Exponentials

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

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 $$\sum _{i=1}^r{n}_i{x}_i,n_i\in \mathbb{Z}(1\le i\le r).$$ 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 $$\begin{array}{ll}{y}_i=\sum _{j=1}^r{a}_{ij}{x}_j(1\le i\le r)(a_{ij}\in \mathbb{Z}).& (Exp\; 1)\end{array}$$ 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 $d=\left|\det A\right|.$

		Proof.
	By elementary transformations on rows and columns we can find $r\times r$ matrices $U,V$ of integers with determinant $\pm 1,$ such that $UAV$ is a diagonal matrix, say $$UAV=\left(\begin{array}{ccc}{d}_1& & \\ & \ddots & \\ & & d_r\end{array}\right)$$ where the $d_i$ are positive integers. (We can even arrange that $d_1\left|d_2\right|\cdots |d_r,$ though we shan't need that fact here.) Let $${y\prime }_i=\sum _j{u}_{ij}{y}_{j}and{x}_k=\sum _l{v}_{kl}{x\prime }_l,$$ so that the ${x\prime }_k$ (resp. the ${y\prime }_i$) are a basis of $L$ (resp. $M$). From (Exp 1) we have $${y\prime }_i=\sum _j{u}_{ij}{y}_j=\sum _{j,k}{u}_{ij}{a}_{jk}{x}_k=\sum _{j,k,l}{u}_{ij}{a}_{jk}{v}_{kl}{x\prime }_l=\sum _l{\left(UAV\right)}_{il}{x\prime }_l$$ and therefore $${y\prime }_i=d_i{x\prime }_i(1\le i\le l).$$ So $L, M$ have bases $${x\prime }_1,...,{x\prime }_r,and{d}_1{x\prime }_1,...,d_r{x\prime }_r,$$ and therefore $L/M\cong \underset{i=1}{\overset{r}{⨁}}\mathbb{Z}/d_i\mathbb{Z}$ is a direct sum of cyclic groups of orders $d_1,...,d_r,$ hence $$|L/M|=d_1,...,d_r=\det \left(UAV\right)=\pm \det A,$$ since $U,V$ have determinant $\pm 1.$ $\square$

Now suppose $V$ carries a positive definite scalar product $⟨x,y⟩,$ and let $L$ be a lattice in $V,$ as before. Let $$L^{*}=\{y\in V|⟨x,y⟩\in \mathbb{Z}for\; allx\in L\}.$$ 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 $$\begin{array}{rclr}y\in L^{*}& \iff & ⟨x_i,y⟩\in \mathbb{Z}& for1\le i\le r\\ & \iff & y=\sum _1^r{n}_i{x}_i^{*},& n_i\in \mathbb{Z},\end{array}$$ 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 $⟨x,y⟩\in \mathbb{Z}$ 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 $\eta \in M^{*}, x\in L$ and define $${\chi }_{\eta }(x+y)=e^{2\pi i⟨x,\eta ⟩}.$$ If $y\in M$ we have $${\chi }_{\eta }(x+y)=e^{2\pi i⟨x+y,\eta ⟩}=e^{2\pi i⟨x,\eta ⟩}{e}^{2\pi i⟨y,\eta ⟩}={\chi }_{\eta }\left(x\right)because\; ⟨y,\eta ⟩\in \mathbb{Z},$$ 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 $⟨\xi ,x⟩\in \mathbb{Z}$ for $x\in L$). Hence $\eta \mapsto {\chi }_{\eta }$ 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 $$y_i=\sum _{y=1}^r{a}_{ij}{x}_j(a_{ij}\in \mathbb{Z})$$ and $a_{ij}=⟨y_i,x_j^{*}⟩,$ so that also $$x_j^{*}=\sum _{i=1}^r{a}_{ij}{y}_i^{*}$$ or, interchanging $i$ and $j,$ $$x_i^{*}=\sum _{j=1}^r{a}_{ji}{y}_j^{*}.$$

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 $\lambda =\sum _{\alpha \in R}{\eta }_{\alpha }\alpha , {\eta }_{\alpha }\in \mathbb{Z}.$ If $B=\{{\alpha }_1,...,{\alpha }_r\}$ 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 ${\eta }_i:$ $$Q\left(R\right)=\underset{i=1}{\overset{r}{⨁}}\mathbb{Z}{\alpha }_i$$ 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 $Q\left(R^{\vee }\right):$ $$P\left(R\right)=Q{\left(R^{\vee }\right)}^{*}$$ hence is also a lattice, called the weight lattice of $R.$ Since $⟨\beta ,{\alpha }^{\vee }⟩\in \mathbb{Z}$ for all $\alpha ,\beta \in R$ (integrality axiom (R1)) it follows that $$Q\left(R\right)\subseteq P\left(R\right)$$ (and likewise $Q\left(R^{\vee }\right)\subseteq P\left(R^{\vee }\right)$).

Let $({\pi }_1,...,{\pi }_r)$ be the basis of $V$ dual to $({\alpha }_1^{\vee },...,{\alpha }_r^{\vee }):$ $⟨{\pi }_i,{\alpha }_j^{\vee }⟩={\delta }_{ij}.$ 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. $$R=\{e_i-e_j|1\le i,j\le r,i\ne j\}type\; A_{n-1}.$$ What are the fundamental weights? $${\alpha }_i={\alpha }_i^{\vee }=e_i-e_{i+1}(1\le i\le n-1).$$ So $⟨{\pi }_i,{\alpha }_j⟩={\delta }_{ij}$ say ${\pi }_i=\sum u_i{e}_i, \sum u_i=0$ $$⟨{\pi }_i,{\alpha }_j⟩=⟨\sum u_i{e}_i,e_j-e_{j+1}⟩=u_j-u_{j+1}.$$ So $$\begin{array}{rcl}{u}_j& =& u_{j+1}ifj\ne i\\ u_i-u_{i+1}& =& 1\\ u_{i+1}& =& \cdots =u_n=usay\\ u_1& =& \cdots =u_i=u+1\\ 0=\sum u_i& =& i(u+1)+(n-i)u=nu+i\end{array}$$ hence $u=-\frac{i}{n}$ $$\phantom{\{}\begin{array}{ccc}{\pi }_i& =& \frac{1}{n}((n-i)(e_1+\cdots +e_i)-i(e_{i+1}+\cdots +e_n))\\ {\pi }_i& =& e_1+\cdots +e_i-ie\\ e& =& \frac{1}{n}\sum _1^n{e}_i\end{array}\}1\le i\le n-1;$$ $${\pi }_n=0.$$

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

We have $${\alpha }_j=\sum _{i=1}^r⟨{\alpha }_i^{\vee },{\alpha }_j⟩{\pi }_{i}i.e.,{\alpha }_j=\sum _{i=1}^r{a}_{ij}{\pi }_i(1\le j\le r)$$ and hence from Proposition 1.1

$$|P\left(R\right)/Q\left(R\right)|=\det A.$$

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

The positive integer $\det A$ is traditionally denoted by $f$ and called the index of connection: $$f=|P\left(R\right)/Q\left(R\right)|=|P\left(R^{\vee }\right)/Q\left(R^{\vee }\right)|$$ 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 $$Q^{+}= {\sum _1^r{\eta }_i{\alpha }_i|{\eta }_i\ge 0} and{P}^{+}= {\sum _1^r{\eta }_i{\pi }_i|{\eta }_i\ge 0} .$$ 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. $\lambda \in P^{+}\iff \lambda -w\lambda \in Q^{+}$ for all $w\in W.$

		Proof.
	We have $\lambda -s_i\lambda =⟨\lambda ,{\alpha }_i^{\vee }⟩{\alpha }_i\in Q,$ because $⟨\lambda ,{\alpha }_i^{\vee }⟩\in \mathbb{Z}.$ Hence $\lambda \equiv s_i\lambda \left(\mathrm{mod}Q\right).$ Since $W$ is generated by the $s_i,$ it follows that $\lambda \equiv w\lambda \left(\mathrm{mod}Q\right)$ for all $w\in W.$ Recall (1.24) the partial order $x\ge y$ on $V: x\ge y$ 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 $\lambda \in P^{+}\iff \lambda \in \stackrel{—}{C}\iff \lambda \ge w\lambda$ for all $w\in W$ (1.24) $\iff \lambda -w\lambda \in \stackrel{—}{C^{*}}\cap Q=Q^{+},$ 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 $⟨\rho ,{\alpha }_i^{\vee }⟩=1(1\le i\le r)$ (1.9). Hence

$$\rho ={\pi }_1+\cdots +{\pi }_r\in P\cap C.$$

Likewise let $$\sigma =\frac{1}{2}\sum _{\alpha \in R^{+}}{\alpha }^{\vee }$$ (warning: ${\rho }^{\vee }$ is not a good notation, because $\sigma \ne \frac{2\rho }{{\left|\rho \right|}^2}$). By (1.9) applied to $R^{\vee }$ we have $⟨\sigma ,{\alpha }_i⟩=1 (1\le i\le r),$ because (1.23) $B^{\vee }=\{{\alpha }_1^{\vee },...,{\alpha }_r^{\vee }\}$ is a basis of $R^{\vee }.$ Hence

For all $\alpha \in R, \mathrm{ht}\left(\alpha \right)=⟨\sigma ,\alpha ⟩.$

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 $$a=\sum _{g\in G}{a}_{g}g$$ 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 $$a+b=\sum _{g\in G}(a_g+b_g)gandab=\sum _{g,h\in G}{a}_g{b}_h\cdot gh$$ 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 $$e^g\cdot e^h=e^{g+h},e^0=1,{\left(e^g\right)}^{-1}=e^{-g}$$ $(g,h\in G).$ The elements of $kG$ are now written $$a=\sum a_g{e}^g$$ 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 $$f=\sum _{\lambda \in P}{f}_{\lambda }{e}^{\lambda }(f_{\lambda }\in ℝ,almost\; all\; zero)$$ where $e^{\lambda }\cdot e^{\mu }=e^{\lambda +\mu }, e^0=1.$ The $e^{\lambda }$ are an $ℝ-$basis of $A.$

The ring $A$ is easily described. The fundamental weights ${\pi }_i (1\le i\le r)$ form a basis of $P,$ so that each $\lambda \in P$ is uniquely of the form $$\lambda =\sum _{i=1}^r{n}_i{\pi }_i(n_i\in \mathbb{Z}).$$ Hence if we put $y_i=e^{{\pi }_i}$ we have $$e^{\lambda }=e^{n_1{\pi }_1}\cdots e^{n_r{\pi }_r}=y_1^{{\eta }_1}\cdots y_r^{{\eta }_r}$$ and therefore $A=ℝ[y_1^{\pm 1},...,y_r^{\pm 1}]$ is the algebra of Laurent polynomials in $y_1,...,y_r$ (i.e. polynomials in the $y\text{'}s$ and their inverses). Since $ℝ[y_1,...,y_r]$ 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 $\alpha ,\beta \in {\mathbb{Z}}^r$ 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 $A:$ $$iff=\sum _{\lambda \in P}{f}_{\lambda }{e}^{\lambda }\in Athenwf=\sum _{\lambda \in P}{f}_{\lambda }{e}^{w\lambda }.$$ $f\in A$ 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 $w\in W.(\iff s_{i}f=-f(1\le i\le r).)$

We shall start by considering the $W-$skew elements of $A.$ Let $\theta :A\to A$ be the linear mapping defined by $$\theta f=\sum _{w\in W}\epsilon \left(w\right)wf(f\in A).$$ We have $(v\in W)\theta v=v\theta =\sum \epsilon \left(w\right)vw=\epsilon \left(v\right)\theta .$

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 $\theta \left(e^{\lambda +\rho }\right),\lambda \in P^{+},$ 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 $$\theta f=\sum _{w\in W}\epsilon \left(w\right)\cdot wf=\left|W\right|fso\; thatf=\frac{\theta f}{\left|W\right|}\in \theta \left(A\right).$$ 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 $$f=\epsilon \left(w\right)wf=\sum _{\mu }\epsilon \left(w\right)f_{\mu }{e}^{w\mu }$$ 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 $$\begin{array}{ll}f=\sum _{\mu \in P^{+}}{f}_{\mu }\left(\sum _{w\in W}\epsilon \left(w\right)e^{w\mu }\right)=\sum _{\mu \in P^{+}}{f}_{\mu }\theta \left(e^{\mu }\right).& (*)\end{array}$$ 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 $W(\lambda +\rho )$ are disjoint). $\square$

Next let $$d=e^{\rho }\prod _{\alpha \in R^{+}}(1-e^{-\alpha })=\prod _{\alpha \in R^{+}}(e^{\frac{\alpha }{2}}-e^{-\frac{\alpha }{2}})\in A.$$

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 $f\mapsto f{d}^{-1}$ 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 $$s_{i}d=(e^{-\frac{{\alpha }_i}{2}}-e^{\frac{{\alpha }_i}{2}})\prod _{\genfrac{}{}{0ex}{}{\alpha \in R^{+}}{\alpha \ne {\alpha }_i}}(e^{\frac{\alpha }{2}}-e^{-\frac{\alpha }{2}})=-d(1\le i\le r)$$ from which it follows that $d\in S.$ Hence (Prop. 3.2) $$\begin{array}{ll}d=\sum _{\lambda \in P^{+}}{d}_{\lambda }\theta \left(e^{\lambda +\rho }\right)(d_{\lambda }\in \mathbb{Z})& (ExpWeyl\; 1)\end{array}$$ where $d_{\lambda }=$ coefficient of $e^{\lambda +\rho }$ in $d.$ Now on expansion we have $$\begin{array}{ll}d=e^{\rho }+\sum _{\mu <\rho }{a}_{\mu }{e}^{\mu }& (ExpWeyl\; 2)\end{array}$$ say; hence $d_0=$ coefficient of $e^{\rho }$ in $d=1,$ and if $\lambda \in P^{+}, \lambda \ne 0$ 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 $\lambda \in P^{+}, \lambda \ne 0,$ 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 $W^{+}=\{w\in W|\epsilon \left(w\right)=1\},$ then $W=W^{+}\cup s_{\alpha }{W}^{+}$ (disjoint union). Then $$\theta \left(e^{\lambda +\rho }\right)=\sum _{w\in W^{+}}\epsilon \left(w\right)(e^{w(\lambda +\rho )}-e^{s_{\alpha }w(\lambda +\rho )}).$$ It is enough to show that each term in this sum is divisible by $1-e^{-\alpha };$ i.e. that $$e^{\mu }-e^{s_{\alpha }\mu }(\mu \in P)$$ is divisible by $1-e^{-\alpha }.$ Now $s_{\alpha }\mu =\mu -⟨\mu ,{\alpha }^{\vee }⟩\alpha =\mu -m\alpha$ say, where $m\in \mathbb{Z},$ hence $$e^{\mu }-e^{s_{\alpha }\mu }=e^{\mu }(1-e^{-m\alpha })=e^{\mu }(1-e^{-\alpha })(1+e^{-\alpha }+\cdots +e^{-(m-1)\alpha })if\; m>0;$$ and if $m<0,$ say $m=-n,$ then $$(1-e^{-m\alpha })=(1-e^{n\alpha })=-e^{n\alpha }(1-e^{-n\alpha })$$ 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 $$e^{\rho }\prod _{\alpha >0}(1-e^{-\alpha })=\sum _{w\in W}\epsilon \left(w\right)e^{w\rho }$$ is Weyl's denominator formula. I will give an alternative, more combinatorial proof.

Example. $R$ of type $A_{n-1},$ roots ${\epsilon }_i-{\epsilon }_j (1\le i,j\le n,i\ne j).$ We have $$\begin{array}{rcl}\rho =\frac{1}{2}\sum _{i<j}({\epsilon }_i-{\epsilon }_j)& =& \frac{1}{2}((n-1){\epsilon }_1+(n-3){\epsilon }_2+\cdots +(1-n){\epsilon }_n)\\ & =& (n-1){\epsilon }_1+(n-2){\epsilon }_2+\cdots +{\epsilon }_{n-1}-\frac{1}{2}(n-1)({\epsilon }_1+\cdots +{\epsilon }_n)\\ & =& \delta -\frac{1}{2}(n-1)\epsilon .\end{array}$$ Let $x_i=e^{{\epsilon }_i},$ then Weyl's formula is in this case $$x^{\delta -\frac{n-1}{2}\epsilon }\cdot \prod _{i<j}(1-x_i^{-1}{x}_j)=\sum _{w\in S_n}\epsilon \left(w\right)x^{w\delta -\frac{n-1}{2}\epsilon }.$$ Multiply both sides by ${(x_1,...,x_n)}^{\frac{n-1}{2}}$ and we get $$\prod _{1\le i<j\le n}(x_i-x_j)=\sum _{w\in S_n}\epsilon \left(w\right)x^{w\delta }(Vandermonde\; identity).$$

Exercise: Write it down explicitly for the other $R$ of classical type $(B_n,C_n,D_n).$

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 $${\sigma }_E=\sum _{\alpha \in E}\alpha .$$ Suppose that $\rho -{\sigma }_E$ is a regular element of $V$ (i.e. does not lie in any hyperplane $H_{\alpha }$). Then $E=R\left(w\right) (=R^{+}\cap w{R}^{-})$ 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 $w^{-1}(\rho -{\sigma }_E)\in C.$ Now $$\begin{array}{ll}\rho -{\sigma }_E=\frac{1}{2}\sum _{\alpha \in R^{+}}{u}_{\alpha }\alpha & (ExpLem\; 1)\end{array}$$ where $$\begin{array}{ll}{u}_{\alpha }=\{\begin{array}{ll}+1,& if\alpha \notin E,\\ -1,& if\alpha \in E.\end{array}\phantom{\}}& (ExpLem\; 2)\end{array}$$ i.e. $\rho -{\sigma }_E$ is of the form $\frac{1}{2}\sum _{\alpha \in R^{+}}\pm \alpha .$ Hence so is $w^{-1}(\rho -{\sigma }_E),$ hence $$\begin{array}{ll}{w}^{-1}(\rho -{\sigma }_E)=\rho -{\sigma }_F& (ExpLem\; 3)\end{array}$$ 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 $⟨{\sigma }_F,{\alpha }_i⟩\le 0 (1\le i\le r),$ 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 $w^{-1}(\rho -{\sigma }_E)=\rho$ it follows from (ExpLem 1) that $$\frac{1}{2}\sum _{\alpha \in R^{+}}{u}_{\alpha }{w}^{-1}\alpha =\frac{1}{2}\sum _{\beta \in R^{+}}\beta .$$ Hence for each $\alpha \in R^{+},$ $$\begin{array}{ll}{u}_{\alpha }=\{\begin{array}{ll}+1,& if{w}^{-1}\alpha \in R^{+}\\ -1,& if{w}^{-1}\alpha \in R^{-}.\end{array}\phantom{\}}& (ExpLem\; 4)\end{array}$$ Hence from (ExpLem 2) and (ExpLem 4) $\alpha \in E \iff w^{-1}\alpha \in R^{-},$ 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 $$d=e^{\rho }\prod _{\alpha \in R^{+}}(1-e^{-\alpha })=\sum _{E\subseteq R^{+}}{(-1)}^{\left|E\right|}{e}^{\rho -{\sigma }_E},$$ on multiplying out. But $d$ is skew, hence $$\left|W\right|\cdot d=\theta d=\sum _{E\subseteq R^{+}}{(-1)}^{\left|E\right|}\theta \left(e^{\rho -{\sigma }_E}\right).$$ 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 $$\left|W\right|\cdot d=\sum _{w\in W}\epsilon \left(w\right)\theta \left(e^{w\rho }\right)$$ since ${(-1)}^{\left|E\right|}={(-1)}^{l\left(w\right)}=\epsilon \left(w\right);$ and $$\epsilon \left(w\right)\theta \left(e^{w\rho }\right)=\epsilon \left(w\right)\theta w\left(e^{\rho }\right)=\theta \left(e^{\rho }\right)$$ so that $\left|W\right|\cdot d=\left|W\right|\theta \left(e^{\rho }\right)$ 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

$$\sum _{w\in W}w\left(\prod _{\alpha \in R^{+}}\frac{1-t_{\alpha }{e}^{-\alpha }}{1-e^{-\alpha }}\right)=\sum _{w\in W}{t}_{R\left(w\right)}.$$

		Proof.
	Let $L$ denote the left-hand side of Proposition 3.5, so that $$L=\sum _{w\in W}w\left(d^{-1}{e}^{\rho }\prod _{\alpha \in R^{+}}(1-t_{\alpha }{e}^{-\alpha })\right)$$ and therefore $$\begin{array}{rcl}Ld& =& \sum _{w\in W}\epsilon \left(w\right)e^{w\rho }\prod _{\alpha \in R^{+}}(1-t_{\alpha }{e}^{-w\alpha })\\ & =& \sum _{w\in W}\epsilon \left(w\right)\sum _{E\subseteq R^{+}}{(-1)}^{\left|E\right|}{t}_E{e}^{w(\rho -{\sigma }_E)}\end{array}$$ on multiplying out the product, i.e., $$Ld=\sum _{E\subseteq R^{+}}{(-1)}^{\left|E\right|}{t}_E\theta \left(e^{\rho -{\sigma }_E}\right).$$ 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 $$E=R\left(w\right)and\rho -{\sigma }_E=w\rho$$ for some $w\in W.$ Hence (since ${(-1)}^{\left|R\left(w\right)\right|}={(-1)}^{l\left(w\right)}=\epsilon \left(w\right)$) $$\begin{array}{ll}Ld=\sum _{w\in W}\epsilon \left(w\right)t_{R\left(w\right)}\theta \left(e^{w\rho }\right)& (ExpId\; 1)\end{array}$$ 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 $$L=\sum _{w\in W}{t}_{R\left(w\right)}.$$ $\square$

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

$$\sum _{w\in W}w\left(\prod _{\alpha \in R^{+}}\frac{1-t{e}^{-\alpha }}{1-e^{-\alpha }}\right)=\sum _{w\in W}{t}^{l\left(w\right)}=W\left(t\right).$$

Here the right hand side is independent of the $e^{-\alpha },$ so we may specialize them in any way we please. Define $\phi :ℝQ\to ℝ[t,t^{-1}]$ by $$\phi \left(e^{{\alpha }_i}\right)=t^{-1}(1\le i\le r)$$ so that $$\phi \left(e^{-\alpha }\right)=t^{\mathrm{ht}\left(\alpha \right)}$$ and therefore $$\begin{array}{ll}\phi \left(\prod _{\alpha \in R^{+}}\frac{1-t{e}^{-\alpha }}{1-e^{-\alpha }}\right)=\prod _{\alpha \in R^{+}}\frac{1-t^{\mathrm{ht}\left(\alpha \right)+1}}{1-t^{\mathrm{ht}\left(\alpha \right)}}.& (Exp\; 2)\end{array}$$ What about $$\begin{array}{ll}\phi \left(\prod _{\alpha \in R^{+}}\frac{1-t{e}^{-w\alpha }}{1-e^{-w\alpha }}\right)& (Exp\; 3)\end{array}$$ 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 $$\phi (1-t{e}^{-w\alpha })=\phi (1-t{e}^{{\alpha }_i})=1-t\cdot t^{-1}=0$$ so that (Exp 3) is zero when $w\ne 1.$ Hence from Proposition 3.6 we obtain

$$\prod _{\alpha \in R^{+}}\frac{1-t^{1+\mathrm{ht}\left(\alpha \right)}}{1-t^{\mathrm{ht}\left(\alpha \right)}}=\sum _{w\in W}{t}^{l\left(w\right)}$$

another multiplicative formula for the Poincaré polynomial $W\left(t\right)=P(\mathscr{H};t)$ of the harmonics. From Proposition 3.7 and (3.31) it follows that $$\begin{array}{rcl}\prod _{i=1}^r\frac{1-t^{d_i}}{1-t}& =& \prod _{\alpha \in R^{+}}\frac{1-t^{1+\mathrm{ht}\left(\alpha \right)}}{1-t^{\mathrm{ht}\left(\alpha \right)}}\\ & =& \prod _{i\ge 1}{\left(\frac{1-t^{i+1}}{1-t^i}\right)}^{h_i}\end{array}$$ where $h_i (i\ge 1)$ 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 $$\begin{array}{ll}\prod _{i=1}^r(1-t^{d_i})=\prod _{i\ge 1}{(1-t^{i+1})}^{h_i-h_{i+1}}& (Exp\; 4)\end{array}$$ so that $h_i\ge h_{i+1}$ for each $i\ge 1,$ i.e. $$\eta =(h_1,h_2,...)$$ is a partition of $N=\mathrm{Card}\left(R^{+}\right);$ $h_i-h_{i+1}=m_i(\eta \prime ),$ the multiplicity of $i$ as part of the conjugate partition $\eta \prime ,$ so that

The partitions $(h_1,h_2,...)$ and $(d_1-1,d_2-1,...,d_r-1)=(e_1,...,e_r)$ where $(d_1\ge \cdots \ge d_r)$ 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 $A^W=\{f\in A|wf=fallw\in W\}$ 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 $$f=wf=\sum _{\mu }{f}_{\mu }{e}^{w\mu }$$ 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 $W\lambda , \lambda \in P^{+},$ so that $$f=\sum _{\lambda \in P^{+}}{f}_{\lambda }\left(\sum _{\mu \in W\lambda }{e}^{\mu }\right)$$ from which it follows that

The orbit-sums $$m_{\lambda }=\sum _{\mu \in W\lambda }{e}^{\mu }(\lambda \in P^{+})$$ are an $ℝ-$basis of $A^W$ (indeed, a $\mathbb{Z}-$basis of ${\left(\mathbb{Z}P\right)}^W$).

Weyl characters

For each $\lambda \in P^{+}$ we define $${\chi }_{\lambda }=\frac{\theta \left(e^{\lambda +\rho }\right)}{d}=\frac{\sum _{w\in W}\epsilon \left(w\right)e^{w(\lambda +\rho )}}{\sum _{w\in W}\epsilon \left(w\right)e^{w\rho }},by\; Proposition\; 3.3.$$ By Proposition 3.3 (ii) we have ${\chi }_{\lambda }\in A^W.$ The ${\chi }_{\lambda }$ are Weyl characters. Since (Prop. 3.2 (iii)) the $\theta (e^{\lambda +\rho )}, \lambda \in P^{+},$ form an $ℝ-$basis of the space $S$ of $W-$skew elements of $A,$ hence by Proposition 3.3 (ii) the ${\chi }_{\lambda }, \lambda \in P^{+}$ are an $ℝ-$basis of $A^W.$

We have $$\begin{array}{rcl}{\chi }_{\lambda }& =& \frac{\sum _{w\in W}\epsilon \left(w\right)e^{w(\lambda +\rho )-\rho }}{\prod _{\alpha \in R^{+}}(1-e^{-\alpha })}\\ & =& e^{\lambda }+ \; lower\; terms,\end{array}$$ where by lower terms is meant a linear combination of $e^{\mu }$ with $\mu <\lambda$ (because if $w\ne 1$ then $w(\lambda +\rho )<\chi +\rho ,$ i.e. $w(\lambda +\rho )-\rho <\lambda$). Hence, as ${\chi }_{\lambda }$ is symmetric, we can write

$${\chi }_{\lambda }=\sum _{\genfrac{}{}{0ex}{}{\mu \in P^{+}}{\mu \le \lambda }}{K}_{\lambda \mu }{m}_{\mu }with{K}_{\lambda \lambda }=1.$$

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. $R=A_{n-1}, \lambda =({\lambda }_1,...,{\lambda }_n)\in P$ (so that $\sum {\lambda }_i=0$). Since $⟨\lambda ,{\alpha }^{\vee }⟩$ is an integer for all $\alpha \in R$ it follows that ${\lambda }_i-{\lambda }_j\in \mathbb{Z}$ 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, $\lambda \in P^{+} \iff {\lambda }_1\ge {\lambda }_2\ge \cdots \ge {\lambda }_n$ (and $\sum {\lambda }_i=0$).

So any partition $\mu =({\mu }_1,...,{\mu }_n)$ of length $\le n$ determines a dominant weight $${\mu }^{*}=({\mu }_1-\frac{1}{n}\left|\mu \right|,...,{\mu }_n-\frac{1}{n}\left|\mu \right|).$$ So we have a surjective (but not bijective) mapping $\mu \mapsto {\mu }^{*}$ 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 $x_i=e^{{\epsilon }_i} (1\le i\le n).$ Then the Weyl characters ${\chi }_{\lambda }$ are essentially Schur functions: to be precise, if $\mu$ is a partition of length $\le n,$ then $${\chi }_{{\mu }^{*}}=\frac{s_{\mu }(x_1,...,x_n)}{{(x_1\cdots x_n)}^{\frac{\left|\mu \right|}{n}}}$$ (because it must have degree 0 as a polynomial in the $x_i^{\pm 1}$). In particular, since $${\pi }_i=({\epsilon }_1+\cdots +{\epsilon }_i)-\frac{i}{n}\sum _1^n{\epsilon }_j,we\; have{\chi }_{{\pi }_i}=\frac{e_i(x_1,...,x_n)}{{(x_1\cdots x_n)}^{\frac{i}{n}}}=m_{{\pi }_i},(1\le i\le n-1)$$ $({\pi }_n=0,{\chi }_{{\pi }_n}={\chi }_0=1).$

Return to the general case

For each $i\in [1,r]$ let $u_i$ be an element of $A^W$ of the form $$\begin{array}{ll}{u}_i=e^{{\pi }_i}+lower\; terms& (Exp\; 5)\end{array}$$

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

$A^W=ℝ[u_1,...,u_r]$ 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 $$\begin{array}{rcl}{u}_{\lambda }=u_1^{n_1}\cdots u_r^{n_r}& =& e^{\sum n_i{\pi }_i}+lower\; terms\\ & =& e^{\lambda }+lower\; terms\\ & =& m_{\lambda }+\sum _{\genfrac{}{}{0ex}{}{\mu M\lambda }{\mu \in P^{+}}}(*)m_{\mu }\end{array}$$ for certain coefficients $(*).$ 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 $\mathbb{Z}$), 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 $${\tilde{P}}_{\lambda }\left(t\right)={\tilde{P}}_{\lambda }=\sum _{w\in W}w\left(e^{\lambda }\prod _{\alpha \in R^{+}}\frac{1-t{e}^{-\alpha }}{1-e^{-\alpha }}\right).$$ Notation $W\left(t\right)=\sum _{w\in W}{t}^{l\left(w\right)};$ so in particular ${\tilde{P}}_0=W\left(t\right)$ by Proposition 3.6. We have $$\begin{array}{rcl}{\tilde{P}}_{\lambda }& =& \sum _{w\in W}w\left(\frac{e^{\lambda +\rho }}{d}\prod _{\alpha \in R^{+}}(1-t{e}^{-\alpha })\right)\\ & =& d^{-1}\sum _{w\in W}\epsilon \left(w\right)\cdot w\left(e^{\lambda +\rho }\prod _{\alpha \in R^{+}}(1-t{e}^{-\alpha })\right)\\ & =& d^{-1}\theta \left(e^{\lambda +\rho }\prod _{\alpha \in R^{+}}(1-t{e}^{-\alpha })\right)\end{array}$$ which shows that ${\tilde{P}}_{\lambda }\in A^W\left[t\right].$

Let $W_{\lambda }=\{w\in W|w\lambda =\lambda \}$ be the subgroup of $W$ that fixes $\lambda .$ Since $\lambda \in P^{+}\subseteq \stackrel{—}{C}, W_{\lambda }$ is generated by the $s_i, i\in I.$ Let $R_{\lambda }$ be the root system with basis ${\alpha }_i (i\in I).$ 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 $${\tilde{P}}_{\lambda }=\sum _{u\in U}u\left(e^{\lambda }\prod _{\alpha \in R^{+}-R_{\lambda }^{+}}\frac{1-t{e}^{-\alpha }}{1-e^{-\alpha }}\cdot \sum _{v\in W_{\lambda }}v\left(\prod _{\alpha \in R_{\lambda }^{+}}\frac{1-t{e}^{-\alpha }}{1-e^{-\alpha }}\right)\right)$$ from what we have just observed. By Proposition 3.6 it follows that $${\tilde{P}}_{\lambda }=W_{\lambda }\left(t\right)\sum _{u\in W/W_{\lambda }}u\left(e^{\lambda }\prod _{\alpha \in R^{+}-R_{\lambda }^{+}}\frac{1-t{e}^{-\alpha }}{1-e^{-\alpha }}\right)$$ and hence all the coefficients of ${\tilde{P}}_{\lambda }$ (which are polynomials in $t$) are divisible by $W_{\lambda }\left(t\right),$ and we can define $$\begin{array}{rcll}{P}_{\lambda }\left(t\right)& =& W_{\lambda }{\left(t\right)}^{-1}\sum _{w\in W}w\left(e^{\lambda }\prod _{\alpha \in R^{+}}\frac{1-t{e}^{-\alpha }}{1-e^{-\alpha }}\right)& (Exp\; 6)\\ & =& \sum _{u\in W/W_{\lambda }}u\left(e^{\lambda }\prod _{\genfrac{}{}{0ex}{}{\alpha \in R^{+}}{-R_{\lambda }^{+}}}\frac{1-t{e}^{-\alpha }}{1-e^{-\alpha }}\right)& (Exp\; 7)\\ & =& e^{\lambda }+lower\; terms& \\ & =& m_{\lambda }+lower\; terms.& \end{array}$$

Particular case:

1. $t=1,$ then $$P_{\lambda }=\sum _{u\in W/W_{\lambda }}{e}^{u\lambda }=m_{\lambda },from(Exp\; 7)$$
2. $t=0,$ then $$\begin{array}{rcl}{P}_{\lambda }& =& \sum _{w\in W}w\left(e^{\lambda }\prod _{\alpha \in R^{+}}\frac{1}{1-e^{-\alpha }}\right)\\ & =& d^{-1}\sum _{w\in W}\epsilon \left(w\right)e^{w(\lambda +\rho )}={\chi }_{\lambda }.\end{array}$$

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 $${\chi }_{\lambda }=\sum _{\genfrac{}{}{0ex}{}{\mu \le \lambda }{\mu \in P^{+}}}{K}_{\lambda \mu }\left(t\right)P_{\mu }(\lambda \in P^{+})$$ 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 $$F(\xi ;t)=\sum _{\left(m_{\alpha }\right)}{t}^{\sum m_{\alpha }}$$ 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 $F(\xi ;t)=0$ unless $\xi \in Q^{+}.$)

1. $K_{\lambda \mu }\left(t\right)=\sum _{w\in W}\epsilon \left(w\right)F(w(\lambda +\rho )-(\mu +\rho );t),(\lambda ,\mu \in P^{+})$ (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 $$\prod _{\alpha \in R^{+}}{(1-t{R}_{\alpha })}^{-1},$$ 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 $\prod _{\alpha \in R}{(1-e^{\alpha })}^k$ is $\prod _{i=1}^r\left(\genfrac{}{}{0ex}{}{k{d}_i}{k}\right).$
2. (C1q) The constant term in $\prod _{\alpha \in R^{+}}\prod _{i=0}^{k-1}(1-q^i{e}^{\alpha })(1-q^{i+1}{e}^{-\alpha })$ is $\prod _{i=1}^r\left[\genfrac{}{}{0ex}{}{k{d}_i}{k}\right]$ where $$\left[\genfrac{}{}{0ex}{}{n}{r}\right]=\frac{(1-q^n)(1-q^{n-1})\cdots (1-q^{n-r+1})}{(1-q)(1-q^2)\cdots (1-q^r)}$$ 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 $$\begin{array}{rcl}\prod _{\alpha \in R}(1-e^{\alpha })& =& \left(e^{\rho }\prod _{\alpha \in R^{+}}(1-e^{-\alpha })\right)\cdot \left(e^{-\rho }\prod _{\alpha \in R^{+}}(1-e^{\alpha })\right)\\ & =& \left(\sum _{w\in W}\epsilon \left(w\right)e^{w\rho }\right)\left(\sum _{\nu \in W}\epsilon \left(\nu \right)e^{-\nu \rho }\right)by\; Proposition\; 3.3.\end{array}$$ Since $\rho \in C$ we have $w\rho =\nu \rho \iff \nu =w,$ so the constant term is $\left|W\right|=d_1{d}_2\cdots d_r,$ which is (C1) for $k=1.$

Next, when $k=1$ we have to consider the constant term $C\left(q\right)$ in $$\prod _{\alpha \in R^{+}}(1-e^{\alpha })(1-q{e}^{-\alpha })=\prod _{\alpha \in R}(1-e^{\alpha })\cdot \prod _{\alpha \in R^{+}}\frac{1-q{e}^{-\alpha }}{1-e^{-\alpha }}.$$ Hence $C\left(q\right)=\frac{1}{\left|W\right|}\times$ constant term in $$\prod _{\alpha \in R}(1-e^{\alpha })\sum _{w\in W}w\left(\prod _{\alpha \in R^{+}}\frac{1-q{e}^{-\alpha }}{1-e^{-\alpha }}\right)=\prod _{\alpha \in R}(1-e^{\alpha })\cdot W\left(q\right)by\; Proposition\; 3.5.$$ But we have just seen that the constant term in $\prod _{\alpha \in R}(1-e^{\alpha })$ is $\left|W\right|,$ hence $$C\left(q\right)=W\left(q\right)=\prod _{i=1}^r\frac{1-q^{d_i}}{1-q}=\prod _{i=1}^r\left[\genfrac{}{}{0ex}{}{d_i}{1}\right].$$

Formal exponentials as functions on $V$

Let $\lambda \in P.$ We may regard $e^{\lambda }$ as a (complex-valued) function on $V,$ as follows $$e^{\lambda }\left(x\right)=e^{2\pi i⟨\lambda ,x⟩}(x\in V)$$ where the $e$ on the right is the usual one: 2.718...

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

Let $\Pi$ be the parallelepiped $$\Pi = {\sum x_i{\alpha }_i^{\vee }|0\le x_i<1(1\le i\le r)}$$ so that the projection $\phi :V\to T$ restricted to $\Pi$ is a bijection $\Pi \tilde{\to }T.$

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

Hence $\tilde{f}$ can be expanded as a convergent Fourier series: $$\tilde{f}=\sum _{\lambda \in P}{f}_{\lambda }{e}^{\lambda }$$ — now in general an infinite series — where $$f_{\lambda }={\int }_{\Pi }\tilde{f}\left(x\right)e^{-\lambda }\left(x\right)dx:={\int }_{T}f\left(\stackrel{\cdot }{x}\right)e^{-\lambda }\left(\stackrel{\cdot }{x}\right)d\stackrel{\cdot }{x}$$ 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 $\tilde{f}$ is $$f_0={\int }_{T}f\left(\stackrel{\cdot }{x}\right)d\stackrel{\cdot }{x}.$$ These considerations apply to the conjectures (C1) and (C1q). Consider the latter. Introduce the notation $${(x;q)}_{\infty }=\prod _{i=1}^{\infty }(1-q^{i}x)$$ whenever the product makes sense. Then if $k$ is an integer $\ge 0$ $$\frac{{(x;q)}_{\infty }}{{(x{q}^k;q)}_{\infty }}=\prod _{i=0}^{k-1}(1-q^{i}x).$$ So in (C1q) we seek the constant term in $$\begin{array}{ll}\prod _{\alpha \in R^{+}}\frac{{(e^{\alpha },q)}_{\infty }{(q{e}^{-\alpha };q)}_{\infty }}{{(q^k{e}^{\alpha };q)}_{\infty }{(q^{k+1}{e}^{-\alpha };q)}_{\infty }}.& (Exp\; 9)\end{array}$$ Think of $q$ now as a real number, $0\le q<1.$ The infinite product $${(e^{\alpha };q)}_{\infty }=\prod _{i\ge 0}(1-q^i{e}^{\alpha })$$ converges uniformly on $T$ to a continuous function (because $\sum q^i{e}^{\alpha }$ converges). So does ${(q^k{e}^{\alpha };q)}_{\infty }$ for any real or complex $k$ $(q^k=e^{k\log q});$ 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 $\Gamma (k+1)=k!$ is the limit as $q\to 1$ of $${\Gamma }_q(k+1)=\frac{(1-q)(1-q^2)\cdots (1-q^k)}{{(1-q)}^k}=\frac{{(q;q)}_{\infty }}{{(q^{k+1};q)}_{\infty }{(1-q)}^k}.$$ 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 $$\left[\genfrac{}{}{0ex}{}{k{d}_i}{k}\right]as\frac{{\Gamma }_q(k{d}_i+1)}{{\Gamma }_q(k+1){\Gamma }_q(k(d_i-1)+1)}$$ (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 $$p\left(\mu \right)=no.\; of\; sequences\; {\left(n_{\alpha }\right)}_{\alpha \in R^{+}} \; such\; that\; n_{\alpha }\in \mathbb{N} \; (all\; \alpha )\; and\; \sum _{\alpha \in R^{+}}{n}_{\alpha }\alpha =\mu .$$ (So $p\left(\mu \right)=0$ unless $\mu \in Q^{+}$.) Clearly $$\begin{array}{ll}\prod _{\alpha \in R^{+}}{(1-e^{-\alpha })}^{-1}=\sum _{\xi \in Q^{+}}p\left(\xi \right)e^{-\xi }.& (Exp\; 10)\end{array}$$ Define $K_{\lambda \mu }$ by $${\chi }_{\lambda }=\sum _{\mu }{K}_{\lambda \mu }{m}_{\mu }$$ and $c_{\lambda \mu }^{\nu }$ by $${\chi }_{\lambda }{\chi }_{\mu }=\sum _{\nu }{c}_{\lambda \mu }^{\nu }{\chi }_{\nu }$$ $(\lambda ,\mu ,\nu \in P^{+}).$

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

		Proof.
	$K_{\lambda \mu }$ is the coefficient of $e^{\mu }$ in $${\chi }_{\lambda }=\left(\sum _{w\in W}\epsilon \left(w\right)e^{w(\lambda +\rho )-\rho }\right)\prod _{\alpha \in R^{+}}{(1-e^{-\alpha })}^{-1}=\sum _{\genfrac{}{}{0ex}{}{w\in W}{\xi \in Q^{+}}}\epsilon \left(w\right)p\left(\xi \right)e^{w(\lambda +\rho )-\xi -\rho }.$$ So we must have $w(\lambda +\rho )-\xi -\rho =\mu ,$ i.e. $\xi =w(\lambda +\rho )-(\mu +\rho )$ and hence we get $$K_{\lambda \mu }=\sum _{w\in W}\epsilon \left(w\right)p(w(\lambda +\rho )-(\mu +\rho )).$$ We have $$\begin{array}{ll}\frac{\theta \left(e^{\lambda +\rho }\right)\theta \left(e^{\mu +\rho }\right)}{d}=\sum _{\nu }{c}_{\lambda \mu }^{\nu }\theta \left(e^{\nu +\rho }\right)& (Exp\; Kos)\end{array}$$ and therefore $c_{\lambda \mu }^{\nu }$ is the coefficient of $e^{\nu +\rho }$ in the left hand side of (Exp Kos), i.e. in $$\left(\sum _{v,w}\epsilon \left(vw\right)e^{v(\lambda +\rho )+w(\mu +\rho )-\rho }\right)\left(\prod _{\alpha \in R^{+}}{(1-e^{-\alpha })}^{-1}\right)=\sum _{\genfrac{}{}{0ex}{}{v,w\in W}{\xi \in Q^{+}}}\epsilon \left(vw\right)p\left(\xi \right)e^{v(\lambda +\rho )+w(\mu +\rho )-\xi -\rho }.$$ So we must have $\nu (\lambda +\rho )+w(\mu +\rho )-\xi -\rho =\nu +\rho$ i.e. $$\xi =v(\lambda +\rho )+w(\mu +\rho )-(\nu +2\rho )$$ 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.

References

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

Version: March 12, 2012

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