## Orthogonal polynomials

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 $f— = ∑λ∈P fλe-λ ( f—(x) = f(x)—, x∈V )$ 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 ${A}^{W}$ $⟨f,g⟩1 = 1|W| [fg—]1 (OP 1)$

Let $\lambda ,\mu \in {P}^{+}.$ Then ${⟨{m}_{\lambda },{m}_{\mu }⟩}_{1}=0$ if $\lambda \ne \mu ,$ and ${⟨{m}_{\lambda },{m}_{\lambda }⟩}_{1}={|{W}_{\lambda }|}^{-1}.$

 Proof. We have $mλ = ∑ν∈Wλeν$ and the number of terms in this sum is $|W\lambda |=\frac{|W|}{|{W}_{\lambda }|}.$ Hence $⟨mλ,mμ⟩ = 1|W| ∑ ν∈Wλ π∈Wμ [eν⋅e-π]1 = 1|W| Card(Wλ∩Wμ).$ If $\lambda \ne \mu$ this is 0, and if $\lambda =\mu$ it is $\frac{1}{|W|}\mathrm{Card}\left(W\lambda \right)=\frac{1}{|W\lambda |}.$ $\square$

$⟨f,g⟩0 = 1|W| [fg—dd—]1 = ⟨fd,gd⟩1 (OP 2)$ where as before $d = eρ ∏α∈R+ (1-e-α) = ∑w∈W ε(w) ewρ$ is the Weyl denominator.

Let $\lambda ,\mu \in {P}^{+}.$ Then ${⟨{\chi }_{\lambda },{\chi }_{\mu }⟩}_{0}={\delta }_{\lambda \mu }.$

 Proof. Since $χλ = d-1 θ(eλ+ρ) = d-1 ∑w∈W ε(w) ew(λ+ρ) ,$ we have $⟨χλ,χμ⟩0 = 1|W| ∑v,w∈W ε(v) ε(w) [ ev(λ+ρ) ⋅ e-w(μ+ρ) ]1$ which is zero if $\lambda \ne \mu ,$ because the orbits do not intersect. If $\lambda =\mu$ we have (because $\lambda +\rho \in C$). Hence we get $\frac{1}{|W|}\sum _{w\in W}1=1.$ $\square$

Let $t$ be a real number, $0\le t\le 1,$ and let $Δ = Δ(t) = ∏α∈R 1-eα 1-teα .$ 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 $⟨f,g⟩t = 1|W| [fg—Δ]1 = ⟨fΔ,g⟩1 = ⟨fΔ+,gΔ+⟩1 = 1|W| ∫Tfg—Δ positive definite, symmetric.$ Notice that when $t=1$ we have $\Delta =1,$ and when $t=0$ we have $Δ = ∏α∈R (1-eα) = dd—$ 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 $P˜λ = ∑w∈W w eλ ∏α∈R+ 1-te-α 1-e-α (λ∈P+).$ Since $\Delta$ is $W-$invariant it follows that $P˜λΔ = ∑w∈W w eλ ∏α∈R+ 1-eα 1-teα$ in which $eλ ∏α∈R+ 1-eα 1-teα = eλ ∏α∈R+ 1+∑r≥1 (tr-tr-1) erα$ a product of convergent series, since $0\le t<1.$ Hence we can multiply them all together and get say $eλ ∏α∈R+ 1-eα 1-teα = ∑ν∈Q+ aνeλ+ν$ with coefficients ${a}_{\nu }\in ℤ\left[t\right],$ and in particular ${a}_{0}=1.$ Hence we have $P˜λΔ = ∑ν∈Q+aν ∑w∈W ew(λ+ν).$ If $\lambda +\nu ={w}_{1}\pi ,$ where $\pi \in {P}^{+}$ and ${w}_{1}\in W,$ then $π≥w1π = λ+ν≥λ$ with equality if and only if $\nu =0.$ Hence $P˜λΔ = |Wλ| mλ + higher terms. (*)$ 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 $Pμ = mμ+ lower terms. (**)$ If $\mu \ne \lambda ,$ the series (*) and (**) have just one term in common, namely ${m}_{\lambda },$ and hence $⟨P˜λ,Pλ⟩t = ⟨P˜λΔ,Pλ⟩1 = |Wλ| ⟨mλ,mλ⟩1 = 1 by Proposition 1.1$ so that ${⟨{P}_{\lambda },{P}_{\lambda }⟩}_{t}={W}_{\lambda }{\left(t\right)}^{-1}.$ $\square$

(So are dual bases.)

 Proof of (4.18). Since $χλ = ∑μ Kλμ(t)Pμ$ we have $Kλμ(t) = ⟨χλ,P˜μ⟩t.$ Now $Δ = ∏α∈R 1-eα 1-teα = dd—Π where Π = ∏α∈R (1-teα)-1.$ Hence $Kλμ(t) = 1|W| [χ—λP˜μΔ]1 = 1|W| [χ—λP˜μdd—Π]1 = ⟨χλ,P˜μΠ⟩0 (†)$ But now $P˜μΠ = ∑w∈W w eμ ∏α∈R+ 1-te-α 1-e-α ⋅ ∏α∈R 1 1-teα = d-1 ∑w∈W ε(w)w eμ+ρ ∏α∈R+ (1-teα)-1 = ∏α∈R+ (1-tRα)-1 χμ.$ From (†) it follows that $Kλμ(t) = coefficient of χλ in ∏α∈R+ (1-tRα)-1 χμ (††)$ which proves (4.18)(ii). Now $∏α∈R+ (1-tRα)-1 = ∏α∈R+ ∑mα=0∞ tmα Rαmα = ∑ξ∈Q+ F(ξ;t) Rξ$ and if $\xi \in {Q}^{+}$ $χμ+ξ = d-1θ( eμ+ξ+ρ ).$ If $\mu +\xi +\rho$ is singular, then ${\chi }_{\mu +\xi }=0;$ if nonsingular, then $μ+ξ+ρ = w(λ+ρ)$ for a unique $w\in W$ and $\lambda \in {P}^{+}:$ and then $ξ = w(λ+ρ)-(μ+ρ) and χμ+ξ = ε(w) = ε(w)χλ.$ So from (††) we have $Kλμ(t) = ∑w∈W ε(w) F( w(λ+ρ)-(μ+ρ) ; t ).$ $\square$

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?

$t=1:$ ${K}_{\lambda \mu }=\sum _{w\in W}\epsilon \left(w\right)F\left(w\left(\lambda +\rho \right)-\left(\mu +\rho \right);1\right)$ is Kostant's formula for weight multiplicities. $χμχν = ∑λ cμνλχλ cμνλ ≥ 0.$

Now let $q,t$ be independent indeterminants [or real variables, ] and let $Δ = Δ(q,t) = ∏α∈R (eα;q)∞ (teα;q)∞ .$ In particular, if $t={q}^{k}$ where $k\in ℕ$ then $Δ = ∏α∈R (eα;q)k ∈ AW.$ We now define a new scalar product on $A:$ if $f,g\in A$ then $⟨f,g⟩q,t = 1|W| [fg—Δ]1 ( = 1|W| ∫T fg—Δ ) = ⟨fΔ+,gΔ+⟩1 .$ Once again, this is symmetric and positive definite. Notice that when $q=0$ we have $Δ(0,t) = ∏α∈R 1-eα 1-teα = Δ(t)$ so that ${⟨f,g⟩}_{0,t}={⟨f,g⟩}_{t}.$ Also if $t=q$ (i.e., $k=1$) we have $Δ(q,q) = ∏α∈R (1-eα) = dd—$ so that $⟨f,g⟩q,q = ⟨f,g⟩0.$ Finally if $t=1$ (i.e., $k=0$) we have $Δ(q,1) = 1$ and hence ${⟨f,g⟩}_{q,1}={⟨f,q⟩}_{1}:$ both independent of $q.$

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

1. ${P}_{\lambda }={m}_{\lambda }+\sum _{\genfrac{}{}{0}{}{\mu \in {P}^{+}}{\mu <\lambda }}{u}_{\lambda \mu }\left(q,t\right){m}_{\mu }\phantom{\rule{2em}{0ex}}\left({u}_{\lambda \mu }\in ℚ\left(q,t\right)\right).$
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 $f,g\in {A}^{W}\phantom{\rule{2em}{0ex}}$ ($E$ is self adjoint),
2. ($E$ is triangular relative to the orbit-sums ${m}_{\lambda }$),
3. ($E$ 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 $Eλ = ∏ μ<λ μ∈P+ E-cμ cλ-cμ$ and then let $Pλ = Eλmλ .$ 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 $∏μ≤λ (X-cμ)$ since the ${c}_{\mu }$ are all distinct. Hence $\prod _{\mu \le \lambda }\left(E-{c}_{\mu }\right)=0$ on ${M}_{\lambda },$ i.e. $\left(E-{c}_{\lambda }\right){E}_{\lambda }=0$ on ${M}_{\lambda },$ and therefore $EPλ = EEλmλ = cλEλmλ = cλPλ$ (so the ${P}_{\lambda }$ are the eigenvectors of $E$). Hence $cλ ⟨Pλ,Pμ⟩ = ⟨EPλ,Pμ⟩ = ⟨Pλ,EPμ⟩ by (a) = cμ ⟨Pλ,Pμ⟩.$ 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 Clearly 0 is always a minuscule weight.

Let $\left({\omega }_{1},...,{\omega }_{r}\right)$ be the basis of $V$ dual to $\left({\alpha }_{1},...,{\alpha }_{r}\right):$ $⟨{\alpha }_{i},{\omega }_{j}⟩={\delta }_{ij}.$ Then ${\omega }_{1},...,{\omega }_{r}$ is a basis of the lattice ${P}^{\vee }\left(={Q}^{*}\right).$ Also let $φ = ∑i=1r miαi$ be the highest root of $R,$ and let $I=\left\{i\in \left[1,r\right]\phantom{\rule{.5em}{0ex}}|\phantom{\rule{.5em}{0ex}}{m}_{i}=1\right\}.$ (It might be empty.)

The minuscule weights of ${R}^{\vee }$ are 0 and

 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 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 (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:

Examples. $R An Bn Cn Dn E6 E7 E8 F4 G2 f n+1 2 2 4 3 2 1 1 1$ 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 some $r\ge 0.$

PICTURE OF DYNKIN DIAGRAMS

## First construction

Suppose $f>1$ (i.e. that $R$ is not of type or ${G}_{2}$) and let $\pi$ be a nonzero minuscule weight of ${R}^{\vee },$ so that $⟨π,α⟩ = 0 or 1, all α∈R+. (OP 3)$ Define ${T}_{\pi }:A\to A$ by $Tπeλ = q⟨λ,π⟩eλ (λ∈P)$ extended by linearity. Clearly $Tπ (eλ+μ) = Tπ(eλ) ⋅ Tπ(eμ)$ so that ${T}_{\pi }$ is an automorphism of the $ℝ-$algebra $A.$ Moreover, ${T}_{\pi }$ is self-adjoint for the scalar product $⟨f,g⟩1 = 1|W| [fg—]1.$ For if then $⟨Tπf,g⟩1 = 1|W| [(Tπf)g—]1 = 1|W| ∑λ,μ q⟨λ,μ⟩ fλgμ eλ-μ 1 = 1|W| ∑λ q⟨λ,π⟩ fλgλ$ 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 — $eλ(x) = q⟨λ,x⟩ (x∈V)$ then $(Tπeλ)(x) = q⟨λ,π⟩ ⋅ q⟨λ,x⟩ = q⟨λ,x+π⟩ = eλ(x+π)$ and hence by linearity $(Tπf) (x) = f(x+π)$ for all $f\in A,$ i.e. ${T}_{\pi }$ is a translation operator.

From above we have $Tπeα = { qeα, if ⟨π,α⟩ = 1, eα if ⟨π,α⟩ = 0, }$ and hence $Tπ (eα;q)∞ (eα;q)∞ = { 1 1-eα , if ⟨π,α⟩ = 1, 1, if ⟨π,α⟩ = 0. } (OP 4)$ Now let $Δ+ = ∏α∈R+ (eα;q)∞ (teα;q)∞ ( ∈A if t=qk, k∈ℕ )$ (so that $\Delta ={\Delta }^{+},{\stackrel{—}{\Delta }}^{+}$) and let $Φπ = TπΔ+ Δ+ = ∏α∈R+ Tπ(eα;q)∞ (eα;q)∞ ( Tπ(teα;q)∞ (teα;q)∞ ) = ∏ α∈R+ ⟨π,α⟩=1 1-teα 1-eα by (OP 4) above$ or equivalently $Φπ = ∏α∈R+ 1-t⟨π,α⟩eα 1-eα . (OP 5)$ We now define, for $f\in A,$ $Eπf = ∑w∈W w(ΦπTπf) = ∑w∈W w( Tπ(Δ+f) Δ+ )$ (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 we have $Δ = wΔ+ ⋅ wΔ+—$ and hence $\left(f,g\in {A}^{W}\right)$ $⟨Eπf,g⟩q,t = |W|-1 [ (Eπf) Δg— ]1 = |W|-1 ∑w∈W [ w(Tπ(Δ+f)) w(Δ+g)— ]1 (wg=g) = [ Tπ(Δ+f) ⋅ Δ+g— ]1 = |W| ⟨ Tπ(Δ+f), Δ+g ⟩1.$ 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 We have from (OP 5) $Φπ = ∏α∈R+ 1-t⟨π,α⟩eα 1-eα = t⟨π,2ρ⟩ ∏α∈R+ 1-t-⟨π,α⟩e-α 1-e-α = t⟨π,2ρ⟩ d ∑X⊆R+ (-1)|X| t-⟨π,σx⟩ eρ-σX$ where as before ${\sigma }_{X}=\sum _{\alpha \in X}\alpha$ for $X\subseteq {R}^{+}.$ Hence $Eπeμ = ∑w∈W w( Φπ ⋅ q⟨π,μ⟩ eμ ) = q⟨π,μ⟩ t⟨π,2ρ⟩ ∑w∈W w d-1 ∑X⊆R+ (-1)|X| t-⟨π,σX⟩ eμ+ρ-σX = q⟨π,μ⟩ t⟨π,2ρ⟩ ∑X&sunseteq;R+ (-1)|X| t-⟨π,σX⟩ χμ-σX$ which shows that ${E}_{\pi }$ maps $A$ into ${A}^{W}.$

Now let $\lambda \in {P}^{+},$ then $Eπ(mλ) = ∑μ∈wλ Eπeμ = t⟨π,2ρ⟩ ∑X⊆R+ (-1)|X| t-⟨π,σX⟩ ∑μ∈wλ q⟨π,μ⟩ χμ-σX.$ 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 $w(μ+ρ-σX) = ν+ρ, and χμ-σX = ε(w) χν.$ Now $w\left(\rho -{\sigma }_{X}\right)=\rho -{\sigma }_{Y}$ for some $Y\subseteq {R}^{+},$ hence $ν = w(μ+ρ-σX)-ρ = (wμ+ρ-σY) - ρ = wμ-σY ≤ wμ ≤ λ$ and it follows that ${E}_{\pi }{m}_{\lambda }$ is a linear combination of the and hence also a linear combination of the

Moreover in the above calculations we shall have $\nu =\lambda$ if and only if $w\mu =\lambda$ and ${\sigma }_{Y}=0,$ i.e. $w\left(\rho -{\sigma }_{X}\right)=\rho ,$ or ${\sigma }_{X}=\rho -{w}^{-1}\rho$ which by (4.11) forces $X=R\left({w}^{-1}\right)$ and hence $|X|=l\left(w\right).$ So the coefficient of ${X}_{\lambda }$ (or equivalently of ${m}_{\lambda }$) in ${E}_{\pi }{m}_{\lambda }$ is $cλ = t⟨π,2ρ⟩ ∑w∈W (-1)l(w) t-⟨π,ρ-w-1ρ⟩ q⟨π,w-1λ⟩ ε(w) = t⟨π,ρ⟩ ∑w∈W q⟨wπ,λ⟩ t⟨wπ,ρ⟩ = t⟨π,ρ⟩ ∑w∈W q⟨wπ,λ+kρ⟩ (t=qk) [ = t⟨π,ρ⟩ m˜π (λ+kρ) ].$ It remains to examine whether ${E}_{\pi }$ satisfies condition (c) of Proposition 1.6, i.e. whether the are all distinct. In fact this is true in all cases except (and where there is no nonzero minuscule weight). With $\pi$ as above let $pr(x) = ∑w∈W ⟨x,wπ⟩r (x∈V).$ Clearly the 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 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 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 ( excluded) $\lambda ,\mu \in {P}^{+}$ are such that ${c}_{\lambda }={c}_{\mu },$ i.e. $∑w∈W q⟨λ+kρ,wπ⟩ = ∑w∈W q⟨μ+kρ,wπ⟩ (all q,k).$ Operate on both sides with ${\left(q\frac{d}{dq}\right)}^{r}$ and then set $q=1.$ We obtain $pr(λ+kρ) = pr(μ+kρ) (r≥1).$ Hence by Proposition 3.2 $f(λ+kρ) = f(μ+kρ)$ 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 $g\left({x}_{i}\right)={c}_{i}&nsbp;\left(1\le i\le n\right).$ To prove this it is enough to construct 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 $\left(i,j\right)$ 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 $\left({x}_{1},...,{x}_{n}\right)$ and $\left({y}_{1},...,{y}_{m}\right)$ be distinct $W-$orbits in $V.$ From above there exists $g\in S\left(V\right)$ such that Let $f=\frac{1}{|W|}\sum _{w\in W}wg.$ 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) $⟨π,α⟩ = { 0 or 1, if α∈R+, α≠π∨, 2, if α=π∨. }$ As before, let ${\Phi }_{\pi }=\frac{{T}_{\pi }{\Delta }^{+}}{{\Delta }^{+}}$ and define now $\left(f\in A\right)$ $Eπf = ∑w∈W ( Φπ(Tπf-f) ).$ 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 $\left({E}_{8},{F}_{4},{G}_{2}\right)$ not covered by the previous construction. So the existence of the ${P}_{\lambda }\left(q,t\right)$ 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 $\left(t={q}^{k}\right)$ $⟨Pλ,Pλ⟩q,t = ∏α∈R+ ∏i=1k-1 1-q ⟨λ+kρ,α∨⟩+i ⟨λ+kρ,α∨⟩-i .$ True if $R$ of type $A.$

Consider the special case $\lambda =0$ of (C2q). Since ${P}_{0}=1$ we have ${⟨{P}_{0},{P}_{0}⟩}_{q,t}=\frac{1}{|W|}{\left[\Delta \right]}_{1}$ and hence (C2q) would give $1|W| [Δ]1 = ∏α∈R+ ∏i=1k-1 = 1-q ⟨kρ,α∨⟩+i ⟨kρ,α∨⟩-i . (OP 6)$ I claim that (OP 6) is equivalent to (C1q). Let $Δ′ = ∏α∈R+ ∏i=0k-1 ( 1-qieα ) ( 1-qi+1e-α )$ then $Δ′Δ = ∏α∈R+ ∏i=0k-1 1-qi+1e-α 1-e-α (t=qk)$ and therefore (since $\Delta$ is $W-$invariant) $Δ-1 ∑w∈W wΔ′ = ∑w∈W w( 1-te-α 1-e-α ) = W(t) by (4.12)$ i.e. $W(t)Δ = ∑w∈W wΔ′$ and therefore $W(t)[Δ]1 = |W|[Δ′]1$ (because have the same constant term). It follows from (OP 6) that $[Δ′]1 = W(t) |W| [Δ]1 = ∏α∈R+ 1-q kht(α∨)+k 1-q kht(α∨) [Δ]1 by (4.13) = ∏α∈R+ ∏i=0k-1 1-q kht(α∨)+i+1 1-q kht(α∨)-i .$ Consider then the polynomial $∑α∈R+ ∑i=0k-1 ( qkht(α∨)+i+1 - qkht(α∨)-i ) = ∑α∈R+ qkht(α∨) ( qk+qk-1+⋯+q-1-q-1-⋯-q1-k ) = ∑α∈R+ qkht(α∨) (qk-1) ( 1+q-1+⋯+q1-k ) .$ Now since $∏α∈R+ 1-tht(α∨)+1 1-tht(α∨) = W(t) = ∏i=1r 1-tdi 1-t$ it follows that $∑α∈R+ ( tht(α∨)+1 - tht(α∨) ) = ∑i=1r tdi-t$ i.e. that $∑α∈R+ qkht(α∨) (qk-1) = ∑i=1r ( qkdi-qk ).$ Hence (OP 6) becomes $∑i=1r (qkdi-qk) ( 1+q-1+⋯+q1-k ) = ∑i=1r { ( qkdi + qkdi-1 +⋯+ qkdi-k+1 ) - ( q+q2+⋯+qk ) }$ and therefore $∏α∈R+ ∏i=0k-1 1-q kht(α∨)+i+1 1-q kht(α∨)-i = ∏i=1r (1-qkdi)⋯(1-qkdi-k+1) (1-q)⋯(1-qk) = ∏i=1r [ kdi k ].$ So (C2q) $\left(\lambda =0\right)=$ (C1q). $\phantom{\rule{5em}{0ex}}$(Tip of the iceberg!)

Consider ${|{P}_{\lambda }|}^{2}$ as $k\to \infty$ (i.e. $t\to 0$). We have $\Delta \left(q,0\right)=\prod _{\alpha \in R}{\left({e}^{\alpha };q\right)}_{\infty }.$ (continuous function on $T=V/{Q}^{\vee }$) $∏i=1k-1 ( 1-q ⟨λ+kρ,α∨⟩+i ) = ( q⟨λ,α∨⟩+1 t⟨ρ,α∨⟩ ; q )∞ ( q⟨λ,α∨⟩+1 t⟨ρ,α∨⟩+1 ; q )∞ →1 ∏i=1k-1 ( 1-q ⟨λ+kρ,α∨⟩-i ) = ( q⟨λ,α∨⟩+1 t⟨ρ,α∨⟩-1 ; q )∞ ( q⟨λ,α∨⟩+1 t⟨ρ,α∨⟩ ; q )∞ →1 → { 1, if α≠αi, ( q⟨λ,α∨⟩+1 ; q )∞, if α=αi. }$ Hence $⟨Pλ,Pλ⟩2 → ∏i=1r ( q⟨λ,αi∨⟩+1 ; q ) ∞-1 = ∏i=1r ( qni+1 ; q )∞-1 if λ = ∑1r niπi (ni≥0).$ True when $\lambda =0.$ (From the denominator formula for affine root systems.)

## The limiting case $q\to 1$

As $q\to 1$ we have $\left(t={q}^{k}\right)$ $Δ(q,t) = ∏α∈R (eα;q)k → Δ(k) = ∏α∈R (1-eα)k$ as kernel for the scalar product. So we have ${P}_{\lambda }^{\left(k\right)}=\underset{q\to 1}{\mathrm{lim}}{P}_{\lambda }\left(q,{q}^{k}\right).$ 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) $|Pλ|(k)2 = ⟨Pλ,Pλ⟩(k) = ∏α∈R+ ∏i=1k-1 ⟨λ+kρ,α∨⟩+i ⟨λ+kρ,α∨⟩-i (λ∈P+, k∈ℕ). (OP 7)$ Proof is now elementary — depends on their notion of shift operators which enables one to compute the quotient $|Pλ|k2 |Pλ+ρ|k-12$ and hence prove (OP 7) (for $k$ a positive integer) by induction on $k$ (when and ${|{\chi }_{\lambda }|}^{2}=1.$)

## 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 $⟨αi,αi+1⟩ < 0 (1≤i≤m-1) and ⟨αm,α1⟩ < 0.$ Let ${\theta }_{i}=$ angle between ${\alpha }_{i}$ and Then ${\theta }_{i}$ is $>\frac{\pi }{2},$ hence $\ge \frac{2\pi }{3}$ and therefore Equivalently, if ${e}_{i}=\frac{{\alpha }_{i}}{|{\alpha }_{i}|},$ we have $⟨ei,ei+1⟩ ≤ -12 (1≤i≤m)$ and if $|i-j|\ge 2.$ Consider the vector $v={e}_{1}+\cdots +{e}_{m}:$ we have $|v|2 = ∑i,j=1m ⟨ei,ej⟩ = m+2∑i So ${|v|}^{2}=0$ 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 $|\alpha |=|\beta |$ 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 $⟨{\alpha }^{\vee },\beta ⟩=⟨\alpha ,{\beta }^{\vee }⟩=±1$ by (1.2)(i) $\left(⟨{\alpha }^{\vee },\beta ⟩⟨\alpha ,{\beta }^{\vee }⟩=1,2\phantom{\rule{.5em}{0ex}}or\phantom{\rule{.5em}{0ex}}3\right).$ Replacing $\alpha$ by $-\alpha ={s}_{\alpha }\alpha$ if necessary, we may assume that $⟨{\alpha }^{\vee },\beta ⟩=⟨\alpha ,{\beta }^{\vee }⟩=+1.$ Then we have $\square$

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 $|{W}_{\phi }|=|W\left({A}_{5}\right)|=6!$ So $|W(E6)| = 72⋅6! = 27⋅34⋅5.$

PICTURE

${E}_{7}:$ 126 roots, forming a single orbit. Here $|{W}_{\phi }|=|W\left({D}_{6}\right)|={2}^{5}\cdot 6!$ So $|W(E7)| = 25⋅6!⋅126 = 210⋅34⋅5⋅7 = 7!⋅4!⋅4!.$

PICTURE

${E}_{8}:$ 240 roots, $|{W}_{\phi }|=|W\left({E}_{7}\right)|.$ So $|W(E8)| = 240⋅|W(E7)| = 8!⋅6!⋅4! = 214⋅35⋅52⋅7.$

${F}_{4}:$ 24 long roots, $|{W}_{\phi }|=|W\left({C}_{3}\right)|$ implies $|W(F4)| = 24⋅|W(C3)| = 24⋅8⋅3! = 1152 = 27⋅32.$

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 $I\subseteq \left[1,r\right]$ 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 $\mathrm{Supp}\left(-\alpha \right)=\mathrm{Supp}\left(\alpha \right)$). 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 $\left({p}_{i}>0\right)$ and hence $∑pi ⟨αi,α⟩ = ⟨α,α⟩ > 0$ 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 $β, β+αi (=α), ..., β+pαi$ say, where $p\ge 1.$ By (1.5) $p=⟨\beta ,{\alpha }_{i}^{\vee }⟩,$ hence $⟨\beta ,{\alpha }_{i}⟩>0$ and since $\beta =\sum _{\genfrac{}{}{0}{}{j\in S}{j\ne i}}{p}_{j}{\alpha }_{j}$ it follows that $⟨{\alpha }_{j},{\alpha }_{i}⟩>0$ for some 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 $|I|.$ OK if $|I|=1,$ so assume $|I|\ge 2.$ Since $D$ contains no cycles, neither does ${D}_{I},$ i.e. ${D}_{I}$ is a tree. Let ${\alpha }_{i}$ be an end vertex of Then ${D}_{J}$ is connected, hence ${\alpha }_{J}\in R,$ and $⟨{\alpha }_{i},{\alpha }_{j}⟩<0$ for some $j\in J,$ so that $⟨αi,αJ⟩ = ∑k∈J ⟨αi,αk⟩ < 0$ 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 $\left[1,r\right].$ Let $BI = {αi | i∈I} VI = subspace of V spanned by BI RI = R∩VI WI = subgroup of W generated by si,i∈I.$ 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 ${B}_{I}$ is a basis of ${V}_{I},$ hence every $\alpha \in {R}_{I}$ is a linear combination of the 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

 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 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 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 the highest root or Let $\stackrel{—}{D}$ be the extended Dynkin diagram of $R,$ with vertices corresponding to and labels ${m}_{i},$ where $∑i=0r miα = 0 (m0=1). (OP 8)$ 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 ${B}_{i}=\left\{{\alpha }_{j}\phantom{\rule{.5em}{0ex}}|\phantom{\rule{.5em}{0ex}}0\le j\le r,j\ne i\right\},$ 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 $miαi = -∑j≠imjαj ∈ Ri. (OP 9)$ Consider the root lattices Since ${R}_{i}\subseteq R$ we have ${Q}_{i}\subseteq Q,$ and $Q/{Q}_{i}$ is a finite group. Let 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 then $m{\alpha }_{i}\in Q\left({R}_{i}\right),$ hence $m{\alpha }_{i}$ is a linear combination of the But (OP 8) is the only linear dependence relation among the 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 ${|\alpha |}^{2}=2$ for each $\alpha \in R,$ so that ${\alpha }^{\vee }=\alpha$ and hence ${R}^{\vee }=R.$ ($R$ is simply laced.) Then the dual lattices of 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 $Pi P Q Qi mi f mi$ 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. $mi2 = fi f .$

Examples:

1. $R$ of type ${E}_{8}:$ the extended Dynkin diagram is $PICTURE$ So 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 ${P}_{8}=P\left({A}_{8}\right):$ $P8⊃3P = Q⊃3Q8.$ This underlies the construction of ${E}_{8}$ that I gave earlier: $Q\left({A}_{8}\right)\subseteq {ℝ}^{9}.$
2. $R$ of type ${E}_{8}:$
3. $R$ of type ${E}_{8}:$ $P4>5P = Q>5Q4.$ Exercise: Figure out the roots of ${E}_{8}$ in this representation.
4. $R$ of type ${E}_{6}:$ $PICTURE$ Here $P3⊃3P⊃3Q⊃3Q3$ 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 ${ℂ}^{3}\left(\simeq {ℝ}^{6}\right)$ and take (${e}_{1},{e}_{2},{e}_{3}$ basis of ${ℂ}^{3}$ as $ℂ-$vector space) $α1=e1, α2=ωe1, α4=ωe2, α5=e2, α6=ωe3, α0=e3,$ where $\omega =\mathrm{exp}\frac{2\pi i}{3}.$
What about ${\alpha }_{3}?$ We must have $3α3 = -( α1+2α2+2α4+α5+2α6+α0 ) = -(1+2ω) (e1+e2+e3)$ so that $α3 = θ(e1+e2+e3)$ where $\theta =-\frac{1}{3}\left(1+2\omega \right)$ (see figure) $PICTURE$ In this representation the roots of ${E}_{6}$ are $±ζei where ζ3=1, ±θ( ζ1e1 + ζ2e2 + ζ3e3 ) where ζ1,ζ2,ζ3 are cube roots of unity.$ For these are certainly roots, and there are $2⋅3⋅3+2⋅33 = 18+54 = 72$ 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 $P7>2P>2Q>2Q7$ ${Q}_{7}$ is easily described; it is the lattice of all $\left({n}_{1},...,{n}_{8}\right)\in {ℤ}^{8}$ such that $\sum {n}_{i}=0.$
Exercise: Figure out the roots of ${E}_{7}$ in this representation $ei-ej, 56 of these, 12 ∑±ei with 4 minus signs, 70 of these.$ What are the roots of ${E}_{7}$ in this description?
Take then $α7 = -12 ( α1+2α2+3α3+4α4+3α5+2α6+α0 ) = -12 ( e1+⋯+e4-e5-⋯-e8 ).$ So certainly all of the vectors are roots. There are $2\left(\genfrac{}{}{0}{}{8}{2}\right)+\left(\genfrac{}{}{0}{}{8}{4}\right)=56+70=126$ of these vectors, hence they are the lot.
6. Again take So we may take and then $α6 = -12 ( α1+2α2+3α3+4α4+3α5+2α7+α0 ) = -12 ( e1+⋯+e6+f7 ).$ So certainly the following vectors are roots: $±ei±ej (1≤i So these are all roots.

## Orders of the Weyl groups of

Recall (Exercise): $R$ irreducible, 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 $|W| = |Wφ|Card(R).$ ${E}_{6}:$ hence $|W| = 72⋅6! = 27⋅34⋅5.$ ${E}_{7}:$ hence $|W| = 126⋅25⋅6! = 210⋅34⋅5⋅7 = 7!⋅4!⋅4!.$ ${E}_{8}:$ hence $|W| = 240⋅210⋅34⋅5⋅7 = 214⋅35⋅52⋅7 = 8!⋅6!⋅4!.$ ${F}_{4}:$ 24 long roots, ${W}_{\phi }=W\left({C}_{3}\right),$ hence $|W| = 24⋅23⋅3! = 1152 = 27⋅32.$

## The Weyl groups of

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 ${|\alpha |}^{2}=2$ 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 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 $|G/B| = W(q) |B| = (q-1)rqN$ ($r$ the rank, $N$ the number of positive roots). So $|G| = qN(q-1)rW(q) = qN ∏i=1r (qdi-1).$ If $G=S{O}_{5}\left({𝔽}_{3}\right)$ then $q=3,$ $R$ is of type ${B}_{2},$ so that $|W| = |SO5(𝔽3)| = 34(33-1)(34-1) = 34⋅8⋅80 = 27⋅34⋅5.$
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 ${|\alpha |}^{2}=2$ for all $\alpha \in R,$ then it induces a nonsingular alternating bilinear form on acts and we have an isomorphism $W/(±1) →˜ Sp6(𝔽2).$ For $S{p}_{6},$ the root system is ${C}_{3},$ so that here $q=2,N=9,{d}_{1}=2,{d}_{2}=4,{d}_{3}=6.$ $|Sp6(𝔽2)| = 29 (22-1) (24-1) (26-1) = 29⋅3⋅15⋅63 = 29⋅34⋅5⋅7 (=12|W|).$
3. Again for is a vector space of dimension 7 over ${𝔽}_{2}.$ Normalize the scalar product so that ${|\alpha |}^{2}=1$ 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 $W/(±1) →˜ O7(𝔽2).$
4. Repeat Exercise 3 for ${E}_{8}.$ Again we find $W/(±1) →˜ O8(𝔽2)$ (check the orders). For ${O}_{8}$ the root system is ${D}_{4},$ so that $q=2,$ the $d\text{'}s$ are $\left(2,4,4,6\right)$ and $N=12:$ $|SO8(𝔽2)| = 212 (22-1) (24-1)2 (26-1) = 212⋅3⋅152⋅63 = 212⋅35⋅52⋅7$

Arun: The next section appears to be out of order/without context. Have typed and left to be fixed/moved. $∏i=1r 1-tdi 1-t = ∏α∈R+ 1-tht(α)+1 1-tht(α) .$ Let ${h}_{i}=$ number of roots of height Then $∏1r 1-tdi 1-t = ∏i≥1 ( 1-ti+1 1-ti )hi = 1 (1-t)r ∏j≥1 ( 1-ti+1 ) hi-hi+1.$ So $Π(1-tdi) = Π( 1-tj+1 ) hj-hj+1.$ Hence ${h}_{j}-{h}_{j+1}=$ number of degrees ${d}_{i}$ equal to $j+1$ so ${h}_{j}\ge {h}_{j+1}$ i.e. $\eta =\left({h}_{1},{h}_{2},...\right)$ is a partition and $\eta \prime =\left({d}_{1}-1,{d}_{2}-1,...,{d}_{r}-1\right).$

Can now check (3) directly case by case. Probably as quick, if not quicker, than a uniform proof $c(αi) = βi = -∑Ujiγj = -∑UjiLkj-1αk γj = ∑Lkj-1α + k L-1U.$