# How is the Eichler-Shimura congruence related to L-functions?

My understanding is that the Eichler-Shimura relation expresses the Hecke operator $T_p$ in terms of the geometric Frobenius map. Specifically, $T_p = Frob + Ver$ for Frobenius map $Frob$ and it's transpose $Ver$.

However, Wikipedia states that "the Eichler–Shimura congruence relation expresses the local L-function of a modular curve at a prime p in terms of the eigenvalues of Hecke operators".

I have no idea how that statement comes from the above definition of the congruence. How does such a decomposition of Hecke operators have anything to with its eigenvalues or L-functions of a modular curve?

EDIT: Will's amazing answer gives the connection between L-functions and eigenvalues, but how is the local L-function of a modular curve restated explicitly using Eichler-Shimura?

Recall that the local $L$-function of the modular curve $X$ is $$\frac{1}{\det \left(1 - p^{-s} \operatorname{Frob}_p, H^1(X, \mathbb Q_\ell)\right)}.$$

The denominator is simply a variant of the characteristic polynomial of Frobenius, o it is sufficient to relate the eigenvalues of Frobenius to the Hecke eigenvalues.

Now because $FV = p$, $F$ and $V$ commute, and so for each $\lambda$ an eigenvalue of $F$, $\lambda+ p/\lambda$ is an eigenvalue of $F+V$.

Using this, one can give formulas for the characteristic polynomial of the Hecke operator in terms of the characteristic polynomial of Frobenius, or vice versa.

• Thank you very much! How can you explicitly restate the local L-function using Eichler-Shimura? – TreFox Apr 22 '18 at 18:27
• @TreFox The formula you end up with is $\prod_f 1/ (1 - a_p(f) p^{-1} + p^{1-2s})$ where $f$ are a basis for the cusp forms and $a_p(f)$ is the $p$th Hecke eigenvalue. This is for the compact modular curve, there are additional terms in the noncompact case (which can be related to the Eisenstein series). – Will Sawin Apr 22 '18 at 18:55
• Huh - so, despite the similarties, the L-function of a modular form is not the same as the zeta function of the Hecke operator? (en.wikipedia.org/wiki/Zeta_function_(operator)) – TreFox Apr 22 '18 at 19:50
• @TreFox Because there are only finitely many Hecke eigenvalues on any given space of holomorphic modular forms, the zeta function of the Hecke operator would not make an interesting L-function. – Will Sawin Apr 22 '18 at 20:03
• Is the L-function of a modular form a special case of the operator-zeta function for any operator? – TreFox Apr 22 '18 at 20:29