Modular forms

This is the modular-form analogue of "a degree-$n$ polynomial has exactly $n$ roots counted with multiplicity": $f$ can be viewed as a section of a line bundle of degree $k/12$ on the compact orbifold $X(1)=\mathbb{H}^*/\mathrm{SL}_2(\mathbb{Z})\cong\mathbb{P}^1(\mathbb{C})$, and the total number of its zeros (weighted) must equal that degree. The weights $\tfrac12,\tfrac13$ appear only at the two points $i$ and $\rho=e^{i\pi/3}$ because those are exactly the points where $\mathrm{SL}_2(\mathbb{Z})$ has extra (order-$2$ and order-$3$) stabilizers, so the quotient map wraps around them $2$ and $3$ times.

Fix the standard fundamental domain $\mathcal{F}=\{\tau\in\mathbb{H} : |\operatorname{Re}(\tau)|\le \tfrac12,\ |\tau|\ge 1\}$, with vertices at $\rho=e^{i\pi/3}$, $\rho+1$, and $i$ on its lower boundary, extending up to a cusp at $\infty$. Truncate it at height $\operatorname{Im}(\tau)=T$ for large $T$, and indent the boundary with small circular arcs around any zeros of $f$ that happen to lie on $\partial\mathcal{F}$ itself. Call the resulting contour $C$. Since $f$ has finitely many zeros in this bounded region, the argument principle gives $\frac{1}{2\pi i}\oint_C \frac{f'(\tau)}{f(\tau)}\,d\tau$ equal to the number of zeros of $f$ strictly inside $C$.

Now evaluate each piece of $\partial\mathcal{F}$. The two vertical sides $\operatorname{Re}(\tau)=-\tfrac12$ and $\operatorname{Re}(\tau)=\tfrac12$ are identified by $T:\tau\mapsto\tau+1$, and since $f(\tau+1)=f(\tau)$, the integrand $f'/f$ is also $1$-periodic; traversed in opposite directions (one side going up, the identified side coming down), these two integrals cancel exactly. This is where the translation symmetry earns its keep: it removes two full sides of the contour for free.

The top edge at height $T$ closes up as $T\to\infty$: writing $f=\sum_{n\ge m}a_nq^n$ with $m=\operatorname{ord}_\infty(f)$ and $a_m\ne0$, we get $f'/f \to 2\pi i\,m$ as $q\to0$, so this edge contributes $-\operatorname{ord}_\infty(f)$ to the total (the minus sign from the contour's orientation, traversed leftward at the top).

What remains is the lower boundary: the arc from $\rho$ to $i$ and the arc from $i$ to $\rho+1$ along $|\tau|=1$, plus the two short vertical segments from $\rho$ up to the top-left corner region and from $\rho+1$ down. The inversion $S:\tau\mapsto-1/\tau$ maps the arc through $i$ to itself (since $S(i)=i$) and swaps the two halves of the circular arc; because $S$ fixes $i$ and rotates a neighborhood of $i$ by angle $\pi$ (the stabilizer of $i$ in $\mathrm{PSL}_2(\mathbb{Z})$ has order $2$), the small indentation around $i$ contributes only $-\tfrac12\operatorname{ord}_i(f)$ instead of a full $-\operatorname{ord}_i(f)$: only "half" of a full residue is actually swept out before $S$-symmetry identifies the rest. Likewise $\rho$ and $\rho+1=S T^{-1}(\rho)$ are identified with stabilizer of order $3$ in $\mathrm{PSL}_2(\mathbb{Z})$ (generated by $ST$, of order $3$), so their combined indentation contributes $-\tfrac13\operatorname{ord}_\rho(f)$. Any other zero $P$ on the remaining, generic part of the arc has trivial stabilizer and contributes the full $-\operatorname{ord}_P(f)$.

Adding every piece: the vertical sides cancel, the top gives $-\operatorname{ord}_\infty(f)$, the corners give $-\tfrac12\operatorname{ord}_i(f)-\tfrac13\operatorname{ord}_\rho(f)$, and any remaining boundary zero $P$ gives $-\operatorname{ord}_P(f)$. But $\frac{1}{2\pi i}\oint_C f'/f\,d\tau$ must also equal $\tfrac{k}{12}$: this constant comes from tracking how the weight-$k$ automorphy factor $(c\tau+d)^k$ forces the total turning of $\arg f$ around the two identified corner arcs (related by $S$, which sends $d\tau\mapsto \tau^{-2}d\tau$ picking up a winding governed by $k$) to contribute exactly $k/12$ net — the same constant that appears because the hyperbolic area of $\mathcal{F}$ is $\pi/3$ and $X(1)$ has exactly one cusp and elliptic points of order $2,3$, which is the Riemann–Hurwitz bookkeeping underlying the $12$ in the formula. Equating the two computations of the same contour integral and moving every zero-order term to one side gives exactly $\operatorname{ord}_\infty(f)+\tfrac12\operatorname{ord}_i(f)+\tfrac13\operatorname{ord}_\rho(f)+\sum_{P}\operatorname{ord}_P(f) = \tfrac{k}{12}$.

This single computation both certifies that the space of weight-$12$ cusp forms is $1$-dimensional (spanned by $\Delta$) and that $\Delta$'s non-vanishing on $\mathbb{H}$ is exactly what makes the $j$-invariant $j=E_4^3/\Delta$ holomorphic on all of $\mathbb{H}$ — the fact that classifies elliptic curves over $\mathbb{C}$ up to isomorphism, the bridge to arithmetic geometry.

First, holomorphy and weight: $E_4,E_6$ are modular forms of weight $4,6$ respectively (standard Eisenstein-series facts, taken as known input here), so $E_4^3$ and $E_6^2$ are both weight $12$. Both have Fourier expansion starting with constant term $1$ (by normalization), so $E_4^3-E_6^2$ has vanishing constant term: it is a cusp form of weight $12$, and $\Delta=(E_4^3-E_6^2)/1728$ is too, in particular $\operatorname{ord}_\infty(\Delta)\ge1$.

Now apply the valence formula (proved above) with $k=12$: the total weighted zero order of $\Delta$ must equal $12/12=1$. Every term on the left — $\operatorname{ord}_\infty(\Delta)$, $\tfrac12\operatorname{ord}_i(\Delta)$, $\tfrac13\operatorname{ord}_\rho(\Delta)$, and any $\operatorname{ord}_P(\Delta)$ for other points — is a non-negative real number, and we already know $\operatorname{ord}_\infty(\Delta)\ge1$. The only way a sum of non-negative terms, one of which is already $\ge1$, can total exactly $1$, is if that term equals exactly $1$ and every other term is exactly $0$. Hence $\operatorname{ord}_\infty(\Delta)=1$ exactly, and $\operatorname{ord}_P(\Delta)=0$ for every $P\in\mathbb{H}$: $\Delta$ never vanishes on $\mathbb{H}$, with a simple zero only at the cusp.

For the dimension claim: let $g$ be any weight-$12$ cusp form. The quotient $g/\Delta$ is $\mathrm{SL}_2(\mathbb{Z})$-invariant (weight $12-12=0$), and holomorphic on $\mathbb{H}$ since $\Delta$ never vanishes there. At the cusp, both $g$ and $\Delta$ vanish to order at least $1$ in $q$, and since $\Delta$'s order there is exactly $1$, the quotient extends holomorphically (no pole) to $q=0$ as well. A weight-$0$ modular function holomorphic everywhere on $\mathbb{H}$ and at the cusp descends to a holomorphic function on the compact Riemann surface $X(1)\cong\mathbb{P}^1(\mathbb{C})$, which by Liouville's theorem (extended to the compactification) must be constant. So $g=c\cdot\Delta$ for a constant $c$: the space of weight-$12$ cusp forms is exactly $1$-dimensional. The identical argument at weight $4$ (total budget $4/12=1/3$, which can only be realized as $\operatorname{ord}_\rho=1$ with every other order $0$, since any integer contribution — such as a nonzero $\operatorname{ord}_\infty$ — would already exceed $1/3$) shows every nonzero weight-$4$ form has its unique zero at $\rho$ with order exactly $1$ and no others, and the same quotient-by-$E_4$ trick forces $\dim M_4(\mathrm{SL}_2(\mathbb{Z}))=1$.

Finally, the $q$-product $\Delta(\tau)=q\prod_{n\ge1}(1-q^n)^{24}$ is the classical identity discovered by Jacobi; we cite it here rather than re-derive it, but note it is fully consistent with what we just proved: the right-hand side visibly vanishes to order exactly $1$ at $q=0$ and is manifestly nonzero for $0<|q|<1$ (a convergent product of nonzero factors), matching the zero-order profile forced by the valence formula above.