Finite-dimensional representations of ${\mathfrak{s}\mathfrak{l}}_2$

The algebra ${\mathfrak{s}\mathfrak{l}}_2$ is semisimple: in the basis $(h,e,f)$ one computes ${\mathrm{ad}}_h=\mathrm{diag}(0,2,-2)$, ${\mathrm{ad}}_e=\left(\begin{array}{ccc}0& 0& 1\\ -2& 0& 0\\ 0& 0& 0\end{array}\right)$ and ${\mathrm{ad}}_f=\left(\begin{array}{ccc}0& -1& 0\\ 0& 0& 0\\ 2& 0& 0\end{array}\right)$, whence $B(h,h)=8$, $B(e,f)=B(f,e)=4$ and the remaining pairings vanish; that matrix has nonzero determinant, so the Killing form is nondegenerate and [L4] makes ${\mathfrak{s}\mathfrak{l}}_2$ semisimple. Consequently [L3] gives (i): every finite-dimensional $V$ is a direct sum of irreducible submodules.

Let $V\mathrm{\ne }0$ be irreducible. Since $h$ is an endomorphism of a nonzero finite-dimensional complex vector space, [L2] gives an eigenvalue $\lambda$ and eigenvector $v\mathrm{\ne }0$ with $hv=\lambda v$; because $h(ev)=e(hv)+[h,e]v=(\lambda +2)ev$ and $h(fv)=(\lambda -2)fv$ by [L1], the sum $W$ of all eigenspaces of $h$ in $V$ is a nonzero submodule, so $W=V$; thus $h$ acts diagonalisably on an irreducible module.

Let $V\mathrm{\ne }0$ be irreducible, choose among its finitely many eigenvalues of $h$ one with maximal real part, say $\lambda$, and choose $0\mathrm{\ne }v\in V$ with $hv=\lambda v$. Then $ev=0$: otherwise $ev$ is an eigenvector of $h$ with eigenvalue $\lambda +2$, contradicting maximality of the real part. Put $v_k=f^{k}v$ for $k\ge 0$; induction on $k$ using $[e,f]=h$ gives $h{v}_k=(\lambda -2k)v_k$ and $e{v}_k=k(\lambda -k+1)v_{k-1}$ for $k\ge 1$.

The vectors $v_k$ of step 2.1 cannot all be nonzero: nonzero $v_k$ are eigenvectors of $h$ with the distinct eigenvalues $\lambda -2k$, hence linearly independent, and $V$ is finite-dimensional. Let $N+1$ be the least index with $v_{N+1}=0$; then $v_0,\dots ,v_N\mathrm{\ne }0$. Applying step 2.1's formula for $e$ at $k=N+1$ gives $0=e{v}_{N+1}=(N+1)(\lambda -N)v_N$, so $\lambda =N\in {\mathbb{Z}}_{\ge 0}$ because the field has characteristic zero. The span of $v_0,\dots ,v_N$ is a nonzero submodule by the same formulas, hence equals $V$ by irreducibility; it has dimension $N+1$ and its $h$-eigenvalues are $N,N-2,\dots ,-N$, each with a one-dimensional eigenspace. This proves (ii).

Finally, an arbitrary nonzero finite-dimensional $V$ is a direct sum of irreducibles by (i), and on each summand $h$ is diagonalisable with the integer eigenvalues of (ii); hence $h$ is diagonalisable on all of $V$ with integer eigenvalues, which is (iii). If $V=0$ all three statements are vacuous.