📂Representation Theory

Special Linear Lie Algebra of Degree 2 sl(2,C)

Definition

The set of $2 \times 2$ complex matrices whose trace is $0$ is the Lie algebra of the special linear group $\operatorname{SL}(2, \mathbb{C})$.

$$ \mathfrak{sl}(2, \mathbb{C}) = \left\{ A \in M_{2}(\mathbb{C}) : \trace(A) = 0 \right\} $$

Explanation

$\mathfrak{sl}(2, \mathbb{C})$ is the case $n = 2$ of the special linear Lie algebra $\mathfrak{sl}(n, \mathbb{C})$, and it is a $3$-dimensional $\mathbb{C}$-vector space. It is also the complexification of the special unitary Lie algebra of degree 2 $\mathfrak{su}(2)$.

Algebraic Structure¹

Basis and Commutation Relations

Let us find out what a basis actually looks like, using the condition for $A \in \mathfrak{sl}(2, \mathbb{C})$. The only condition is that the trace is $0$, so $a_{22} = -a_{11}$. Hence an element of $\mathfrak{sl}(2)$ is expressed as follows for complex numbers $\alpha, \beta, \gamma$, and therefore $\mathfrak{sl}(2, \mathbb{C})$ is a 3-dimensional $\mathbb{C}$-vector space.

$$ A = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} = \alpha \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} + \beta\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} + \gamma\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} $$

$$ H = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}, \quad X = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \quad Y = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} $$

Therefore a basis of $\mathfrak{sl}(2, \mathbb{C})$ is $\left\{ H, X, Y \right\}$, and computing their structure constants gives the following.

$$ [H, X] = 2X, \quad [H, Y] = -2Y, \quad [X, Y] = H \tag{1} $$

Quantization of the Representation Space

Suppose we are given a finite-dimensional $\mathbb{C}$-vector space $V$ and an arbitrary representation $\pi$ from $\mathfrak{sl}(2, \mathbb{C})$ to $V$.

$$ \pi : \mathfrak{sl}(2, \mathbb{C}) \to \operatorname{gl}(V) $$

Since a representation is a homomorphism, the bracket is preserved.

$$ \left[ \pi(H), \pi(X) \right] = \pi([H, X]) = 2\pi(X) $$

What the commutation relations $(1)$ mean is that $\pi(X)$ and $\pi(Y)$ are eigenvectors of the map $\ad_{H}$.

$$ \ad_{H}(X) = 2X, \quad \ad_{H}(Y) = -2Y $$

They also mean that $\pi(X)$ and $\pi(Y)$ are ladder operators for $\pi(H)$ on the vector space $V$. Let $v \in V$ be an eigenvector of $\pi(H)$ corresponding to the eigenvalue $\lambda$. Then the following holds.

$$ \begin{align*} \pi(H)(\pi(X)v) = \left( \pi(X)\pi(H) + 2\pi(X) \right)v = (\lambda + 2)\pi(X)v \\[1em] \pi(H)(\pi(Y)v) = \left( \pi(Y)\pi(H) - 2\pi(Y) \right)v = (\lambda - 2)\pi(Y)v \end{align*} \tag{2} $$

Hence $\pi(X)$ is a raising operator that maps an eigenvector corresponding to the eigenvalue $\lambda$ to an eigenvector corresponding to $\lambda+2$, and $\pi(Y)$ is a lowering operator acting in the opposite direction. Denote the eigenspace corresponding to the eigenvalue $\lambda$ by $V_{\lambda}$.

$$ \pi(X) : V_{\lambda} \to V_{\lambda+2}, \quad \pi(Y) : V_{\lambda+2} \to V_{\lambda} $$

The eigenvalues of $\pi(H)$ are quantized by $\pi(X)$ and $\pi(Y)$ so that they all differ in steps of $2$.

Structure of the Representation Space

By $(2)$, $\pi(H)\pi(X)^{k}v = (\lambda + 2k)\pi(X)^{k}v$ holds. This says that for each $k$, $\pi(X)^{k}v$ is an eigenvector of $\pi(H)$ corresponding to the distinct eigenvalue $(\lambda + 2k)$, which in turn means that $\left\{ \pi(X)^{k}v \right\}$ is linearly independent. Since $V$ is finite-dimensional, the number of linearly independent eigenvectors is also finite. In other words, there must exist some $N \ge 0$ satisfying the following.

$$ \pi(X)^{N}v \ne 0, \quad \pi(X)^{N+1}v = 0 $$

Now denote by $v_{0}$ the vector just before it becomes $0$ under $\pi(X)$. This is an eigenvector of $\pi(H)$ with eigenvalue $\mu = \lambda + 2N$.

$$ \begin{align*} v_{0} &= \pi(X)^{N}v \\ \pi(H) v_{0} &= \mu v_{0} \\ \pi(X) v_{0} &= 0 \\ \end{align*} $$

Let $v_{k} = \pi(Y)^{k}v_{0}$. By $(2)$ we obtain the following.

$$ \pi(H) v_{k} = \pi(H)(\pi(Y)^{k}v_{0}) = (\mu - 2k)(\pi(Y)^{k}v_{0}) = (\mu - 2k) v_{k} $$

We can also obtain the following recurrence relation.

$$ \begin{align*} \pi(X) v_{1} &= \pi(X)\pi(Y) v_{0} = \left[ \pi(Y)\pi(X) + \pi(H) \right] v_{0} \\ &= \mu v_{0} \\ &= 1(\mu - 0) v_{0} \\ & \\ \pi(X) v_{2} &= \pi(X)\pi(Y)\pi(Y) v_{0} = \left[ \pi(Y)\pi(X) + \pi(H) \right]\pi(Y) v_{0} \\ &= \left[ \pi(Y)\pi(X)\pi(Y) + \pi(H)\pi(Y) \right] v_{0} \\ &= \left[ \pi(Y)\pi(Y)\pi(X) + \pi(Y)\pi(H) + \pi(H)\pi(Y) \right] v_{0} \\ &= \left[ \mu + (\mu - 2) \right]\pi(Y)v_{0} \\ &= 2(\mu - 1) v_{1} \end{align*} $$

$$ \begin{align*} \pi(X) v_{3} &= \pi(X)\pi(Y)\pi(Y)\pi(Y) v_{0} = \left[ \pi(Y)\pi(X) + \pi(H) \right]\pi(Y)\pi(Y) v_{0} \\ &= \left[ \pi(Y)\pi(X)\pi(Y)\pi(Y) + \pi(H)\pi(Y)\pi(Y) \right] v_{0} \\ &= \left[ \pi(Y)\left( \pi(Y)\pi(X) + \pi(H) \right)\pi(Y) + \pi(H)\pi(Y)\pi(Y) \right] v_{0} \\ &= \left[ \pi(Y)\pi(Y)\pi(X)\pi(Y) + \pi(Y)\pi(H)\pi(Y) + \pi(H)\pi(Y)\pi(Y) \right] v_{0} \\ &= \left[ \pi(Y)\pi(Y)\pi(Y)\pi(X) + \pi(Y)\pi(Y)\pi(H) + \pi(Y)\pi(H)\pi(Y) + \pi(H)\pi(Y)\pi(Y) \right] v_{0} \\ &= \left[ \mu + (\mu - 2) + (\mu - 4) \right] \pi(Y)\pi(Y)v_{0} \\ &= 3(\mu - 2) v_{2} \\ & \qquad \vdots \end{align*} $$

$$ \pi(X)v_{k} = k(\mu - (k-1))v_{k-1} \quad (k \ge 1) \tag{3} $$

And for the same reason as in the case of $\pi(X)$, there must exist some $m \ge 0$ satisfying the following.

$$ v_{m} = \pi(Y)^{m}v_{0} \ne 0, \quad v_{m+1} = \pi(Y)^{m+1}v_{0} = 0 $$

Then $\pi(X) v_{m+1} = 0$, and by the recurrence relation $(3)$ we obtain the following.

$$ \pi(X) v_{m+1} = (m+1)(\mu - m)v_{m} = 0 \\[1em] $$

This means that $\mu$, the eigenvalue of the last eigenvector before it becomes the zero vector under $\pi(X)$, is some positive integer $m$. Now, putting the above results together, for the representation $\pi$ there exist eigenvectors $v_{0}$, $\dots$, $v_{m}$ such that the following holds.

$$ \begin{align*} \pi(H) v_{k} &= (m -2k)v_{k} \\ \pi(Y) v_{k} &= \begin{cases} v_{k+1} & k \lt m \\ 0 & k = m \end{cases} \\ \pi(X)v_{k} &= \begin{cases} k(m - (k-1))v_{k-1} & k \gt 0 \\ 0 & k = 0 \end{cases} \end{align*} \tag{4} $$

Since their eigenvalues are distinct, $\left\{ v_{k} \right\}$ is linearly independent. By $(4)$, $W = \span\left\{ v_{0}, \dots, v_{m} \right\}$ is an invariant subspace for $\pi(H)$, $\pi(X)$, and $\pi(Y)$. Moreover, since $\left\{ H, X, Y \right\}$ is a basis of $\mathfrak{sl}(2, \mathbb{C})$, $W$ is also invariant under any operator $\pi(Z)$ $(Z \in \mathfrak{sl}(2, \mathbb{C}))$. Here, if $\pi$ is an irreducible representation, it has no nontrivial invariant subspace, so $W = V$ and $\left\{ v_{k} \right\}$ becomes a basis of $V$. Also, although we only defined $V$ to be finite-dimensional, we learn that $m+1$ is the dimension of $V$.

$$ V = \span \left\{ v_{0}, \dots, v_{m} \right\} \text{ and } \quad \dim V = m+1, \quad \text{if $\pi$ is irreducible} $$

Conversely, if we define $\pi(H)$, $\pi(X)$, $\pi(Y)$ so that they satisfy $(4)$, they satisfy the commutation relations $(1)$ and $\pi$ becomes an irreducible representation of $\mathfrak{sl}(2, \mathbb{C})$.

Furthermore, even if we take another representation $\pi^{\prime}$ different from $\pi$, $\pi^{\prime}$ satisfies $(4)$ by the same argument as above, and $\pi$ and $\pi^{\prime}$ are isomorphic.

Properties

(a) $\mathfrak{sl}(2, \mathbb{C})$ is the complexification of $\mathfrak{su}(2)$. $$ \mathfrak{su}(2)_{\mathbb{C}} \cong \mathfrak{sl}(2, \mathbb{C}) $$

(b) Irreducible representations of $\mathfrak{sl}(2, \mathbb{C})$ exist.

(c) Every eigenvalue of $\pi(H)$ is an integer. Moreover, if an integer $k$ is an eigenvalue of $\pi(H)$, then all of the following are also eigenvalues.

$$ -|k|, -|k|+2, \dots, |k|-2, |k| $$

(d) The operators $\pi(X)$ and $\pi(Y)$ are nilpotent transformations.