6
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ There is an (unanswered) question about this being true for any $p$ in level 1: mathoverflow.net/q/105713/6518 $\endgroup$
    – Kimball
    Jul 4, 2015 at 2:29
  • 1
    $\begingroup$ 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. $\endgroup$ 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
    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. $\endgroup$ 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
    Jul 22, 2015 at 13:55

1 Answer 1

6
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Ah, right. I was originally thinking about "minimal" modular forms (say prime level). Do you know anything about this case. $\endgroup$
    – Kimball
    Jul 4, 2015 at 4:33
  • 1
    $\begingroup$ 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. $\endgroup$ 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
    Jul 22, 2015 at 13:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.