# How does one see Hecke Operators as helping to generalize Quadratic Reciprocity?

I posted this question on math stackexchange: http://math.stackexchange.com/questions/56040/how-does-one-see-hecke-operators-as-helping-to-generalize-quadratic-reciprocity

and got 10 upvotes but no answers. I interpret this as evidence that maybe I've matured to mathoverflow. Here is what I wrote:

My question is really about how to think of Hecke operators as helping to generalize quadratic reciprocity.

Quadratic reciprocity can be stated like this: Let $\rho: Gal(\mathbb{Q})\rightarrow GL_1(\mathbb{C})$ be a $1$-dimensional representation that factors through $Gal(\mathbb{Q}(\sqrt{W})/\mathbb{Q})$. Then for any $\sigma \in Gal(\mathbb{Q})$, $\sigma(\sqrt{W})=\rho(\sigma)\sqrt{W}$. Define for each prime number $p$ an operator on the space of functions from $(\mathbb{Z}/4|W|\mathbb{Z})^{\times}$ to $\mathbb{C}^{\times}$ by $T(p)$ takes the function $\alpha$ to the function that takes $x$ to $\alpha(\frac{x}{p})$. Then there is a simultaneous eigenfunction $\alpha$, with eigenvalue $a_p$ for $T(p)$, such that for all $p\not|4|W|$ $\rho(Frob_p)=a_p$. (and to relate it to the undergraduate-textbook-version of quadratic reciprocity, one need only note that $\rho(Frob_p)$ is just the Legendre symbol $\left( \frac{W}{p}\right)$.)

Now I'm trying to understand how people think of generalizations of this. First, still in the one dimensional case, let's say we are not working over a quadratic field. What would the generalization be? What would take the place of $4|W|$? Would the space of functions that the $T(p)$'s work on still thes space of functions from $(\mathbb{Z}/N\mathbb{Z})^{\times}$ to $\mathbb{C}^{\times}$? What is this $N$?

Now let's jump to the $2$-dimensional case. Here we have the actual theory of Hecke operators. However, as I understand it, there is a basis of simultaneous eigenvalues only for the cusp forms. Now I'm finding it hard to match everything up: are we dealing just with irreducible $2$-dimensional representations? Instead of $\rho$ do we take the character? Would we say that for each representation there's a cusp form such that it's a simultaneous eigenfunction and such that $\xi(Frob_p)=a_p$ (the eigenvalues) where $\xi$ is the character of $\rho$? This should probably be for all $p$ that don't divide some $N$. What is this $N$? Does it relate to the cusp forms somehow? Is it their weight? Their level?

In other words:

### Questions

$1$. What is the precise statement of the generalization (in the terminology above) of quadratic reciprocity for the $1$-dimensional case?

$2$. What is the precise statement of the generalization (in the terminology above) of quadratic reciprocity for the $2$-dimensional case?

### Edit

Actually, now that I have the attention of experts, let me add two more questions:

$3$. Does Langlands predict anything for $1$-dimensional representations with infinite image?

and

$4$. I very much want to understand Hecke operators better. For example, why are the $T(p)$'s that I gave above the $1$-dimensional analogue of the usual Hecke operators? I've heard something about Hecke correspondences, and I wonder what is a good reference I can sink my teeth into about that.

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There are no continuous 1-dimensional representations with infinite image, because the Galois group is compact. If you want Frobenius eigenvalues that are not roots of unity, you need to consider representations of a different group, e.g., the Weil group. – S. Carnahan Aug 8 '11 at 3:46
For a way of stating quadratic reciprocity different from Ash - Gross see rzuser.uni-heidelberg.de/~hb3/publ/hecke-op.pdf – Franz Lemmermeyer Aug 8 '11 at 11:10

This is what Langlands reciprocity is about, you can read about it in many surveys. Very briefly it says that every $d$-dimensional (continuous) Galois representation is associated to an automorphic form on $\mathrm{GL}_d$ so that their $L$-functions agree. For a Galois representation the $L$-function is given in terms of the characteristic polynomials of the Frobenius elements, while for an automorphic form the $L$-function is given in terms of Hecke eigenvalues (more precisely of Langlands-Satake parameters that can be read off from Hecke eigenvalues or vice versa). So Langlands reciprocity says that for any $d$-dimensional Galois representation there is a great harmony of the images of the Frobenius elements: they yield a function on $\mathrm{GL}_d$ with fantastic properties (namely an automorphic form). Your case is $d=2$. And yes, $N$ is the level.