Representation theory of semisimple Lie algebras

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 2 September 2013

Weights

Let $C={\Vert c_{ij}\Vert }_{1\le i,j\le n}$ be a Cartan matrix corresponding to a simple Lie algebra. The dimension, $n,$ of the Cartan matrix is called the rank (note that this has nothing to do with the rank of the Cartan matrix). Let ${\omega }_1,{\omega }_2,\dots ,{\omega }_n$ be linearly independent vectors spanning a vector space denoted by ${𝔥}^{*}\text{.}$ The elements of ${𝔥}^{*}$ are called weights. The ${\omega }_i$ are called the fundamental weights. If $\lambda \in {\sum }_{i=1}^n{\lambda }_i{\omega }_i\in {𝔥}^{*}$ we use the notation $⟨\lambda ,{\alpha }_i⟩$ for ${\lambda }_i,$ i.e. $$\lambda =\sum _{i=1}^n⟨\lambda ,{\alpha }_i⟩{\omega }_i\text{.}$$ The weight lattice is the lattice $$P=\sum _{i=1}^n\mathbb{Z}{\omega }_i\text{.}$$ The elements of $P$ are called integral weights and the dominant integral weights are the weights in the set $$P^{+}=\sum _{i=1}^n\mathbb{N}{\omega }_i,$$ where $\mathbb{N}$ denotes the nonegative integers. Define vectors ${\alpha }_1,{\alpha }_2,\dots ,{\alpha }_n,$ the simple roots, by $${\alpha }_i=\sum _{j=1}^n{c}_{ij}{\omega }_j\text{.}$$

There is a partial ordering on the weights $\lambda \in {𝔥}^{*}$ given by defining $\mu \le \lambda$ if $$\mu =\lambda -\sum _i{a}_i{\alpha }_i,$$ for nonnegative integers $a_i\text{.}$ This is called the dominance partial order.

For each $1\le i\le n$ define a linear transformation $s_i:{𝔥}^{*}\to {𝔥}^{*}$ by $$s_i\left(\lambda \right)=\lambda -⟨\lambda ,{\alpha }_i⟩{\alpha }_i,$$ for each $\lambda \in {𝔥}^{*}\text{.}$ The Weyl group $W$ is the group generated by the $s_i\text{.}$ There is a partial order on the Weyl group induced by the dominance ordering on weights and given by $$\tau \ge \kappa \hspace{0.17em}\text{when}\hspace{0.17em}\tau \rho \le \kappa \rho \hspace{0.17em}\text{in dominance.}$$ Here $\tau ,\kappa \in W$ and $\rho ={\sum }_i{\omega }_i\text{.}$ This partial order on $W$ is called the Bruhat order. The weight $\rho$ is called the half-sum of the positive roots.

The enveloping algebra $𝔘\left(C\right)$

Fix a (simple) Cartan matrix $C$ and define $𝔘\left(C\right)$ to be the associative algebra (over $ℂ\text{)}$ with 1 generated by $x_i,y_i,h_i,$ $1\le i\le n$ with relations (Serre relations, see [Hum1972]) $$\begin{array}{ccc}& [h_i,h_j]=0(1\le i,j\le n),& \text{(}S1\text{)}\\ & [x_i,y_i]=h_i,[x_i,y_j]=0\hspace{0.17em}\text{if}\hspace{0.17em}i\ne j,& \text{(}S2\text{)}\\ & [h_i,x_j]=⟨{\alpha }_j,{\alpha }_i⟩x_j,[h_i,y_j]=-⟨{\alpha }_j,{\alpha }_i⟩y_j,& \text{(}S3\text{)}\\ & {\left(\text{ad}\hspace{0.17em}{x}_i\right)}^{-⟨{\alpha }_j,{\alpha }_i⟩+1}\left(x_j\right)=0(i\ne j),& \text{(}{S}_{ij}^{+}\text{)}\\ & {\left(\text{ad}\hspace{0.17em}{y}_i\right)}^{-⟨{\alpha }_j,{\alpha }_i⟩+1}\left(y_j\right)=0(i\ne j)\text{.}& \text{(}{S}_{ij}^{-}\text{)}\end{array}$$ Here $[a,b]=ab-ba$ and ${\left(\text{ad}\hspace{0.17em}a\right)}^k\left(b\right)=[a,[a,[a,\cdots ,[a,b]]\cdots ]]\text{.}$ One shows easily that $${\left(\text{ad}\hspace{0.17em}a\right)}^k\left(b\right)=\sum _{i=0}^k{(-1)}^i\left(\genfrac{}{}{0ex}{}{k}{i}\right)a^{k-i}b{a}^i\text{.}$$ We will use the following notations: $$\begin{array}{ccc}𝔘& =& 𝔘\left(C\right)\text{;}\\ {𝔘}^{+}& =& \text{subalgebra of}\hspace{0.17em}𝔘\hspace{0.17em}\text{generated by the}\hspace{0.17em}{x}_i\text{;}\\ 𝔥& =& \text{subalgebra of}\hspace{0.17em}𝔘\hspace{0.17em}\text{generated by the}\hspace{0.17em}{h}_i\text{;}\\ {𝔘}^{-}& =& \text{subalgebra of}\hspace{0.17em}𝔘\hspace{0.17em}\text{generated by the}\hspace{0.17em}{y}_i\text{.}\end{array}$$

Remark. Note that we could have defined the algebra $𝔘\left(C\right)$ using only the generators $x_i$ and $y_i,$ $1\le i\le n\text{.}$

Representations of $𝔘$

Let $V$ be a $𝔘$ module. A vector $v\in V$ is called a weight vector if, for each $i,$ $h_{i}v={\lambda }_{i}v,$ for some constant ${\lambda }_i\in ℂ\text{.}$ We associate to $v$ the weight $\lambda ={\sum }_i{\lambda }_i{\omega }_i\in {𝔥}^{*}$ and write $$h_{i}v=⟨\lambda ,{\alpha }_i⟩v\text{.}$$ Let $wt\left(v\right)$ denote the weight of a weight vector $v\text{.}$ Given a weight $\lambda \in {𝔥}^{*}$ the weight space $V_{\lambda }$ corresponding to $\lambda$ is the subspace of $V$ given by $$V_{\lambda }=\{v\in V\hspace{0.17em}|\hspace{0.17em}v\hspace{0.17em}\text{is a weight vector and}\hspace{0.17em}wt\left(v\right)=\lambda \}\text{.}$$ A vector in $v^{+}\in V$ is a highest weight vector if $v^{+}$ is a weight vector and if, for each $i,$ $x_i{v}^{+}=0\text{.}$

If $V$ is a finite dimensional $𝔘$ module then $V$ contains a weight vector.

		Proof.
	The proof is by induction. Since $V$ is finite dimensional it contains an eigenvector $v_1$ for $h_1\text{.}$ The induction step is as follows. Suppose that $v_k\in V$ is a weight vector for $h_1,h_2,\dots ,h_k,$ i.e. there are constants ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_k\in ℂ$ such that $h_{i}v={\lambda }_{i}v\text{.}$ Let $W=\text{span}\{v_k,h_{k+1}v,h_{k+1}^{2}v,\dots \}\text{.}$ Since $$h_j{h}_{k+1}^{s}v=h_{k+1}^s{h}_{j}v={\lambda }_j{h}_{k+1}^{s}v,$$ for each $1\le j\le k,$ and all $s,$ $W$ is an eigenspace for each of the $h_j\text{.}$ But $h_{k+1}$ takes $W$ into itself. Thus, since $W$ is finite dimensional, there is an eigenvector $v_{k+1}$ of $h_{k+1}$ in $W\text{.}$ Since $v_{k+1}$ is in $W$ it is also an eigenvector for all the $h_j,$ $1\le j\le k\text{.}$ $\square$

If $v$ is a weight vector then

(1)	$y_{i}v$ is a weight vector and $wt\left(y_{i}v\right)=wt\left(v\right)-{\alpha }_i,$
(2)	$x_{i}v$ is a weight vector and $wt\left(x_{i}v\right)=wt\left(v\right)+{\alpha }_i\text{.}$
(3)	$y^{M}v=y_{i_k}^{m_k}\cdots y_{i_1}^{m_1}v$ is a weight vector, and $wt\left(y^{M}v\right)=wt\left(v\right)-{\sum }_j{m}_j{\alpha }_{i_j}\text{.}$
(4)	$x^{M}v=x_{i_k}^{m_k}\cdots x_{i_1}^{m_1}v$ is a weight vector, and $wt\left(x^{M}v\right)=wt\left(v\right)+{\sum }_j{m}_j{\alpha }_{i_j}\text{.}$

		Proof.
	We have $$\begin{array}{ccc}{h}_j{y}_{i}v& =& y_i{h}_{j}v+[h_j,y_i]v\\ & =& y_i⟨\lambda ,{\alpha }_j⟩v-⟨{\alpha }_i,{\alpha }_j⟩y_{i}v\\ & =& ⟨\lambda -{\alpha }_i,{\alpha }_j⟩y_{i}v,\end{array}$$ giving (1). (2) is proved similarly. (3) and (4) follow from (1) and (2) respectively, by induction. $\square$

If $V$ is a finite dimensional $𝔘$ module then $V$ contains a highest weight vector.

		Proof.
	We know $V$ contains a weight vector $v_0\text{.}$ If there exists $x_i$ such that $x_{i}v\ne 0$ then let $v_1=x_{i}v\text{.}$ Repeating this process constructs a sequence of weight vectors $v_0,v_1,\dots \text{.}$ Since these vectors have different weights, i.e. different “eigenvalues” they must be linearly independent. Since $V$ is finite dimensional we know that this sequence must be finite, i.e. for some $v_k$ we have that $x_i{v}_k=0$ for all $i\text{.}$ Then $w_k$ is a highest weight vector. $\square$

Let $V$ be a $𝔘$ module and let $v\in V$ be a weight vector. Let $wt\left(v\right)=\lambda ={\sum }_i{\lambda }_i{\omega }_i\text{.}$ Then for each $m\ge 1$ we have $$x_i\frac{y_i^m}{m!}v=\frac{y_i^m}{m!}{x}_{i}v+({\lambda }_i-m+1)\frac{y_i^{m-1}}{(m-1)!}\text{.}$$

		Proof.
	The proof is by induction. The statement is easy for $m=1\text{.}$ The induction step is as follows. Using Lemma (3.2) and the fact that $⟨{\alpha }_i,{\alpha }_i⟩=2$ we have that $$\begin{array}{ccc}m{x}_i\frac{y_i^m}{m!}v& =& y_i{x}_i\frac{y_i^{m-1}}{(m-1)!}+h_i\frac{y_i^{m-1}}{(m-1)!}v\\ & =& y_i(\frac{y_i^{m-1}}{(m-1)!}xv+({\lambda }_i-(m-1)+1)\frac{y_i^{m-2}}{(m-2)!}v)+⟨\lambda -(m-1){\alpha }_i,{\alpha }_i⟩\frac{y_i^{m-1}}{(m-1)!}v\\ & =& \frac{y_i^m}{(m-1)!}xv+({\lambda }_i-m+2)(m-1)\frac{y_i^{m-1}}{(m-1)!}+({\lambda }_i-2(m-1))\frac{y_i^{m-1}}{(m-1)!}v\\ & =& \frac{y_i^m}{(m-1)!}+m({\lambda }_i-m+1)\frac{y_i^{m-1}}{(m-1)!}v\text{.}\end{array}$$ $\square$

If $V$ is finite dimensional and $v^{+}$ is a highest weight vector then $\lambda =wt\left(v^{+}\right)\in P^{+}={\sum }_i\mathbb{N}{\omega }_i\text{.}$

		Proof.
	Since $V$ is finite dimensional we know that $\frac{y_i^{m+1}}{(m+1)!}{v}^{+}=0$ for some $m\in \mathbb{N}\text{.}$ Let $m$ be minimal such that this is true. By the previous lemma we have that $$\begin{array}{ccc}0& =& x_i\frac{y_i^{m+1}}{(m+1)!}{v}^{+}\\ & =& ({\lambda }_i-m)\frac{y_i^m}{m!}{v}^{+}\text{.}\end{array}$$ Since $(y_i^m/m!)v^{+}\ne 0$ we must have that ${\lambda }_i=m\text{.}$ $\square$

Let $V$ be a $𝔘$ module (not necessarily assumed to be finite dimensional). Suppose that $v$ contains a highest weight vector $v^{+}\text{.}$ Note that $𝔘v^{+}$ is a submodule of $V\text{.}$ If $V=𝔘v^{+}$ (in particular, when $V$ is irreducible) then

(1)	$$V={𝔘}^{-}{v}^{+}=\text{span}\hspace{0.17em}\left\{y^M{v}^{+}\right\}$$ where $y^M$ denotes a monomial in the $y\text{'s}$ of the form $y_{i_1}^{m_1}{y}_{i_2}^{m_2}\cdots y_{i_k}^{m_k}=y^M\text{.}$
(2)	$V$ is a direct sum of its weight spaces, $$\begin{array}{c}V=\underset{\mu \in P,\mu \le \lambda }{⨁}{V}_{\mu },\text{and}\\ V_{\mu }=\text{span}\left\{y^M{v}^{+}\hspace{0.17em}\right|\hspace{0.17em}wt\left(y^M{v}^{+}\right)=\mu \}\text{.}\end{array}$$

		Proof.
	(1) For each $i$ we have that $x_i{v}^{+}=0$ and that $h_i{v}^{+}=c{v}^{+}$ for some constant $c\in C\text{.}$ Using Lemma (3.6) we have that $h_i{y}_{i_k}^{m_k}\cdots y_{i_1}^{m_1}{v}^{+}=c{y}_{i_k}^{m_k}\cdots y_{i_1}^{m_1}{v}^{+}$ for some $c\in ℂ\text{.}$ $$\begin{array}{ccc}{x}_i{y}_{i_k}^{m_k}\cdots y_{i_1}^{m_1}{v}^{+}& =& (y_{i_k}{x}_i{y}_{i_k}^{m_k-1}\cdots y_{i_1}^{m_1}+\left[x_i{y}_{i_k}\right]y_{i_k}^{m_k-1}\cdots y_{i_1}^{m_1})v^{+}\\ & =& (y_{i_k}{x}_i{y}_{i_k}^{m_k-1}\cdots y_{i_1}^{m_1}+c{h}_i{y}_{i_k}^{m_k-1}\cdots y_{i_1}^{m_1})v^{+}\end{array}$$ for some constant $c\in ℂ\text{.}$ Thus by induction we have that $x_i{y}^M{v}^{+}\in {𝔘}^{-}{v}^{+}$ and $h_i{y}^M{v}^{+}\in {𝔘}^{-}{v}^{+}$ for each $i$ and each monomial $y_{i_1}^{m_1}{y}_{i_2}^{m_2}\cdots y_{i_k}^{m_k}=y^M\text{.}$ (2) The $V_{\mu }$ are the weight spaces under the action of $𝔥,$ i.e. they are eigenspaces with different eigenvalues. So they must decompose as a direct sum. The monomials $y^M{v}^{+}$ have weight $\mu$ exactly when $\mu =\lambda -{\sum }_j{m}_j{a}_{i_j}\le \lambda \text{.}$ Since $\lambda \in P^{+}\subseteq P$ and each ${\alpha }_i\in P$ we have that $\mu \in P\text{.}$ $\square$

If $V$ is an irreducible $𝔘$ module then $V$ can have at most one highest weight vector (up to multiplication by a constant).

		Proof.
	Suppose that $V$ is an irreducible module that contains highest weight vectors $v^{+}$ and $w^{+}\text{.}$ Since $V$ is irreducible $w^{+}\in V={𝔘}^{-}{v}^{+}\text{.}$ By () $wt\left(w^{+}\right)\le wt\left(v^{+}\right)\text{.}$ But we also have that $wt\left(w^{+}\right)\le wt\left(v^{+}\right)\text{.}$ So $w^{+}\in \text{span}\left\{y^0{v}^{+}\right\},$ i.e., $w^{+}=c{v}^{+}$ for some $c\in ℂ\text{.}$ $\square$

We did not prove the following three results in class.

There is a unique finite dimensional irreducible representation $V^{\lambda }$ of $𝔘$ corresponding to each dominant weight $\lambda \in P^{+}\text{.}$

Let $V^{\lambda }$ be the irreducible module with highest weight vector $v^{+}$ of weight $\lambda ={\sum }_i{\lambda }_i{\omega }_i\in P^{+}\text{.}$ Then the defining relations for $𝔘$ and the relations $$y_i^{{\lambda }_i+1}{v}^{+}=0$$ are a complete set of relations determining the irreducible module $V^{\lambda }={𝔘}^{-}{v}^{+}\text{.}$

(Weyl) Every finite dimensional representation $V$ of $𝔘$ is a direct sum of irreducible representations.

Symmetry

Let $V$ be a finite dimensional $𝔘$ module and define the following operator on $V,$ $$\begin{array}{c}{\stackrel{\sim }{s}}_i=\text{exp}\left(x_i\right)\text{exp}(-y_i)\text{exp}\left(x_i\right),\text{where}\\ \text{exp}\left(a\right)=\sum _{k\ge 0}\frac{x^k}{k!}\text{.}\end{array}$$

$$\left(\text{exp}\hspace{0.17em}a\right)b\left(\text{exp}(-a)\right)=\left(\text{exp}\left(\text{ad}\hspace{0.17em}a\right)\right)\left(b\right)$$

		Proof.
	$$\begin{array}{ccc}\left(\text{exp}\hspace{0.17em}a\right)b\left(\text{exp}(-a)\right)& =& \sum _{k\ge 0}\sum _{s=0}^k\frac{{(-1)}^s}{s!}\frac{1}{(k-s)!}{a}^{k-s}b{a}^s\\ & =& \sum _{k\ge 0}\frac{1}{k!}\sum _{s=0}^k{(-1)}^s\left(\genfrac{}{}{0ex}{}{k}{s}\right)a^{k-s}b{a}^s\\ & =& \sum _{k\ge 0}\frac{1}{k!}{\left(\text{ad}\hspace{0.17em}a\right)}^{k}b\\ & =& \left(\text{exp}\left(\text{ad}\hspace{0.17em}a\right)\right)b\text{.}\end{array}$$ $\square$

For each $i,{\stackrel{\sim }{s}}_i,$ is a bijection from the weight space $V_{\mu }$ to the weight space $V_{s_i\mu }\text{.}$

		Proof.
	${\stackrel{\sim }{s}}_i$ is a bijection since it is invertible with inverse ${\stackrel{\sim }{s}}_i^{-1}=\text{exp}(-x_i)\text{exp}\left(y_i\right)\text{exp}(-x_i)\text{.}$ Using Lemma (3.11) we have that $${\stackrel{\sim }{s}}_i^{-1}{h}_i{\stackrel{\sim }{s}}_i=\text{exp}\left(\text{ad}\hspace{0.17em}{y}_i\right)\text{exp}\left(\text{ad}(-x_i)\right)\text{exp}\left(\text{ad}\hspace{0.17em}{y}_i\right)h_i\text{.}$$ Using the relations $$\begin{array}{c}[y_i,x_i]=-h_i,[y_i,h_i]=2y_i,[y_i,y_i]=0,\\ [x_i,x_i]=0,[x_i,h_i]=-2x_i,[x_i,y_i]=h_i,\end{array}$$ we can express the action of $\text{ad}(-x_i)$ and $\text{ad}\hspace{0.17em}{y}_i$ on the subspace $\text{span}\{x_i,y_i,h_i\}$ in terms of the matrices $$\left(\text{ad}(-x_i)\right)=\left(\begin{array}{ccc}0& 0& 2\\ 0& 0& 0\\ 0& -1& 0\end{array}\right),\left(\text{ad}\hspace{0.17em}{y}_i\right)=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 2\\ -1& 0& 0\end{array}\right)\text{.}$$ From this one easily computes that $$\text{exp}\left(\text{ad}\hspace{0.17em}{y}_i\right)\text{exp}\left(\text{ad}(-x_i)\right)\text{exp}\left(\text{ad}\hspace{0.17em}{y}_i\right)=\left(\begin{array}{ccc}0& -1& 0\\ -1& 0& 0\\ 0& 0& -1\end{array}\right)\text{.}$$ This gives that ${\stackrel{\sim }{s}}_i^{-1}{h}_i{\stackrel{\sim }{s}}_i=-h_i\text{.}$ So we have that if $v\in V_{\mu }$ then $$\begin{array}{ccc}{h}_i\left({\stackrel{\sim }{s}}_{i}v\right)& =& -{\stackrel{\sim }{s}}_i{h}_{i}v\\ & =& -{\mu }_i\left({\stackrel{\sim }{s}}_{i}v\right)\\ & =& (⟨\mu ,{\alpha }_i⟩-2⟨\mu ,{\alpha }_i⟩){\stackrel{\sim }{s}}_{i}v\\ & =& ⟨\mu -⟨\mu ,{\alpha }_i⟩{\alpha }_i,{\alpha }_i⟩{\stackrel{\sim }{s}}_{i}v\\ & =& ⟨s_i\mu ,{\alpha }_i⟩{\stackrel{\sim }{s}}_{i}v\text{.}\end{array}$$ Then ${\stackrel{\sim }{s}}_{i}v\in V_{s_i\mu }\text{.}$ $\square$

Suggested References

[Bou1968] N. Bourbaki, Groupes et algèbres de Lie, Chapitres 4, 5 et 6, Elements de Mathématique, Hermann, Paris, 1968. MR 39:1590

[Hum1972] J.E. Humphreys, Introduction to Lie Algebras and Representation Theory, Springer-Verlag, New York 1972.

[Kac1985] V. Kac, Infinite dimensional Lie algebras, Cambridge University Press 1985.

[Mac1991] I.G. Macdonald, Lecture Notes, University of California, San Diego, Spring 1991.

[Mac1979] I.G. Macdonald, Hall Polynomials and Symmetric Functions, Oxford Univ. Press, 1979.

Roots and the Weyl group

Positive and negative roots

The roots are the elements of the set $$\Phi =\{\alpha \in {𝔥}^{*}\hspace{0.17em}|\hspace{0.17em}\alpha =w{\alpha }_i\hspace{0.17em}\text{for some}\hspace{0.17em}w\in W\hspace{0.17em}\text{and some}\hspace{0.17em}{\alpha }_i\}\text{.}$$ The positive roots and the negative roots are the roots in the sets $$\begin{array}{ccc}{\Phi }^{+}& =& \{\alpha \in \Phi \hspace{0.17em}|\hspace{0.17em}\alpha \ge 0\},\\ {\Phi }^{-}& =& \{\alpha \in \Phi \hspace{0.17em}|\hspace{0.17em}\alpha \le 0\},\end{array}$$ respectively.

Let $V$ be a finite dimensional $𝔘$ module.

1)	If $\mu$ is a weight of $V$ and $s_i\mu >\mu$ then $\mu +{\alpha }_i$ is a weight of $V\text{.}$
2)	If $\mu$ is a weight of $V$ and $s_i\mu <\mu$ then $\mu -{\alpha }_i$ is a weight of $V\text{.}$

		Proof.
	Let $v\in V_{\mu }\text{.}$ Using the defining relations for $𝔘$ we can expand to write the action of $s_i$ on $v$ in the form $$\begin{array}{ccc}{s}_{i}v& =& \text{exp}\left(x_i\right)\text{exp}(-y_i)\text{exp}\left(x_i\right)v\\ & =& av+\sum _m{b}_m{y}_i^{m}v+\sum _{m,m\prime }{c}_{m,m\prime }{y}_i^m{x}_i^{m\prime }v\text{.}\end{array}$$ Since the weight of $s_{i}v$ is $s_i\mu >\mu$ we get that $a=0$ and $b_m=0\text{.}$ Since ${\stackrel{\sim }{s}}_i$ is a bijection ${\stackrel{\sim }{s}}_{i}v\ne 0$ so $x_{i}v\ne 0\text{.}$ Thus $wt\left(x_{i}v\right)=\mu +{\alpha }_i$ is a weight of $V,$ proving 1). 2) is proved similarly. $\square$

Let $V^{\lambda }$ be the irreducible $𝔘$ module corresponding to the weight $\lambda \in P^{+}\text{.}$

1)	If $w\lambda -{\alpha }_i$ is a weight of $V^{\lambda }$ then $w^{-1}{\alpha }_i\ge 0\text{.}$
2)	If $w\lambda +{\alpha }_i$ is a weight of $V^{\lambda }$ then $w^{-1}{\alpha }_i\le 0\text{.}$
3)	$w\lambda +{\alpha }_i$ and $w\lambda -{\alpha }_i$ cannot both be weights of $V^{\lambda }\text{.}$
4)	$s_{i}w\lambda =w\lambda$ if and only if neither $w\lambda +{\alpha }_i$ nor $w\lambda -{\alpha }_i$ is a weight of $V^{\lambda }\text{.}$
5)	$s_{i}w\lambda =w\lambda$ cannot hold for all $\lambda \in P^{+}\text{.}$

		Proof.
	If $w\lambda -{\alpha }_i$ is a weight so is $w^{-1}(w\lambda -{\alpha }_i)=\lambda -w^{-1}{\alpha }_i\text{.}$ But we know that all weights of $V^{\lambda }$ are $\le \lambda \text{.}$ So $w^{-1}{\alpha }_i\ge 0,$ proving 1). 2) is proved similarly. To prove 3) we see from 1) and 2) that $w^{-1}{\alpha }_i\le 0$ and $w^{-1}{\alpha }_i\ge 0$ giving that $w^{-1}{\alpha }_i=0\text{.}$ But this implies that ${\alpha }_i=0\text{.}$ This is a contradiction (provided that the $i\text{th}$ row of the Cartan matrix is not all zeros). Assume that $s_{i}w\lambda =w\lambda \text{.}$ If $w\lambda +{\alpha }_i$ is a weight then $s_i(w\lambda +{\alpha }_i)=w\lambda -{\alpha }_i$ is also a weight which is a contradiction to 3). The same argument shows that $w\lambda -{\alpha }_i$ can not be a weight. This proves the forward implication of 4). Conversely, if neither $w\lambda +{\alpha }_i$ nor $w\lambda -{\alpha }_i$ are weights then $x_{i}v=0$ and $y_{i}v=0$ for any $v\in {\left(V^{\lambda }\right)}_{w\lambda }\text{.}$ This shows that ${\stackrel{\sim }{s}}_{i}v=\text{exp}\left(y_i\right)\text{exp}(-x_i)\text{exp}\left(y_i\right)v=1v=v,$ i.e., that $s_{i}w\lambda =wt\left({\stackrel{\sim }{s}}_{i}v\right)=wt\left(v\right)=w\lambda \text{.}$ Prove 5) by contradiction. If $s_{i}w\lambda =w\lambda$ for all $\lambda \in P^{+}$ then $s_{i}w{\omega }_j=w{\omega }_j$ for all $j\text{.}$ Since this is true on a basis we have that $s_{i}w=w\text{.}$ But this implies that $s_i=1$ which is a contradiction. $\square$

All roots are either positive or negative, i.e., $$\Phi ={\Phi }^{+}\cup {\Phi }^{-}\text{.}$$

		Proof.
	Let $w\in W$ and let $1\le i\le n,$ then by Lemma(5.2) 5),we know that there exists $\lambda \in P^{+}$ such that $s_{i}w\lambda \ne w\lambda \text{.}$ Then by Lemma (5.2) 1) and 2) we have that $w^{-1}{\alpha }_i$ is either positive or negative. $\square$

Let $m$ be a positive integer greater than 1. $m{\alpha }_i$ is not a root for any ${\alpha }_i\text{.}$

		Proof.
	Suppose that $m{\alpha }_i$ is a root and let $w\in W$ and ${\alpha }_j$ be such that $m{\alpha }_i=w{\alpha }_j\text{.}$ Consider the irreducible $𝔘\text{-module}$ $V^{\rho }$ where $\rho ={\sum }_i{\omega }_i\text{.}$ Then $s_i\rho =\rho -{\alpha }_i$ is a weight of $V^{\rho }\text{.}$ So $w^{-1}(\rho -{\alpha }_i)=w^{-1}\rho -(1/m){\alpha }_j$ is a root of $V^{\rho }\text{.}$ But we know that all weights of $V^{\rho }$ are of the form $\rho -{\sum }_i{a}_i{\alpha }_i$ where the $a_i$ are positive integers. Thus we have a contradiction since $w^{-1}\rho -{\alpha }_i$ is not of this form. $\square$

1)	$s_i$ permutes the set ${\Phi }^{+}-\left\{{\alpha }_i\right\}\text{.}$
2)	$\rho ={\sum }_i{\omega }_i=\frac{1}{2}{\sum }_{\alpha \in {\Phi }^{+}}\alpha \text{.}$

		Proof.
	1) Let $\beta \in {\Phi }^{+},$ $\beta \ne {\alpha }_i\text{.}$ Then $\beta =\sum m_j{\alpha }_j$ for nonnegative integers $m_j\text{.}$ Furthermore we know that since $\beta \ne {\alpha }_i$ we must have $m_j\ne 0$ for some $j\ne i\text{.}$ Since $$s_i\beta =\beta -⟨\beta ,{\alpha }_i⟩{\alpha }_i$$ we have that $s_i\beta$ also has $m_j>0\text{.}$ Thus, since $s_i\beta$ is a root it must be a positive root. Let $\rho \prime =\frac{1}{2}{\sum }_{\alpha \in {\Phi }^{+}}\alpha \text{.}$ We show that $\rho \prime =\rho \text{.}$ $$\begin{array}{ccc}{s}_i\rho \prime & =& s_i\left(\frac{1}{2}\sum _{\alpha \in {\Phi }^{+}-\left\{{\alpha }_i\right\}}\alpha \right)+\frac{1}{2}{s}_i{\alpha }_i\\ & =& -{\alpha }_i+\frac{1}{2}\sum _{\alpha \in {\Phi }^{+}}\alpha \\ & =& \rho \prime -{\alpha }_i\text{.}\end{array}$$ This implies that $⟨\rho \prime ,{\alpha }_i⟩=1\text{.}$ So $\rho \prime ={\sum }_i{\omega }_i=\rho \text{.}$ $\square$

The Weyl chamber

The Weyl chamber is the subset of ${𝔥}^{*}$ given by $$\bar{C}=\{\mu \in {𝔥}^{*}\hspace{0.17em}|\hspace{0.17em}⟨\mu ,{\alpha }_i⟩\ge 0\hspace{0.17em}\text{for all}\hspace{0.17em}i\}\text{.}$$ Note that $P^{+}=P\cap \bar{C}\text{.}$

Every $\mu \in P$ has a unique image in the Weyl chamber, i.e., there exists $w\in W$ such that $w\mu \in P^{+}=\bar{C}\cap P$ and that if $w\prime \mu \in P^{+}$ then $w\prime \mu =w\mu \text{.}$

		Proof.
	If $w\mu \notin P^{+}$ then there exists $i$ such that $⟨w\mu ,{\alpha }_i⟩<0$ and so $s_{i}w\mu >\mu$ in dominance. In this way we can construct a sequence $\mu ={\mu }^{\left(0\right)}<{\mu }^{\left(1\right)}<\cdots$ of weights where in each case we have that ${\mu }^{(j+1)}=s_i{\mu }^{\left(j\right)}$ for some $i\text{.}$ If the Weyl group is finite this must stop, call the last element $\lambda \text{.}$ Then $\lambda =w\mu \in P^{+}\text{.}$ Now suppose that $\lambda \prime =w\prime \mu$ is also in $P^{+}\text{.}$ Then $\lambda =w{\left(w\prime \right)}^{-1}\lambda \prime =\bar{w}\lambda \text{.}$ Let $V^{\lambda }$ be the irreducible $𝔘$ module corresponding to $\lambda$ let $v^{+}$ be a highest weight vector and let $v=\bar{w}{v}^{+}\text{.}$ Since $\lambda \prime \in P^{+}$ we have that $⟨\lambda \prime ,{\alpha }_i⟩\ge 0$ for all $i\text{.}$ This implies that $s_i\lambda \prime \le \lambda \prime$ for all $i\text{.}$ Then by Lemma (5.1) 2) and Lemma (5.2) 3) and 4) we have that $\lambda \prime +{\alpha }_i$ is not a weight of $V^{\lambda }\text{.}$ This implies that $x_{i}v=0$ for all $i\text{.}$ Thus $v$ is a highest weight vector. As $V^{\lambda }$ has only one highest weight $\lambda \prime =\lambda \text{.}$ $\square$

Length in the Weyl group

Since the Weyl group$$W is generated by the simple transpositions $s_i$ every element $w\in W$ can be written in the form $$w=s_{i_1}{s}_{i_2}\cdots s_{i_p}\text{.}$$ $s_{i_1}\cdots s_{i_p}$ is a reduced word for $w$ if $p$ is minimal. In this case $p$ is called the length of $w$ and denoted $\ell \left(w\right)\text{.}$ The sign of $w$ is defined to be $$\epsilon \left(w\right)={(-1)}^{\ell \left(w\right)}\text{.}$$ One can also view $\epsilon$ as the homomorphism $$\begin{array}{cccc}\epsilon :& W& ⟶& W\\ & s_i& ⟼& -1\end{array}$$

Let $w\in W$ and let $s_i$ be a simple reflection corresponding to the simple root ${\alpha }_i\text{.}$

1)	$\ell \left(s_{i}w\right)>\ell \left(w\right)\iff \ell \left(s_{i}w\right)=\ell \left(w\right)+1\iff w^{-1}{\alpha }_i>0\text{.}$
2)	$\ell \left(s_{i}w\right)<\ell \left(w\right)\iff \ell \left(s_{i}w\right)=\ell \left(w\right)-1\iff w^{-1}{\alpha }_i<0\text{.}$

		Proof.
	2) follows from 1) by replacing $w$ by $s_{i}w\text{.}$ $\square$

If $\ell \left(w\right)>\ell \left(s_{i}w\right)$ then $$\Phi \left(w\right)=s_i\Phi \left(s_{i}w\right)\cup \left\{{\alpha }_i\right\}\text{.}$$

		Proof.
	$\square$

For each $w\in W,$ $$\ell \left(w\right)=\text{Card}\left(\Phi \left(w\right)\right)\text{.}$$

		Proof.
	By Lemmas () and () and induction on the length of $w\text{.}$ $\square$

Let $w=s_{i_1}{s}_{i_2}\cdots s_{i_p}$ be a reduced decomposition of $w\in W\text{.}$ Set $${\theta }_j=s_{i_1}{s}_{i_2}\cdots s_{i_{j-1}}\left({\alpha }_{i_j}\right)$$ for each $j=1,2,\dots ,p\text{.}$ Then the roots ${\theta }_j$ are $>0,$ distinct, and $w^{-1}{\theta }_j<0\text{.}$ Furthermore every root $\alpha \in {\Phi }^{+}$ such that $w^{-1}\alpha <0$ is one of the ${\theta }_j\text{.}$

		Proof.
	If $\alpha \in \Phi \left(w\right)=\{\alpha \in {\Phi }^{+}\hspace{0.17em}|\hspace{0.17em}{w}^{-1}\alpha <0\}$ then there exists $1\le j\le p$ such that $$s_{i_{j-1}}\cdots s_{i_1}\alpha >0\text{and}{s}_{i_j}\cdots s_{i_1}\alpha <0\text{.}$$ Since $s_{i_j}$ permutes the set ${\Phi }^{+}-\left\{{\alpha }_{i_j}\right\}$ we have that $$s_{i_{j-1}}\cdots s_{i_1}\alpha ={\alpha }_{i_j}\text{.}$$ So $\alpha =s_{i_1}\cdots s_{i_{j-1}}\left({\alpha }_{i_j}\right)\text{.}$ This shows that $\Phi \left(w\right)\subseteq \left\{{\theta }_j\right\}\text{.}$ But by Lemma () the cardinalities of these two sets are the same. $\square$

Schubert Images

Let $V^{\lambda }$ be a finite dimensional irreducible $𝔘$ module of highest weight $\lambda$ and let $v^{+}$ be a highest weight vector.

Let $w\in W$ and let $s_{i_1}{s}_{i_2}\cdots s_{i_p}=w$ be a reduced decomposition of $w\text{.}$ Let $w{v}^{+}$ denote ${\stackrel{\sim }{s}}_{i_1}{\stackrel{\sim }{s}}_{i_2}\cdots {\stackrel{\sim }{s}}_{i_p}{v}^{+}\text{.}$ $w{v}^{+}$ is well defined up to scalar since we know that $wt\left(w{v}^{+}\right)=w\lambda$ and that $\text{dim}{\left(V^{\lambda }\right)}_{w\lambda }=1\text{.}$ i.e., up to scalar multiples $w{v}^{+}$ does not depend on the reduced decomposition of $w\text{.}$ The Schubert image is the ${𝔘}^{+}$ module $$V^{\lambda }\left(w\right)={𝔘}^{+}w{v}^{+}\text{.}$$

Let $1$ denote the identity element of $W,$ and let $w\in W\text{.}$

1)	$V^{\lambda }\left(1\right)=ℂv^{+}\text{.}$
2)	If $s_{i}w\lambda \le w\lambda ,$ then $V^{\lambda }\left(s_{i}w\right)\supseteq V^{\lambda }\left(w\right)\text{.}$ Furthermore $V^{\lambda }\left(s_{i}w\right)\supset V^{\lambda }\left(w\right)$ if $s_{i}w\lambda <w\lambda \text{.}$
3)	If $w_0$ is such that $s_i{w}_0\lambda \ge w_0\lambda$ for all $i,$ then $V^{\lambda }\left(w_0\right)=V^{\lambda }\text{.}$

		Proof.
	Since $v^{+}$ is a highest weight vector $V^{\lambda }\left(1\right)={𝔘}^{+}{v}^{+}=ℂv^{+}$ giving 1). To prove 2), suppose that $s_{i}w\lambda =w\lambda -⟨w\lambda ,{\alpha }_i⟩\le w\lambda$ and let $m_i=⟨w\lambda ,{\alpha }_i⟩\ge 0\text{.}$ Then $x_i^{m_i}{s}_{i}w{v}^{+}\in {\left(V^{\lambda }\right)}_{w\lambda }\text{.}$ Since $\text{dim}{\left(V^{\lambda }\right)}_{w\lambda }=1,$ $x_i^{m_i}{s}_{i}w{v}^{+}=cw{v}^{+}$ for some constant $c\in ℂ\text{.}$ Thus $$V^{\lambda }\left(w\right)={𝔘}^{+}w{v}^{+}={𝔘}^{+}{x}_i^{m_i}{s}_{i}w{v}^{+}={𝔘}^{+}{x}_i^{m_i}{s}_{i}w{v}^{+}\subseteq {𝔘}^{+}{s}_{i}w{v}^{+}=V^{\lambda }\left(s_{i}w\right)\text{.}$$ The fact that the inclusion is proper if $s_{i}w\lambda <w\lambda$ follows from the fact that all weights of $V^{\lambda }\left(w\right)$ are $\ge w\lambda \text{.}$ To show 3) we see that since $s_i{w}_0\lambda \ge w_0\lambda$ for all $i$ then it follows from Lemma (5.2) 3) and 4) that then $w_0\lambda -{\alpha }_i$ is not a weight of $V^{\lambda }$ for any $i\text{.}$ This means that $y_i{w}_0{v}^{+}=0$ for all $i\text{.}$ So $w_0{v}^{+}$ is a lowest weight vector, giving that $V^{\lambda }=𝔘w_0{v}^{+}={𝔘}^{+}{w}_0{v}^{+}=V^{\lambda }\left(w_0\right)\text{.}$ $\square$

Let $V^{\lambda }$ be an irreducible $𝔘$ module of highest weight $\lambda \text{.}$ Then there is a filtration of $V^{\lambda }$ by Schubert images in the form $$V^{\lambda }=V^{\lambda }\left(w_0\right)\supset \cdots \supset V^{\lambda }\left(w_j\right)\supset \cdots \supset V^{\lambda }\left(1\right)=ℂv^{+}\text{.}$$ Furthermore this filtration can be chosen such that for each $j$ there is no Schubert image $V^{\lambda }\left(w\right)$ with $V^{\lambda }\left(w_j\right)\supset V^{\lambda }\left(w\right)\supset V^{\lambda }\left(w_{j-1}\right)\text{.}$

		Proof.
	Construct a chain of elements $w_j\in W$ in the following fashion. Let $w_1=1\text{.}$ Inductively $w_{j+1}$ is defined to be $s_i{w}_j$ where $i$ is such that $⟨w_j\lambda ,{\alpha }_i⟩>0\text{.}$ In this way $w_{j+1}\lambda <w_j\lambda$ in dominance. This process must stop since $V^{\lambda }$ is finite dimensional. We shall call the last element in this chain $w_0\text{.}$ The various parts of Lemma (6.1) show that we have a filtration $$ℂv^{+}=V^{\lambda }\left(1\right)\supset V^{\lambda }\left(w_2\right)\subset \cdots \subset V^{\lambda }\left(w_0\right)=V^{\lambda }\text{.}$$ If there existed some Schubert image $V^{\lambda }\left(w\right)$ such that $V^{\lambda }\left(w_{j+1}\right)\supset V^{\lambda }\left(w\right)\supset V^{\lambda }\left(w_j\right)$ then we must have that $w_{j+1}\lambda <w\lambda <w_j\lambda$ as these are the lowest weights in this chain of Schubert images. Assume that $w\lambda =w_{j+1}\lambda +{\sum }_i{m}_i{\alpha }_i$ and that $w_j\lambda =w\lambda +{\sum }_i{m}_i^{\prime }{\alpha }_i\text{.}$ Let $k$ be such that $w_{j+1}=s_k{w}_j$ and let $m=⟨w_j\lambda ,{\alpha }_k⟩,$ so that we get $$w_j\lambda w_{j+1}\lambda +m{\alpha }_k=w_{j+1}\lambda +\sum _i(m_i+m_i^{\prime }){\alpha }_i\text{.}$$ Since the ${\alpha }_i$ form a basis we have that $m_i+m_i^{\prime }=0$ for $i\ne k$ and $m_k+m_k^{\prime }=m\text{.}$ This shows that $w\lambda =w_{j+1}\lambda +m_k{\alpha }_k$ where $0<m_k<m\text{.}$ Then we would have that both $w\lambda +{\alpha }_i$ and $w\lambda -{\alpha }_i$ are weights of $V^{\lambda }$ which is a contradiction to Lemma (5.2) 3). $\square$

The character ring

Corresponding to each $\mu \in {𝔥}^{*}$ we write formally $e^{\mu }$ so that we have $$e^{\lambda }{e}^{\mu }=e^{\lambda +\mu }\text{.}$$ (Essentially we are just working with the group ${𝔥}^{*}$ just using multiplication as our operation instead of addition.) Define $$w{e}^{\lambda }=e^{w\lambda },$$ for each element $w\in W\text{.}$ Define $$A=ℝ[e^{\pm {\omega }_1},e^{\pm {\omega }_2},\dots ,e^{\pm {\omega }_n}]\text{.}$$ The ring $$A^W=\{f\in A\hspace{0.17em}|\hspace{0.17em}wf=f\hspace{0.17em}\text{for all}\hspace{0.17em}w\in W\}$$ is called the ring of Weyl group symmetric functions or the character ring. Define the character of a representation $V$ of $𝔘$ to be $$\chi =\sum _{\mu \in {𝔥}^{*}}\left(\text{dim}\hspace{0.17em}{V}_{\mu }\right)e^{\mu },$$ where $V_{\mu }$ is the weight space of $V$ corresponding to $\mu \text{.}$ Note that $\chi$ is an element of $A^W$ since, by Proposition (3.12), $\text{dim}\hspace{0.17em}{V}_{w\mu }=\text{dim}\hspace{0.17em}{V}_{\mu }$ for every $w\in W$ and $\mu \in P\text{.}$

If $f={\sum }_{\mu \in P}{f}_{\mu }{e}^{\mu }\in A^W$ then for all $w\in W$ we have $$f=wf=\sum _{\mu }{f}_{\mu }{e}^{w\mu }$$ so that $f_{\mu }=f_{w\mu },$ i.e. the coefficients $f_{\mu }$ are constant on each $W\text{-orbit}$ in $P\text{.}$ By Proposition (5.6) each $W\text{-orbit}$ is of the form $W\nu$ for a unique $\nu \in P^{+},$ so that $$f=\sum _{\nu \in P^{+}}{f}_{\lambda }\left(\sum _{\mu \in W\nu }{e}^{\mu }\right),$$ from which it follows that the orbit sums $$\begin{array}{cc}{m}_{\nu }=\sum _{\mu \in W\nu }{e}^{\mu },(\nu \in P^{+})& \text{(7.1)}\end{array}$$ are an $ℝ$ basis of $A^W$ (indeed a $\mathbb{Z}$ basis of $\mathbb{Z}{\left[e^{\pm {\omega }_i}\right]}^W\text{).}$

Let us use ${\chi }^{\lambda }$ to denote the character of the irreducible $𝔘\text{-module}$ $V^{\lambda }\text{.}$ ${\chi }^{\lambda }$ is in a sense a generating function for the values $K_{\lambda \mu }=\text{dim}{\left(V^{\lambda }\right)}_{\mu },$ $\lambda \in P^{+},$ $\mu \in P$ since by definition $${\chi }^{\lambda }=\sum _{\mu \in P}{K}_{\lambda \mu }{e}^{\mu }\text{.}$$ The values $K_{\lambda \mu },\lambda \in P^{+},$ $\mu \in P$ are completely determined by the values $K_{\lambda \nu },\lambda ,\nu \in P^{+}$ and that $${\chi }^{\lambda }=\sum _{\nu \in P^{+}}{K}_{\lambda \nu }{m}_{\nu }\text{.}$$

(Weyl character formula) Let $\lambda \in P^{+}$ be a dominant weight. The character of the irreducible representation $V^{\lambda }$ corresponding to $\lambda$ is given by $${\chi }^{\lambda }=\frac{{\sum }_{w\in W}\epsilon \left(w\right)e^{w(\lambda +\rho )}}{{\sum }_{w\in W}\epsilon \left(w\right)w^{w\rho }},$$ where $\rho ={\sum }_i{\omega }_i\text{.}$

Alternating symmetric functions $\text{(}W\text{-skew}$ functions)

A polynomial $f\in A$ is $W\text{-skew,}$ or alternating, if $$wf=\epsilon \left(w\right)f,\text{for all}\hspace{0.17em}w\in W\text{.}$$ Let $\theta :A\to A$ be the linear mapping defined by $$\theta f=\sum _{w\in W}\epsilon \left(w\right)wf,$$ for all $f\in A\text{.}$ If $v\in W$ then $$\begin{array}{cc}\theta v=v\theta =\sum \epsilon \left(w\right)vw=\epsilon \left(v\right)\theta \text{.}& \text{(7.3)}\end{array}$$

1)	$S=\theta \left(A\right)$ is the space of $W\text{-skew}$ elements of $A\text{.}$
2)	If $f\in A$ and $s_{i}f=f$ for some $i,$ then $\theta \left(f\right)=0\text{.}$
3)	The elements $\theta \left(e^{\lambda +\rho }\right),$ $\lambda \in P^{+},$ form a basis of $S\text{.}$

		Proof.
	1) Let $f\in A\text{.}$ Then $\theta f$ is $W\text{-skew,}$ because if $v\in W$ we have $v\theta f=\epsilon \left(v\right)\theta f$ by (7.3). Conversely if $f$ is $W\text{-skew}$ then $$\theta f=\sum _{w\in W}\epsilon \left(w\right)wf=\left|W\right|f,$$ so that $f=\theta \left(\frac{f}{\left|W\right|}\right)\in \theta \left(A\right)\text{.}$ 2) If $s_{i}f=f$ then $\theta f=\theta s_{i}f=\epsilon \left(s_i\right)\theta f=-\theta f$ and so $\theta f=0\text{.}$ 3) Suppose $f={\sum }_{\mu }{f}_{\mu }{e}^{\mu }\in S\text{.}$ Then $$f=\epsilon \left(w\right)wf=\sum _{\mu }\epsilon \left(w\right)f_{\mu }{e}^{w\mu },$$ and therefore $f_{w\mu }=\epsilon \left(w\right)f_{\mu }\text{.}$ Since each $W\text{-orbit}$ in $P$ meets $P^{+}$ in just one point it follows that $$\begin{array}{cc}f=\sum _{\mu \in P^{+}}{f}_{\mu }\left(\sum _{w\in W}\epsilon \left(w\right)e^{w\mu }\right)=\sum _{\mu \in P^{+}}{f}_{\mu }\theta \left(e^{\mu }\right)\text{.}& (*)\end{array}$$ Now $\mu \in P^{+}$ is of the form $\mu =\sum m_i{\omega }_i$ with $m_i\ge 0\text{.}$ If $m_i=0$ for some $i$ then $⟨\mu ,{\alpha }_i⟩=0$ and hence $s_i\mu =\mu ,$ so that $\theta \left(e^{\mu }\right)=0$ by 2). Hence in $(*)$ the sum is restricted to $\mu =\sum m_i{\omega }_i$ with each $m_i\ge 1,$ hence if we define $\lambda ={\sum }_i(m_i-1){\omega }_i$ then $\mu =\lambda +\rho$ with $\lambda \in P^{+}\text{.}$ So $f$ is a linear combination of the $\theta \left(e^{\lambda +\rho }\right)$ with $\lambda \in P^{+},$ and these are linearly independent since the orbits $W(\lambda +\rho )$ are disjoint (Proposition (5.6)). $\square$

If $\lambda \in P^{+}$ and $\lambda \le 0$ then $\lambda =0\text{.}$

		Proof.
	We will use the assumption that the Cartan matrix is symmetrizable and let $(,)$ be the bilinear form on ${𝔥}^{*}$ given by $$⟨\lambda ,{\alpha }_i⟩=\frac{(\lambda ,{\alpha }_i)}{({\alpha }_i,{\alpha }_i)}\text{.}$$ Then the fact that $\lambda \in P^{+}$ means that $(\lambda ,{\alpha }_i)\ge 0$ for all $i\text{.}$ The fact that $\lambda \le 0$ means that $\lambda ={\sum }_i{a}_i{\alpha }_i$ where the $a_i$ are integers $\le 0\text{.}$ Then $$\begin{array}{ccc}(\lambda ,\lambda )& =& \sum _i(\lambda ,a_i{\alpha }_i)\\ & =& \sum _i{a}_i(\lambda ,{\alpha }_i)\\ & \le & 0\end{array}$$ which implies that $\lambda =0\text{.}$ $\square$

(Weyl denominator formula) $$\sum _{w\in W}\epsilon \left(w\right)w{e}^{\rho }=\prod _{\alpha \in {\Phi }^{+}}(e^{\alpha /2}-e^{-\alpha /2})$$

		Proof.
	Let $d={\prod }_{\alpha \in {\Phi }^{+}}(e^{\alpha /2}-e^{-\alpha /2})\text{.}$ Then $$\begin{array}{ccc}{s}_{i}d& =& \left[s_i\prod _{\alpha \in {\Phi }^{+}-\left\{{\alpha }_i\right\}}(e^{\alpha /2}-e^{-\alpha /2})\right]\left[s_i(e^{{\alpha }_i/2}-e^{-{\alpha }_i/2})\right]\\ & =& \left[\prod _{\alpha \in {\Phi }^{+}-\left\{{\alpha }_i\right\}}(e^{\alpha /2}-e^{-\alpha /2})\right]\left[(e^{-{\alpha }_i/2}-e^{{\alpha }_i/2})\right]\\ & =& -d\text{.}\end{array}$$ So $d$ is $W\text{-skew.}$ Thus, by Proposition (7.4) 3), $d$ can be written in the form $$d=\sum _{\lambda \in P^{+}}{d}_{\lambda }\theta \left(e^{\lambda +\rho }\right),$$ where $d_{\lambda }$ is the coefficient of $e^{\lambda +\rho }$ in $d\text{.}$ On expansion of $d$ we have that $d$ is in the form $$d=e^{\rho }+\sum _{\mu <\rho }{a}_{\mu }{e}^{\mu },$$ giving that $d_0=1$ (the coefficient of $e^{\rho }\text{),}$ furthermore if $d_{\lambda }\ne 0$ for $\lambda \in P^{+},$ $\lambda \ne 0$ then we must have $\lambda +\rho <\rho ,$ i.e. that $\lambda <0\text{.}$ By Lemma (7.5) this cannot be. Thus we have that $d=\theta \left(e^{\rho }\right)\text{.}$ $\square$

If $\lambda ,\mu \in P$ are not proportional then $1-e^{\lambda }$ and $1-e^{\mu }$ are coprime in $A,$ i.e. $1-e^{\lambda }$ and $1-e^{\mu }$ have no common factors that are not units in $A\text{.}$

		Proof.
	This is proved in Bourbaki [Bou1968] Ch. VI, §3, Lemma 1. $\square$

Each $f\in S$ is divisible in $A$ by $$\begin{array}{ccc}d& =& \theta \left(e^{\rho }\right)\\ & =& e^{\rho }\prod _{\alpha \in {\Phi }^{+}}(1-e^{-\alpha })\end{array}$$ and $f{d}^{-1}$ is $W\text{-symmetric.}$ Furthermore the map $$\begin{array}{ccc}S& ⟶& A^W\\ f& ⟼& f{d}^{-1}\end{array}$$ is a bijection of $S$ onto the $W\text{-invariants.}$

		Proof.
	First show that for each $\alpha \in {\Phi }^{+},$ $(1-e^{-\alpha })$ divides $\theta \left(e^{\lambda +\rho }\right)$ for each $\lambda \in P^{+}\text{.}$ Suppose that $\alpha =w{\alpha }_i\text{.}$ Define $$s_{\alpha }=w{s}_i{w}^{-1}\text{.}$$ Note that $s_{\alpha }^2=1,$ $\epsilon \left(s_{\alpha }\right)=-1$ and that $$s_{\alpha }\mu =\mu -⟨w^{-1}\mu ,{\alpha }_i⟩\alpha \text{.}$$ Let $M$ be a system of representatives of the left cosets of the subgroup $\{1,s_{\alpha }\}\text{.}$ Then $$\theta \left(e^{\lambda +\rho }\right)=\sum _{w\in M}\epsilon \left(w\right)(e^{w(\lambda +\rho )}-e^{s_{\alpha }w(\lambda +\rho )})\text{.}$$ Thus it is sufficient to show that $e^{\mu }-e^{s_{\alpha }\mu }$ is divisible by $1-e^{-\alpha }$ for all $\mu \in P\text{.}$ Let $m=⟨w^{-1}\mu ,{\alpha }_i⟩$ so that we have that $$\begin{array}{ccc}{e}^{\mu }-e^{s_{\alpha }\mu }& =& e^{\mu }-e^{\mu -m\alpha }\\ & =& e^{\mu }(1-e^{-m\alpha })\\ & =& \{\begin{array}{cc}{e}^{\mu }(1-e^{-\alpha })(1+e^{-\alpha }+\cdots +e^{-(m-1)\alpha }),& \text{if}\hspace{0.17em}m>0\text{;}\\ 0& \text{if}\hspace{0.17em}m=0\text{;}\\ e^{\mu }(1-e^{-\alpha })(-e^{-m\alpha }-e^{-(m+1)\alpha }-\cdots -e^{\alpha }),& \text{if}\hspace{0.17em}m<0\text{.}\end{array}\end{array}$$ This shows that $(1-e^{-\alpha })$ divides $\theta \left(e^{\lambda +\rho }\right)$ for all $\alpha \in {\Phi }^{+}\text{.}$ By Lemma (5.4) and Lemma (7.7) we know that the factors $(1-e^{-\alpha })$ are pairwise coprime giving that $e^{\rho }{\prod }_{\alpha \in {\Phi }^{+}}(1-e^{-\alpha })=d$ divides $\theta \left(e^{\lambda +\rho }\right)\text{.}$ The map $f\mapsto f{d}^{-1}$ is a bijection as it is invertible with inverse given by $g\mapsto gd\text{.}$ It is clear that if $g\in A^W$ then $gd\in S$ since $w\left(gd\right)=\left(wg\right)\left(wd\right)=g\left(\epsilon \left(w\right)d\right)\text{.}$ $\square$

Since, (7.4) 3), the $\theta \left(e^{\lambda +\rho }\right),$ $\lambda \in P^{+},$ form an $ℝ\text{-basis}$ of the space $S$ of $W\text{-skew}$ elements of $A,$ hence by (7.8) the ${\chi }^{\lambda },$ $\lambda \in P^{+},$ given by the Weyl character formula (7.2) are an $ℝ\text{-basis}$ of $A^W\text{.}$

The Demazure operator

The proof of Proposition (7.8) encourages us to define the following operator. For each simple root ${\alpha }_i,$ define the Demazure operator ${\Delta }_i$ on the ring $A={\sum }_{\mu \in P}ℂe^{\mu }$ by $${\Delta }_i\left(e^{\lambda }\right)=\frac{e^{\lambda }-s_i{e}^{\lambda }}{1-e^{-{\alpha }_i}}\text{.}$$ Let $w\in W\text{.}$ Define $${\Delta }_w={\Delta }_{i_1}{\Delta }_{i_2}\cdots {\Delta }_{i_p},$$ where $s_{i_1}{s}_{i_2}\cdots s_{i_p}=w$ is a reduced word for $w\text{.}$ It is sometimes helpful to view ${\Delta }_i$ as an element of the group ring $𝒜\left[W\right]$ of the Weyl group over the ring of fractions $𝒜$ of $A\text{.}$

${\Delta }_w$ is well defined and does not depend on the reduced decomposition of $w\text{.}$

		Proof.
	I do not know an easy proof of this. It follows from Theorem () below (which I will not prove either). $\square$

1)	${\Delta }_i{\Delta }_i={\Delta }_i\text{.}$
2)	${\Delta }_i{\Delta }_w=\{\begin{array}{cc}{\Delta }_{s_{i}w}& \text{if}\hspace{0.17em}\ell \left(s_{i}w\right)>\ell \left(w\right)\text{;}\\ {\Delta }_w& \text{if}\hspace{0.17em}\ell \left(s_{i}w\right)<\ell \left(w\right)\text{.}\end{array}$

		Proof.
	The computation $$\begin{array}{ccc}{\Delta }_i{\Delta }_i{e}^{\mu }& =& {(1-e^{-{\alpha }_i})}^{-1}(1-s_i)\left[{(1-e^{-{\alpha }_i})}^{-1}(1-s_i)e^{\mu }\right]\\ & =& {(1-e^{-{\alpha }_i})}^{-1}(1-s_i)\left[{(1-e^{-{\alpha }_i})}^{-1}(e^{\mu }-e^{s_i\mu })\right]\\ & =& {(1-e^{-{\alpha }_i})}^{-1}[{(1-e^{-{\alpha }_i})}^{-1}(e^{\mu }-e^{s_i\mu })-(1-e^{{\alpha }_i})(e^{s_i\mu }-e^{\mu })]\\ & =& {(1-e^{-{\alpha }_i})}^{-1}{(1-e^{-{\alpha }_i})}^{-1}[e^{\mu }-e^{s_i\mu }-e^{-{\alpha }_i}(e^{\mu }-e^{s_i\mu })]\\ & =& {\Delta }_i{e}^{\mu }\end{array}$$ gives 1). 2) follows from the fact that if $\ell \left(s_{i}w\right)>\ell \left(w\right)$ and $s_{i_1}\cdots s_{i_p}$ is a reduced decomposition of $w$ then $s_i{s}_{i_1}\cdots s_{i_p}$ is a reduced decomposition of $s_{i}w\text{.}$ The second part follows similarly after interchanging $w$ with $s_{i}w\text{.}$ $\square$

Let $w_0\in W$ be such that $s_i{w}_0<w_0$ for all $i\text{.}$ Then

1)	For each $i,$ $s_i{\Delta }_{w_0}=e^{-{\alpha }_i}{\Delta }_{w_0}\text{.}$
2)	$${\Delta }_{w_0}=\epsilon \left(w_0\right)\prod _{\alpha \in {\Phi }^{+}}{(1-e^{-\alpha })}^{-1}{w}_0+\sum _{w\prime \le w_0}a\left(w\prime \right)w\prime ,$$ where the coefficients $a\left(w\prime \right)$ are elements in the field of fractions of $A\text{.}$

		Proof.
	By Lemma () 2) we have that ${\Delta }_i{\Delta }_{w_0}={\Delta }_{w_0}$ for all $i\text{.}$ So $(1-s_i){\Delta }_{w_0}=(1-e^{-{\alpha }_i}){\Delta }_{w_0}$ and subtracting ${\Delta }_{w_0}$ from each side gives 1). Let $s_{i_1}{s}_{i_2}\cdots s_{i_p}=w_0$ be a reduced decomposition of $w_0\text{.}$ Let $$a={(1-e^{-{\alpha }_{i_1}})}^{-1}{s}_{i_1}\cdots {(1-e^{-{\alpha }_{i_p}})}^{-1}{s}_{i_p}\text{.}$$ Then one gets easily by multiplying $$\begin{array}{ccc}{\Delta }_{w_0}& =& {(1-e^{-{\alpha }_{i_1}})}^{-1}(1-s_{i_1})\cdots {(1-e^{-{\alpha }_{i_p}})}^{-1}(1-s_{i_p})\\ & =& \epsilon \left(w_0\right)a+\sum _{w\prime <w_0}a\left(w\prime \right)w\prime ,\end{array}$$ where the coefficients $a\left(w\prime \right)$ are rational functions in $e^{{\omega }_i}\text{.}$ Re-expressing $a$ we get $$a=\frac{s_{i_1}{s}_{i_2}\cdots s_{i_p}}{(1-e^{-{\alpha }_{i_1}})\left[s_{i_1}(1-e^{-{\alpha }_{i_2}})\right]\cdots \left[s_{i_1}{s}_{i_2}\cdots s_{i_{p-1}}(1-e^{-{\alpha }_{i_p}})\right]}\text{.}$$ It follows from Proposition () that $$a=\frac{w_0}{{\prod }_{\alpha \in {\Phi }^{+}}(1-e^{-\alpha })}\text{.}$$ $\square$

Let $\theta ={\sum }_{w\in W}\epsilon \left(w\right)w$ and let $$D=e^{-\rho }\theta \left(e^{\rho }\right){\Delta }_{w_0},$$ where $\rho ={\sum }_i{\omega }_i\text{.}$ Then $$D=\theta$$

		Proof.
	Let $s_i$ be a simple reflection. Then using Lemma () ) $$\begin{array}{ccc}{s}_{i}D& =& \left(s_i{e}^{-\rho }\right)\left(s_i\theta \left(e^{\rho }\right)\right)\left(s_i{\Delta }_{w_0}\right)\\ & =& e^{-(\rho -{\alpha }_i)}\epsilon \left(s_i\right)\theta \left(e^{\rho }\right){\Delta }_{w_0}\\ & =& -D\text{.}\end{array}$$ This shows that $wD=\epsilon \left(w\right)D$ for all $w\in W,$ giving that $\theta D=D\theta =\left|W\right|D\text{.}$ This means that $$\frac{\theta }{\left|W\right|}D\frac{\theta }{\left|W\right|}=D\text{.}$$ But since $\theta /\left|W\right|$ is a minimal idempotent in the group algebra of $\left|W\right|$ we have that $D$ must be a multiple of $\theta \text{.}$ $D=\theta$ follows by using Lemma () to compare coefficients of $w_0\text{.}$ $\square$

For each $w\in W$ define another operator on $A$ by $${\bar{\Delta }}_w=e^{-\rho }{\Delta }_w{e}^{\rho }\text{.}$$ Note that one can define ${\bar{\Delta }}_w$ by defining ${\bar{\Delta }}_i=e^{-\rho }{\Delta }_i{e}^{\rho }$ for each $i$ and then defining ${\bar{\Delta }}_w={\bar{\Delta }}_{i_1}{\bar{\Delta }}_{i_2}\dots {\bar{\Delta }}_{i_p}$ where $s_{i_1}{s}_{i_2}\cdots s_{i_p}$ is a reduced word for $w\text{.}$

Let $\lambda \in P^{+},$ $w_{0}W$ be such that $\ell \left(s_i{w}_0\right)<\ell \left(w_0\right)$ for all $i,$ and recall that ${\chi }^{\lambda }$ denotes the Weyl character given by the Weyl character formula. Then one has $${\bar{\Delta }}_{w_0}\left(e^{\lambda }\right)={\chi }^{\lambda }\text{.}$$

		Proof.
	Using Lemma () we have that $$\begin{array}{ccc}{\bar{\Delta }}_{w_0}{e}^{\lambda }& =& e^{-\rho }{\Delta }_{w_0}{e}^{\lambda +\rho }\\ & =& \frac{D{e}^{\lambda +\rho }}{\theta \left(e^{\rho }\right)}\\ & =& \frac{\theta \left(e^{\lambda +\rho }\right)}{\theta \left(e^{\rho }\right)}\text{.}\end{array}$$ The result follows from the Weyl character formula. $\square$

We shall finish with the following deep and amazing theorem.

(Demazure character formula) For any subspace $M$ of a finite dimensional $𝔘\text{-module}$ $V$ define $$\text{char}\hspace{0.17em}M=\sum _{\mu \in P}\text{dim}\left(M_{\mu }\right)e^{\mu }\in A$$ where $M_{\mu }=M\cap V_{\mu }\text{.}$ Then for every Schubert image $V^{\lambda }\left(W\right),$ $\lambda \in P^{+},$ $w\in W,$ $${\bar{\Delta }}_w{e}^{\lambda }=\text{char}\hspace{0.17em}{V}^{\lambda }\left(w\right)\text{.}$$

		Proof.
	We shall not prove this as this is much too deep a theorem to prove in a short time. The most accessible proof I know of is in [Jos1985]. $\square$

Suggested References

Roots and weights

[Bou1968] N. Bourbaki, Groupes et algèbres de Lie, Chapitres 4, 5 et 6, Elements de Mathématique, Hermann, Paris, 1968. MR 39:1590

[Mac1991] I.G. Macdonald, Lecture Notes, University of California, San Diego, Spring 1991.

Enveloping algebra and representations

[Hum1972] J.E. Humphreys, Introduction to Lie Algebras and Representation Theory, Springer-Verlag, New York 1972.

[Kac1985] V. Kac, Infinite dimensional Lie algebras, Cambridge University Press 1985.

The character ring

[Bou1968] N. Bourbaki, Groupes et algèbres de Lie, Chapitres 4, 5 et 6, Elements de Mathématique, Hermann, Paris, 1968. MR 39:1590

[Mac1979] I.G. Macdonald, Hall Polynomials and Symmetric Functions, Oxford Univ. Press, 1979.

Schubert images and Demazure operators

[Dem1974] M. Demazure, Désingularisation des variétés de Schubert généralisées, Ann. Sci. École Norm. Sup. 7 (1974), 53–88.

[Dem1974-2] M. Demazure, Une nouvelle formule des characteres, Bull. Sc. Math. 98, (1974) 163-172.

[Jos1985] A. Joseph, On the Demazure character formula, Ann. Scient. Ec. Norm. Sup. 18, (1985) 389-419.

[LSe1986] V. Lakshmibai and C.S. Seshadri, Geometry of $G/P-V$, J. of Algebra 100 (1986), 462-557.

Notes and References

This is a copy of lectures notes for 18.318 Topics in Combinatorics, MIT Spring 1992, Prof. G.-C. Rota, given by Arun Ram.

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