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Consider the space of modular forms $M_k(N)$. Any modular form $f \in M_k(N)$ is determined by a finite number of Fourier coefficients (e.g., Sturm's bound), thus there is a finite set of Hecke operators that lets us distinguish eigenforms from each other. In fact this is true for $T_p$'s rather than $T_n$'s

Question: Does there always exist a Hecke operator $T_p$ that distinguishes eigenforms? I.e., is there always some $T_p$ acting on $M_k(N)$ with distinct eigenvalues?

This is not true for all $p$ certainly (e.g., this question), but I want to know if you can have strange situations like $f_1, f_2, f_3$ are distinct eigenforms with $a_p(f_1) = a_p(f_2)$ for $p \equiv 1, 2$ mod $4$ and $a_p(f_1) = a_p(f_3)$ for $p \equiv 3$ mod $4$, say.

I would also be interested in partial results, e.g., cuspidal newforms in weight 2.

Edit: As pointed out in a comment and an answer, it's easy to come up with counterexamples using quadratic twists. I would still like to know what happens if one restricts to "minimal" modular forms, say newforms of prime level.

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  • $\begingroup$ There is an (unanswered) question about this being true for any $p$ in level 1: mathoverflow.net/q/105713/6518 $\endgroup$
    – Kimball
    Commented Jul 4, 2015 at 2:29
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    $\begingroup$ To construct a counterexample, use three forms $f_1,f_2,f_3$ in the same space (they can even be weight-2 newforms) that are quadratic twists of each other. For each $p$ at least two of the $f_i$ must have the same $T_p$ eigenvalue. $\endgroup$
    – Noam D. Elkies
    Commented Jul 4, 2015 at 2:53
  • $\begingroup$ @NoamD.Elkies Thanks. I was originally thinking of the weight 2 prime or squarefree level case, so I wasn't thinking about quadratic twists. Do you know what happens if you restrict to prime or squarefree level? $\endgroup$
    – Kimball
    Commented Jul 4, 2015 at 4:29
  • $\begingroup$ There is a paper by Koopa Koo, William Stein, and Gabor Wiese that is about exactly this question. $\endgroup$
    – David Loeffler
    Commented Jul 4, 2015 at 15:16
  • $\begingroup$ @DavidLoeffler Thanks for the reference. I finally had time to look at it, but I don't quite see the connection. What they prove says that (say when N is squarefree) in each Galois conjugacy class the Fourier coefficients are distinct at a density 1 set of primes. But there can be many Galois orbits with the same coefficient field. Can you please clarify? $\endgroup$
    – Kimball
    Commented Jul 22, 2015 at 13:55

1 Answer 1

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I think that this is false for Dirichlet characters (since you can take, say, $\chi_1, \chi_2, \chi_1 \chi_2$ when $\chi_1 \chi_2$ are quadratic -- at least one will take the value $1$ on a prime) and then you can just twist your favorite modular form by these guys.

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  • $\begingroup$ Ah, right. I was originally thinking about "minimal" modular forms (say prime level). Do you know anything about this case. $\endgroup$
    – Kimball
    Commented Jul 4, 2015 at 4:33
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    $\begingroup$ Without thinking too carefully, it probably follows from Chebotarev that the density of places where Hecke eigenvalues coincide is zero for twist inequivalent forms, so that case should be OK. $\endgroup$
    – FriendlyWendy
    Commented Jul 4, 2015 at 4:44
  • $\begingroup$ Just thinking about the weight 1 case, is it true that for a finite group $G$ there is a conjugacy class that distinguishes quadratic-twist classes of characters? If not, it's not obvious to me how to use Chebotarev. $\endgroup$
    – Kimball
    Commented Jul 22, 2015 at 13:47

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