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

Last update: 30 May 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.


R a root system in V; W the Weyl group, acting on V as a group of isometries. (W could be any finite Coxeter group). Each vV (in particular, each αR) may be regarded as a linear function on V: v(x) = v,x (xV). So if v1,...,vkV we may define v1vk as a polydomial function on V: (v1vk)(x) = v1(x) vk(x) (xV). Let 𝒮 denote the -algebra of polynomial functions on V. If x1,...,xr is a basis of V then 𝒮 = [x1,...,xr].

  1. 𝒮 is graded, i.e. 𝒮 = n0𝒮n, 𝒮n the polynomials of degree n.
  2. W acts on 𝒮 as a group of automorphisms: if wW, f𝒮 we define (wf)(x) = f(w-1x), (xV). We have w(f+g) = wf+wg, w(fg) = wfwg (f,g𝒮).
Moreover W respects the grading, i.e. W acts on each 𝒮n.

Let αR, f𝒮. Then sαf-f is divisible by α𝒮.

Let g = sα-f, so that g(x) = (sαf)(x) - f(x) = f(sαx)-f(x) for xV. Hence g(x) = 0 if sαx=x, i.e. if α(x) (= α,x ) = 0. Hence α divides g𝒮.

Δα = 1-sα α : 𝒮𝒮 (αR). Now let = 𝒮W = {f𝒮 | wf=f, all wW}. is the ring of polynomial invariants of W. is a graded ring: = n0 n, n = 𝒮n, (0=).


  1. W=Sn acting on V=n by permuting the coordinates. Then W acts on 𝒮 = [x1,...,xn] by permuting the xi, and is the subring of symmetric polynomials: = [e1,...,en] where ei are the elementary symmetric functions.
  2. W=Sn acting on V= hyperplane x1++xn = 0 in n. This time 𝒮 = [x1,...,xn] but now the xi are not independent: they are related by x1++xn=0. So now = [e2,...,en] (e1=0).
  3. W= Weyl group of type Bn, acting on n by signed permutations. If f(x1,...,xn), then f is unchanged by replacing xi by -xi, hence is a symmetric polynomial in the xi2. So this time = [f1,...,fn] where fi=ith elementary symmetric function of x12,..., xn2.
  4. Any R: αR(t+α) = αR+(t2-α2) is W-invariant, hence the elementary symmetric functions of the α2 (αR+) are W-invariants.


Define a:𝒮𝒮 by a(f) = 1|W| wW wf.

  1. a(f) for all f𝒮.
  2. a(f)=f if f.
  3. a(fg)=a(f)g if f𝒮, g.

Our first aim is to prove Chevalley's theorem.

There exist homogeneous elements I1,...,Ir (r=dim(V)) such that = [I1,...,Ir] and I1,...,Ir are algebraically independent over .

Let + = n1 n, so that an element g lies in + if and only if g(0)=0. Let 𝒥=𝒮+ be the ideal in 𝒮 generated by +, so that each element of 𝒥 can be written as a finite sum gihi with gi𝒮 and hi+. Since 𝒮 is a polynomial ring, it follows from Hilbert's basis theorem that there is a finite set I1,...,In of homogeneous elements of + such that (a) I1,...,In generate the ideal 𝒥; (b) no proper subset does. I claim that these Ij do what we want. So we have three things to prove:

  1. every element of is a polynomial in the Ij (with real coefficients);
  2. I1,...,In are algebraically independent over ;
  3. n=r.

Proof of 1.
Let f, without loss of generality homogeneous of degree d. Induction on d. d=0 is OK; suppose d>0, then f+𝒥, hence f = j=1n pjIj, with pj𝒮. Since f and the Ij are homogeneous, we may assume pj homogeneous of degree d-dj<d (where dj=degIj). Now apply the averaging operator (Prop. 2.1): f = j=1n a(pj) Ij. By (Prop. 2.1), a(pj) and has degree <d, hence, by the inductive hypothesis, is a polynomial in I1,...,In. Hence, so is f.

For the proof of 2. we require a lemma:

Let F1,...Fm be such that F1 is not in ideal 𝔞 of generated by F2,...,Fm. Suppose p1,...,pm are homogeneous elements of 𝒮 such that i=1m piFi = 0. Then p1 𝒮+ = 𝒥.

Induction on d=degp1.
  1. d=0, then p1: by averaging the relation i=1m = 0 over W we get p1F1 = a(p1F1) = - i=2m a(pi) Fi 𝔞, by (Prop. 2.1). If p10 this shows that F1𝔞, a contradiction. Hence p1=0𝒥.
  2. d>0. Let αR. By (Prop. 1.1) we have pi-sαpi = αqi say qi = Δαpi where qi is homogeneous of degree degpi-1. From the relation piFi=0 we deduce sα(pi)Fi =0 and hence on subtraction i=1m qiFi = 0. Each qi𝒮 is homogeneous, and degq1 = degp1-1 = d-1, so by the inductive hypothesis q1𝒥, i.e. p1-sαp1 𝒥, or p1 sαp1 mod 𝒥. Since the sα generate W and 𝒥 is W-stable it follows that p1 wp1 mod 𝒥 for all wW, or p1-wp1𝒥. Averaging over W we get p1-a(p1) 𝒥, hence p1𝒥 (because a(p1)+).

Proof of 2.
Suppose F is a nonzero polynomial in n variables such that F(I1,...,In)=0. We may assume that all the monomials in the Ij that occur in F have the same degree d as elements of 𝒮, and that d is minimal. Let Fi=FIi   (1in), then degFi<d and so Fi0.

Consider the ideal in generated by F1,...,Fn. We may choose the numbering so that F1,...,Fm, but no proper subset, generate this ideal. Then there exist gij such that Fi = j=1m gijFj, (i<m). (Inv 1) Fi is homogeneous of degree d-di ( di = degIi ). Hence we may assume gij homogeneous of degree dj-di (so zero if dj<di).

Since F(I1,...,In) = 0 we have Fxk = 0 (1kr), i.e. i=1n Fi Ii xk = 0 (1kr). Using (Inv 1) this becomes i=1m Fi Ii xk + i=m+1n j=1m gijFj Ii xk = 0 for i=1m Fi Ii xk + j=m+1n gji Ij xk = 0. Now apply the Lemma 2.3: F1 is not in the ideal of generated by F2,...,Fm, and each of the polynomials Ii xk + jgji Ij xk is homogeneous (of degree di-1). Hence I1 xk + i=m+1n gj1 Ij xk lies in 𝒥, hence is of the form i=1n hikIi with hik𝒮 homogeneous of degree (d1-1)-di. In particular, h1k=0, so that I1 xk + i=m+1n gj1 Ij xk = i=2n hikIi, (1kr). Now multiply by xk and sum over k, remembering Euler's formula: we get d1I1 + m+1n gj1djIj = i=2n k=1r hikxk Ii, showing (since d1>0) that I1 lies in the ideal of 𝒮 generated by I2,...,In: a contradiction.

It remains to prove that n=r.

Proof of 3.
Let K = (f1,...,fn) = field of fractions of , L = (x1,...,xr) = field of fractions of 𝒮. Then tr.deg.K/=n, tr.deg.L/=r. Now transcendence degree is additive in towers L>K>: tr.deg.L/ = tr.deg.L/K + tr.deg.K/. So we have to prove that tr.deg.L/K=0, i.e. that L/K is an algebraic extension. Since L=K(x1,...,xr) it is enough to show that each xi is algebraic over K, i.e. is a root of an equation with coefficients in K. Consider the equation in t wW (t-wxi) = 0. It has a root, and its coefficients are clearly W-invariant, i.e. K. This completes the proof of Theorem 2.2.

In Theorem 2.2 the invariants I1,...,Ir are not uniquely determined, but their degrees di=degIi are. To see why this is so we introduce Poincaré series.

Poincaré Series

Let M = n0Mn be a graded vector space (over ), with each Mn finite dimensional. We define the Poincaré series of M to be P(M,t) = n0 (dimMn) tn [[t]]. If N = n0Nn is another graded vector space (with dimNn<, all n) then MN and MN are graded. (MN)n = MnNn, (MN)n = p+q=n MpNq, from which it follows that

P(MN,t) = P(M,t) + P(N,t), P(MN,t) = P(M,t) P(N,t).

Suppose M=[y] where y𝒮 is homogeneous, degy=d1. Then dimMn = { 0, if dn, 1, if dn, } so that P(M,t) = m0 tmd = 1 1-td . From this and (Prop. 3.1) we deduce

  1. P(𝒮,t) = 1 (1-t)r ,
  2. P(,t) = i=1r 1 1-tdi .

𝒮 = [x1,...,xr] [x1][xr], (degxi=1), from which i. follows. Likewise = [I1,...,Ir] [I1][Ir], (degIi=di) by Chevalley's Theorem (2.2), which gives ii.

Now suppose that also = [I1,...,Ir], where I1,...,Ir + are homogeneous and algebraically independent over . If di = degIi, then from (Prop. 3.2) we have P(,t) = i=1r (1-tdi)-1 and hence i=1r (1-tdi) = i=1r (1-tdi), from which it follows that (after renumbering if necessary) di = di (1ir). The numbers d1,...,dr are called the degrees of W.

(Molien's formula) P(,t) = 1|W| wW 1 det(1-wt) .

In general, if U is a finite dimensional vector space (over a field of characteristic 0) and φ:UU is an idempotent linear transformation (i.e. φ2=φ) then trφ = rkφ ( =dimφ(U) ) (for each eigenvalue of φ is 0 or 1, hence with respect to a suitable basis of U the matrix of φ is 1r 0 0 0 . )

Apply this to the averaging operator a acting on 𝒮n: dimn = dima(𝒮n) = tr(a,𝒮n) = 1|W| wW tr(w,𝒮n), so that P(,t) = n0 (dimn)tn = 1|W| wW n0 tr(w,𝒮n)tn . (Inv 2) Suppose w acting on V has eigenvalues λ1,...,λr (roots of unity, since w has finite order). Then the eigenvalues of w acting on 𝒮n=Sn(V) will all be monomials λ1a1 λrar with ai = n, as one sees by diagonalising w. Hence tr(w,𝒮n) = hn(λ1,...,λr) and therefore n0 tr(w,𝒮n)tn = n0 hn( λ1,...,λr )tr = i=1r 1 1-λit = 1 det(1-wt) .

We shall use (Prop. 3.3) to show that the degrees di satisfy

  1. d1dr = |W|.
  2. d1++dr = r+N, where N = |R+| = number of positive roots.

From (Prop. 3.2) and (Prop. 3.3) we have i=1r 1-t 1-tdi = 1|W| wW (1-t)r det(1-wt) . (Inv 3)
  1. Set t=1. Then all terms on the right vanish except that for w=1, so the right hand side is equal to 1|W|, and the left hand side is equal to 1r 1di. This proves i.
  2. Suppose wW is a reflection. I claim that w=sα, some αR+. For if not, then wx=x for some x not orthogonal to any root, i.e. xVreg. But then wx=x implies w=1 by (1.21) (ref. to Root Systems?), contradiction. Now w is a reflection if and only if det(1-wt) = (1-t)r-1 (1+t). Hence the right hand side of (Inv 3) is of the form 1|W| ( 1+ N(1-t) 1+t + (1-t)2F(t) ), where F(t) is a rational function of t with no pole at t=1. Now put t=1+u in (Inv 3) . The left hand side becomes i=1r u (1+u)di-1 = i=1r 1 di+ (di2)u+ = 1|W| i=1r ( 1+ di-1 2 u + )-1 = 1|W| i=1r ( 1- di-1 2 u + ) = 1|W| ( 1- i=1r di-1 2 u + O(u2) ). On the right hand side, we get 1|W| ( 1- Nu 2+u + O(u2) ). Comparing coefficients of u now gives i=1r di-1 2 = N2.

We shall later improve on (Prop. 3.4).

Suppose W is any finite group of linear transformations of V. Define 𝒮=S(V),   =𝒮W,   𝒥=𝒮+ as before. Then is generated as an -algebra by I1,...,In. (Same proof as before.)

Moreover if (x,y) is any positive definite scalar product on V then x,y = 1|W| wW (wx,wy) is a W-invariant positive definite scalar product (positive definite because x,x = 1|W| wW (wx,wx)0, and vanishes if and only if x=0). So we may assume W is a group of isometries (or orthogonal transformations) of V (relative to a suitably chosen scalar product).

Hence it makes sense to say that W is generated by reflections. Reflection: r-1 eigenvalues equal to 1.

Consider the following four statements:

  1. W is generated by reflections.
  2. Lemma 2.3 holds.
  3. I1,...,In are algebraically independent. (C') n=r.
  4. P(,t) = i=1n (1-tdi) for suitable positive integers d1,...,dn.
Then we have seen that A ⇒ B ⇒ C ⇒ D, and C ⇔ C'. Moreover it is easy to see that conversely D ⇒ C, for D shows that the dimension of each n is equal to the number of monomials I1a1 Irar it contains.

In fact it is also true that C ⇒ A (Shephard and Todd). Hence A, B, C, D are all equivalent. In particular the weird Lemma 2.3 which appears to have no geometric content, is equivalent to W being a reflection group.

Moreover Molien's formula (Prop. 3.3) is valid for any finite W<GL(V). Hence we can avoid the appeal to field theory to prove that n=r. We have i=1n 1 1-tdi = 1|W| wW 1 det(1-tw) . The left hand side has a pole of order n at t=1; the right hand side has a pole of order r=dimV, arising from the term 1 det(1-tw) = 1 (1-t)r when   w=1 (all other terms have poles of order <r at t=1).

Finally we can improve on (Prop. 3.4). For each wW let d(w) = dimVw = multiplicity of 1 as an eigenvalue of   w. Then

(Shephard and Todd, Solomon) wW td(w) = i=1r (t+di-1).

This implies (Prop. 3.4.i) (set t=1) and (Prop. 3.4.ii): d(w)=r-1 if and only if w is a reflection, hence N = no. of reflections   = i=1r (di-1).

Exercise: Verify Theorem 3.7 when W=Sn acting on V=n by permuting the coordinates.

Decompose wSn into disjoint cycles. Then each cycle gives a single w-fixes vector (e.g. the cycle (1 2 r) gives x1++xr) and hence d(w) = l(τ(w)), where τ(w) is the cycle type of w. Hence 1n! wSn Xd(w) = ρn zρ-1 Xl(ρ). (Inv 4) Let φX: Λ[X] be defined by φX(pr) = X, all r1. Then the right hand side of (Inv 4) is φX ρn zρ-1 pρ = φX(hn). Now if X replaced by a positive integer N, then φN(pr) = N for all r1, so that φn(f) = f (1,...,1) n for all fΛ. Hence φN(hn) = hn(1,...,1) = coefficient of   tn   in   (1-t)-N = N(N+1)(N+n-1) n! . Since this is true for all positive integers N, it is true identically: 1n! wSn Xd(w) = φX(hn) = X(X+1)(X+n-1) n! giving wSn Xd(w) = i=1n (X+i-1) in agreement with Theorem 3.7, since the di are (1,2,...,n).

In fact there is a fancier version of Theorem 3.7. Let m be a positive integer, ζ a primitive mth root of unity, e.g. ζ = e2πim. For each wW, let dm(w) = multiplicity of  ζ  as as eigenvalue of  w (this is independent of which primitive mth root of unity we choose).

wW tdm(w) = i=1r ( tχ(m|di) + di-1 ), where χ is the truth function, i.e. χ(mdi) = { 1, if mdi, 0, if mdi. }

m=1 gives Theorem 3.7: d1(w)=d(w). (So the degrees di determine the eigenvalue distributions in W.)

Proposition 3.8 is a consequence of a fancier version of Molien's formula:

1|W| wW det(1+qw) det(1-tw) = i=1r 1+qtdi-1 1-tdi , where t,q are independent variables.

I won't prove Proposition 3.9 yet, but I will show you how to deduce Proposition 3.8 (and hence in particular Theorem 3.7) from it.

Proof of Proposition 3.8, assuming Proposition 3.9.
First replace q,t by qζ, tζ: then (Prop. 3.9) becomes 1|W| wW det(ζ+qw) det(ζ-tw) = i=1r ζdi+qtdi-1 ζdi-tdi . (Inv 5) Now set q=(1-t)X-1, where X is another variable, and then let t1.

Consider first the left hand side of (Inv 5). Let wW and let λ1,...,λr be the eigenvalues of w (acting on V as usual). Then det(ζ+qw) det(ζ-tw) = i=1r ζ+λiq ζ-λit = i=1r ζ+λi((1-t)X-1) ζ-λit . Suppose first λiζ. On putting t=1 we get ζ-λi ζ-λi = 1. If on the other hand λi=ζ, the corresponding factor in the product is 1+(1-t)X-1 1-t = X, so when we put q=(1-t)X-1, the limit as t1 of det(ζ+qw) det(ζ-tw) is just Xdm(w), and hence the left hand side of (Inv 5) gives 1|W| wW Xdm(w). Now consider the factors on the right hand side: ζdi + qtdi-1 ζdi-tdi = ζdi + ( (1-t)X-1 ) tdi-1 ζdi - tdi . Suppose first that mdi, so that ζdi1. We can then safely put t=1 and get ζdi-1 ζdi-1 = 1. If on the other hand mdi so that ζdi=1, we have to compute the limit as t1 of 1+ ( (1-t)X-1 )tdi-1 1-tdi = 1+(X-1)tdi-1-Xtdi 1-tdi , which we can do by l'Hôpital's rule: differentiate numerator and denominator with respect to t, and then set t=1. This gives (di-1)(X-1)-diX -di = X+di-1 di . So the limit as t1 of the right hand side of (Inv 5) is mdi X+di-1 di = i=1r Xχ(mdi)+di-1 di = 1|W| i=1r ( Xχ(mdi) +di-1 ), since we know (Prop. 3.4) that d1dr = |W|. This completes the proof of Proposition 3.8 (modulo Proposition 3.9 which I will prove later, perhaps).


  1. Take m=2, i.e. ζ=-1. Then -1W if and only if the right-hand side of (Prop. 3.8) has degree rd1,...,dr are even.
  2. If m does not divide any di then dm(w)=0 for all wW. So all eigenvalues ζ of wW satisfy ζdi=1 for some i. (Is this otherwise obvious?)
  3. Reference: C.L. Morgan, Can. J. Math. (1979) vol. 31, 252-254.

W-skew polynomials

Each wW (acting on V) is an orthogonal transformation, hence ε(w) = det(w) = ±1, ε:W{±1} is the sign character of W. Clearly ε(w1w2) = det(w1w2) = det(w1) det(w2) = ε(w1) ε(w2), i.e. ε is a homomorphism of W into the 2-elements group {±1}. In particular, ε(sα)=-1 for a reflection sα. Hence if l(w)=p and w=t1tp is a reduced expression for w, where each ti is an sj, we have ε(w) = ε(t1)ε(tp) = (-1)p, i.e.,

ε(w) = (-1)l(w).

A polynomial f𝒮 is W-skew if wf = ε(w)f, for all wW.

Let p = αR+α, the product of the positive roots (if R is of type An then p is the Vandermonde determinant).

The set of W-skew polynomials in 𝒮 is p.

Suppose f𝒮 is skew. Then for each αR+ we have sαf=-f, so that f = 12( f-sαf ) is divisible by α by (Prop. 1.1). Hence f is divisible by p in 𝒮, say f=pg.

Moreover p itself is skew, because by (1.9) (Ref to Root Systems page?) sip = αR+ si(α) = -αi αR+ ααi α = -p and therefore wp = (-1)l(w)p = ε(w)p by (Prop. 4.1). Hence in the factorization f=pg both f and p are skew, hence g is symmetric, i.e. g, and conversely each pg, g is W-skew.

As before let I1,...,Ir be a set of fundamental polynomial invariants.

det( Ii xj ) = λp   for some   λ,  λ0.

In general, if φ:VV is a mapping, say φ(x) = ( φ1(x) ,..., φr(x) ), x = (x1,...,xr), where the φi have partial derivatives, the Jacobian of φ is Jφ = det( φi xj ): V. In particular, if φ is linear, say φi(x) = j=1r aijxj, then φi xj = aij and so Jφ=detφ (constant).

If θ:VV is another mapping, then by the chain rule we have Jθφ = Jθ(φx) Jφ(x) . (Inv 6) Define θ by θ(x) = ( I1(x) ,..., Ir(x) ) and let φ=wW. Then (θφ)(x) = θ(φ(x)) = θ(w(x)) = θ(x) (because the Ij are W-invariant), i.e. θφ=θ. So (Inv 6) gives Jθ(x) = Jθ(wx) det(w), i.e. Jθ = ε(w) w-1Jθ for all wW, whence J=Jθ is W-skew. But also J is a homogeneous polynomial of degree i=1r (di-1) = N = degp by (Prop. 3.4.ii), hence by (Prop. 4.2) we have J=λp for some λ, and it remains to show that λ0.

Consider as before the fields of fractions K,L of ,𝒮 respectively. We have L = K(x1,...,xr) and each xi is algebraic over K. Hence for each i there is a polynomial Fi(t) [t] of minimal degree such that Fi(xi)=0. Differentiate with respect to xk: j=1r Fi Ij (xi) + δik Fi(xi) = 0, (1kr) where Fi = dFi dt . Write these equations as a matrix identity, say AB=C where A = ( Fi Ij (xi) ), B = ( Ij xk ), C = -diag( Fi(xi) ). Taking determinants we get (detA) (detB) = (-1)r i=1r Fi(xi) 0, by the minimality of the Fi.

Hence J=detB0 in 𝒮.

The last part of this proof uses only that I1,...,Ir are algebraically independent over . Indeed, if k is a field of characteristic 0, and I1,...,Ir k[x1,...,xr], then I1,...,Ir are algebraically independend over k if and only if the Jacobian det( Ii xj ) is not identically zero.

(If I1,...,Ir are algebraically independent over k, let f(y1,...,yr) be a nonzero polynomial of minimal degree such that f(I1,...,Ir) = 0. Since f is not constant, not all the partial derivatives f yi vanish, hence (as they have smaller degree than f), not all the polynomials gi = f yi (I1,...,Ir) are zero. But by differentiating f(I1,...,Ir)=0 with respect to xj we get i=1r gi Ii xj = 0, which shows that det( Ii xj ) =0. )

(Shephard and Todd) Let W be a finite subgroup of GL(V), and suppose that is generated as an -algebra by r independent homogeneous elements I1,...,Ir. Then W is generated by reflections.

Let W be the subgroup of W generated by the reflections in W, and let =𝒮W be its ring of invariants. By Chevalley's theorem we have = [ I1,...,Ir ] with I1,...,Ir algebraically independent. We have , hence each Ii is (uniquely) a polynomial in the Ij, and Ii xk = j=1r Ii Ij Ij xk so that det( Ii xk ) = det( Ii Ij ) det( Ij xk ). By the remark above it follows that Ii Ij 0. Expanding, some product Iσ(1) I1 Iσ(r) Ir 0, (σSr). Renumbering the I's we may assume that I1 I1 Ir Ir 0. Let di = degIi,  di = degIi. Then we must have didi   (1ir). On the other hand, from (Prop. 3.4.ii), if N is the number of reflections in W (or W) N = (di-1) = (di-1). So 1r (di-di) = 0 and therefore di=di. But by (Prop. 3.4.i) |W| = di and |W| = di. Hence |W|=|W| and finally W=W.

Differential operators

Let x1,...,xr be coordinates relative to an (orthonormal) basis of V, so that 𝒮 = [x1,...,xr]. Let End𝒮 denote the -algebra of linear maps 𝒮𝒮. Define an -algebra homomorphism : 𝒮End𝒮 by (xi) = xi . Explicitly, if f𝒮, say f = α cαxα   ( α = ( α1,...,αr ) r;   cα;   xα = x1α1 xrαr ) then (f) = αcα ( x1 )α1 ( xr )αr. We shall write f in place of (f). The operators f:𝒮𝒮 are linear partial differential operators with constant coefficients.

The action of W is given by

Let f𝒮, wW. Then wf = wfw-1.

If this is true for f and g, then it is true for fg: w(fg) = wfwg = wf wg = ( wfw-1 ) ( wgw-1 ) = wfgw-1 = wfgw-1. So it is enough to prove Proposition 5.1 when f is linear, say f=ξV. The easiest thing to do is to go back to the definition: if g𝒮 then (ξg)(x) = limt0 g(x+tξ)-g(x) t and therefore (wξg)(x) = limt0 g(x+tξ)-g(x) t = limt0 ( w-1g )( w-1x+tξ ) - ( w-1g )( w-1x ) t = ( ξ(w-1g) )(w-1x) = (wξ (w-1g) )(x) giving wξg = wξw-1g, i.e. wξ = wξw-1.

Scalar product on 𝒮

The scalar product on V induces one on 𝒮, which may be defined as follows: if f,g𝒮 then f,g = ( fg )(0). If wW we have wf,g = f,w-1g, because wf,g = (wfg)(0) = ( wf( w-1g ) ) (0) by (Prop. 5.1) = f( w-1g )(0) ( because w-1(0)=0 ) = f,w-1g. As before let x1,...,xr be coordinates relative to an orthonormal basis in V and let α,βr be multi-indices. Then the definition gives xα,xβ = i=1r ( xi )αi xiβi evaluated at 0, which is zero if αβ and is equal to α! = αi! if α=β: xα,xβ = α!δαβ. It follows that the scalar product is symmetric (not immediately obvious from the definition) and positive definite, and that the monomials xα form an orthogonal (but not orthonormal) basis of 𝒮. The homogeneous components 𝒮n of 𝒮 are pairwise orthogonal, and on 𝒮1=V we have our original scalar product.

Example. Let ui,vi V   (1in), then f = u1un,   g = v1vn 𝒮n and f,g = u1 un ( v1vn ). Since ui(vj) = ui,vj it follows that f,g = wSn i=1r ui,vw(i) = per( ui,vj ), the permanent of the n×n matrix (ui,vj). (This is, as far as I am aware, the only place the permanent occurs in nature.)

Let f𝒮 and let μf: 𝒮𝒮 be multiplication by f:  μfg = fg.

f is the adjoint of μf.

Let gh𝒮. Then g,fh = g (fh) (0) = fg(h)(0) = fg,h = μfg,h.

Harmonic polynomials

A polynomial h𝒮 is said to be W-harmonic if fh=0 for all f+. Since is generated by I1,...,Ir (Thm. 2.2) it is enough to require that Iih = 0   (1ir).

  1. If x1,...,xr are coordinates in V relative to an orthonormal basis, then |x|2 = x12++xr2, hence certainly a W-harmonic polynomial h must satisfy 1r 2h xi2 = 0, i.e. it is a solution to Laplace's equation, hence is harmonic in the usual sense of the word.
  2. 1 is harmonic; so is p, because Iip is skew of degree <N, hence zero.

Let 𝒮 be the vector space of W-harmonic polynomials. It is a graded vector space: = n0 n, n = 𝒮n. Moreover each n is W-stable, because if hn and wW, f+ we have f(wh) = wf(wh) = wfh = 0 using (Prop. 5.1). So is a graded W-module.

Recall that 𝒥=𝒮+ is the ideal in 𝒮 generated by I1,...,Ir.

is the orthogonal complement of 𝒥 in 𝒮, i.e. =𝒥.

We have only to apply the definitions: h fh = 0 all f+ g,fh = 0 all g𝒮, f+ fg,h = 0 all g𝒮, f+ by (Prop. 6.1) h𝒥.

So 𝒮=𝒥. Next we have

The mapping φ:𝒮 defined by φ(hf)=hf is an isomorphism of W-modules (i.e. φ commutes with the actions of W).

We have to show that φ is (a) surjective, (b) injective.
  1. We show by induction on n that 𝒮n. Clear for n=0 (since 1). From (Prop. 7.2) we have 𝒮n = n+𝒥n = n + p=1n 𝒮n-p p. By the inductive hypothesis, 𝒮n-p, hence 𝒮n. Hence 𝒮𝒮, i.e. =𝒮.
  2. For each α𝒩r, let Iα = I1α1 Irαr . The Iα form an -basis of . Let (hn)n0 be a basis of consisting of homogeneous elements. Then the hnIα form a basis of and we have to show that their images under φ, namely hnIα, are linearly independent over . So we suppose we have a nontrivial relations n,α cnα hnIα = 0, cnα, cnα 0. (Inv 7) By homogeneity we may assume that deghn + degIα is constant for each term in the sum. Consider the monomials Iα occuring in (Inv 7), and choose one of least degree (N.B. degIα = αidegIi = αidi ). This is not in the ideal of generated by the other Iβ occurring in (Inv 7), otherwise the Iα would be linearly dependent over , contradicting (Thm. 2.2). Hence by the Lemma 2.3 we have h = ncnαhn 𝒥. But h, hence by (Prop. 7.2) h=0, i.e. cnα=0 for each n: contradiction.

  1. P(,t) = i=1r 1-tdi 1-t .
  2. dim = dim(𝒮/𝒥) = |W|.

  1. From (Prop. 3.1) and (Prop. 7.4) we have P(,t) P(,t) = P(𝒮,t) hence P(,t) = P(𝒮,t) / P(,t) = i=1r 1-tdi 1-t by (Prop. 3.2).
  2. From (Prop. 7.2) we have 𝒮𝒥, hence 𝒮/𝒥 as vector spaces, so that dim(𝒮/𝒥) = dim = dimn = P(,1) = i=1r di = |W|, by (Prop. 3.4).

(It follows that n=0 if n > (di-1) = N and that dimN=1. Since p we have N=p.)

Suppose M=n0Mn is a graded vector space over such that W acts on each Mn (and each Mn is finite dimensional). (W here could be any finite group.) Then we can define the "graded character" of M: if wW and t is an indeterminate, we define χM(w;t) = n0 trace(w,Mn)tn [[t]]. If N=n0Nn is another such, then W acts on each MpNq (by the rule w(xy)=wxwy,   xMp,   yNq) and we have trace(w,MpNq) = trace(w,Mp) trace(w,Nq). Hence, as in (Prop. 3.1) we have

χMN(w;t) = χM(w;t) χN(w;t) (wW).

We shall use this to prove

(or equivalently 𝒮/𝒥) affords the regular representation of W.

Since a representation is determined by its character, it is enough to prove that trace(w;) = { |W|, if w=1, 0, if w1. } By (Prop. 7.5) and (Prop. 7.3) we have χ(w;t) = χ𝒮(w;t) χ(w;t) and χ𝒮(w;t) = 1 det(1-tw) as in the proof of Molien's formula (Prop. 3.3). Moreover w acts as the identity on , hence tr(w,n) = dimn for each n and therefore χ(w;t) = n0 (dimn)tn = P(,t) = 1 1-tdi , by (Prop. 3.2). Hence χ(w;t) = (1-tdi) det(1-tw) , (Inv 8) and trace(w,) = n0 trace(w,n) = χ(w;1).

If w=1 we get |W|, and if w1 we get 0, because then χ(w;t) vanishes at t=1 (the numerator of (Inv 8) is divisible by (1-t)r and the denominator by at most (1-t)r-1.

The usual proof of Proposition 7.6 is to invoke the normal basis theorem of Galois theory. I will say something about this later.

From (Prop. 7.6) it follows that the regular representation of W carries a natural grading: =n0n (or on 𝒮/𝒥). For each irreducible representation ρ of W, let mult(ρ,n) denote the number of times ρ occurs in n. Then we can form the generating function Fρ(W;t) = n0 mult(ρ,n)tn 𝒩[[t]]. This polynomial is given by

Fρ(W;t) = 1|W| i=1r (1-tdi) wW χρ(w) det(1-tw) , where χρ is the character of the representation ρ.

The multiplicity of ρ in n is the scalar product (on W) of χρ with the character of n, i.e., mult(ρ,n) = 1|W| wW χρ(w) trace(w,n) . Hence Fρ(W;t) = 1|W| wW χρ(w) n0 trace(w,n)tn = 1|W| wW χρ(w) χ(w;t) = 1|W| i=1r (1-tdi) wW χρ(w) det(1-tw) .

Setting t=1 in Proposition 7.8 gives Fρ(W;1) = χρ(1) = deg(ρ) in accordance with the fact that the regular representation of a finite group contains each irreducible representation ρ as many times as its degree.


  1. W=Sn acting on V=n by permuting the coordinates. Here P(,t) = i=1n (1-ti)-1. The irreducible characters of Sn are χλ, λn; if wSn has cycle type ρ, then det(1-tw) = (1-tρi) so that 1 det(1-tw) = i=1l(ρ) 1 1-tρi = i=1l(ρ) ( 1+tρi+t2ρi+ ) = pρ(1,t,t2,...), and therefore by (Prop. 7.8) Fλ(Sn;t) = i=1n (1-ti) ρn zρ-1 χρλ pρ(1,t,t2,...) = i=1n (1-ti) sλ(1,t,t2,...) = tn(λ) i=1n (1-ti) xλ (1-th(x)) Check: Fλ(Sn;1) = n! xλh(x) = degree of   χλ, as it should.
  2. If ρ is an irreducible representation, so is ερ, with character εχρ. We have Fερ(W;t) = tNFρ(W;t-1) from (Prop. 7.8) (use di = N+r, (Prop. 3.4)). In particular, F1(W;t)=1 and Fε(W;t)=tN.

Recall: K = (I1,...,Ir) = field of fractions of , L = (x1,...,xr) = field of fractions of 𝒮. W acts on 𝒮, hence on L: w(fg) = wfwg,   ( f,g𝒮 ).

  1. Every element of L is of the form fg where f𝒮, g.
  2. K=LW.

  1. Let uL, say u=fg (f,g𝒮). Multiply numerator and denominator by w1 wg, then u=f*g* say with g* = wW wg .
  2. Let uL;   u=fg   (f𝒮, g). Then u is fixed by W if and only if f is, i.e. if and only if f. So uLW uK.

  1. 𝒮 is a free -module of rank |W|.
  2. And -basis of is an -basis of 𝒮.
  3. Any -basis of 𝒮 is a K-basis of L.

Let (hi) be an -basis of (1i|W|), so that = ihi and hence = i hi. It now follows from (Prop. 7.3) that 𝒮 = i hi. This proves ii. and hence i. As to iii., let (fi) be any -basis of 𝒮, so that every f𝒮 has a unique expression f = figi, with gi. Hence if g (g0) we have fg = fi gig uniquely. By (Prop. 7.10(i)) this shows that (fi) is a basis of L as a K-vector space.

If L is any field, W any finite group of automorphisms of L, K=LW the subfield of invariants, then there exists xL such that the transforms wx   (wW) of x are all distinct and are a K-basis of L (so that (L:K) = |W| ). This is the "normal basis theorem" of Galois theory. We have managed to avoid it by consideration of Poincaré polynomials.

𝒮/𝒥 = H*(G/B); G/B   flag manifold.

Divided differences

Let αR,  f𝒮. Recall (Prop. 1.1) that f-sαf is divisible by α in 𝒮. Hence we may define an operator Δα: 𝒮𝒮 by Δα = α-1 (1-sα) (i.e., Δαf = α-1 ( f-sαf ) ). Δα has degree -1, i.e. it maps 𝒮n into 𝒮n-1 (n0). Clearly it kills .

  1. Δα2 = 0.
  2. Δα(fg) = (Δαf)g + (sαf)(Δαg) (f,g𝒮).
  3. Δα(fg) = (Δαf)g   if   f𝒮,g ( i.e. Δα is -linear ).
  4. Δαf = f,α   if   f   is linear   ( i.e.   f𝒮1 = V ).

  1. We have sαΔα = (sαα)-1 ( sα-sα2 ) = (-α)-1(sα-1) = Δα, hence Δα2 = α-1(Δα-sαΔα) = 0.
  2. Δα(fg) = α-1( fg- (sαf) (sαg) ) = α-1{ (f-sαf)g + sαf (g-sαg) } = (Δαf)g + (sαf)Δαg.
  3. If g then sαg=g, hence Δαg=0.
  4. If fV then sαf = f-f,αα, so Δαf=f,α.

We shall use these operators only when α is a simple root of αi, and we shall write Δi=Δαi   (1ir). For any sequence a_ = (a1,...,ap) with each ai[1,r], let Δa_ = Δa1Δap. Say a_=(a1,...,ap) is reduced if w = sa1 sap is a reduced expression, i.e. if l(w)=p.

  1. If a_ is reduced, then Δa_ depends only on w = sa1 sap (write Δw=Δa1Δap).
  2. If a_ is not reduced, then Δa_=0.

It is tempting to attempt to prove (i) by observing that it is only necessary to verify that the Δi satisfy the Coxeter relations ΔiΔjΔi = ΔjΔiΔj (Inv 9) (mij terms on either side, where mij= order of sisj in W). However it seems not to be easy to verify (Inv 9) by brute force, and we therefore adopt another approach (due to Bernstein, Gelfand and Gelfand).

We shall prove (i) and (ii) by induction on the length of the sequence a_. p=0 is OK, so assume p1.
  1. Let a_ be reduced, w=sa1sap. We shall compute Δa_ (fg) when f,g𝒮 and f is linear, using Proposition 8.1.ii: Δa_ (fg) = Δa1 Δap (fg) = sa1 sap(f) Δa1 Δap (fg) + i=1p ( sa1 sai-1 Δai ( sai+1 sapf ) ) Δa1 Δ^ai Δapg. (Inv 10) Now by Proposition 8.1.iv we have Δai sai+1 sapf = sai+1 sapf, αai = f,βi , where βi = sap sai+1 (αai). By (Root Systems, 1.16) the roots β1,...,βp are precisely the positive roots β such that wβ is negative. Also, by the inductive hypothesis, Δa1 Δ^ai Δap = 0 unless sa1 s^ai sap is reduced, i.e. of length p-1; moveover sa1 s^ai sap = wsap sai sap = wsβi. So we may rewrite (Inv 10) as follows: Δa_ (fg) = (wf) Δa_ (g) = β f,β Δwsβ(g) (Inv 11) summed over βR+ such that (1) wβ<0 and (2) l(wsβ) = l(w)-1. Thus the right hand side of (Inv 11) depends only on w (and f,g), not on the reduced word a_ for w.

    Now let b_=(b1,...,bp) be another reduced word for w, and let E = Δa_ - Δb_. This is an operator of degree -p, so that in particular E(1)=0. From (Inv 11) it follows that E(fg) = (wf)E(g) (g𝒮) (Inv 12) if f𝒮 is linear. But then by iteration we see that if f1,...,fn are linear then E(f1,...,fng) = (wf1) E(f2fng) == w(f1fn) E(g) and hence (Inv 12) is true for all f𝒮. Taking now g=1 we have E(f) = (wf)E(1) = 0 i.e., E=0 and therefore Δa_ = Δb_. This completes the induction step for Proposition 8.2.i.
  2. Suppose now a_=(a1,...,ap) is not reduced, and let a_ = ( a1,...,ap-1 ). If a_ is not reduced then Δa_ = 0 by the inductive hypothesis, hence Δa_ = Δa_ Δap = 0. If a_ is reduced, let w = sa1 sap,  v = sa1 sap-1 so that l(v)=p-1 and v = wsap; since l(w)p-1 it follows that l(w)=p-2, and hence Δv = Δw Δap. Consequently Δa_ = Δv Δap = Δw Δap2 = 0 by Proposition 8.1.i.

From Proposition 8.2 it follows that

Let v,wW. Then Δv Δw = { Δvw, if l(vw) = l(v) + l(w), 0, otherwise. }

(Consider reduced words for v,w and apply Proposition 8.2.)

Let wW. Then Δw is an operator of the form Δw = v avwv, where avwL (the field of fractions of 𝒮) and in particular aww = ε(w) αR(w) α-1, where R(w) = R+ wR-. [In fact avw=0 unless vw (Bruhat order) ... ]

Let w = sa1 sap   ( p=l(w) ). Then Δw = Δa1 Δap = αa1-1 (1-sa1) αa2-1 (1-sa2) αap-1 (1-sap) in which the coefficient of w = sa1 sap is (-1)p ( αa1 sa1 (αa2) sa1 sa2 (αa3) sa1ap-1 (αap) )-1 = ε(w) αR(w) α-1 by (Root Systems, 1.16).

HW: Show that pavw is a polynomial, i.e. pavw𝒮 for all v,w.

Let w0 be the longest element of W. Then Δw0 = p-1 wW ε(w)w.

Say Δw0 = wW aww, with coefficients aw   (=aw,w0)   L. Since l(siw0) < l(w0) for 1ir, it follows from Proposition 8.4 that ΔiΔw0 = 0, i.e. Δw0si = Δw0. Since the si generate W, we conclude that Δw0 = vΔw0 for all vW, and hence Δw0 = wW v(aw) vw. Comparing this with the previous expression, we see that v(aw) = avw for v,wW*. Hence it is enough to determine one of the coefficients aw; now by Proposition 8.5 we have aw0 = ε(w0)p-1 (since R(w0)=R+). Hence aw = ww0(aw0) = w(p)-1 = ε(w)p-1.

The wW are linearly independent over L. (Dedekind's lemma)

Dedekind's lemma

Dedekind's lemma. Distinct automorphisms of a field L are linearly independent over L.

Suppose we have a linear dependence relation 1n fiαi = 0 with fiL,  αiAutL. May assume n minimal (hence the fi0). Let g,hL. Then we have 1n fiαi(g) = 0, (Inv 13) 1n fi αi(g) αi(h) = 0. (Inv 14) Multiply (Inv 13) by α1(h) and subtract from (Inv 14). 2n fi( αi(h) - α1(h) ) αi(g) = 0, for all  g. Since α2α1 there exists h such that α2(h) α1(h): but then 2n fi( αi(h) - α1(h) ) αi = 0 is a nontrivial linear dependence relation of length <n.

Now let Δi* be the adjoint of Δi: Δi*,g = f,Δig . Since Δi has degree -1, Δi* has degree +1. To get an idea of what Δi* looks like, we prove (although we shan't use it)

Let f𝒮. Then there exists g𝒮 (not unique) such that αig=f (indefinite integral: g=fdαi). For any such g we have Δi*f = g-sig.

Let h𝒮. Then g-sig,h = g,h-sih = g, αiΔih = αig, Δih = f,Δih = Δi*f,h . Since h was arbitrary it follows that g-sig = Δi*f.

Now let wW, w = sa1 sap a reduced expression and define Δw* = Δap* Δa1* of degree   p=l(w). Δw* is the adjoint of Δw and we have, from Proposition 8.4:

Δv* Δw* = { Δwv*, if l(v)+l(w) = l(vw), 0, otherwise. }

(Take adjoints: Δv*Δw* is the adjoint of ΔwΔv (in that order).)

  1. Δw maps 𝒥 into 𝒥.
  2. Δw* maps into .

  1. Let f𝒮, g+. Then by Proposition 8.1.iii Δi(fg) = (Δif)g so that Δw(fg) = (Δwf)g, which proves (i).
  2. Recall Proposition 7.2 that is the orthogonal complement of 𝒥 in 𝒮. Hence if h,  f𝒥 we have Δw*h, f = h, Δwf = 0, because Δwf𝒥. Hence Δw*h 𝒥=.

Now let (wW) hw = Δw*(1). By Proposition 9.4, hw and is homogeneous of degree l(w).

(BGG) { hw | wW } is an -basis of .

Let us first establish that hw00. We have hw0,p = Δw0* (1),p = 1, Δw0p and by Proposition 8.6 and skew symmetry of p we have Δw0p = p-1 wW ε(w) wp = |W|. So hw0,p = |W| 0 and therefore hw00. (It must be a scalar multiple of p.)

Since dim=|W| (Proposition 7.4) it is enough to show that the hw are linearly independent over . So suppose we have a linear dependence relation cwhw = 0 (cw). (Inv 15) Choose vW of minimal length such that cv0. Let u=v-1w0 and apply Δu* to (Inv 15). We have by Proposition 9.3 Δu*hv = Δu* Δv* (1) = Δw0*(1) = hw0 ( because vu=w0 ) and if wv Δu* hw = Δu* Δw* (1) = 0 by Proposition 9.3 because l(u) + l(w) = N-l(v)+l(w) N, and uww0 if wv. So we obtain cvhw0 = 0, hence cv=0.

From Theorem 9.5 it follows that for each n0 dimn = Card{ wW | l(w)=n } and hence P(;t) = wW tl(w). Comparing this with Proposition 8.6 we have the polynomial identity wW tl(w) = i=1r 1-tdi 1-t (Inv 16) valid for any finite reflection group.

Instead of appealing to Proposition 8.6 we can prove directly that the hw span . Indeed let 0 be the span of the hw. Then 0 so that 0 =𝒥, and 0=    0=𝒥. So we need to prove that if   f𝒮   and   f,hw = 0   for all   wW,  then   f𝒥. (Inv 17)

Proof of (Inv 17).
We may assume f𝒮d and proceed by induction on d. If d=0 we have f = f,1 = f,h1 = 0, so f=0𝒥. If d>0 consider hw, Δif = Δi*hw,f = Δi* Δw* (1),f and Δi*Δw* is either Δwsi* or 0, so that hw, Δif if either 0 or equal to hwsi,f = 0. So in both cases hw, Δif = 0 for all wW, hence (induction hypothesis) Δif𝒥, i.e. f sif   (mod 𝒥) for all wW. Hence finally fa(f)   =1|W| wW wf mod𝒥 , and a(f)𝒥+ (because d>0) so finally f𝒥 as required.

(The last part of this argument we have encountered before, in the proof of Chevalley's Lemma 2.3.)

We have hsi=αi because if f𝒮 is linear hsi,f = Δi*(1),f = 1, Δif = αi,f by Proposition 8.1.iv.

Schubert polynomials

As before let p = αR+α and define (wW) Sw = 1|W| Δw-1w0(p) 𝒮. Sw is homogeneous of degree N-l(w-1w0) = l(w) (since l(w-1w0) = l(w0) - l(w-1) = N-l(w) ).

  1. hv,Sw = δv,w (v,wW).
  2. Let Sw_ = Sw+𝒥 𝒮/𝒥. Then ( Sw_ ) wW is an -basis of 𝒮/𝒥.

  1. We have hv,Sw = 1|W| Δv*(1), Δw-1w0(p) = 1|W| 1, ΔvΔw-1w0(p) . Suppose this is 0. Then for reasons of degree we must have l(v) + l(w-1w0) = N i.e., l(v)=l(w), and by Proposition 8.4 l(v) + l(w-1w0) = l(vw-1w0) which forces v=w. So hv,Sw = 0 unless v=w, and hw,Sw = 1|W| 1, Δw0(p) and since by Proposition 8.6 Δw0(p) = 1p wW ε(w) wp = |W| we have hw,Sw = 1.
  2. Suppose wW awSw 𝒥 (aw).

    Take the scalar produce of each side with hv: since hv𝒥 we get w aw hv,Sw = 0 and hence av=0 by (i) above. Hence the Sw_ are linearly independent over , and since dim𝒮/𝒥 = dim = |W|, they are a basis.

Suppose W=Sn acting on 𝒮 = [x1,...,xn] by permuting the xi. Then the monomials xα = x1α1 xnαn with αi n-i form a basis of 𝒮mod𝒥. In particular xδ = x1n-1 x2n-2 xn-1 is a generator of 𝒮Nmod𝒥 ( N = 12n (n-1) ). Since W=Sn acts on (𝒮/𝒥)N as the sign representation, it follows that xwδ = w(xδ) ε(w) xδ mod𝒥 and hence p = i<j (xi-xj) = wSn ε(w) xwδ n!xδ mod𝒥. Hence Sw = 1n! Δw-1w0(p) Δw-1w0(xδ) mod𝒥 which ties up the Sw with the Schubert polynomials of Lascoux and Schutzenberger.

Coxeter elements

Let X be a tree, G a group and S:XG a mapping xsxG. Suppose that sx,sy commute whenever x,y are not joined by an edge in X. Then all the products xX sx are conjugate in G.

Remark first that if g1,...,gnG then g2g3gng1 = g1-1 ( g1g2gn )g1 is conjugate to g1gn. So cyclic permutation of a product produces conjugate elements.

We prove the lemma by induction on the number n of vertices in X. When n=1 there is nothing to prove. Suppose n>1. Since X is a tree it has an end vertex a, say, joined to just one other vertex b. Consider a product π of the sx,xG. Up to conjugacy in G we may assume that sa comes first: π = sasb = sausbv say. Now sa commutes with every sx except sb, hence commutes with u: π = sausbv = usasbv, hence π is conjugate to sasbvu. So it remains to show that all products sasv are conjugate in G, and to do this we use the inductive hypothesis. Let X is a tree with n-1 vertices. Define s: XG by sb = sasb,   sx = sx if xa,b. Now apply the inductive hypothesis to X, s.

Now let W be an ???? Weyl group (more generally, a finite reflection group), s1,...,sr the generators of W. The Dynkin diagram D of W is a tree hence (X,G) = (D,W) satisfies the conditions of the lemma. Consequently all the products sπ(1) sπ(r)   ( πSr ) of the generators s1,...,sr are conjugate in W. Moreover if we had taken a different basis B=wB of R the generators would be si = wsiw-1   (wW), so all the products (for all bases) are conjugate in W. These are the Coxeter elements of W; they form a distinguished conjugacy class.

Example. W=Sn: si = (i,i+1),   1in-1. Then s1s2sn-1 is an n-cycle, so the Coxeter elements of Sn are the n-cycles.

Let cW be a Coxeter element (we are assuming R irreducible). Since conjugate elements of a finite group have the same order, we can define h = order of   c   in   W. h is the Coxeter number of W.

Now consider the characteristic polynomial P(t) = det(1-tc) [t] (if  W  is a Weyl group) whose root are the eigenvalues of c. Again they do not depend on which c we take, and they are hth roots of unity, say ζj = exp 2πiej h (1jr) where 1e1 e2 er<h. (One shows that 1 is not an eigenvalue of c, i.e. that c has no fixed points 0 in V). Since the eigenvalues are also ζj_ (because they are the roots of a real polynomial) it follows that the sequence (h-ej) is the sequence (ej) in reverse order, i.e. ej+ek = h if jk = r+1. (Inv 18) Adding up these equations, we get e1++er = 12hr. (Inv 19) The relevance of this to polynomial invariants is that if d1dr are the degrees of W then di = ei+1 (1ir) ( dj+dk = h+2 if j+k = r+1 ). (Inv 20) Since by Proposition 3.4 1r (di-1) = N it follows that N = ei = 12hr: hr = 2N = Card(R). so h = Card(R) rank(R) = top degree . (Inv 21) Again, since there is a polynomial invariant I2 = |x|2 of degree 2, we have e1=1, hence ζ1 = e2πih is an eigenvalue of c, i.e. a root of P(t) [t]. Hence all primitive hth roots of unity are eigenvalues of c, i.e. each positive integer less than h and prime to h is an exponent. ( e1=1 er = h-1 dr = h so h = max( d1,...,dr ). )


  1. W=Sn acting irreducibly on the hyperplane V in n with equation x1++xn=0. c is an n-cycle, so its eigenvalues are the nth roots of unity other than 1. So h=n and (e1,...,en-1) = (1,2,...,n-1). This agrees with the above, because we have seen earlier that (d1,...,dn-1) = (2,3,...,n).
  2. W=E8. Here r=8 and Card(R)=240, so that h=30. The positive integers less than 30 and prime to 30 are 1,7,11,13, 17,19,23,29. There are 8 of them, hence they are precisely the exponents of E8.
  3. W=E7: r=7,  Card(R) = 126, so h=18. The positive integers less than 18 and prime to 18 are 1,5,7,11,13,17. So these are 6 of the exponents, and the other one must be 12h=9, by (1) above.
  4. W=E6: r=6, Card(R) = 72, so h=12. Hence 1, 5, 7, 11 are four of the exponents. In fact, the other two are 4, 8, but one needs some extra information to deduce this. For example if one knew that |W| = 27345 = di = (ei+1) = 26812(e+1)(e+1) where e,e are the two unknown exponents, then since 26812=2732 we have (e+1) (e+1) = 325 = 45 which together with e+e=12 gives ee=45-1-12=32 and hence e,e=4,8.
  5. W=F4:r=4, Card(R)=48, so h=12 again. Hence the exponents are 1, 5, 7, 11.
In fact the eigenvalues of the Cartan matrix A = ( αi, αj ) 1i,jr are 4sin2 πei 2h   (1ir), by an ingenious argument due to Coxeter [Bourbaki, p.140 Ex. 3,4].

G= simple (or almost simple) compact Lie group, R the root system of G relative to a maximal torus T. Then (over the reals) G has the same cohomology as a product of spheres of dimensions 2ei+1, i.e., H*(G;) is an exterior algebra with generators of degrees 2ei+1   (1ir).

G= Chevalley group over 𝔽q: |G| = |G/B| |B|. |G/B| = wW ql(w) |B| = (q-1)rqN, so |G| = qN(q-1)r wql(w) = qN(qdi-1) = q2N+r+.

Let c=s1sr be a Coxeter element of W and put αik = s1sk(αi), 0kr, 1ir. Thus αi0=αi, αir = c(αi) = βi. Let γi = αi,i-1. We have αi,k-1 - αik = s1sk-1 ( αi-skαi ) = αk, αi s1sk-1(αk) = akiγk, where aki = αk,αi is the (k,i) element of the Cartan matrix A. Thus αi,k-1 = αik + akiγk and therefore, summing from k=1 to i-1, αi = γi + k<i akiγk. (Inv 23) Likewise, summing from k=i to k=r, βi = γi - ki akiγk = - γi + k>i aki γk (Inv 24) since aii=2.

Let U = 1 aij 0 1 , L = 1 aij 1 , so that A=U+L. Then (Inv 23), (Inv 24) show that the matrix expressing the β's in terms of the γ's is -L, and that expressing the α's in terms of the γ's is U-1. Hence if C is the matrix of the Coxeter element c we have C=-U-1L and hence det(λ2-c) = det(λ21r-C) = det( λ2+U-1L ) = det( λ2U+L ) ( because detU=1 ) = λ2+1 λ2aij aij λ2+1 . Now in general if X=(xij) is an r×r matrix, its determinant is a sum of terms ±x1σ(1) xrσ(r) σSr. (Inv 25) Express the permutation σ as a product of disjoint cycles: each cycle (i1,...,ip) in σ will contribute xi1i2 xi2i3 xipi1 to the monomial (Inv 25). Apply this observation to the determinant above: the Dynkin diagram contains no cycles, hence any σSr containing cycles of length >2 will give zero. In other words, any term in the expansion of this determinant that contains λ2aij (i<j) must also contain aji (below the diagonal). In other words we have det(λ2-c) = λ2+1 λaij λaij λ2+1 = det( (λ-1)2 + λA ) since aii=2 = λr det( λ+λ-1 -2+A ). This vanishes when λ2 = exp( 2πiej h ) ( 1jr ), i.e. when λ = ±exp( πiej h ), i.e. when X = λ+λ-1-2 = ±cos πej h -2. Since -cos πej h = cos( π- πej h ) = cosπh (h-ej) = cosπh er+1-j it follows that det(X+A) = i=1r ( X+2-cos πej h ) and hence the eigenvalues of A are 2( 1-cos πej h ) = 4sin2 πej 2h .

The roots γi = sisi-1(αi) are precisely the positive roots γ such that c-1γ is negative (Root systems, 1.18). The formula (Inv 23) shows that they are a basis of V. Let Ωi be the c-orbit of γi: Ωi = { γi,cγi,c2γi,... }. Then one can show that the r orbits Ωi are pairwise disjoint, that each contains h elements, and that R = 1r Ωi.


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

Version: September 20, 2001

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