[proofplan] The proof separates odd and even weights. Odd weights vanish because the matrix $-I \in \mathrm{SL}_2(\mathbb{Z})$ fixes every point of the upper half-plane but acts on weight-$k$ modular forms by the scalar $(-1)^k$. For even weights, the [structure theorem for level-one modular forms](/theorems/4235) identifies a basis of $M_k(\mathrm{SL}_2(\mathbb{Z}))$ with the monomials $E_4^a E_6^b$ of total weight $k$, so the dimension is reduced to counting nonnegative integer solutions of $4a+6b=k$. [/proofplan]

[step:Use the action of $-I$ to eliminate odd weights]Let $\mathbb{H} := \{z \in \mathbb{C} : \operatorname{Im}(z) > 0\}$ denote the complex upper half-plane. Let \begin{align*} f: \mathbb{H} &\to \mathbb{C} \end{align*} be an element of $M_k(\mathrm{SL}_2(\mathbb{Z}))$. The defining transformation law for a modular form of weight $k$ says that, for every matrix \begin{align*} \gamma = \begin{pmatrix} a & b\\ c & d \end{pmatrix} \in \mathrm{SL}_2(\mathbb{Z}) \end{align*} and every $z \in \mathbb{H}$, \begin{align*} f\left(\frac{az+b}{cz+d}\right) = (cz+d)^k f(z). \end{align*} Apply this with $\gamma = -I$, so $a=d=-1$ and $b=c=0$. Then $\gamma z=z$ and $cz+d=-1$, hence \begin{align*} f(z) = (-1)^k f(z) \end{align*} for every $z \in \mathbb{H}$. If $k$ is odd, then $(-1)^k=-1$, so \begin{align*} f(z) = -f(z) \end{align*} for every $z \in \mathbb{H}$. Therefore $f(z)=0$ for every $z \in \mathbb{H}$, and hence \begin{align*} M_k(\mathrm{SL}_2(\mathbb{Z})) = \{0\}. \end{align*}[/step]

[guided]The point of this step is that level-one modular forms transform under every element of $\mathrm{SL}_2(\mathbb{Z})$, including the central element $-I$. Let \begin{align*} f: \mathbb{H} &\to \mathbb{C} \end{align*} be a modular form of weight $k$, where $\mathbb{H} := \{z \in \mathbb{C} : \operatorname{Im}(z)>0\}$. The transformation law gives \begin{align*} f\left(\frac{az+b}{cz+d}\right) = (cz+d)^k f(z) \end{align*} for every \begin{align*} \begin{pmatrix} a & b\\ c & d \end{pmatrix} \in \mathrm{SL}_2(\mathbb{Z}) \end{align*} and every $z \in \mathbb{H}$. Now take the particular matrix $-I$. It acts on $\mathbb{H}$ by \begin{align*} z \mapsto \frac{-z+0}{0z-1} = z, \end{align*} so it fixes every point of $\mathbb{H}$. However, the automorphy factor is $cz+d=-1$. Therefore the transformation law becomes \begin{align*} f(z)=(-1)^k f(z). \end{align*} When $k$ is odd, this is \begin{align*} f(z)=-f(z), \end{align*} so $2f(z)=0$ for every $z \in \mathbb{H}$. Since the functions are complex-valued and $2 \neq 0$ in $\mathbb{C}$, this forces $f(z)=0$ for every $z$. Thus the only modular form of odd weight is the zero form.[/guided]

[step:Reduce the even-weight dimension to a lattice-point count]Assume now that $k$ is even, and define the integer $n := k/2$. By the structure theorem for level-one modular forms (citing a result not yet in the wiki: Structure Theorem for Level-One Modular Forms), the graded $\mathbb{C}$-algebra of level-one modular forms is \begin{align*} \bigoplus_{j \ge 0} M_j(\mathrm{SL}_2(\mathbb{Z})) \cong \mathbb{C}[E_4,E_6], \end{align*} where $E_4 \in M_4(\mathrm{SL}_2(\mathbb{Z}))$ and $E_6 \in M_6(\mathrm{SL}_2(\mathbb{Z}))$ are algebraically independent. Therefore the monomials \begin{align*} E_4^a E_6^b \end{align*} with $a,b \in \mathbb{Z}_{\ge 0}$ and \begin{align*} 4a+6b=k \end{align*} form a $\mathbb{C}$-basis of $M_k(\mathrm{SL}_2(\mathbb{Z}))$. Dividing the weight equation by $2$, we obtain the equivalent Diophantine equation \begin{align*} 2a+3b=n. \end{align*} Hence \begin{align*} \dim_{\mathbb{C}} M_k(\mathrm{SL}_2(\mathbb{Z})) = \#\{(a,b)\in \mathbb{Z}_{\ge 0}^2 : 2a+3b=n\}. \end{align*}[/step]

[guided]For even $k$, the structure theorem for level-one modular forms converts the analytic problem into a finite counting problem. The theorem says that the graded ring of level-one modular forms is the polynomial algebra generated by the Eisenstein series $E_4$ and $E_6$: \begin{align*} \bigoplus_{j \ge 0} M_j(\mathrm{SL}_2(\mathbb{Z})) \cong \mathbb{C}[E_4,E_6], \end{align*} where $E_4$ has weight $4$ and $E_6$ has weight $6$ (citing a result not yet in the wiki: Structure Theorem for Level-One Modular Forms). Because this is a polynomial algebra, the distinct monomials \begin{align*} E_4^a E_6^b \end{align*} are linearly independent over $\mathbb{C}$. The weight of such a monomial is \begin{align*} 4a+6b. \end{align*} Therefore the weight-$k$ piece has a basis indexed exactly by pairs $(a,b)\in \mathbb{Z}_{\ge 0}^2$ satisfying \begin{align*} 4a+6b=k. \end{align*} Since $k$ is even, define $n:=k/2$. Dividing the equation by $2$ gives \begin{align*} 2a+3b=n. \end{align*} Thus computing the dimension of $M_k(\mathrm{SL}_2(\mathbb{Z}))$ is the same as counting the nonnegative integer solutions of this equation: \begin{align*} \dim_{\mathbb{C}} M_k(\mathrm{SL}_2(\mathbb{Z})) = \#\{(a,b)\in \mathbb{Z}_{\ge 0}^2 : 2a+3b=n\}. \end{align*}[/guided]

[step:Count the nonnegative solutions of $2a+3b=n$]For a fixed $b \in \mathbb{Z}_{\ge 0}$, the equation $2a+3b=n$ determines \begin{align*} a = \frac{n-3b}{2}. \end{align*} Thus $a$ is a nonnegative integer precisely when \begin{align*} 0 \le b \le \left\lfloor \frac{n}{3}\right\rfloor \end{align*} and $n-3b$ is even. Since $3b \equiv b \pmod{2}$, the parity condition is \begin{align*} b \equiv n \pmod{2}. \end{align*} Therefore the desired dimension is the number of integers $b$ satisfying \begin{align*} 0 \le b \le \left\lfloor \frac{n}{3}\right\rfloor, \qquad b \equiv n \pmod{2}. \end{align*} Write \begin{align*} k = 12q+r, \end{align*} where $q=\left\lfloor k/12\right\rfloor$ and $r\in\{0,2,4,6,8,10\}$, since $k$ is even. Then \begin{align*} n = \frac{k}{2} = 6q+\frac{r}{2}. \end{align*} We count the admissible values of $b$ in each residue class. If $r=0$, then $n=6q$ and $\lfloor n/3\rfloor=2q$. The admissible $b$ are the even integers in $\{0,1,\dots,2q\}$, namely $0,2,\dots,2q$, giving $q+1$ solutions. If $r=2$, then $n=6q+1$ and $\lfloor n/3\rfloor=2q$. The admissible $b$ are the odd integers in $\{0,1,\dots,2q\}$, namely $1,3,\dots,2q-1$, giving $q$ solutions. If $r=4$, then $n=6q+2$ and $\lfloor n/3\rfloor=2q$. The admissible $b$ are the even integers in $\{0,1,\dots,2q\}$, giving $q+1$ solutions. If $r=6$, then $n=6q+3$ and $\lfloor n/3\rfloor=2q+1$. The admissible $b$ are the odd integers in $\{0,1,\dots,2q+1\}$, giving $q+1$ solutions. If $r=8$, then $n=6q+4$ and $\lfloor n/3\rfloor=2q+1$. The admissible $b$ are the even integers in $\{0,1,\dots,2q+1\}$, giving $q+1$ solutions. If $r=10$, then $n=6q+5$ and $\lfloor n/3\rfloor=2q+1$. The admissible $b$ are the odd integers in $\{0,1,\dots,2q+1\}$, giving $q+1$ solutions. Hence the count is $q$ exactly when $r=2$, and it is $q+1$ for all other even residues $r$ modulo $12$.[/step]

[guided]We now solve the counting problem completely. Fix $b \in \mathbb{Z}_{\ge 0}$. The equation \begin{align*} 2a+3b=n \end{align*} forces \begin{align*} a=\frac{n-3b}{2}. \end{align*} Thus, after choosing $b$, there is at most one possible value of $a$. This value is allowed exactly when two things happen: first $a\ge 0$, and second $a$ is an integer. The inequality $a\ge 0$ is \begin{align*} n-3b \ge 0, \end{align*} which is equivalent to \begin{align*} 0 \le b \le \left\lfloor \frac{n}{3}\right\rfloor. \end{align*} The integrality condition says that $n-3b$ is even. Since $3b$ has the same parity as $b$, this is equivalent to \begin{align*} b \equiv n \pmod{2}. \end{align*} So the problem is reduced to counting the integers $b$ in the interval \begin{align*} 0 \le b \le \left\lfloor \frac{n}{3}\right\rfloor \end{align*} whose parity agrees with $n$. Now write \begin{align*} k=12q+r, \end{align*} where $q=\lfloor k/12\rfloor$ and, because $k$ is even, \begin{align*} r\in\{0,2,4,6,8,10\}. \end{align*} Since $n=k/2$, we have \begin{align*} n=6q+\frac{r}{2}. \end{align*} We inspect the six possible even residues modulo $12$. For $r=0$, the integer $n=6q$ is even and $\lfloor n/3\rfloor=2q$. Thus $b$ must be even in $\{0,1,\dots,2q\}$, giving \begin{align*} b=0,2,\dots,2q, \end{align*} so there are $q+1$ choices. For $r=2$, the integer $n=6q+1$ is odd and $\lfloor n/3\rfloor=2q$. Thus $b$ must be odd in $\{0,1,\dots,2q\}$, giving \begin{align*} b=1,3,\dots,2q-1. \end{align*} There are $q$ choices. This is the exceptional residue class. For $r=4$, the integer $n=6q+2$ is even and $\lfloor n/3\rfloor=2q$. Thus the admissible $b$ are again the even integers from $0$ to $2q$, so there are $q+1$ choices. For $r=6$, the integer $n=6q+3$ is odd and $\lfloor n/3\rfloor=2q+1$. The admissible $b$ are the odd integers from $1$ to $2q+1$, so there are $q+1$ choices. For $r=8$, the integer $n=6q+4$ is even and $\lfloor n/3\rfloor=2q+1$. The admissible $b$ are the even integers from $0$ to $2q$, so there are $q+1$ choices. For $r=10$, the integer $n=6q+5$ is odd and $\lfloor n/3\rfloor=2q+1$. The admissible $b$ are the odd integers from $1$ to $2q+1$, so there are $q+1$ choices. Therefore the number of nonnegative solutions of $2a+3b=n$ is $q$ when $k\equiv 2 \pmod{12}$ and is $q+1$ otherwise.[/guided]

[step:Translate the count back into the stated dimension formula] Since $q=\left\lfloor k/12\right\rfloor$, the preceding count gives \begin{align*} \dim_{\mathbb{C}} M_k(\mathrm{SL}_2(\mathbb{Z})) = \begin{cases} \left\lfloor \frac{k}{12}\right\rfloor, & k \equiv 2 \pmod{12},\\ \left\lfloor \frac{k}{12}\right\rfloor + 1, & k \not\equiv 2 \pmod{12}, \end{cases} \end{align*} for every even $k \ge 0$. Together with the odd-weight vanishing proved above, this proves the theorem. [/step]