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The valence formula for the modular group shows that $\Delta =(E_4^3-E_6^2)/1728$ has no zeros in the half-plane. For every $a\in \mathbb{C}$, the weight-twelve form $E_4^3-a\Delta$ has total weighted zero order one. At $a=0$ its sole zero is the order-three zero at the cubic elliptic orbit; at $a=1728$ its sole zero is the order-two zero at the quadratic elliptic orbit. Otherwise it has one simple ordinary zero orbit. Thus each value of the Klein j-invariant occurs on exactly one modular orbit and classifies one homothety of complex lattices.

The modular group is $S{L}_2(\mathbb{Z})$ acting on the complex upper half-plane byIts central element $-I$ acts trivially on points, so the effective group is $PS{L}_2(\mathbb{Z})$. It is generated by $S:\tau \mapsto -1/\tau$ and $T:\tau \mapsto \tau +1$: the Euclidean algorithm reduces the bottom row of a matrix using these operations until it is a translation. The relations $S^2=(ST{)}^3=1$ hold in the projective group.

A standard fundamental domain of the modular group isTo see that each orbit meets it, use $\mathrm{Im}(\gamma \tau )=\mathrm{Im}\tau /|c\tau +d{|}^2$. Choose a coprime integer pair $(c,d)$ minimizing the nonzero lattice modulus $|c\tau +d|$; a modular matrix with that bottom row maximizes the imaginary part. Translate to the central strip. If the result had modulus below one, inversion would further increase its imaginary part, a contradiction. Interior representatives are unique: for a point in the interior, $|c\tau +d|>1$ when $c\ne 0$. For $|c|\ge 2$ use $\mathrm{Im}\tau >\sqrt{3}/2$; for $|c|=1$ use the strip and the strict unit-circle bound. Thus such a matrix lowers the imaginary part, which would contradict the same fact applied to its inverse if both representatives were in the interior. Matrices with $c=0$ are translations, and cannot move one interior strip point to another.

The vertical sides are identified by $T$, and the circular halves by $S$. The exceptional points with elliptic stabilizers of the modular group are represented by $i$ and $\rho =e^{2\pi i/3}$, of projective stabilizer orders two and three. The rational boundary points together with infinity form one modular cusp orbit. In the hyperbolic metric the region has area ${\int }_{-1/2}^{1/2}(1-x^2{)}^{-1/2}dx=\pi /3$.

A useful subgroup is the level-two principal congruence subgroup,Reduction modulo two is onto $S{L}_2({\mathbb{F}}_2)$, which has six elements and permutes the three nonzero vectors. Hence its kernel has index six, and the quotient is $S_3$. In the projective action, $\Gamma (2)$ is torsion-free. Indeed finite-order nonidentity modular elements have trace zero or $\pm 1$; matrices in $\Gamma (2)$ have even trace, and trace zero is impossible because $a,d$ are odd while $b,c$ are even: $ad-bc=1$ with $d=-a$ would imply $bc=-a^2-1\equiv 2(\mathrm{mod}4)$. The subgroup has three modular cusp classes, represented by $\infty ,0,1$, each of width two. Parity of a primitive pair distinguishes these classes, and elementary level-two row operations reduce each class to its representative. Six translates of $\mathcal{F}$ supply a fundamental region. Its compactified level-two modular curve is obtained by adjoining these three modular cusp points.

Invariant holomorphic functions arise naturally from theta constants. In the radian convention of Question 4, put ${\Theta }_j(\tau )={\theta }_j(0,\tau )$, and defineHere ${\Theta }_2={\sum }_{n\in \mathbb{Z}}{q}^{(n+1/2{)}^2}$. The theta-constant inversion and translation laws, obtained by Gaussian Poisson summation and by reindexing, interchange ${\Theta }_2,{\Theta }_4$ under $S$ with the common factor $\sqrt{-i\tau }$, leave ${\Theta }_3$ fixed up to that factor, and under $T$ interchange ${\Theta }_3,{\Theta }_4$ while multiplying ${\Theta }_2$ by $e^{\pi i/4}$. Together with the Jacobi abstruse identity ${\Theta }_3^4={\Theta }_2^4+{\Theta }_4^4$, this givesThese transformations generate the six permutations of $0,1,\infty$. In particular $T^2$ and $S{T}^2{S}^{-1}$ fix $\lambda$; with $-I$, these generate $\Gamma (2)$. Thus the modular lambda function is invariant under this subgroup, but not under the whole modular group. Its even theta constants do not vanish on $\mathbb{H}$, so it is holomorphic there and avoids $0,1,\infty$. Near infinity, $\lambda =16q+O(q^2)$, with $q=e^{\pi i\tau }$, which is the width-two modular cusp coordinate. The two transformation laws give modular cusp values one at zero and infinity at one, with the same simple local orders. Thus $\lambda$ has exactly one simple pole on the compactified level-two quotient and no pole in the half-plane. A meromorphic map with one simple pole has degree one, so it identifies that quotient with the sphere and the open quotient with $\mathbb{C}\setminus \{0,1\}$. This proves the coordinate assertion and describes all three modular cusps.

Symmetrizing this subgroup invariant produces a full modular-invariant function:Substitution shows that this rational expression is unchanged by both $\lambda \mapsto 1-\lambda$ and $\lambda \mapsto \lambda /(\lambda -1)$, hence by $S$ and $T$. It is the Klein j-invariant. Its denominator does not vanish on $\mathbb{H}$, so it is holomorphic there; at the modular cusp it has $j\sim q^{-2}$, a simple pole in $Q=e^{2\pi i\tau }$. It therefore remains nonconstant without being a bounded holomorphic function on the compactified quotient. In fact a holomorphic invariant extending holomorphically to the compactified modular curve would be constant by the maximum principle.

This also connects invariance to elliptic curves: changing an oriented basis of the lattice $\mathbb{Z}+\tau \mathbb{Z}$ gives precisely the modular action after a rescaling. On the compact full modular quotient, $j$ has just the one simple modular cusp pole, so its map to the sphere has degree one. It therefore classifies the resulting homothety of complex lattices classes, while the modular lambda function retains a level-two labeling of the branch points in the Legendre model $y^2=x(x-1)(x-\lambda )$. Finally, a modular form of nonzero weight obeys a transformation law with a factor $(c\tau +d{)}^k$, not plain invariance; ratios of equal-weight forms instead give weight-zero modular functions. The distinction between forms, invariant holomorphic functions on the open half-plane, and their modular cusp behavior is essential to this construction.

Let $\Lambda =\mathbb{Z}{\omega }_1+\mathbb{Z}{\omega }_2$ be a complex lattice. Orient its basis so that $\mathrm{Im}({\omega }_1/{\omega }_2)>0$. Multiplication by ${\omega }_2^{-1}$ gives the normalized complex lattice $\mathbb{Z}\tau +\mathbb{Z}$ with $\tau ={\omega }_1/{\omega }_2\in \mathbb{H}$. Thus every homothety of complex lattices class has a normalized representative.

If $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in S{L}_2(\mathbb{Z})$, thenThe last equality follows because the two integer vectors are an integral basis: their determinant is one. Hence points in the same modular orbit give homothetic complex lattices.

Conversely, suppose ${\Lambda }_{{\tau }^{\prime }}=\alpha {\Lambda }_{\tau }$ for $\alpha \ne 0$. There is an integer basis-change matrix of determinant $\pm 1$ such thatBoth normalized bases have positive orientation, and multiplication by $\alpha$ preserves real orientation because its real determinant is $|\alpha {|}^2>0$. The basis-change determinant is therefore $+1$. Dividing the two equalities gives ${\tau }^{\prime }=(a\tau +b)/(c\tau +d)$. Thus the modular orbits and homothety of complex lattices classes correspond bijectively.

For a complex lattice, put $g_2=60G_4$ and $g_3=140G_6$. The complex lattice form of the Klein j-invariant isThe denominator is nonzero for a complex lattice: after normalization it is a nonzero scalar multiple of $E_4^3-E_6^2$, whose absence of zeros was proved in part (b). The equal lattice Eisenstein sums therefore give equal $j$ values, so parts (a) and (b) give ${\Lambda }^{\prime }=\alpha \Lambda$ for some $\alpha \ne 0$. Scaling the absolutely convergent lattice Eisenstein sums givesIf both sums are nonzero, ${\alpha }^4={\alpha }^6=1$, so ${\alpha }^2=1$ and $\alpha \Lambda =\Lambda$.

If $G_6=0$, then $G_4\ne 0$ and $j=1728$. The complex lattice is homothetic to $\mathbb{Z}i+\mathbb{Z}$, which is preserved by multiplication by $i$. The constraint ${\alpha }^4=1$ allows only powers of $i$, each already preserving the complex lattice. If $G_4=0$, then $G_6\ne 0$ and $j=0$. The complex lattice is homothetic to $\mathbb{Z}\rho +\mathbb{Z}$. Multiplication by $\rho$ preserves this complex lattice because ${\rho }^2=-1-\rho$, and multiplication by $-1$ always preserves it. The constraint ${\alpha }^6=1$ allows exactly the sixth roots generated by these two multiplications. Again $\alpha \Lambda =\Lambda$. Both sums cannot be zero simultaneously because the discriminant is nonzero. Hence in every caseThe exceptional cases are essential in a complex lattice is determined by its fourth and sixth Eisenstein sums; merely deducing homothety of complex lattices would not prove equality.