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 action on $V$ is $H\cdot e_i=H_{ii}{e}_i$ and $E_{ij}\cdot e_l={\delta }_{jl}{e}_i$; hence the weights of $V$ are ${\epsilon }_1,\dots ,{\epsilon }_n$, each with one-dimensional weight space, and $e_1$ is killed by every positive root vector $E_{ij}$ with $i<j$, since $j>i\ge 1$ forces $j\mathrm{\ne }1$.

$V$ is irreducible: if $0\mathrm{\ne }v={\sum }_i{v}_i{e}_i$ and $v_l\mathrm{\ne }0$, then $E_{il}\cdot v=v_l{e}_i$ for every $i\mathrm{\ne }l$. Choose one such $i$ (possible since $n\ge 2$); applying $E_{li}$ to the resulting $e_i$ gives $e_l$. All matrix units used are off-diagonal and belong to ${\mathfrak{s}\mathfrak{l}}_n$. Hence every basis vector belongs to the submodule generated by $v$, which equals $V$.

For the dual module the action is $(X\cdot \phi )(v)=-\phi (X\cdot v)$, so the weights of $V^{\ast }$ are $-{\epsilon }_1,\dots ,-{\epsilon }_n$ on the dual basis vectors $e_i^{\ast }$; the vector $e_n^{\ast }$ is killed by every positive root vector: $(E_{ij}\cdot e_n^{\ast })(v)=-e_n^{\ast }(E_{ij}v)=-e_n^{\ast }(v_j{e}_i)=-v_j{\delta }_{in}=0$ because $i<j\le n$ gives $i\mathrm{\ne }n$.

By [L2] and [L3], ${\epsilon }_1(h_{{\alpha }_j})={\epsilon }_1(E_{jj}-E_{j+1,j+1})={\delta }_{1j}={\omega }_1(h_{{\alpha }_j})$ for every $j$, and the simple coroots span $\mathfrak{h}$; hence ${\epsilon }_1={\omega }_1$, so $V$ is irreducible with highest weight ${\omega }_1$ by steps 1.1, 1.2 and [L4].

$V^{\ast }$ is irreducible: if $0\mathrm{\ne }W\subseteq V^{\ast }$ is a submodule, then its annihilator $W^{\perp }\subseteq V$ is a submodule: for $v\in W^{\perp }$ and $\phi \in W$, $\phi (Xv)=-(X\phi )(v)=0$. Its dimension is $\dim W^{\perp }=n-\dim W<n$, hence $W^{\perp }=0$ by step 1.2 and $W=V^{\ast }$.

Finally $-{\epsilon }_n(h_{{\alpha }_j})=-(E_{jj}-E_{j+1,j+1}{)}_{nn}={\delta }_{j,n-1}={\omega }_{n-1}(h_{{\alpha }_j})$ for every $j$, so $-{\epsilon }_n={\omega }_{n-1}$ by [L3]; hence $V^{\ast }$ is irreducible of highest weight ${\omega }_{n-1}$ by steps 1.3, 2.2 and [L4].

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 $e_{\{1,\dots ,k\}}=e_1\wedge \cdots \wedge e_k$ has weight ${\epsilon }_1+\cdots +{\epsilon }_k={\omega }_k$ by [L1] and [L2], and it is killed by every positive root vector $E_{ij}$ with $i<j$: if $j\le k$ then $i<j\le k$ gives $i\in \{1,\dots ,k\}$ and the replacement repeats an index, while if $j>k$ then $e_j$ is not a factor at all; either way $E_{ij}\cdot e_{\{1,\dots ,k\}}=0$ by [L2].

From any basis wedge $e_I\mathrm{\ne }{e}_{\{1,\dots ,k\}}$ one reaches $e_{\{1,\dots ,k\}}$ by positive root vectors: let $j$ be the smallest positive integer not in $I$, so $j\le k$, and choose $i\in I$ with $i>j$, which exists since $I$ has $k$ elements; then $E_{ji}\cdot e_I=\pm e_{I^{\mathrm{\prime }}}$ with $I^{\mathrm{\prime }}=I\setminus \{i\}\cup \{j\}$ a nonvanishing basis wedge whose index sum is strictly smaller, and repeating finitely many times reaches $\{1,\dots ,k\}$.

From $e_{\{1,\dots ,k\}}$ one reaches every basis wedge by negative root vectors: for $a\in \{1,\dots ,k\}$ and $b>k$ the operator $E_{ba}$ replaces the factor $e_a$ by $e_b$ without repeated indices (as $b\mathrm{\notin }\{1,\dots ,k\}$), giving $\pm e_{I^{\mathrm{\prime }\mathrm{\prime }}}$ with $I^{\mathrm{\prime }\mathrm{\prime }}=I_0\setminus \{a\}\cup \{b\}$; successive replacements of this kind produce every increasing index set.

$W$ is irreducible: if $0\mathrm{\ne }U\subseteq W$ is a submodule, then by [L3] some basis wedge $e_I$ lies in $U$, so by step 2.1 the highest vector $e_{\{1,\dots ,k\}}$ lies in $U$, and by step 2.2 every basis wedge lies in $U$; hence $U=W$ by [L4].

By step 1.1 the vector $e_{\{1,\dots ,k\}}$ is a highest weight vector of weight ${\omega }_k$, and by step 3.1 the module is irreducible; hence ${\mathrm{\Lambda }}^k({\mathbb{C}}^n)$ has highest weight ${\omega }_k$, as asserted.

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.