Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I have two "vague questions" which are the following: if you have two $n$-dimensional $\ell$-adic Galois representations of a number field $K$ (with the standard ramification conditions) that have the same char. polynomial in a set of primes of density $1$, then they are isomorphic. The same is true if I replace Galois representations by automorphic representations and I ask the local representations to be the same in a set of density $1$. So the question is: "what is the best bound for such result"? (assume $n=2$ and $K= \mathbb Q$ if you want).

So for example, is it true that given $\epsilon >0$ there exist two representations that are not isomorphic but whose Frobenius/local components are the same in a set of density $1-\epsilon$? (at least in the case $n=1$?)

The second somehow related question (but has nothing to do, just in spirit) is if I have a restricted tensor product of local automorphic representations (say $K=\mathbb Q$, $n=2$) then there is no reason for it to be modular, but is there any result saying "there exists $\delta <1$ such that if you change the local component appropiately in a set of primes of density $\delta$ then it is an automorphic form"?

Of course, if one starts with a Galois representation and construct a restricted tensor product via the local langlands correspondence, one would like to say that such representation $\Pi$ is modular (under the appropiate assumptions on the Galois representation), so my question can be viewed as "how far (in terms of density) are we from proving modularity in general"

share|improve this question
    
Here's a silly comment: In the n = 1 case, you can't have epsilon any smaller than 1/2. If chi is the quotient between two l-adic Galois characters under comparison, then the density of primes p for which chi(p) = 1 is 1/[G_Q^ab : ker(chi)]. If chi is not injective then this density is no more than 1/2. Certainly the bound 1/2 can be attained by choosing chi to be a nontrivial quadratic character. I think the question becomes significantly more interesting when n = 2! (If you restrict to irreducible Galois representations, can you attain epsilon = 1/2?) –  Jared Weinstein Mar 17 '10 at 4:33
add comment

1 Answer

up vote 8 down vote accepted

Firstly, at the beginning of the question you are missing irreducibility/cuspidality assumptions. If $\rho_1$ and $\rho_2$ are $\ell$-adic Galois reps with the same char poly in a set of primes of density 1, then you can only deduce their semisimplifications are isomorphic. A counterexample to your statement would be given by $\rho_1=1+\omega$ ($\omega$ the cyclotomic character) and $\rho_2$ some non-split extension coming from Kummer theory (taking $\ell$-poower roots of some prime number $p$ for example). Of course you knew that already.

On the automorphic side you make "the same slip", well, a related slip. If $\chi$ and $\psi$ are two Grossencharacters and their ratio is the norm character at some place $v$ (or possibly even infinitely many, or even all $v$), then the induction of $\chi+\psi$ from the Borel to $GL_2(\mathbf{A})$ is reducible, and any Jordan-Hoelder factor (which by definition means a tensor product of J-H factors of the local inductions) ramified at all but finitely many places is an automorphic representation. So now you can build two non-cuspidal automorphic representations which are isomorphic at all but one place (and such that the local components don't even have the same dimension at the bad place) easily. Of course for cuspidal representations you have strong multiplicity 1 theorems.

Passing remark: it is a source of some confusion to me as to why these errors are "similar" but on the other hand they don't "biject". The problem is that on the Kummer theory side, if you allow ramification at two primes, you get a whole projective line of non-isomorphic Galois representations (all with the same semisimplificiation). On the automorphic side the amount of control you have is much more combinatorial (you can change a finite set of places from trivial to Steinberg and that's it). So $\pi$ and $\rho$s don't match up (so Toby Gee and I only conjecture that given any (EDIT: algebraic) $\pi$ one expects a semi-simple $\rho$, and nothing more, and for $GL_n$ this observation is no doubt much older).

OK so after these pedantic remarks, that you no doubt knew anyway, but could have avoided if you'd put "irreducible" and "cuspidal" in the appropriate places, we move on to the far more interesting question of whether one can do better than multiplicity 1. And your hunch is correct. First you should do the following exercise on the Galois side: if $\rho_1$ and $\rho_2$ are two irreducible 2-dimensional representations of a finite group $G$, and their traces agree on a set $S$ in $G$ of density greater than 7/8, then $\rho_1$ and $\rho_2$ are isomorphic. Big hint: orthogonality relations. Next you should convince yourself that 7/8 is optimal in this result Big hint: D_8 x D_8. (EDIT: slightly simpler is D_8 x C_2---here D_8 has 8 elements). So now for silly Artin representation reasons you can't expect to beat 7/8 (like you can't expect to beat 1/2 in the GL_1 case as in Jared's comment).

So now the great news is that Dinakar Ramakrishnan proved an analogue on the automorphic side! At least in some cases. See the appendix to "$l$-adic representations associated to modular forms over imaginary quadratic fields. II." by Richard Taylor (Inventiones 116).

As for your second question though, it's completely hopeless. Even for $GL_1$ your hope is (provably) way out, so for $GL_2$ one can perhaps construct "Eisenstein" counterexamples (induced from non-automorphic characters of $GL_1$). The first objection is that $\pi_p$ can be ramified for infinitely many $p$, making you dead in the water. But even if $\pi_p$ is unramified for all $p$ you've still got no chance. Let me stick to $GL_1/\mathbf{Q}$. The point is that every Grossencharacter for $GL_1/\mathbf{Q}$ is the product of a finite order character $\chi$ (a Dirichlet character) and $||.||^s$, so you can recursively construct characters of $\mathbf{Q}_p^\times$ each of which is totally at odds with everything that came before. For example lets send $Frob_2$ to 1. Now let's write down all the solutions to $2^s.\zeta=1$, with $\zeta$ a root of unity and $s$ a complex number. There are only countably many values of $s$ in this list. So choose $s$ not in the list and send $Frob_3$ to $3^s$. Now knock off countably many more $s$ and send $Frob_5$ to $5^s$ for $s$ in neither list. We are making a collection of representations here that are pairwise completely incompatible! This is only one of the many obstructions. For example, for cuspidal automorphic representations there are Weil bound obstructions (Ramanujan conjecture) and arithmeticity obstructions in the holomorphic case---all deep theorems or conjectures about automorphic forms in some cases that can easily be violated if we're allowed to build $\pi$ locally. These arguments trivially show that the set of automorphic $\pi$s have density zero (and "a very small zero" at that) amongst all the $\pi$s. In fact here's a much cleaner objection for $GL_2$: if you stick to cuspidal automorphic representations with a fixed central character then there are only countably many! But you have uncountably many choices at each local place! So it's completely hopeless I think.

share|improve this answer
    
My question was a little vague, but I had in mind the irreducibility conditions. For n=1, my guess was that $1/2$ was the minimum ammount but my argument was not so clean as Jared´s one. For bigger dimensions, I didn´t have a guess, and I might being too sleepy but I cannot see how the orthogonality relations can give you the bound you mentioned! BTW, I am quite impressed with the $D_4 x C_2$ example ($D_4$ it is the easiest group with a $2$-dimensional representation and many trace zero elements, but I don´t think you start looking at each group of small order), may I ask how to came to it? –  A. Pacetti Mar 17 '10 at 23:26
    
My supervisor asked me the question about finite groups in the 1990s, presumably when he was thinking about the paper he was writing that I mention in the answer. Orthogonality relns say sum_g(chi_1(g)chi_2-bar(g))=0 and sum_g(chi_1(g)chi_1-bar(g))=|G|. If chi_1=chi_2 on a set S of density > 7/8 then then taking the difference we get sum_{g not in S}(chi_1(g)(chi_1-bar(g)-chi_2-bar(g))=|G| but this is a contradiction because |chi(g)|<=2 and so the sum isn't over enough elements. –  Kevin Buzzard Mar 17 '10 at 23:38
    
The proof shows that to obtain 7/8 you need to find chi_1 and chi_2 with chi_1(g)=chi_2(g) for all g in a set of density 7/8, and then chi_1(g)=-chi_2(g) for the rest, and the rest must all be contained in the centre, because |chi(g)|=2 implies g is central (at least in the image of rho_1). You try all groups of order 8 and they don't seem to work, and then you realise that G x C_2 is a good trick because there's a natural way of coming up with two reps with chi_1(g)=+-chi_2(g) and then you're home. –  Kevin Buzzard Mar 17 '10 at 23:42
    
Ah, you are saying that 7/8 works for dimension 2! I understood you said that this was for any dimension, and the inequality $|\chi(g)| \le 2$ is not true (replace the $2$ by $n$). So for dimension $n$ the bound is $1 - \frac{1}{2n^2}$. Is this optimal? (I will play around with your example in small dimensions). Also the other natural questions is if Ramakrishnan result hilds in higher dimensions as well? –  A. Pacetti Mar 18 '10 at 7:53
    
Yes, I was only making assertions about 2-dimensional representations when I was talking about 7/8. Generalising Ramakrishnan's theorem---one might conjecture that some variant (where 7/8 is replaced by some other number) holds in higher dimensions but proving it will be another matter. –  Kevin Buzzard Mar 20 '10 at 9:31
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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