Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 29 / 5 / b / Solution

Reduction modulo $N$ has kernel $\Gamma (N)$ and is onto $S{L}_2(\mathbb{Z}/N\mathbb{Z})$. To see surjectivity, use the Chinese remainder theorem to reduce to $\mathbb{Z}/p^a\mathbb{Z}$. In that local ring a unimodular column has a unit entry. Elementary row operations move a unit to the upper position and clear the lower one; diagonal matrices $\mathrm{diag}(u,u^{-1})$ are products of elementary matrices as well. Explicitly, with $E_{12},E_{21}$ denoting elementary shears,and multiplying this by the same matrix with $u=-1$ gives $\mathrm{diag}(u,u^{-1})$. Thus elementary matrices generate the special linear group over each local ring. Chinese remaindering their parameters gives generation over the product ring, and every such parameter lifts to an integer. This proves the reduction is onto.

Over ${\mathbb{F}}_p$, there are $p^2-1$ choices for a nonzero first column and $p$ choices for a second column with determinant one, so $|S{L}_2({\mathbb{F}}_p)|=p(p^2-1)$. Each reduction from modulus $p^{a+1}$ to $p^a$ has kernel represented by $I+p^{a}B$, with $B$ modulo $p$ and $\mathrm{tr}B=0$, of size $p^3$. Hence, for $N>1$,For $N=1$ the index is one. Since $-I\in \Gamma (N)$ exactly when $N=1,2$,

For $N\ge 2$ there are no nontrivial effective elliptic stabilizers. Indeed $A=I+NB\in \Gamma (N)$ with determinant one has $\mathrm{tr}B=-N\det B$, and thereforeA noncentral elliptic element of $S{L}_2(\mathbb{Z})$ has trace $0$ or $\pm 1$, none congruent to $2$ modulo $N^2$ for $N\ge 2$. The central element $-I$ at level two acts trivially and is not an elliptic point of the effective quotient. Thus ${\nu }_2={\nu }_3=0$ for $N\ge 2$, while at level one both numbers are one.

The subgroup $\Gamma (N)$ is normal, so all its modular cusps have the same effective width as infinity. That width is $N$: the shear $T^h$ belongs to $\pm \Gamma (N)$ exactly when $N\mid h$; for $N\ge 3$ its diagonal entries exclude the minus sign, and for $N=1,2$ that sign changes nothing. Since modular cusp widths sum to $\mu$,For $N\ge 3$ the resulting principal congruence modular curve genus isFor clarity, the small levels, including all exceptional terms, areFor $N>6$, every factor in the genus correction is positive, so $g>1$. Therefore the genus-zero levels are exactly $N=1,2,3,4,5$.