= Lie algebra of $SL(2)$
{c}
= Lie algebra of the special linear group of degree 2
{c}
{synonym}
This is a good first concrete example of a Lie algebra. Shown at <Lie Groups, Physics, and Geometry by Robert Gilmore (2008)> Chapter 4.2 "How to linearize a Lie Group" has an example.
We can use use the following parametrization of the <special linear group> on variables $x$, $y$ and $z$:
$$
M =
\begin{bmatrix}
1 + x & y \\
z & (1 + yz)/(1 + x) \\
\end{bmatrix}
$$
Every element with this parametrization has <determinant> 1:
$$
det(M) = (1 + x)(1 + yz)/(1 + x) - yz = 1
$$
Furthermore, any element can be reached, because by independently settting $x$, $y$ and $z$, $M_{00}$, $M_{01}$ and $M_{10}$ can have any value, and once those three are set, $M_{11}$ is fixed by the determinant.
To find the elements of the <Lie algebra>, we evaluate the derivative on each parameter at 0:
$$
\begin{aligned}
M_x
&=
\evalat{\dv{M}{x}}{(x,y,z) = (0,0,0)}
&=
\evalat{
\begin{bmatrix}
1 & 0 \\
0 & -(1 + yz)/(1 + x)^2 \\
\end{bmatrix}
}{(x,y,z) = (0,0,0)}
&=
\begin{bmatrix}
1 & 0 \\
0 & -1 \\
\end{bmatrix}
\\
M_y
&=
\evalat{\dv{M}{y}}{(x,y,z) = (0,0,0)}
&=
\evalat{
\begin{bmatrix}
0 & 1 \\
0 & z/(1 + x) \\
\end{bmatrix}
}{(x,y,z) = (0,0,0)}
&=
\begin{bmatrix}
0 & 1 \\
0 & 0 \\
\end{bmatrix}
\\
M_z
&=
\evalat{\dv{M}{z}}{(x,y,z) = (0,0,0)}
&=
\evalat{
\begin{bmatrix}
0 & 0 \\
1 & y/(1 + x) \\
\end{bmatrix}
}{(x,y,z) = (0,0,0)}
&=
\begin{bmatrix}
0 & 0 \\
1 & 0 \\
\end{bmatrix}
\\
\end{aligned}
$$
Remembering that the <Lie bracket of a matrix Lie group> is really simple, we can then observe the following <Lie bracket> relations between them:
$$
\begin{aligned}
[M_x, M_y] &= M_xM_y - M_yM_x &= \begin{bmatrix}0 & 1 \\ 0 & 0 \\\end{bmatrix} &- \begin{bmatrix}0 & -1 \\ 0 & 0 \\\end{bmatrix} &= \begin{bmatrix}0 & 2 \\ 0 & 0 \\\end{bmatrix} &= 2M_y\\
[M_x, M_z] &= M_xM_z - M_zM_x &= \begin{bmatrix}0 & 0 \\ -1 & 0 \\\end{bmatrix} &- \begin{bmatrix}0 & 0 \\ 1 & 0 \\\end{bmatrix} &= \begin{bmatrix}0 & 0 \\ -2 & 0 \\\end{bmatrix} &= -2M_z\\
[M_y, M_z] &= M_yM_z - M_zM_y &= \begin{bmatrix}1 & 0 \\ 0 & 0 \\\end{bmatrix} &- \begin{bmatrix}0 & 0 \\ 0 & 1 \\\end{bmatrix} &= \begin{bmatrix}1 & 0 \\ 0 & -1 \\\end{bmatrix} &= M_x\\
\end{aligned}
$$
One key thing to note is that the specific matrices $M_x$, $M_y$ and $M_z$ are not really fundamental: we could easily have had different matrices if we had chosen any other parametrization of the group.
TODO confirm: however, no matter which parametrization we choose, the <Lie bracket> relations between the three elements would always be the same, since it is the number of elements, and the definition of the <Lie bracket>, that is truly fundamental.
<Lie Groups, Physics, and Geometry by Robert Gilmore (2008)> Chapter 4.2 "How to linearize a Lie Group" then calculates the <exponential map> of the vector $xM_x + yM_y + zM_z$ as:
$$
I cosh(\theta) + M_x sinh(\theta)/\theta
$$
with:
$$
\theta^2 = x^2 + bc
$$
TODO now the natural question is: can we cover the entire Lie group with this exponential? <Lie Groups, Physics, and Geometry by Robert Gilmore (2008)> Chapter 7 "EXPonentiation" explains why not.