# Why can Hecke operators be regarded as finite flat cohomological correspondence?

I'm reading the paper "Higher Hida and Coleman theories on the modular curve" by G.Boxer and V.Pilloni. But I'm confused with the different views towards Hecke operators.

$$N$$ is an integer and $$p$$ is a prime such that $$(p,N)=1$$. Let $$X\rightarrow \mathrm{Spec} \mathbb{Z}_p$$ be the compactified modular of level $$\Gamma_1(N)$$ defined by P.Deligne and M.Rapoport. And let $$X_0(p)$$ be the modular curve of level $$\Gamma_1(N)\cap \Gamma_0(p)$$. We have two projections:

$$p_1\colon X_0(p)\rightarrow X,~(E,H)\mapsto E$$

and

$$p_2\colon X_0(p)\rightarrow X,~(E,H)\mapsto E/H,$$ where $$H\subset E[p]$$ is a subgroup of order $$p$$.

I have two problems:

1. Why are those two maps finite flat? It's quite easy to check for $$p_1$$. But I have no ideas with $$p_2$$.
2. Let $$\omega$$ be the sheaf of invariant diffrential. What is the map $$p_2^{\ast} \omega^k\rightarrow p_1^{\ast} \omega^k$$ and how it relates to Hecke operators?
• I don't know if this is the best way to prove this, but there is an automorphism of $X_0(p)$ that exchanges $p_1$ and $p_2$ by sending $(E, H)$ to $(E/H, E[p]/H)$, so if one is finite flat then the other is. To check this is an automorphism you just need to check that its composition with itself is the identity. May 23, 2023 at 9:57
• @WillSawin That's right. But I guess the composition is actually the diamond operator $\langle p \rangle$. May 23, 2023 at 12:57
• Of course, yes. May 23, 2023 at 16:07

We want to define Hecke operators on the complex $$R\Gamma(X, \omega^k)$$, because this is the complex whose $$H^0$$ is modular forms for $$k \ge 1$$ (and whose $$H^1$$ is dual to cusp forms for $$k \le 1$$).
The definition of $$T_p$$ is going to be, essentially, the composite $$R\Gamma(X, \omega^k) \xrightarrow{(p_2)^*} R\Gamma(X_0(p), p_2^* \omega^k) \xrightarrow{???} R\Gamma(X_0(p), p_1^* \omega^k) \xrightarrow{(p_1)_*} R\Gamma(X, \omega^k).$$ (Some of these $$*$$-s should be $$!$$-s, I don't remember exactly which but the paper will tell you.) The middle arrow $$???$$ needs to be there: you can't just compose $$(p_1)_*$$ and $$(p_2)^*$$, the composition doesn't make sense. So this is why we want to make a map between $$p_2^* \omega^k$$ and $$p_1^* \omega^k$$.
As for how we get a map: there is a canonical isogeny $$p_1^* E \to p_2^* E$$, because this is what $$X_0(p)$$ parameterises; so we get a "naive $$T_p$$" by setting $$???$$ to be the pullback of differential forms along this isogeny (or the pushforward along the dual isogeny, if you prefer). The subtle thing is normalising this by the "correct" power of $$p$$, so that the resulting correspondence is defined integrally and its ordinary part is not trivially 0.
• How to understand $X_0(p)$ parametrises $p_1^{\ast}E \rightarrow p_2^{\ast}E$? I know $X_0(p)$ classifies triples $(E,H,P)$. May 24, 2023 at 2:31