Orthogonal polynomials

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

Last update: 11 August 2012

Abstract.
This is a typed version of I.G. Macdonald's lecture notes from lectures at the University of California San Diego from January to March of 1991.

Introduction

Notation etc. as in the previous chapter. If $f=\sum _{\lambda \in P}{f}_{\lambda }{e}^{\lambda }\in A,$ let $$\stackrel{—}{f}=\sum _{\lambda \in P}{f}_{\lambda }{e}^{-\lambda }(\stackrel{—}{f}\left(x\right)=\stackrel{—}{f\left(x\right)},x\in V)$$ and let ${\left[f\right]}_1$ denote the constant term $f_0$ of $f$ (the coefficient of $e^0=1$). I am going to define various scalar products on $A^W$ $$\begin{array}{ll}{⟨f,g⟩}_1=\frac{1}{\left|W\right|}{\left[f\stackrel{—}{g}\right]}_1& (OP\; 1)\end{array}$$

Let $\lambda ,\mu \in P^{+}.$ Then ${⟨m_{\lambda },m_{\mu }⟩}_1=0$ if $\lambda \ne \mu ,$ and ${⟨m_{\lambda },m_{\lambda }⟩}_1={\left|W_{\lambda }\right|}^{-1}.$

		Proof.
	We have $$m_{\lambda }=\sum _{\nu \in W\lambda }{e}^{\nu }$$ and the number of terms in this sum is $\left|W\lambda \right|=\frac{\left|W\right|}{\left|W_{\lambda }\right|}.$ Hence $$⟨m_{\lambda },m_{\mu }⟩=\frac{1}{\left|W\right|}\sum _{\genfrac{}{}{0ex}{}{\nu \in W\lambda }{\pi \in W\mu }}{[e^{\nu }\cdot e^{-\pi }]}_1=\frac{1}{\left|W\right|}\mathrm{Card}(W\lambda \cap W\mu ).$$ If $\lambda \ne \mu$ this is 0, and if $\lambda =\mu$ it is $\frac{1}{\left|W\right|}\mathrm{Card}\left(W\lambda \right)=\frac{1}{\left|W\lambda \right|}.$ $\square$

$$\begin{array}{ll}{⟨f,g⟩}_0=\frac{1}{\left|W\right|}{\left[f\stackrel{—}{g}d\stackrel{—}{d}\right]}_1={⟨fd,gd⟩}_1& (OP\; 2)\end{array}$$ where as before $$d=e^{\rho }\prod _{\alpha \in R^{+}}(1-e^{-\alpha })=\sum _{w\in W}\epsilon \left(w\right)e^{w\rho }$$ is the Weyl denominator.

Let $\lambda ,\mu \in P^{+}.$ Then ${⟨{\chi }_{\lambda },{\chi }_{\mu }⟩}_0={\delta }_{\lambda \mu }.$

		Proof.
	Since $${\chi }_{\lambda }=d^{-1}\theta \left(e^{\lambda +\rho }\right)=d^{-1}\sum _{w\in W}\epsilon \left(w\right)e^{w(\lambda +\rho )},$$ we have $${⟨{\chi }_{\lambda },{\chi }_{\mu }⟩}_0=\frac{1}{\left|W\right|}\sum _{v,w\in W}\epsilon \left(v\right)\epsilon \left(w\right){[e^{v(\lambda +\rho )}\cdot e^{-w(\mu +\rho )}]}_1$$ which is zero if $\lambda \ne \mu ,$ because the orbits $W(\lambda +\rho ), W(\mu +\rho )$ do not intersect. If $\lambda =\mu$ we have $v(\lambda +\rho )=w(\lambda +\rho ) \iff v=w$ (because $\lambda +\rho \in C$). Hence we get $\frac{1}{\left|W\right|}\sum _{w\in W}1=1.$ $\square$

Let $t$ be a real number, $0\le t\le 1,$ and let $$\Delta =\Delta \left(t\right)=\prod _{\alpha \in R}\frac{1-e^{\alpha }}{1-t{e}^{\alpha }}.$$ As explained earlier, we may regard $\Delta$ as a continuous function on the torus $T=V/Q^{\vee }.$ For $f,g\in A$ we define $$\begin{array}{rcl}{⟨f,g⟩}_t& =& \frac{1}{\left|W\right|}{\left[f\stackrel{—}{g}\Delta \right]}_1\\ & =& {⟨f\Delta ,g⟩}_1\\ & =& {⟨f{\Delta }^{+},g{\Delta }^{+}⟩}_1\\ & =& \frac{1}{\left|W\right|}{\int }_{T}f\stackrel{—}{g}\Delta positive\; definite,\; symmetric.\end{array}$$ Notice that when $t=1$ we have $\Delta =1,$ and when $t=0$ we have $$\Delta =\prod _{\alpha \in R}(1-e^{\alpha })=d\stackrel{—}{d}$$ so that when $t=1,0$ we get the two previous scalar products.

Let $\lambda ,\mu \in P^{+}.$ Then ${⟨P_{\lambda },P_{\mu }⟩}_t=0$ if $\lambda \ne \mu ,$ and ${⟨P_{\lambda },P_{\lambda }⟩}_t=W_{\lambda }{\left(t\right)}^{-1}.$

		Proof.
	Recall that $${\tilde{P}}_{\lambda }=\sum _{w\in W}w\left(e^{\lambda }\prod _{\alpha \in R^{+}}\frac{1-t{e}^{-\alpha }}{1-e^{-\alpha }}\right)(\lambda \in P^{+}).$$ Since $\Delta$ is $W-$invariant it follows that $${\tilde{P}}_{\lambda }\Delta =\sum _{w\in W}w\left(e^{\lambda }\prod _{\alpha \in R^{+}}\frac{1-e^{\alpha }}{1-t{e}^{\alpha }}\right)$$ in which $$e^{\lambda }\prod _{\alpha \in R^{+}}\frac{1-e^{\alpha }}{1-t{e}^{\alpha }}=e^{\lambda }\prod _{\alpha \in R^{+}}\left(1+\sum _{r\ge 1}(t^r-t^{r-1})e^{r\alpha }\right)$$ a product of convergent series, since $0\le t<1.$ Hence we can multiply them all together and get say $$e^{\lambda }\prod _{\alpha \in R^{+}}\frac{1-e^{\alpha }}{1-t{e}^{\alpha }}=\sum _{\nu \in Q^{+}}{a}_{\nu }{e}^{\lambda +\nu }$$ with coefficients $a_{\nu }\in \mathbb{Z}\left[t\right],$ and in particular $a_0=1.$ Hence we have $${\tilde{P}}_{\lambda }\Delta =\sum _{\nu \in Q^{+}}{a}_{\nu }\sum _{w\in W}{e}^{w(\lambda +\nu )}.$$ If $\lambda +\nu =w_1\pi ,$ where $\pi \in P^{+}$ and $w_1\in W,$ then $$\pi \ge w_1\pi =\lambda +\nu \ge \lambda$$ with equality if and only if $\nu =0.$ Hence $$\begin{array}{ll}{\tilde{P}}_{\lambda }\Delta =\left|W_{\lambda }\right|m_{\lambda }+higher\; terms.& (*)\end{array}$$ Now let $\mu \in P^{+}.$ We cannot have $\mu >\lambda$ and $\lambda >\mu ,$ hence (since the scalar product is symmetric) we may assume $\mu \ngtr \lambda .$ We have $$\begin{array}{ll}{P}_{\mu }=m_{\mu }+lower\; terms.& (**)\end{array}$$ If $\mu \ne \lambda ,$ the series (*) and (**) have just one term in common, namely $m_{\lambda },$ and hence $$\begin{array}{rcl}{⟨{\tilde{P}}_{\lambda },P_{\lambda }⟩}_t& =& {⟨{\tilde{P}}_{\lambda }\Delta ,P_{\lambda }⟩}_1\\ & =& \left|W_{\lambda }\right|{⟨m_{\lambda },m_{\lambda }⟩}_1\\ & =& 1by\; Proposition\; 1.1\end{array}$$ so that ${⟨P_{\lambda },P_{\lambda }⟩}_t=W_{\lambda }{\left(t\right)}^{-1}.$ $\square$

(So $\left(P_{\lambda }\right), \left({\tilde{P}}_{\lambda }\right)$ are dual bases.)

		Proof of (4.18).
	Since $${\chi }_{\lambda }=\sum _{\mu }{K}_{\lambda \mu }\left(t\right)P_{\mu }$$ we have $$K_{\lambda \mu }\left(t\right)={⟨{\chi }_{\lambda },{\tilde{P}}_{\mu }⟩}_t.$$ Now $$\Delta =\prod _{\alpha \in R}\frac{1-e^{\alpha }}{1-t{e}^{\alpha }}=d\stackrel{—}{d}\Pi where\Pi =\prod _{\alpha \in R}{(1-t{e}^{\alpha })}^{-1}.$$ Hence $$\begin{array}{ll}\begin{array}{rcl}{K}_{\lambda \mu }\left(t\right)& =& \frac{1}{\left|W\right|}{\left[{\stackrel{—}{\chi }}_{\lambda }{\tilde{P}}_{\mu }\Delta \right]}_1\\ & =& \frac{1}{\left|W\right|}{\left[{\stackrel{—}{\chi }}_{\lambda }{\tilde{P}}_{\mu }d\stackrel{—}{d}\Pi \right]}_1\\ & =& {⟨{\chi }_{\lambda },{\tilde{P}}_{\mu }\Pi ⟩}_0\end{array}& (†)\end{array}$$ But now $$\begin{array}{rcl}{\tilde{P}}_{\mu }\Pi & =& \sum _{w\in W}w\left(e^{\mu }\prod _{\alpha \in R^{+}}\frac{1-t{e}^{-\alpha }}{1-e^{-\alpha }}\cdot \prod _{\alpha \in R}\frac{1}{1-t{e}^{\alpha }}\right)\\ & =& d^{-1}\sum _{w\in W}\epsilon \left(w\right)w\left(e^{\mu +\rho }\prod _{\alpha \in R^{+}}{(1-t{e}^{\alpha })}^{-1}\right)\\ & =& \prod _{\alpha \in R^{+}}{(1-t{R}_{\alpha })}^{-1}{\chi }_{\mu }.\end{array}$$ From (†) it follows that $$\begin{array}{ll}{K}_{\lambda \mu }\left(t\right)=coefficient\; of{\chi }_{\lambda }in\prod _{\alpha \in R^{+}}{(1-t{R}_{\alpha })}^{-1}{\chi }_{\mu }& (††)\end{array}$$ which proves (4.18)(ii). Now $$\begin{array}{rcl}\prod _{\alpha \in R^{+}}{(1-t{R}_{\alpha })}^{-1}& =& \prod _{\alpha \in R^{+}}\sum _{m_{\alpha }=0}^{\infty }{t}^{m_{\alpha }}{R}_{\alpha }^{m_{\alpha }}\\ & =& \sum _{\xi \in Q^{+}}F(\xi ;t)R_{\xi }\end{array}$$ and if $\xi \in Q^{+}$ $${\chi }_{\mu +\xi }=d^{-1}\theta \left(e^{\mu +\xi +\rho }\right).$$ If $\mu +\xi +\rho$ is singular, then ${\chi }_{\mu +\xi }=0;$ if nonsingular, then $$\mu +\xi +\rho =w(\lambda +\rho )$$ for a unique $w\in W$ and $\lambda \in P^{+}:$ and then $$\xi =w(\lambda +\rho )-(\mu +\rho )and{\chi }_{\mu +\xi }=\epsilon \left(w\right)=\epsilon \left(w\right){\chi }_{\lambda }.$$ So from (††) we have $$K_{\lambda \mu }\left(t\right)=\sum _{w\in W}\epsilon \left(w\right)F(w(\lambda +\rho )-(\mu +\rho );t).$$ $\square$

In fact the $K_{\lambda \mu }\left(t\right)$ are polynomials in $t$ with positive coefficients (Lusztig: he shows that they are Kazhdan-Lusztig polynomials for the affine Weyl group). Elementary proof?

$t=1:$ $K_{\lambda \mu }=\sum _{w\in W}\epsilon \left(w\right)F(w(\lambda +\rho )-(\mu +\rho );1)$ is Kostant's formula for weight multiplicities. $${\chi }_{\mu }{\chi }_{\nu }=\sum _{\lambda }{c}_{\mu \nu }^{\lambda }{\chi }_{\lambda }{c}_{\mu \nu }^{\lambda }\ge 0.$$

Now let $q,t$ be independent indeterminants [or real variables, $0\le q<1, t\ge 0$] and let $$\Delta =\Delta (q,t)=\prod _{\alpha \in R}\frac{{(e^{\alpha };q)}_{\infty }}{{(t{e}^{\alpha };q)}_{\infty }}.$$ In particular, if $t=q^k$ where $k\in \mathbb{N}$ then $$\Delta =\prod _{\alpha \in R}{(e^{\alpha };q)}_k\in A^W.$$ We now define a new scalar product on $A:$ if $f,g\in A$ then $$\begin{array}{rcl}{⟨f,g⟩}_{q,t}& =& \frac{1}{\left|W\right|}{\left[f\stackrel{—}{g}\Delta \right]}_1(=\frac{1}{\left|W\right|}{\int }_{T}f\stackrel{—}{g}\Delta )\\ & =& {⟨f{\Delta }^{+},g{\Delta }^{+}⟩}_1.\end{array}$$ Once again, this is symmetric and positive definite. Notice that when $q=0$ we have $$\Delta (0,t)=\prod _{\alpha \in R}\frac{1-e^{\alpha }}{1-t{e}^{\alpha }}=\Delta \left(t\right)$$ so that ${⟨f,g⟩}_{0,t}={⟨f,g⟩}_t.$ Also if $t=q$ (i.e., $k=1$) we have $$\Delta (q,q)=\prod _{\alpha \in R}(1-e^{\alpha })=d\stackrel{—}{d}$$ so that $${⟨f,g⟩}_{q,q}={⟨f,g⟩}_0.$$ Finally if $t=1$ (i.e., $k=0$) we have $$\Delta (q,1)=1$$ and hence ${⟨f,g⟩}_{q,1}={⟨f,q⟩}_1:$ both independent of $q.$

For each $\lambda \in P^{+}$ there exists $P_{\lambda }=P_{\lambda }(q,t)\in A^W$ such that

1. $P_{\lambda }=m_{\lambda }+\sum _{\genfrac{}{}{0ex}{}{\mu \in P^{+}}{\mu <\lambda }}{u}_{\lambda \mu }(q,t)m_{\mu }(u_{\lambda \mu }\in \mathbb{Q}(q,t)).$
2. ${⟨P_{\lambda },P_{\mu }⟩}_{q,t}=0$ if $\lambda \ne \mu .$

This will be a consequence of the following proposition:

There exists a linear operator $E:A^W\to A^W$ satisfying the following three conditions:

1. ${⟨Ef,g⟩}_{q,t}={⟨f,Eg⟩}_{q,t}$ for all $f,g\in A^W$ ($E$ is self adjoint),
2. $E{m}_{\lambda }=c_{\lambda }{m}_{\lambda }+ \; lower\; terms$ ($E$ is triangular relative to the orbit-sums $m_{\lambda }$),
3. $\lambda \ne \mu \Rightarrow c_{\lambda }\ne c_{\mu }$ ($E$ has distinct eigenvalues).

		Proof.
	Let us first show that Proposition 1.6 implies Proposition 1.5, and then address the problem of constructing a suitable $E$ for each root system $R.$ For each $\lambda \in P^{+}$ define $$E_{\lambda }=\prod _{\genfrac{}{}{0ex}{}{\mu <\lambda }{\mu \in P^{+}}}\frac{E-c_{\mu }}{c_{\lambda }-c_{\mu }}$$ and then let $$P_{\lambda }=E_{\lambda }{m}_{\lambda }.$$ Then the $P_{\lambda }$ so defined have the desired properties. Indeed, it is clear from the definition and (b) above that $P_{\lambda }$ satisfies Proposition 1.5 (i). Let $M_{\lambda }$ be the (finite-dimensional) subspace of $A^W$ spanned by the $m_{\mu }$ such that $\mu \le \lambda$ (and $\mu \in P^{+}$). Then $M_{\lambda }$ is stable under $E,$ and the minimal (characteristic polynomial of $E{|}_{M_{\lambda }}$ is $$\prod _{\mu \le \lambda }(X-c_{\mu })$$ since the $c_{\mu }$ are all distinct. Hence $\prod _{\mu \le \lambda }(E-c_{\mu })=0$ on $M_{\lambda },$ i.e. $(E-c_{\lambda })E_{\lambda }=0$ on $M_{\lambda },$ and therefore $$E{P}_{\lambda }=E{E}_{\lambda }{m}_{\lambda }=c_{\lambda }{E}_{\lambda }{m}_{\lambda }=c_{\lambda }{P}_{\lambda }$$ (so the $P_{\lambda }$ are the eigenvectors of $E$). Hence $$\begin{array}{rcl}{c}_{\lambda }⟨P_{\lambda },P_{\mu }⟩& =& ⟨E{P}_{\lambda },P_{\mu }⟩\\ & =& ⟨P_{\lambda },E{P}_{\mu }⟩by\; (a)\\ & =& c_{\mu }⟨P_{\lambda },P_{\mu }⟩.\end{array}$$ Since $c_{\lambda }\ne c_{\mu }$ if $\lambda \ne \mu ,$ it follows that ${⟨P_{\lambda },P_{\mu }⟩}_{q,t}=0.$ $\square$

So it remains to construct a linear operator $E$ on $A^W$ for each reduced irreducible root system $R,$ satisfying the conditions (a), (b), (c) of Proposition 1.6. In fact we need two constructions, according as $f>1$ or $f=1.$

Minuscule weights (of $R^{\vee }$)

Assume $R$ reduced, irreducible. Let $\pi \in P^{\vee }=P\left(R^{\vee }\right).$ $\pi$ is a minuscule weight of $R^{\vee }$ if $$⟨\pi ,\alpha ⟩=0 \; or\; 1,all\; \alpha \in R^{+}.$$ Clearly 0 is always a minuscule weight.

Let $({\omega }_1,...,{\omega }_r)$ be the basis of $V$ dual to $({\alpha }_1,...,{\alpha }_r):$ $⟨{\alpha }_i,{\omega }_j⟩={\delta }_{ij}.$ Then ${\omega }_1,...,{\omega }_r$ is a basis of the lattice $P^{\vee }(=Q^{*}).$ Also let $$\phi =\sum _{i=1}^r{m}_i{\alpha }_i$$ be the highest root of $R,$ and let $I=\{i\in [1,r]|m_i=1\}.$ (It might be empty.)

The minuscule weights of $R^{\vee }$ are 0 and ${\omega }_i (i\in I).$

		Proof.
	Suppose $\pi =\sum _1^r{n}_i{\omega }_i$ is a minuscule weight. Then $n_i=⟨\pi ,{\alpha }_i⟩=0$ or 1 for each $i,$ and $$\sum _1^r{m}_i{n}_i=⟨\pi ,\phi ⟩=0 \; or\; 1.$$ If this is zero, then all the $n_i$ are 0, and $\pi =0.$ If it is 1, then for some $i$ we have $m_i{n}_i=1$ (whence $m_i=n_i=1$) and $m_j{n}_j=0$ for $j\ne i$ (whence $n_j=0,$ since $m_j>0$). So $\pi ={\omega }_i$ and $m_i=1,$ i.e. $i\in I.$ Conversely, if $\pi ={\omega }_i$ and $\alpha =\sum p_j{\alpha }_j\in R^{+},$ we have $0\le p_j\le m_j$ for all $j,$ because $\alpha \le \phi ;$ hence $⟨\pi ,\alpha ⟩=⟨{\omega }_i,\alpha ⟩=p_i=0 \; or\; 1$ (since $m_i=1$). $\square$

The minuscule weights have other nice properties:

1. They are coset representatives of $Q^{\vee }$ in $P^{\vee },$ hence there are $f$ of them (i.e. $f-1$ nonzero minuscule weights).
2. Let $\lambda \in P^{+}.$ Then $\lambda$ is a minuscule weight (for $R,$ not $R^{\vee }$) if and only if ${\chi }_{\lambda }=m_{\lambda }$ (i.e. the weights form a single $W-$orbit).
3. They are the minimal elements of $P^{+}$ with respect to: $\lambda \ge \mu \iff \lambda -\mu \in Q^{+}.$

Examples. $$\begin{array}{cccccccccc}R& A_n& B_n& C_n& D_n& E_6& E_7& E_8& F_4& G_2\\ f& n+1& 2& 2& 4& 3& 2& 1& 1& 1\end{array}$$ So there are nonzero weights except in types $E_8,F_4,G_2.$
$A_n:$ all fundamental weights are minuscule and property (2) above translate into the fact that $s_{\lambda }=m_{\lambda } \iff \lambda =\left(1^r\right),$ some $r\ge 0.$

First construction

Suppose $f>1$ (i.e. that $R$ is not of type $E_8, F_4$ or $G_2$) and let $\pi$ be a nonzero minuscule weight of $R^{\vee },$ so that $$\begin{array}{ll}⟨\pi ,\alpha ⟩=0or1,all\alpha \in R^{+}.& (OP\; 3)\end{array}$$ Define $T_{\pi }:A\to A$ by $$T_{\pi }{e}^{\lambda }=q^{⟨\lambda ,\pi ⟩}{e}^{\lambda }(\lambda \in P)$$ extended by linearity. Clearly $$T_{\pi }\left(e^{\lambda +\mu }\right)=T_{\pi }\left(e^{\lambda }\right)\cdot T_{\pi }\left(e^{\mu }\right)$$ so that $T_{\pi }$ is an automorphism of the $ℝ-$algebra $A.$ Moreover, $T_{\pi }$ is self-adjoint for the scalar product $${⟨f,g⟩}_1=\frac{1}{\left|W\right|}{\left[f\stackrel{—}{g}\right]}_1.$$ For if $f=\sum f_{\lambda }{e}^{\lambda }, g=\sum g_{\mu }{e}^{\mu }$ then $${⟨T_{\pi }f,g⟩}_1=\frac{1}{\left|W\right|}{\left[\left(T_{\pi }f\right)\stackrel{—}{g}\right]}_1=\frac{1}{\left|W\right|}{ [\sum _{\lambda ,\mu }{q}^{⟨\lambda ,\mu ⟩}{f}_{\lambda }{g}_{\mu }{e}^{\lambda -\mu }] }_1=\frac{1}{\left|W\right|}\sum _{\lambda }{q}^{⟨\lambda ,\pi ⟩}{f}_{\lambda }{g}_{\lambda }$$ which is symmetrical in $f$ and $g,$ hence equal to ${⟨T_{\pi }g,f⟩}_1={⟨f,T_{\pi }g⟩}_1.$

If we interpret $e^{\lambda }$ as a function on $V$ by the rule — different from the previous one — $$e^{\lambda }\left(x\right)=q^{⟨\lambda ,x⟩}(x\in V)$$ then $$\left(T_{\pi }{e}^{\lambda }\right)\left(x\right)=q^{⟨\lambda ,\pi ⟩}\cdot q^{⟨\lambda ,x⟩}=q^{⟨\lambda ,x+\pi ⟩}=e^{\lambda }(x+\pi )$$ and hence by linearity $$\left(T_{\pi }f\right)\left(x\right)=f(x+\pi )$$ for all $f\in A,$ i.e. $T_{\pi }$ is a translation operator.

From above we have $$T_{\pi }{e}^{\alpha }=\{\begin{array}{ll}q{e}^{\alpha },& if⟨\pi ,\alpha ⟩=1,\\ e^{\alpha }& if⟨\pi ,\alpha ⟩=0,\end{array}\phantom{\}}$$ and hence $$\begin{array}{ll}\frac{T_{\pi }{(e^{\alpha };q)}_{\infty }}{{(e^{\alpha };q)}_{\infty }}=\{\begin{array}{ll}\frac{1}{1-e^{\alpha }},& if⟨\pi ,\alpha ⟩=1,\\ 1,& if⟨\pi ,\alpha ⟩=0.\end{array}\phantom{\}}& (OP\; 4)\end{array}$$ Now let $${\Delta }^{+}=\prod _{\alpha \in R^{+}}\frac{{(e^{\alpha };q)}_{\infty }}{{(t{e}^{\alpha };q)}_{\infty }}(\in Aift=q^k,k\in \mathbb{N})$$ (so that $\Delta ={\Delta }^{+},{\stackrel{—}{\Delta }}^{+}$) and let $$\begin{array}{rcl}{\Phi }_{\pi }& =& \frac{T_{\pi }{\Delta }^{+}}{{\Delta }^{+}}\\ & =& \frac{\left(\prod _{\alpha \in R^{+}}\frac{T_{\pi }{(e^{\alpha };q)}_{\infty }}{{(e^{\alpha };q)}_{\infty }}\right)}{\left(\frac{T_{\pi }{(t{e}^{\alpha };q)}_{\infty }}{{(t{e}^{\alpha };q)}_{\infty }}\right)}\\ & =& \prod _{\genfrac{}{}{0ex}{}{\alpha \in R^{+}}{⟨\pi ,\alpha ⟩=1}}\frac{1-t{e}^{\alpha }}{1-e^{\alpha }}by(OP\; 4)above\end{array}$$ or equivalently $$\begin{array}{ll}{\Phi }_{\pi }=\prod _{\alpha \in R^{+}}\frac{1-t^{⟨\pi ,\alpha ⟩}{e}^{\alpha }}{1-e^{\alpha }}.& (OP\; 5)\end{array}$$ We now define, for $f\in A,$ $$E_{\pi }f=\sum _{w\in W}w\left({\Phi }_{\pi }{T}_{\pi }f\right)=\sum _{w\in W}w\left(\frac{T_{\pi }\left({\Delta }^{+}f\right)}{{\Delta }^{+}}\right)$$ (conjugate $T_{\pi }$ by ${\Delta }^{+}$ and then sum over the Weyl group). We shall show that $E_{\pi }$ maps $A^W$ into $A^W$ and satisfies conditions (a) and (b) (self adjointness and triangularity) of Proposition 1.6, but not always (c) (distinct eigenvalues)).

Let us first verify that $E_{\pi }$ is self-adjoint. Since $\Delta ={\Delta }^{+}\cdot \stackrel{—}{{\Delta }^{+}}$ and $\Delta =w\Delta (w\in W)$ we have $$\Delta =w{\Delta }^{+}\cdot \stackrel{—}{w{\Delta }^{+}}$$ and hence $(f,g\in A^W)$ $$\begin{array}{rcl}{⟨E_{\pi }f,g⟩}_{q,t}& =& {\left|W\right|}^{-1}{\left[\left(E_{\pi }f\right)\Delta \stackrel{—}{g}\right]}_1\\ & =& {\left|W\right|}^{-1}\sum _{w\in W}{\left[w\left(T_{\pi }\left({\Delta }^{+}f\right)\right)\stackrel{—}{w\left({\Delta }^{+}g\right)}\right]}_1(wg=g)\\ & =& {[T_{\pi }\left({\Delta }^{+}f\right)\cdot \stackrel{—}{{\Delta }^{+}g}]}_1\\ & =& \left|W\right|{⟨T_{\pi }\left({\Delta }^{+}f\right),{\Delta }^{+}g⟩}_1.\end{array}$$ Since ${⟨T_{\pi }u,v⟩}_1={⟨u,T_{\pi }v⟩}_1$ as we saw earlier, it follows that ${⟨E_{\pi }f,g⟩}_{q,t}={⟨f,E_{\pi }g⟩}_{q,t}.$

Next we have to compute $E_{\pi }{e}^{\mu } (\mu \in P).$ We have from (OP 5) $$\begin{array}{rcl}{\Phi }_{\pi }& =& \prod _{\alpha \in R^{+}}\frac{1-t^{⟨\pi ,\alpha ⟩}{e}^{\alpha }}{1-e^{\alpha }}\\ & =& t^{⟨\pi ,2\rho ⟩}\prod _{\alpha \in R^{+}}\frac{1-t^{-⟨\pi ,\alpha ⟩}{e}^{-\alpha }}{1-e^{-\alpha }}\\ & =& \frac{t^{⟨\pi ,2\rho ⟩}}{d}\sum _{X\subseteq R^{+}}{(-1)}^{\left|X\right|}{t}^{-⟨\pi ,{\sigma }_x⟩}{e}^{\rho -{\sigma }_X}\end{array}$$ where as before ${\sigma }_X=\sum _{\alpha \in X}\alpha$ for $X\subseteq R^{+}.$ Hence $$\begin{array}{rcl}{E}_{\pi }{e}^{\mu }& =& \sum _{w\in W}w({\Phi }_{\pi }\cdot q^{⟨\pi ,\mu ⟩}{e}^{\mu })\\ & =& q^{⟨\pi ,\mu ⟩}{t}^{⟨\pi ,2\rho ⟩}\sum _{w\in W}w\left(d^{-1}\sum _{X\subseteq R^{+}}{(-1)}^{\left|X\right|}{t}^{-⟨\pi ,{\sigma }_X⟩}{e}^{\mu +\rho -{\sigma }_X}\right)\\ & =& q^{⟨\pi ,\mu ⟩}{t}^{⟨\pi ,2\rho ⟩}\sum _{X\&sunseteq;R^{+}}{(-1)}^{\left|X\right|}{t}^{-⟨\pi ,{\sigma }_X⟩}{\chi }_{\mu -{\sigma }_X}\end{array}$$ which shows that $E_{\pi }$ maps $A$ into $A^W.$

Now let $\lambda \in P^{+},$ then $$\begin{array}{rcl}{E}_{\pi }\left(m_{\lambda }\right)& =& \sum _{\mu \in w\lambda }{E}_{\pi }{e}^{\mu }\\ & =& t^{⟨\pi ,2\rho ⟩}\sum _{X\subseteq R^{+}}{(-1)}^{\left|X\right|}{t}^{-⟨\pi ,{\sigma }_X⟩}\sum _{\mu \in w\lambda }{q}^{⟨\pi ,\mu ⟩}{\chi }_{\mu -{\sigma }_X}.\end{array}$$ Consider ${\chi }_{\mu -{\sigma }_X}:$ if $\mu +\rho -{\sigma }_X$ is singular (i.e. not regular) then ${\chi }_{\mu -{\sigma }_X}=0.$ If $\mu +\rho -{\sigma }_X$ is regular, then there exists $w\in W$ and $\nu \in P^{+}$ such that $$w(\mu +\rho -{\sigma }_X)=\nu +\rho ,and{\chi }_{\mu -{\sigma }_X}=\epsilon \left(w\right){\chi }_{\nu }.$$ Now $w(\rho -{\sigma }_X)=\rho -{\sigma }_Y$ for some $Y\subseteq R^{+},$ hence $$\begin{array}{rcl}\nu & =& w(\mu +\rho -{\sigma }_X)-\rho \\ & =& (w\mu +\rho -{\sigma }_Y)-\rho \\ & =& w\mu -{\sigma }_Y\\ & \le & w\mu \le \lambda \end{array}$$ and it follows that $E_{\pi }{m}_{\lambda }$ is a linear combination of the ${\chi }_{\nu }, \nu \in P^{+}, \nu \le \lambda ,$ and hence also a linear combination of the $m_{\nu } (\nu \in P^{+},\nu \le \lambda ).$

Moreover in the above calculations we shall have $\nu =\lambda$ if and only if $w\mu =\lambda$ and ${\sigma }_Y=0,$ i.e. $w(\rho -{\sigma }_X)=\rho ,$ or ${\sigma }_X=\rho -w^{-1}\rho$ which by (4.11) forces $X=R\left(w^{-1}\right)$ and hence $\left|X\right|=l\left(w\right).$ So the coefficient of $X_{\lambda }$ (or equivalently of $m_{\lambda }$) in $E_{\pi }{m}_{\lambda }$ is $$\begin{array}{rcl}{c}_{\lambda }& =& t^{⟨\pi ,2\rho ⟩}\sum _{w\in W}{(-1)}^{l\left(w\right)}{t}^{-⟨\pi ,\rho -w^{-1}\rho ⟩}{q}^{⟨\pi ,w^{-1}\lambda ⟩}\epsilon \left(w\right)\\ & =& t^{⟨\pi ,\rho ⟩}\sum _{w\in W}{q}^{⟨w\pi ,\lambda ⟩}{t}^{⟨w\pi ,\rho ⟩}\\ & =& t^{⟨\pi ,\rho ⟩}\sum _{w\in W}{q}^{⟨w\pi ,\lambda +k\rho ⟩}(t=q^k)\\ & & [=t^{⟨\pi ,\rho ⟩}{\tilde{m}}_{\pi }(\lambda +k\rho )].\end{array}$$ It remains to examine whether $E_{\pi }$ satisfies condition (c) of Proposition 1.6, i.e. whether the $c_{\lambda }, \lambda \in P^{+},$ are all distinct. In fact this is true in all cases except $D_n (n\ge 4)$ (and $E_8, F_4, G_2$ where there is no nonzero minuscule weight). With $\pi$ as above let $$p_r\left(x\right)=\sum _{w\in W}{⟨x,w\pi ⟩}^r(x\in V).$$ Clearly the $p_r, r\ge 1,$ are $W-$invariant polynomial functions on $V,$ i.e. $p_r\in ℐ$ in the notation of Chapter III. (Also $p_1=0.$)

Suppose that $R$ is not of type $D_n (n\ge 4) E_8, F_4$ or $G_2,$ and that if $R$ is of type $A_n$ the minuscule weight $\pi$ is the fundamental weight corresponding to an end node of the Dynkin diagram. Then the $p_r (r\ge 2)$ generate the $ℝ-$algebra $ℐ.$

This is easily checked for $R$ of type $A,B$ or $C.$ For $E_6$ and $E_7$ we refer to M.L. Mehta (Communications in Algebra 16 (1988) 1083-1098).

Suppose now ($D_n, n\ge 4$ excluded) $\lambda ,\mu \in P^{+}$ are such that $c_{\lambda }=c_{\mu },$ i.e. $$\sum _{w\in W}{q}^{⟨\lambda +k\rho ,w\pi ⟩}=\sum _{w\in W}{q}^{⟨\mu +k\rho ,w\pi ⟩}(allq,k).$$ Operate on both sides with ${\left(q\frac{d}{dq}\right)}^r$ and then set $q=1.$ We obtain $$p_r(\lambda +k\rho )=p_r(\mu +k\rho )(r\ge 1).$$ Hence by Proposition 3.2 $$f(\lambda +k\rho )=f(\mu +k\rho )$$ for all $f\in ℐ,$ from which we can conclude that $\lambda =\mu ,$ thanks to the following general lemma:

Let $W$ be a finite group of automorphisms of $V,$ and let $ℐ=S{\left(V\right)}^W$ be the $ℝ-$algebra of $W-$invariant polynomial functions on $V.$ Suppose $x,y\in V$ are such that $f\left(x\right)=f\left(y\right)$ for all $f\in ℐ.$ Then $x,y$ are in the same $W-$orbit in $V$ (i.e. $ℐ$ separates the $W-$orbits in $V.$)

		Proof.
	Let $x_1,...,x_n$ be distinct points of $V$ and let $c_1,...,c_n\in ℝ.$ Then I claim that there exists $g\in S\left(V\right)$ such that To prove this it is enough to construct $g_i\in S\left(V\right) (1\le i\le n)$ such that $g_i\left(x_j\right)={\delta }_{ij},$ for we can then take $g=\sum c_i{g}_i.$ For each ordered pair $(i,j)$ with $1\le i,j\le n$ and $i\ne j$ we can find $u_{ij}\in S\left(V\right)$ such that $u_{ij}\left(x_i\right)=1$ and $u_{ij}\left(x_j\right)=0$ (we may take $u_{ij}$ linear, nonhomogeneous). Then $g_i=\prod _{k\ne i}{u}_{ik}$ does the trick. Now let $(x_1,...,x_n)$ and $(y_1,...,y_m)$ be distinct $W-$orbits in $V.$ From above there exists $g\in S\left(V\right)$ such that $g\left(x_i\right)=1 (1\le i\le n), g\left(y_j\right)=0 (1\le j\le m).$ Let $f=\frac{1}{\left|W\right|}\sum _{w\in W}wg.$ Then $f\in ℐ$ and $f$ takes the value 1 (resp. 0) at each $x_i$ (resp. $y_j$). So $ℐ$ separates the $W-$orbits in $V,$ which is equivalent to the result stated. $\square$

What about $D_n$? Here there are 3 minuscule weights, hence 3 operators $E_{\pi }.$ It turns out that no one of them has distinct eigenvalues, but that a suitable linear combination $E$ of all three does. We omit the details. So we have constructed for each $R$ with $f>1$ an operator $E$ satisfying the three conditions of Proposition 1.6.

Second construction

This time we take $\pi =$ highest root of $R^{\vee },$ so that (2.4) $$⟨\pi ,\alpha ⟩=\{\begin{array}{ll}0or1,& if\alpha \in R^{+},\alpha \ne {\pi }^{\vee },\\ 2,& if\alpha ={\pi }^{\vee }.\end{array}\phantom{\}}$$ As before, let ${\Phi }_{\pi }=\frac{T_{\pi }{\Delta }^{+}}{{\Delta }^{+}}$ and define now $(f\in A)$ $$E_{\pi }f=\sum _{w\in W}\left({\Phi }_{\pi }(T_{\pi }f-f)\right).$$ Then it turns out, as a result of calculations analogous to those we have already done, that $E_{\pi }$ has the 3 desired properties in the cases $(E_8,F_4,G_2)$ not covered by the previous construction. So the existence of the $P_{\lambda }(q,t)$ claimed by Proposition 1.5 is established in all cases.

What about ${⟨P_{\lambda },P_{\lambda }⟩}_{q,t}?$ Here I can only conjecture

1. (C2q) Let $\lambda \in P^{+}.$ Then $(t=q^k)$ $${⟨P_{\lambda },P_{\lambda }⟩}_{q,t}=\prod _{\alpha \in R^{+}}\prod _{i=1}^{k-1}\frac{1-q^{⟨\lambda +k\rho ,{\alpha }^{\vee }⟩+i}}{⟨\lambda +k\rho ,{\alpha }^{\vee }⟩-i}.$$ True if $R$ of type $A.$

Consider the special case $\lambda =0$ of (C2q). Since $P_0=1$ we have ${⟨P_0,P_0⟩}_{q,t}=\frac{1}{\left|W\right|}{\left[\Delta \right]}_1$ and hence (C2q) would give $$\begin{array}{ll}\frac{1}{\left|W\right|}{\left[\Delta \right]}_1=\prod _{\alpha \in R^{+}}\prod _{i=1}^{k-1}=\frac{1-q^{⟨k\rho ,{\alpha }^{\vee }⟩+i}}{⟨k\rho ,{\alpha }^{\vee }⟩-i}.& (OP\; 6)\end{array}$$ I claim that (OP 6) is equivalent to (C1q). Let $$\Delta \prime =\prod _{\alpha \in R^{+}}\prod _{i=0}^{k-1}(1-q^i{e}^{\alpha })(1-q^{i+1}{e}^{-\alpha })$$ then $$\begin{array}{rcl}\frac{\Delta \prime }{\Delta }& =& \prod _{\alpha \in R^{+}}\prod _{i=0}^{k-1}\frac{1-q^{i+1}{e}^{-\alpha }}{1-e^{-\alpha }}(t=q^k)\end{array}$$ and therefore (since $\Delta$ is $W-$invariant) $$\begin{array}{rcl}{\Delta }^{-1}\sum _{w\in W}w\Delta \prime & =& \sum _{w\in W}w\left(\frac{1-t{e}^{-\alpha }}{1-e^{-\alpha }}\right)\\ & =& W\left(t\right)by\; (4.12)\end{array}$$ i.e. $$W\left(t\right)\Delta =\sum _{w\in W}w\Delta \prime$$ and therefore $$W\left(t\right){\left[\Delta \right]}_1=\left|W\right|{[\Delta \prime ]}_1$$ (because $\Delta \prime , w\Delta \prime$ have the same constant term). It follows from (OP 6) that $$\begin{array}{rcl}{[\Delta \prime ]}_1& =& \frac{W\left(t\right)}{\left|W\right|}{\left[\Delta \right]}_1=\prod _{\alpha \in R^{+}}\frac{1-q^{k\mathrm{ht}\left({\alpha }^{\vee }\right)+k}}{1-q^{k\mathrm{ht}\left({\alpha }^{\vee }\right)}}{\left[\Delta \right]}_{1}by\; (4.13)\\ & =& \prod _{\alpha \in R^{+}}\prod _{i=0}^{k-1}\frac{1-q^{k\mathrm{ht}\left({\alpha }^{\vee }\right)+i+1}}{1-q^{k\mathrm{ht}\left({\alpha }^{\vee }\right)-i}}.\end{array}$$ Consider then the polynomial $$\begin{array}{rcl}\sum _{\alpha \in R^{+}}\sum _{i=0}^{k-1}(q^{k\mathrm{ht}\left({\alpha }^{\vee }\right)+i+1}-q^{k\mathrm{ht}\left({\alpha }^{\vee }\right)-i})& =& \sum _{\alpha \in R^{+}}{q}^{k\mathrm{ht}\left({\alpha }^{\vee }\right)}(q^k+q^{k-1}+\cdots +q-1-q^{-1}-\cdots -q^{1-k})\\ & =& \sum _{\alpha \in R^{+}}{q}^{k\mathrm{ht}\left({\alpha }^{\vee }\right)}(q^k-1)(1+q^{-1}+\cdots +q^{1-k}).\end{array}$$ Now since $$\prod _{\alpha \in R^{+}}\frac{1-t^{\mathrm{ht}\left({\alpha }^{\vee }\right)+1}}{1-t^{\mathrm{ht}\left({\alpha }^{\vee }\right)}}=W\left(t\right)=\prod _{i=1}^r\frac{1-t^{d_i}}{1-t}$$ it follows that $$\sum _{\alpha \in R^{+}}(t^{\mathrm{ht}\left({\alpha }^{\vee }\right)+1}-t^{\mathrm{ht}\left({\alpha }^{\vee }\right)})=\sum _{i=1}^r{t}^{d_i}-t$$ i.e. that $$\sum _{\alpha \in R^{+}}{q}^{k\mathrm{ht}\left({\alpha }^{\vee }\right)}(q^k-1)=\sum _{i=1}^r(q^{k{d}_i}-q^k).$$ Hence (OP 6) becomes $$\sum _{i=1}^r(q^{k{d}_i}-q^k)(1+q^{-1}+\cdots +q^{1-k})=\sum _{i=1}^r\{(q^{k{d}_i}+q^{k{d}_i-1}+\cdots +q^{k{d}_i-k+1})-(q+q^2+\cdots +q^k)\}$$ and therefore $$\prod _{\alpha \in R^{+}}\prod _{i=0}^{k-1}\frac{1-q^{k\mathrm{ht}\left({\alpha }^{\vee }\right)+i+1}}{1-q^{k\mathrm{ht}\left({\alpha }^{\vee }\right)-i}}=\prod _{i=1}^r\frac{(1-q^{k{d}_i})\cdots (1-q^{k{d}_i-k+1})}{(1-q)\cdots (1-q^k)}=\prod _{i=1}^r\left[\genfrac{}{}{0ex}{}{k{d}_i}{k}\right].$$ So (C2q) $(\lambda =0)=$ (C1q). $$(Tip of the iceberg!)

Consider ${\left|P_{\lambda }\right|}^2$ as $k\to \infty$ (i.e. $t\to 0$). We have $\Delta (q,0)=\prod _{\alpha \in R}{(e^{\alpha };q)}_{\infty }.$ (continuous function on $T=V/Q^{\vee }$) $$\begin{array}{rcl}\prod _{i=1}^{k-1}(1-q^{⟨\lambda +k\rho ,{\alpha }^{\vee }⟩+i})& =& \frac{{(q^{⟨\lambda ,{\alpha }^{\vee }⟩+1}{t}^{⟨\rho ,{\alpha }^{\vee }⟩};q)}_{\infty }}{{(q^{⟨\lambda ,{\alpha }^{\vee }⟩+1}{t}^{⟨\rho ,{\alpha }^{\vee }⟩+1};q)}_{\infty }}\to 1\\ \prod _{i=1}^{k-1}(1-q^{⟨\lambda +k\rho ,{\alpha }^{\vee }⟩-i})& =& \frac{{(q^{⟨\lambda ,{\alpha }^{\vee }⟩+1}{t}^{⟨\rho ,{\alpha }^{\vee }⟩-1};q)}_{\infty }}{{(q^{⟨\lambda ,{\alpha }^{\vee }⟩+1}{t}^{⟨\rho ,{\alpha }^{\vee }⟩};q)}_{\infty }}\to 1\\ & \to & \{\begin{array}{ll}1,& if\alpha \ne {\alpha }_i,\\ {(q^{⟨\lambda ,{\alpha }^{\vee }⟩+1};q)}_{\infty },& if\alpha ={\alpha }_i.\end{array}\phantom{\}}\end{array}$$ Hence $$\begin{array}{rcl}{⟨P_{\lambda },P_{\lambda }⟩}^2& \to & \prod _{i=1}^r{(q^{⟨\lambda ,{\alpha }_i^{\vee }⟩+1};q)}_{\infty }^{-1}\\ & & =\prod _{i=1}^r{(q^{n_i+1};q)}_{\infty }^{-1}if\lambda =\sum _1^r{n}_i{\pi }_i(n_i\ge 0).\end{array}$$ True when $\lambda =0.$ (From the denominator formula for affine root systems.)

The limiting case $q\to 1$

As $q\to 1$ we have $(t=q^k)$ $$\Delta (q,t)=\prod _{\alpha \in R}{(e^{\alpha };q)}_k\to \Delta \left(k\right)=\prod _{\alpha \in R}{(1-e^{\alpha })}^k$$ as kernel for the scalar product. So we have $P_{\lambda }^{\left(k\right)}=\underset{q\to 1}{\lim }{P}_{\lambda }(q,q^k).$ When $R$ is of type $A$ these are Jack's symmetric functions (suitably normalized) with parameter $\alpha =\frac{1}{k}.$ When $q\to 1,$ (C2q) is in fact true: (Heckman, Opdam) $$\begin{array}{ll}{\left|P_{\lambda }\right|}_{\left(k\right)}^2={⟨P_{\lambda },P_{\lambda }⟩}_{\left(k\right)}=\prod _{\alpha \in R^{+}}\prod _{i=1}^{k-1}\frac{⟨\lambda +k\rho ,{\alpha }^{\vee }⟩+i}{⟨\lambda +k\rho ,{\alpha }^{\vee }⟩-i}(\lambda \in P^{+},k\in \mathbb{N}).& (OP\; 7)\end{array}$$ Proof is now elementary — depends on their notion of shift operators which enables one to compute the quotient $$\frac{{\left|P_{\lambda }\right|}_k^2}{{\left|P_{\lambda +\rho }\right|}_{k-1}^2}$$ and hence prove (OP 7) (for $k$ a positive integer) by induction on $k$ (when $k=1, P_{\lambda }^{\left(k\right)}={\chi }_{\lambda }$ and ${\left|{\chi }_{\lambda }\right|}^2=1.$)

Exercises

$D$ contains no cycles (hence is a tree if $R$ is irreducible).

		Proof.
	Suppose $D$ contains a cycle of length $m\ge 3.$ Then with a suitable numbering of the simple roots ${\alpha }_i$ we have $$⟨{\alpha }_i,{\alpha }_{i+1}⟩<0(1\le i\le m-1)and⟨{\alpha }_m,{\alpha }_1⟩<0.$$ Let ${\theta }_i=$ angle between ${\alpha }_i$ and ${\alpha }_{i+1} ({\alpha }_{m+1}={\alpha }_1).$ Then ${\theta }_i$ is $>\frac{\pi }{2},$ hence $\ge \frac{2\pi }{3}$ and therefore $\cos {\theta }_i\le -\frac{1}{2} (1\le i\le m).$ Equivalently, if $e_i=\frac{{\alpha }_i}{\left|{\alpha }_i\right|},$ we have $$⟨e_i,e_{i+1}⟩\le -\frac{1}{2}(1\le i\le m)$$ and $⟨e_i,e_i⟩=1, ⟨e_i,e_j⟩\le 0$ if $|i-j|\ge 2.$ Consider the vector $v=e_1+\cdots +e_m:$ we have $$\begin{array}{rcl}{\left|v\right|}^2& =& \sum _{i,j=1}^m⟨e_i,e_j⟩\\ & =& m+2\sum _{i<j}⟨e_i,e_j⟩\\ & \le & m+2\sum _{i=1}^m⟨e_i,e_{i+1}⟩(e_{m+1},e_1)\\ & \le & m+2m-\frac{1}{2}=0.\end{array}$$ So ${\left|v\right|}^2=0$ and hence $v=0,$ impossible as the $c_i$ are linearly independent. $\square$

Let $R$ be an irreducible root system; $\alpha ,\beta \in R.$ If $\left|\alpha \right|=\left|\beta \right|$ then $\beta =w\alpha$ for some $w\in W.$

		Proof.
	Since $R$ is irreducible, $V$ is an irreducible $W-$module (2.1). Hence $W\alpha$ spans $V,$ hence there exists $w\in W$ such that $⟨w\alpha ,\beta ⟩\ne 0.$ Replace $\alpha$ by $w\alpha ,$ we may assume $⟨\alpha ,\beta ⟩\ne 0.$ If $\alpha =\beta$ we are done; if $\alpha \ne \beta$ we have $⟨{\alpha }^{\vee },\beta ⟩=⟨\alpha ,{\beta }^{\vee }⟩=\pm 1$ by (1.2)(i) $(⟨{\alpha }^{\vee },\beta ⟩⟨\alpha ,{\beta }^{\vee }⟩=1,2or3).$ Replacing $\alpha$ by $-\alpha =s_{\alpha }\alpha$ if necessary, we may assume that $⟨{\alpha }^{\vee },\beta ⟩=⟨\alpha ,{\beta }^{\vee }⟩=+1.$ Then we have $s_{\beta }\left(\alpha \right)=\alpha -\beta , s_{\alpha }{s}_{\beta }\left(\alpha \right)=s_{\alpha }(\alpha -\beta )=-\alpha -(\beta -\alpha )=-\beta , s_{\beta }{s}_{\alpha }{s}_{\beta }\left(\alpha \right)=\beta .$ $\square$

Can use this, together with (1.21a), to find the orders of the Weyl groups of the exceptional root systems. (Use $\phi \in \stackrel{—}{C}$.)

$E_6:$ 72 roots, forming a single orbit. The extended Dynkin diagram shows that the highest root $\phi$ is orthogonal to all ${\alpha }_i$ except ${\alpha }_6,$ hence $\left|W_{\phi }\right|=\left|W\left(A_5\right)\right|=6!$ So $$\left|W\left(E_6\right)\right|=72\cdot 6!=2^7\cdot 3^4\cdot 5.$$

$E_7:$ 126 roots, forming a single orbit. Here $\left|W_{\phi }\right|=\left|W\left(D_6\right)\right|=2^5\cdot 6!$ So $$\left|W\left(E_7\right)\right|=2^5\cdot 6!\cdot 126=2^{10}\cdot 3^4\cdot 5\cdot 7=7!\cdot 4!\cdot 4!.$$

$E_8:$ 240 roots, $\left|W_{\phi }\right|=\left|W\left(E_7\right)\right|.$ So $$\left|W\left(E_8\right)\right|=240\cdot \left|W\left(E_7\right)\right|=8!\cdot 6!\cdot 4!=2^{14}\cdot 3^5\cdot 5^2\cdot 7.$$

$F_4:$ 24 long roots, $\left|W_{\phi }\right|=\left|W\left(C_3\right)\right|$ implies $$\left|W\left(F_4\right)\right|=24\cdot \left|W\left(C_3\right)\right|=24\cdot 8\cdot 3!=1152=2^7\cdot 3^2.$$

Let $\alpha \in R$ (reduced), say $\alpha =\sum _{i=1}^r{p}_i{\alpha }_i.$ The support of $\alpha$ is the subgraph $S=\mathrm{Supp}\left(\alpha \right)$ of $D$ (= Dynkin diagram of $R$) obtained by deleting the vertices ${\alpha }_i$ of $D$ for which $m_i=0.$

1. Show that $\mathrm{Supp}\left(\alpha \right)$ is connected, for each $\alpha \in R.$
2. Let $I\subseteq [1,r]$ be such that $D_I$ is connected ($D_I:$ delete verticies outside $I$). Show that $\sum _{i\in I}{\alpha }_i\in R.$

		Proof.
	May assume that $\alpha \in R^{+}$ (because clearly $\mathrm{Supp}(-\alpha )=\mathrm{Supp}\left(\alpha \right)$). Induction on $\mathrm{ht}\left(\alpha \right)=\sum p_i.$ If $\mathrm{ht}\left(\alpha \right)=1$ then $\alpha ={\alpha }_i$ and $S$ consists of one point. So assume $\mathrm{ht}\left(\alpha \right)>1,$ then $S$ has at least two elements. We have $\alpha =\sum _{i\in S}{p}_i{\alpha }_i$ say $(p_i>0)$ and hence $$\sum p_i⟨{\alpha }_i,\alpha ⟩=⟨\alpha ,\alpha ⟩>0$$ so that $⟨{\alpha }_i,\alpha ⟩>0$ for some $i\in S$ and hence (1.4) $\beta =\alpha -{\alpha }_i\in R:$ indeed $\beta \in R^{+},$ because the coefficients of $\beta$ in terms of the ${\alpha }_i$ are $\ge 0.$ If $p_i\ge 2$ then $\mathrm{Supp}\left(\beta \right)=\mathrm{Supp}\left(\alpha \right)=S,$ and $\mathrm{ht}\left(\beta \right)<\mathrm{ht}\left(\alpha \right),$ hence by the inductive hypothesis $S$ is connected. If $p_i=1$ then $\beta -{\alpha }_i\notin R,$ hence the ${\alpha }_i-$string through $\beta$ is $$\beta ,\beta +{\alpha }_i(=\alpha ),...,\beta +p{\alpha }_i$$ say, where $p\ge 1.$ By (1.5) $p=⟨\beta ,{\alpha }_i^{\vee }⟩,$ hence $⟨\beta ,{\alpha }_i⟩>0$ and since $\beta =\sum _{\genfrac{}{}{0ex}{}{j\in S}{j\ne i}}{p}_j{\alpha }_j$ it follows that $⟨{\alpha }_j,{\alpha }_i⟩>0$ for some $j\in S, j\ne i.$ Now $\mathrm{Supp}\left(\beta \right)=S-\left\{i\right\}$ is connected, by the induction hypothesis; and $i$ is joined to some $j\in S.$ Hence $S$ is connected. Induction on $\left|I\right|.$ OK if $\left|I\right|=1,$ so assume $\left|I\right|\ge 2.$ Since $D$ contains no cycles, neither does $D_I,$ i.e. $D_I$ is a tree. Let ${\alpha }_i$ be an end vertex of $D_I, J=I-\left\{i\right\}.$ Then $D_J$ is connected, hence ${\alpha }_J\in R,$ and $⟨{\alpha }_i,{\alpha }_j⟩<0$ for some $j\in J,$ so that $$⟨{\alpha }_i,{\alpha }_J⟩=\sum _{k\in J}⟨{\alpha }_i,{\alpha }_k⟩<0$$ and therefore (1.4) ${\alpha }_I={\alpha }_i+{\alpha }_J\in R.$ $\square$

$R$ of type $A_n,$ then all positive roots are as in (ii). For the other root systems this is not so.

(Parabolic subgroups of $W$). Let $I$ be a subset of $[1,r].$ Let $$\begin{array}{rcl}{B}_I& =& \left\{{\alpha }_i\right|i\in I\}\\ V_I& =& subspace\; ofVspanned\; by{B}_I\\ R_I& =& R\cap V_I\\ W_I& =& subgroup\; ofWgenerated\; by{s}_i,i\in I.\end{array}$$ Then

1. $R_I$ is a root system in $V_I$ with basis $B_I$ and Weyl group $W_I,$ if $I\ne 0.$
2. $W_I\ne W_J$ if $I\ne J.$
3. The subgroup of $W$ generated by $W_I$ and $W_J$ is $W_{I\cup J}.$
4. $W_I\cap W_J=W_{I\cap J}.$

Thus the $W_I$ form a lattice of $2^r$ subgroups of $W$ (parabolic subgroups).

		Proof of (i).
	(i) Check the axioms: $R_I$ clearly satisfies (R1), also (R2) because if $\alpha ,\beta \in R_I$ then $s_{\alpha }\left(\beta \right)=\beta -⟨{\alpha }^{\vee },\beta ⟩\alpha \in R\cap V_I (because\alpha ,\beta \in V_I)=R_I.$ $B_I$ is a basis of $V_I,$ hence every $\alpha \in R_I$ is a linear combination of the ${\alpha }_i, i\in I.$ The coefficients must be integers all of the same sign, hence $B_I$ is a basis of $R_I.$ Hence $W_I$ is the Weyl group of $R_I.$ $\square$

Claim. Let $\left({\pi }_i\right)$ be the basis of $V$ dual to the basis $\left({\alpha }_i\right),$ so that $⟨{\alpha }_i,{\pi }_j⟩={\delta }_{ij}.$ If $w\in W$ then $$w\in W_I\iff w{\pi }_j={\pi }_{j}all\; j\notin I.$$

		Proof of claim.
	⇒) Let $j\notin I.$ If $i\in I$ then $s_i{\pi }_j={\pi }_j$ (because ${\alpha }_i\perp {\pi }_j$). Since $W_I$ is generated by the $s_i (i\in I)$ it follows that $w{\pi }_j={\pi }_j$ for all $w\in W_I.$ ⇐) Let $x=\sum _{j\notin I}{\pi }_j.$ If $w$ fixes ${\pi }_j$ for all $j\notin I$ then $w$ fixes $x,$ hence (1.21a) $w$ is a product of reflections $s_{\alpha }$ fixing $x.$ So suppose $\alpha \in R^{+},$ say $\alpha =\sum _{i=1}^r{m}_i{\alpha }_i.$ If $s_{\alpha }$ fixes $x,$ then $⟨\alpha ,x⟩=0,$ but $⟨\alpha ,x⟩=\sum _{j\notin I}{m}_j,$ hence $m_j=0$ for each $j\notin I,$ so that $\alpha \in R\cap V_I=R_I$ and $s_{\alpha }\in W_I.$ Hence $w\in W_I.$ $\square$

Now do (ii)-(iv):

		Proof of (ii)-(iv).
	(ii) If $I\ne J$ we may assume there exists $j\in J, j\notin I.$ We have $s_j\in W_J$ but $s_j\notin W_I.$ (Otherwise $s_j$ would fix ${\pi }_j,$ hence $⟨{\alpha }_j,{\pi }_j⟩=0.$) So $W_I\ne W_J.$ (iv) Suppose $w\in W_I\cap W_J,$ then $w$ fixes ${\pi }_j$ whenever $j\notin I$ or $j\notin J$ hence whenever $j\notin I\cap J.$ Hence $w\in W_{I\cap J}.$ $\square$

Descriptions of the exceptional root systems

Let $R$ be a reduced irreducible root system with basis ${\alpha }_1,...,{\alpha }_r, \phi$ the highest root or $R, {\alpha }_0=-\phi .$ Let $\stackrel{—}{D}$ be the extended Dynkin diagram of $R,$ with vertices corresponding to ${\alpha }_i (0\le i\le r)$ and labels $m_i,$ where $$\begin{array}{ll}\sum _{i=0}^r{m}_i\alpha =0(m_0=1).& (OP\; 8)\end{array}$$ If we remove the vertex ${\alpha }_i$ of $\stackrel{—}{D}$ we get a (usually disconnected) diagram $D_i$ of a root system $R_i$ in $V$ with basis $B_i=\left\{{\alpha }_j\right|0\le j\le r,j\ne i\},$ and Weyl group $W_i,$ say. (In particular $R_0=R.$) Since $W_i\subseteq W$ and $B_i\subseteq R$ it follows that $R_i=W_i{B}_i\subseteq R.$ From (OP 8) we have $$\begin{array}{ll}{m}_i{\alpha }_i=-\sum _{j\ne i}{m}_j{\alpha }_j\in R_i.& (OP\; 9)\end{array}$$ Consider the root lattices $Q=Q\left(R\right), Q_i=Q\left(R_i\right).$ Since $R_i\subseteq R$ we have $Q_i\subseteq Q,$ and $Q/Q_i$ is a finite group. Let ${\stackrel{—}{\alpha }}_j (1\le j\le r)$ be the image of ${\alpha }_j$ in $Q/Q_i.$ Then ${\stackrel{—}{\alpha }}_j=0$ if $j\ne i,$ and (OP 9) shows that $m_i{\stackrel{—}{\alpha }}_i=0.$ Moreover if $m{\stackrel{—}{\alpha }}_i=0 (m\in \mathbb{Z})$ then $m{\alpha }_i\in Q\left(R_i\right),$ hence $m{\alpha }_i$ is a linear combination of the ${\alpha }_j, j\ne i.$ But (OP 8) is the only linear dependence relation among the ${\alpha }_j (0\le j\le r),$ hence $m$ is a multiple of $m_i.$ It follows that $Q/Q_i$ is a cyclic group of order $m_i,$ generated by ${\stackrel{—}{\alpha }}_i.$

Assume now that ${\left|\alpha \right|}^2=2$ for each $\alpha \in R,$ so that ${\alpha }^{\vee }=\alpha$ and hence $R^{\vee }=R.$ ($R$ is simply laced.) Then the dual lattices of $Q, Q_i$ are $P=P\left(R\right)$ and $P_i=P\left(R_i\right);$ we have $P_i\supseteq P,$ and $P_i/P$ is cyclic of order $m_i$ (4.1), (4.2). So we have the tower of lattices $$\begin{array}{c} \[ P_i \] \[ P \] \[ Q \] \[ Q_i \] \[ m_i \] \[ f \] \[ m_i \]\end{array}$$ and hence if $f_i=f\left(R_i\right)$ is the index of connection of $R_i$ we have $f_i=f{m}_i^2,$ i.e. $$m_i^2=\frac{f_i}{f}.$$

Examples:

1. $R$ of type $E_8:$ the extended Dynkin diagram is $$PICTURE$$ So $R_8=A_8, f_8=8+1=9: f=1$ and hence the root lattice $Q$ of $E_8$ lies between the root lattice $Q_8=Q\left(A_8\right)$ and the weight lattice $P_8=P\left(A_8\right):$ $$P_8{\supset }_{3}P=Q{\supset }_3{Q}_8.$$ This underlies the construction of $E_8$ that I gave earlier: $Q\left(A_8\right)\subseteq {ℝ}^9.$
2. $R$ of type $E_8:$ $R_7=D_8, f_7=f\left(D_8\right)=4,$ $$P_7{\supset }_{2}P=Q{\supset }_2{Q}_7(Bourbaki\text{'}s\; description\; of\; E_8).$$
3. $R$ of type $E_8:$ $R_4=A_4\times A_4, f_4=5\times 5$ $$P_4{>}_{5}P=Q{>}_5{Q}_4.$$ Exercise: Figure out the roots of $E_8$ in this representation.
4. $R$ of type $E_6:$ $$PICTURE$$ Here $R_3=A_2\times A_2\times A_2, f_3=3^3=27, f=|P/Q|=3:$ $$P_3{\supset }_{3}P{\supset }_{3}Q{\supset }_3{Q}_3$$ So we have 3 pairwise orthogonal copies of $A_2$ sitting inside $E_6:$ $A_2$ we can think of as the sixth roots of unity in $ℂ.$ So we may work in ${ℂ}^3(\simeq {ℝ}^6)$ and take ($e_1,e_2,e_3$ basis of ${ℂ}^3$ as $ℂ-$vector space) $${\alpha }_1=e_1,{\alpha }_2=\omega e_1,{\alpha }_4=\omega e_2,{\alpha }_5=e_2,{\alpha }_6=\omega e_3,{\alpha }_0=e_3,$$ where $\omega =\exp \frac{2\pi i}{3}.$
   What about ${\alpha }_3?$ We must have $$\begin{array}{rcl}3{\alpha }_3& =& -({\alpha }_1+2{\alpha }_2+2{\alpha }_4+{\alpha }_5+2{\alpha }_6+{\alpha }_0)\\ & =& -(1+2\omega )(e_1+e_2+e_3)\end{array}$$ so that $${\alpha }_3=\theta (e_1+e_2+e_3)$$ where $\theta =-\frac{1}{3}(1+2\omega )$ (see figure) $$PICTURE$$ In this representation the roots of $E_6$ are $$\begin{array}{cl}\pm \zeta e_i& where{\zeta }^3=1,\\ \pm \theta ({\zeta }_1{e}_1+{\zeta }_2{e}_2+{\zeta }_3{e}_3)& where{\zeta }_1,{\zeta }_2,{\zeta }_{3}are\; cube\; roots\; of\; unity.\end{array}$$ For these are certainly roots, and there are $$2\cdot 3\cdot 3+2\cdot 3^3=18+54=72$$ of them. So they are all the roots of $E_6.$
   
   Exercise: Figure out the action of the Weyl group in this description.
5. $R$ of the type $E_7:$ $$PICTURE$$ Here $R_7=A_7, f_7=7+1=8:$ $$P_7{>}_{2}P{>}_{2}Q{>}_2{Q}_7$$ $Q_7$ is easily described; it is the lattice of all $(n_1,...,n_8)\in {\mathbb{Z}}^8$ such that $\sum n_i=0.$
   Exercise: Figure out the roots of $E_7$ in this representation $$\begin{array}{ll}{e}_i-e_j,& 56\; of\; these,\\ \frac{1}{2}\sum \pm e_{i}with\; 4\; minus\; signs,& 70\; of\; these.\end{array}$$ What are the roots of $E_7$ in this description?
   Take ${\alpha }_i=e_i-e_{i+1} (1\le i\le 6), {\alpha }_0=e_7-e_8,$ then $$\begin{array}{rcl}{\alpha }_7& =& -\frac{1}{2}({\alpha }_1+2{\alpha }_2+3{\alpha }_3+4{\alpha }_4+3{\alpha }_5+2{\alpha }_6+{\alpha }_0)\\ & =& -\frac{1}{2}(e_1+\cdots +e_4-e_5-\cdots -e_8).\end{array}$$ So certainly all of the vectors $$\pm (e_i-e_j) (1\le i<j\le 8)and\frac{1}{2}\sum _{i=1}^8\pm e_i$$ are roots. There are $2\left(\genfrac{}{}{0ex}{}{8}{2}\right)+\left(\genfrac{}{}{0ex}{}{8}{4}\right)=56+70=126$ of these vectors, hence they are the lot.
6. Again take $R=E_7: R_6=D_6\times A_1.$ So we may take $${\alpha }_i=e_i-e_{i+1} (1\le i\le 5),{\alpha }_7=e_5+e_6,{\alpha }_0=f_7=e_7\sqrt{2},$$ and then $$\begin{array}{rcl}{\alpha }_6& =& -\frac{1}{2}({\alpha }_1+2{\alpha }_2+3{\alpha }_3+4{\alpha }_4+3{\alpha }_5+2{\alpha }_7+{\alpha }_0)\\ & =& -\frac{1}{2}(e_1+\cdots +e_6+f_7).\end{array}$$ So certainly the following vectors are roots: $$\begin{array}{lr}\pm e_i\pm e_j(1\le i<j\le 6)& 4\left(\genfrac{}{}{0ex}{}{6}{2}\right)=60of\; these\\ \pm f_7& 2\; of\; these\\ \frac{1}{2}\left(\sum _{i=1}^6\pm e_i\right)\pm \frac{1}{2}{f}_7& 2^5\cdot 2=64of\; these\\ & 126\; total\end{array}$$ So these are all roots.

Orders of the Weyl groups of $E_6, E_7, E_8$

Recall (Exercise): $R$ irreducible, $\alpha ,\beta \in R, \left|\alpha \right|=\left|\beta \right|.$ Then $\alpha ,\beta$ are in the same $W-$orbit. Apply this when $R=E_6,E_7,E_8.$ The roots form a single $W-$orbit. Consider in particular the highest root $\phi =-{\alpha }_0.$ Since $\phi \in \stackrel{—}{C},$ the stabilizer $W_{\phi }$ is generated by the $s_i$ such that ${\alpha }_i\perp \phi .$ So we can read off $W_{\phi }$ from the extended Dynkin diagram, and we have $\mathrm{Card}\left(R\right)=|W/W_{\phi }|,$ hence $$\left|W\right|=\left|W_{\phi }\right|\mathrm{Card}\left(R\right).$$ $E_6:$ $\mathrm{Card}\left(R\right)=72, W_{\phi }=W\left(A_5\right),$ hence $$\left|W\right|=72\cdot 6!=2^7\cdot 3^4\cdot 5.$$ $E_7:$ $\mathrm{Card}\left(R\right)=126, W_{\phi }=W\left(D_6\right),$ hence $$\left|W\right|=126\cdot 2^5\cdot 6!=2^{10}\cdot 3^4\cdot 5\cdot 7=7!\cdot 4!\cdot 4!.$$ $E_8:$ $\mathrm{Card}\left(R\right)=240, W_{\phi }=W\left(E_7\right),$ hence $$\begin{array}{rcl}\left|W\right|& =& 240\cdot 2^{10}\cdot 3^4\cdot 5\cdot 7\\ & =& 2^{14}\cdot 3^5\cdot 5^2\cdot 7=8!\cdot 6!\cdot 4!.\end{array}$$ $F_4:$ 24 long roots, $W_{\phi }=W\left(C_3\right),$ hence $$\left|W\right|=24\cdot 2^3\cdot 3!=1152=2^7\cdot 3^2.$$

The Weyl groups of $E_6, E_7, E_8$

1. For $E_6$ we have $|P/Q|=f=3,$ hence $$P>Q>3P.$$ Since $|P/3P|=3^6$ it follows that $Q/3P$ has order $3^5,$ hence is a 5-dimensional vector space over the field ${𝔽}_3$ of 3 elements. If the scalar product is normalized so that ${\left|\alpha \right|}^2=2$ for all $\alpha \in R,$ it induces a nonsingular scalar product on $Q/3P.$ So $W$ acts on $Q/3P$ as a group of orthogonal transformations, i.e. $$W<O_5\left({𝔽}_3\right).$$ In fact $W$ has index 2 in this group (but it is not $S{O}_5\left({𝔽}_3\right),$ but another subgroup of index 2). Now in general for a Chevalley group $G=G\left({𝔽}_q\right)$ we have $$\begin{array}{rcl}|G/B|& =& W\left(q\right)\\ \left|B\right|& =& {(q-1)}^r{q}^N\end{array}$$ ($r$ the rank, $N$ the number of positive roots). So $$\begin{array}{rcl}\left|G\right|& =& q^N{(q-1)}^{r}W\left(q\right)\\ & =& q^N\prod _{i=1}^r(q^{d_i}-1).\end{array}$$ If $G=S{O}_5\left({𝔽}_3\right)$ then $q=3,$ $R$ is of type $B_2,$ so that $N=4, d_1=2, d_2=4.$ $$\begin{array}{rcl}\left|W\right|=\left|S{O}_5\left({𝔽}_3\right)\right|& =& 3^4(3^3-1)(3^4-1)\\ & =& 3^4\cdot 8\cdot 80=2^7\cdot 3^4\cdot 5.\end{array}$$
2. For $E_7$ we have $|P/Q|=f=2,$ hence $$P>Q>2P.$$ Since $|P/2P|=2^7$ it follows that $Q/2P$ has order $2^6,$ hence is a 6 dimensional space over ${𝔽}_2.$ Normalize the scalar product so that ${\left|\alpha \right|}^2=2$ for all $\alpha \in R,$ then it induces a nonsingular alternating bilinear form on $Q/2P; W$ acts and we have an isomorphism $$W/(\pm 1)\tilde{\to }S{p}_6\left({𝔽}_2\right).$$ For $S{p}_6,$ the root system is $C_3,$ so that here $q=2,N=9,d_1=2,d_2=4,d_3=6.$ $$\begin{array}{rcl}\left|S{p}_6\left({𝔽}_2\right)\right|& =& 2^9(2^2-1)(2^4-1)(2^6-1)\\ & =& 2^9\cdot 3\cdot 15\cdot 63\\ & =& 2^9\cdot 3^4\cdot 5\cdot 7(=\frac{1}{2}\left|W\right|).\end{array}$$
3. Again for $E_7, Q/2Q$ is a vector space of dimension 7 over ${𝔽}_2.$ Normalize the scalar product so that ${\left|\alpha \right|}^2=1$ this time, for all $\alpha \in R.$ Then we have a nondegenerate quadratic form on $Q/2Q,$ and $W$ acts as a group of orthogonal transformations: we have an isomorphism $$W/(\pm 1)\tilde{\to }{O}_7\left({𝔽}_2\right).$$
4. Repeat Exercise 3 for $E_8.$ Again we find $$W/(\pm 1)\tilde{\to }{O}_8\left({𝔽}_2\right)$$ (check the orders). For $O_8$ the root system is $D_4,$ so that $q=2,$ the $d\text{'}s$ are $(2,4,4,6)$ and $N=12:$ $$\begin{array}{rcl}\left|S{O}_8\left({𝔽}_2\right)\right|& =& 2^{12}(2^2-1){(2^4-1)}^2(2^6-1)\\ & =& 2^{12}\cdot 3\cdot {15}^2\cdot 63\\ & =& 2^{12}\cdot 3^5\cdot 5^2\cdot 7\end{array}$$

Arun: The next section appears to be out of order/without context. Have typed and left to be fixed/moved. $$\prod _{i=1}^r\frac{1-t^{d_i}}{1-t}=\prod _{\alpha \in R^{+}}\frac{1-t^{\mathrm{ht}\left(\alpha \right)+1}}{1-t^{\mathrm{ht}\left(\alpha \right)}}.$$ Let $h_i=$ number of roots of height $i (h_1=r).$ Then $$\begin{array}{rcl}\prod _1^r\frac{1-t^{d_i}}{1-t}& =& \prod _{i\ge 1}{\left(\frac{1-t^{i+1}}{1-t^i}\right)}^{h_i}\\ & =& \frac{1}{{(1-t)}^r}\prod _{j\ge 1}{(1-t^{i+1})}^{h_i-h_{i+1}}.\end{array}$$ So $$\Pi (1-t^{d_i})=\Pi {(1-t^{j+1})}^{h_j-h_{j+1}}.$$ Hence $h_j-h_{j+1}=$ number of degrees $d_i$ equal to $j+1$ so $h_j\ge h_{j+1}$ i.e. $\eta =(h_1,h_2,...)$ is a partition and $\eta \prime =(d_1-1,d_2-1,...,d_r-1).$

Can now check (3) directly case by case. Probably as quick, if not quicker, than a uniform proof $$\begin{array}{rcl}c\left({\alpha }_i\right)={\beta }_i& =& -\sum U_{ji}{\gamma }_j=-\sum U_{ji}{L}_{kj}^{-1}{\alpha }_k\\ {\gamma }_j& =& \sum L_{kj}^{-1}\alpha +k{L}^{-1}U.\end{array}$$

References

I.G. Macdonald
Issac Newton Institute for the Mathematical Sciences
20 Clarkson Road
Cambridge CB3 OEH U.K.

Version: March 12, 2012

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