## Exponentials

Last update: 04 August 2012

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

## Lattices

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 ${x}_{1},...,{x}_{r}$ (which are therefore a basis of $V$). So the elements of $L$ are all linear combinations $∑i=1rnixi, ni∈ℤ (1≤i≤r).$ Let $M$ be another lattice in $V,$ with basis ${y}_{1},...,{y}_{r},$ and suppose that $M\subseteq L.$ Then each ${y}_{i}\in L,$ hence we have equations $yi = ∑j=1raijxj (1≤i≤r) (aij∈ℤ). (Exp 1)$ The matrix $A=\left({a}_{ij}\right)$ is a nonsingular matrix of integers, because it transforms one basis of $V$ into another.

$L/M$ is a finite abelian group of order $d=|\mathrm{det}A|.$

 Proof. 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 ${d}_{i}$ are positive integers. (We can even arrange that ${d}_{1}|{d}_{2}|\cdots |{d}_{r},$ though we shan't need that fact here.) Let $y′i = ∑j uijyj and xk = ∑l vkl x′l,$ so that the ${x\prime }_{k}$ (resp. the ${y\prime }_{i}$) are a basis of $L$ (resp. $M$). From (Exp 1) we have $y′i = ∑j uijyj = ∑j,k uij ajk xk = ∑j,k,l uij ajk vkl x′l = ∑l (UAV)il x′l$ and therefore $y′i = dix′i (1≤i≤l).$ So have bases $x′1,...,x′r, and d1x′1,...,drx′r,$ and therefore $L/M\cong \underset{i=1}{\overset{r}{⨁}}ℤ/{d}_{i}ℤ$ is a direct sum of cyclic groups of orders ${d}_{1},...,{d}_{r},$ hence $|L/M| = d1,...,dr = det(UAV) = ±detA,$ since $U,V$ have determinant $±1.$ $\square$

Now suppose $V$ carries a positive definite scalar product $⟨x,y⟩,$ and let $L$ be a lattice in $V,$ as before. Let $L* = {y∈V | ⟨x,y⟩∈ℤ for all x∈L}.$ Let ${x}_{1},...,{x}_{r}$ be a basis of $L$ and let ${x}_{1}^{*},...,{x}_{r}^{*}$ be the dual basis of $V:$ $⟨{x}_{i},{x}_{j}^{*}⟩={\delta }_{ij}.$ Then $y∈L* ⇔ ⟨xi,y⟩∈ℤ for 1≤i≤r ⇔ y = ∑1r nixi*, ni∈ℤ,$ so that ${L}^{*}$ is a lattice in $V$ with basis ${x}_{1}^{*},...,{x}_{r}^{*}.$ ${L}^{*}$ is called the dual of $L.$ We have ${L}^{**}=L$ (because ${L}^{**}$ has basis ${x}_{1},...,{x}_{r}$).

If $M\subseteq L$ we have ${L}^{*}\subseteq {M}^{*},$ because if $y\in {L}^{*}$ we have $⟨x,y⟩\in ℤ$ for all $x\in L$ and a fortiori for all $x\in M.$

${M}^{*}/{L}^{*}$ is the dual (or character group) of $L/M.$

(In particular $L/M\cong {M}^{*}/{L}^{*}$ — but not canonically.)

 Proof. Let and define $χη(x+y) = e2πi⟨x,η⟩ .$ If $y\in M$ we have i.e. ${\chi }_{\eta }$ is constant on the cosets of $M$ in $L$ and therefore defines a character of the finite abelian group $L/M.$ Moreover if $\xi \in {L}^{*}$ then ${\chi }_{\eta +\xi }={\chi }_{\eta }$ (because $⟨\xi ,x⟩\in ℤ$ for $x\in L$). Hence $\eta ↦{\chi }_{\eta }$ defines a mapping of ${M}^{*}/{L}^{*}$ into the character group of $L/M.$ Check that it is an isomorphism. $\square$

If ${y}_{1},...,{y}_{r}$ is a basis of $M,$ ${y}_{1}^{*},...,{y}_{r}^{*}$ the dual basis we have $yi = ∑y=1r aijxj (aij∈ℤ)$ and ${a}_{ij}=⟨{y}_{i},{x}_{j}^{*}⟩,$ 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 $Q=Q\left(R\right)$ be the subgroup of $V$ generated by the roots: so the elements of $Q$ are all linear combinations If $B=\left\{{\alpha }_{1},...,{\alpha }_{r}\right\}$ is a basis of $R,$ every $\alpha \in R,$ and hence every $\lambda \in Q,$ is uniquely of the form $\sum _{1}^{r}{\eta }_{i}{\alpha }_{i}$ with integer coefficients ${\eta }_{i}:$ $Q(R) = ⨁i=1rℤαi$ so that $Q$ is a lattice in $V,$ called the root lattice. Likewise $Q\left({R}^{\vee }\right).$ From above, $P\left(R\right)$ is the dual of $Q\left({R}^{\vee }\right):$ $P(R) = Q(R∨)*$ hence is also a lattice, called the weight lattice of $R.$ Since $⟨\beta ,{\alpha }^{\vee }⟩\in ℤ$ for all $\alpha ,\beta \in R$ (integrality axiom (R1)) it follows that $Q(R) ⊆ P(R)$ (and likewise $Q\left({R}^{\vee }\right)\subseteq P\left({R}^{\vee }\right)$).

Let $\left({\pi }_{1},...,{\pi }_{r}\right)$ be the basis of $V$ dual to $\left({\alpha }_{1}^{\vee },...,{\alpha }_{r}^{\vee }\right):$ $⟨{\pi }_{i},{\alpha }_{j}^{\vee }⟩={\delta }_{ij}.$ The ${\pi }_{i}$ are a basis of the weight lattice $P\left(R\right):$ they are called the fundamental weights. They span the Weyl chamber $\stackrel{—}{C}.$

Recall that $A=\left({a}_{ij}\right)$ is the Cartan matrix, where ${a}_{ij}=⟨{\alpha }_{i}^{\vee },{\alpha }_{j}⟩.$

Example. What are the fundamental weights? $αi = αi∨ = ei-ei+1 (1≤i≤n-1).$ So $⟨{\pi }_{i},{\alpha }_{j}⟩={\delta }_{ij}$ say $⟨πi,αj⟩ = ⟨∑uiei,ej-ej+1⟩ = uj-uj+1.$ So $uj = uj+1 if j≠i ui-ui+1 = 1 ui+1 = ⋯=un=u say u1 = ⋯=ui=u+1 0=∑ui = i(u+1) + (n-i)u = nu+i$ hence $u=-\frac{i}{n}$ ${ πi = 1n( (n-i) (e1+⋯+ei) - i(ei+1+⋯+en) ) πi = e1+⋯+ei-ie e = 1n ∑1nei } 1≤i≤n-1;$ $πn=0.$

Exercise: Calculate the ${\pi }_{i}$ for types $B,C,D.$

We have $αj = ∑i=1r ⟨αi∨,αj⟩πi i.e., αj = ∑i=1r aijπi (1≤j≤r)$ and hence from Proposition 1.1

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

(We know from before that $\mathrm{det}A$ is positive.)

The positive integer $\mathrm{det}A$ 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 $P\left({R}^{\vee }\right)$ and $Q\left({R}^{\vee }\right)$ are respectively the duals of $Q\left(R\right)$ and $P\left(R\right).$

Finally, $W$ acts on everything: it acts on $R,$ and on ${R}^{\vee },$ hence on $Q\left(R\right)$ and $Q\left({R}^{\vee }\right),$ hence on $P\left(R\right)$ and $P\left({R}^{\vee }\right).$ Let $Q+ = ∑1rηiαi ηi≥0 and P+ = ∑1rηiπi ηi≥0 .$ Then ${P}^{+}=P\cap \stackrel{—}{C},$ hence (1.21).

Every $W-$orbit in $P$ meets ${P}^{+}$ in exactly one point.

The $\lambda \in {P}^{+}$ are called the dominant weights.

Let $\lambda \in P.$

1. $\lambda -w\lambda \in Q,$ for all $w\in W.$
2. $\lambda \in {P}^{+}⇔\lambda -w\lambda \in {Q}^{+}$ for all $w\in W.$

 Proof. We have $\lambda -{s}_{i}\lambda =⟨\lambda ,{\alpha }_{i}^{\vee }⟩{\alpha }_{i}\in Q,$ because $⟨\lambda ,{\alpha }_{i}^{\vee }⟩\in ℤ.$ Hence Since $W$ is generated by the ${s}_{i},$ it follows that for all $w\in W.$ Recall (1.24) the partial order $x\ge y$ on iff $x-y\in \stackrel{—}{{C}^{*}},$ where $\stackrel{—}{{C}^{*}}$ is the cone spanned by the simple roots ${\alpha }_{i}.$ We have ${Q}^{+}=Q\cap \stackrel{—}{{C}^{*}}.$ Now if $\lambda \in P$ we have $\lambda \in {P}^{+}\phantom{\rule{.5em}{0ex}}⇔\phantom{\rule{.5em}{0ex}}\lambda \in \stackrel{—}{C}\phantom{\rule{.5em}{0ex}}⇔\phantom{\rule{.5em}{0ex}}\lambda \ge w\lambda$ for all $w\in W$ (1.24) $⇔\lambda -w\lambda \in \stackrel{—}{{C}^{*}}\cap Q={Q}^{+},$ using (i). $\square$

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

Recall $\rho =\frac{1}{2}\sum _{\alpha \in {R}^{+}}\alpha$ is such that $⟨\rho ,{\alpha }_{i}^{\vee }⟩=1\phantom{\rule{.5em}{0ex}}\left(1\le i\le r\right)$ (1.9). Hence

$ρ = π1+⋯+πr ∈ P∩C.$

Likewise let $σ = 12 ∑α∈R+ α∨$ (warning: ${\rho }^{\vee }$ is not a good notation, because $\sigma \ne \frac{2\rho }{{|\rho |}^{2}}$). By (1.9) applied to ${R}^{\vee }$ we have because (1.23) ${B}^{\vee }=\left\{{\alpha }_{1}^{\vee },...,{\alpha }_{r}^{\vee }\right\}$ is a basis of ${R}^{\vee }.$ Hence

For all

For if $\alpha =\sum _{1}^{r}{n}_{i}{\alpha }_{i},$ then $⟨\sigma ,\alpha ⟩=\sum {n}_{i}⟨\sigma ,{\alpha }_{i}⟩=\sum {n}_{i}=\mathrm{ht}\left(\alpha \right).$ $\sigma$ is the "altimeter".

## Exponentials

If $G$ is a group (finite or infinite) and $k$ is a commutative ring, the group algebra $kG$ consists of all formal sums $a = ∑g∈G agg$ with coefficients ${a}_{g}\in k,$ only finitely many non-zero (i.e. $kG$ is the free $k-$module with $G$ as basis). So in particular each $g\in G$ is an element of $kG.$ Add and multiply in the obvious way: if $b=\sum _{g\in G}{b}_{g}g$ is another element of $kG,$ then $a+b = ∑g∈G (ag+bg)g and ab = ∑g,h∈G agbh⋅gh$ 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. ${e}^{g}$ for the element of $kG$ corresponding to $g:$ the $e\text{'}s$ are "formal exponentials" and satisfy $eg⋅eh = eg+h, e0=1, (eg)-1 = e-g$ $\left(g,h\in G\right).$ 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 $A=ℝP$ denote the group algebra of $P$ over $k=ℝ.$ So the elements of $A$ are finite sums $f = ∑λ∈P fλeλ (fλ∈ℝ, almost all zero )$ where The ${e}^{\lambda }$ are an $ℝ-$basis of $A.$

The ring $A$ is easily described. The fundamental weights form a basis of $P,$ so that each $\lambda \in P$ is uniquely of the form $λ = ∑i=1r niπi (ni∈ℤ).$ Hence if we put ${y}_{i}={e}^{{\pi }_{i}}$ we have $eλ = en1π1 ⋯ enrπr = y1η1 ⋯ yrηr$ and therefore $A=ℝ\left[{y}_{1}^{±1},...,{y}_{r}^{±1}\right]$ is the algebra of Laurent polynomials in ${y}_{1},...,{y}_{r}$ (i.e. polynomials in the $y\text{'}s$ and their inverses). Since $ℝ\left[{y}_{1},...,{y}_{r}\right]$ is a unique factorization domain, so is $A.$

Let $\lambda ,\mu \in P$ be linearly independent (i.e. not proportional). Then $1-{e}^{\lambda }$ and $1-{e}^{\mu }$ are coprime in $A.$

 Proof. This amounts to showing that if $\alpha ,\beta \in {ℤ}^{r}$ are not proportional, then $1-{y}^{\alpha }$ and $1-{y}^{\beta }$ have no common factor. Left as exercise. $\square$

The Weyl group $W$ acts on $P,$ hence on $A:$ $if f = ∑λ∈P fλeλ ∈ A then wf = ∑λ∈P fλewλ.$ $f\in A$ is

1. $W-$invariant (or $W-$symmetric) if $wf=f,$ all $w\in W.$
2. $W-$skew (or alternating) if $wf=\epsilon \left(w\right)f,$ all $w\in W.\phantom{\rule{2em}{0ex}}\left(⇔{s}_{i}f=-f\phantom{\rule{.5em}{0ex}}\left(1\le i\le r\right).\right)$

We shall start by considering the $W-$skew elements of $A.$ Let $\theta :A\to A$ be the linear mapping defined by $θf = ∑w∈W ε(w)wf (f∈A).$ We have $\left(v\in W\right)\phantom{\rule{.5em}{0ex}}\theta v=v\theta =\sum \epsilon \left(w\right)vw=\epsilon \left(v\right)\theta .$

1. $S=\theta \left(A\right)$ is the space of $W-$skew elements of $A.$
2. If $f\in A$ and ${s}_{\alpha }f=f$ for some $\alpha \in R,$ then $\theta f=0.$
3. The $\theta \left({e}^{\lambda +\rho }\right),\phantom{\rule{.5em}{0ex}}\lambda \in {P}^{+},$ form an $ℝ-$basis of $S.$

 Proof. Let $f\in A.$ Then $\theta f$ is $W-$skew, because if $v\in W$ we have $v\theta f=\epsilon \left(v\right)\theta f.$ Conversely if $f$ is $W-$skew then $θf = ∑w∈W ε(w)⋅wf = |W|f so that f = θf |W| ∈ θ(A).$ We have ${\theta }_{{s}_{\alpha }}=\epsilon \left({s}_{\alpha }\right)\theta =-\theta ,$ hence $\theta f=\theta {s}_{\alpha }f=-\theta f$ and therefore $\theta f=0.$ Suppose $f=\sum _{\mu }{f}_{\mu }{e}^{\mu }\in S.$ Then $f = ε(w)wf = ∑με(w) fμewμ$ and therefore ${f}_{w\mu }=\epsilon \left(w\right){f}_{\mu }.$ Since each $W-$orbit in $P$ meets ${P}^{+}$ in just one point (Prop. 2.2) it follows that $f = ∑μ∈P+fμ ∑w∈W ε(w) ewμ = ∑μ∈P+ fμθ(eμ). (*)$ Now $\mu \in {P}^{+}$ is of the form $\mu =\sum {m}_{i}{\pi }_{i}$ with ${m}_{i}&he;0.$ If ${m}_{i}=0$ for some $i$ then $⟨\mu ,{\alpha }_{i}^{\vee }⟩=0$ and hence ${s}_{i}\mu =\mu ,$ so that $\theta \left({e}^{\mu }\right)=0$ by (ii). Hence in (*) the sum is restricted to $\mu =\sum {m}_{i}{\pi }_{i}$ with each ${m}_{i}\ge 1,$ hence (Prop. 2.5) $\mu =\lambda +\rho$ with $\lambda \in {P}^{+}.$ So $f$ is a linear combination of the $\theta \left({e}^{\lambda +\rho }\right)$ with $\lambda \in {P}^{+},$ and these are clearly linearly independent (the orbits $W\left(\lambda +\rho \right)$ are disjoint). $\square$

Next let $d = eρ ∏α∈R+ (1-e-α) = ∏α∈R+ ( eα2 - e-α2 ) ∈A.$

1. $d=\theta \left({e}^{\rho }\right)$ and $d$ is $W-$skew (H. Weyl).
2. Each $f\in S$ is divisible in $A$ by $d$ and $f{d}^{-1}$ is invariant. Moreover $f↦f{d}^{-1}$ is a bijection of $S$ onto the $W-$invariants.
(So $d$ plays the same role here as $p=\prod _{\alpha \in {R}^{+}}\alpha$ did for polynomials.)

 Proof. ${s}_{i}$ permutes the positive roots other than ${\alpha }_{i},$ and ${s}_{i}{\alpha }_{i}=-{\alpha }_{i}$ (1.9). Hence $sid = ( e- αi 2 - e αi 2 ) ∏ α∈R+ α≠αi ( eα2 - e-α2 ) = -d (1≤i≤r)$ from which it follows that $d\in S.$ Hence (Prop. 3.2) $d = ∑λ∈P+ dλ θ(eλ+ρ) (dλ∈ℤ) (ExpWeyl 1)$ where ${d}_{\lambda }=$ coefficient of ${e}^{\lambda +\rho }$ in $d.$ Now on expansion we have $d = eρ + ∑μ<ρ aμeμ (ExpWeyl 2)$ say; hence ${d}_{0}=$ coefficient of ${e}^{\rho }$ in $d=1,$ and if we have $\lambda +\rho >\rho ,$ so that from (ExpWeyl 2) the coefficient of ${e}^{\lambda +\rho }$ in $d$ is zero. So ${d}_{\lambda }=0$ if and hence (ExpWeyl 1) shows that $d=\theta \left({e}^{\rho }\right).$ By Prop. 3.1 and Prop. 3.2 it is enough to show that $\theta \left({e}^{\lambda +\rho }\right)$ is divisible by $1-{e}^{-\alpha }$ for each $\alpha \in {R}^{+}.$ Let ${W}^{+}=\left\{w\in W\phantom{\rule{.5em}{0ex}}|\phantom{\rule{.5em}{0ex}}\epsilon \left(w\right)=1\right\},$ then $W={W}^{+}\cup {s}_{\alpha }{W}^{+}$ (disjoint union). Then $θ(eλ+ρ) = ∑w∈W+ ε(w) ( ew(λ+ρ) - esαw(λ+ρ) ).$ It is enough to show that each term in this sum is divisible by $1-{e}^{-\alpha };$ i.e. that $eμ-esαμ (μ∈P)$ is divisible by $1-{e}^{-\alpha }.$ Now ${s}_{\alpha }\mu =\mu -⟨\mu ,{\alpha }^{\vee }⟩\alpha =\mu -m\alpha$ say, where $m\in ℤ,$ hence 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 ${e}^{\mu }-{e}^{{s}_{\alpha }\mu }=0$ if $m=0.$) Finally, if $f$ is skew then $f{d}^{-1}$ is $W-$symmetric. Conversely if $g\in {A}^{W}$ then $gd\in S,$ which proves the last statement of Proposition 3.3 (ii). $\square$

The formula $d=\theta \left({e}^{\rho }\right),$ or explicitly $eρ ∏α>0 (1-e-α) = ∑w∈W ε(w) ewρ$ is Weyl's denominator formula. I will give an alternative, more combinatorial proof.

Example. $R$ of type ${A}_{n-1},$ roots We have $ρ = 12 ∑i Let ${x}_{i}={e}^{{\epsilon }_{i}},$ then Weyl's formula is in this case $x δ-n-12ε ⋅ ∏i Multiply both sides by ${\left({x}_{1},...,{x}_{n}\right)}^{\frac{n-1}{2}}$ and we get $∏1≤i

Exercise: Write it down explicitly for the other $R$ of classical type $\left({B}_{n},{C}_{n},{D}_{n}\right).$

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

For each subset $E\subseteq {R}^{+}$ let $σE = ∑α∈Eα.$ Suppose that $\rho -{\sigma }_{E}$ is a regular element of $V$ (i.e. does not lie in any hyperplane ${H}_{\alpha }$). Then for some $w\in W$ and ${\sigma }_{E}=\rho -w\rho .$

 Proof. We have $\rho -{\sigma }_{E}\in wC$ for some $w\in W$ ($C$ the Weyl chamber), hence ${w}^{-1}\left(\rho -{\sigma }_{E}\right)\in C.$ Now $ρ-σE = 12 ∑α∈R+ uαα (ExpLem 1)$ where $uα = { +1, if α∉E, -1, if α∈E. } (ExpLem 2)$ i.e. $\rho -{\sigma }_{E}$ is of the form $\frac{1}{2}\sum _{\alpha \in {R}^{+}}±\alpha .$ Hence so is ${w}^{-1}\left(\rho -{\sigma }_{E}\right),$ hence $w-1 (ρ-σE) = ρ-σF (ExpLem 3)$ for some $F\subseteq {R}^{+},$ depending on $E$ and $w,$ and which we could describe explicitly if we needed. So we have $\rho -{\sigma }_{F}\in C,$ hence $⟨\rho -{\sigma }_{F},{\alpha }_{i}^{\vee }⟩$ is $>0$ and hence $\ge 1$ for each $i=1,2,...,r.$ But $⟨\rho ,{\alpha }_{i}^{\vee }⟩=1,$ hence $⟨{\sigma }_{F},{\alpha }_{i}^{\vee }⟩\le 0$ and therefore also so that $⟨{\sigma }_{F},\mu ⟩\le 0$ for all $\mu \in {Q}^{+}.$ But ${\sigma }_{F}\in {Q}^{+},$ hence $⟨{\sigma }_{F},{\sigma }_{F}⟩\le 0,$ hence ${\sigma }_{F}=0$ (and so $F=\phi$). From (ExpLem 3) it now follows that $\rho -{\sigma }_{E}=w\rho ,$ i.e., ${\sigma }_{E}=\rho -w\rho .$ Moreover, since ${w}^{-1}\left(\rho -{\sigma }_{E}\right)=\rho$ it follows from (ExpLem 1) that $12 ∑α∈R+ uαw-1α = 12 ∑β∈R+β.$ Hence for each $\alpha \in {R}^{+},$ $uα = { +1, if w-1α∈R+ -1, if w-1α∈R-. } (ExpLem 4)$ Hence from (ExpLem 2) and (ExpLem 4) i.e. $\alpha \in w{R}^{-},$ so $E={R}^{+}\cap w{R}^{-}=R\left(w\right).$ $\square$

 Second proof of Proposition 3.3 (i). Now consider $d = eρ ∏α∈R+ (1-e-α) = ∑E⊆R+ (-1)|E| eρ-σE,$ on multiplying out. But $d$ is skew, hence $|W|⋅d = θd = ∑E⊆R+ (-1)|E| θ(eρ-σE).$ If $\rho -{\sigma }_{E}$ is not regular, $\theta \left({e}^{\rho -{\sigma }_{E}}\right)=0$ by Proposition 3.2 (ii). If $\rho -{\sigma }_{E}$ is regular, then $E=R\left(w\right)$ and $\rho -{\sigma }_{E}=w\rho$ for some $w\in W.$ So $|W|⋅d = ∑w∈W ε(w) θ(ewρ)$ since ${\left(-1\right)}^{|E|}={\left(-1\right)}^{l\left(w\right)}=\epsilon \left(w\right);$ and $ε(w) θ(ewρ) = ε(w)θw(eρ) = θ(eρ)$ so that $|W|\cdot d=|W|\theta \left({e}^{\rho }\right)$ and hence $d=\theta \left({e}^{\rho }\right)$ as required. $\square$

The same technique can be used to prove the following identity. For each $\alpha \in {R}^{+},$ let ${t}_{\alpha }$ be an indeterminate, and for each subset $E\subseteq {R}^{+}$ let ${t}_{E}=\prod _{\alpha \in E}{t}_{\alpha }.$ Then we have

$∑w∈W w ∏α∈R+ 1-tαe-α 1-e-α = ∑w∈W tR(w).$

 Proof. Let $L$ denote the left-hand side of Proposition 3.5, so that $L = ∑w∈W w d-1eρ ∏α∈R+ (1-tαe-α)$ and therefore $Ld = ∑w∈W ε(w) ewρ ∏α∈R+ (1-tαe-wα) = ∑w∈W ε(w) ∑E⊆R+ (-1)|E| tE ew(ρ-σE)$ on multiplying out the product, i.e., $Ld = ∑E⊆R+ (-1)|E| tE θ(eρ-σE).$ Now by Proposition 3.2 we have $\theta \left({e}^{\rho -{\sigma }_{E}}\right)=0$ if $\rho -{\sigma }_{E}$ is not regular, and if $\rho -{\sigma }_{E}$ is regular then by Proposition 3.4 $E=R(w) and ρ-σE = wρ$ for some $w\in W.$ Hence (since ${\left(-1\right)}^{|R\left(w\right)|}={\left(-1\right)}^{l\left(w\right)}=\epsilon \left(w\right)$) $Ld = ∑w∈W ε(w) tR(w) θ(ewρ) (ExpId 1)$ and $\theta \left({e}^{w\rho }\right)=\theta w\left({e}^{\rho }\right)=\epsilon \left(w\right)\theta \left({e}^{\rho }\right)=\epsilon \left(w\right)d.$ So we can cancel $d$ from either side of (ExpId 1) and obtain $L = ∑w∈W tR(w).$ $\square$

In particular, if ${t}_{\alpha }=t$ for all $\alpha \in {R}^{+}:$

$∑w∈W w ∏α∈R+ 1-te-α 1-e-α = ∑w∈W tl(w) = W(t).$

Here the right hand side is independent of the ${e}^{-\alpha },$ so we may specialize them in any way we please. Define $\phi :ℝQ\to ℝ\left[t,{t}^{-1}\right]$ by $φ(eαi) = t-1 (1≤i≤r)$ so that $φ(e-α) = tht(α)$ and therefore $φ ∏α∈R+ 1-te-α 1-e-α = ∏α∈R+ 1-tht(α)+1 1-tht(α) . (Exp 2)$ What about $φ ∏α∈R+ 1-te-wα 1-e-wα (Exp 3)$ when $w\ne 1?$ In that case some ${w}^{-1}{\alpha }_{i}$ is a negative root $-\alpha$ say, so that ${\alpha }_{i}=w\alpha ,$ some $\alpha \in {R}^{+},$ and then $φ( 1-te-wα ) = φ( 1-teαi ) = 1-t⋅t-1 =0$ so that (Exp 3) is zero when $w\ne 1.$ Hence from Proposition 3.6 we obtain

$∏α∈R+ 1-t1+ht(α) 1-tht(α) = ∑w∈W tl(w)$

another multiplicative formula for the Poincaré polynomial $W\left(t\right)=P\left(ℋ;t\right)$ of the harmonics. From Proposition 3.7 and (3.31) it follows that $∏i=1r 1-tdi 1-t = ∏α∈R+ 1-t1+ht(α) 1-tht(α) = ∏i≥1 ( 1-ti+1 1-ti ) hi$ where is the number of roots $\alpha \in {R}^{+}$ of height $i.$ Since ${h}_{1}=r$ (the roots of height 1 being the ${\alpha }_{i}$) it follows that $∏i=1r (1-tdi) = ∏i≥1 (1-ti+1) hi-hi+1 (Exp 4)$ so that ${h}_{i}\ge {h}_{i+1}$ for each $i\ge 1,$ i.e. $η = (h1,h2,...)$ is a partition of $N=\mathrm{Card}\left({R}^{+}\right);$ ${h}_{i}-{h}_{i+1}={m}_{i}\left(\eta \prime \right),$ the multiplicity of $i$ as part of the conjugate partition $\eta \prime ,$ so that

The partitions $\left({h}_{1},{h}_{2},...\right)$ and $\left({d}_{1}-1,{d}_{2}-1,...,{d}_{r}-1\right)=\left({e}_{1},...,{e}_{r}\right)$ where $\left({d}_{1}\ge \cdots \ge {d}_{r}\right)$ are conjugate (Shapiro, Kostant).

So the degrees ${d}_{i}$ (or the exponents ${e}_{i}$) can be read off from the root system.

Exercise: Check Proposition 3.8 for the classical root systems (i.e. compute the ${h}_{i}$).

## Bases of ${A}^{W}$

Let ${A}^{W}=\left\{f\in A\phantom{\rule{.5em}{0ex}}|\phantom{\rule{.5em}{0ex}}wf=f\phantom{\rule{.5em}{0ex}}all\phantom{\rule{.5em}{0ex}}w\in W\right\}$ be the subring of $W-$invariants of $A.$ I will list several $ℝ-$bases of ${A}^{W},$ generalizing familiar objects from the theory of symmetric functions.

### Orbit-sums (monomial symmetric functions)

Suppose $f=\sum _{\mu \in P}{f}_{\mu }{e}^{\mu }\in {A}^{W}.$ Then for all $w\in W$ we have $f = wf = ∑μ fμewμ$ so that ${f}_{\mu }={f}_{w\mu },$ i.e. the coefficients ${f}_{\mu }$ are constant on each $W-$orbit in $P.$ By Proposition 2.2 each such $W-$orbit is of the form so that $f = ∑λ∈P+ fλ ∑μ∈Wλ eμ$ from which it follows that

The orbit-sums $mλ = ∑μ∈Wλeμ (λ∈P+)$ are an $ℝ-$basis of ${A}^{W}$ (indeed, a $ℤ-$basis of ${\left(ℤP\right)}^{W}$).

### Weyl characters

For each $\lambda \in {P}^{+}$ we define $χλ = θ(eλ+ρ) d = ∑w∈W ε(w) ew(λ+ρ) ∑w∈W ε(w) ewρ , by Proposition 3.3.$ By Proposition 3.3 (ii) we have ${\chi }_{\lambda }\in {A}^{W}.$ The ${\chi }_{\lambda }$ are Weyl characters. Since (Prop. 3.2 (iii)) the form an $ℝ-$basis of the space $S$ of $W-$skew elements of $A,$ hence by Proposition 3.3 (ii) the are an $ℝ-$basis of ${A}^{W}.$

We have where by lower terms is meant a linear combination of ${e}^{\mu }$ with $\mu <\lambda$ (because if $w\ne 1$ then $w\left(\lambda +\rho \right)<\chi +\rho ,$ i.e. $w\left(\lambda +\rho \right)-\rho <\lambda$). Hence, as ${\chi }_{\lambda }$ is symmetric, we can write

$χλ = ∑ μ∈P+ μ≤λ Kλμ mμ with Kλλ = 1.$

In fact the coefficients ${K}_{\lambda \mu }$ are non-negative integers, (because they are the multiplicities of the weights in the simple $𝔤-$module indexed by $\lambda$).

Problem. Prove directly (i.e. without recourse to representation theory of Lie groups or Lie algebras) that the ${K}_{\lambda \mu }$ are $\ge 0$).

Example. (so that $\sum {\lambda }_{i}=0$). Since $⟨\lambda ,{\alpha }^{\vee }⟩$ is an integer for all $\alpha \in R$ it follows that ${\lambda }_{i}-{\lambda }_{j}\in ℤ$ for all $i,j,$ but the individual ${\lambda }_{i}$ 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, (and $\sum {\lambda }_{i}=0$).

So any partition $\mu =\left({\mu }_{1},...,{\mu }_{n}\right)$ of length $\le n$ determines a dominant weight $μ* = ( μ1-1n|μ| ,..., μn-1n|μ| ).$ So we have a surjective (but not bijective) mapping $\mu ↦{\mu }^{*}$ of the set ${𝒫}_{n}$ of partitions of length $\le n$ onto the cone ${P}^{+}$ of dominant weights for the root system ${A}_{n-1}.$

As before put Then the Weyl characters ${\chi }_{\lambda }$ are essentially Schur functions: to be precise, if $\mu$ is a partition of length $\le n,$ then $χμ* = sμ(x1,...,xn) (x1⋯xn) |μ|n$ (because it must have degree 0 as a polynomial in the ${x}_{i}^{±1}$). In particular, since $πi = (ε1+⋯+εi) - in ∑1n εj, we have χπi = ei(x1,...,xn) (x1⋯xn) in = mπi, (1≤i≤n-1)$ $\left({\pi }_{n}=0,\phantom{\rule{.5em}{0ex}}{\chi }_{{\pi }_{n}}={\chi }_{0}=1\right).$

For each $i\in \left[1,r\right]$ let ${u}_{i}$ be an element of ${A}^{W}$ of the form $ui = eπi + lower terms (Exp 5)$

Examples.

1. ${u}_{i}={m}_{{\pi }_{i}},$
2. ${u}_{i}={\chi }_{{\pi }_{i}}.$
[When $R$ is of type ${A}_{n-1}$ these are the same, but in general they aren't.]

Then we have

${A}^{W}=ℝ\left[{u}_{1},...,{u}_{r}\right]$ and the ${u}_{i}$ are algebraically independent over $ℝ.$

 Proof. Consider a monomial ${u}_{1}^{{n}_{1}},...,{u}_{r}^{{n}_{r}}$ (${n}_{1},...,{n}_{r}$ integers $\ge 0$). From (Exp 5) it follows that if $\lambda =\sum {n}_{i}{\pi }_{i}\in {P}^{+}$ we have $uλ = u1n1 ⋯ urnr = e∑niπi + lower terms = eλ + lower terms = mλ + ∑ μMλ μ∈P+ (*)mμ$ for certain coefficients $\left(*\right).$ Hence the transition matrix between the $u\text{'}s$ and the orbit sums is unitriangular, hence invertible. So by Proposition 4.1 the ${u}_{\lambda }$ are a basis of ${A}^{W}$ (even over $ℤ$), which proves Proposition 4.3. $\square$

Note that the ${\chi }_{\lambda }$ are defined for all $\lambda \in P,$ not just $\lambda \in {P}^{+}.$

### Hall-Littlewood analogs

Next, we have analogs of the Hall-Littlewood symmetric functions. Let $t$ be an indeterminate, let $\lambda \in {P}^{+}$ and define $P˜λ(t) = P˜λ = ∑w∈W w eλ ∏α∈R+ 1-te-α 1-e-α .$ Notation $W\left(t\right)=\sum _{w\in W}{t}^{l\left(w\right)};$ so in particular ${\stackrel{˜}{P}}_{0}=W\left(t\right)$ by Proposition 3.6. We have $P˜λ = ∑w∈W w eλ+ρ d ∏α∈R+ (1-te-α) = d-1 ∑w∈W ε(w) ⋅ w eλ+ρ ∏α∈R+ (1-te-α) = d-1θ eλ+ρ ∏α∈R+ (1-te-α)$ which shows that ${\stackrel{˜}{P}}_{\lambda }\in {A}^{W}\left[t\right].$

Let ${W}_{\lambda }=\left\{w\in W\phantom{\rule{.5em}{0ex}}|\phantom{\rule{.5em}{0ex}}w\lambda =\lambda \right\}$ be the subgroup of $W$ that fixes $\lambda .$ Since is generated by the Let ${R}_{\lambda }$ be the root system with basis Then ${R}_{\lambda }^{+}\subseteq {R}^{+}.$ Now ${s}_{i}$ permutes ${R}^{+}-\left\{{\alpha }_{i}\right\}$ and also permutes ${R}_{\lambda }^{+}-\left\{{\alpha }_{i}\right\}$ if $i\in I.$ Hence ${s}_{i}$ permutes ${R}^{+}-{R}_{\lambda }^{+}$ if $i\in I,$ and hence ${W}_{\lambda }$ permutes ${R}^{+}-{R}_{\lambda }^{+}.$

Now let $U$ be a set of left coset representatives of ${W}_{\lambda }$ in $W,$ so that each $w\in W$ is uniquely $w=uv$ with $u\in U$ and $v\in {W}_{\lambda }.$ Then $P˜λ = ∑u∈U u eλ ∏ α∈R+-Rλ+ 1-te-α 1-e-α ⋅ ∑v∈Wλ v ∏α∈Rλ+ 1-te-α 1-e-α$ from what we have just observed. By Proposition 3.6 it follows that $P˜λ = Wλ(t) ∑u∈W/Wλ u eλ ∏α∈R+-Rλ+ 1-te-α 1-e-α$ and hence all the coefficients of ${\stackrel{˜}{P}}_{\lambda }$ (which are polynomials in $t$) are divisible by ${W}_{\lambda }\left(t\right),$ and we can define $Pλ(t) = Wλ(t)-1 ∑w∈W w eλ ∏α∈R+ 1-te-α 1-e-α (Exp 6) = ∑u∈W/Wλ u eλ ∏ α∈R+ -Rλ+ 1-te-α 1-e-α (Exp 7) = eλ + lower terms = mλ + lower terms.$

Particular case:

1. $t=1,$ then $Pλ = ∑u∈W/Wλ euλ = mλ, from (Exp 7)$
2. $t=0,$ then $Pλ = ∑w∈W w eλ ∏α∈R+ 1 1-e-α = d-1 ∑w∈W ε(w) ew(λ+ρ) = χλ.$
So the ${P}_{\lambda }$ include the orbit sums ${m}_{\lambda }$ and the Weyl characters ${\chi }_{\lambda }$ as particular cases.

## Kostka polynomials

These are defined by $χλ = ∑ μ≤λ μ∈P+ Kλμ(t)Pμ (λ∈P+)$ Then ${K}_{\lambda \mu }\left(0\right)={\delta }_{\lambda \mu }$ and ${K}_{\lambda \mu }\left(1\right)={K}_{\lambda \mu }$ and the ${K}_{\lambda \mu }\left(t\right)$ are $q-$analogs of weight multiplicities. For $\xi \in P$ let $F(ξ;t) = ∑(mα) t∑mα$ summed over all families ${\left({m}_{\alpha }\right)}_{\alpha \in {R}^{+}}$ of non-negative integers such that $\xi =\sum _{\alpha \in {R}^{+}}{m}_{\alpha }\alpha .$

(So $F\left(\xi ;t\right)=0$ unless $\xi \in {Q}^{+}.$)

1. ${K}_{\lambda \mu }\left(t\right)=\sum _{w\in W}\epsilon \left(w\right)F\left(w\left(\lambda +\rho \right)-\left(\mu +\rho \right);t\right),\phantom{\rule{2em}{0ex}}\left(\lambda ,\mu \in {P}^{+}\right)\phantom{\rule{2em}{0ex}}$ (Kato)
2. Let $K\left(t\right)={\left({K}_{\lambda \mu }\left(t\right)\right)}_{\lambda ,\mu \in {P}^{+}}.$ Then the transpose of $K\left(t\right)$ is the matrix of the operator $∏α∈R+ (1-tRα)-1,$ where the ${R}_{\alpha }$ are raising operators: ${R}_{\alpha }{\chi }_{\lambda }={\chi }_{\lambda +\alpha }.$

### Constant term conjectures

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

Let $R$ be a reduced root system, $k$ an integer $\ge 0;$ ${d}_{1},...,{d}_{r}$ the degrees of the Weyl group $W$ of $R.$ If $f=\sum _{\lambda \in P}{f}_{\lambda }{e}^{\lambda }\in A,$ the constant term of $f$ is ${f}_{0}.$

1. (C1) The constant term in $\prod _{\alpha \in R}{\left(1-{e}^{\alpha }\right)}^{k}$ is $\prod _{i=1}^{r}\left(\genfrac{}{}{0}{}{k{d}_{i}}{k}\right).$
2. (C1q) The constant term in $\prod _{\alpha \in {R}^{+}}\prod _{i=0}^{k-1}\left(1-{q}^{i}{e}^{\alpha }\right)\left(1-{q}^{i+1}{e}^{-\alpha }\right)$ is $\prod _{i=1}^{r}\left[\genfrac{}{}{0}{}{k{d}_{i}}{k}\right]$ 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 $∏α∈R (1-eα) = eρ ∏α∈R+ (1-e-α) ⋅ e-ρ ∏α∈R+ (1-eα) = ∑w∈W ε(w) ewρ ∑ν∈W ε(ν) e-νρ by Proposition 3.3.$ Since $\rho \in C$ we have so the constant term is $|W|={d}_{1}{d}_{2}\cdots {d}_{r},$ which is (C1) for $k=1.$

Next, when $k=1$ we have to consider the constant term $C\left(q\right)$ in $∏α∈R+ (1-eα) (1-qe-α) = ∏α∈R (1-eα) ⋅ ∏α∈R+ 1-qe-α 1-e-α .$ Hence $C\left(q\right)=\frac{1}{|W|}×$ constant term in $∏α∈R (1-eα) ∑w∈W w ∏α∈R+ 1-qe-α 1-e-α = ∏α∈R (1-eα) ⋅ W(q) by Proposition 3.5.$ But we have just seen that the constant term in $\prod _{\alpha \in R}\left(1-{e}^{\alpha }\right)$ is $|W|,$ hence $C(q) = W(q) = ∏i=1r 1-qdi 1-q = ∏i=1r [di1].$

## Formal exponentials as functions on $V$

Let $\lambda \in P.$ We may regard ${e}^{\lambda }$ as a (complex-valued) function on $V,$ as follows $eλ(x) = e2πi⟨λ,x⟩ (x∈V)$ where the $e$ on the right is the usual one: 2.718...

Now let $\mu \in {Q}^{\vee }=Q\left({R}^{\vee }\right).$ Then $eλ(x+μ) = e2πi⟨λ,x+μ⟩$ and $⟨\lambda ,x+\mu ⟩=⟨\lambda ,x⟩+⟨\lambda ,\mu ⟩\equiv ⟨\lambda ,x⟩$ module integers, so that $eλ(x+μ) = eλ(x) (x∈V, μ∈Q∨). (Exp 8)$ Let $T=V/{Q}^{\vee }$ and let $\phi :x↦\stackrel{\cdot }{x}=x+{Q}^{\vee }$ be the canonical mapping $V\to T.$ Then (Exp 8) shows that ${e}^{\lambda }$ defines a function on $T,$ $eλ(x⋅) = eλ(x).$ Since it follows that $T$ is a direct sum (or direct product) of $r$ copies of $ℝ/ℤ,$ the circle group: $\left(x↦{e}^{2\pi i}\right)$ i.e., $T$ is an $r-$dimensional torus: a compact commutative topological group.

Let $\Pi$ be the parallelepiped $Π = ∑xiαi∨ 0≤xi<1 (1≤i≤r)$ so that the projection $\phi :V\to T$ restricted to $\Pi$ is a bijection $\Pi \stackrel{˜}{\to }T.$

Now let $f$ be a complex-valued continuous function on $T.$ Lifting it back to $V,$ we get a continuous function $\stackrel{˜}{f}$ on $V:$ $f˜(x) = f(x⋅) = f(φ(x))$ i.e. $\stackrel{˜}{f}=f\circ \phi ;$

$\stackrel{˜}{f}$ is periodic: $\phantom{\rule{.5em}{0ex}}\stackrel{˜}{f}\left(x+\mu \right)=\stackrel{˜}{f}\left(x\right)\phantom{\rule{2em}{0ex}}\left(\mu \in {Q}^{\vee }\right).$
Hence $\stackrel{˜}{f}$ can be expanded as a convergent Fourier series: $f˜ = ∑λ∈P fλeλ$ — now in general an infinite series — where $fλ = ∫Π f˜(x) e-λ(x)dx := ∫T f(x⋅) e-λ(x⋅)dx⋅$ and $dx=d{x}_{1}\cdots d{x}_{r}$ — integration from 0 to 1 in each ${x}_{i}.$

So in particular the constant term of $\stackrel{˜}{f}$ 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 $\ge 0$ $(x;q)∞ (xqk;q)∞ = ∏i=0k-1 (1-qix).$ So in (C1q) we seek the constant term in $∏α∈R+ (eα,q)∞ (qe-α;q)∞ (qkeα;q)∞ (qk+1e-α;q)∞ . (Exp 9)$ Think of $q$ now as a real number, $0\le q<1.$ The infinite product $(eα;q)∞ = ∏i≥0 (1-qieα)$ converges uniformly on $T$ to a continuous function (because $\sum {q}^{i}{e}^{\alpha }$ converges). So does ${\left({q}^{k}{e}^{\alpha };q\right)}_{\infty }$ for any real or complex $k$ $\left({q}^{k}={e}^{k\mathrm{log}q}\right);$ moreover it does not vanish anywhere on $T$ unless one of the factors $1-{q}^{k+i}{e}^{\alpha }$ does, and this can happen only if $k$ is an integer $\le 0.$ 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 $\ge 0,$ we introduce the $q-$gamma function. Suppose first that $k$ is an integer $\ne 0,$ then $\Gamma \left(k+1\right)=k!$ is the limit as $q\to 1$ 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 $k\in ℂ$ 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 $\mu \in P$ let (So $p\left(\mu \right)=0$ unless $\mu \in {Q}^{+}$.) Clearly $∏α∈R+ (1-e-α)-1 = ∑ξ∈Q+ p(ξ) e-ξ. (Exp 10)$ Define ${K}_{\lambda \mu }$ by $χλ = ∑μ Kλμ mμ$ and ${c}_{\lambda \mu }^{\nu }$ by $χλχμ = ∑ν cλμνχν$ $\left(\lambda ,\mu ,\nu \in {P}^{+}\right).$

1. ${K}_{\lambda \mu }=\sum _{w\in W}\epsilon \left(w\right)p\left(w\left(\lambda +\rho \right)-\left(\mu +\rho \right)\right).$
2. ${c}_{\lambda \mu }^{\nu }=\sum _{v,w\in W}\epsilon \left(vw\right)p\left(v\left(\lambda +\rho \right)+w\left(\mu +\rho \right)-\left(\nu +2\rho \right)\right).$

 Proof. ${K}_{\lambda \mu }$ is the coefficient of ${e}^{\mu }$ in $χλ = ∑w∈W ε(w) ew(λ+ρ)-ρ ∏α∈R+ (1-e-α)-1 = ∑ w∈W ξ∈Q+ ε(w) p(ξ) ew(λ+ρ)-ξ-ρ.$ So we must have $w\left(\lambda +\rho \right)-\xi -\rho =\mu ,$ i.e. $\xi =w\left(\lambda +\rho \right)-\left(\mu +\rho \right)$ and hence we get $Kλμ = ∑w∈W ε(w) p( w(λ+ρ) - (μ+ρ) ).$ We have $θ(eλ+ρ) θ(eμ+ρ) d = ∑ν cλμν θ(eν+ρ) (Exp Kos)$ and therefore ${c}_{\lambda \mu }^{\nu }$ is the coefficient of ${e}^{\nu +\rho }$ in the left hand side of (Exp Kos), i.e. in $∑v,w ε(vw) e v(λ+ρ) + w(μ+ρ) - ρ ∏α∈R+ (1-e-α)-1 = ∑ v,w∈W ξ∈Q+ ε(vw) p(ξ) e v(λ+ρ) + w(μ+ρ) - ξ-ρ .$ So we must have $\nu \left(\lambda +\rho \right)+w\left(\mu +\rho \right)-\xi -\rho =\nu +\rho$ i.e. $ξ = v(λ+ρ) + w(μ+ρ) - (ν+2ρ)$ and hence we get the formula (ii). $\square$

Both ${K}_{\lambda \mu }$ and ${c}_{\lambda \mu }^{\nu }$ are integers $\ge 0.$ This is not apparent from these formulas.