Lie algebra of sl(2) – Random Thought

The “S” in $SL(2,\mathbb{C})$ stands for special, meaning that $A \in SL(2,\mathbb{C})$ , then $det(A)=1$ .

$sl(2,\mathbb{C})=\left\{\begin{pmatrix}a&b\\c&d\end{pmatrix}: \det\exp \left(t\begin{pmatrix}a&b\\c&d\end{pmatrix} \right)= 1,\forall t \right\}$

$\det\exp \left(t\begin{pmatrix}a&b\\c&d\end{pmatrix} \right)=\det \left(t\begin{pmatrix}1+ta&tb\\tc&1+td\end{pmatrix} + O(t^2) \right)$

$= 1+t(a+d)+O(t^2)=1, \forall t$ .

The condition obviously requires $a+d=0$ . It turns out that the converse is true as well and so $sl(2,\mathbb{C}) =\left\{\begin{pmatrix}a&b\\c&d\end{pmatrix}: a+d=0\right\}$ as shown in the following.

Let $H=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ ,
$X=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ , and
$Y=\begin{pmatrix}0&0\\1&0\end{pmatrix}$ .

Then any matrix in $\left\{\begin{pmatrix}a&b\\c&d\end{pmatrix}: a+d=0\right\}$ can be represented by $aH+bX+cY$ . By the linearity of $sl(2,\mathbb{C})$ . We can show that $aH+bX+cY\in sl(2,\mathbb{C})$ and thus $sl(2,\mathbb{C}) =\left\{\begin{pmatrix}a&b\\c&d\end{pmatrix}: a+d=0\right\}$ as long as we can show

$H,X,Y \in sl(2,\mathbb{C})$ .

Note that $\forall t$ , $\exp(t H) = \exp\begin{pmatrix}t& 0\\0&-t\end{pmatrix}=\begin{pmatrix}1+t+\cdots& 0\\0&1-t+\cdots\end{pmatrix}= \begin{pmatrix}e^t& 0\\0&e^{-t}\end{pmatrix}$ .

Thus, $\det(\exp(tH))=e^t e^{-t}=1 \Rightarrow H\in sl(2,\mathbb{C})$

and for $X$ , $\exp(t X)= \exp\begin{pmatrix}0& t\\0&0\end{pmatrix}=I+\begin{pmatrix}0& t\\0&0\end{pmatrix}+\frac{1}{2}\begin{pmatrix}0& t\\0&0\end{pmatrix}^2+\cdots=\begin{pmatrix}1& t\\0&1\end{pmatrix}$