Highest Weight Theory for Complex Semisimple Lie Algebras — Examples

The operators define a representation: on each basis vector, $HF{v}_k-FH{v}_k=(n-2k-2)v_{k+1}-(n-2k)v_{k+1}=-2v_{k+1}=-2F{v}_k$, and similarly $HE{v}_k-EH{v}_k=2E{v}_k$; moreover $EF{v}_k-FE{v}_k=(k+1)(n-k)v_k-k(n-k+1)v_k=(n-2k)v_k=H{v}_k$, with both sides zero for $k=n$ and $k=0$ respectively. This verifies the three bracket relations on every basis vector, hence (i).

The eigenvalues $n-2k$, $k=0,\dots ,n$, of $H$ are pairwise distinct, so every $H$-eigenspace of $V(n)$ is one-dimensional, spanned by the corresponding $v_k$.

$V(n)$ is irreducible: if $0\mathrm{\ne }W\subseteq V(n)$ is a submodule, then $W$ is $H$-stable and contains a nonzero $H$-eigenvector, hence some $v_k$; applying $E$ exactly $k$ times gives $E^k{v}_k=k!(n-k+1)(n-k+2)\cdots n{v}_0\mathrm{\ne }0$ because each coefficient $j(n-j+1)$ with $1\le j\le n$ is nonzero, so $v_0\in W$; applying $F$ repeatedly then gives $v_1,\dots ,v_n\in W$; hence $W=V(n)$ by [L2].

Every finite-dimensional irreducible ${\mathfrak{s}\mathfrak{l}}_2$-module $W$ is isomorphic to some $V(n)$: by [L1] its top weight is an integer $n\ge 0$, and it has a highest weight vector $w_n$ with $E{w}_n=0$ and $H{w}_n=n{w}_n$; the commutation identity $[E,F^j]=j{F}^{j-1}(H-(j-1))$, proved by induction, gives $E{F}^j{w}_n=j(n-j+1)F^{j-1}{w}_n$ and $H{F}^j{w}_n=(n-2j)F^j{w}_n$; the span of $F^j{w}_n$, $j=0,\dots ,n$, is nonzero and stable under $E,F,H$, hence equals $W$ by irreducibility, and the assignment $F^j{w}_n\mapsto v_j$ is an isomorphism $W\to V(n)$.

Steps 1.1, 3.1 and 4.1 establish (i), (ii) and (iii), and the modules $V(n)$ for distinct $n$ are non-isomorphic because $H$ has different eigenvalue sets.

The relations $h\mapsto \lambda (h)$ and $e\mapsto 0$ define a character ${\chi }_{\lambda }:U(\mathbb{C}h\oplus \mathbb{C}e)\to \mathbb{C}$. By PBW, multiplication is a linear isomorphism $U(\mathbb{C}f)\otimes U(\mathbb{C}h\oplus \mathbb{C}e)⟶U({\mathfrak{s}\mathfrak{l}}_2)\mathrm{.}$ Tensoring this factorisation over $U(\mathbb{C}h\oplus \mathbb{C}e)$ with the one-dimensional module ${\mathbb{C}}_{\lambda }$ identifies $M(\lambda )$ linearly with $U(\mathbb{C}f)$. Thus the classes of $f^k$, $k\ge 0$, are a basis of $M(\lambda )$.

Therefore $M(\lambda )$ is infinite-dimensional, and the action on $v_k=f^k\cdot v_0$ is $h\cdot v_k=(\lambda (h)-2k)v_k$ (eigenvalue computation), $f\cdot v_k=v_{k+1}$, and $e\cdot v_k=[e,f^k]\cdot v_0=k(\lambda (h)-k+1)v_{k-1}$ by [L3]; in particular $e\cdot v_k=0$ exactly when $k=0$ or $\lambda (h)=k-1$.

If $\lambda (h)\mathrm{\notin }{\mathbb{Z}}_{\ge 0}$, then $e\cdot v_k\mathrm{\ne }0$ for every $k\ge 1$. Given $0\mathrm{\ne }w={\sum }_{k=0}^N{c}_k{v}_k$, choose the largest $K$ with $c_K\mathrm{\ne }0$. For $k<K$ one has $e^K{v}_k=0$, whereas $e^K{v}_K=K!{\prod }_{j=1}^K(\lambda (h)-j+1)v_0\mathrm{\ne }0.$ Thus $e^{K}w$ is a nonzero multiple of $v_0$. Hence $v_0\in U({\mathfrak{s}\mathfrak{l}}_2)w$, so every nonzero submodule is all of $M(\lambda )$. The module is therefore simple and infinite-dimensional and has no nonzero finite-dimensional quotient.

If $\lambda (h)=n$ is a nonnegative integer, then $e\cdot v_{n+1}=0$ by step 2.1, and $K={\sum }_{k\ge n+1}\mathbb{C}{v}_k$ is a nonzero proper submodule. Let $N$ be a submodule not contained in $K$, and choose a finite nonzero sum $w=\sum c_k{v}_k\in N\setminus K$. Since the $h$-eigenvalues $n-2k$ are pairwise distinct, a polynomial in $h$ isolates from this finite sum a nonzero term $c_j{v}_j\in N$ with $j\le n$. Then $e^j{v}_j=j!{\prod }_{r=1}^j(n-r+1)v_0\mathrm{\ne }0$, so $v_0\in N$ and $N=M(\lambda )$. Consequently every proper submodule lies in $K$; hence $K$ is the unique maximal proper submodule and $M(\lambda )\mathrm{/}K$ is the unique simple quotient.

On $M(\lambda )\mathrm{/}K$ the classes of $v_0,\dots ,v_n$ satisfy exactly the relations of $V(n)$ from All irreducible finite-dimensional sl2 modules, with $h$-eigenvalues $n-2k$ and operators $f,e$ acting by $v_k\mapsto v_{k+1}$ and $v_k\mapsto k(n-k+1)v_{k-1}$; hence the quotient is the finite-dimensional simple module $V(n)$ by [L4].

Collecting steps 1.1–4.1, $M(\lambda )$ is an infinite-dimensional highest weight module of highest weight $\lambda$, and it has a finite-dimensional simple quotient exactly for $\lambda (h)\in {\mathbb{Z}}_{\ge 0}$, in which case that quotient is $V(\lambda (h))$; this proves all the assertions.

The vector $x_1^m$ is a highest weight vector: $H\cdot x_1^m=m{H}_{11}{x}_1^m$, so its weight is $m{\epsilon }_1=m{\omega }_1$; and $E_{ij}\cdot x_1=0$ for $i<j$ because $j\mathrm{\ne }1$, so $E_{ij}\cdot x_1^m=0$ for every positive root vector.

Let $0\mathrm{\ne }U\subseteq W$ be a submodule; writing a nonzero element as a sum of distinct-weight monomials, [L2] produces an element of $U(\mathfrak{h})$ that projects onto one of them, so $U$ contains a monomial $x_1^{a_1}\cdots x_n^{a_n}$ with $a_j>0$ for some $j>1$ unless it already contains $x_1^m$.

From any monomial $x_1^{a_1}\cdots x_n^{a_n}$ with $a_j>0$, $j>1$, applying $E_{1j}$ exactly $a_j$ times replaces all $x_j$-factors by $x_1$-factors with nonzero coefficient $a_j!$, and repeating for $j=2,\dots ,n$ reaches a nonzero multiple of $x_1^m$; hence $x_1^m\in U$ by step 1.2.

Conversely, from $x_1^m$ the operators $E_{l1}$ with $l>1$ replace $x_1$-factors by $x_l$-factors, and applying them $a_l$ times successively for $l=2,\dots ,n$ produces a nonzero multiple of $x_1^{a_1}\cdots x_n^{a_n}$; hence $U$ contains the whole monomial basis and $U=W$.

Therefore every nonzero submodule of $W$ is $W$, so $W$ is irreducible, and by step 1.1 its highest weight is $m{\omega }_1=$ the weight of $x_1^m$; this proves the assertion.

The vector $E_{1n}$ has weight ${\epsilon }_1-{\epsilon }_n$: $[H,E_{1n}]=(H_{11}-H_{nn})E_{1n}$ by [L2].

$E_{1n}$ is killed by every positive root vector: for $i<j$ we have $[E_{ij},E_{1n}]={\delta }_{j1}{E}_{in}-{\delta }_{in}{E}_{1j}$, and ${\delta }_{j1}=0$ because $i<j$ with $i\ge 1$, while ${\delta }_{in}=0$ because $i<j\le n$ forces $i<n$; hence $[E_{ij},E_{1n}]=0$.

The adjoint submodule generated by $E_{1n}$ is all of ${\mathfrak{s}\mathfrak{l}}_n$. It contains $H=[E_{n1},E_{1n}]=E_{nn}-E_{11}$; then $[E_{nk},H]$ is a nonzero scalar multiple of $E_{nk}$ for every $k<n$. It also contains $E_{jn}=[E_{j1},E_{1n}]$ for $1<j<n$, as well as the original $E_{1n}$. Finally, $[E_{nk},E_{jn}]={\delta }_{kj}{E}_{nn}-E_{jk}$ supplies every off-diagonal $E_{jk}$ with $j,k<n$ and every diagonal difference $E_{nn}-E_{jj}$. These matrices span ${\mathfrak{s}\mathfrak{l}}_n$.

In the Euclidean model of the roots take the traceless vector $v=(n-1,n-3,\dots ,1-n)$: $(v,{\epsilon }_i-{\epsilon }_j)=2(j-i)$, so it is regular and selects exactly $i<j$. Each positive root has the expansion ${\epsilon }_i-{\epsilon }_j={\alpha }_i+\cdots +{\alpha }_{j-1}$. The $n-1$ vectors ${\alpha }_j$ are independent: comparing successive coordinates in $\sum c_j({\epsilon }_j-{\epsilon }_{j+1})=0$ gives every $c_j=0$. If $j>i+1$ the root splits at $i+1$ into two positive roots; an adjacent root cannot so split because the two nonempty interval expansions would have to sum to its single coefficient one. Thus these adjacent differences are exactly the simple roots.

For every positive root ${\epsilon }_i-{\epsilon }_j$ with $i<j$, one has $({\epsilon }_1-{\epsilon }_n)-({\epsilon }_i-{\epsilon }_j)=({\alpha }_1+\cdots +{\alpha }_{i-1})+({\alpha }_j+\cdots +{\alpha }_{n-1}),$ a nonnegative integral combination of simple roots. Thus ${\epsilon }_1-{\epsilon }_n$ is the highest root. By steps 1.1–1.3, $E_{1n}$ has that weight, is killed by all positive root spaces, and generates the adjoint module, so it is a highest weight vector and the adjoint module has highest weight ${\epsilon }_1-{\epsilon }_n$. (Height and highest root, Highest-weight vectors and modules, step 1.1, step 1.2, step 1.3, step 1.4, L4) ∎

By [L1] the sum ${\chi }_n(z)={\sum }_{k=0}^n{z}^{n-2k}$ is the sum of $z^m$ over the $h$-eigenvalues $m$ of $V(n)$, each counted with its multiplicity, so it is the rank-one formal character under the convention fixed in the given data.

The telescoping identity $(z-z^{-1}){\chi }_n(z)={\sum }_{k=0}^n(z^{n-2k+1}-z^{n-2k-1})=z^{n+1}-z^{-(n+1)}$ holds as an identity of Laurent polynomials.

For $z\mathrm{\ne }0$ with $z-z^{-1}\mathrm{\ne }0$, that is for $z\mathrm{\ne }\pm 1$, division gives ${\chi }_n(z)=\frac{z^{n+1}-z^{-(n+1)}}{z-z^{-1}}$, which is the displayed formula on the regular set.

The identity of step 1.2 is the algebraic cancellation $z^{n+1}-z^{-(n+1)}=(z-z^{-1}){\chi }_n(z)$ in the Laurent polynomial ring; it exhibits ${\chi }_n$ as the quotient after cancelling the common factor $z-z^{-1}$, and evaluating that Laurent polynomial at $z=1$ gives ${\chi }_n(1)=n+1$, matching $\dim V(n)=n+1$ by [L1].

Hence the character identity on the regular set and the dimension formula both hold, as asserted.

By [L2] the adjoint module is $\mathfrak{g}=\mathfrak{h}\oplus {⨁}_{i\mathrm{\ne }j}\mathbb{C}{E}_{ij}$, so $\dim \mathfrak{g}=2+6=8$; the zero-weight space is $\mathfrak{h}$ of dimension $2$, and each of the six root spaces $\mathbb{C}{E}_{ij}$ is a one-dimensional weight space of weight ${\epsilon }_i-{\epsilon }_j$.

By [L1] the adjoint module has highest vector $E_{13}$ and highest weight ${\epsilon }_1-{\epsilon }_3$; by [L2] and [L3] that weight is the highest root ${\alpha }_1+{\alpha }_2$ and equals ${\omega }_1+{\omega }_2$.

Collecting the dimensions and weights, the adjoint representation of ${\mathfrak{s}\mathfrak{l}}_3(\mathbb{C})$ has dimension $8$, six one-dimensional root-weight spaces, a two-dimensional zero-weight space, and highest weight ${\alpha }_1+{\alpha }_2={\omega }_1+{\omega }_2$, as asserted.

The weight multiplicities of the tensor product are $w_m=\mathrm{\#}\{(r,s):0\le r\le a,0\le s\le b,a+b-2(r+s)=m\}$ by [L1]: the sum $m$ of the two weights $a-2r$ and $b-2s$ occurs once for each such pair.

For the right-hand side ${⨁}_{i=0}^{\min (a,b)}V(a+b-2i)$ the same weight $m$ occurs in the summand $V(a+b-2i)$ exactly when $\mathrm{\mid }m\mathrm{\mid }\le a+b-2i$ and $m\equiv a+b$ modulo $2$, so its multiplicity is $w_m^{\mathrm{\prime }}=\max (0,\min (\min (a,b),\lfloor (a+b-\mathrm{\mid }m\mathrm{\mid })\mathrm{/}2\rfloor )+1)$ in that parity case and $0$ otherwise.

The counts agree, $w_m=w_m^{\mathrm{\prime }}$ for every integer $m$. If $m\equiv ̸a+b(\mathrm{m}\mathrm{o}\mathrm{d}2)$ or $\mathrm{\mid }m\mathrm{\mid }>a+b$, both counts are zero. Otherwise put $j=(a+b-m)\mathrm{/}2$. For $m\ge 0$ one has $0\le j\le (a+b)\mathrm{/}2$, and the tensor count is $c_j=\max (0,\min (j,a)-\max (0,j-b)+1)\mathrm{.}$ A direct case check for $a\le b$ and $a>b$ identifies this with the right-hand count of step 1.2. The identity for $m<0$ follows from the symmetries $w_{-m}=w_m$ and $w_{-m}^{\mathrm{\prime }}=w_m^{\mathrm{\prime }}$ obtained by reflecting the weight strings.

Both sides are direct sums of irreducibles and the left side is completely reducible by [L2]; moreover in a completely reducible ${\mathfrak{s}\mathfrak{l}}_2$-module the multiplicity $c_n$ of $V(n)$ is determined by the weight multiplicities through $w_n={\sum }_{m\ge n,m\equiv n(2)}{c}_m$, so that $c_n=w_n-{\sum }_{m>n,m\equiv n(2)}{c}_m$ is recovered by downward induction on $n$.

Applying the recovery of step 2.2 to the two modules, whose weight multiplicities agree by step 2.1, gives equal multiplicities of every $V(n)$ and hence the multiplicity-one decomposition $V(a)\otimes V(b)\cong {⨁}_{i=0}^{\min (a,b)}V(a+b-2i)$.

The functional $\lambda$ with $\lambda (h)=-1$ is integral by [L1], so it satisfies the hypothesis of the refuted statement.

By [L2] the highest weight module $M(\lambda )$ for this $\lambda$ is infinite-dimensional; the conclusion of the refuted statement ("$M(\lambda )$ is finite-dimensional") thus fails.

The witness is explicit: the integral but nondominant functional $\lambda$ with $\lambda (h)=-1$ and the module $M(\lambda )$ with its infinite basis $v_k=f^k\cdot v_0$, $k\ge 0$; the failed conclusion is the implication from integrality to finite-dimensionality.

The element $-I$ acts on a homogeneous polynomial of degree $m$ by $(-I)\cdot p(x)=p((-I{)}^{-1}x)=p(-x)=(-1{)}^{m}p(x)$, so the operator ${\rho }_m(-I)$ on $W_m$ is the scalar $(-1{)}^m$.

A homomorphism $\rho :SU(2)\to GL(V)$ factors as $\bar{\rho }\circ \pi$ for a homomorphism $\bar{\rho }:SO(3)\to GL(V)$ if and only if $\rho (-I)={\mathrm{id}}_V$: if $\rho =\bar{\rho }\circ \pi$ then $\pi (-I)=I$ gives $\rho (-I)=\mathrm{id}$, while conversely $\rho (-g)=\rho (g)\rho (-I)=\rho (g)$ whenever $\rho (-I)=\mathrm{id}$, so $\rho$ is constant on the fibers $\{g,-g\}$ of the surjective $\pi$ from [L1] and descends uniquely to the quotient $SO(3)\cong SU(2)\mathrm{/}\{\pm I\}$.