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By definition, an eigenform is a simultaneous eigenvector for all the Hecke operators $T_n$. Suppose I know that $f$ is an eigenvector of a particular Hecke operator, say $T_N$, does it follow that $f$ is also an eigenvector for all the other Hecke operators? Or are there counter-examples to this?

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If I'm not mistaken, to find a counterexample, it suffices to find a (non-isomorphic) pair of elliptic curves over $\mathbb{Q}$ and a prime $p$ at which both curves have good reduction and the same number of $\mathbb{F}_p$ points. Then you can take a convex combination of the corresponding eigenforms. – S. Carnahan Dec 16 '12 at 11:26
@S. Carnahan. You're right. And it suffices to take two elliptic curves supersingular at the same prime $p$ to ensure that the have the same number of $\mathbb F_p$-points, namely $p+1$. To be sure that two such curves exist without looking at a table, just take two CM curves by imaginary quadratic fields $K$ and $K'$, with a prime $p$ that is both inert in $K$ and $K'$, which is really easy to find. This answer the question by the negative for some level $N>1$. I think the answer is also no in level $1$ (which might be what the OP had in mind), bud I'd like to see a simple argument – Joël Dec 16 '12 at 14:37
I would take a common eigenform and make $U_p$ act on it: that wouldn't change the fact that it's an eigenform for most Hecke operators, but would probably break things for $T_p$. – Julien Puydt Dec 16 '12 at 15:14
An eigenvector of $T_N$ may not be an eigenform if and only if $T_N$ has a multiple eigenvalue. Whether this happen for some $N$, I don't know and I leave the question to specialists. – Denis Serre Dec 16 '12 at 15:41
If $N=1$, it is possible that the answer be not known. If I remember correctly, one does not know any example of a $T_N$ acting on a space of cuspidal modular forms of weight $k$ with a non-irreducible characteristic polynomial over $\mathbb Q$. If this was always true, that would implies simple eigenvalues for $T_N$ and settle the question. So certainly it is either not known or false that the answer to the OP's question is "no". On the other hand, if the answer is yes, I don't know how one would prove such a result. – Joël Dec 16 '12 at 18:44
up vote 8 down vote accepted

Following the comments, here is perhaps the simplest counterexample (once you know the Breuil-Conrad-Diamond-Taylor modularity theorem). The curves $y^2 + y = x^3$ and $y^2 + y = x^3 + 2x$ both reduce mod 2 to the same smooth curve, so their Hecke eigenforms have equal $T_2$ eigenvalues.

However, the curves over $\mathbb{Q}$ are non-isomorphic, since only the first curve has vanishing $j$-invariant. This means the Hecke eigenforms have different coefficients for some prime $p$, and hence different eigenvalues for $T_p$ (in fact, by counting the mod 3 solutions, you can see that $p=3$ works).

If you add the eigenforms, you get a (non-normalized) cusp form that is an eigenfunction for $T_2$, but not for $T_p$.

Edit I might as well say what I know about the level 1 case. It is "classically" known that Hecke operators are self-adjoint with respect to the Petersson inner product on cusp forms, and they commute with each other. By the spectral theorem, the Hecke operators are therefore simultaneously diagonalizable, i.e., there is a basis of the space of cusp forms made out of eigenforms.

Now, let us assume all of the eigenvalues of all of the Hecke operators are distinct (in contrast to the counterexample above). If you have a form $f$ for which some $T_n$ acts by a scalar, then $f$ cannot be a non-trivial linear combination of some basis of eigenforms, i.e., it is necessarily an eigenform. This multiplicity one property is not known in level one in general, but it is known to be true for $T_2$, as I learned in Cardinal Wolsey's brilliant answer to your previous question (the answer was deleted, it seems, because you didn't accept it within 4 days).

More generally, the multiplicity one property is implied by a conjecture of Maeda, which asserts that Hecke operators act irreducibly on $S_k(SL_2(\mathbb{Z}))$ for each $k$. In fact, David Loeffler mentioned, in a comment on the previous question, that there is a lot of computational evidence behind an even stronger assertion, namely that the eigenvalues generate as-big-as-possible Galois extensions over $\mathbb{Q}$.

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Thank you for the great answer! Just to clarify, when you say "all of the eigenvalues of all of the Hecke operators are distinct," do you mean that given a particular Hecke operator we want its eigenvalues to be distinct (which is the case if Maeda's conjecture holds), or do you mean that if we collect all the eigenvalues of all the Hecke operators we want nobody to be equal. The second statement implies the first but I don't know if they are equivalent. – Flip Dec 19 '12 at 14:55
I meant the weaker statement: for each $n \geq 1$ and each sufficiently large even $k$, the eigenvalues of $T_n$ on $S_k(SL_2(\mathbb{Z}))$ have multiplicity one. – S. Carnahan Dec 20 '12 at 0:15

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