From MAILER-DAEMON Sat Jul 24 15:08:13 1999 Date: Sat, 24 Jul 1999 15:08:13 -0700 (PDT) From: Mail System Internal Data Subject: DON'T DELETE THIS MESSAGE -- FOLDER INTERNAL DATA X-IMAP: 0932854093 0000000000 Status: RO This text is part of the internal format of your mail folder, and is not a real message. It is created automatically by the mail system software. If deleted, important folder data will be lost, and it will be re-created with the data reset to initial values. From was@math.berkeley.edu Sat Jul 24 15:07:48 1999 -0700 Status: X-Status: X-Keywords: Date: Sat, 24 Jul 1999 15:07:48 -0700 (PDT) From: William Arthur Stein To: Amod Agashe Subject: new "almost-everywhere" congruences? In-Reply-To: Message-ID: MIME-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII I think I told you that Ken's level raising gives congruence at all but fin many primes. But it looks like that may still be enough for our purposed. I'm not sure about this. What one needs is A_f intersect A_g to contain some nontrivial p-torsion. When f and g are both new, this is guaranteed when (1) _all_ a_p are congruent (2) one knows multiplicity one (so the residue characteristic should not be 2). I don't yet understand what happens when say f is old and g is new. What I am wondering is, can one have an almost-everywhere but not everywhere congruence at the SAME level? I think that is not possible, though I am not sure why. Any idea? eg at level 1091, I want to be sure that there is no other form congruent mod 7 to the one associated to the winding quotient, at almost every prime but not all. SHORT ANSWER: I suspect your intuition isn't quite right here, especially when the level is divisible by small primes (but see below for prime level). For example, take N=2*3*113=678. Consider the eigenforms f and g which I call, 678I2 and 678B1. The q-expansion of f to 11 terms is: f = q + q^2 + q^3 + q^4 + 2*q^5 + q^6 + a*q^7 + q^8 + q^9 + 2*q^10 + (-a-1)*q^11 + O(q^12), where a^2-a-15 = 0. The expansion of g is: g = q + -1*q^2 + q^3 + q^4 + -1*q^5 + -1*q^6 + -3*q^7 + -1*q^8 + q^9 + q^10 + 2*q^11 + O(q^12) Notice that the a5, a7, and, for the choice a=0 of root, the a11 are congruent modulo 3, though the a2's are not. Going a little farther, f-g (mod 3) = -q^2 - q^6 - q^8 + q^10 - q^18 + q^22 - q^24 - q^26 + q^30 -q^32 - q^34 + O(q^38), so the ap for p<=37 *prime* and prime to N are cong modulo 3. One other point: the mod 3 representation *is* irreducible, as 678B1 is isolated in its isogeny class. I've checked for this sort of thing much farther in the course of our visibility computations; in fact, that's where I found this example. There are other examples like this at levels 765, 766, 855, 894, 902, 913, 946, 966, 980, 998, and the list goes on. (Those listed are all mod 3 congruences). LONG ANSWER: But there are conditions under which it doesn't happen. You ask whether one can have two newforms f and g, each of level N, maximal ideals m and n of residue characteristic p, and some choice of isomorphism of residue fields so that (a) a_p(f) mod m == a_p(g) mod n for all p not dividing N, and (b) but there is a prime p dividing N so that a_p(f) mod m =/= a_p(g) mod n. (1) This question is almost discussed in Cremona-Mazur in the section titled "The relation of m-congruence". But, somehow, it is not addressed. (2) Here's an argument which, unless I'm mistaken, works for your p=1091 example. By the Sturm bound, the Hecke algebra is generated as a Z-module by the Hecke operators T1, T2, ..., Tb where b=(1092)/6 = 182. This means, in particular, that T1091 is a Z-linear combination of T1,...,Tb. Further, each of the T1,...,Tb is in the ring generated over Z by the Tell for prime ell<182. In particular, it is possible to write T1091 as a polynomial, over Z, in terms of T2, T3, T5, ..., T181. For any particular eigenform, this means that a_{1091} can be written as a polynomial over Z in terms of a_2, a_3,..., a_{181}. If we assume that the a_2, a_3, ..., a_{181} for two distinct eigenforms are congruent, then the a_{1091}'s must also be congruent. (3) Step (2) suggests the following criterion: The prime to N Hecke algebra T' generated by all T_n with n not dividing N, is equal to the full Hecke algebra T generatedy by ALL T_n. In the case of large prime level, T must equal T' because of the Sturm bound. Best, William