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

Last update: 04 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.


Let V be a real vector space of dimension r>0. A lattice in V is a subgroup L of V (hence closed under addition and subtraction) that spans V and is generated by r elements x1,...,xr (which are therefore a basis of V). So the elements of L are all linear combinations i=1rnixi, ni (1ir). Let M be another lattice in V, with basis y1,...,yr, and suppose that ML. Then each yiL, hence we have equations yi = j=1raijxj (1ir) (aij). (Exp 1) The matrix A=(aij) is a nonsingular matrix of integers, because it transforms one basis of V into another.

L/M is a finite abelian group of order d = |detA|.

By elementary transformations on rows and columns we can find r×r matrices U,V of integers with determinant ±1, such that UAV is a diagonal matrix, say UAV = d1 dr where the di are positive integers. (We can even arrange that d1|d2||dr, though we shan't need that fact here.)

Let yi = j uijyj and xk = l vkl xl, so that the xk (resp. the yi) are a basis of L (resp. M). From (Exp 1) we have yi = j uijyj = j,k uij ajk xk = j,k,l uij ajk vkl xl = l (UAV)il xl and therefore yi = dixi (1il). So L, M have bases x1,...,xr, and d1x1,...,drxr, and therefore L/M i=1r /di is a direct sum of cyclic groups of orders d1,...,dr, hence |L/M| = d1,...,dr = det(UAV) = ±detA, since U,V have determinant ±1.

Now suppose V carries a positive definite scalar product x,y, and let L be a lattice in V, as before. Let L* = {yV | x,y for all xL}. Let x1,...,xr be a basis of L and let x1*,...,xr* be the dual basis of V: xi,xj* = δij. Then yL* xi,y for 1ir y = 1r nixi*, ni, so that L* is a lattice in V with basis x1*,...,xr*. L* is called the dual of L. We have L**=L (because L** has basis x1,...,xr).

If ML we have L*M*, because if yL* we have x,y for all xL and a fortiori for all xM.

M*/L* is the dual (or character group) of L/M.

(In particular L/M M*/L* — but not canonically.)

Let ηM*, xL and define χη(x+y) = e2πix,η . If yM we have χη(x+y) = e2πix+y,η = e2πix,η e2πiy,η = χη(x) because   y,η , i.e. χη is constant on the cosets of M in L and therefore defines a character of the finite abelian group L/M. Moreover if ξL* then χη+ξ = χη (because ξ,x for xL). Hence ηχη defines a mapping of M*/L* into the character group of L/M. Check that it is an isomorphism.

If y1,...,yr is a basis of M, y1*,...,yr* the dual basis we have yi = y=1r aijxj (aij) and aij = yi,xj* , so that also xj* = i=1raijyi* or, interchanging i and j, xi* = j=1rajiyj*.

The root lattice and the weight lattice

Let R be a reduced root system, spanning V, and let Q=Q(R) be the subgroup of V generated by the roots: so the elements of Q are all linear combinations λ = αR ηαα,   ηα. If B={α1,...,αr} is a basis of R, every αR, and hence every λQ, is uniquely of the form 1rηiαi with integer coefficients ηi: Q(R) = i=1rαi so that Q is a lattice in V, called the root lattice. Likewise Q(R). From above, P(R) is the dual of Q(R): P(R) = Q(R)* hence is also a lattice, called the weight lattice of R. Since β,α for all α,βR (integrality axiom (R1)) it follows that Q(R) P(R) (and likewise Q(R) P(R) ).

Let ( π1,...,πr ) be the basis of V dual to ( α1,...,αr ): πi,αj = δij. The πi are a basis of the weight lattice P(R): they are called the fundamental weights. They span the Weyl chamber C.

Recall that A=(aij) is the Cartan matrix, where aij = αi,αj .

Example. R = {ei-ej | 1i,jr,ij} type   An-1. What are the fundamental weights? αi = αi = ei-ei+1 (1in-1). So πi,αj = δij say πi = uiei,   ui=0 πi,αj = uiei,ej-ej+1 = uj-uj+1. So uj = uj+1 if ji ui-ui+1 = 1 ui+1 = =un=u say u1 = =ui=u+1 0=ui = i(u+1) + (n-i)u = nu+i hence u=-in { πi = 1n( (n-i) (e1++ei) - i(ei+1++en) ) πi = e1++ei-ie e = 1n 1nei } 1in-1; πn=0.

Exercise: Calculate the πi for types B,C,D.

We have αj = i=1r αi,αjπi i.e., αj = i=1r aijπi (1jr) and hence from Proposition 1.1

|P(R)/Q(R)| = detA.

(We know from before that detA is positive.)

The positive integer detA is traditionally denoted by f and called the index of connection: f = |P(R)/Q(R)| = |P(R)/Q(R)| by Proposition 1.2, since P(R) and Q(R) are respectively the duals of Q(R) and P(R).

Finally, W acts on everything: it acts on R, and on R, hence on Q(R) and Q(R), hence on P(R) and P(R). Let Q+ = 1rηiαi ηi0 and P+ = 1rηiπi ηi0 . Then P+ = PC, hence (1.21).

Every W-orbit in P meets P+ in exactly one point.

The λP+ are called the dominant weights.

Let λP.

  1. λ-wλQ, for all wW.
  2. λP+ λ-wλ Q+ for all wW.

  1. We have λ-siλ = λ,αi αi Q, because λ,αi . Hence λsiλ   (modQ). Since W is generated by the si, it follows that λwλ   (modQ) for all wW.
  2. Recall (1.24) the partial order xy on V:   xy iff x-yC*, where C* is the cone spanned by the simple roots αi. We have Q+ = QC*. Now if λP we have λP+ λC λwλ for all wW (1.24) λ-wλ C*Q = Q+, using (i).

Proposition 2.3 (i) says that the Weyl group W acts trivially on the finite group P/Q.

Recall ρ = 12 αR+α is such that ρ,αi = 1 (1ir) (1.9). Hence

ρ = π1++πr PC.

Likewise let σ = 12 αR+ α (warning: ρ is not a good notation, because σ 2ρ |ρ|2 ). By (1.9) applied to R we have σ,αi = 1   (1ir), because (1.23) B = {α1,...,αr} is a basis of R. Hence

For all αR, ht(α)=σ,α.

For if α = 1r niαi, then σ,α = niσ,αi = ni = ht(α). σ is the "altimeter".


If G is a group (finite or infinite) and k is a commutative ring, the group algebra kG consists of all formal sums a = gG agg with coefficients agk, only finitely many non-zero (i.e. kG is the free k-module with G as basis). So in particular each gG is an element of kG. Add and multiply in the obvious way: if b = gG bgg is another element of kG, then a+b = gG (ag+bg)g and ab = g,hG agbhgh i.e. the multiplication in kG is just the multiplication in G, extended by linearity.

If however the group operation in G is addition (g+h rather than gh) we have to change our notation and write e.g. eg for the element of kG corresponding to g: the e's are "formal exponentials" and satisfy egeh = eg+h, e0=1, (eg)-1 = e-g (g,hG). The elements of kG are now written a = ageg and add and multiply as indicated.

This applies in particular to the group P, the weight lattice. Let A=P denote the group algebra of P over k=. So the elements of A are finite sums f = λP fλeλ (fλ, almost all zero ) where eλeμ = eλ+μ,   e0=1. The eλ are an -basis of A.

The ring A is easily described. The fundamental weights πi   (1ir) form a basis of P, so that each λP is uniquely of the form λ = i=1r niπi (ni). Hence if we put yi=eπi we have eλ = en1π1 enrπr = y1η1 yrηr and therefore A = [ y1±1,...,yr±1 ] is the algebra of Laurent polynomials in y1,...,yr (i.e. polynomials in the y's and their inverses). Since [y1,...,yr] is a unique factorization domain, so is A.

Let λ,μP be linearly independent (i.e. not proportional). Then 1-eλ and 1-eμ are coprime in A.

This amounts to showing that if α,βr are not proportional, then 1-yα and 1-yβ have no common factor. Left as exercise.

The Weyl group W acts on P, hence on A: if f = λP fλeλ A then wf = λP fλewλ. fA is

  1. W-invariant (or W-symmetric) if wf=f, all wW.
  2. W-skew (or alternating) if wf=ε(w)f, all wW. ( sif = -f (1ir). )

We shall start by considering the W-skew elements of A. Let θ:AA be the linear mapping defined by θf = wW ε(w)wf (fA). We have (vW) θv = vθ = ε(w)vw = ε(v)θ.

  1. S=θ(A) is the space of W-skew elements of A.
  2. If fA and sαf=f for some αR, then θf=0.
  3. The θ(eλ+ρ), λP+, form an -basis of S.

  1. Let fA. Then θf is W-skew, because if vW we have vθf = ε(v)θf. Conversely if f is W-skew then θf = wW ε(w)wf = |W|f so that f = θf |W| θ(A).
  2. We have θsα = ε(sα)θ = -θ, hence θf = θsαf = -θf and therefore θf=0.
  3. Suppose f = μ fμeμ S. Then f = ε(w)wf = με(w) fμewμ and therefore fwμ = ε(w)fμ. Since each W-orbit in P meets P+ in just one point (Prop. 2.2) it follows that f = μP+fμ wW ε(w) ewμ = μP+ fμθ(eμ). (*) Now μP+ is of the form μ = miπi with mi&he;0. If mi=0 for some i then μ,αi = 0 and hence siμ=μ, so that θ(eμ) = 0 by (ii). Hence in (*) the sum is restricted to μ = miπi with each mi1, hence (Prop. 2.5) μ=λ+ρ with λP+. So f is a linear combination of the θ(eλ+ρ) with λP+, and these are clearly linearly independent (the orbits W(λ+ρ) are disjoint).

Next let d = eρ αR+ (1-e-α) = αR+ ( eα2 - e-α2 ) A.

  1. d=θ(eρ) and d is W-skew (H. Weyl).
  2. Each fS is divisible in A by d and fd-1 is invariant. Moreover ffd-1 is a bijection of S onto the W-invariants.
(So d plays the same role here as p = αR+α did for polynomials.)

  1. si permutes the positive roots other than αi, and siαi=-αi (1.9). Hence sid = ( e- αi 2 - e αi 2 ) αR+ ααi ( eα2 - e-α2 ) = -d (1ir) from which it follows that dS. Hence (Prop. 3.2) d = λP+ dλ θ(eλ+ρ) (dλ) (ExpWeyl 1) where dλ= coefficient of eλ+ρ in d. Now on expansion we have d = eρ + μ<ρ aμeμ (ExpWeyl 2) say; hence d0= coefficient of eρ in d=1, and if λP+, λ0 we have λ+ρ>ρ, so that from (ExpWeyl 2) the coefficient of eλ+ρ in d is zero. So dλ=0 if λP+, λ0, and hence (ExpWeyl 1) shows that d=θ(eρ).
  2. By Prop. 3.1 and Prop. 3.2 it is enough to show that θ(eλ+ρ) is divisible by 1-e-α for each αR+. Let W+ = {wW | ε(w)=1}, then W = W+sαW+ (disjoint union). Then θ(eλ+ρ) = wW+ ε(w) ( ew(λ+ρ) - esαw(λ+ρ) ). It is enough to show that each term in this sum is divisible by 1-e-α; i.e. that eμ-esαμ (μP) is divisible by 1-e-α. Now sαμ = μ-μ,αα = μ-mα say, where m, hence eμ-esαμ = eμ (1-e-mα) = eμ (1-e-α)( 1+e-α++e-(m-1)α ) if   m>0; and if m<0, say m=-n, then (1-e-mα) = (1-enα) = -enα (1-e-nα) and the conclusion is the same as before. (Finally eμ-esαμ=0 if m=0.)

    Finally, if f is skew then fd-1 is W-symmetric. Conversely if gAW then gdS, which proves the last statement of Proposition 3.3 (ii).

The formula d = θ(eρ), or explicitly eρ α>0 (1-e-α) = wW ε(w) ewρ is Weyl's denominator formula. I will give an alternative, more combinatorial proof.

Example. R of type An-1, roots εi-εj   (1i,jn, ij). We have ρ = 12 i<j (εi-εj) = 12( (n-1)ε1 + (n-3)ε2 + + (1-n)εn ) = (n-1)ε1 + (n-2)ε2 + + εn-1 - 12 (n-1) (ε1++εn) = δ-12(n-1)ε. Let xi=eεi, then Weyl's formula is in this case x δ-n-12ε i<j ( 1-xi-1xj ) = wSn ε(w) x wδ-n-12ε . Multiply both sides by (x1,...,xn) n-1 2 and we get 1i<jn (xi-xj) = wSn ε(w) xwδ (Vandermonde identity).

Exercise: Write it down explicitly for the other R of classical type (Bn,Cn,Dn).

For the second proof of Proposition 3.3 (i) we need a lemma, which will come in useful later:

For each subset ER+ let σE = αEα. Suppose that ρ-σE is a regular element of V (i.e. does not lie in any hyperplane Hα). Then E = R(w)   ( =R+wR- ) for some wW and σE=ρ-wρ.

We have ρ-σE wC for some wW (C the Weyl chamber), hence w-1 (ρ-σE) C. Now ρ-σE = 12 αR+ uαα (ExpLem 1) where uα = { +1, if αE, -1, if αE. } (ExpLem 2) i.e. ρ-σE is of the form 12 αR+ ±α. Hence so is w-1 (ρ-σE), hence w-1 (ρ-σE) = ρ-σF (ExpLem 3) for some FR+, depending on E and w, and which we could describe explicitly if we needed. So we have ρ-σF C, hence ρ-σF,αi is >0 and hence 1 for each i=1,2,...,r. But ρ,αi = 1, hence σF,αi 0 and therefore also σF,αi 0   (1ir), so that σF,μ 0 for all μQ+. But σFQ+, hence σF,σF 0, hence σF=0 (and so F=φ). From (ExpLem 3) it now follows that ρ-σE = wρ, i.e., σE=ρ-wρ.

Moreover, since w-1 (ρ-σE) = ρ it follows from (ExpLem 1) that 12 αR+ uαw-1α = 12 βR+β. Hence for each αR+, uα = { +1, if w-1αR+ -1, if w-1αR-. } (ExpLem 4) Hence from (ExpLem 2) and (ExpLem 4) αE    w-1α R-, i.e. αwR-, so E = R+wR- = R(w).

Second proof of Proposition 3.3 (i).
Now consider d = eρ αR+ (1-e-α) = ER+ (-1)|E| eρ-σE, on multiplying out. But d is skew, hence |W|d = θd = ER+ (-1)|E| θ(eρ-σE). If ρ-σE is not regular, θ(eρ-σE) = 0 by Proposition 3.2 (ii). If ρ-σE is regular, then E=R(w) and ρ-σE = wρ for some wW. So |W|d = wW ε(w) θ(ewρ) since (-1)|E| = (-1)l(w) = ε(w); and ε(w) θ(ewρ) = ε(w)θw(eρ) = θ(eρ) so that |W|d = |W|θ(eρ) and hence d=θ(eρ) as required.

The same technique can be used to prove the following identity. For each αR+, let tα be an indeterminate, and for each subset ER+ let tE = αEtα. Then we have

wW w αR+ 1-tαe-α 1-e-α = wW tR(w).

Let L denote the left-hand side of Proposition 3.5, so that L = wW w d-1eρ αR+ (1-tαe-α) and therefore Ld = wW ε(w) ewρ αR+ (1-tαe-wα) = wW ε(w) ER+ (-1)|E| tE ew(ρ-σE) on multiplying out the product, i.e., Ld = ER+ (-1)|E| tE θ(eρ-σE). Now by Proposition 3.2 we have θ(eρ-σE) = 0 if ρ-σE is not regular, and if ρ-σE is regular then by Proposition 3.4 E=R(w) and ρ-σE = wρ for some wW. Hence (since (-1)|R(w)| = (-1)l(w) = ε(w) ) Ld = wW ε(w) tR(w) θ(ewρ) (ExpId 1) and θ(ewρ) = θw(eρ) = ε(w) θ(eρ) = ε(w)d. So we can cancel d from either side of (ExpId 1) and obtain L = wW tR(w).

In particular, if tα=t for all αR+:

wW w αR+ 1-te-α 1-e-α = wW tl(w) = W(t).

Here the right hand side is independent of the e-α, so we may specialize them in any way we please. Define φ:Q[t,t-1] by φ(eαi) = t-1 (1ir) so that φ(e-α) = tht(α) and therefore φ αR+ 1-te-α 1-e-α = αR+ 1-tht(α)+1 1-tht(α) . (Exp 2) What about φ αR+ 1-te-wα 1-e-wα (Exp 3) when w1? In that case some w-1αi is a negative root -α say, so that αi=wα, some αR+, and then φ( 1-te-wα ) = φ( 1-teαi ) = 1-tt-1 =0 so that (Exp 3) is zero when w1. Hence from Proposition 3.6 we obtain

αR+ 1-t1+ht(α) 1-tht(α) = wW tl(w)

another multiplicative formula for the Poincaré polynomial W(t) = P(;t) of the harmonics. From Proposition 3.7 and (3.31) it follows that i=1r 1-tdi 1-t = αR+ 1-t1+ht(α) 1-tht(α) = i1 ( 1-ti+1 1-ti ) hi where hi (i1) is the number of roots αR+ of height i. Since h1=r (the roots of height 1 being the αi) it follows that i=1r (1-tdi) = i1 (1-ti+1) hi-hi+1 (Exp 4) so that hi hi+1 for each i1, i.e. η = (h1,h2,...) is a partition of N = Card(R+) ; hi-hi+1 = mi(η), the multiplicity of i as part of the conjugate partition η, so that

The partitions (h1,h2,...) and (d1-1, d2-1,..., dr-1) = (e1,...,er) where (d1dr) are conjugate (Shapiro, Kostant).

So the degrees di (or the exponents ei) can be read off from the root system.

Exercise: Check Proposition 3.8 for the classical root systems (i.e. compute the hi).

Bases of AW

Let AW = {fA | wf=f all wW} be the subring of W-invariants of A. I will list several -bases of AW, generalizing familiar objects from the theory of symmetric functions.

Orbit-sums (monomial symmetric functions)

Suppose f = μP fμeμ AW. Then for all wW we have f = wf = μ fμewμ so that fμ = fwμ, i.e. the coefficients fμ are constant on each W-orbit in P. By Proposition 2.2 each such W-orbit is of the form Wλ,   λP+, so that f = λP+ fλ μWλ eμ from which it follows that

The orbit-sums mλ = μWλeμ (λP+) are an -basis of AW (indeed, a -basis of (P)W ).

Weyl characters

For each λP+ we define χλ = θ(eλ+ρ) d = wW ε(w) ew(λ+ρ) wW ε(w) ewρ , by Proposition 3.3. By Proposition 3.3 (ii) we have χλ AW. The χλ are Weyl characters. Since (Prop. 3.2 (iii)) the θ(eλ+ρ),   λP+, form an -basis of the space S of W-skew elements of A, hence by Proposition 3.3 (ii) the χλ,   λP+ are an -basis of AW.

We have χλ = wW ε(w) ew(λ+ρ)-ρ αR+ (1-e-α) = eλ +   lower terms, where by lower terms is meant a linear combination of eμ with μ<λ (because if w1 then w(λ+ρ) < χ+ρ, i.e. w(λ+ρ)-ρ<λ). Hence, as χλ is symmetric, we can write

χλ = μP+ μλ Kλμ mμ with Kλλ = 1.

In fact the coefficients Kλμ are non-negative integers, (because they are the multiplicities of the weights in the simple 𝔤-module indexed by λ).

Problem. Prove directly (i.e. without recourse to representation theory of Lie groups or Lie algebras) that the Kλμ are 0).

Example. R = An-1,   λ = (λ1,...,λn) P (so that λi=0). Since λ,α is an integer for all αR it follows that λi-λj for all i,j, but the individual λi need not be integers: in general they will be rational numbers with denominators dividing n, as we say last time from computing the fundamental weights. In particular, λP+    λ1λ2λn (and λi=0).

So any partition μ = (μ1,...,μn) of length n determines a dominant weight μ* = ( μ1-1n|μ| ,..., μn-1n|μ| ). So we have a surjective (but not bijective) mapping μμ* of the set 𝒫n of partitions of length n onto the cone P+ of dominant weights for the root system An-1.

As before put xi = eεi   (1in). Then the Weyl characters χλ are essentially Schur functions: to be precise, if μ is a partition of length n, then χμ* = sμ(x1,...,xn) (x1xn) |μ|n (because it must have degree 0 as a polynomial in the xi±1). In particular, since πi = (ε1++εi) - in 1n εj, we have χπi = ei(x1,...,xn) (x1xn) in = mπi, (1in-1) ( πn = 0, χπn = χ0 = 1 ).

Return to the general case

For each i[1,r] let ui be an element of AW of the form ui = eπi + lower terms (Exp 5)


  1. ui=mπi,
  2. ui=χπi.
[When R is of type An-1 these are the same, but in general they aren't.]

Then we have

AW = [u1,...,ur] and the ui are algebraically independent over .

Consider a monomial u1n1 ,..., urnr (n1,...,nr integers 0). From (Exp 5) it follows that if λ = niπi P+ we have uλ = u1n1 urnr = eniπi + lower terms = eλ + lower terms = mλ + μMλ μP+ (*)mμ for certain coefficients (*). Hence the transition matrix between the u's and the orbit sums is unitriangular, hence invertible. So by Proposition 4.1 the uλ are a basis of AW (even over ), which proves Proposition 4.3.

Note that the χλ are defined for all λP, not just λP+.

Hall-Littlewood analogs

Next, we have analogs of the Hall-Littlewood symmetric functions. Let t be an indeterminate, let λP+ and define P˜λ(t) = P˜λ = wW w eλ αR+ 1-te-α 1-e-α . Notation W(t) = wW tl(w); so in particular P˜0 = W(t) by Proposition 3.6. We have P˜λ = wW w eλ+ρ d αR+ (1-te-α) = d-1 wW ε(w) w eλ+ρ αR+ (1-te-α) = d-1θ eλ+ρ αR+ (1-te-α) which shows that P˜λ AW[t].

Let Wλ = {wW | wλ=λ} be the subgroup of W that fixes λ. Since λP+C,   Wλ is generated by the si,   iI. Let Rλ be the root system with basis αi   (iI). Then Rλ+ R+. Now si permutes R+-{αi} and also permutes Rλ+-{αi} if iI. Hence si permutes R+-Rλ+ if iI, and hence Wλ permutes R+-Rλ+.

Now let U be a set of left coset representatives of Wλ in W, so that each wW is uniquely w=uv with uU and vWλ. Then P˜λ = uU u eλ αR+-Rλ+ 1-te-α 1-e-α vWλ v αRλ+ 1-te-α 1-e-α from what we have just observed. By Proposition 3.6 it follows that P˜λ = Wλ(t) uW/Wλ u eλ αR+-Rλ+ 1-te-α 1-e-α and hence all the coefficients of P˜λ (which are polynomials in t) are divisible by Wλ(t), and we can define Pλ(t) = Wλ(t)-1 wW w eλ αR+ 1-te-α 1-e-α (Exp 6) = uW/Wλ u eλ αR+ -Rλ+ 1-te-α 1-e-α (Exp 7) = eλ + lower terms = mλ + lower terms.

Particular case:

  1. t=1, then Pλ = uW/Wλ euλ = mλ, from (Exp 7)
  2. t=0, then Pλ = wW w eλ αR+ 1 1-e-α = d-1 wW ε(w) ew(λ+ρ) = χλ.
So the Pλ include the orbit sums mλ and the Weyl characters χλ as particular cases.

Kostka polynomials

These are defined by χλ = μλ μP+ Kλμ(t)Pμ (λP+) Then Kλμ(0) = δλμ and Kλμ(1) = Kλμ and the Kλμ(t) are q-analogs of weight multiplicities. For ξP let F(ξ;t) = (mα) tmα summed over all families (mα)αR+ of non-negative integers such that ξ = αR+mαα.

(So F(ξ;t)=0 unless ξQ+.)

  1. Kλμ(t) = wW ε(w) F( w(λ+ρ) - (μ+ρ) ; t ), (λ,μP+) (Kato)
  2. Let K(t) = (Kλμ(t)) λ,μP+. Then the transpose of K(t) is the matrix of the operator αR+ (1-tRα)-1, where the Rα are raising operators: Rαχλ = χλ+α.

Constant term conjectures

(SIAM J. Math. Analysis 13 (1982) 988-1007)

Let R be a reduced root system, k an integer 0; d1,...,dr the degrees of the Weyl group W of R. If f = λP fλeλ A, the constant term of f is f0.

  1. (C1) The constant term in αR (1-eα)k is i=1r ( kdi k ).
  2. (C1q) The constant term in αR+ i=0k-1 (1-qieα) (1-qi+1e-α) is i=1r [ kdi k ] where [ n r ] = (1-qn) (1-qn-1)(1-qn-r+1) (1-q)(1-q2)(1-qr) is the q-binomial coefficient or Gaussian polynomial.

Clearly (C1q)⇒(C1) by setting q=1. Consider the simplest nontrivial case, i.e. k=1. Then αR (1-eα) = eρ αR+ (1-e-α) e-ρ αR+ (1-eα) = wW ε(w) ewρ νW ε(ν) e-νρ by Proposition 3.3. Since ρC we have wρ=νρ    ν=w, so the constant term is |W| = d1d2dr, which is (C1) for k=1.

Next, when k=1 we have to consider the constant term C(q) in αR+ (1-eα) (1-qe-α) = αR (1-eα) αR+ 1-qe-α 1-e-α . Hence C(q) = 1|W| × constant term in αR (1-eα) wW w αR+ 1-qe-α 1-e-α = αR (1-eα) W(q) by Proposition 3.5. But we have just seen that the constant term in αR (1-eα) is |W|, hence C(q) = W(q) = i=1r 1-qdi 1-q = i=1r [di1].

Formal exponentials as functions on V

Let λP. We may regard eλ as a (complex-valued) function on V, as follows eλ(x) = e2πiλ,x (xV) where the e on the right is the usual one: 2.718...

Now let μQ = Q(R) . Then eλ(x+μ) = e2πiλ,x+μ and λ,x+μ = λ,x+λ,μ λ,x module integers, so that eλ(x+μ) = eλ(x) (xV, μQ). (Exp 8) Let T=V/Q and let φ:xx = x+Q be the canonical mapping VT. Then (Exp 8) shows that eλ defines a function on T, eλ(x) = eλ(x). Since V = αi,   Q = αi it follows that T is a direct sum (or direct product) of r copies of /, the circle group: (xe2πi) i.e., T is an r-dimensional torus: a compact commutative topological group.

Let Π be the parallelepiped Π = xiαi 0xi<1 (1ir) so that the projection φ:VT restricted to Π is a bijection Π˜T.

Now let f be a complex-valued continuous function on T. Lifting it back to V, we get a continuous function f˜ on V: f˜(x) = f(x) = f(φ(x)) i.e. f˜=fφ;

f˜ is periodic: f˜(x+μ) = f˜(x) (μQ).
Hence f˜ can be expanded as a convergent Fourier series: f˜ = λP fλeλ — now in general an infinite series — where fλ = Π f˜(x) e-λ(x)dx := T f(x) e-λ(x)dx and dx = dx1dxr — integration from 0 to 1 in each xi.

So in particular the constant term of f˜ is f0 = T f(x)dx. These considerations apply to the conjectures (C1) and (C1q). Consider the latter. Introduce the notation (x;q) = i=1 (1-qix) whenever the product makes sense. Then if k is an integer 0 (x;q) (xqk;q) = i=0k-1 (1-qix). So in (C1q) we seek the constant term in αR+ (eα,q) (qe-α;q) (qkeα;q) (qk+1e-α;q) . (Exp 9) Think of q now as a real number, 0q<1. The infinite product (eα;q) = i0 (1-qieα) converges uniformly on T to a continuous function (because qieα converges). So does (qkeα;q) for any real or complex k (qk = eklogq); moreover it does not vanish anywhere on T unless one of the factors 1-qk+ieα does, and this can happen only if k is an integer 0. So if k is not an integer <0, the expression (Exp 9) is a continuous function on the torus T, and its constant term is therefore well-defined.

To make sense of the other side when k is not an integer 0, we introduce the q-gamma function. Suppose first that k is an integer 0, then Γ(k+1) = k! is the limit as q1 of Γq(k+1) = (1-q)(1-q2)(1-qk) (1-q)k = (q;q) (qk+1;q) (1-q)k . This makes sense whenever k is not a negative integer (it has poles at the negative integers). We now interpret the Gaussian polynomial [ kdi k ] as Γq(kdi+1) Γq(k+1) Γq(k(di-1)+1) (The +1s in the notation are a nuisance, but it is 200 years too late to change.)

Kostant's partition function

If μP let p(μ) = no. of sequences   (nα)αR+   such that   nα   (all  α) and   αR+ nαα = μ. (So p(μ)=0 unless μQ+.) Clearly αR+ (1-e-α)-1 = ξQ+ p(ξ) e-ξ. (Exp 10) Define Kλμ by χλ = μ Kλμ mμ and cλμν by χλχμ = ν cλμνχν (λ,μ,νP+).

  1. Kλμ = wW ε(w) p( w(λ+ρ)-(μ+ρ) ).
  2. cλμν = v,wW ε(vw) p( v(λ+ρ) + w(μ+ρ) - (ν+2ρ) ).

  1. Kλμ is the coefficient of eμ in χλ = wW ε(w) ew(λ+ρ)-ρ αR+ (1-e-α)-1 = wW ξQ+ ε(w) p(ξ) ew(λ+ρ)-ξ-ρ. So we must have w(λ+ρ)-ξ-ρ = μ, i.e. ξ = w(λ+ρ)-(μ+ρ) and hence we get Kλμ = wW ε(w) p( w(λ+ρ) - (μ+ρ) ).
  2. We have θ(eλ+ρ) θ(eμ+ρ) d = ν cλμν θ(eν+ρ) (Exp Kos) and therefore cλμν is the coefficient of eν+ρ in the left hand side of (Exp Kos), i.e. in v,w ε(vw) e v(λ+ρ) + w(μ+ρ) - ρ αR+ (1-e-α)-1 = v,wW ξQ+ ε(vw) p(ξ) e v(λ+ρ) + w(μ+ρ) - ξ-ρ . So we must have ν(λ+ρ) + w(μ+ρ) - ξ-ρ = ν+ρ i.e. ξ = v(λ+ρ) + w(μ+ρ) - (ν+2ρ) and hence we get the formula (ii).

Both Kλμ and cλμν are integers 0. This is not apparent from these formulas.


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

Version: March 12, 2012

page history