Classification of finite-dimensional sl2 representations

For every $n\ge 0$ there is one irreducible ${\mathfrak{sl}}_2(\mathbb{C})$-module $V(n)$ of dimension $n+1$, with weights $n,n-2,\dots ,-n$, each of multiplicity one. Every finite-dimensional representation is a direct sum of these modules.

For the defining sl2 Lie algebra module with basis $x,y$, put $A_k=x\otimes x^{3-k}{y}^k$ and $B_k=y\otimes x^{3-k}{y}^k$. The raising operator has $e{A}_k=k{A}_{k-1}$ and $e{B}_k=A_k+k{B}_{k-1}$; the lowering operator has $f{A}_k=B_k+(3-k)A_{k+1}$ and $f{B}_k=(3-k)B_{k+1}$. The highest-weight vectors $A_0$ and $B_0-A_1$ generate irreducible submodules of highest weights four and two. Bases are respectively $(A_0,3A_1+B_0,A_2+B_1,A_3+3B_2,B_3)$ and $(B_0-A_1,B_1-A_2,B_2-A_3)$. Their vectors are independent in every weight space, proving the displayed direct sum without only comparing dimensions.

For a finite-dimensional sl2 Lie algebra representation, write $\chi (q)={\sum }_j{c}_j{q}^j$. Each irreducible summand has character $q^m+q^{m-2}+\cdots +q^{-m}$, so $c_j=c_{-j}$ and $c_j\ge c_{j+2}$ for nonnegative $j$. Thus each parity subsequence is symmetric and unimodal. If all weights have one parity, removing the unused exponents gives an ordinarily unimodal coefficient sequence. The full sequence including every integer exponent need not be unimodal: $q+q^{-1}$ has coefficients $1,0,1$.

In the sl2 Lie algebra, if $Xv=0$ and $Hv=mv$, then the commutator relations give $H{Y}^{j}v=(m-2j)Y^{j}v$ and $X{Y}^{k+1}v=(k+1)(m-k)Y^{k}v$. Induct using $XY=YX+H$. This controls the raising operator on a string formed by the lowering operator and yields the classification of finite-dimensional sl2 representations.

For a highest-weight vector $v$ of weight $k$ in an irreducible finite-dimensional sl2 Lie algebra module, the basis $v,Yv,\dots ,Y^{k}v$ diagonalizes $XY$, with eigenvalue $(j+1)(k-j)$ on $Y^{j}v$. Summing those eigenvalues gives $\mathrm{tr}(XY)=k(k+1)(k+2)/6$. The formula $X{Y}^{n}v=n(k-n+1)Y^{n-1}v$ requires $v$ to be a highest-weight vector, not an arbitrary vector of the module.

In the irreducible sl2 Lie algebra module $V(m)$, use $v_j=f^j{v}_0$, with $0\le j\le m$. The Lie-invariant bilinear form $B(v_j,v_k)=(-1{)}^j{1}_{\{j+k=m\}}$ is nondegenerate: its anti-diagonal is nonzero. Invariance follows from $h{v}_j=(m-2j)v_j$, $f{v}_j=v_{j+1}$ and $e{v}_j=j(m-j+1)v_{j-1}$. Its transpose equals $(-1{)}^{m}B$, so it is a symmetric bilinear form when $m$ is an even number, and an alternating bilinear form when $m$ is an odd integer. The Weyl complete reducibility theorem gives a nondegenerate invariant form on any finite-dimensional module by taking the orthogonal direct sum of these forms.

On any finite-dimensional ${\mathfrak{sl}}_2$-module, the lowering operator $f$ is injective from the $m$-weight space to the $(m-2)$-weight space whenever $m>0$. This follows on each irreducible summand from its standard weight string.