Orthogonal polynomials

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last update: 11 August 2012

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.


Notation etc. as in the previous chapter. If f = λP fλeλ A, let f = λP fλe-λ ( f(x) = f(x), xV ) and let [f]1 denote the constant term f0 of f (the coefficient of e0=1). I am going to define various scalar products on AW f,g1 = 1|W| [fg]1 (OP 1)

Let λ,μP+. Then mλ,mμ1 = 0 if λμ, and mλ,mλ1 = |Wλ|-1.

We have mλ = νWλeν and the number of terms in this sum is |Wλ| = |W| |Wλ| . Hence mλ,mμ = 1|W| νWλ πWμ [eνe-π]1 = 1|W| Card(WλWμ). If λμ this is 0, and if λ=μ it is 1|W| Card(Wλ) = 1|Wλ|.

f,g0 = 1|W| [fgdd]1 = fd,gd1 (OP 2) where as before d = eρ αR+ (1-e-α) = wW ε(w) ewρ is the Weyl denominator.

Let λ,μP+. Then χλ,χμ0 = δλμ.

Since χλ = d-1 θ(eλ+ρ) = d-1 wW ε(w) ew(λ+ρ) , we have χλ,χμ0 = 1|W| v,wW ε(v) ε(w) [ ev(λ+ρ) e-w(μ+ρ) ]1 which is zero if λμ, because the orbits W(λ+ρ),   W(μ+ρ) do not intersect. If λ=μ we have v(λ+ρ) = w(λ+ρ)    v=w (because λ+ρC). Hence we get 1|W| wW1 = 1.

Let t be a real number, 0t1, and let Δ = Δ(t) = αR 1-eα 1-teα . As explained earlier, we may regard Δ as a continuous function on the torus T=V/Q. For f,gA we define f,gt = 1|W| [fgΔ]1 = fΔ,g1 = fΔ+,gΔ+1 = 1|W| TfgΔ positive definite, symmetric. Notice that when t=1 we have Δ=1, and when t=0 we have Δ = αR (1-eα) = dd so that when t=1,0 we get the two previous scalar products.

Let λ,μP+. Then Pλ,Pμt = 0 if λμ, and Pλ,Pλt = Wλ(t)-1.

Recall that P˜λ = wW w eλ αR+ 1-te-α 1-e-α (λP+). Since Δ is W-invariant it follows that P˜λΔ = wW w eλ αR+ 1-eα 1-teα in which eλ αR+ 1-eα 1-teα = eλ αR+ 1+r1 (tr-tr-1) erα a product of convergent series, since 0t<1. Hence we can multiply them all together and get say eλ αR+ 1-eα 1-teα = νQ+ aνeλ+ν with coefficients aν [t], and in particular a0=1. Hence we have P˜λΔ = νQ+aν wW ew(λ+ν). If λ+ν = w1π, where πP+ and w1W, then πw1π = λ+νλ with equality if and only if ν=0. Hence P˜λΔ = |Wλ| mλ + higher terms. (*) Now let μP+. We cannot have μ>λ and λ>μ, hence (since the scalar product is symmetric) we may assume μλ. We have Pμ = mμ+ lower terms. (**) If μλ, the series (*) and (**) have just one term in common, namely mλ, and hence P˜λ,Pλt = P˜λΔ,Pλ1 = |Wλ| mλ,mλ1 = 1 by Proposition 1.1 so that Pλ,Pλt = Wλ(t)-1.

(So (Pλ), (P˜λ) are dual bases.)

Proof of (4.18).
Since χλ = μ Kλμ(t)Pμ we have Kλμ(t) = χλ,P˜μt. Now Δ = αR 1-eα 1-teα = ddΠ where Π = αR (1-teα)-1. Hence Kλμ(t) = 1|W| [χλP˜μΔ]1 = 1|W| [χλP˜μddΠ]1 = χλ,P˜μΠ0 (†) But now P˜μΠ = wW w eμ αR+ 1-te-α 1-e-α αR 1 1-teα = d-1 wW ε(w)w eμ+ρ αR+ (1-teα)-1 = αR+ (1-tRα)-1 χμ. From (†) it follows that Kλμ(t) = coefficient of χλ in αR+ (1-tRα)-1 χμ (††) which proves (4.18)(ii). Now αR+ (1-tRα)-1 = αR+ mα=0 tmα Rαmα = ξQ+ F(ξ;t) Rξ and if ξQ+ χμ+ξ = d-1θ( eμ+ξ+ρ ). If μ+ξ+ρ is singular, then χμ+ξ = 0; if nonsingular, then μ+ξ+ρ = w(λ+ρ) for a unique wW and λP+: and then ξ = w(λ+ρ)-(μ+ρ) and χμ+ξ = ε(w) = ε(w)χλ. So from (††) we have Kλμ(t) = wW ε(w) F( w(λ+ρ)-(μ+ρ) ; t ).

In fact the Kλμ(t) 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: Kλμ = wW ε(w) F( w(λ+ρ)-(μ+ρ) ; 1 ) is Kostant's formula for weight multiplicities. χμχν = λ cμνλχλ cμνλ 0.

Now let q,t be independent indeterminants [or real variables, 0q<1, t0] and let Δ = Δ(q,t) = αR (eα;q) (teα;q) . In particular, if t=qk where k then Δ = αR (eα;q)k AW. We now define a new scalar product on A: if f,gA then f,gq,t = 1|W| [fgΔ]1 ( = 1|W| T fgΔ ) = fΔ+,gΔ+1 . Once again, this is symmetric and positive definite. Notice that when q=0 we have Δ(0,t) = αR 1-eα 1-teα = Δ(t) so that f,g0,t = f,gt . Also if t=q (i.e., k=1) we have Δ(q,q) = αR (1-eα) = dd so that f,gq,q = f,g0. Finally if t=1 (i.e., k=0) we have Δ(q,1) = 1 and hence f,gq,1 = f,q1 : both independent of q.

For each λP+ there exists Pλ = Pλ(q,t) AW such that

  1. Pλ = mλ+ μP+ μ<λ uλμ(q,t)mμ ( uλμ (q,t) ).
  2. Pλ,Pμq,t = 0 if λμ.

This will be a consequence of the following proposition:

There exists a linear operator E:AWAW satisfying the following three conditions:

  1. Ef,gq,t = f,Egq,t for all f,gAW (E is self adjoint),
  2. Emλ = cλmλ +   lower terms (E is triangular relative to the orbit-sums mλ),
  3. λμ    cλcμ (E has distinct eigenvalues).

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 λP+ define Eλ = μ<λ μP+ E-cμ cλ-cμ and then let Pλ = Eλmλ . Then the Pλ so defined have the desired properties. Indeed, it is clear from the definition and (b) above that Pλ satisfies Proposition 1.5 (i).

Let Mλ be the (finite-dimensional) subspace of AW spanned by the mμ such that μλ (and μP+). Then Mλ is stable under E, and the minimal (characteristic polynomial of E|Mλ is μλ (X-cμ) since the cμ are all distinct. Hence μλ (E-cμ) = 0 on Mλ, i.e. (E-cλ)Eλ = 0 on Mλ, and therefore EPλ = EEλmλ = cλEλmλ = cλPλ (so the Pλ are the eigenvectors of E). Hence cλ Pλ,Pμ = EPλ,Pμ = Pλ,EPμ by (a) = cμ Pλ,Pμ. Since cλcμ if λμ, it follows that Pλ,Pμq,t = 0.

So it remains to construct a linear operator E on AW 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)

Assume R reduced, irreducible. Let πP = P(R) . π is a minuscule weight of R if π,α = 0   or   1, all   αR+. Clearly 0 is always a minuscule weight.

Let (ω1,...,ωr) be the basis of V dual to (α1,...,αr): αi,ωj = δij. Then ω1,...,ωr is a basis of the lattice P (=Q*) . Also let φ = i=1r miαi be the highest root of R, and let I = {i[1,r] | mi=1} . (It might be empty.)

The minuscule weights of R are 0 and ωi   (iI).

Suppose π = 1r niωi is a minuscule weight. Then ni = π,αi = 0 or 1 for each i, and 1r mini = π,φ = 0   or   1. If this is zero, then all the ni are 0, and π=0. If it is 1, then for some i we have mini=1 (whence mi=ni=1) and mjnj=0 for ji (whence nj=0, since mj>0). So π=ωi and mi=1, i.e. iI.

Conversely, if π=ωi and α = pjαj R+, we have 0pjmj for all j, because αφ; hence π,α = ωi,α = pi = 0  or  1 (since mi=1).

The minuscule weights have other nice properties:

  1. They are coset representatives of Q in P, hence there are f of them (i.e. f-1 nonzero minuscule weights).
  2. Let λP+. Then λ is a minuscule weight (for R, not R) if and only if χλ = mλ (i.e. the weights form a single W-orbit).
  3. They are the minimal elements of P+ with respect to: λμ    λ-μQ+.

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 E8,F4,G2.
An: all fundamental weights are minuscule and property (2) above translate into the fact that sλ=mλ    λ=(1r), some r0.


First construction

Suppose f>1 (i.e. that R is not of type E8, F4 or G2) and let π be a nonzero minuscule weight of R, so that π,α = 0 or 1, all αR+. (OP 3) Define Tπ:AA by Tπeλ = qλ,πeλ (λP) extended by linearity. Clearly Tπ (eλ+μ) = Tπ(eλ) Tπ(eμ) so that Tπ is an automorphism of the -algebra A. Moreover, Tπ is self-adjoint for the scalar product f,g1 = 1|W| [fg]1. For if f = fλeλ ,  g = gμeμ then Tπf,g1 = 1|W| [(Tπf)g]1 = 1|W| λ,μ qλ,μ fλgμ eλ-μ 1 = 1|W| λ qλ,π fλgλ which is symmetrical in f and g, hence equal to Tπg,f1 = f,Tπg1.

If we interpret eλ as a function on V by the rule — different from the previous one — eλ(x) = qλ,x (xV) then (Tπeλ)(x) = qλ,π qλ,x = qλ,x+π = eλ(x+π) and hence by linearity (Tπf) (x) = f(x+π) for all fA, i.e. Tπ 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+ = αR+ (eα;q) (teα;q) ( A if t=qk, k ) (so that Δ=Δ+,Δ+) and let Φπ = TπΔ+ Δ+ = αR+ Tπ(eα;q) (eα;q) ( Tπ(teα;q) (teα;q) ) = αR+ π,α=1 1-teα 1-eα by (OP 4) above or equivalently Φπ = αR+ 1-tπ,αeα 1-eα . (OP 5) We now define, for fA, Eπf = wW w(ΦπTπf) = wW w( Tπ(Δ+f) Δ+ ) (conjugate Tπ by Δ+ and then sum over the Weyl group). We shall show that Eπ maps AW into AW 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 Eπ is self-adjoint. Since Δ = Δ+Δ+ and Δ=wΔ   (wW) we have Δ = wΔ+ wΔ+ and hence (f,gAW) Eπf,gq,t = |W|-1 [ (Eπf) Δg ]1 = |W|-1 wW [ w(Tπ(Δ+f)) w(Δ+g) ]1 (wg=g) = [ Tπ(Δ+f) Δ+g ]1 = |W| Tπ(Δ+f), Δ+g 1. Since Tπu,v1 = u,Tπv1 as we saw earlier, it follows that Eπf,gq,t = f,Eπgq,t .

Next we have to compute Eπeμ   (μP). We have from (OP 5) Φπ = αR+ 1-tπ,αeα 1-eα = tπ,2ρ αR+ 1-t-π,αe-α 1-e-α = tπ,2ρ d XR+ (-1)|X| t-π,σx eρ-σX where as before σX = αXα for XR+. Hence Eπeμ = wW w( Φπ qπ,μ eμ ) = qπ,μ tπ,2ρ wW w d-1 XR+ (-1)|X| t-π,σX eμ+ρ-σX = qπ,μ tπ,2ρ X&sunseteq;R+ (-1)|X| t-π,σX χμ-σX which shows that Eπ maps A into AW.

Now let λP+, then Eπ(mλ) = μwλ Eπeμ = tπ,2ρ XR+ (-1)|X| t-π,σX μwλ qπ,μ χμ-σX. Consider χμ-σX: if μ+ρ-σX is singular (i.e. not regular) then χμ-σX = 0. If μ+ρ-σX is regular, then there exists wW and νP+ such that w(μ+ρ-σX) = ν+ρ, and χμ-σX = ε(w) χν. Now w(ρ-σX) = ρ-σY for some YR+, hence ν = w(μ+ρ-σX)-ρ = (wμ+ρ-σY) - ρ = wμ-σY wμ λ and it follows that Eπmλ is a linear combination of the χν,   νP+,   νλ, and hence also a linear combination of the mν   (νP+, νλ).

Moreover in the above calculations we shall have ν=λ if and only if wμ=λ and σY=0, i.e. w(ρ-σX) = ρ, or σX = ρ-w-1ρ which by (4.11) forces X = R(w-1) and hence |X| = l(w). So the coefficient of Xλ (or equivalently of mλ) in Eπmλ is cλ = tπ,2ρ wW (-1)l(w) t-π,ρ-w-1ρ qπ,w-1λ ε(w) = tπ,ρ wW qwπ,λ twπ,ρ = tπ,ρ wW qwπ,λ+kρ (t=qk) [ = tπ,ρ m˜π (λ+kρ) ]. It remains to examine whether Eπ satisfies condition (c) of Proposition 1.6, i.e. whether the cλ,   λP+, are all distinct. In fact this is true in all cases except Dn   (n4) (and E8, F4, G2 where there is no nonzero minuscule weight). With π as above let pr(x) = wW x,wπr (xV). Clearly the pr, r1, are W-invariant polynomial functions on V, i.e. pr in the notation of Chapter III. (Also p1=0.)

Suppose that R is not of type Dn (n4) E8, F4 or G2, and that if R is of type An the minuscule weight π is the fundamental weight corresponding to an end node of the Dynkin diagram. Then the pr (r2) generate the -algebra .

This is easily checked for R of type A,B or C. For E6 and E7 we refer to M.L. Mehta (Communications in Algebra 16 (1988) 1083-1098).

Suppose now (Dn, n4 excluded) λ,μP+ are such that cλ=cμ, i.e. wW qλ+kρ,wπ = wW qμ+kρ,wπ (all q,k). Operate on both sides with (qddq)r and then set q=1. We obtain pr(λ+kρ) = pr(μ+kρ) (r1). Hence by Proposition 3.2 f(λ+kρ) = f(μ+kρ) for all f, from which we can conclude that λ=μ, thanks to the following general lemma:

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

Let x1,...,xn be distinct points of V and let c1,...,cn. Then I claim that there exists gS(V) such that g(xi) = ci &nsbp; (1in).

To prove this it is enough to construct giS(V)   (1in) such that gi(xj) = δij, for we can then take g = cigi.

For each ordered pair (i,j) with 1i,jn and ij we can find uijS(V) such that uij(xi) = 1 and uij(xj) = 0 (we may take uij linear, nonhomogeneous). Then gi = ki uik does the trick.

Now let (x1,...,xn) and (y1,...,ym) be distinct W-orbits in V. From above there exists gS(V) such that g(xi) = 1   (1in),   g(yj) = 0   (1jm). Let f = 1|W| wWwg. Then f and f takes the value 1 (resp. 0) at each xi (resp. yj). So separates the W-orbits in V, which is equivalent to the result stated.

What about Dn? Here there are 3 minuscule weights, hence 3 operators Eπ. 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 π= highest root of R, so that (2.4) π,α = { 0 or 1, if αR+, απ, 2, if α=π. } As before, let Φπ = TπΔ+ Δ+ and define now (fA) Eπf = wW ( Φπ(Tπf-f) ). Then it turns out, as a result of calculations analogous to those we have already done, that Eπ has the 3 desired properties in the cases (E8,F4,G2) not covered by the previous construction. So the existence of the Pλ(q,t) claimed by Proposition 1.5 is established in all cases.

What about Pλ,Pλq,t? Here I can only conjecture

  1. (C2q) Let λP+. Then (t=qk) Pλ,Pλq,t = αR+ i=1k-1 1-q λ+kρ,α+i λ+kρ,α-i . True if R of type A.

Consider the special case λ=0 of (C2q). Since P0=1 we have P0,P0q,t = 1|W| [Δ]1 and hence (C2q) would give 1|W| [Δ]1 = αR+ i=1k-1 = 1-q kρ,α+i kρ,α-i . (OP 6) I claim that (OP 6) is equivalent to (C1q). Let Δ = αR+ i=0k-1 ( 1-qieα ) ( 1-qi+1e-α ) then ΔΔ = αR+ i=0k-1 1-qi+1e-α 1-e-α (t=qk) and therefore (since Δ is W-invariant) Δ-1 wW wΔ = wW w( 1-te-α 1-e-α ) = W(t) by (4.12) i.e. W(t)Δ = wW wΔ and therefore W(t)[Δ]1 = |W|[Δ]1 (because Δ, wΔ have the same constant term). It follows from (OP 6) that [Δ]1 = W(t) |W| [Δ]1 = αR+ 1-q kht(α)+k 1-q kht(α) [Δ]1 by (4.13) = αR+ i=0k-1 1-q kht(α)+i+1 1-q kht(α)-i . Consider then the polynomial αR+ i=0k-1 ( qkht(α)+i+1 - qkht(α)-i ) = αR+ qkht(α) ( qk+qk-1++q-1-q-1--q1-k ) = αR+ qkht(α) (qk-1) ( 1+q-1++q1-k ) . Now since αR+ 1-tht(α)+1 1-tht(α) = W(t) = i=1r 1-tdi 1-t it follows that αR+ ( tht(α)+1 - tht(α) ) = i=1r tdi-t i.e. that αR+ 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 αR+ 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) (λ=0)= (C1q). (Tip of the iceberg!)

Consider |Pλ|2 as k (i.e. t0). We have Δ(q,0) = αR (eα;q) . (continuous function on T=V/Q) i=1k-1 ( 1-q λ+kρ,α+i ) = ( qλ,α+1 tρ,α ; q ) ( qλ,α+1 tρ,α+1 ; q ) 1 i=1k-1 ( 1-q λ+kρ,α-i ) = ( qλ,α+1 tρ,α-1 ; q ) ( qλ,α+1 tρ,α ; q ) 1 { 1, if ααi, ( qλ,α+1 ; q ), if α=αi. } Hence Pλ,Pλ2 i=1r ( qλ,αi+1 ; q ) -1 = i=1r ( qni+1 ; q )-1 if λ = 1r niπi (ni0). True when λ=0. (From the denominator formula for affine root systems.)

The limiting case q1

As q1 we have (t=qk) Δ(q,t) = αR (eα;q)k Δ(k) = αR (1-eα)k as kernel for the scalar product. So we have Pλ(k) = limq1 Pλ (q,qk). When R is of type A these are Jack's symmetric functions (suitably normalized) with parameter α=1k. When q1, (C2q) is in fact true: (Heckman, Opdam) |Pλ|(k)2 = Pλ,Pλ(k) = αR+ i=1k-1 λ+kρ,α+i λ+kρ,α-i (λP+, k). (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 k=1,   Pλ(k) = χλ and |χλ|2=1.)


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

Suppose D contains a cycle of length m3. Then with a suitable numbering of the simple roots αi we have αi,αi+1 < 0 (1im-1) and αm,α1 < 0. Let θi= angle between αi and αi+1   (αm+1=α1). Then θi is >π2, hence 2π3 and therefore cosθi -12   (1im). Equivalently, if ei = αi |αi| , we have ei,ei+1 -12 (1im) and ei,ei = 1,  ei,ej 0 if |i-j| 2. Consider the vector v = e1++em : we have |v|2 = i,j=1m ei,ej = m+2i<j ei,ej m+2 i=1m ei,ei+1 (em+1,e1) m+2m-12 = 0. So |v|2=0 and hence v=0, impossible as the ci are linearly independent.

Let R be an irreducible root system; α,βR. If |α| = |β| then β=wα for some wW.

Since R is irreducible, V is an irreducible W-module (2.1). Hence Wα spans V, hence there exists wW such that wα,β 0. Replace α by wα, we may assume α,β 0. If α=β we are done; if αβ we have α,β = α,β = ±1 by (1.2)(i) ( α,β α,β = 1,2 or 3 ). Replacing α by -α=sαα if necessary, we may assume that α,β = α,β = +1. Then we have sβ(α) = α-β,   sαsβ(α) = sα(α-β) = -α-(β-α) = -β,   sβsαsβ(α) = β.

Can use this, together with (1.21a), to find the orders of the Weyl groups of the exceptional root systems. (Use φC.)

E6: 72 roots, forming a single orbit. The extended Dynkin diagram shows that the highest root φ is orthogonal to all αi except α6, hence |Wφ| = |W(A5)| = 6! So |W(E6)| = 726! = 27345.


E7: 126 roots, forming a single orbit. Here |Wφ| = |W(D6)| = 256! So |W(E7)| = 256!126 = 2103457 = 7!4!4!.


E8: 240 roots, |Wφ| = |W(E7)|. So |W(E8)| = 240|W(E7)| = 8!6!4! = 21435527.

F4: 24 long roots, |Wφ| = |W(C3)| implies |W(F4)| = 24|W(C3)| = 2483! = 1152 = 2732.

Let αR (reduced), say α = i=1r piαi. The support of α is the subgraph S=Supp(α) of D (= Dynkin diagram of R) obtained by deleting the vertices αi of D for which mi=0.

  1. Show that Supp(α) is connected, for each αR.
  2. Let I[1,r] be such that DI is connected (DI: delete verticies outside I). Show that iI αi R.

  1. May assume that αR+ (because clearly Supp(-α)=Supp(α)). Induction on ht(α) = pi. If ht(α)=1 then α=αi and S consists of one point. So assume ht(α)>1, then S has at least two elements. We have α = iS piαi say (pi>0) and hence pi αi,α = α,α > 0 so that αi,α>0 for some iS and hence (1.4) β=α-αiR: indeed βR+, because the coefficients of β in terms of the αi are 0.
    1. If pi2 then Supp(β)=Supp(α)=S, and ht(β) < ht(α), hence by the inductive hypothesis S is connected.
    2. If pi=1 then β-αiR, hence the αi-string through β is β, β+αi (=α), ..., β+pαi say, where p1. By (1.5) p = β,αi, hence β,αi > 0 and since β = jS ji pjαj it follows that αj,αi > 0 for some jS, ji. Now Supp(β)=S-{i} is connected, by the induction hypothesis; and i is joined to some jS. Hence S is connected.
  2. Induction on |I|. OK if |I|=1, so assume |I|2. Since D contains no cycles, neither does DI, i.e. DI is a tree. Let αi be an end vertex of DI, J=I-{i}. Then DJ is connected, hence αJR, and αi,αj < 0 for some jJ, so that αi,αJ = kJ αi,αk < 0 and therefore (1.4) αI = αi+αJR.

R of type An, 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 BI = {αi | iI} VI = subspace of V spanned by BI RI = RVI WI = subgroup of W generated by si,iI. Then

  1. RI is a root system in VI with basis BI and Weyl group WI, if I0.
  2. WIWJ if IJ.
  3. The subgroup of W generated by WI and WJ is WIJ.
  4. WIWJ = WIJ.
Thus the WI form a lattice of 2r subgroups of W (parabolic subgroups).

Proof of (i).
(i) Check the axioms: RI clearly satisfies (R1), also (R2) because if α,βRI then sα(β) = β-α,βα RVI   (because α,βVI) = RI. BI is a basis of VI, hence every αRI is a linear combination of the αi, iI. The coefficients must be integers all of the same sign, hence BI is a basis of RI. Hence WI is the Weyl group of RI.

Claim. Let (πi) be the basis of V dual to the basis (αi), so that αi,πj = δij. If wW then wWI wπj = πj all   jI.

Proof of claim.
⇒) Let jI. If iI then siπj=πj (because αiπj). Since WI is generated by the si (iI) it follows that wπj=πj for all wWI.

⇐) Let x = jI πj. If w fixes πj for all jI then w fixes x, hence (1.21a) w is a product of reflections sα fixing x. So suppose αR+, say α = i=1rmiαi. If sα fixes x, then α,x = 0, but α,x = jI mj, hence mj=0 for each jI, so that αRVI=RI and sαWI. Hence wWI.

Now do (ii)-(iv):

Proof of (ii)-(iv).
(ii) If IJ we may assume there exists jJ, jI. We have sjWJ but sjWI. (Otherwise sj would fix πj, hence αj,πj=0.) So WIWJ.

(iv) Suppose wWIWJ, then w fixes πj whenever jI or jJ hence whenever jIJ. Hence wWIJ.

Descriptions of the exceptional root systems

Let R be a reduced irreducible root system with basis α1,...,αr, φ the highest root or R,  α0 = -φ. Let D be the extended Dynkin diagram of R, with vertices corresponding to αi   (0ir) and labels mi, where i=0r miα = 0 (m0=1). (OP 8) If we remove the vertex αi of D we get a (usually disconnected) diagram Di of a root system Ri in V with basis Bi = {αj | 0jr, ji}, and Weyl group Wi, say. (In particular R0=R.) Since WiW and BiR it follows that Ri = WiBi R. From (OP 8) we have miαi = -jimjαj Ri. (OP 9) Consider the root lattices Q=Q(R) ,   Qi=Q(Ri). Since RiR we have QiQ, and Q/Qi is a finite group. Let αj   (1jr) be the image of αj in Q/Qi. Then αj = 0 if ji, and (OP 9) shows that miαi=0. Moreover if mαi = 0   (m) then mαi Q(Ri) , hence mαi is a linear combination of the αj, ji. But (OP 8) is the only linear dependence relation among the αj   (0jr), hence m is a multiple of mi. It follows that Q/Qi is a cyclic group of order mi, generated by αi.

Assume now that |α|2=2 for each αR, so that α=α and hence R=R. (R is simply laced.) Then the dual lattices of Q, Qi are P=P(R) and Pi=P(Ri); we have PiP, and Pi/P is cyclic of order mi (4.1), (4.2). So we have the tower of lattices Pi P Q Qi mi f mi and hence if fi=f(Ri) is the index of connection of Ri we have fi = fmi2, i.e. mi2 = fi f .


  1. R of type E8: the extended Dynkin diagram is PICTURE So R8=A8 ,  f8=8+1=9:   f=1 and hence the root lattice Q of E8 lies between the root lattice Q8 = Q(A8) and the weight lattice P8=P(A8): P83P = Q3Q8. This underlies the construction of E8 that I gave earlier: Q(A8)9.
  2. R of type E8: R7=D8 ,   f7=f(D8)=4, P72P = Q2Q7 (Bourbaki's description of  E8).
  3. R of type E8: R4=A4×A4 ,   f4=5×5 P4>5P = Q>5Q4. Exercise: Figure out the roots of E8 in this representation.
  4. R of type E6: PICTURE Here R3=A2×A2×A2,   f3=33=27,   f=|P/Q|=3: P33P3Q3Q3 So we have 3 pairwise orthogonal copies of A2 sitting inside E6: A2 we can think of as the sixth roots of unity in . So we may work in 3(6) and take (e1,e2,e3 basis of 3 as -vector space) α1=e1, α2=ωe1, α4=ωe2, α5=e2, α6=ωe3, α0=e3, where ω = exp2πi3.
    What about α3? We must have 3α3 = -( α1+2α2+2α4+α5+2α6+α0 ) = -(1+2ω) (e1+e2+e3) so that α3 = θ(e1+e2+e3) where θ = -13 (1+2ω) (see figure) PICTURE In this representation the roots of E6 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 233+233 = 18+54 = 72 of them. So they are all the roots of E6.

    Exercise: Figure out the action of the Weyl group in this description.
  5. R of the type E7: PICTURE Here R7=A7,   f7=7+1=8: P7>2P>2Q>2Q7 Q7 is easily described; it is the lattice of all (n1,...,n8) 8 such that ni=0.
    Exercise: Figure out the roots of E7 in this representation ei-ej, 56 of these, 12 ±ei with 4 minus signs, 70 of these. What are the roots of E7 in this description?
    Take αi = ei-ei+1   (1i6),   α0 = e7-e8, then α7 = -12 ( α1+2α2+3α3+4α4+3α5+2α6+α0 ) = -12 ( e1++e4-e5--e8 ). So certainly all of the vectors ±(ei-ej)   (1i<j8) and 12 i=18 ±ei are roots. There are 2(82)+(84) = 56+70 = 126 of these vectors, hence they are the lot.
  6. Again take R=E7: R6=D6×A1. So we may take αi=ei-ei+1   (1i5), α7 = e5+e6, α0 = f7 = e72, and then α6 = -12 ( α1+2α2+3α3+4α4+3α5+2α7+α0 ) = -12 ( e1++e6+f7 ). So certainly the following vectors are roots: ±ei±ej (1i<j6) 4( 6 2 ) =60 of these ±f7 2 of these 12 i=16±ei ±12f7 252=64 of these 126 total So these are all roots.

Orders of the Weyl groups of E6, E7, E8

Recall (Exercise): R irreducible, α,βR,   |α|=|β|. Then α,β are in the same W-orbit. Apply this when R = E6,E7,E8. The roots form a single W-orbit. Consider in particular the highest root φ=-α0. Since φC, the stabilizer Wφ is generated by the si such that αiφ. So we can read off Wφ from the extended Dynkin diagram, and we have Card(R) = |W/Wφ|, hence |W| = |Wφ|Card(R). E6: Card(R) = 72,   Wφ = W(A5), hence |W| = 726! = 27345. E7: Card(R) = 126,   Wφ = W(D6), hence |W| = 126256! = 2103457 = 7!4!4!. E8: Card(R) = 240,   Wφ = W(E7), hence |W| = 2402103457 = 21435527 = 8!6!4!. F4: 24 long roots, Wφ=W(C3), hence |W| = 24233! = 1152 = 2732.

The Weyl groups of E6, E7, E8

  1. For E6 we have |P/Q| = f = 3, hence P>Q>3P. Since |P/3P| = 36 it follows that Q/3P has order 35, hence is a 5-dimensional vector space over the field 𝔽3 of 3 elements. If the scalar product is normalized so that |α|2 = 2 for all αR, it induces a nonsingular scalar product on Q/3P. So W acts on Q/3P as a group of orthogonal transformations, i.e. W<O5(𝔽3). In fact W has index 2 in this group (but it is not SO5(𝔽3), but another subgroup of index 2). Now in general for a Chevalley group G=G(𝔽q) 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 = SO5(𝔽3) then q=3, R is of type B2, so that N=4, d1=2, d2=4. |W| = |SO5(𝔽3)| = 34(33-1)(34-1) = 34880 = 27345.
  2. For E7 we have |P/Q| = f = 2, hence P>Q>2P. Since |P/2P| = 27 it follows that Q/2P has order 26, hence is a 6 dimensional space over 𝔽2. Normalize the scalar product so that |α|2 = 2 for all αR, then it induces a nonsingular alternating bilinear form on Q/2P; W acts and we have an isomorphism W/(±1) ˜ Sp6(𝔽2). For Sp6, the root system is C3, so that here q=2, N=9, d1=2, d2=4, d3=6. |Sp6(𝔽2)| = 29 (22-1) (24-1) (26-1) = 2931563 = 293457 (=12|W|).
  3. Again for E7, Q/2Q is a vector space of dimension 7 over 𝔽2. Normalize the scalar product so that |α|2=1 this time, for all αR. 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 E8. Again we find W/(±1) ˜ O8(𝔽2) (check the orders). For O8 the root system is D4, so that q=2, the d's are (2,4,4,6) and N=12: |SO8(𝔽2)| = 212 (22-1) (24-1)2 (26-1) = 212315263 = 21235527

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 = αR+ 1-tht(α)+1 1-tht(α) . Let hi= number of roots of height i (h1=r). Then 1r 1-tdi 1-t = i1 ( 1-ti+1 1-ti )hi = 1 (1-t)r j1 ( 1-ti+1 ) hi-hi+1. So Π(1-tdi) = Π( 1-tj+1 ) hj-hj+1. Hence hj-hj+1= number of degrees di equal to j+1 so hj hj+1 i.e. η=(h1,h2,...) is a partition and η = ( d1-1,d2-1,...,dr-1 ).

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.


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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