---- > [!definition] Definition. ([[special linear group]]) > The **special linear group** ~~$\text{SL}_{n}(R)$~~ is the [[group]] of $n \times n$ [[matrix|matrices]] with [[determinant of a matrix|determinant]] $1$. > [!basicexample] > $\text{SL}_{2}(\mathbb{Z})$ is the set of $2 \times 2$ [[matrix|matrices]] with integer entries and [[determinant]] $1$. It is [[generating set of a group|generated by]] $s=\begin{bmatrix} > 0 & -1 \\ > 1 & 0 > \end{bmatrix} \text{ and } t= \begin{bmatrix} > 1 & 1 \\ 0 & 1 > \end{bmatrix}.$ > Indeed, let $H$ be the [[subgroup]] [[generating set of a group|generated by]] $s$ and $t$. Given a [[matrix]] $m=\begin{bmatrix}a & b \\ c & d\end{bmatrix} \in \text{SL}_{2}(\mathbb{Z})$, it suffices to show the [[identity matrix]] can be obtained by multiplying $m$ by suitably chosen elements of $H$. We choose the following elements: given $q \in \mathbb{Z}$, it is clear that $\begin{bmatrix} > 1 & -q \\ > 0 & 1 > \end{bmatrix} \text{ and } \begin{bmatrix} > 1 & 0 \\ -q & 1 > \end{bmatrix}$ are in $H$, as one may check (or visualize) $t^{-1}=\begin{bmatrix}1 & -1 \\ 0 & 1\end{bmatrix}$ and in turn $t^{-q}=\begin{bmatrix}1 & -q \\ 0 & 1\end{bmatrix}$, (larger $q$ implies sharper shear) as well as $s t^{-q}s ^{-1}=\begin{bmatrix}1 & 0 \\ -q & 1\end{bmatrix}$. ![[CleanShot 2024-06-17 at [email protected]]] > $\begin{bmatrix} > a & b \\ c & d > \end{bmatrix} \begin{bmatrix} > 1 & -q \\ 0&1 > \end{bmatrix}=\begin{bmatrix} > a & b-qa \\ > c & d-qc > \end{bmatrix} \text{ and } \begin{bmatrix} > a & b \\ > c & d > \end{bmatrix} \begin{bmatrix} > 1 & 0 \\ > -q & 1 > \end{bmatrix}=\begin{bmatrix} > a-qb & b \\ c-qd & d > \end{bmatrix}.$ > Now, if $c=0$ then to maintain unit determinant we must have $d=1$, in which case $\begin{bmatrix}a & b \\ c & d\end{bmatrix}=\begin{bmatrix}1 & b \\ 0 & 1\end{bmatrix}=t^{b} \in H$. If $d=0$ then to maintain unit determinant we must have $bc=-1$, so either we have $\begin{bmatrix} > a & b\\ c & d > \end{bmatrix}=\begin{bmatrix} > 1 & 1 \\ > -1 & 0 > \end{bmatrix}= ts ^{-1} \text{ or } \begin{bmatrix} > a & b \\ c & d > \end{bmatrix}=\begin{bmatrix} > 1 & -1 \\ > 1 & 0 > \end{bmatrix}=ts$ > both of which are in $H$. So assume $c$ and $d$ are both nonzero. First assume $c \leq d$. ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```