We have the following basis elements for $\sl(2,\RR)\cong\su(2,\CC')\cong\so(2,1)$, a real form of $\su(2)$: \begin{equation} \sigma_0=\frac12\sigma_z, \quad \sigma_\pm=\frac12(\sigma_x\mp s_y)=\frac12(\sigma_x\pm i\sigma_y), \end{equation} with commutation relations \begin{equation} [\sigma_0,\sigma_\pm]=\pm\sigma_\pm, \quad [\sigma_+,\sigma_-]=2\sigma_0. \label{su2comm} \end{equation} These basis elements also form a basis of the complexified Lie algebra $\su(2)\otimes\CC$.

Figure 1: The root diagram of $\su(2)$.

We can thus represent $\sl(2,\RR)$ graphically as the points $0,\pm1\in\RR$, representing $\sigma_z$ acting on itself and $\sigma_\pm$, respectively, connected by oriented arrows representing the action of $\sigma_\pm$, as shown in Figure 1. This diagram fully captures the algebraic description $\sl(2,\RR)$ acting on itself, the so-called adjoint representation of $\sl(2,\RR)$. Each of these statements can be reinterpreted as being about $\su(2)\otimes\CC$; Figure 1 is normally called the root diagram of $\su(2)$.

We can now ask about more general representations of $\su(2)$, with $\rho(\su(2))$ acting on some vector space $V$. The commutation relations (\ref{su2comm}) show that $\sigma_0$ is diagonal in the given basis. It turns out that $L_z=\rho(\sigma_0)$ is diagonalizable in any representation $\rho$, 1) so we can choose a basis for $V$ consisting entirely of eigenvectors of $L_z$. If $w\ne0$ is one such eigenvector, we have \begin{equation} L_z w = \lambda w \end{equation} for some $w\in\CC$. Letting $L_\pm=\rho(\sigma_\pm)$, we have \begin{equation} L_z L_\pm w = [L_z,L_\pm] w + L_\pm L_z w = \pm L_\pm w + L_\pm \lambda w = (\lambda\pm1) L_\pm w \end{equation} Thus, $L_\pm w$ is also an eigenvector of $L_z$, with eigenvalue $\lambda\pm1$.

We want $V$ to be an irreducible representation of $\su(2)$, by which we mean that there should be no (nonzero, proper) subrepresentations of $\su(2)$ in $V$. Thus, acting repeatedly on $w$ with $L_\pm$ must generate a basis for $V$, as any vector not contained in the resulting span would itself generate a disjoint subrepresentation.

We also want $V$ to be finite. Since we are changing the eigenvalue at each step, this can only happen if there is a “biggest” eigenvalue. That is, we can assume without loss of generality that \begin{equation} L_+ w = 0 \end{equation} and that the remaining basis vectors are obtained by repeated action of $L_-$.

We now compute \begin{align} L_+L_- w &= [L_+,L_-] w + L_-L_+ w = 2L_z w = 2\lambda w \\ L_+L_- L_-w &= [L_+,L_-] L_-w + L_-L_+L_- w \nonumber\\ &= 2 L_zL_-w + 2\lambda L_-w = 2(2\lambda-1) L_w \\ \vdots\qquad &= \qquad\qquad\vdots \nonumber\\ L_+(L_-)^k w &= … = \bigl(2k\lambda-k(k-1)\bigr) (L_-)^{k-1} w \label{Lmax} \end{align} But for $V$ to be finite, $(L_-)^k w$ must be zero for some positive integer $k$. Assume that $k$ is the smallest such integer. Then $(L_-)^{k-1}$ is not zero, and therefore \begin{equation} 2k\lambda-k(k-1)=0 \end{equation} by (\ref{Lmax}). Since $k\ne0$, we conclude first of all that \begin{equation} \lambda = \frac{k-1}{2} \end{equation} is an integer or half-integer, so that there are $k=2\lambda+1$ basis vectors, with eigenvalues \begin{equation} \lambda, \lambda-1, …, \lambda-2\lambda=-\lambda . \end{equation}

We conclude that there is exactly one (irreducible) representation of $\su(2)$ for each dimension $k\ge2$, with eigenvalues $\{-\frac{k-1}{2},…,\frac{k-1}{2}\}$. Put differently, we can reproduce the commutation relations (\ref{su2comm}) using $n\times n$ matrices for any $n\ge2$, and can do so in essentially just one way (up to change of basis).