# Example: SL(2,C)

## Example: SL(2,C)

### Definition

The group $SL(2,\mathbf{C})$ is defined to be the group of 2-by-2 complex matrices with determinant 1 ("S" in the name stands for "special", which means determinant 1). This is a topologically closed subgroup of $GL(2,\mathbf{C})$ because it's cut out by a continuous equation $\det(M)=1$ .

### Lie algebra

The Lie algebra of $SL(2,\mathbf{C})$ is $\mathfrak{sl}(2,\mathbf{C})=\left\{\begin{pmatrix}a&b\\ c&d\end{pmatrix}\ :\ \det\exp t\begin{pmatrix}a&b\\ c&d\end{pmatrix}=1\ \forall t\in\mathbf{R}\right\}$ To first order in $t$ , $\det\exp t\begin{pmatrix}a&b\\ c&d\end{pmatrix}=\det\begin{pmatrix}1+ta&tb\\ tc&1+td\end{pmatrix}+\mathcal{O}(t^{2})=(1+ta)(1+td)-t^{2}bc+\mathcal{O}(t^{2}% )=1+t(a+d)+\mathcal{O}(t^{2}).$ Differentiating with respect to $t$ at $t=0$ gives: $\left.\frac{d}{dt}\right|_{t=0}\det\exp t\begin{pmatrix}a&b\\ c&d\end{pmatrix}=a+d=\left.\frac{d}{dt}\right|_{t=0}1=0.$ Therefore, if $M=\begin{pmatrix}a&b\\ c&d\end{pmatrix}$ is in $\mathfrak{sl}(2,\mathbf{C})$ , then the trace of $M$ $\mathrm{Tr}(M)=a+d$ is zero. It turns out that this is "if and only if" (we'll see this later) so that $\mathfrak{sl}(2,\mathbf{C})=\{M\ :\ \mathrm{Tr}(M)=0\}$ The general element of $\mathfrak{sl}(2,\mathbf{C})$ is therefore $\begin{pmatrix}a&b\\ c&-a\end{pmatrix}$ and we see that the Lie algebra is a 3-dimensional (complex) vector space with coordinates $a,b,c$ . By contrast, it's trickier to write down a formula for the general element of $SL(2,\mathbf{C})$ .

A basis for this vector space is given by setting one coordinate at a time equal to 1 (and the others equal to zero): $H=\begin{pmatrix}1&0\\ 0&-1\end{pmatrix},\quad X=\begin{pmatrix}0&1\\ 0&0\end{pmatrix},\quad Y=\begin{pmatrix}0&0\\ 1&0\end{pmatrix}$ That is, any element of $\mathfrak{sl}(2,\mathbf{C})$ can be written in a unique way as $aH+bX+cY$ for some $a,b,c\in\mathbf{C}$ .

### Checking that sl(2,C) is a Lie algebra

Since $\mathfrak{sl}(2,\mathbf{C})$ is a Lie algebra, it's supposed to be preserved under taking commutator bracket. Let's check: $[H,X]=HX-XH=\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}\begin{pmatrix}0&1\\ 0&0\end{pmatrix}-\begin{pmatrix}0&1\\ 0&0\end{pmatrix}\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}=\begin{pmatrix}0&2\\ 0&0\end{pmatrix}=2X.$ $[H,Y]=HY-YH=\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}\begin{pmatrix}0&0\\ 1&0\end{pmatrix}-\begin{pmatrix}0&0\\ 1&0\end{pmatrix}\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}=\begin{pmatrix}0&0\\ -2&0\end{pmatrix}=-2Y.$ $[X,Y]=XY-YX=\begin{pmatrix}0&1\\ 0&0\end{pmatrix}\begin{pmatrix}0&0\\ 1&0\end{pmatrix}-\begin{pmatrix}0&0\\ 1&0\end{pmatrix}\begin{pmatrix}0&1\\ 0&0\end{pmatrix}=\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}=H.$ You can see that the bracket of two things in $\mathfrak{sl}(2,\mathbf{C})$ still lives in $\mathfrak{sl}(2,\mathbf{C})$ . All the other bracket computations follow from antisymmetry. For example: $[X,X]=[Y,Y]=[H,H]=0$ and $[Y,X]=-[X,Y]=-H,\quad[X,H]=-[H,X]=-2X,\mbox{ etc.}$ This example is of such fundamental importance that I'll restate the brackets we just found: $[H,X]=2X,\quad[H,Y]=-2Y,\quad[X,Y]=H.$ We will use these relations all the time later in the course.

### Trace(M) = 0 implies det(exp(M)) = 1

There's one loose end to tie up, namely showing that if $\mathrm{Tr}(M)=0$ then $M\in\mathfrak{sl}(2,\mathbf{C})$ . We have a basis ($H,X,Y$ ) for the space of trace-zero matrices. Because we know that $\mathfrak{sl}(2,\mathbf{C})$ is a vector space, it suffices to show that $H,X,Y$ are in $\mathfrak{sl}(2,\mathbf{C})$ because then any linear combination of them will live in $\mathfrak{sl}(2,\mathbf{C})$ .

Lemma:

$H\in\mathfrak{sl}(2,\mathbf{C})$ .

Proof:

We need to check that $\det\exp(tH)=1$ for all $t$ . We have $\exp(tH)=\exp\begin{pmatrix}t&0\\ 0&-t\end{pmatrix}=\begin{pmatrix}1+t+t^{2}/2+\cdots&0\\ 0&1-t+t^{2}/2-\cdots\end{pmatrix}=\begin{pmatrix}e^{t}&0\\ 0&e^{-t}\end{pmatrix}$ which has determinant 1.

## Pre-class exercise

Exercise:

Show that $X$ and $Y$ are in $\mathfrak{sl}(2,\mathbf{C})$ .