[proofplan] We use the double-coset definition of the Hecke operator. The modular transformation law follows because right multiplication by an element of $\Gamma$ permutes the left cosets in $\Gamma \backslash \Delta_n$, and the weight-$k$ slash action is invariant under left multiplication by $\Gamma$ when applied to a modular form. Holomorphy on the upper half-plane is immediate from the fact that matrices in $\Delta_n$ preserve $\mathbb{H}$. Finally, holomorphy at the cusp is checked from the standard triangular representatives, which give an explicit Fourier coefficient formula with no negative powers of $q$. [/proofplan]

[step:Declare the slash action and the Hecke operator]Let $\mathbb{C}$ denote the field of complex numbers, let $\mathbb{R}\subset\mathbb{C}$ denote the real line, let $\mathbb{Z}$ denote the ring of integers, and let \begin{align*} \mathbb{H}:=\{z\in\mathbb{C}:\operatorname{Im}z>0\} \end{align*} denote the complex upper half-plane. Let $\Gamma:=SL_2(\mathbb{Z})$, and define \begin{align*} \Delta_n := \{\alpha \in M_2(\mathbb{Z}) : \det \alpha = n\}. \end{align*} For a matrix \begin{align*} \alpha = \begin{pmatrix} a & b\\ c & d \end{pmatrix} \in \Delta_n \end{align*} and a [holomorphic function](/page/Holomorphic%20Function) $f:\mathbb{H}\to \mathbb{C}$, define the weight-$k$ slash action \begin{align*} (f|_k \alpha): \mathbb{H} &\to \mathbb{C}\\ z &\mapsto \det(\alpha)^{k-1}(cz+d)^{-k} f\left(\frac{az+b}{cz+d}\right). \end{align*} The denominator $cz+d$ is nonzero for $z \in \mathbb{H}$. Indeed, if $cz+d=0$ and $c\ne 0$, then \begin{align*} z=-\frac{d}{c}\in\mathbb{R}, \end{align*} contradicting $z\in\mathbb{H}$. If $c=0$, then $d\ne 0$ because $\det \alpha=n>0$. Choose a finite set $R_n \subset \Delta_n$ of representatives for the left cosets $\Gamma\backslash \Delta_n$. Define \begin{align*} (T_n f): \mathbb{H} &\to \mathbb{C}\\ z &\mapsto \sum_{\alpha \in R_n} (f|_k \alpha)(z). \end{align*} Here $M_k(\Gamma)$ denotes the space of holomorphic functions $f:\mathbb{H}\to\mathbb{C}$ satisfying $f|_k\gamma=f$ for every $\gamma\in\Gamma$ and holomorphicity at the cusp, meaning that $f$ has a Fourier expansion in nonnegative powers of $q=e^{2\pi i z}$ for $\operatorname{Im}z$ sufficiently large. This definition is independent of the chosen representatives: if $\alpha'=\gamma\alpha$ with $\gamma\in\Gamma$, then \begin{align*} f|_k\alpha' = f|_k(\gamma\alpha) = (f|_k\gamma)|_k\alpha = f|_k\alpha, \end{align*} because $f\in M_k(\Gamma)$ satisfies $f|_k\gamma=f$ for every $\gamma\in\Gamma$.[/step]

[guided]We first fix the precise normalization of the operator and the ambient notation. Let $\mathbb{C}$ denote the field of complex numbers, let $\mathbb{R}\subset\mathbb{C}$ denote the real line, let $\mathbb{Z}$ denote the ring of integers, and let \begin{align*} \mathbb{H}:=\{z\in\mathbb{C}:\operatorname{Im}z>0\} \end{align*} denote the complex upper half-plane. Let $\Gamma:=SL_2(\mathbb{Z})$. The relevant matrices are \begin{align*} \Delta_n := \{\alpha \in M_2(\mathbb{Z}) : \det \alpha = n\}. \end{align*} Each such matrix has positive determinant, so it acts on the upper half-plane $\mathbb{H}$ by the usual fractional linear transformation. If \begin{align*} \alpha = \begin{pmatrix} a & b\\ c & d \end{pmatrix}, \end{align*} we define \begin{align*} (f|_k \alpha): \mathbb{H} &\to \mathbb{C}\\ z &\mapsto \det(\alpha)^{k-1}(cz+d)^{-k} f\left(\frac{az+b}{cz+d}\right). \end{align*} The expression is well-defined on $\mathbb{H}$. Indeed, if $cz+d=0$ and $c\ne 0$, then \begin{align*} z=-\frac{d}{c}\in\mathbb{R}, \end{align*} contradicting $z\in\mathbb{H}$. If $c=0$, then $d\ne 0$ because $\det\alpha=ad=n>0$. Now choose a finite set $R_n\subset\Delta_n$ representing the left cosets in $\Gamma\backslash\Delta_n$, and define \begin{align*} (T_n f): \mathbb{H} &\to \mathbb{C}\\ z &\mapsto \sum_{\alpha\in R_n}(f|_k\alpha)(z). \end{align*} Here $M_k(\Gamma)$ denotes the space of holomorphic functions $f:\mathbb{H}\to\mathbb{C}$ satisfying $f|_k\gamma=f$ for every $\gamma\in\Gamma$ and holomorphicity at the cusp, meaning that $f$ has a Fourier expansion in nonnegative powers of $q=e^{2\pi i z}$ for $\operatorname{Im}z$ sufficiently large. Why does this not depend on the choice of representative? If $\alpha'$ is another representative of the same left coset, then $\alpha'=\gamma\alpha$ for some $\gamma\in\Gamma$. The slash action satisfies associativity: \begin{align*} f|_k(\gamma\alpha)=(f|_k\gamma)|_k\alpha. \end{align*} Since $f$ is a modular form of weight $k$ for $\Gamma$, it satisfies $f|_k\gamma=f$ for every $\gamma\in\Gamma$. Therefore \begin{align*} f|_k\alpha'=f|_k(\gamma\alpha)=(f|_k\gamma)|_k\alpha=f|_k\alpha. \end{align*} Thus the coset sum is intrinsic.[/guided]

[step:Show the Hecke sum has the modular transformation law]Let $\gamma\in\Gamma$. Since right multiplication by $\gamma$ maps $\Delta_n$ bijectively to itself, it induces a permutation of the finite set $\Gamma\backslash\Delta_n$. Hence for each $\alpha\in R_n$ there are unique $\beta(\alpha)\in R_n$ and some $\delta_\alpha\in\Gamma$ such that \begin{align*} \alpha\gamma=\delta_\alpha \beta(\alpha). \end{align*} Using associativity of the slash action and the $\Gamma$-invariance of $f$, we obtain \begin{align*} (T_n f)|_k\gamma &= \sum_{\alpha\in R_n} f|_k(\alpha\gamma)\\ &= \sum_{\alpha\in R_n} f|_k(\delta_\alpha\beta(\alpha))\\ &= \sum_{\alpha\in R_n} (f|_k\delta_\alpha)|_k\beta(\alpha)\\ &= \sum_{\alpha\in R_n} f|_k\beta(\alpha)\\ &= \sum_{\beta\in R_n} f|_k\beta\\ &= T_n f. \end{align*} Thus $T_n f$ transforms with weight $k$ under every $\gamma\in\Gamma$.[/step]

[guided]We now prove the central invariance statement. Fix $\gamma\in\Gamma$. Since $\det\gamma=1$, right multiplication by $\gamma$ preserves determinants: \begin{align*} \det(\alpha\gamma)=\det(\alpha)\det(\gamma)=n. \end{align*} Thus $\alpha\mapsto \alpha\gamma$ is a bijection $\Delta_n\to\Delta_n$, with inverse $\alpha\mapsto \alpha\gamma^{-1}$. Consequently it permutes the left cosets in $\Gamma\backslash\Delta_n$. Because $R_n$ contains exactly one representative from each left coset, for every $\alpha\in R_n$ there are a unique representative $\beta(\alpha)\in R_n$ and an element $\delta_\alpha\in\Gamma$ such that \begin{align*} \alpha\gamma=\delta_\alpha\beta(\alpha). \end{align*} The map $\alpha\mapsto\beta(\alpha)$ is a permutation of $R_n$. Now compute using the slash action. First, \begin{align*} (T_n f)|_k\gamma = \left(\sum_{\alpha\in R_n} f|_k\alpha\right)|_k\gamma = \sum_{\alpha\in R_n} f|_k(\alpha\gamma), \end{align*} where the last equality uses linearity of the slash action and associativity. Substitute the coset relation $\alpha\gamma=\delta_\alpha\beta(\alpha)$: \begin{align*} (T_n f)|_k\gamma = \sum_{\alpha\in R_n} f|_k(\delta_\alpha\beta(\alpha)) = \sum_{\alpha\in R_n} (f|_k\delta_\alpha)|_k\beta(\alpha). \end{align*} Since $\delta_\alpha\in\Gamma$ and $f\in M_k(\Gamma)$, the modular transformation law gives $f|_k\delta_\alpha=f$. Therefore \begin{align*} (T_n f)|_k\gamma = \sum_{\alpha\in R_n} f|_k\beta(\alpha). \end{align*} Because $\alpha\mapsto\beta(\alpha)$ is a permutation of $R_n$, the final sum is just the original Hecke sum: \begin{align*} (T_n f)|_k\gamma = \sum_{\beta\in R_n} f|_k\beta = T_n f. \end{align*} This proves that $T_n f$ satisfies the same weight-$k$ modular transformation law as $f$.[/guided]

[step:Verify holomorphy on the upper half-plane] For each $\alpha\in R_n$, the map \begin{align*} \varphi_\alpha:\mathbb{H}&\to\mathbb{H}\\ z&\mapsto \frac{az+b}{cz+d} \end{align*} is holomorphic, and the factor $z\mapsto \det(\alpha)^{k-1}(cz+d)^{-k}$ is holomorphic on $\mathbb{H}$. Since $f:\mathbb{H}\to\mathbb{C}$ is holomorphic, each summand $f|_k\alpha$ is holomorphic on $\mathbb{H}$. The sum defining $T_n f$ is finite, so $T_n f$ is holomorphic on $\mathbb{H}$. [/step]

[step:Use triangular representatives to compute the Fourier expansion at the cusp]We first justify the triangular representatives. Let \begin{align*} \alpha= \begin{pmatrix} A & B\\ C & D \end{pmatrix} \in\Delta_n. \end{align*} Choose $a:=\gcd(A,C)>0$ and integers $r,s\in\mathbb{Z}$ with $rA+sC=a$. Then \begin{align*} \gamma_0:= \begin{pmatrix} r & s\\ -C/a & A/a \end{pmatrix} \in SL_2(\mathbb{Z}), \end{align*} because $\det\gamma_0=(rA+sC)/a=1$, and \begin{align*} \gamma_0\alpha= \begin{pmatrix} a & b_0\\ 0 & d \end{pmatrix} \end{align*} with $d=n/a>0$. Multiplying on the left by \begin{align*} \begin{pmatrix} 1 & t\\ 0 & 1 \end{pmatrix} \in\Gamma \end{align*} replaces $b_0$ by $b_0+td$, so a unique choice of $t\in\mathbb{Z}$ gives $0\le b<d$. Conversely, if two such triangular matrices lie in the same left $\Gamma$-coset, then comparing the bottom-left, determinant, and top-right entries forces the same $a$, the same $d$, and the same residue $b$ modulo $d$; the condition $0\le b<d$ gives equality. Hence the matrices \begin{align*} \alpha_{a,b,d}:= \begin{pmatrix} a & b\\ 0 & d \end{pmatrix}, \qquad ad=n,\quad d>0,\quad 0\le b<d, \end{align*} are exactly one representative for each class in $\Gamma\backslash\Delta_n$. With this choice, \begin{align*} (T_n f)(z) = \sum_{\substack{ad=n\\ d>0}} \sum_{b=0}^{d-1} n^{k-1}d^{-k} f\left(\frac{az+b}{d}\right). \end{align*} Since $f$ is holomorphic at the cusp, there are coefficients $(c_m)_{m\ge 0}\subset\mathbb{C}$ such that, for $q=e^{2\pi i z}$ and $\operatorname{Im}z$ sufficiently large, \begin{align*} f(z)=\sum_{m=0}^{\infty} c_m q^m. \end{align*} For each fixed pair $(a,d)$ with $ad=n$ and $d>0$, the substitution $w=(az+b)/d$ satisfies $\operatorname{Im}w=a\operatorname{Im}z/d$. Thus, after increasing the lower bound on $\operatorname{Im}z$ if necessary, all finitely many points $(az+b)/d$ lie in the half-plane where the [Fourier series](/page/Fourier%20Series) of $f$ converges locally uniformly. Since the sums over $(a,d)$ and $b$ are finite, substitution and rearrangement with these finite sums are justified by local [uniform convergence](/page/Uniform%20Convergence) of the Fourier series on that half-plane. Substituting this expansion gives \begin{align*} (T_n f)(z) &= \sum_{\substack{ad=n\\ d>0}} n^{k-1}d^{-k} \sum_{b=0}^{d-1} \sum_{m=0}^{\infty} c_m \exp\left(2\pi i m\frac{az+b}{d}\right)\\ &= \sum_{\substack{ad=n\\ d>0}} n^{k-1}d^{-k} \sum_{m=0}^{\infty} c_m q^{ma/d} \sum_{b=0}^{d-1}\exp\left(\frac{2\pi i mb}{d}\right). \end{align*} The finite geometric sum satisfies \begin{align*} \sum_{b=0}^{d-1}\exp\left(\frac{2\pi i mb}{d}\right) = \begin{cases} d, & d\mid m,\\ 0, & d\nmid m. \end{cases} \end{align*} Writing $m=d\ell$ in the surviving terms, with $\ell\in\mathbb{Z}_{\ge 0}$, we obtain \begin{align*} (T_n f)(z) &= \sum_{\substack{ad=n\\ d>0}} n^{k-1}d^{1-k} \sum_{\ell=0}^{\infty} c_{d\ell}q^{a\ell}. \end{align*} Therefore the coefficient of $q^r$ is \begin{align*} \sum_{a\mid \gcd(n,r)} a^{k-1}c_{nr/a^2}, \end{align*} with the usual interpretation that, for $r=0$, every $a\mid n$ contributes the constant coefficient $a^{k-1}c_0$. In particular, no negative powers of $q$ occur.[/step]

[guided]The remaining condition for being a modular form is holomorphy at the cusp. We check it by using representatives adapted to Fourier expansions, and first verify why those representatives are valid. Let \begin{align*} \alpha= \begin{pmatrix} A & B\\ C & D \end{pmatrix} \in\Delta_n. \end{align*} Set $a:=\gcd(A,C)>0$. Choose $r,s\in\mathbb{Z}$ such that $rA+sC=a$, and define \begin{align*} \gamma_0:= \begin{pmatrix} r & s\\ -C/a & A/a \end{pmatrix}. \end{align*} Then $\gamma_0\in SL_2(\mathbb{Z})$ because $\det\gamma_0=(rA+sC)/a=1$, and left multiplication gives \begin{align*} \gamma_0\alpha= \begin{pmatrix} a & b_0\\ 0 & d \end{pmatrix} \end{align*} with $d=n/a>0$. Multiplying on the left by $\begin{pmatrix}1&t\\0&1\end{pmatrix}\in\Gamma$ changes $b_0$ to $b_0+td$, so we may choose the unique residue $b$ satisfying $0\le b<d$. Uniqueness follows because if $\gamma\begin{pmatrix}a&b\\0&d\end{pmatrix}=\begin{pmatrix}a'&b'\\0&d'\end{pmatrix}$ with $\gamma=\begin{pmatrix}p&q\\r&s\end{pmatrix}\in\Gamma$, then the bottom-left entry gives $ra=0$, hence $r=0$; from $ps=1$ and $d,d'>0$ we get $p=s=1$, so $a'=a$, $d'=d$, and $b'=b+qd$. Since both $b$ and $b'$ lie in $\{0,\dots,d-1\}$, $b=b'$. Therefore a representative set for $\Gamma\backslash\Delta_n$ is \begin{align*} \alpha_{a,b,d}:= \begin{pmatrix} a & b\\ 0 & d \end{pmatrix}, \qquad ad=n,\quad d>0,\quad 0\le b<d. \end{align*} For such a matrix, the slash action becomes \begin{align*} (f|_k\alpha_{a,b,d})(z) = n^{k-1}d^{-k}f\left(\frac{az+b}{d}\right), \end{align*} because the lower-left entry is $0$ and the lower-right entry is $d$. Hence \begin{align*} (T_n f)(z) = \sum_{\substack{ad=n\\ d>0}} \sum_{b=0}^{d-1} n^{k-1}d^{-k} f\left(\frac{az+b}{d}\right). \end{align*} Since $f$ is holomorphic at the cusp, it has a Fourier expansion with no negative powers: \begin{align*} f(z)=\sum_{m=0}^{\infty} c_m q^m, \qquad q=e^{2\pi i z}, \end{align*} for all $z\in\mathbb{H}$ with $\operatorname{Im}z$ sufficiently large. For each fixed divisor pair $(a,d)$ with $ad=n$ and $d>0$, the imaginary part of $w=(az+b)/d$ is $a\operatorname{Im}z/d$. Because there are only finitely many such pairs and finitely many values of $b$ for each $d$, we can require $\operatorname{Im}z$ to be large enough that every point $(az+b)/d$ lies in the half-plane where the Fourier expansion of $f$ converges locally uniformly. This local uniform convergence justifies substituting the expansion into each summand and then moving the finite sums over $a,d,b$ past the infinite Fourier series. Substituting $w=(az+b)/d$ gives \begin{align*} f\left(\frac{az+b}{d}\right) = \sum_{m=0}^{\infty} c_m \exp\left(2\pi i m\frac{az+b}{d}\right) = \sum_{m=0}^{\infty} c_m q^{ma/d}\exp\left(\frac{2\pi i mb}{d}\right). \end{align*} Therefore \begin{align*} (T_n f)(z) &= \sum_{\substack{ad=n\\ d>0}} n^{k-1}d^{-k} \sum_{b=0}^{d-1} \sum_{m=0}^{\infty} c_m q^{ma/d}\exp\left(\frac{2\pi i mb}{d}\right)\\ &= \sum_{\substack{ad=n\\ d>0}} n^{k-1}d^{-k} \sum_{m=0}^{\infty} c_m q^{ma/d} \sum_{b=0}^{d-1}\exp\left(\frac{2\pi i mb}{d}\right). \end{align*} The inner sum over $b$ is a finite geometric sum over the $d$th roots of unity. It equals $d$ exactly when $d$ divides $m$, and it equals $0$ otherwise: \begin{align*} \sum_{b=0}^{d-1}\exp\left(\frac{2\pi i mb}{d}\right) = \begin{cases} d, & d\mid m,\\ 0, & d\nmid m. \end{cases} \end{align*} Thus only the terms $m=d\ell$, with $\ell\in\mathbb{Z}_{\ge 0}$, remain. Substituting $m=d\ell$ gives \begin{align*} (T_n f)(z) &= \sum_{\substack{ad=n\\ d>0}} n^{k-1}d^{1-k} \sum_{\ell=0}^{\infty} c_{d\ell}q^{a\ell}. \end{align*} This is a Fourier series in nonnegative integral powers of $q$, because $a\ell\ge 0$. Equivalently, if $r\ge 0$, the coefficient of $q^r$ is obtained by requiring $r=a\ell$, so $a\mid r$ and $d=n/a$. The coefficient is then \begin{align*} \sum_{a\mid \gcd(n,r)} a^{k-1}c_{nr/a^2}. \end{align*} For $r=0$, the same formula means that every divisor $a$ of $n$ contributes to the constant term. The important conclusion is that no negative powers of $q$ appear, so $T_n f$ is holomorphic at the cusp.[/guided]

[step:Conclude that the Hecke operator preserves $M_k(\Gamma)$] We have shown that $T_n f$ is holomorphic on $\mathbb{H}$, satisfies $(T_n f)|_k\gamma=T_n f$ for every $\gamma\in\Gamma$, and has a Fourier expansion at the cusp with no negative powers of $q$. Hence $T_n f\in M_k(\Gamma)$. Since $f\in M_k(\Gamma)$ was arbitrary, $T_n(M_k(\Gamma))\subseteq M_k(\Gamma)$. [/step]