# Relation between Hecke operators and coefficient of L-functions

This question has its seed in this one by Gory, which found an enlightening answer but one of the comments kept me wondering. I am beginning to discover Hecke operators, and there appears to be an ubiquitous relation between Hecke eigenvalues and coefficients of L-functions that I do not get at all. I will try to state everything in details.

Hecke operators. Let us fix a place $p$ and consider an unramified local component $\pi_p$ of an automorphic representation of $GL_2$ over $F$. Let $K_p$ denote $GL_2(\mathcal{O}_p)$. We define the Hecke operator $T_{p^i}$ as the convolution action of the characteristic function of $$\bigcup_{\substack{a+b = i \\ a \geqslant b}} K_p \left( \begin{array}{cc} p^a & \\ & p^b \end{array} \right) K_p$$

L-functions. The automorphic representation $\pi$ also has an attached $L$-function (built on the Satake parameters at unramified places and a specific completion defining the remaining factors) which can be written as (and this defines the $\lambda_\pi(n)$) $$L(s, \pi) = \sum_{n \geqslant 1} \lambda_\pi(n) n^{-s}$$

Coefficients as eigenvalues. With all those definitions in hand, if $\phi$ is a function in the (one-dimensional) subspace of $K_p$-invariant vectors of $\pi_p$, do we have that $$T_{p^i} \star \phi = p^{1/2} \lambda_\pi(p^i) \phi \quad ?$$

Questions. More precisely, I would like to ask both following (maybe elementary) questions:

• I know it for $i=1$ (for instance Gelbart), however does it remain for $i \geqslant 2$, and do you have a proof of that?
• in the case where $\pi_p$ is ramified, those convolutions always give zero because there is no $K_p$ invariant vector in $\pi_p$ but the convolution creates such invariant vectors. In order to get the coefficient $\lambda_\pi(p^i)$ is this case, can I do exactly the same construction replacing $K_p$ by $K_1(p^f)$ where $f$ is the (additive) arithmetic conductor of $\pi_p$? (in that case the vector space of vectors fixed by it is one-dimensional)

I would appreciate any details or good reference for this matters, thanks in advance!

• Yes we have, but this is something we could not capture at the same time. I think things we could capture is the moment such like the order 1 moment, it is the expectation. And higher order moments, such like order 2 moments is just variation. And what you write is just the binichi identity for curvatures. – Hu xiyu Nov 7 '17 at 9:37
• @Huxiyu Thanks for the different point of view. It is fine for me if I have a way to capture the coefficients for a fixed representation, and in any case we can sieve by the set of ramified places and by arithmetic conductor to ensure a common Hecke algebra. So my question remains on details about the proof and whether or not this hold (with the normalization I wrote or with some other modifications) :) – Desiderius Severus Nov 7 '17 at 9:40
• This could hold in my opinion but I could not proved it, In fact I try to solve a similar thing 2 months ago when I try to get some information of primes distribution in polynomial partition. But this rely on the underlying L-function which I consider should not vanish at 0. So I need informations on the coefficients. But due to all the nontrivial theorem in this type, the dimension of representation is infinite, this is the key obstacle for us to get the information on coefficient on the underly L-function. – Hu xiyu Nov 7 '17 at 9:51
• To settle this difficult I try to use some multi-scale teiqenicle but fail due to I could not get some estimate to make it coverage. Maybe you could try this method, I do not know what will happen in your setting. – Hu xiyu Nov 7 '17 at 9:52

A normalized version of your guess is right. First note that the $T_{p^n}$'s satisfy the relation $$T_{p^{n+1}} = T_p T_{p^n} - p T_{p^{n-1}}$$ (e.g., Bump Prop 4.6.4). This gives you a recursion relation among Hecke eigenvalues of $\phi$. E.g., $T_{p^2} = T_p T_p - p T_1$ says if the eigenvalue for $T_p$ is $p^{1/2} a_p$, then the eigenvalue for $T_{p^2}$ is $p(a_p^2 - 1)$.
Now you want to compare with coefficients of the Dirichlet series. Say the Satake parameters of $\pi$ are $\alpha=\alpha_p$ and $\alpha^{-1}$, so $a_p = \alpha + \alpha^{-1}$ (e.g., Bump Prop 4.6.6--here I'm assuming trivial central character for simplicity). By the Euler product, you only need to look at the coefficients of the Dirichlet series for the factor at $p$, which is defined to be $$L_p(s,\pi) = \frac 1{1-\alpha p^{-s}} \frac 1{1-\alpha^{-1} p^{-s}} = (\sum \alpha^i p^{-is} )( \sum \alpha^{-j} p^{-js} )$$ Here the coefficient of $p^{ns}$ is $$c_n := \lambda_\pi(p^n) = \sum_{i+j=n} \frac{\alpha^i}{\alpha^j}.$$ Now it is not hard to see the $c_n$'s satisfy the relation $$c_{n+1} = c_1 c_n - c_{n-1}.$$ For instance, for $n=2$, we have $c_2 = c_1 c_1 - c_0 = a_p^2 - 1$. Comparing with the Hecke recurrence, one gets $$a_{p^n} = p^{\frac n2} c_n = p^{\frac n2} \lambda_\pi(p^n),$$ where $a_{p^n}$ is the eigenvalue of $T_{p^n}$.
For ramified representations, yes you can define Hecke operators using appropriate congruence subgroups $K_0(p^n)$ or $K_1(p^n)$, but you should do this a little differently if you want to get coefficients of Dirichlet series of newforms as Hecke eigenvalues. That is, you shouldn't look at all double cosets in $GL_2(\mathbb Z_p$). See e.g. the book of Knightly and Li.