Special linear group ($SL(n)$)

This Lie group consists of real two-by-two matrices of determinant one. Its Lie algebra is the space of traceless real two-by-two matrices. Its Exponential map of a Lie group is not onto: by the Cayley-Hamilton theorem, $A^2=-\det (A)I$ for traceless $A$, and its exponential has trace either $2\cosh s$ or $2\cos s$, hence never less than $-2$. The determinant-one matrix $\mathrm{diag}(-2,-1/2)$ is therefore outside its exponential image.

The upper and lower unipotent matrices generate SL2R. For $a\ne 0$, let $w(a)=U(a)L(-a^{-1})U(a)$; multiplication gives $w(a)=\left(\begin{array}{cc}0& a\\ -a^{-1}& 0\end{array}\right)$ and $w(a)w(-1)=\mathrm{diag}(a,a^{-1})$. A determinant-one matrix with upper-left entry $a\ne 0$ factors as $L(c/a)\mathrm{diag}(a,a^{-1})U(b/a)$. If that entry vanishes, first add the second row to the first by $U(1)$. Lower unipotents are conjugates of upper unipotents, so the latter also generate the whole group normally.

The quotient of the special linear group by its scalar center, acting faithfully on projective space. It is a simple group for $n\ge 3$, or for $n=2$ and $|F|>3$. Elementary transvection matrices give generation and perfectness, and the Iwasawa simplicity lemma applies to the abelian subgroup of transvections with a fixed center.

A linear transformation fixing every one-dimensional subspace is scalar. Fixing the coordinate lines makes it diagonal, and fixing the lines spanned by $e_i+e_j$ makes all diagonal entries equal. Within determinant-one matrices the scalar satisfies ${\lambda }^n=1$. Conversely scalars fix all projective points. Commutation with all elementary transvections shows these scalars are exactly the center of the special linear group, so its projective quotient acts faithfully.

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 $(p-5)(p-7)/24$.