[proofplan] We verify the defining conditions for a modular form of weight $k+\ell$ on $SL_2(\mathbb{Z})$. The product of holomorphic functions is holomorphic on $\mathbb{H}$, and the two automorphy factors multiply to give the correct transformation law. At the cusp $\infty$, we use the $q$-expansions of $f$ and $g$; their product has a convergent $q$-expansion with non-negative powers, and if one original constant term is zero then the product constant term is zero. [/proofplan] [step:Multiply the transformation laws to obtain weight $k+\ell$] Let \begin{align*} h: \mathbb{H} &\to \mathbb{C} \\ z &\mapsto f(z)g(z) \end{align*} be the pointwise product. Since $f$ and $g$ are modular forms of weights $k$ and $\ell$ on $SL_2(\mathbb{Z})$, for every matrix \begin{align*} \gamma = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in SL_2(\mathbb{Z}) \end{align*} and every $z \in \mathbb{H}$, we have \begin{align*} f(\gamma z) &= (cz+d)^k f(z), \\ g(\gamma z) &= (cz+d)^\ell g(z), \end{align*} where $\gamma z := \frac{az+b}{cz+d}$. Multiplying these two identities gives \begin{align*} h(\gamma z) &= f(\gamma z)g(\gamma z) \\ &= (cz+d)^k f(z)(cz+d)^\ell g(z) \\ &= (cz+d)^{k+\ell}h(z). \end{align*} Thus $h$ satisfies the modular transformation law of weight $k+\ell$ for $SL_2(\mathbb{Z})$. [guided] The transformation law is the algebraic part of the definition of a modular form. We must prove it for the new function $h$, not merely state that products behave well. Define the product map \begin{align*} h: \mathbb{H} &\to \mathbb{C} \\ z &\mapsto f(z)g(z). \end{align*} Let \begin{align*} \gamma = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in SL_2(\mathbb{Z}) \end{align*} and let $z \in \mathbb{H}$. Since $f$ has weight $k$, its transformation law gives \begin{align*} f(\gamma z) = (cz+d)^k f(z). \end{align*} Since $g$ has weight $\ell$, its transformation law gives \begin{align*} g(\gamma z) = (cz+d)^\ell g(z). \end{align*} Multiplying the two equations is legitimate because all quantities are complex numbers. We obtain \begin{align*} h(\gamma z) &= f(\gamma z)g(\gamma z) \\ &= (cz+d)^k f(z)(cz+d)^\ell g(z) \\ &= (cz+d)^{k+\ell}f(z)g(z) \\ &= (cz+d)^{k+\ell}h(z). \end{align*} This is exactly the automorphy relation required for weight $k+\ell$. [/guided] [/step] [step:Use holomorphy of products on the upper half-plane] The functions $f: \mathbb{H} \to \mathbb{C}$ and $g: \mathbb{H} \to \mathbb{C}$ are holomorphic by hypothesis. Since pointwise products of holomorphic complex-valued functions are holomorphic, the map $h: \mathbb{H} \to \mathbb{C}$, $z \mapsto f(z)g(z)$, is holomorphic on $\mathbb{H}$. [/step] [step:Multiply the $q$-expansions to prove holomorphy at the cusp] Set \begin{align*} q: \mathbb{H} &\to \mathbb{C} \\ z &\mapsto e^{2\pi i z}. \end{align*} Since $f$ and $g$ are holomorphic at the cusp $\infty$, there exist complex sequences $(a_n)_{n=0}^{\infty}$ and $(b_n)_{n=0}^{\infty}$ such that, for all $z \in \mathbb{H}$ with $|q(z)|$ sufficiently small, \begin{align*} f(z) &= \sum_{n=0}^{\infty} a_n q(z)^n, \\ g(z) &= \sum_{n=0}^{\infty} b_n q(z)^n, \end{align*} and both [power series](/page/Power%20Series) converge in a neighbourhood of $q=0$. Define the sequence $(c_n)_{n=0}^{\infty}$ by \begin{align*} c_n := \sum_{r=0}^{n} a_r b_{n-r}. \end{align*} The product of two convergent power series has the Cauchy product expansion in a possibly smaller neighbourhood of $q=0$, so \begin{align*} h(z) &= f(z)g(z) \\ &= \left(\sum_{r=0}^{\infty} a_r q(z)^r\right) \left(\sum_{s=0}^{\infty} b_s q(z)^s\right) \\ &= \sum_{n=0}^{\infty} c_n q(z)^n \end{align*} for all $z \in \mathbb{H}$ with $|q(z)|$ sufficiently small. This expansion contains only non-negative powers of $q$, so $h$ is holomorphic at the cusp $\infty$. [guided] The cusp condition is best checked in the local coordinate $q=e^{2\pi i z}$. Define \begin{align*} q: \mathbb{H} &\to \mathbb{C} \\ z &\mapsto e^{2\pi i z}. \end{align*} The condition that $f$ is holomorphic at $\infty$ means that near $q=0$ it has a convergent power series with no negative powers: \begin{align*} f(z) = \sum_{n=0}^{\infty} a_n q(z)^n \end{align*} for a complex sequence $(a_n)_{n=0}^{\infty}$. Similarly, the condition that $g$ is holomorphic at $\infty$ gives a complex sequence $(b_n)_{n=0}^{\infty}$ such that \begin{align*} g(z) = \sum_{n=0}^{\infty} b_n q(z)^n \end{align*} near $q=0$. We now multiply these two expansions. For each integer $n \geq 0$, define the coefficient \begin{align*} c_n := \sum_{r=0}^{n} a_r b_{n-r}. \end{align*} This is the Cauchy product coefficient of total degree $n$. Since both original power series converge in a neighbourhood of $q=0$, their product is again represented by the Cauchy product in a possibly smaller neighbourhood of $q=0$. Hence, for all $z \in \mathbb{H}$ with $|q(z)|$ sufficiently small, \begin{align*} h(z) &= f(z)g(z) \\ &= \left(\sum_{r=0}^{\infty} a_r q(z)^r\right) \left(\sum_{s=0}^{\infty} b_s q(z)^s\right) \\ &= \sum_{n=0}^{\infty} c_n q(z)^n. \end{align*} No negative powers of $q$ appear in this expansion. Therefore the product $h$ is holomorphic at the cusp $\infty$. [/guided] [/step] [step:Read off the constant term to prove cuspidality] The constant term of the $q$-expansion of $h$ is \begin{align*} c_0 = a_0 b_0. \end{align*} If $f \in S_k(SL_2(\mathbb{Z}))$, then $a_0=0$, and hence $c_0=0$. If $g \in S_\ell(SL_2(\mathbb{Z}))$, then $b_0=0$, and hence again $c_0=0$. Thus, whenever at least one of $f$ and $g$ is cuspidal, the product $h$ has zero constant term at $\infty$. Therefore $h \in S_{k+\ell}(SL_2(\mathbb{Z}))$. Combining the transformation law, holomorphy on $\mathbb{H}$, and holomorphy at the cusp proves $h \in M_{k+\ell}(SL_2(\mathbb{Z}))$, and the preceding paragraph proves the asserted cusp-form statement. [/step]