $g$ of level $M$ dividing $N$ (and high weight), where $M$ is prime to $p$. In particular, by induction, all forms $f$ are connected to a form $g$ of level $1$ in at most $d$ steps, where $d$ is the number of distinct prime divisors of $N$. Yet all level one forms are congruent to $\Delta$ modulo $2$. This is actually related to ideas behind the proof of Serre's conjecture:

http://en.wikipedia.org/wiki/Serre%27s_modularity_conjecture

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Talking to yourself is the first sign of madness... – Lavender Honey Jan 10 2012 at 4:25
@Wanax: you know, there is an edit history where you could check who actually said what. – quid Jan 10 2012 at 17:57
@quid: where is this edit history you speak of? – Marty Jan 10 2012 at 21:01
This answer was written by The Hamburglar (who, by the way, I think is the same person behind the users Frictionless Jellyfish and Trust Me, and I think I know who it is in real life, but this margin won't contain it.) I edited a trivial misprint; did that automatically convert it to a CW status? – David Hansen Jan 11 2012 at 19:35