[proofplan] We prove that multiplication by $\Delta$ sends modular forms of weight $k-12$ to cusp forms of weight $k$, then prove injectivity from the fact that $\Delta$ is not identically zero. For surjectivity, given a cusp form $f$ of weight $k$, we form the quotient $f/\Delta$ on $\mathbb H$. The product expansion of $\Delta$ shows that it has no zeros on $\mathbb H$, and the $q$-expansions at infinity show that the quotient is holomorphic at the cusp and has weight $k-12$. [/proofplan] [step:Verify that multiplication by $\Delta$ lands in cusp forms of weight $k$] Fix an integer $k \geq 12$. Define \begin{align*} \Phi_k: M_{k-12}(SL_2(\mathbb Z)) &\to \{\text{holomorphic functions } \mathbb H \to \mathbb C\} \\ g &\mapsto \Delta g. \end{align*} Let $g \in M_{k-12}(SL_2(\mathbb Z))$. Since $\Delta \in S_{12}(SL_2(\mathbb Z))$, the product $\Delta g: \mathbb H \to \mathbb C$ is holomorphic and satisfies the weight-$k$ transformation law under $SL_2(\mathbb Z)$: \begin{align*} (\Delta g)\left(\frac{az+b}{cz+d}\right) &= \Delta\left(\frac{az+b}{cz+d}\right) g\left(\frac{az+b}{cz+d}\right) \\ &= (cz+d)^{12}\Delta(z)\,(cz+d)^{k-12}g(z) \\ &= (cz+d)^k(\Delta g)(z) \end{align*} for every $\begin{pmatrix}a & b \\ c & d\end{pmatrix} \in SL_2(\mathbb Z)$ and every $z \in \mathbb H$. It remains to check the cusp condition. Let \begin{align*} q: \mathbb H &\to \{w \in \mathbb C : 0 < |w| < 1\} \\ z &\mapsto e^{2\pi i z}. \end{align*} Since $g$ is holomorphic at infinity, there is a [holomorphic function](/page/Holomorphic%20Function) \begin{align*} G: \{w \in \mathbb C : |w| < 1\} &\to \mathbb C \end{align*} such that $g(z)=G(q(z))$. Since \begin{align*} \Delta(z)=q(z)\prod_{n=1}^{\infty}(1-q(z)^n)^{24}, \end{align*} there is a holomorphic function \begin{align*} D: \{w \in \mathbb C : |w| < 1\} &\to \mathbb C \end{align*} with $D(0)=0$ and $D(w)=w+O(w^2)$ as $w \to 0$, such that $\Delta(z)=D(q(z))$. Thus $(\Delta g)(z)=D(q(z))G(q(z))$, and $D(w)G(w)$ vanishes at $w=0$. Hence $\Delta g \in S_k(SL_2(\mathbb Z))$. The map $\Phi_k$ is complex-linear because multiplication by the fixed function $\Delta$ distributes over addition and scalar multiplication. [/step] [step:Prove injectivity using nonvanishing of $\Delta$ as a function] Suppose $g \in M_{k-12}(SL_2(\mathbb Z))$ and $\Phi_k(g)=0$. Then \begin{align*} \Delta(z)g(z)=0 \end{align*} for every $z \in \mathbb H$. The function $\Delta$ is not identically zero because its $q$-expansion begins with $q$. Therefore the set \begin{align*} U := \{z \in \mathbb H : \Delta(z) \neq 0\} \end{align*} is a nonempty open subset of $\mathbb H$. On $U$, the equality $\Delta g=0$ gives $g(z)=0$. We justify the [analytic continuation](/page/Analytic%20Continuation) step directly. Let \begin{align*} Z := \{z \in \mathbb H : g(z)=0\}. \end{align*} The set $Z$ is closed in $\mathbb H$ by continuity of $g$, and $U \subset Z$, so the interior of $Z$ is nonempty. If $z_0$ is a boundary point in $\mathbb H$ of the interior of $Z$, then the Taylor expansion of $g$ at $z_0$ has every coefficient equal to $0$, because $g$ vanishes on points of the interior of $Z$ accumulating at $z_0$. Hence $g$ vanishes on a neighbourhood of $z_0$, contradicting that $z_0$ is a boundary point. Thus the interior of $Z$ is both open and closed in the connected domain $\mathbb H$, and therefore $Z=\mathbb H$. Hence $\ker \Phi_k=\{0\}$, so $\Phi_k$ is injective. [/step] [step:Show that $\Delta$ has no zeros on the upper half-plane] Let $z \in \mathbb H$, and set $q=e^{2\pi i z}$. Since $\operatorname{Im}(z)>0$, we have \begin{align*} |q|=|e^{2\pi i z}|=e^{-2\pi \operatorname{Im}(z)}<1. \end{align*} The product formula for the discriminant gives \begin{align*} \Delta(z)=q\prod_{n=1}^{\infty}(1-q^n)^{24}. \end{align*} Since $q \neq 0$, it is enough to show that the infinite product is nonzero. For every $n \geq 1$, $|q^n|<1$, so $1-q^n \neq 0$. Moreover, \begin{align*} \sum_{n=1}^{\infty} |q^n| = \frac{|q|}{1-|q|}<\infty. \end{align*} We now record the product argument in this case. Since $\sum_{n=1}^{\infty}|q^n|<\infty$, the series $\sum_{n=1}^{\infty}\log(1-q^n)$ converges absolutely after choosing the holomorphic branch of $\log(1+w)$ on $|w|<1$ and using the estimate $|\log(1+w)|\leq 2|w|$ for $|w|\leq 1/2$, with the finitely many remaining factors handled separately. Exponentiating the convergent sum gives \begin{align*} \prod_{n=1}^{\infty}(1-q^n) =\exp\left(\sum_{n=1}^{\infty}\log(1-q^n)\right)\neq 0. \end{align*} Therefore $\Delta(z)\neq 0$ for every $z \in \mathbb H$. [guided] We need this step because the surjectivity argument divides by $\Delta$. Division is legitimate on $\mathbb H$ only if $\Delta$ has no zeros there. Fix $z \in \mathbb H$ and define $q=e^{2\pi i z}$. Since $\operatorname{Im}(z)>0$, we have \begin{align*} |q| = |e^{2\pi i z}| = e^{-2\pi \operatorname{Im}(z)} < 1. \end{align*} The product expansion of the discriminant is \begin{align*} \Delta(z)=q\prod_{n=1}^{\infty}(1-q^n)^{24}. \end{align*} The factor $q$ is nonzero. For each integer $n \geq 1$, the inequality $|q^n|<1$ implies $q^n \neq 1$, hence $1-q^n \neq 0$. It remains to rule out the possibility that the infinite product converges to zero. The absolute convergence estimate \begin{align*} \sum_{n=1}^{\infty}|q^n|=\frac{|q|}{1-|q|}<\infty \end{align*} lets us prove this directly. Choose $N \in \mathbb N$ such that $|q|^n \leq 1/2$ for every $n \geq N$. On the disk $|w|<1$, take the holomorphic branch of $\log(1+w)$ given near $0$ by its [power series](/page/Power%20Series); for $|w|\leq 1/2$ it satisfies $|\log(1+w)|\leq 2|w|$. Therefore \begin{align*} \sum_{n=N}^{\infty}|\log(1-q^n)|\leq 2\sum_{n=N}^{\infty}|q|^n<\infty. \end{align*} The finitely many factors with $1\leq n<N$ are nonzero, and the tail product satisfies \begin{align*} \prod_{n=N}^{\infty}(1-q^n) =\exp\left(\sum_{n=N}^{\infty}\log(1-q^n)\right)\neq 0. \end{align*} Multiplying by the finitely many preceding nonzero factors gives \begin{align*} \prod_{n=1}^{\infty}(1-q^n)\neq 0. \end{align*} Raising this nonzero number to the twenty-fourth power preserves nonvanishing. Thus $\Delta(z)\neq 0$. [/guided] [/step] [step:Construct the inverse image of a cusp form by dividing by $\Delta$] Let $f \in S_k(SL_2(\mathbb Z))$. Define \begin{align*} h: \mathbb H &\to \mathbb C \\ z &\mapsto \frac{f(z)}{\Delta(z)}. \end{align*} The preceding step shows that $\Delta(z)\neq 0$ for every $z \in \mathbb H$, so $h$ is a holomorphic function on $\mathbb H$. We verify the transformation law. For every $\gamma=\begin{pmatrix}a & b \\ c & d\end{pmatrix} \in SL_2(\mathbb Z)$ and every $z \in \mathbb H$, \begin{align*} h\left(\frac{az+b}{cz+d}\right) &= \frac{f\left(\frac{az+b}{cz+d}\right)}{\Delta\left(\frac{az+b}{cz+d}\right)} \\ &= \frac{(cz+d)^k f(z)}{(cz+d)^{12}\Delta(z)} \\ &= (cz+d)^{k-12}h(z). \end{align*} Thus $h$ satisfies the weight-$(k-12)$ modular transformation law. It remains to prove holomorphy at infinity. Since $f$ is a cusp form, there is a holomorphic function \begin{align*} F: \{w \in \mathbb C : |w|<1\} &\to \mathbb C \end{align*} such that $f(z)=F(q(z))$ and $F(0)=0$. Hence $F(w)=wF_1(w)$ for a holomorphic function \begin{align*} F_1: \{w \in \mathbb C : |w|<1\} &\to \mathbb C. \end{align*} The discriminant has the form \begin{align*} \Delta(z)=q(z)D_1(q(z)), \end{align*} where \begin{align*} D_1: \{w \in \mathbb C : |w|<1\} &\to \mathbb C \\ w &\mapsto \prod_{n=1}^{\infty}(1-w^n)^{24}. \end{align*} We justify that $D_1$ is holomorphic on the unit disk. If $0<r<1$ and $|w|\leq r$, then \begin{align*} \sum_{n=1}^{\infty}|w|^n\leq \sum_{n=1}^{\infty}r^n<\infty. \end{align*} The same logarithmic product argument as above gives [uniform convergence](/page/Uniform%20Convergence) of the product on $\{w\in\mathbb C:|w|\leq r\}$, and each finite partial product is holomorphic. Hence $D_1$ is holomorphic by locally uniform convergence of holomorphic functions. Also $D_1(0)=1$. Therefore $D_1$ is nonzero in a neighbourhood of $0$, and \begin{align*} \frac{F(w)}{wD_1(w)}=\frac{F_1(w)}{D_1(w)} \end{align*} is holomorphic near $w=0$. This proves that $h$ is holomorphic at infinity. Hence $h \in M_{k-12}(SL_2(\mathbb Z))$. [guided] The only delicate point in defining $h=f/\Delta$ is holomorphy. On $\mathbb H$, holomorphy follows from the previous step: $\Delta$ has no zeros on $\mathbb H$. The quotient of two holomorphic functions is holomorphic wherever the denominator is nonzero, so \begin{align*} h: \mathbb H &\to \mathbb C \\ z &\mapsto \frac{f(z)}{\Delta(z)} \end{align*} is holomorphic. Next we check the modular transformation law. Since $f$ has weight $k$ and $\Delta$ has weight $12$, for every $\gamma=\begin{pmatrix}a & b \\ c & d\end{pmatrix} \in SL_2(\mathbb Z)$, \begin{align*} h\left(\frac{az+b}{cz+d}\right) &= \frac{f\left(\frac{az+b}{cz+d}\right)}{\Delta\left(\frac{az+b}{cz+d}\right)} \\ &= \frac{(cz+d)^k f(z)}{(cz+d)^{12}\Delta(z)} \\ &= (cz+d)^{k-12}h(z). \end{align*} Thus the quotient has exactly the expected weight: dividing by a weight-$12$ modular form lowers the weight by $12$. Finally we verify holomorphy at the cusp. Let \begin{align*} q: \mathbb H &\to \{w \in \mathbb C : 0<|w|<1\} \\ z &\mapsto e^{2\pi i z}. \end{align*} Because $f$ is a cusp form, its $q$-expansion has no constant term. Equivalently, there is a holomorphic function \begin{align*} F: \{w \in \mathbb C : |w|<1\} &\to \mathbb C \end{align*} with $F(0)=0$ and $f(z)=F(q(z))$. Since $F(0)=0$, the removable zero [factor theorem](/theorems/3235) gives a holomorphic function \begin{align*} F_1: \{w \in \mathbb C : |w|<1\} &\to \mathbb C \end{align*} such that $F(w)=wF_1(w)$. For $\Delta$, the product expansion gives \begin{align*} \Delta(z)=q(z)D_1(q(z)), \end{align*} where \begin{align*} D_1: \{w \in \mathbb C : |w|<1\} &\to \mathbb C \\ w &\mapsto \prod_{n=1}^{\infty}(1-w^n)^{24}. \end{align*} We need $D_1$ to be holomorphic near $0$, not merely formally defined by a product. Fix $0<r<1$. For every $w$ with $|w|\leq r$, \begin{align*} \sum_{n=1}^{\infty}|w|^n\leq \sum_{n=1}^{\infty}r^n<\infty. \end{align*} Using the logarithmic estimate for $\log(1-u)$ on $|u|\leq 1/2$, after discarding finitely many initial factors uniformly in $|w|\leq r$, the series $\sum_{n=1}^{\infty}\log(1-w^n)$ converges uniformly on $|w|\leq r$. Hence the product defining $D_1$ converges uniformly on compact subsets of the unit disk. Since the finite partial products are holomorphic, $D_1$ is holomorphic on the unit disk. Also $D_1(0)=1$, so continuity gives a neighbourhood of $0$ on which $D_1$ is nonzero. Therefore near $w=0$, \begin{align*} \frac{F(w)}{wD_1(w)}=\frac{wF_1(w)}{wD_1(w)}=\frac{F_1(w)}{D_1(w)} \end{align*} is holomorphic. This proves that $h=f/\Delta$ has a holomorphic $q$-expansion at infinity, so $h \in M_{k-12}(SL_2(\mathbb Z))$. [/guided] [/step] [step:Conclude that multiplication by $\Delta$ is an isomorphism] For the cusp form $f \in S_k(SL_2(\mathbb Z))$ chosen above, the function $h=f/\Delta$ belongs to $M_{k-12}(SL_2(\mathbb Z))$ and satisfies \begin{align*} \Phi_k(h)=\Delta h=f. \end{align*} Thus $\Phi_k$ is surjective. Since we already proved that $\Phi_k$ is injective and complex-linear, $\Phi_k$ is an isomorphism of complex vector spaces. [/step]