Characteristic polynomials of Hecke operators $T_\ell$, with $\ell$ prime, acting on cusp forms $S_k$ of level one and weight $k$ "appear to be" squarefree (even irreducible!).

This can be interpreted as follows: if $p$ is any prime, then the $p$-adic Galois representations $\rho_i$, where $1\leq i\leq \dim(S_k)$, attached to eigenforms in $S_k$ "appear to be" pairwise non isomorphic ${\it locally}$ at primes $\ell\neq p$.

This is completely false for other levels. For example in the two dimensional space $S_2(\Gamma_0(37))$, I learn from Magma and Cremona's tables that there are exactly two normalized eigenforms, $f_1$ and $f_2$, with rational coefficients, corresponding to two elliptic curves $E_1$ and $E_2$ defined over $\mathbf{Q}$, of conductor $37$, and uniquely determined up to $Q$-isogeny.

Looking at some Hecke operators on this space, one easily finds examples of $T_\ell$ acting diagonally on $S_2(\Gamma_0(37))$, i.e., examples of primes $\ell\neq 37$ for which the two elliptic curves have $p$-adic Tate modules isomorphic as local Galois modules at $\ell$ (some of the $\ell$'s for which this happens are $7$, $31$, $41$, $101$, $137$, $173$, $179$,..$39769$).

$Q1$: Is it reasonable to suspect that $E_1$ and $E_2$ become isogenous over an extension $F$ of $Q$? If this were the case, then one should see the phenomenon described above for primes $\ell$ that are split in $F$, right?

$Q2$: On the other hand, given an elliptic curve $E$ over $\mathbf{Q}$, what are the known ways to construct more elliptic curves $A$, defined over $\mathbf{Q}$, possibly of the same conductor as $E$, which are not $\mathbf{Q}$-isogenous to $E$ but such that they become so over a non-trivial extension of $Q$?

$Q3$: Can we say why we do not see the above phenomenon in level one?

  • 1
    $\begingroup$ Interesting. Can you tell me more about the phenomenon in level 1? Did you observe it yourself? is tab a well-known observation? Do you have references? $\endgroup$ – Joël Oct 7 '11 at 14:07
  • 1
    $\begingroup$ Re. Q2: Twisting is the only way to do this. Use Faltings's proof of the Tate conjecture plus Chebotarev: if $E_1,E_2$ are defined over $\mathbf{Q}$ and $Hom_{G_K}(\rho_1,\rho_2)$ is nonempty for some $K/\mathbf{Q}$ of finite degree, then $\rho_1$ and $\rho_2$ are twist-equivalent by Frobenius reciprocity. $\endgroup$ – David Hansen Oct 7 '11 at 14:55
  • 1
    $\begingroup$ I think that people do not have any example where $T_p$ acting on level ONE and weight k has a reducible characteristic polynomial (let alone its squarefreeness). I've always heard of computations aimed to check this, but never did it myself. Google points me towards this paper: www.math.clemson.edu/~kevja/PAPERS/FarmerJames2000.pdf. I am not sure about any kind of theoretical evidence. Maeda's conjecture, saying that in level one there is only one galois orbit of forms, addresses a related issue. $\endgroup$ – Tommaso Centeleghe Oct 7 '11 at 15:02
  • $\begingroup$ @David: thanks, that's helpful. The twist is a quadratic one, right? $\endgroup$ – Tommaso Centeleghe Oct 7 '11 at 15:06
  • $\begingroup$ @Tommaso: Yup! $\endgroup$ – David Hansen Oct 7 '11 at 15:07

I imagine this is a weight issue, not a level issue. Let f and g be the two weight-2 newforms in $S_2(\Gamma_0(37))$. Then a "random" coefficient of f-g is going to have size about $p^{1/2}$, so there should be about $X^{1/2-\epsilon}$ primes $p$ less than $X$ such that $a_p(f) = a_p(g)$, just by chance. When the weight is larger, the Fourier coefficients are bigger, and it is much more surprising to see coincidences of Fourier coefficients.

Try other weight 2 cases, and try some higher weight cases in level 37, and I'll bet you'll see that your phenomenon happens in weight 2 and not in weight 4.

  • 1
    $\begingroup$ By the way, this is somewhat badly written: I should have written "E_1 and E_2" since I'm really using the fact that the coefficients of f and g are rational integers. $\endgroup$ – JSE Oct 7 '11 at 15:21
  • $\begingroup$ Ok, thanks. I guess your argument can be adapted to the case where there is no form with rational coefficients in $S_2(\gamma_0(N))$. At least one can observe a similar phenomenon as that above even for values of $N$ (e.s $23$) for which there is not elliptic curve of that conductor. $\endgroup$ – Tommaso Centeleghe Oct 7 '11 at 15:30
  • $\begingroup$ Try it for a case where the space has dimension higher than 2 and the forms are all Galois conjugate! $\endgroup$ – JSE Oct 7 '11 at 16:12
  • $\begingroup$ Nice answer, but does it mean that you guys think that the observation that in level 1 it never happens that two eigenforms have the same T_l eigenvalues for any l is false? But then it would be nice to have a counter-example. $\endgroup$ – Joël Oct 7 '11 at 16:22
  • $\begingroup$ @JSE: wait, do you have a proof of your last assertion, or this is just something with empirical evidence? $\endgroup$ – Tommaso Centeleghe Oct 7 '11 at 17:31

There is an easier way to answer Q2 at least if E1 does not have CM. Let phi be the isogeny from E1 to E2 defined over some extension, of degree d>1, and let sigma be a Galois automorphism. Then the composite of phi and dual(phi^sigma) is an isogeny from E1 to itself of degree d^2 so must be [d] or -[d]; hence phi^sigma = either +sigma or -sigma. If the latter occurs at all then there is a quadratic extension L of K such that phi is defined over L. Now compose with the L/K quadratic twist of E2 to get a curve E3 defined over K; the composite of phi and the twisting isomorphism is an isogeny from E1 to E3 of degree d and defined over K. So (1) phi had to be defined over a quadratic extension of K, and (2) E1 is isogenous over K to a quadratic twist of E2.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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