The dimension of the space of level-one modular forms

Upper bound. Choose $d:=\lfloor k\mathrm{/}12\rfloor +1$ distinct non-elliptic classes of points of $\mathfrak{H}$ if $k\equiv ̸2(\mathrm{m}\mathrm{o}\mathrm{d}12)$, and $d:=\lfloor k\mathrm{/}12\rfloor$ such classes if $k\equiv 2(\mathrm{m}\mathrm{o}\mathrm{d}12)$; such classes exist because non-elliptic classes are infinite. Suppose $0\mathrm{\ne }f\in M_k$ vanished at all chosen classes. Their total contribution to the valence sum is $N\ge d$ (each has ${\nu }_P=1$). If $k\equiv ̸2(\mathrm{m}\mathrm{o}\mathrm{d}12)$ then $d=\lfloor k\mathrm{/}12\rfloor +1>k\mathrm{/}12\ge k\mathrm{/}12-{\mathrm{ord}}_{\mathrm{\infty }}(f)$, contradicting [F1]. If $k\equiv 2(\mathrm{m}\mathrm{o}\mathrm{d}12)$, write $k=12q+2$, so $k\mathrm{/}12=q+1\mathrm{/}6$ and $d=q$. If ${\mathrm{ord}}_{\mathrm{\infty }}(f)\ge 1$ then the valence sum is at most $q+1\mathrm{/}6-1=q-5\mathrm{/}6<q\le N$, a contradiction. If ${\mathrm{ord}}_{\mathrm{\infty }}(f)=0$, let $m={\mathrm{ord}}_i(f)$, $r={\mathrm{ord}}_{\omega }(f)$ and $N\ge q$ the contribution of the remaining classes; multiplying the valence identity by $6$ gives $6N+3m+2r=6q+1$, so $m$ is odd and $r\equiv 2(\mathrm{m}\mathrm{o}\mathrm{d}3)$; hence $m\ge 1$, $r\ge 2$ and $N+m\mathrm{/}2+r\mathrm{/}3\ge q+1\mathrm{/}2+2\mathrm{/}3=q+7\mathrm{/}6>q+1\mathrm{/}6=k\mathrm{/}12$, again contradicting [F1]. Therefore evaluation at the $d$ chosen classes is injective on $M_k$, so $\dim M_k\le d$.

Lower bound. If $4a+6b=4a^{\mathrm{\prime }}+6b^{\mathrm{\prime }}=k$ with $a<a^{\mathrm{\prime }}$ then ${\mathrm{ord}}_{\omega }(E_4^a{E}_6^b)=a<a^{\mathrm{\prime }}={\mathrm{ord}}_{\omega }(E_4^{a^{\mathrm{\prime }}}{E}_6^{b^{\mathrm{\prime }}})$ by [F3], so in a linear relation ${\sum }_a{c}_a{E}_4^a{E}_6^{b(a)}=0$ the term with least $a$, if its coefficient were nonzero, would give the sum the finite order $a$ at the class of $\omega$; since the sum is identically zero its order is infinite, so $c_a=0$ for the least $a$, and induction gives that all coefficients vanish. Hence the monomials are linearly independent, so $\dim M_k$ is at least their number. The pairs $(a,b)$ with $4a+6b=k$, $a,b\ge 0$, are indexed by the integers $a\ge 0$ with $a\equiv k(\mathrm{m}\mathrm{o}\mathrm{d}3)$ and $a\le k\mathrm{/}4$ (then $b=(k-4a)\mathrm{/}6\ge 0$ is a nonnegative integer). Writing $k=12q+s$ with $s\in \{0,2,4,6,8,10\}$, put $a_0\in \{0,1,2\}$ for the least nonnegative residue of $s$ modulo $3$. The solutions are $a=a_0+3j$ for $0\le j\le \lfloor (k\mathrm{/}4-a_0)\mathrm{/}3\rfloor$, so their number is $\max (0,\lfloor (k\mathrm{/}4-a_0)\mathrm{/}3\rfloor +1)=q+1$ for $s=0,4,6,8,10$ and $q$ for $s=2$; this is exactly $\lfloor k\mathrm{/}12\rfloor +1$ for $k\equiv ̸2(\mathrm{m}\mathrm{o}\mathrm{d}12)$ and $\lfloor k\mathrm{/}12\rfloor$ for $k\equiv 2(\mathrm{m}\mathrm{o}\mathrm{d}12)$. With 1.1 this proves the dimension formula.

Cusp forms and examples. For $k\ge 4$, ${\epsilon }_{\mathrm{\infty }}$ is nonzero and surjective onto $\mathbb{C}$, so by [F2] $\dim S_k=\dim M_k-1$; for $0\le k<12$ the formula gives $\dim M_k=1$ for $k=0,4,6,8,10$ and $\dim M_2=0$, so $S_k=0$ there (for $k=0,4,6,8,10$ because the kernel of a nonzero functional on a one-dimensional space is zero, for $k=2$ because $M_2=0$, and $S_0\subseteq M_0$). In particular $M_0=\mathbb{C}$ (the constants lie in $M_0$ and it is one-dimensional), $M_2=0$, $M_4,M_6,M_8,M_{10}$ are one-dimensional, and $M_{12}$ is two-dimensional. All the listed monomial counts are covered by 2.1, which also gives the asserted linear independence.