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

I posted this question on math stackexchange: https://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.

• 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. Aug 8, 2011 at 3:46
• For a way of stating quadratic reciprocity different from Ash - Gross see rzuser.uni-heidelberg.de/~hb3/publ/hecke-op.pdf Aug 8, 2011 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.