Adjoint representation of Sl(2,C)

Question

What is the adjoint representation of Sl(2,C). Can anyone help explain this please?

Answer 1

From your comment I understand that you are talking about the Lie algebra $\mathfrak{sl}(2, \mathbb{C})$ not the group $SL(2, \mathbb{C})$.

Let's zoom out a bit and talk about Lie algebra representations in general.

Lie algebra representations involve (obviously) a Lie algebra $\mathfrak{g}$, but equally importantly a vector space $V$.

The third and final ingredient is a map $\phi \colon \mathfrak{g} \to End(V)$, where the latter is the space of all linear maps from $V$ to itself (equivalently: the space of all $n \times n$-matrices, with $n = \dim V$), viewed as a Lie algebra with bracket $[A, B] = AB - BA$.

So the map $\phi$ assigns to each element $X \in \mathfrak{g}$ a matrix (or linear map, depending on your perspective). This is in line with your comment.

Also in line with your comment we can say that in the special case of adjoint representations the map $\phi$ is called 'ad'. But before we can say what is this map we must first answer the more important question: what is the space $V$?

This is the important part: the vector space $V$ on which the Lie algebra acts is the lie algebra $\mathfrak{g}$ itself! We forget some of its structure, and just see it as a vector space.

So in the case of $\mathfrak{g} = \mathfrak{sl}(2, \mathbb{C})$ we have that the representation space $V$ is three dimensional and hence for each $X \in \mathfrak{sl}(2, \mathbb{C})$ the corresponding matrix $ad(X)$ is a three-by-three (not two by two) matrix!

However, in this case things are a bit easier if you think about $ad(X)$ not so much as a matrix but more as a linear transformation, i.e. a linear map sending elements of the vector space $\mathfrak{g}$ to elements of the vector space $\mathfrak{g}$. In this setting the map $ad(X)$ (for any Lie algebra, not just $\mathfrak{sl}(2, \mathbb{C})$) is easily described:

$ad(X)$ is the map that takes an element $Y$ and sends it to the element $[X, Y]$

It is a nice exercise to compute the corresponding matrix with respect to a vector space basis of $\mathfrak{g}$. Elements of $\mathfrak{g}$ now appear in two ways: as vectors in the representation space $V$ (which just happens to be $\mathfrak{g}$ itself), and as matrices acting on that space (their images under ad).

For instance, if I take the standard basis of $\mathfrak{sl}(2, \mathbb{C})$ where $E$ is the two-by-two matrix with a 1 in the top right corner, and otherwise zeroes, $F$ is its transpose, and $H$ is the diagonal matrix with 1 in the top left and -1 in the lower right corner then, viewed as an element of the space $V$ the element $E$ looks like the vector $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$.

The element $ad(E)$ on the other hand looks like $\begin{pmatrix} 0 & 0 & - 2 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}$.

It is easy to see that if you multiply matrix $ad(E)$ with vector $E$ you get the zero-vector. This is correct because the outcome should be the vector corresponding to $[E, E]$, which is zero. Similarly since $[E, F] = H$ we know that $ad(E)$ maps $F$ to $H$ and hence that multiplying the matrix $ad(E)$ given above with the vector $\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$ we will get the vector $\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$ corresponding to $H$ in the $\{E, F, H\}$ basis.

See if you can compute the matrices corresponding to $ad(F)$ and $ad(H)$ yourself!