[proofplan] The proof has two parts. First, the valence formula forces every nonzero cusp form of full level and weight $k$ to have $k \geq 12$, so all weights below $12$ contribute no cusp forms. Second, multiplication by the modular discriminant $\Delta \in S_{12}(SL_2(\mathbb{Z}))$ identifies $M_{k-12}(SL_2(\mathbb{Z}))$ with $S_k(SL_2(\mathbb{Z}))$ for $k \geq 12$. The explicit formula follows by substituting $k-12$ into the known dimension formula for full-level modular forms. [/proofplan]

[step:Use the valence formula to rule out cusp forms of weight below $12$]Let $k \in \mathbb{Z}$ with $k < 12$. Suppose, toward a contradiction, that there exists a nonzero cusp form \begin{align*} f: \mathfrak{H} \to \mathbb{C} \end{align*} with $f \in S_k(SL_2(\mathbb{Z}))$, where $\mathfrak{H} := \{z \in \mathbb{C} : \operatorname{Im}(z) > 0\}$ is the complex upper half-plane. If $k$ is odd, then the modular transformation law for the matrix $-I \in SL_2(\mathbb{Z})$ gives \begin{align*} f(z) = (-1)^k f(z) = -f(z) \end{align*} for every $z \in \mathfrak{H}$, hence $f = 0$, a contradiction. Thus a nonzero cusp form has even weight. Assume now that $k$ is even. By the valence formula for full-level modular forms (citing a result not yet in the wiki: Valence Formula for Modular Forms of Full Level), a nonzero modular form $f \in M_k(SL_2(\mathbb{Z}))$ satisfies \begin{align*} \operatorname{ord}_{\infty}(f) + \frac{1}{2}\operatorname{ord}_{i}(f) + \frac{1}{3}\operatorname{ord}_{\rho}(f) + \sum_{p \in SL_2(\mathbb{Z}) \backslash \mathfrak{H},\, p \ne i,\rho} \operatorname{ord}_{p}(f) = \frac{k}{12}, \end{align*} where $\rho := e^{2\pi i/3}$, $\operatorname{ord}_{p}(f)$ denotes the vanishing order of $f$ at the orbit of $p$, and $\operatorname{ord}_{\infty}(f)$ denotes the vanishing order at the cusp. Because $f$ is holomorphic on $\mathfrak{H}$, every finite vanishing order on the left-hand side is nonnegative. Because $f$ is a cusp form, $\operatorname{ord}_{\infty}(f) \geq 1$. Therefore \begin{align*} \frac{k}{12} = \operatorname{ord}_{\infty}(f) + \frac{1}{2}\operatorname{ord}_{i}(f) + \frac{1}{3}\operatorname{ord}_{\rho}(f) + \sum_{p \in SL_2(\mathbb{Z}) \backslash \mathfrak{H},\, p \ne i,\rho} \operatorname{ord}_{p}(f) \geq 1. \end{align*} Hence $k \geq 12$, contradicting $k < 12$. Thus $S_k(SL_2(\mathbb{Z})) = \{0\}$ for every $k < 12$, and \begin{align*} \dim_{\mathbb{C}} S_k(SL_2(\mathbb{Z})) = 0. \end{align*}[/step]

[guided]We first show that a nonzero cusp form cannot have weight below $12$. Let $k \in \mathbb{Z}$ with $k < 12$, and suppose that \begin{align*} f: \mathfrak{H} \to \mathbb{C} \end{align*} is a nonzero element of $S_k(SL_2(\mathbb{Z}))$, where $\mathfrak{H} := \{z \in \mathbb{C} : \operatorname{Im}(z) > 0\}$. There is a preliminary parity obstruction. Since $-I \in SL_2(\mathbb{Z})$, the modular transformation law gives \begin{align*} f(z) = (-1)^k f(z) \end{align*} for every $z \in \mathfrak{H}$. If $k$ is odd, then $(-1)^k = -1$, so $f(z) = -f(z)$ for every $z$, hence $f = 0$. This contradicts the assumption that $f$ is nonzero. Therefore any nonzero full-level modular form must have even weight. Now assume $k$ is even. The key input is the valence formula for full-level modular forms (citing a result not yet in the wiki: Valence Formula for Modular Forms of Full Level). It applies because $f$ is a nonzero holomorphic modular form of weight $k$ for $SL_2(\mathbb{Z})$. The formula states that \begin{align*} \operatorname{ord}_{\infty}(f) + \frac{1}{2}\operatorname{ord}_{i}(f) + \frac{1}{3}\operatorname{ord}_{\rho}(f) + \sum_{p \in SL_2(\mathbb{Z}) \backslash \mathfrak{H},\, p \ne i,\rho} \operatorname{ord}_{p}(f) = \frac{k}{12}, \end{align*} where $\rho := e^{2\pi i/3}$, $\operatorname{ord}_{p}(f)$ is the vanishing order at the orbit of $p$, and $\operatorname{ord}_{\infty}(f)$ is the vanishing order at the cusp. The point of using the valence formula is that every term on the left is nonnegative, while a cusp form has at least one zero at the cusp. Since $f$ is holomorphic on $\mathfrak{H}$, all finite vanishing orders are nonnegative. Since $f$ is a cusp form, its Fourier expansion at infinity has zero constant term, so $\operatorname{ord}_{\infty}(f) \geq 1$. Therefore \begin{align*} \frac{k}{12} = \operatorname{ord}_{\infty}(f) + \frac{1}{2}\operatorname{ord}_{i}(f) + \frac{1}{3}\operatorname{ord}_{\rho}(f) + \sum_{p \in SL_2(\mathbb{Z}) \backslash \mathfrak{H},\, p \ne i,\rho} \operatorname{ord}_{p}(f) \geq 1. \end{align*} Thus $k \geq 12$, contradicting $k < 12$. Hence no nonzero cusp form of weight $k < 12$ exists, so \begin{align*} S_k(SL_2(\mathbb{Z})) = \{0\}, \qquad \dim_{\mathbb{C}} S_k(SL_2(\mathbb{Z})) = 0. \end{align*}[/guided]

[step:Divide by the discriminant to identify cusp forms with lower-weight modular forms]Let $k \geq 12$. Let \begin{align*} \Delta: \mathfrak{H} \to \mathbb{C} \end{align*} denote the modular discriminant, so $\Delta \in S_{12}(SL_2(\mathbb{Z}))$, $\Delta$ has a simple zero at the cusp, and $\Delta$ has no zeros on $\mathfrak{H}$. Define the complex-[linear map](/page/Linear%20Map) \begin{align*} T_k: M_{k-12}(SL_2(\mathbb{Z})) &\to S_k(SL_2(\mathbb{Z})) \\ g &\mapsto \Delta g. \end{align*} The product $\Delta g$ is a modular form of weight $12 + (k-12) = k$. Since $\Delta$ vanishes at the cusp, $\Delta g$ is a cusp form. Thus $T_k$ is well-defined. The map $T_k$ is injective because $\Delta$ is not identically zero and $\mathfrak{H}$ is connected: if $\Delta g = 0$, then $g = 0$ on the nonempty [open set](/page/Open%20Set) where $\Delta \ne 0$, hence $g = 0$ by holomorphicity. To prove surjectivity, let $f \in S_k(SL_2(\mathbb{Z}))$. Define \begin{align*} h: \mathfrak{H} &\to \mathbb{C} \\ z &\mapsto \frac{f(z)}{\Delta(z)}. \end{align*} Because $\Delta$ has no zeros on $\mathfrak{H}$, the function $h$ is holomorphic on $\mathfrak{H}$. The quotient of the weight-$k$ transformation law for $f$ by the weight-$12$ transformation law for $\Delta$ gives that $h$ transforms with weight $k-12$ under $SL_2(\mathbb{Z})$. At the cusp, both $f$ and $\Delta$ vanish, and $\Delta$ has a simple zero. Since $f$ is cuspidal, the quotient $h = f/\Delta$ is holomorphic at the cusp. Therefore $h \in M_{k-12}(SL_2(\mathbb{Z}))$, and $T_k(h) = f$. Thus $T_k$ is a complex-vector-space isomorphism, so \begin{align*} \dim_{\mathbb{C}} S_k(SL_2(\mathbb{Z})) = \dim_{\mathbb{C}} M_{k-12}(SL_2(\mathbb{Z})). \end{align*}[/step]

[guided]We now construct the promised isomorphism. Let $k \geq 12$, and let \begin{align*} \Delta: \mathfrak{H} \to \mathbb{C} \end{align*} be the modular discriminant. We use the standard facts that $\Delta \in S_{12}(SL_2(\mathbb{Z}))$, that $\Delta$ has a simple zero at the cusp, and that $\Delta$ has no zeros on $\mathfrak{H}$ (citing a result not yet in the wiki: Discriminant Division Isomorphism for Full-Level Cusp Forms). Define \begin{align*} T_k: M_{k-12}(SL_2(\mathbb{Z})) &\to S_k(SL_2(\mathbb{Z})) \\ g &\mapsto \Delta g. \end{align*} This is the natural map to try because multiplying by $\Delta$ increases weight by $12$ and introduces vanishing at the cusp. We verify that $T_k$ is well-defined. If $g \in M_{k-12}(SL_2(\mathbb{Z}))$, then the product $\Delta g$ is holomorphic on $\mathfrak{H}$. Its transformation law has weight \begin{align*} 12 + (k-12) = k. \end{align*} Moreover, $\Delta$ vanishes at the cusp and $g$ is holomorphic at the cusp, so $\Delta g$ vanishes at the cusp. Hence $\Delta g \in S_k(SL_2(\mathbb{Z}))$. Next, $T_k$ is injective. Suppose $T_k(g) = 0$, so $\Delta g = 0$ on $\mathfrak{H}$. Since $\Delta$ is holomorphic and not identically zero, the set on which $\Delta \ne 0$ is a nonempty open subset of $\mathfrak{H}$. On that open set, $g = 0$. Because $g$ is holomorphic on the connected domain $\mathfrak{H}$, the identity theorem gives $g = 0$ on all of $\mathfrak{H}$. Thus $\ker T_k = \{0\}$. Finally, $T_k$ is surjective. Let $f \in S_k(SL_2(\mathbb{Z}))$. Since $\Delta$ has no zeros on $\mathfrak{H}$, the quotient \begin{align*} h: \mathfrak{H} &\to \mathbb{C} \\ z &\mapsto \frac{f(z)}{\Delta(z)} \end{align*} is holomorphic on $\mathfrak{H}$. Dividing the weight-$k$ modular transformation law of $f$ by the weight-$12$ transformation law of $\Delta$ shows that $h$ transforms with weight $k-12$ under $SL_2(\mathbb{Z})$. At the cusp, $f$ has vanishing order at least $1$ because it is cuspidal, while $\Delta$ has vanishing order exactly $1$. Therefore the quotient $f/\Delta$ has no pole at the cusp and is holomorphic there. Hence $h \in M_{k-12}(SL_2(\mathbb{Z}))$, and $T_k(h) = f$. Thus $T_k$ is a complex-linear bijection. Therefore \begin{align*} \dim_{\mathbb{C}} S_k(SL_2(\mathbb{Z})) = \dim_{\mathbb{C}} M_{k-12}(SL_2(\mathbb{Z})). \end{align*}[/guided]

[step:Substitute into the full-level modular form dimension formula] Let $k \geq 12$ be even, and define the integer \begin{align*} \ell := k - 12. \end{align*} Then $\ell \geq 0$ is even and $\ell \equiv k \pmod{12}$. By the dimension formula for full-level modular forms (citing a result not yet in the wiki: [Dimension Formula for Modular Forms of Full Level](/theorems/4238)), for even $\ell \geq 0$, \begin{align*} \dim_{\mathbb{C}} M_{\ell}(SL_2(\mathbb{Z})) = \begin{cases} \left\lfloor \frac{\ell}{12}\right\rfloor, & \ell \equiv 2 \pmod{12},\\ \left\lfloor \frac{\ell}{12}\right\rfloor + 1, & \ell \not\equiv 2 \pmod{12}. \end{cases} \end{align*} Since $\ell = k - 12$, \begin{align*} \left\lfloor \frac{\ell}{12}\right\rfloor = \left\lfloor \frac{k-12}{12}\right\rfloor = \left\lfloor \frac{k}{12}\right\rfloor - 1. \end{align*} Also $\ell \equiv k \pmod{12}$. Combining this formula with the isomorphism from the previous step gives \begin{align*} \dim_{\mathbb{C}} S_k(SL_2(\mathbb{Z})) = \begin{cases} \left\lfloor \frac{k}{12}\right\rfloor - 1, & k \equiv 2 \pmod{12},\\ \left\lfloor \frac{k}{12}\right\rfloor, & k \not\equiv 2 \pmod{12}. \end{cases} \end{align*} This proves the stated dimension formula. [/step]