[proofplan] We compute the normalized geometric correspondence on the Tate curve at the cusp, using the complex Tate uniformization $\operatorname{Tate}(q)(\mathbb{C}) \cong \mathbb{C}^{\times}/q^{\mathbb{Z}}$ and writing $\mu_p:=\{\zeta\in\mathbb{C}^{\times}:\zeta^p=1\}$ for the group of $p$th roots of unity. The hypothesis $p \nmid N$ ensures that quotienting by a cyclic subgroup of order $p$ preserves the $\Gamma_0(N)$ level structure, so the degree-$p$ correspondence is defined on level $\Gamma_0(N)$. The $p+1$ cyclic subgroups split into the $p$ subgroups generated by $\zeta q^{1/p}$, giving quotient parameter $\zeta q^{1/p}$ after the normalized trace, and the subgroup $\mu_p$, giving quotient parameter $q^p$. The root-of-unity average extracts exactly the coefficients with index divisible by $p$, while the $\mu_p$ quotient contributes the second term with the factor $p^{k-1}$ coming from the pullback relation on invariant differentials and the global trace factor $p^{-1}$. [/proofplan]

[step:Evaluate the correspondence on the Tate curve at the cusp]Let $D^{\times}$ denote the formal punctured disc over $\mathbb{C}$ with coordinate $q$, and let $\mu_p:=\{\zeta\in\mathbb{C}^{\times}:\zeta^p=1\}$ denote the group of $p$th roots of unity. Let \begin{align*} \operatorname{Tate}: D^{\times} &\to \{\text{generalized elliptic curves over }D^{\times}\} \\ q &\mapsto \operatorname{Tate}(q) \end{align*} denote the [Tate curve](/page/Tate%20Curve), and let $\omega_{\mathrm{Tate}}$ denote its standard invariant differential, corresponding analytically to the logarithmic differential \begin{align*} \frac{d z}{z} \end{align*} on $\mathbb{C}^{\times}$. By the [q-expansion principle](/page/q-Expansion%20Principle), evaluating $f \in M_k(\Gamma_0(N);\mathbb{C})$ on this pair gives \begin{align*} f(\operatorname{Tate}(q),\omega_{\mathrm{Tate}}) = \sum_{m=0}^{\infty} a_m q^m. \end{align*} Let $A_N \subset \operatorname{Tate}(q)$ denote the cyclic subgroup of order $N$ specifying the $\Gamma_0(N)$ level structure. Because $p \nmid N$, every cyclic subgroup $C \subset \operatorname{Tate}(q)[p]$ of order $p$ satisfies $C\cap A_N=\{0\}$: any point in the intersection has order dividing both $p$ and $N$. Hence the quotient isogeny $\pi_C:\operatorname{Tate}(q)\to \operatorname{Tate}(q)/C$ restricts injectively to $A_N$, and $\pi_C(A_N)$ is again a cyclic subgroup of order $N$. Therefore $\operatorname{Tate}(q)/C$ inherits a $\Gamma_0(N)$ level structure. For each order-$p$ subgroup $C\subset \operatorname{Tate}(q)[p]$, define the normalized quotient differential $\widetilde{\omega}_C$ on $\operatorname{Tate}(q)/C$ as follows. If $C=C_{\zeta}$ is complementary to $\mu_p$, then $\widetilde{\omega}_{C_{\zeta}}$ is the standard invariant differential on $\operatorname{Tate}(\zeta q^{1/p})$. If $C=\mu_p$, then \begin{align*} \widetilde{\omega}_{\mu_p}:=p^{-1}\omega_{\mathrm{Tate}(q^p)}. \end{align*} With this convention, the normalized geometric operator is the correspondence \begin{align*} T_p^{\mathrm{geom}} f(\operatorname{Tate}(q),\omega_{\mathrm{Tate}}) := \frac{1}{p}\left( \sum_{C \subset \operatorname{Tate}(q)[p] \atop C \cap \mu_p = \{1\}} f(\operatorname{Tate}(q)/C,\widetilde{\omega}_C) + f(\operatorname{Tate}(q)/\mu_p,\widetilde{\omega}_{\mu_p}) \right), \end{align*} where the sum ranges over the $p$ cyclic subgroups complementary to $\mu_p$. [claim:The cyclic order-$p$ quotients of the Tate curve have the stated Tate parameters] With the analytic uniformization $\operatorname{Tate}(q)(\mathbb{C}) \cong \mathbb{C}^{\times}/q^{\mathbb{Z}}$, the order-$p$ cyclic subgroups are $\mu_p$ and \begin{align*} C_{\zeta}:=\langle \zeta q^{1/p}\rangle \quad\text{for }\zeta \in \mu_p. \end{align*} The quotient by $C_{\zeta}$ is analytically $\operatorname{Tate}(\zeta q^{1/p})$, and the quotient by $\mu_p$ is analytically $\operatorname{Tate}(q^p)$. [/claim] [proof] In $\mathbb{C}^{\times}/q^{\mathbb{Z}}$, the $p$-torsion consists of classes represented by elements $u \in \mathbb{C}^{\times}$ satisfying $u^p \in q^{\mathbb{Z}}$. Thus every class has a representative $\zeta q^{r/p}$ with $\zeta \in \mu_p$ and $r \in \{0,1,\dots,p-1\}$. The subgroup with $r=0$ is $\mu_p$. For $r \ne 0$, replacing the generator by a power gives a generator of the form $\zeta q^{1/p}$, so the remaining $p$ cyclic subgroups are precisely $C_{\zeta}=\langle \zeta q^{1/p}\rangle$ for $\zeta \in \mu_p$. For $C_{\zeta}$, quotienting $\mathbb{C}^{\times}/q^{\mathbb{Z}}$ by the class of $\zeta q^{1/p}$ enlarges the period subgroup from $q^{\mathbb{Z}}$ to $(\zeta q^{1/p})^{\mathbb{Z}}$, which gives \begin{align*} (\mathbb{C}^{\times}/q^{\mathbb{Z}})/C_{\zeta} \cong \mathbb{C}^{\times}/(\zeta q^{1/p})^{\mathbb{Z}} = \operatorname{Tate}(\zeta q^{1/p})(\mathbb{C}). \end{align*} For $\mu_p$, the quotient map is induced by \begin{align*} \varphi_p: \mathbb{C}^{\times} &\to \mathbb{C}^{\times} \\ z &\mapsto z^p. \end{align*} It sends $q^{\mathbb{Z}}$ to $(q^p)^{\mathbb{Z}}$ and has kernel $\mu_p$, hence it induces \begin{align*} (\mathbb{C}^{\times}/q^{\mathbb{Z}})/\mu_p \cong \mathbb{C}^{\times}/(q^p)^{\mathbb{Z}} = \operatorname{Tate}(q^p)(\mathbb{C}). \end{align*} [/proof] For the quotients by $C_{\zeta}$, the normalized quotient differential is the standard differential on $\operatorname{Tate}(\zeta q^{1/p})$. For the quotient by $\mu_p$, let \begin{align*} \pi_{\mu_p}:\operatorname{Tate}(q)&\to\operatorname{Tate}(q^p) \end{align*} denote the isogeny induced by $\varphi_p(z)=z^p$. Since \begin{align*} \varphi_p^*\left(\frac{d z}{z}\right)=\frac{d(z^p)}{z^p}=p\frac{d z}{z}, \end{align*} we have \begin{align*} \pi_{\mu_p}^*\omega_{\mathrm{Tate}(q^p)}=p\omega_{\mathrm{Tate}}. \end{align*} The chosen normalized quotient differential is therefore $\widetilde{\omega}_{\mu_p}=p^{-1}\omega_{\mathrm{Tate}(q^p)}$. Homogeneity of a weight-$k$ modular form gives \begin{align*} f(\operatorname{Tate}(q^p),p^{-1}\omega_{\mathrm{Tate}(q^p)}) = p^k f(\operatorname{Tate}(q^p),\omega_{\mathrm{Tate}(q^p)}). \end{align*} Multiplying by the global prefactor $p^{-1}$ in the normalized degree-$p$ correspondence produces the coefficient $p^{k-1}$. Therefore the evaluation of the normalized correspondence at the Tate curve is \begin{align*} (T_p^{\mathrm{geom}} f)(q) = \frac{1}{p}\sum_{\zeta \in \mu_p} f(\operatorname{Tate}(\zeta q^{1/p}),\omega_{\mathrm{Tate}}) + p^{k-1}f(\operatorname{Tate}(q^p),\omega_{\mathrm{Tate}(q^p)}). \end{align*}[/step]

[guided]The comparison is made at the cusp because the $q$-expansion is, by definition, the value of the modular form on the Tate curve. We work over the complex punctured disc $D^{\times}$ with coordinate $q$ and use the analytic model \begin{align*} \operatorname{Tate}(q)(\mathbb{C}) \cong \mathbb{C}^{\times}/q^{\mathbb{Z}}. \end{align*} The standard invariant differential is represented by $d z / z$ on $\mathbb{C}^{\times}$. Thus the [q-expansion principle](/page/q-Expansion%20Principle) identifies \begin{align*} f(\operatorname{Tate}(q),\omega_{\mathrm{Tate}}) = \sum_{m=0}^{\infty} a_m q^m. \end{align*} Why is the condition $p \nmid N$ needed? Let $A_N \subset \operatorname{Tate}(q)$ be the cyclic subgroup of order $N$ defining the $\Gamma_0(N)$ level structure. If $C\subset \operatorname{Tate}(q)[p]$ is cyclic of order $p$, then $C\cap A_N=\{0\}$ because every point in the intersection has order dividing both $p$ and $N$, and $\gcd(p,N)=1$. Therefore the quotient isogeny $\pi_C:\operatorname{Tate}(q)\to \operatorname{Tate}(q)/C$ is injective on $A_N$, so $\pi_C(A_N)$ is cyclic of order $N$. This gives the quotient curve a $\Gamma_0(N)$ level structure and makes the degree-$p$ Hecke correspondence well-defined on $M_k(\Gamma_0(N);\mathbb{C})$. We now identify the cyclic subgroups. A point of order dividing $p$ in $\mathbb{C}^{\times}/q^{\mathbb{Z}}$ is represented by an element $u \in \mathbb{C}^{\times}$ such that $u^p \in q^{\mathbb{Z}}$. Therefore $u$ can be represented as $\zeta q^{r/p}$ with $\zeta \in \mu_p$ and $r \in \{0,1,\dots,p-1\}$. The case $r=0$ gives the subgroup $\mu_p$. Every case $r \ne 0$ generates one of the subgroups \begin{align*} C_{\zeta}:=\langle \zeta q^{1/p}\rangle, \quad \zeta \in \mu_p. \end{align*} These are exactly the $p+1$ cyclic subgroups of order $p$. For $C_{\zeta}$, quotienting by the class of $\zeta q^{1/p}$ enlarges the period subgroup from $q^{\mathbb{Z}}$ to $(\zeta q^{1/p})^{\mathbb{Z}}$. Hence \begin{align*} (\mathbb{C}^{\times}/q^{\mathbb{Z}})/C_{\zeta} \cong \mathbb{C}^{\times}/(\zeta q^{1/p})^{\mathbb{Z}} = \operatorname{Tate}(\zeta q^{1/p})(\mathbb{C}). \end{align*} For $\mu_p$, the quotient map is induced by the degree-$p$ map \begin{align*} \varphi_p: \mathbb{C}^{\times} &\to \mathbb{C}^{\times} \\ z &\mapsto z^p. \end{align*} Its kernel is $\mu_p$, and it sends the period subgroup $q^{\mathbb{Z}}$ to $(q^p)^{\mathbb{Z}}$. Hence \begin{align*} (\mathbb{C}^{\times}/q^{\mathbb{Z}})/\mu_p \cong \mathbb{C}^{\times}/(q^p)^{\mathbb{Z}} = \operatorname{Tate}(q^p)(\mathbb{C}). \end{align*} It remains to record the normalization of the differential, because this is where the factor $p^{k-1}$ enters. The pullback of the standard invariant differential under $\varphi_p$ is \begin{align*} \varphi_p^*\left(\frac{d z}{z}\right)=\frac{d(z^p)}{z^p}=p\frac{d z}{z}. \end{align*} Equivalently, for the induced isogeny \begin{align*} \pi_{\mu_p}:\operatorname{Tate}(q)&\to\operatorname{Tate}(q^p), \end{align*} we have \begin{align*} \pi_{\mu_p}^*\omega_{\mathrm{Tate}(q^p)}=p\omega_{\mathrm{Tate}}. \end{align*} The normalized quotient differential for the canonical subgroup is defined to be \begin{align*} \widetilde{\omega}_{\mu_p}:=p^{-1}\omega_{\mathrm{Tate}(q^p)}. \end{align*} A weight-$k$ modular form is homogeneous of degree $-k$ in the invariant differential: replacing $\omega$ by $\lambda\omega$ multiplies the value by $\lambda^{-k}$. Therefore evaluating at $\widetilde{\omega}_{\mu_p}$ multiplies the standard value on $\operatorname{Tate}(q^p)$ by $p^k$. The normalized correspondence has the global prefactor $p^{-1}$ on the full degree-$p$ trace, including the canonical quotient, so the canonical quotient contributes with total scalar $p^{-1}p^k=p^{k-1}$. Thus the normalized geometric Hecke operator evaluates on the Tate curve as \begin{align*} (T_p^{\mathrm{geom}} f)(q) = \frac{1}{p}\sum_{\zeta \in \mu_p} f(\operatorname{Tate}(\zeta q^{1/p}),\omega_{\mathrm{Tate}}) + p^{k-1}f(\operatorname{Tate}(q^p),\omega_{\mathrm{Tate}(q^p)}). \end{align*}[/guided]

[step:Average over the étale quotients to extract the coefficients $a_{pm}$]For each $\zeta \in \mu_p$, substitute $\zeta q^{1/p}$ into the $q$-expansion of $f$: \begin{align*} f(\operatorname{Tate}(\zeta q^{1/p}),\omega_{\mathrm{Tate}}) = \sum_{m=0}^{\infty} a_m(\zeta q^{1/p})^m = \sum_{m=0}^{\infty} a_m \zeta^m q^{m/p}. \end{align*} Averaging over $\mu_p$ gives \begin{align*} \frac{1}{p}\sum_{\zeta \in \mu_p} f(\operatorname{Tate}(\zeta q^{1/p}),\omega_{\mathrm{Tate}}) &= \sum_{m=0}^{\infty} a_m q^{m/p} \left(\frac{1}{p}\sum_{\zeta \in \mu_p}\zeta^m\right). \end{align*} The root-of-unity sum satisfies \begin{align*} \frac{1}{p}\sum_{\zeta \in \mu_p}\zeta^m = \begin{cases} 1, & p \mid m,\\ 0, & p \nmid m. \end{cases} \end{align*} Therefore only the terms with $m=pn$ survive, and the étale contribution equals \begin{align*} \sum_{n=0}^{\infty} a_{pn}q^n. \end{align*}[/step]

[guided]We now compute the normalized trace over the $p$ étale quotients. For a fixed $\zeta \in \mu_p$, replacing $q$ by $\zeta q^{1/p}$ in the $q$-expansion gives \begin{align*} f(\operatorname{Tate}(\zeta q^{1/p}),\omega_{\mathrm{Tate}}) = \sum_{m=0}^{\infty} a_m(\zeta q^{1/p})^m = \sum_{m=0}^{\infty} a_m \zeta^m q^{m/p}. \end{align*} The average over all $p$ roots of unity is \begin{align*} \frac{1}{p}\sum_{\zeta \in \mu_p} f(\operatorname{Tate}(\zeta q^{1/p}),\omega_{\mathrm{Tate}}) &= \frac{1}{p}\sum_{\zeta \in \mu_p} \sum_{m=0}^{\infty} a_m \zeta^m q^{m/p} \\ &= \sum_{m=0}^{\infty} a_m q^{m/p} \left(\frac{1}{p}\sum_{\zeta \in \mu_p}\zeta^m\right). \end{align*} The inner sum is the standard orthogonality relation for characters of the cyclic group $\mu_p$: \begin{align*} \frac{1}{p}\sum_{\zeta \in \mu_p}\zeta^m = \begin{cases} 1, & p \mid m,\\ 0, & p \nmid m. \end{cases} \end{align*} Thus every term with non-integral exponent $m/p$ is killed by the averaging, and exactly the terms with $m=pn$ remain. Hence the étale part contributes \begin{align*} \sum_{n=0}^{\infty} a_{pn}q^n. \end{align*}[/guided]

[step:Compute the canonical subgroup contribution] For the subgroup $\mu_p \subset \operatorname{Tate}(q)[p]$, the preceding quotient calculation gives the quotient curve $\operatorname{Tate}(q^p)$. Substituting $q^p$ into the $q$-expansion of $f$ gives \begin{align*} f(\operatorname{Tate}(q^p),\omega_{\mathrm{Tate}(q^p)}) = \sum_{m=0}^{\infty} a_m q^{pm}. \end{align*} The isogeny $\pi_{\mu_p}:\operatorname{Tate}(q)\to\operatorname{Tate}(q^p)$ satisfies $\pi_{\mu_p}^*\omega_{\mathrm{Tate}(q^p)}=p\omega_{\mathrm{Tate}}$, as computed from $z\mapsto z^p$. Since $f$ has weight $k$, evaluating at the normalized quotient differential $p^{-1}\omega_{\mathrm{Tate}(q^p)}$ multiplies the standard value of $f$ by $p^k$, and the normalized correspondence contributes the global trace factor $p^{-1}$. Hence the total scalar multiplying the $q^p$ substitution is $p^{-1}p^k=p^{k-1}$, and the canonical contribution is \begin{align*} p^{k-1}\sum_{m=0}^{\infty} a_m q^{pm}. \end{align*} [/step]

[step:Combine the two contributions] Adding the étale contribution and the canonical contribution, we obtain \begin{align*} (T_p^{\mathrm{geom}} f)(q) &= \sum_{m=0}^{\infty} a_{pm}q^m + p^{k-1}\sum_{m=0}^{\infty} a_m q^{pm}. \end{align*} This is exactly the classical $q$-expansion formula for the Hecke operator $T_p$ at a prime $p \nmid N$, so the geometric correspondence agrees with the classical operator on $q$-expansions. [/step]