Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 29 / 1

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.

Use the normalized Eisenstein series $E_4=1+240q+O(q^2)$ and $E_6=1-504q+O(q^2)$, and putThe valence formula for the modular group applied to this nonzero weight-twelve cusp form gives weighted zero order one. Its order at infinity is already one, so $\Delta$ has no zeros in $\mathbb{H}$. Consequently the Klein j-invariant is holomorphic there and invariant under $S{L}_2(\mathbb{Z})$.

Set $\rho =e^{2\pi i/3}$. The stabilizer of $\rho$ forces $E_4(\rho )=0$, since the weight-four automorphy factor for $\left(\begin{array}{cc}0& -1\\ 1& 1\end{array}\right)$ is $(\rho +1{)}^4\ne 1$. The stabilizer of $i$ forces $E_6(i)=0$, since $i^6=-1$. The valence formula for the modular group giveswith no other zero orbits for either form. In particular $j(\rho )=0$ and $j(i)=1728$.

For any $A\in \mathbb{C}$, the weight-twelve formis nonzero, with constant Fourier coefficient one and hence order zero at infinity. Its total weighted zero order is one. If $A=0$, the triple zero at $\rho$ accounts for this entire order. If $A=1728$, then $h_A=E_6^2$, and the double zero at $i$ accounts for it. For all other $A$, neither elliptic orbit is a zero: $E_4(\rho )=0$ and $\Delta (\rho )\ne 0$ exclude $\rho$, while $E_6(i)=0$ gives $E_4(i{)}^3=1728\Delta (i)$ and excludes $i$. Therefore there is precisely one ordinary zero orbit, of multiplicity one. Since $h_A(\tau )=0$ is exactly $j(\tau )=A$, every complex number occurs at exactly one modular orbit. ThusThis proves Klein j-invariant classifies complex lattice homothety using the weighted zeros, including the two exceptional stabilizers.

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.