[proofplan] Set $m := k-12$. The monomial basis theorem for modular forms identifies the monomials $E_4^aE_6^b$ with $4a+6b=m$ as a basis of $M_m(SL_2(\mathbb{Z}))$. Multiplication by the discriminant cusp form $\Delta$ gives a linear isomorphism from $M_m(SL_2(\mathbb{Z}))$ onto $S_k(SL_2(\mathbb{Z}))$. Therefore the image of the monomial basis under this isomorphism is exactly the displayed family, and hence it is a basis. [/proofplan] [step:Reduce the assertion to the modular form space of weight $k-12$] Define \begin{align*} m := k-12. \end{align*} Since $k \geq 12$ is even, $m$ is a nonnegative even integer. Consider the $\mathbb{C}$-[vector space](/page/Vector%20Space) $M_m(SL_2(\mathbb{Z}))$ of modular forms of weight $m$ on $SL_2(\mathbb{Z})$. By the monomial basis theorem for modular forms on $SL_2(\mathbb{Z})$ (citing a result not yet in the wiki: Monomial Basis Theorem for Modular Forms on $SL_2(\mathbb{Z})$), the set \begin{align*} \mathcal{B}_m := \left\{ E_4^a E_6^b \in M_m(SL_2(\mathbb{Z})) : a,b \in \mathbb{Z}_{\geq 0} \text{ and } 4a+6b=m \right\} \end{align*} is a $\mathbb{C}$-basis of $M_m(SL_2(\mathbb{Z}))$. [/step] [step:Use multiplication by $\Delta$ to identify modular forms with cusp forms] Define the multiplication map \begin{align*} T_\Delta: M_m(SL_2(\mathbb{Z})) &\to S_{m+12}(SL_2(\mathbb{Z})) \\ f &\mapsto \Delta f. \end{align*} Since $\Delta \in S_{12}(SL_2(\mathbb{Z}))$, multiplication by $\Delta$ sends a modular form of weight $m$ to a cusp form of weight $m+12=k$. By the $\Delta$-multiplication isomorphism for level-one modular forms (citing a result not yet in the wiki: Multiplication by $\Delta$ Isomorphism), the map $T_\Delta$ is a $\mathbb{C}$-linear isomorphism \begin{align*} T_\Delta: M_m(SL_2(\mathbb{Z})) \xrightarrow{\cong} S_k(SL_2(\mathbb{Z})). \end{align*} [/step] [step:Take the image of the monomial basis under the isomorphism] Because $T_\Delta$ is a vector space isomorphism and $\mathcal{B}_m$ is a basis of $M_m(SL_2(\mathbb{Z}))$, the image $T_\Delta(\mathcal{B}_m)$ is a basis of $S_k(SL_2(\mathbb{Z}))$. Indeed, if \begin{align*} \sum_{(a,b)} c_{a,b}\, T_\Delta(E_4^aE_6^b)=0 \end{align*} with finitely many coefficients $c_{a,b} \in \mathbb{C}$, then injectivity of $T_\Delta$ gives \begin{align*} \sum_{(a,b)} c_{a,b}\, E_4^aE_6^b=0. \end{align*} Since $\mathcal{B}_m$ is linearly independent, every coefficient $c_{a,b}$ is zero. Thus $T_\Delta(\mathcal{B}_m)$ is linearly independent. For spanning, let $g \in S_k(SL_2(\mathbb{Z}))$. Since $T_\Delta$ is surjective, there exists $f \in M_m(SL_2(\mathbb{Z}))$ such that $T_\Delta(f)=g$. Since $\mathcal{B}_m$ spans $M_m(SL_2(\mathbb{Z}))$, there are coefficients $c_{a,b} \in \mathbb{C}$ such that \begin{align*} f = \sum_{\substack{a,b \in \mathbb{Z}_{\geq 0}\\4a+6b=m}} c_{a,b} E_4^aE_6^b. \end{align*} Applying $T_\Delta$ gives \begin{align*} g = T_\Delta(f) = \sum_{\substack{a,b \in \mathbb{Z}_{\geq 0}\\4a+6b=m}} c_{a,b}\Delta E_4^aE_6^b. \end{align*} Therefore $T_\Delta(\mathcal{B}_m)$ spans $S_k(SL_2(\mathbb{Z}))$. [/step] [step:Identify the image with the displayed family] For each pair $a,b \in \mathbb{Z}_{\geq 0}$ satisfying $4a+6b=m=k-12$, one has \begin{align*} T_\Delta(E_4^aE_6^b)=\Delta E_4^aE_6^b. \end{align*} Hence \begin{align*} T_\Delta(\mathcal{B}_m) = \left\{ \Delta E_4^a E_6^b : a,b \in \mathbb{Z}_{\geq 0} \text{ and } 4a+6b=k-12 \right\}. \end{align*} Since this image is a $\mathbb{C}$-basis of $S_k(SL_2(\mathbb{Z}))$, the displayed family is a $\mathbb{C}$-basis of $S_k(SL_2(\mathbb{Z}))$. [/step]