[proofplan] We use the standard construction $\Delta=(E_4^3-E_6^2)/1728$. The transformation laws for $E_4$ and $E_6$ imply that both $E_4^3$ and $E_6^2$ have weight $12$, so their normalized difference is a modular form of weight $12$. Its first Fourier coefficients are obtained by expanding the known $q$-expansions of $E_4$ and $E_6$. Finally, the valence formula for modular forms on $SL_2(\mathbb Z)$ shows that the single zero at the cusp accounts for the full weighted zero count, hence there are no zeros in $\mathbb H$. [/proofplan] [step:Show that the normalized difference is a modular form of weight $12$] The normalized Eisenstein series \begin{align*} E_4: \mathbb H &\to \mathbb C, \\ E_6: \mathbb H &\to \mathbb C \end{align*} are holomorphic modular forms of weights $4$ and $6$ for $SL_2(\mathbb Z)$, respectively. Thus $E_4^3$ and $E_6^2$ are holomorphic modular forms of weight $12$, because multiplication of modular forms adds weights and preserves holomorphy at $\mathbb H$ and at the cusp $\infty$ (citing a result not yet in the wiki: [Product of Modular Forms](/theorems/4229)). Therefore \begin{align*} \Delta=\frac{E_4^3-E_6^2}{1728} \end{align*} is a holomorphic modular form of weight $12$ for $SL_2(\mathbb Z)$. [guided] We start from the definition \begin{align*} \Delta: \mathbb H &\to \mathbb C \\ \tau &\mapsto \frac{E_4(\tau)^3-E_6(\tau)^2}{1728}. \end{align*} The functions $E_4$ and $E_6$ are the normalized Eisenstein series of weights $4$ and $6$. Their standard modularity theorem says that $E_4$ is a holomorphic modular form of weight $4$ and $E_6$ is a holomorphic modular form of weight $6$ for $SL_2(\mathbb Z)$. We now use the product rule for modular forms: if $f$ has weight $k$ and $g$ has weight $\ell$, then $fg$ has weight $k+\ell$, and holomorphy on $\mathbb H$ and holomorphy at the cusp are preserved. Applying this to the threefold product $E_4^3$ gives weight \begin{align*} 4+4+4=12. \end{align*} Applying it to the twofold product $E_6^2$ gives weight \begin{align*} 6+6=12. \end{align*} Since the space of modular forms of a fixed weight is closed under linear combinations, the difference $E_4^3-E_6^2$ is again a modular form of weight $12$. Multiplication by the scalar $1/1728$ does not change the weight or holomorphy. Hence $\Delta$ is a holomorphic modular form of weight $12$ for $SL_2(\mathbb Z)$. [/guided] [/step] [step:Compute the first Fourier coefficients and identify the cusp order] The normalized Eisenstein series have Fourier expansions \begin{align*} E_4(\tau)&=1+240q(\tau)+2160q(\tau)^2+6720q(\tau)^3+17520q(\tau)^4+\cdots,\\ E_6(\tau)&=1-504q(\tau)-16632q(\tau)^2-122976q(\tau)^3-532728q(\tau)^4+\cdots. \end{align*} Expanding and collecting powers of $q(\tau)$ gives \begin{align*} E_4(\tau)^3 &=1+720q(\tau)+179280q(\tau)^2+16954560q(\tau)^3+396974160q(\tau)^4+\cdots,\\ E_6(\tau)^2 &=1-1008q(\tau)+220752q(\tau)^2+16519104q(\tau)^3+399517776q(\tau)^4+\cdots. \end{align*} Therefore \begin{align*} E_4(\tau)^3-E_6(\tau)^2 &=1728q(\tau)-41472q(\tau)^2+435456q(\tau)^3-2543616q(\tau)^4+\cdots. \end{align*} Dividing by $1728$ yields \begin{align*} \Delta(\tau) &=q(\tau)-24q(\tau)^2+252q(\tau)^3-1472q(\tau)^4+\cdots. \end{align*} The constant term is $0$ and the coefficient of $q(\tau)$ is $1$, so $\Delta$ vanishes at the cusp $\infty$ to order exactly $1$: \begin{align*} \operatorname{ord}_{\infty}(\Delta)=1. \end{align*} In particular, $\Delta \in S_{12}(SL_2(\mathbb Z))$. [/step] [step:Apply the valence formula over orbit representatives to rule out zeros in $\mathbb H$] Let $\rho := e^{2\pi i/3} \in \mathbb H$, and let $\mathcal F \subset \mathbb H$ denote the closed standard fundamental domain for the action of $SL_2(\mathbb Z)$ on $\mathbb H$. Choose a set $\mathcal R \subset \mathcal F \setminus \{i,\rho\}$ containing exactly one representative from each non-elliptic $SL_2(\mathbb Z)$-orbit in $\mathbb H$. The valence formula for a nonzero modular form $f$ of weight $k$ on $SL_2(\mathbb Z)$ states that \begin{align*} \operatorname{ord}_{\infty}(f) +\frac{1}{2}\operatorname{ord}_{i}(f) +\frac{1}{3}\operatorname{ord}_{\rho}(f) +\sum_{z \in \mathcal R}\operatorname{ord}_{z}(f) =\frac{k}{12}, \end{align*} where the sum counts one representative for each non-elliptic orbit, including boundary orbits, rather than only interior points (citing a result not yet in the wiki: Valence Formula for Modular Forms). Apply this formula to $f=\Delta$ and $k=12$. The previous step shows that $\Delta$ is nonzero because its first nonzero Fourier coefficient is $1$, and it also shows that $\operatorname{ord}_{\infty}(\Delta)=1$. Hence \begin{align*} 1 +\frac{1}{2}\operatorname{ord}_{i}(\Delta) +\frac{1}{3}\operatorname{ord}_{\rho}(\Delta) +\sum_{z \in \mathcal R}\operatorname{ord}_{z}(\Delta) =1. \end{align*} Subtracting $1$ from both sides gives \begin{align*} \frac{1}{2}\operatorname{ord}_{i}(\Delta) +\frac{1}{3}\operatorname{ord}_{\rho}(\Delta) +\sum_{z \in \mathcal R}\operatorname{ord}_{z}(\Delta) =0. \end{align*} Each order appearing in this sum is a nonnegative integer, and each coefficient is positive. Hence every term is zero, so $\Delta$ has no zero at $i$, no zero at $\rho$, and no zero at any representative in $\mathcal R$. Since every point of $\mathbb H$ is $SL_2(\mathbb Z)$-equivalent to exactly one of these types of representatives and zeros are preserved by the modular transformation law, $\Delta$ has no zero anywhere in $\mathbb H$. [guided] The last point is to show that the zero at the cusp is the only zero. The correct tool is the valence formula for modular forms on $SL_2(\mathbb Z)$, but we must state it in a form that includes boundary orbits of the fundamental domain. Let $\rho := e^{2\pi i/3} \in \mathbb H$, and let $\mathcal F \subset \mathbb H$ be the closed standard fundamental domain for $SL_2(\mathbb Z)$. Choose a set $\mathcal R \subset \mathcal F \setminus \{i,\rho\}$ containing exactly one representative from each non-elliptic $SL_2(\mathbb Z)$-orbit in $\mathbb H$. This choice is important: boundary points of $\mathcal F$ may be identified with other boundary points under $SL_2(\mathbb Z)$, so the valence formula counts orbits, not merely interior points. The valence formula says that if $f$ is a nonzero modular form of weight $k$ on $SL_2(\mathbb Z)$, then \begin{align*} \operatorname{ord}_{\infty}(f) +\frac{1}{2}\operatorname{ord}_{i}(f) +\frac{1}{3}\operatorname{ord}_{\rho}(f) +\sum_{z \in \mathcal R}\operatorname{ord}_{z}(f) =\frac{k}{12}. \end{align*} The factors $1/2$ and $1/3$ occur because $i$ and $\rho$ are the elliptic representatives with nontrivial stabilizers in $SL_2(\mathbb Z)$. The sum over $\mathcal R$ covers every remaining orbit in $\mathbb H$, including non-elliptic boundary orbits. We verify the hypotheses of the formula. From the previous steps, $\Delta$ is a modular form of weight $12$ for $SL_2(\mathbb Z)$. It is nonzero because its Fourier expansion begins with $q(\tau)$ with coefficient $1$. Therefore the valence formula applies with $f=\Delta$ and $k=12$, giving \begin{align*} \operatorname{ord}_{\infty}(\Delta) +\frac{1}{2}\operatorname{ord}_{i}(\Delta) +\frac{1}{3}\operatorname{ord}_{\rho}(\Delta) +\sum_{z \in \mathcal R}\operatorname{ord}_{z}(\Delta) =1. \end{align*} The Fourier expansion computed above gives $\operatorname{ord}_{\infty}(\Delta)=1$, so substitution yields \begin{align*} 1 +\frac{1}{2}\operatorname{ord}_{i}(\Delta) +\frac{1}{3}\operatorname{ord}_{\rho}(\Delta) +\sum_{z \in \mathcal R}\operatorname{ord}_{z}(\Delta) =1. \end{align*} Subtracting the cusp contribution gives \begin{align*} \frac{1}{2}\operatorname{ord}_{i}(\Delta) +\frac{1}{3}\operatorname{ord}_{\rho}(\Delta) +\sum_{z \in \mathcal R}\operatorname{ord}_{z}(\Delta) =0. \end{align*} Every term on the left is nonnegative: orders of zeros of a [holomorphic function](/page/Holomorphic%20Function) are nonnegative integers, and the coefficients $1/2$, $1/3$, and $1$ are positive. A sum of nonnegative [real numbers](/page/Real%20Numbers) can be zero only when each summand is zero. Hence $\Delta$ has no zero at $i$, no zero at $\rho$, and no zero at any representative in $\mathcal R$. It remains only to pass from orbit representatives to all of $\mathbb H$. Every point of $\mathbb H$ is $SL_2(\mathbb Z)$-equivalent to one of the representatives just considered: either $i$, $\rho$, or a point of $\mathcal R$. The modular transformation law for a weight-$12$ form has the shape \begin{align*} \Delta(\gamma \tau)=(c\tau+d)^{12}\Delta(\tau) \end{align*} for $\gamma=\begin{pmatrix}a&b\\ c&d\end{pmatrix}\in SL_2(\mathbb Z)$. Since $c\tau+d\neq 0$ for $\tau \in \mathbb H$, multiplication by $(c\tau+d)^{12}$ cannot create or remove a zero. Thus zeros are preserved along $SL_2(\mathbb Z)$-orbits, and the absence of zeros at all orbit representatives implies the absence of zeros on all of $\mathbb H$. [/guided] [/step]