Special linear group ($SL(n)$)

Complex two-by-two matrices of determinant one form a complex three-dimensional Lie group, or a real six-dimensional Lie group. Their congruence action on Hermitian two-by-two matrices gives the Lorentz spinor double cover with kernel $\{I,-I\}$. The two Weyl spinor representations are $A$ and $(A^{†}{)}^{-1}$. Treating the group as real is essential for the second, antiholomorphic representation.

Every $g\in S{L}_2(\mathbb{R})$ factors into an upper triangular determinant-one matrix and a rotation matrix. This follows because the triangular subgroup acts transitively on the complex upper half-plane, a group orbit of the full group, while $SO(2)$ is the stabilizer subgroup of $i$. The intersection of the two subgroups is $\{\pm I\}$, giving exactly two matrix pairs $(h,k)$ and $(-h,-k)$. Requiring the diagonal of $h$ to be positive gives uniqueness. For $g=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ and $q=\sqrt{c^2+d^2}$, the positive branch is $h=\left(\begin{array}{cc}1/q& (ac+bd)/q\\ 0& q\end{array}\right)$ and $k=q^{-1}\left(\begin{array}{cc}d& -c\\ c& d\end{array}\right)$.

A congruence subgroup of $S{L}_n(\mathbb{Z})$ contains the kernel of reduction modulo some positive integer.

The principal congruence subgroup of level $N$ isIt has finite index because the target is finite.

The Gamma 0 congruence subgroup is

The subgroupis a congruence subgroup of the modular group.

For an odd prime $p$, the group ${\Gamma }_1(p)$ has $p-1$ cusps. The general primitive-vector count givesThe exceptional group ${\Gamma }_1(2)={\Gamma }_0(2)$ has two cusps, represented by infinity and zero.

For $p\ge 5$, the Gamma 1 congruence subgroup has $(p-1)/2$ modular cusps of cusp width one and $(p-1)/2$ of cusp width $p$. Primitive columns modulo $p$, taken up to sign, are acted on by $(a,c)\mapsto (a+bc,c)$. The classes with $c=0$ have cusp width one, the others cusp width $p$. There are no effective elliptic stabilizers. The genus formula for a modular curve gives genus $(p-5)(p-7)/24$.