sl2 Lie algebra (${\mathfrak{sl}}_2$)

The Lie algebra ${\mathfrak{sl}}_2$ has generators $e,h,f$ with $[h,e]=2e$, $[h,f]=-2f$, and $[e,f]=h$.

This is the $n=2$ instance of the special linear Lie algebra. Classifying its modules is a separate representation-theory problem.

In a finite-dimensional sl2 Lie algebra representation, raising and lowering operators are nilpotent operators, so their exponentials are finite sums. In the defining two-dimensional representation, the displayed product is $\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$. It sends $h$ to $-h$ by conjugation, lifting the nontrivial Weyl group reflection. Its square acts on the irreducible highest-weight module $L(n)={\mathrm{Sym}}^n{\mathbb{C}}^2$ by $(-1{)}^n$. Thus a Weyl reflection can have a representative of order four on a representation.

// Target: linear-operator-theory.bigb

Let $e_i=x^{4-i}{y}^i$ and $w_{ij}=e_i\wedge e_j$ for $i<j$. For the sl2 Lie algebra, the raising and lowering operators are $E=x{\partial }_y$, $F=y{\partial }_x$, and $w_{ij}$ has weight $8-2(i+j)$. A weight basis of the irreducible representation $L(6)$ isA weight basis of the complementary $L(2)$ isThe first vector in each list is a highest-weight vector. Applying $F$ along the first list gives successive coefficients $3,2,1,12,5,2$; along the second it gives $1,2$. The final vectors are killed by $F$. At the overlapping weights $2,0,-2$, the two listed vectors are independent, proving the direct sum.

// Target: semisimple-lie-algebra.bigb

The real traceless two-by-two matrices form a three-dimensional Lie algebra. A basis adapted to conformal mechanics obeys $\{D,E\}=-E$, $\{D,K\}=K$, $\{E,K\}=2D$.

The SU(2) Lie algebra is the compact real form of ${\mathfrak{sl}}_2(\mathbb{C})$. In the Hermitian physics convention its generators satisfy $[J_i,J_j]=i{ϵ}_{ijk}{J}_k$.

In the anti-Hermitian basis $T^a=-i{\sigma }^a/2$, the SU(2) Lie algebra has $[T^a,T^b]={\epsilon }^{abc}{T}^c$ and Killing form $\kappa (T^a,T^b)=-2{\delta }^{ab}$. This is negative definite; the positive internal form for compact Yang-Mills theory is its negative.

With Hermitian angular-momentum generators $[J_a,J_b]=i{ϵ}_{abc}{J}_c$, the Killing form is $2{\delta }_{ab}$ and its inverse contraction is $C_{\kappa }=\frac{1}{2}{\sum }_a{J}_a^2$. Equivalently the anti-Hermitian basis $T_a=-i{J}_a$ has Killing form $-2{\delta }_{ab}$ and gives the same operator. In the complex highest-weight basis, $C_{\kappa }=H^2/8+(E^{+}{E}^{-}+E^{-}{E}^{+})/4$. It acts on spin $j$ by $j(j+1)/2$, so it differs in normalization from the usual angular-momentum quadratic Casimir operator.

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.

The Verma module $M_{\lambda }$ has basis $(f^r{v}_{\lambda }{)}_{r\ge 0}$ andIt is reducible exactly when $\lambda =m\in {\mathbb{Z}}_{\ge 0}$; then its unique proper nonzero submodule is generated by $f^{m+1}{v}_{\lambda }$ and is isomorphic to $M_{-m-2}$.

The Verma module $M(0)$ has basis $v_n=f^n{v}_0$, with $h{v}_n=-2n{v}_n$, $f{v}_n=v_{n+1}$ and $e{v}_n=n(1-n)v_{n-1}$. Its submodule spanned by $v_n$ for $n\ge 1$ is $M(-2)$, and the quotient is the trivial one-dimensional module. A complementary submodule would be trivial and hence lie in the zero-weight space $\mathbb{C}{v}_0$, but $f{v}_0=v_1\ne 0$. Therefore the displayed extension does not split. This explains why the finite-dimensional hypothesis in the Weyl complete reducibility theorem is essential.

An intermediate-series ${\mathfrak{sl}}_2$-module has one-dimensional weight spaces indexed by all integers. In the basis $(w_i{)}_{i\in \mathbb{Z}}$, one useful normalization isIt is irreducible exactly when every raising coefficient is nonzero.