Representation theory of semisimple Lie algebras

Last update: 2 September 2013

Weights

Let $C={‖{c}_{ij}‖}_{1\le i,j\le n}$ 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. $λ=∑i=1n ⟨λ,αi⟩ ωi.$ The weight lattice is the lattice $P=∑i=1n ℤωi.$ The elements of $P$ are called integral weights and the dominant integral weights are the weights in the set $P+=∑i=1n ℕωi,$ where $ℕ$ denotes the nonegative integers. Define vectors ${\alpha }_{1},{\alpha }_{2},\dots ,{\alpha }_{n},$ the simple roots, by $αi=∑j=1n cijωj.$

There is a partial ordering on the weights $\lambda \in {𝔥}^{*}$ given by defining $\mu \le \lambda$ if $μ=λ-∑i aiαi,$ 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 $si(λ)=λ- ⟨λ,αi⟩ αi,$ 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 $τ≥κ when τρ≤ κρ in dominance.$ 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]) $[hi,hj]=0 (1≤i,j≤n), (S1) [xi,yi]= hi, [xi,yj]= 0 if i≠j, (S2) [hi,xj]= ⟨αj,αi⟩ xj, [hi,yj]=- ⟨αj,αi⟩ yj, (S3) (ad xi)-⟨αj,αi⟩+1 (xj)=0 (i≠j), (Sij+) (ad yi)-⟨αj,αi⟩+1 (yj)=0 (i≠j). (Sij-)$ Here $\left[a,b\right]=ab-ba$ and ${\left(\text{ad} a\right)}^{k}\left(b\right)=\left[a,\left[a,\left[a,\cdots ,\left[a,b\right]\right]\cdots \right]\right]\text{.}$ One shows easily that $(ad a)k(b)= ∑i=0k (-1)i (ki)ak-i bai.$ We will use the following notations: $𝔘 = 𝔘(C); 𝔘+ = subalgebra of 𝔘 generated by the xi; 𝔥 = subalgebra of 𝔘 generated by the hi; 𝔘- = subalgebra of 𝔘 generated by the yi.$

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 $hiv= ⟨λ,αi⟩ v.$ 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 $Vλ= { v∈V | v is a weight vector and wt(v) =λ } .$ 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 $W=\text{span}\left\{{v}_{k},{h}_{k+1}v,{h}_{k+1}^{2}v,\dots \right\}\text{.}$ Since $hjhk+1sv= hk+1shjv= λjhk+1sv,$ 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 $hjyiv = yihjv+ [hj,yi]v = yi ⟨λ,αj⟩ v-⟨αi,αj⟩ yiv = ⟨λ-αi,αj⟩ yiv,$ 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 $xiyimm!v= yimm!xiv+ (λi-m+1) yim-1(m-1)!.$

 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 $mxi yimm!v = yixi yim-1(m-1)! +hiyim-1(m-1)! v = yi ( yim-1(m-1)!xv +(λi-(m-1)+1) yim-2(m-2)!v ) +⟨λ-(m-1)αi,αi⟩ yim-1(m-1)!v = yim(m-1)!xv +(λi-m+2)(m-1) yim-1(m-1)! +(λi-2(m-1)) yim-1(m-1)!v = yim(m-1)! +m(λi-m+1) yim-1(m-1)!v.$ $\square$

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

 Proof. Since $V$ is finite dimensional we know that $\frac{{y}_{i}^{m+1}}{\left(m+1\right)!}{v}^{+}=0$ for some $m\in ℕ\text{.}$ Let $m$ be minimal such that this is true. By the previous lemma we have that $0 = xiyim+1(m+1)! v+ = (λi-m) yimm!v+.$ Since $\left({y}_{i}^{m}/m!\right){v}^{+}\ne 0$ 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 $𝔘{v}^{+}$ is a submodule of $V\text{.}$ If $V=𝔘{v}^{+}$ (in particular, when $V$ is irreducible) then

 (1) $V=𝔘-v+= span {yMv+}$ 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, $V=⨁μ∈P,μ≤λ Vμ,and Vμ=span { yMv+ | wt(yMv+) =μ } .$

 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 $c\in ℂ\text{.}$ $xi yikmk⋯ yi1m1 v+ = ( yikxi yikmk-1 ⋯yi1m1+ [xiyik] yikmk-1 ⋯yi1m1 ) v+ = ( yikxi yikmk-1 ⋯yi1m1+ chi yikmk-1 ⋯yi1m1 ) v+$ 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 $yiλi+1v+=0$ 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 $V,$ $s∼i=exp (xi)exp (-yi)exp (xi),where exp(a)= ∑k≥0 xkk!.$

$(exp a)b (exp(-a))= (exp(ad a)) (b)$

 Proof. $(exp a)b (exp(-a)) = ∑k≥0 ∑s=0k (-1)ss! 1(k-s)! ak-sbas = ∑k≥01k! ∑s=0k (-1)s (ks) ak-sbas = ∑k≥01k! (ad a)kb = (exp(ad a)) b.$ $\square$

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 ${\stackrel{\sim }{s}}_{i}^{-1}=\text{exp}\left(-{x}_{i}\right)\text{exp}\left({y}_{i}\right)\text{exp}\left(-{x}_{i}\right)\text{.}$ Using Lemma (3.11) we have that $s∼i-1hi s∼i= exp(ad yi) exp(ad(-xi)) exp(ad yi) hi.$ Using the relations $[yi,xi]=-hi ,[yi,hi]=2 yi,[yi,yi] =0, [xi,xi]=0, [xi,hi]=-2xi ,[xi,yi]=hi,$ we can express the action of $\text{ad}\left(-{x}_{i}\right)$ and $\text{ad} {y}_{i}$ on the subspace $\text{span}\left\{{x}_{i},{y}_{i},{h}_{i}\right\}$ in terms of the matrices $(ad(-xi))= ( 002 000 0-10 ) , (ad yi)= ( 000 002 -100 ) .$ From this one easily computes that $exp(ad yi) exp(ad(-xi)) exp(ad yi)= ( 0-10 -100 00-1 ) .$ 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 $hi(s∼iv) = -s∼ihiv = -μi (s∼iv) = ( ⟨μ,αi⟩ -2⟨μ,αi⟩ ) s∼iv = ⟨μ-⟨μ,αi⟩αi,αi⟩ s∼iv = ⟨siμ,αi⟩ s∼iv.$ 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 $Φ= { α∈𝔥* | α=wαi for some w∈W and some αi } .$ The positive roots and the negative roots are the roots in the sets $Φ+ = {α∈Φ | α≥0}, Φ- = {α∈Φ | α≤0},$ 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 $siv = exp(xi) exp(-yi) exp(xi)v = av+∑mbm yimv+ ∑m,m′ cm,m′ yimxim′ v.$ 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 ${w}^{-1}\left(w\lambda -{\alpha }_{i}\right)=\lambda -{w}^{-1}{\alpha }_{i}\text{.}$ 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 ${s}_{i}\left(w\lambda +{\alpha }_{i}\right)=w\lambda -{\alpha }_{i}$ 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 ${\stackrel{\sim }{s}}_{i}v=\text{exp}\left({y}_{i}\right)\text{exp}\left(-{x}_{i}\right)\text{exp}\left({y}_{i}\right)v=1v=v,$ 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., $Φ=Φ+ ∪Φ-.$

 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 ${w}^{-1}\left(\rho -{\alpha }_{i}\right)={w}^{-1}\rho -\left(1/m\right){\alpha }_{j}$ 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 $siβ=β- ⟨β,αi⟩ αi$ 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 $\rho \prime =\rho \text{.}$ $siρ′ = si ( 12 ∑α∈Φ+-{αi} α ) +12siαi = -αi+12 ∑α∈Φ+α = ρ′-αi.$ 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 $C‾= { μ∈𝔥* | ⟨μ,αi⟩ ≥0 for all i } .$ Note that ${P}^{+}=P\cap \stackrel{‾}{C}\text{.}$

Every $\mu \in P$ has a unique image in the Weyl chamber, i.e., there exists $w\in W$ such that $w\mu \in {P}^{+}=\stackrel{‾}{C}\cap P$ 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 ${\mu }^{\left(j+1\right)}={s}_{i}{\mu }^{\left(j\right)}$ 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 $\lambda =w{\left(w\prime \right)}^{-1}\lambda \prime =\stackrel{‾}{w}\lambda \text{.}$ Let ${V}^{\lambda }$ be the irreducible $𝔘$ module corresponding to $\lambda$ let ${v}^{+}$ be a highest weight vector and let $v=\stackrel{‾}{w}{v}^{+}\text{.}$ 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 groupW is generated by the simple transpositions ${s}_{i}$ every element $w\in W$ can be written in the form $w=si1si2⋯ sip.$ ${s}_{{i}_{1}}\cdots {s}_{{i}_{p}}$ 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 $ε(w)= (-1)ℓ(w).$ One can also view $\epsilon$ as the homomorphism $ε: W ⟶ W si ⟼ -1$

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

 1) $\ell \left({s}_{i}w\right)>\ell \left(w\right)⇔\ell \left({s}_{i}w\right)=\ell \left(w\right)+1⇔{w}^{-1}{\alpha }_{i}>0\text{.}$ 2) $\ell \left({s}_{i}w\right)<\ell \left(w\right)⇔\ell \left({s}_{i}w\right)=\ell \left(w\right)-1⇔{w}^{-1}{\alpha }_{i}<0\text{.}$

 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 $Φ(w)=si Φ(siw)∪ {αi}.$

 Proof. $\square$

For each $w\in W,$ $ℓ(w)=Card (Φ(w)).$

 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 $θj=si1si2 ⋯sij-1 (αij)$ 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 $\alpha \in \Phi \left(w\right)=\left\{\alpha \in {\Phi }^{+} | {w}^{-1}\alpha <0\right\}$ then there exists $1\le j\le p$ such that $sij-1⋯ si1α>0 andsij⋯ si1α<0.$ Since ${s}_{{i}_{j}}$ permutes the set ${\Phi }^{+}-\left\{{\alpha }_{{i}_{j}}\right\}$ we have that $sij-1⋯ si1α= αij.$ 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 $Vλ(w)= 𝔘+wv+.$

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

 1) ${V}^{\lambda }\left(1\right)=ℂ{v}^{+}\text{.}$ 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 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 ${V}^{\lambda }\left(1\right)={𝔘}^{+}{v}^{+}=ℂ{v}^{+}$ 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 $Vλ(w)=𝔘+ wv+=𝔘+ ximisiwv+ =𝔘+ximisi wv+⊆𝔘+si wv+=Vλ(siw) .$ The fact that the inclusion is proper if ${s}_{i}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 ${V}^{\lambda }=𝔘{w}_{0}{v}^{+}={𝔘}^{+}{w}_{0}{v}^{+}={V}^{\lambda }\left({w}_{0}\right)\text{.}$ $\square$

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 $Vλ=Vλ(w0) ⊃⋯⊃Vλ(wj) ⊃⋯⊃Vλ(1)= ℂv+.$ 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 $ℂv+=Vλ(1) ⊃Vλ(w2)⊂⋯ ⊂Vλ(w0)= Vλ.$ 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 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 $wjλwj+1λ+m αk=wj+1λ+ ∑i(mi+mi′) αi.$ 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} 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 $eλeμ=eλ+μ.$ (Essentially we are just working with the group ${𝔥}^{*}$ just using multiplication as our operation instead of addition.) Define $weλ=ewλ,$ for each element $w\in W\text{.}$ Define $A=ℝ [ e±ω1, e±ω2,…, e±ωn ] .$ The ring $AW= { f∈A | wf =f for all w∈W }$ is called the ring of Weyl group symmetric functions or the character ring. Define the character of a representation $V$ of $𝔘$ to be $χ=∑μ∈𝔥* (dim Vμ)eμ,$ 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), $\text{dim} {V}_{w\mu }=\text{dim} {V}_{\mu }$ 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 $f=wf=∑μ fμewμ$ 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 $f=∑ν∈P+ fλ(∑μ∈Wνeμ),$ from which it follows that the orbit sums $mν=∑μ∈Wν eμ,(ν∈P+) (7.1)$ are an $ℝ$ basis of ${A}^{W}$ (indeed a $ℤ$ basis of $ℤ{\left[{e}^{±{\omega }_{i}}\right]}^{W}\text{).}$

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 $χλ=∑μ∈P Kλμeμ.$ 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 $χλ=∑ν∈P+ Kλνmν.$

(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 $χλ= ∑w∈W ε(w) ew(λ+ρ) ∑w∈W ε(w) wwρ ,$ 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 $wf=ε(w)f, for all w∈W.$ Let $\theta :A\to A$ be the linear mapping defined by $θf=∑w∈W ε(w)wf,$ for all $f\in A\text{.}$ If $v\in W$ then $θv=vθ=∑ ε(w)vw= ε(v)θ. (7.3)$

 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 $θf=∑w∈W ε(w)wf= |W|f,$ so that $f=\theta \left(\frac{f}{|W|}\right)\in \theta \left(A\right)\text{.}$ 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 $f=ε(w)wf= ∑με(w) fμewμ,$ 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 $f=∑μ∈P+ fμ ( ∑w∈Wε(w) ewμ ) =∑μ∈P+ fμθ(eμ). (*)$ 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 $\left(*\right)$ the sum is restricted to $\mu =\sum {m}_{i}{\omega }_{i}$ with each ${m}_{i}\ge 1,$ hence if we define $\lambda ={\sum }_{i}\left({m}_{i}-1\right){\omega }_{i}$ 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 $W\left(\lambda +\rho \right)$ 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 $\left(,\right)$ be the bilinear form on ${𝔥}^{*}$ given by $⟨λ,αi⟩= (λ,αi) (αi,αi) .$ Then the fact that $\lambda \in {P}^{+}$ means that $\left(\lambda ,{\alpha }_{i}\right)\ge 0$ 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 $(λ,λ) = ∑i (λ,aiαi) = ∑iai (λ,αi) ≤ 0$ which implies that $\lambda =0\text{.}$ $\square$

(Weyl denominator formula) $∑w∈W ε(w)w eρ=∏α∈Φ+ (eα/2-e-α/2)$

 Proof. Let $d={\prod }_{\alpha \in {\Phi }^{+}}\left({e}^{\alpha /2}-{e}^{-\alpha /2}\right)\text{.}$ Then $sid = [ si ∏α∈Φ+-{αi} (eα/2-e-α/2) ] [ si(eαi/2-e-αi/2) ] = [ ∏α∈Φ+-{αi} (eα/2-e-α/2) ] [ (e-αi/2-eαi/2) ] = -d.$ So $d$ is $W\text{-skew.}$ Thus, by Proposition (7.4) 3), $d$ can be written in the form $d=∑λ∈P+ dλθ (eλ+ρ),$ 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 $d=eρ+ ∑μ<ρ aμeμ,$ 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 $d = θ(eρ) = eρ∏α∈Φ+ (1-e-α)$ and $f{d}^{-1}$ is $W\text{-symmetric.}$ Furthermore the map $S⟶AW f⟼fd-1$ is a bijection of $S$ onto the $W\text{-invariants.}$

 Proof. First show that for each $\alpha \in {\Phi }^{+},$ $\left(1-{e}^{-\alpha }\right)$ divides $\theta \left({e}^{\lambda +\rho }\right)$ for each $\lambda \in {P}^{+}\text{.}$ Suppose that $\alpha =w{\alpha }_{i}\text{.}$ Define $sα=wsi w-1.$ Note that ${s}_{\alpha }^{2}=1,$ $\epsilon \left({s}_{\alpha }\right)=-1$ and that $sαμ=μ- ⟨w-1μ,αi⟩ α.$ Let $M$ be a system of representatives of the left cosets of the subgroup $\left\{1,{s}_{\alpha }\right\}\text{.}$ Then $θ(eλ+ρ)= ∑w∈Mε(w) ( ew(λ+ρ)- esαw(λ+ρ) ) .$ 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 $eμ-esαμ = eμ- eμ-mα = eμ (1-e-mα) = { eμ (1-e-α) ( 1+e-α+ ⋯+ e-(m-1)α ) , if m>0; 0 if m=0; eμ (1-e-α) ( -e-mα -e-(m+1)α -⋯-eα ) , if m<0.$ This shows that $\left(1-{e}^{-\alpha }\right)$ 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 $\left(1-{e}^{-\alpha }\right)$ are pairwise coprime giving that ${e}^{\rho }{\prod }_{\alpha \in {\Phi }^{+}}\left(1-{e}^{-\alpha }\right)=d$ divides $\theta \left({e}^{\lambda +\rho }\right)\text{.}$ The map $f↦f{d}^{-1}$ is a bijection as it is invertible with inverse given by $g↦gd\text{.}$ 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 $A={\sum }_{\mu \in P}ℂ{e}^{\mu }$ by $Δi(eλ)= eλ-sieλ 1-e-αi .$ Let $w\in W\text{.}$ Define $Δw= Δi1 Δi2⋯ Δip,$ 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) ${\Delta }_{i}{\Delta }_{w}=\left\{\begin{array}{cc}{\Delta }_{{s}_{i}w}& \text{if} \ell \left({s}_{i}w\right)>\ell \left(w\right)\text{;}\\ {\Delta }_{w}& \text{if} \ell \left({s}_{i}w\right)<\ell \left(w\right)\text{.}\end{array}$

 Proof. The computation $ΔiΔi eμ = (1-e-αi)-1 (1-si) [ (1-e-αi)-1 (1-si)eμ ] = (1-e-αi)-1 (1-si) [ (1-e-αi)-1 (eμ-esiμ) ] = (1-e-αi)-1 [ (1-e-αi)-1 (eμ-esiμ)- (1-eαi) (esiμ-eμ) ] = (1-e-αi)-1 (1-e-αi)-1 [ eμ-esiμ- e-αi (eμ-esiμ) ] = Δieμ$ 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) $Δw0= ε(w0) ∏α∈Φ+ (1-e-α)-1 w0+∑w′≤w0 a(w′)w′,$ 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 $\left(1-{s}_{i}\right){\Delta }_{{w}_{0}}=\left(1-{e}^{-{\alpha }_{i}}\right){\Delta }_{{w}_{0}}$ 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 $a= (1-e-αi1)-1si1⋯ (1-e-αip)-1sip.$ Then one gets easily by multiplying $Δw0 = (1-e-αi1)-1 (1-si1)⋯ (1-e-αip)-1 (1-sip) = ε(w0)a+ ∑w′ where the coefficients $a\left(w\prime \right)$ are rational functions in ${e}^{{\omega }_{i}}\text{.}$ Re-expressing $a$ we get $a= si1si2⋯sip (1-e-αi1) [si1(1-e-αi2)]⋯ [si1si2⋯sip-1(1-e-αip)] .$ It follows from Proposition () that $a= w0 ∏α∈Φ+ (1-e-α) .$ $\square$

Let $\theta ={\sum }_{w\in W}\epsilon \left(w\right)w$ and let $D=e-ρθ (eρ)Δw0,$ where $\rho ={\sum }_{i}{\omega }_{i}\text{.}$ Then $D=θ$

 Proof. Let ${s}_{i}$ be a simple reflection. Then using Lemma () ) $siD = (sie-ρ) (siθ(eρ)) (siΔw0) = e-(ρ-αi) ε(si)θ (eρ)Δw0 = -D.$ This shows that $wD=\epsilon \left(w\right)D$ for all $w\in W,$ giving that $\theta D=D\theta =|W|D\text{.}$ This means that $θ|W|D θ|W|=D.$ But since $\theta /|W|$ is a minimal idempotent in the group algebra of $|W|$ 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 $Δ‾w= e-ρΔw eρ.$ Note that one can define ${\stackrel{‾}{\Delta }}_{w}$ by defining ${\stackrel{‾}{\Delta }}_{i}={e}^{-\rho }{\Delta }_{i}{e}^{\rho }$ for each $i$ and then defining ${\stackrel{‾}{\Delta }}_{w}={\stackrel{‾}{\Delta }}_{{i}_{1}}{\stackrel{‾}{\Delta }}_{{i}_{2}}\dots {\stackrel{‾}{\Delta }}_{{i}_{p}}$ 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 $Δ‾w0 (eλ)= χλ.$

 Proof. Using Lemma () we have that $Δ‾w0 eλ = e-ρΔw0 eλ+ρ = Deλ+ρ θ(eρ) = θ(eλ+ρ) θ(eρ) .$ 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 $char M=∑μ∈P dim(Mμ)eμ∈A$ where ${M}_{\mu }=M\cap {V}_{\mu }\text{.}$ Then for every Schubert image ${V}^{\lambda }\left(W\right),$ $\lambda \in {P}^{+},$ $w\in W,$ $Δ‾weλ =char Vλ(w).$

 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.