Invariants

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

Last update: 30 May 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.

Introduction

$R$ a root system in $V;$ $W$ the Weyl group, acting on $V$ as a group of isometries. ($W$ could be any finite Coxeter group). Each $v\in V$ (in particular, each $\alpha \in R$) may be regarded as a linear function on $V:v\left(x\right)=⟨v,x⟩(x\in V).$ So if $v_1,...,v_k\in V$ we may define $v_1\cdots v_k$ as a polydomial function on $V:$ $$(v_1\cdots v_k)\left(x\right)=v_1\left(x\right)\cdots v_k\left(x\right)(x\in V).$$ Let $𝒮$ denote the $ℝ-$algebra of polynomial functions on $V.$ If $x_1,...,x_r$ is a basis of $V$ then $$𝒮=ℝ[x_1,...,x_r].$$

1. $𝒮$ is graded, i.e. $𝒮=\underset{n\ge 0}{⨁}{𝒮}^n,$ ${𝒮}^n$ the polynomials of degree $n.$
2. $W$ acts on $𝒮$ as a group of automorphisms: if $w\in W,$ $f\in 𝒮$ we define $$\left(wf\right)\left(x\right)=f\left(w^{-1}x\right),(x\in V).$$ We have $w(f+g)=wf+wg,w\left(fg\right)=wf\cdot wg(f,g\in 𝒮).$

Moreover $W$ respects the grading, i.e. $W$ acts on each ${𝒮}^n.$

Let $\alpha \in R,$ $f\in 𝒮.$ Then $s_{\alpha }f-f$ is divisible by $\alpha \in 𝒮.$

		Proof.
	Let $g=s_{\alpha }-f,$ so that $g\left(x\right)=\left(s_{\alpha }f\right)\left(x\right)-f\left(x\right)=f\left(s_{\alpha }x\right)-f\left(x\right)$ for $x\in V.$ Hence $g\left(x\right)=0$ if $s_{\alpha }x=x,$ i.e. if $\alpha \left(x\right)(=⟨\alpha ,x⟩)=0.$ Hence $\alpha$ divides $g\in 𝒮.$ $\square$

$${\Delta }_{\alpha }=\frac{1-s_{\alpha }}{\alpha }:𝒮\to 𝒮(\alpha \in R).$$ Now let $$ℐ={𝒮}^W=\{f\in 𝒮|wf=f,allw\in W\}.$$ $ℐ$ is the ring of polynomial invariants of $W.$ $ℐ$ is a graded ring: $$ℐ=\underset{n\ge 0}{⨁}{ℐ}^n,{ℐ}^n=ℐ\cap {𝒮}^n,({ℐ}^0=ℝ).$$

Examples.

1. $W=S_n$ acting on $V={ℝ}^n$ by permuting the coordinates. Then $W$ acts on $𝒮=ℝ[x_1,...,x_n]$ by permuting the $x_i,$ and $ℐ$ is the subring of symmetric polynomials: $ℐ=ℝ[e_1,...,e_n]$ where $e_i$ are the elementary symmetric functions.
2. $W=S_n$ acting on $V=$ hyperplane $x_1+\cdots +x_n=0$ in ${ℝ}^n.$ This time $𝒮=ℝ[x_1,...,x_n]$ but now the $x_i$ are not independent: they are related by $x_1+\cdots +x_n=0.$ So now $$ℐ=ℝ[e_2,...,e_n](e_1=0).$$
3. $W=$ Weyl group of type $B_n,$ acting on ${ℝ}^n$ by signed permutations. If $f(x_1,...,x_n)\in ℐ,$ then $f$ is unchanged by replacing $x_i$ by $-x_i,$ hence is a symmetric polynomial in the $x_i^2.$ So this time $ℐ=ℝ[f_1,...,f_n]$ where $f_i=i^{\mathrm{th}}$ elementary symmetric function of $x_1^2,...,x_n^2.$
4. Any $R:$ $$\prod _{\alpha \in R}(t+\alpha )=\prod _{\alpha \in R^{+}}(t^2-{\alpha }^2)$$ is $W-$invariant, hence the elementary symmetric functions of the ${\alpha }^2$ ($\alpha \in R^{+}$) are $W-$invariants.

Averaging

Define $a:𝒮\to 𝒮$ by $$a\left(f\right)=\frac{1}{\left|W\right|}\sum _{w\in W}wf.$$

1. $a\left(f\right)\in ℐ$ for all $f\in 𝒮.$
2. $a\left(f\right)=f$ if $f\in ℐ.$
3. $a\left(fg\right)=a\left(f\right)g$ if $f\in 𝒮,$ $g\in ℐ.$

Our first aim is to prove Chevalley's theorem.

There exist homogeneous elements $I_1,...,I_r\in ℐ(r=\dim \left(V\right))$ such that $$ℐ=ℝ[I_1,...,I_r]$$ and $I_1,...,I_r$ are algebraically independent over $ℝ.$

Let ${ℐ}^{+}=\underset{n\ge 1}{⨁}{ℐ}^n,$ so that an element $g\in ℐ$ lies in ${ℐ}^{+}$ if and only if $g\left(0\right)=0.$ Let $𝒥=𝒮{ℐ}^{+}$ be the ideal in $𝒮$ generated by ${ℐ}^{+},$ so that each element of $𝒥$ can be written as a finite sum $\sum g_i{h}_i$ with $g_i\in 𝒮$ and $h_i\in {ℐ}^{+}.$ Since $𝒮$ is a polynomial ring, it follows from Hilbert's basis theorem that there is a finite set $I_1,...,I_n$ of homogeneous elements of ${ℐ}^{+}$ such that (a) $I_1,...,I_n$ generate the ideal $𝒥;$ (b) no proper subset does. I claim that these $I_j$ do what we want. So we have three things to prove:

1. every element of $ℐ$ is a polynomial in the $Ij$ (with real coefficients);
2. $I_1,...,I_n$ are algebraically independent over $ℝ;$
3. $n=r.$

		Proof of 1.
	Let $f\in ℐ,$ without loss of generality homogeneous of degree $d.$ Induction on $d.$ $d=0$ is OK; suppose $d>0,$ then $f\in {ℐ}^{+}\subseteq 𝒥,$ hence $$f=\sum _{j=1}^n{p}_j{I}_j,$$ with $p_j\in 𝒮.$ Since $f$ and the $I_j$ are homogeneous, we may assume $p_j$ homogeneous of degree $d-d_j<d$ (where $d_j=\mathrm{deg}{I}_j$). Now apply the averaging operator (Prop. 2.1): $$f=\sum _{j=1}^{n}a\left(p_j\right)I_j.$$ By (Prop. 2.1), $a\left(p_j\right)\in ℐ$ and has degree $<d,$ hence, by the inductive hypothesis, is a polynomial in $I_1,...,I_n.$ Hence, so is $f.$ $\square$

For the proof of 2. we require a lemma:

Let $F_1,...F_m\in ℐ$ be such that $F_1$ is not in ideal $𝔞$ of $ℐ$ generated by $F_2,...,F_m.$ Suppose $p_1,...,p_m$ are homogeneous elements of $𝒮$ such that $\sum _{i=1}^m{p}_i{F}_i=0.$ Then $p_1\in 𝒮{ℐ}^{+}=𝒥.$

		Proof.
	Induction on $d=\mathrm{deg}{p}_1.$ $d=0,$ then $p_1\in ℝ:$ by averaging the relation $\sum _{i=1}^m=0$ over $W$ we get $$p_1{F}_1=a\left(p_1{F}_1\right)=-\sum _{i=2}^{m}a\left(p_i\right)F_i\in 𝔞,$$ by (Prop. 2.1). If $p_1\ne 0$ this shows that $F_1\in 𝔞,$ a contradiction. Hence $p_1=0\in 𝒥.$ $d>0.$ Let $\alpha \in R.$ By (Prop. 1.1) we have $$p_i-s_{\alpha }{p}_i=\alpha q_{i}say{q}_i={\Delta }_{\alpha }{p}_i$$ where $q_i$ is homogeneous of degree $\mathrm{deg}{p}_i-1.$ From the relation $\sum p_i{F}_i=0$ we deduce $\sum s_{\alpha }\left(p_i\right)F_i=0$ and hence on subtraction $$\sum _{i=1}^m{q}_i{F}_i=0.$$ Each $q_i\in 𝒮$ is homogeneous, and $\mathrm{deg}{q}_1={\mathrm{deg}}_{p_1}-1=d-1,$ so by the inductive hypothesis $q_1\in 𝒥,$ i.e. $p_1-s_{\alpha }{p}_1\in 𝒥,$ or $p_1\equiv s_{\alpha }{p}_1\mathrm{mod}𝒥.$ Since the $s_{\alpha }$ generate $W$ and $𝒥$ is $W-$stable it follows that $p_1\equiv w{p}_1\mathrm{mod}𝒥$ for all $w\in W,$ or $p_1-w{p}_1\in 𝒥.$ Averaging over $W$ we get $p_1-a\left(p_1\right)\in 𝒥,$ hence $p_1\in 𝒥$ (because $a\left(p_1\right)\in {ℐ}^{+}$). $\square$

		Proof of 2.
	Suppose $F$ is a nonzero polynomial in $n$ variables such that $F(I_1,...,I_n)=0.$ We may assume that all the monomials in the $I_j$ that occur in $F$ have the same degree $d$ as elements of $𝒮,$ and that $d$ is minimal. Let $F_i=\frac{\partial F}{\partial I_i} (1\le i\le n),$ then $\mathrm{deg}{F}_i<d$ and so $F_i\ne 0.$ Consider the ideal in $ℐ$ generated by $F_1,...,F_n.$ We may choose the numbering so that $F_1,...,F_m,$ but no proper subset, generate this ideal. Then there exist $g_{ij}\in ℐ$ such that $$\begin{array}{ll}{F}_i=\sum _{j=1}^m{g}_{ij}{F}_j,(i<m).& (Inv\; 1)\end{array}$$ $F_i$ is homogeneous of degree $d-d_i(d_i=\mathrm{deg}{I}_i).$ Hence we may assume $g_{ij}$ homogeneous of degree $d_j-d_i$ (so zero if $d_j<d_i$). Since $F(I_1,...,I_n)=0$ we have $\frac{\partial F}{\partial x_k}=0(1\le k\le r),$ i.e. $$\sum _{i=1}^n{F}_i\frac{\partial I_i}{\partial x_k}=0(1\le k\le r).$$ Using (Inv 1) this becomes $$\sum _{i=1}^m{F}_i\frac{\partial I_i}{\partial x_k}+\sum _{i=m+1}^n\left(\sum _{j=1}^m{g}_{ij}{F}_j\right)\frac{\partial I_i}{\partial x_k}=0$$ for $$\sum _{i=1}^m{F}_i\left(\frac{\partial I_i}{\partial x_k}+\sum _{j=m+1}^n{g}_{ji}\frac{\partial I_j}{\partial x_k}\right)=0.$$ Now apply the Lemma 2.3: $F_1$ is not in the ideal of $ℐ$ generated by $F_2,...,F_m,$ and each of the polynomials $\frac{\partial I_i}{\partial x_k}+\sum _j{g}_{ji}\frac{\partial I_j}{\partial x_k}$ is homogeneous (of degree $d_i-1$). Hence $$\frac{\partial I_1}{\partial x_k}+\sum _{i=m+1}^n{g}_{j1}\frac{\partial I_j}{\partial x_k}$$ lies in $𝒥,$ hence is of the form $\sum _{i=1}^n{h}_{ik}{I}_i$ with $h_{ik}\in 𝒮$ homogeneous of degree $(d_1-1)-d_i.$ In particular, $h_{1k}=0,$ so that $$\frac{\partial I_1}{\partial x_k}+\sum _{i=m+1}^n{g}_{j1}\frac{\partial I_j}{\partial x_k}=\sum _{i=2}^n{h}_{ik}{I}_i,(1\le k\le r).$$ Now multiply by $x_k$ and sum over $k,$ remembering Euler's formula: we get $$d_1{I}_1+\sum _{m+1}^n{g}_{j1}{d}_j{I}_j=\sum _{i=2}^n\left(\sum _{k=1}^r{h}_{ik}{x}_k\right)I_i,$$ showing (since $d_1>0$) that $I_1$ lies in the ideal of $𝒮$ generated by $I_2,...,I_n:$ a contradiction. $\square$

It remains to prove that $n=r.$

		Proof of 3.
	Let $\begin{array}{rcl}K& =& ℝ(f_1,...,f_n)=field\; of\; fractions\; ofℐ,\\ L& =& ℝ(x_1,...,x_r)=field\; of\; fractions\; of𝒮.\end{array}$ Then $\mathrm{tr.deg.}K/ℝ=n,$ $\mathrm{tr.deg.}L/ℝ=r.$ Now transcendence degree is additive in towers $L>K>ℝ:$ $$\mathrm{tr.deg.}L/ℝ=\mathrm{tr.deg.}L/K+\mathrm{tr.deg.}K/ℝ.$$ So we have to prove that $\mathrm{tr.deg.}L/K=0,$ i.e. that $L/K$ is an algebraic extension. Since $L=K(x_1,...,x_r)$ it is enough to show that each $x_i$ is algebraic over $K,$ i.e. is a root of an equation with coefficients in $K.$ Consider the equation in $t$ $$\prod _{w\in W}(t-w{x}_i)=0.$$ It has a root, and its coefficients are clearly $W-$invariant, i.e. $ℐ\subseteq K.$ This completes the proof of Theorem 2.2. $\square$

In Theorem 2.2 the invariants $I_1,...,I_r$ are not uniquely determined, but their degrees $d_i=\mathrm{deg}{I}_i$ are. To see why this is so we introduce Poincaré series.

Poincaré Series

Let $M=\underset{n\ge 0}{⨁}{M}_n$ be a graded vector space (over $ℝ$), with each $M_n$ finite dimensional. We define the Poincaré series of $M$ to be $$P(M,t)=\sum _{n\ge 0}\left(\dim M_n\right)t^n\in \mathbb{Z}\left[[t]\right].$$ If $N=\underset{n\ge 0}{⨁}{N}_n$ is another graded vector space (with $\dim N_n<\infty ,$ all $n$) then $M\oplus N$ and $M\otimes N$ are graded. $${(M\oplus N)}_n=M_n\oplus N_n,{(M\otimes N)}_n=\underset{p+q=n}{⨁}{M}_p\otimes N_q,$$ from which it follows that

$$P(M\oplus N,t)=P(M,t)+P(N,t),P(M\otimes N,t)=P(M,t)P(N,t).$$

Suppose $M=ℝ\left[y\right]$ where $y\in 𝒮$ is homogeneous, $\mathrm{deg}y=d\ge 1.$ Then $$\dim M_n=\{\begin{array}{ll}0,& ifd\nmid n,\\ 1,& ifd\mid n,\end{array}\phantom{\}}$$ so that $$P(M,t)=\sum _{m\ge 0}{t}^{md}=\frac{1}{1-t^d}.$$ From this and (Prop. 3.1) we deduce

1. $$P(𝒮,t)=\frac{1}{{(1-t)}^r},$$
2. $$P(ℐ,t)=\prod _{i=1}^r\frac{1}{1-t^{d_i}}.$$

		Proof.
	$$𝒮=ℝ[x_1,...,x_r]\cong ℝ\left[x_1\right]\otimes \cdots \otimes ℝ\left[x_r\right],(\mathrm{deg}{x}_i=1),$$ from which i. follows. Likewise $$ℐ=ℝ[I_1,...,I_r]\cong ℝ\left[I_1\right]\otimes \cdots \otimes ℝ\left[I_r\right],(\mathrm{deg}{I}_i=d_i)$$ by Chevalley's Theorem (2.2), which gives ii. $\square$

Now suppose that also $$ℐ=ℝ[{I\prime }_1,...,{I\prime }_r],$$ where ${I\prime }_1,...,{I\prime }_r\in {ℐ}^{+}$ are homogeneous and algebraically independent over $ℝ.$ If ${d\prime }_i=\mathrm{deg}{I\prime }_i,$ then from (Prop. 3.2) we have $$P(ℐ,t)=\prod _{i=1}^r{(1-t^{{d\prime }_i})}^{-1}and\; hence\prod _{i=1}^r(1-t^{d_i})=\prod _{i=1}^r(1-t^{{d\prime }_i}),$$ from which it follows that (after renumbering if necessary) $d_i={d\prime }_i(1\le i\le r).$ The numbers $d_1,...,d_r$ are called the degrees of $W.$

(Molien's formula) $$P(ℐ,t)=\frac{1}{\left|W\right|}\sum _{w\in W}\frac{1}{\det (1-wt)}.$$

		Proof.
	In general, if $U$ is a finite dimensional vector space (over a field of characteristic 0) and $\phi :U\to U$ is an idempotent linear transformation (i.e. ${\phi }^2=\phi$) then $$\mathrm{tr}\phi =\mathrm{rk}\phi (=\dim \phi \left(U\right))$$ (for each eigenvalue of $\phi$ is 0 or 1, hence with respect to a suitable basis of $U$ the matrix of $\phi$ is $\left(\begin{array}{cc}{1}_r& 0\\ 0& 0\end{array}\right).$) Apply this to the averaging operator $a$ acting on ${𝒮}^n:$ $$\dim {ℐ}^n=\dim a\left({𝒮}^n\right)=\mathrm{tr}(a,{𝒮}^n)=\frac{1}{\left|W\right|}\sum _{w\in W}\mathrm{tr}(w,{𝒮}^n),$$ so that $$\begin{array}{ll}P(ℐ,t)=\sum _{n\ge 0}\left(\dim {ℐ}^n\right)t^n=\frac{1}{\left|W\right|}\sum _{w\in W}\left(\sum _{n\ge 0}\mathrm{tr}(w,{𝒮}^n)t^n\right).& (Inv\; 2)\end{array}$$ Suppose $w$ acting on $V$ has eigenvalues ${\lambda }_1,...,{\lambda }_r$ (roots of unity, since $w$ has finite order). Then the eigenvalues of $w$ acting on ${𝒮}^n=S^n\left(V\right)$ will all be monomials ${\lambda }_1^{a_1}\cdots {\lambda }_r^{a_r}$ with $\sum a_i=n,$ as one sees by diagonalising $w.$ Hence $$\mathrm{tr}(w,{𝒮}^n)=h_n({\lambda }_1,...,{\lambda }_r)$$ and therefore $$\sum _{n\ge 0}\mathrm{tr}(w,{𝒮}^n)t^n=\sum _{n\ge 0}{h}_n({\lambda }_1,...,{\lambda }_r)t^r=\prod _{i=1}^r\frac{1}{1-{\lambda }_{i}t}=\frac{1}{\det (1-wt)}.$$ $\square$

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

1. $d_1\cdots d_r=\left|W\right|.$
2. $d_1+\cdots +d_r=r+N,$ where $N=\left|R^{+}\right|=number\; of\; positive\; roots.$

		Proof.
	From (Prop. 3.2) and (Prop. 3.3) we have $$\begin{array}{ll}\prod _{i=1}^r\frac{1-t}{1-t^{d_i}}=\frac{1}{\left|W\right|}\sum _{w\in W}\frac{{(1-t)}^r}{\det (1-wt)}.& (Inv\; 3)\end{array}$$ Set $t=1.$ Then all terms on the right vanish except that for $w=1,$ so the right hand side is equal to $\frac{1}{\left|W\right|},$ and the left hand side is equal to $\prod _1^r\frac{1}{d_i}.$ This proves i. Suppose $w\in W$ is a reflection. I claim that $w=s_{\alpha },$ some $\alpha \in R^{+}.$ For if not, then $wx=x$ for some $x$ not orthogonal to any root, i.e. $x\in V_{\mathrm{reg}}.$ But then $wx=x$ implies $w=1$ by (1.21) (ref. to Root Systems?), contradiction. Now $w$ is a reflection if and only if $\det (1-wt)={(1-t)}^{r-1}(1+t).$ Hence the right hand side of (Inv 3) is of the form $$\frac{1}{\left|W\right|}(1+\frac{N(1-t)}{1+t}+{(1-t)}^{2}F\left(t\right)),$$ where $F\left(t\right)$ is a rational function of $t$ with no pole at $t=1.$ Now put $t=1+u$ in (Inv 3) . The left hand side becomes $$\begin{array}{rcl}\prod _{i=1}^r\frac{u}{{(1+u)}^{d_i}-1}=\prod _{i=1}^r\frac{1}{d_i+\left(\genfrac{}{}{0ex}{}{d_i}{2}\right)u+\cdots }& =& \frac{1}{\left|W\right|}\prod _{i=1}^r{(1+\frac{d_i-1}{2}u+\cdots )}^{-1}\\ & =& \frac{1}{\left|W\right|}\prod _{i=1}^r(1-\frac{d_i-1}{2}u+\cdots )\\ & =& \frac{1}{\left|W\right|}(1-\sum _{i=1}^r\frac{d_i-1}{2}u+O\left(u^2\right)).\end{array}$$ On the right hand side, we get $$\frac{1}{\left|W\right|}(1-\frac{Nu}{2+u}+O\left(u^2\right)).$$ Comparing coefficients of $u$ now gives $$\sum _{i=1}^r\frac{d_i-1}{2}=\frac{N}{2}.$$ $\square$

We shall later improve on (Prop. 3.4).

Suppose $W$ is any finite group of linear transformations of $V.$ Define $𝒮=S\left(V\right), ℐ={𝒮}^W, 𝒥=𝒮{ℐ}^{+}$ as before. Then $ℐ$ is generated as an $ℝ-$algebra by $I_1,\mathrm{...},I_n.$ (Same proof as before.)

Moreover if $(x,y)$ is any positive definite scalar product on $V$ then $$⟨x,y⟩=\frac{1}{\left|W\right|}\sum _{w\in W}(wx,wy)$$ is a $W-$invariant positive definite scalar product (positive definite because $$⟨x,x⟩=\frac{1}{\left|W\right|}\sum _{w\in W}(wx,wx)\ge 0,$$ and vanishes if and only if $x=0$). So we may assume $W$ is a group of isometries (or orthogonal transformations) of $V$ (relative to a suitably chosen scalar product).

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

Consider the following four statements:

1. $W$ is generated by reflections.
2. Lemma 2.3 holds.
3. $I_1,...,I_n$ are algebraically independent. (C') $n=r.$
4. $P(ℐ,t)=\prod _{i=1}^n(1-t^{d_i})$ for suitable positive integers $d_1,...,d_n.$

Then we have seen that A ⇒ B ⇒ C ⇒ D, and C ⇔ C'. Moreover it is easy to see that conversely D ⇒ C, for D shows that the dimension of each ${ℐ}^n$ is equal to the number of monomials $I_1^{a_1}\cdots I_r^{a_r}$ it contains.

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

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

Finally we can improve on (Prop. 3.4). For each $w\in W$ let $$d\left(w\right)=\dim V^w=multiplicity\; of\; 1\; as\; an\; eigenvalue\; of\; w.$$ Then

(Shephard and Todd, Solomon) $$\sum _{w\in W}{t}^{d\left(w\right)}=\prod _{i=1}^r(t+d_i-1).$$

This implies (Prop. 3.4.i) (set $t=1$) and (Prop. 3.4.ii): $d\left(w\right)=r-1$ if and only if $w$ is a reflection, hence $$N=no.\; of\; reflections\; =\sum _{i=1}^r(d_i-1).$$

Exercise: Verify Theorem 3.7 when $W=S_n$ acting on $V={ℝ}^n$ by permuting the coordinates.

Decompose $w\in S_n$ into disjoint cycles. Then each cycle gives a single $w-$fixes vector (e.g. the cycle $(12\cdots r)$ gives $x_1+\cdots +x_r$) and hence $$d\left(w\right)=l\left(\tau \left(w\right)\right),$$ where $\tau \left(w\right)$ is the cycle type of $w.$ Hence $$\begin{array}{ll}\frac{1}{n!}\sum _{w\in S_n}{X}^{d\left(w\right)}=\sum _{\rho ⊢n}{z}_{\rho }^{-1}{X}^{l\left(\rho \right)}.& (Inv\; 4)\end{array}$$ Let ${\phi }_X:{\Lambda }_{\mathbb{Q}}\to \mathbb{Q}\left[X\right]$ be defined by ${\phi }_X\left(p_r\right)=X,$ all $r\ge 1.$ Then the right hand side of (Inv 4) is $${\phi }_X\left(\sum _{\rho ⊢n}{z}_{\rho }^{-1}{p}_{\rho }\right)={\phi }_X\left(h_n\right).$$ Now if $X$ replaced by a positive integer $N,$ then ${\phi }_N\left(p_r\right)=N$ for all $r\ge 1,$ so that ${\phi }_n\left(f\right)=f{\displaystyle \underset{n}{\underbrace{(1,...,1)}}}$ for all $f\in \Lambda .$ Hence $${\phi }_N\left(h_n\right)=h_n(1,...,1)=coefficient\; of\; t^n \; in\; {(1-t)}^{-N}=\frac{N(N+1)\cdots (N+n-1)}{n!}.$$ Since this is true for all positive integers $N,$ it is true identically: $$\frac{1}{n!}\sum _{w\in S_n}{X}^{d\left(w\right)}={\phi }_X\left(h_n\right)=\frac{X(X+1)\cdots (X+n-1)}{n!}$$ giving $$\sum _{w\in S_n}{X}^{d\left(w\right)}=\prod _{i=1}^n(X+i-1)$$ in agreement with Theorem 3.7, since the $d_i$ are $(1,2,...,n).$

In fact there is a fancier version of Theorem 3.7. Let $m$ be a positive integer, $\zeta$ a primitive $m^{\mathrm{th}}$ root of unity, e.g. $\zeta =e^{\frac{2\pi i}{m}}.$ For each $w\in W,$ let $$d_m\left(w\right)=multiplicity\; of\; \zeta \; as\; as\; eigenvalue\; of\; w$$ (this is independent of which primitive $m^{\mathrm{th}}$ root of unity we choose).

$$\sum _{w\in W}{t}^{d_m\left(w\right)}=\prod _{i=1}^r(t^{\chi \left(m\right|d_i)}+d_i-1),$$ where $\chi$ is the truth function, i.e. $$\chi (m\mid d_i)=\{\begin{array}{ll}1,& ifm\mid d_i,\\ 0,& ifm\nmid d_i.\end{array}\phantom{\}}$$

$m=1$ gives Theorem 3.7: $d_1\left(w\right)=d\left(w\right).$ (So the degrees $d_i$ determine the eigenvalue distributions in $W.$)

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

$$\frac{1}{\left|W\right|}\sum _{w\in W}\frac{\det (1+qw)}{\det (1-tw)}=\prod _{i=1}^r\frac{1+q{t}^{d_i-1}}{1-t^{d_i}},$$ where $t,q$ are independent variables.

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

		Proof of Proposition 3.8, assuming Proposition 3.9.
	First replace $q,t$ by $\frac{q}{\zeta },\frac{t}{\zeta }:$ then (Prop. 3.9) becomes $$\begin{array}{ll}\frac{1}{\left|W\right|}\sum _{w\in W}\frac{\det (\zeta +qw)}{\det (\zeta -tw)}=\prod _{i=1}^r\frac{{\zeta }^{d_i}+q{t}^{d_i-1}}{{\zeta }^{d_i}-t^{d_i}}.& (Inv\; 5)\end{array}$$ Now set $q=(1-t)X-1,$ where $X$ is another variable, and then let $t\to 1.$ Consider first the left hand side of (Inv 5). Let $w\in W$ and let ${\lambda }_1,...,{\lambda }_r$ be the eigenvalues of $w$ (acting on $V$ as usual). Then $$\frac{\det (\zeta +qw)}{\det (\zeta -tw)}=\prod _{i=1}^r\frac{\zeta +{\lambda }_{i}q}{\zeta -{\lambda }_{i}t}=\prod _{i=1}^r\frac{\zeta +{\lambda }_i((1-t)X-1)}{\zeta -{\lambda }_{i}t}.$$ Suppose first ${\lambda }_i\ne \zeta .$ On putting $t=1$ we get $\frac{\zeta -{\lambda }_i}{\zeta -{\lambda }_i}=1.$ If on the other hand ${\lambda }_i=\zeta ,$ the corresponding factor in the product is $$\frac{1+(1-t)X-1}{1-t}=X,$$ so when we put $q=(1-t)X-1,$ the limit as $t\to 1$ of $\frac{\det (\zeta +qw)}{\det (\zeta -tw)}$ is just $X^{d_m\left(w\right)},$ and hence the left hand side of (Inv 5) gives $$\frac{1}{\left|W\right|}\sum _{w\in W}{X}^{d_m\left(w\right)}.$$ Now consider the factors on the right hand side: $$\frac{{\zeta }^{d_i}+q{t}^{d_i-1}}{{\zeta }^{d_i}-t^{d_i}}=\frac{{\zeta }^{d_i}+((1-t)X-1)t^{d_i-1}}{{\zeta }^{d_i}-t^{d_i}}.$$ Suppose first that $m\nmid d_i,$ so that ${\zeta }^{d_i}\ne 1.$ We can then safely put $t=1$ and get $\frac{{\zeta }^{d_i}-1}{{\zeta }^{d_i}-1}=1.$ If on the other hand $m\mid d_i$ so that ${\zeta }^{d_i}=1,$ we have to compute the limit as $t\to 1$ of $$\frac{1+((1-t)X-1)t^{d_i-1}}{1-t^{d_i}}=\frac{1+(X-1)t^{d_i-1}-X{t}^{d_i}}{1-t^{d_i}},$$ which we can do by l'Hôpital's rule: differentiate numerator and denominator with respect to $t,$ and then set $t=1.$ This gives $$\frac{(d_i-1)(X-1)-d_{i}X}{-d_i}=\frac{X+d_i-1}{d_i}.$$ So the limit as $t\to 1$ of the right hand side of (Inv 5) is $$\prod _{m\mid d_i}\frac{X+d_i-1}{d_i}=\prod _{i=1}^r\frac{X^{\chi (m\mid d_i)}+d_i-1}{d_i}=\frac{1}{\left|W\right|}\prod _{i=1}^r(X^{\chi (m\mid d_i)}+d_i-1),$$ since we know (Prop. 3.4) that $d_1\cdots d_r=\left|W\right|.$ This completes the proof of Proposition 3.8 (modulo Proposition 3.9 which I will prove later, perhaps). $\square$

Examples.

1. Take $m=2,$ i.e. $\zeta =-1.$ Then $-1\in W$ if and only if the right-hand side of (Prop. 3.8) has degree $r\iff d_1,...,d_r$ are even.
2. If $m$ does not divide any $d_i$ then $d_m\left(w\right)=0$ for all $w\in W.$ So all eigenvalues $\zeta$ of $w\in W$ satisfy ${\zeta }^{d_i}=1$ for some $i.$ (Is this otherwise obvious?)
3. Reference: C.L. Morgan, Can. J. Math. (1979) vol. 31, 252-254.

$W-$skew polynomials

Each $w\in W$ (acting on $V$) is an orthogonal transformation, hence $$\epsilon \left(w\right)=\det \left(w\right)=\pm 1,$$ $\epsilon :W\to \{\pm 1\}$ is the sign character of $W.$ Clearly $$\epsilon \left(w_1{w}_2\right)=\det \left(w_1{w}_2\right)=\det \left(w_1\right)\det \left(w_2\right)=\epsilon \left(w_1\right)\epsilon \left(w_2\right),$$ i.e. $\epsilon$ is a homomorphism of $W$ into the 2-elements group $\{\pm 1\}.$ In particular, $\epsilon \left(s_{\alpha }\right)=-1$ for a reflection $s_{\alpha }.$ Hence if $l\left(w\right)=p$ and $w=t_1\cdots t_p$ is a reduced expression for $w,$ where each $t_i$ is an $s_j,$ we have $\epsilon \left(w\right)=\epsilon \left(t_1\right)\cdots \epsilon \left(t_p\right)={(-1)}^p,$ i.e.,

$$\epsilon \left(w\right)={(-1)}^{l\left(w\right)}.$$

A polynomial $f\in 𝒮$ is $W-$skew if $$wf=\epsilon \left(w\right)f,$$ for all $w\in W.$

Let $$p=\prod _{\alpha \in R^{+}}\alpha ,$$ the product of the positive roots (if $R$ is of type $A_n$ then $p$ is the Vandermonde determinant).

The set of $W-$skew polynomials in $𝒮$ is $pℐ.$

		Proof.
	Suppose $f\in 𝒮$ is skew. Then for each $\alpha \in R^{+}$ we have $s_{\alpha }f=-f,$ so that $f=\frac{1}{2}(f-s_{\alpha }f)$ is divisible by $\alpha$ by (Prop. 1.1). Hence $f$ is divisible by $p$ in $𝒮,$ say $f=pg.$ Moreover $p$ itself is skew, because by (1.9) (Ref to Root Systems page?) $$s_{i}p=\prod _{\alpha \in R^{+}}{s}_i\left(\alpha \right)=-{\alpha }_i\cdot \prod _{\genfrac{}{}{0ex}{}{\alpha \in R^{+}}{\alpha \ne {\alpha }_i}}\alpha =-p$$ and therefore $wp={(-1)}^{l\left(w\right)}p=\epsilon \left(w\right)p$ by (Prop. 4.1). Hence in the factorization $f=pg$ both $f$ and $p$ are skew, hence $g$ is symmetric, i.e. $g\in ℐ,$ and conversely each $pg,$ $g\in ℐ$ is $W-$skew. $\square$

As before let $I_1,...,I_r$ be a set of fundamental polynomial invariants.

$$\det \left(\frac{\partial I_i}{\partial x_j}\right)=\lambda p \; for\; some\; \lambda \in ℝ, \lambda \ne 0.$$

		Proof.
	In general, if $\phi :V\to V$ is a mapping, say $$\phi \left(x\right)=({\phi }_1\left(x\right),...,{\phi }_r\left(x\right)),x=(x_1,...,x_r),$$ where the ${\phi }_i$ have partial derivatives, the Jacobian of $\phi$ is $$J_{\phi }=\det \left(\frac{\partial {\phi }_i}{\partial x_j}\right):V\to ℝ.$$ In particular, if $\phi$ is linear, say ${\phi }_i\left(x\right)=\sum _{j=1}^r{a}_{ij}{x}_j,$ then $\frac{\partial {\phi }_i}{\partial x_j}=a_{ij}$ and so $J_{\phi }=\det \phi$ (constant). If $\theta :V\to V$ is another mapping, then by the chain rule we have $$\begin{array}{ll}{J}_{\theta \circ \phi }=J_{\theta }\left(\phi x\right)J_{\phi }\left(x\right).& (Inv\; 6)\end{array}$$ Define $\theta$ by $$\theta \left(x\right)=(I_1\left(x\right),...,I_r\left(x\right))$$ and let $\phi =w\in W.$ Then $(\theta \circ \phi )\left(x\right)=\theta \left(\phi \left(x\right)\right)=\theta \left(w\left(x\right)\right)=\theta \left(x\right)$ (because the $I_j$ are $W-$invariant), i.e. $\theta \circ \phi =\theta .$ So (Inv 6) gives $J_{\theta }\left(x\right)=J_{\theta }\left(wx\right)\cdot \det \left(w\right),$ i.e. $$J_{\theta }=\epsilon \left(w\right)\cdot w^{-1}{J}_{\theta }$$ for all $w\in W,$ whence $J=J_{\theta }$ is $W-$skew. But also $J$ is a homogeneous polynomial of degree $\sum _{i=1}^r(d_i-1)=N=\mathrm{deg}p$ by (Prop. 3.4.ii), hence by (Prop. 4.2) we have $J=\lambda p$ for some $\lambda \in ℝ,$ and it remains to show that $\lambda \ne 0.$ Consider as before the fields of fractions $K,L$ of $ℐ,𝒮$ respectively. We have $L=K(x_1,...,x_r)$ and each $x_i$ is algebraic over $K.$ Hence for each $i$ there is a polynomial $F_i\left(t\right)\in ℐ\left[t\right]$ of minimal degree such that $F_i\left(x_i\right)=0.$ Differentiate with respect to $x_k:$ $$\sum _{j=1}^r\frac{\partial F_i}{\partial I_j}\left(x_i\right)+{\delta }_{ik}{F}_i\prime \left(x_i\right)=0,(1\le k\le r)$$ where $F_i\prime =\frac{d{F}_i}{dt}.$ Write these equations as a matrix identity, say $AB=C$ where $$A=\left(\frac{\partial F_i}{\partial I_j}\left(x_i\right)\right),B=\left(\frac{\partial I_j}{\partial x_k}\right),C=-\mathrm{diag}(F_i\prime \left(x_i\right)).$$ Taking determinants we get $\left(\det A\right)\left(\det B\right)={(-1)}^r\prod _{i=1}^r{F}_i\prime \left(x_i\right)\ne 0,$ by the minimality of the $F_i.$ Hence $J=\det B\ne 0$ in $𝒮.$ $\square$

The last part of this proof uses only that $I_1,...,I_r$ are algebraically independent over $ℝ.$ Indeed, if $k$ is a field of characteristic 0, and $I_1,...,I_r\in k[x_1,...,x_r],$ then $I_1,...,I_r$ are algebraically independend over $k$ if and only if the Jacobian $\det \left(\frac{\partial I_i}{\partial x_j}\right)$ is not identically zero.

(If $I_1,...,I_r$ are algebraically independent over $k,$ let $f(y_1,...,y_r)$ be a nonzero polynomial of minimal degree such that $f(I_1,...,I_r)=0.$ Since $f$ is not constant, not all the partial derivatives $\frac{\partial f}{\partial y_i}$ vanish, hence (as they have smaller degree than $f$), not all the polynomials $$g_i=\frac{\partial f}{\partial y_i}(I_1,...,I_r)$$ are zero. But by differentiating $f(I_1,...,I_r)=0$ with respect to $x_j$ we get $$\sum _{i=1}^r{g}_i\frac{\partial I_i}{\partial x_j}=0,$$ which shows that $\det \left(\frac{\partial I_i}{\partial x_j}\right)=0.$)

(Shephard and Todd) Let $W$ be a finite subgroup of $GL\left(V\right),$ and suppose that $ℐ$ is generated as an $ℝ-$algebra by $r$ independent homogeneous elements $I_1,...,I_r.$ Then $W$ is generated by reflections.

		Proof.
	Let $W\prime$ be the subgroup of $W$ generated by the reflections in $W,$ and let $ℐ\prime ={𝒮}^{W\prime }$ be its ring of invariants. By Chevalley's theorem we have $ℐ\prime =ℝ[I_1\prime ,...,I_r\prime ]$ with $I_1\prime ,...,I_r\prime$ algebraically independent. We have $ℐ\subseteq ℐ\prime ,$ hence each $I_i$ is (uniquely) a polynomial in the $I_j\prime ,$ and $$\frac{\partial I_i}{\partial x_k}=\sum _{j=1}^r\frac{\partial I_i}{\partial I_j\prime }\frac{\partial I_j\prime }{\partial x_k}$$ so that $$\det \left(\frac{\partial I_i}{\partial x_k}\right)=\det \left(\frac{\partial I_i}{\partial I_j\prime }\right)\cdot \det \left(\frac{\partial I_j\prime }{\partial x_k}\right).$$ By the remark above it follows that $$\frac{\partial I_i}{\partial I_j\prime }\ne 0.$$ Expanding, some product $$\frac{\partial I_{\sigma \left(1\right)}}{\partial I_1\prime }\cdots \frac{\partial I_{\sigma \left(r\right)}}{\partial I_r\prime }\ne 0,(\sigma \in S_r).$$ Renumbering the $I\text{'}s$ we may assume that $$\frac{\partial I_1}{\partial I_1\prime }\cdots \frac{\partial I_r}{\partial I_r\prime }\ne 0.$$ Let $d_i=\mathrm{deg}{I}_i, d_i\prime =\mathrm{deg}{I}_i\prime .$ Then we must have $d_i\ge d_i\prime (1\le i\le r).$ On the other hand, from (Prop. 3.4.ii), if $N$ is the number of reflections in $W$ (or $W\prime$) $$N=\sum (d_i-1)=\sum (d_i\prime -1).$$ So $\sum _1^r(d_i-d_i\prime )=0$ and therefore $d_i=d_i\prime .$ But by (Prop. 3.4.i) $\left|W\right|=\prod d_i$ and $|W\prime |=\prod d_i\prime .$ Hence $|W\prime |=\left|W\right|$ and finally $W\prime =W.$ $\square$

Differential operators

Let $x_1,...,x_r$ be coordinates relative to an (orthonormal) basis of $V,$ so that $𝒮=ℝ[x_1,...,x_r].$ Let $\mathrm{End}𝒮$ denote the $ℝ-$algebra of linear maps $𝒮\to 𝒮.$ Define an $ℝ-$algebra homomorphism $$\partial :𝒮\to \mathrm{End}𝒮by\partial \left(x_i\right)=\frac{\partial }{\partial x_i}.$$ Explicitly, if $f\in 𝒮,$ say $f=\sum _{\alpha }{c}_{\alpha }{x}^{\alpha } (\alpha =({\alpha }_1,...,{\alpha }_r)\in {\mathbb{N}}^r; c_{\alpha }\in ℝ; x^{\alpha }=x_1^{{\alpha }_1}\cdots x_r^{{\alpha }_r})$ then $$\partial \left(f\right)=\sum _{\alpha }{c}_{\alpha }{\left(\frac{\partial }{\partial x_1}\right)}^{{\alpha }_1}\cdots {\left(\frac{\partial }{\partial x_r}\right)}^{{\alpha }_r}.$$ We shall write ${\partial }_f$ in place of $\partial \left(f\right).$ The operators ${\partial }_f:𝒮\to 𝒮$ are linear partial differential operators with constant coefficients.

The action of $W$ is given by

Let $f\in 𝒮, w\in W.$ Then $${\partial }_{wf}=w{\partial }_f{w}^{-1}.$$

		Proof.
	If this is true for $f$ and $g,$ then it is true for $fg:$ $$\begin{array}{rcl}{\partial }_{w\left(fg\right)}& =& {\partial }_{wf\cdot wg}={\partial }_{wf}{\partial }_{wg}=\left(w{\partial }_f{w}^{-1}\right)\left(w{\partial }_g{w}^{-1}\right)\\ & =& w{\partial }_f{\partial }_g{w}^{-1}=w{\partial }_{fg}{w}^{-1}.\end{array}$$ So it is enough to prove Proposition 5.1 when $f$ is linear, say $f=\xi \in V.$ The easiest thing to do is to go back to the definition: if $g\in 𝒮$ then $\left({\partial }_{\xi }g\right)\left(x\right)=\underset{t\to 0}{\lim }\frac{g(x+t\xi )-g\left(x\right)}{t}$ and therefore $$\begin{array}{rcl}\left({\partial }_{w\xi }g\right)\left(x\right)& =& \underset{t\to 0}{\lim }\frac{g(x+t\xi )-g\left(x\right)}{t}\\ & =& \underset{t\to 0}{\lim }\frac{\left(w^{-1}g\right)(w^{-1}x+t\xi )-\left(w^{-1}g\right)\left(w^{-1}x\right)}{t}\\ & =& \left({\partial }_{\xi }\left(w^{-1}g\right)\right)\left(w^{-1}x\right)\\ & =& \left(w{\partial }_{\xi }\left(w^{-1}g\right)\right)\left(x\right)\end{array}$$ giving ${\partial }_{w\xi }g=w{\partial }_{\xi }{w}^{-1}g,$ i.e. ${\partial }_{w\xi }=w{\partial }_{\xi }{w}^{-1}.$ $\square$

Scalar product on $𝒮$

The scalar product on $V$ induces one on $𝒮,$ which may be defined as follows: if $f,g\in 𝒮$ then $$⟨f,g⟩=\left({\partial }_{f}g\right)\left(0\right).$$ If $w\in W$ we have $⟨wf,g⟩=⟨f,w^{-1}g⟩,$ because $$\begin{array}{rcl}⟨wf,g⟩& =& \left({\partial }_{wf}g\right)\left(0\right)\\ & =& \left(w{\partial }_f\left(w^{-1}g\right)\right)\left(0\right)by\; (Prop.\; 5.1)\\ & =& {\partial }_f\left(w^{-1}g\right)\left(0\right)(because{w}^{-1}\left(0\right)=0)\\ & =& ⟨f,w^{-1}g⟩.\end{array}$$ As before let $x_1,...,x_r$ be coordinates relative to an orthonormal basis in $V$ and let $\alpha ,\beta \in {\mathbb{N}}^r$ be multi-indices. Then the definition gives $$⟨x^{\alpha },x^{\beta }⟩=\prod _{i=1}^r{\left(\frac{\partial }{\partial x_i}\right)}^{{\alpha }_i}{x}_i^{{\beta }_i}evaluated\; at\; 0,$$ which is zero if $\alpha \ne \beta$ and is equal to $\alpha !=\prod {\alpha }_i!$ if $\alpha =\beta :$ $$⟨x^{\alpha },x^{\beta }⟩=\alpha !{\delta }_{\alpha \beta }.$$ It follows that the scalar product is symmetric (not immediately obvious from the definition) and positive definite, and that the monomials $x^{\alpha }$ form an orthogonal (but not orthonormal) basis of $𝒮.$ The homogeneous components ${𝒮}^n$ of $𝒮$ are pairwise orthogonal, and on ${𝒮}^1=V$ we have our original scalar product.

Example. Let $u_i,v_i\in V (1\le i\le n),$ then $f=u_1\cdots u_n, g=v_1\cdots v_n\in {𝒮}^n$ and $$⟨f,g⟩={\partial }_{u_1}\cdots {\partial }_{u_n}(v_1\cdots v_n).$$ Since ${\partial }_{u_i}\left(v_j\right)=⟨u_i,v_j⟩$ it follows that $$⟨f,g⟩=\sum _{w\in S_n}\prod _{i=1}^r⟨u_i,v_{w\left(i\right)}⟩=\mathrm{per}\left(⟨u_i,v_j⟩\right),$$ the permanent of the $n\times n$ matrix $\left(⟨u_i,v_j⟩\right).$ (This is, as far as I am aware, the only place the permanent occurs in nature.)

Let $f\in 𝒮$ and let ${\mu }_f:𝒮\to 𝒮$ be multiplication by $f: {\mu }_{f}g=fg.$

${\partial }_f$ is the adjoint of ${\mu }_f.$

		Proof.
	Let $gh\in 𝒮.$ Then $$⟨g,{\partial }_{f}h⟩={\partial }_g\left({\partial }_{f}h\right)\left(0\right)={\partial }_{fg}\left(h\right)\left(0\right)=⟨fg,h⟩=⟨{\mu }_{f}g,h⟩.$$ $\square$

Harmonic polynomials

A polynomial $h\in 𝒮$ is said to be $W-$harmonic if ${\partial }_{f}h=0$ for all $f\in {ℐ}^{+}.$ Since $ℐ$ is generated by $I_1,...,I_r$ (Thm. 2.2) it is enough to require that ${\partial }_{I_i}h=0 (1\le i\le r).$

1. If $x_1,...,x_r$ are coordinates in $V$ relative to an orthonormal basis, then ${\left|x\right|}^2=x_1^2+\cdots +x_r^2\in ℐ,$ hence certainly a $W-$harmonic polynomial $h$ must satisfy $$\sum _1^r\frac{{\partial }^{2}h}{\partial x_i^2}=0,$$ i.e. it is a solution to Laplace's equation, hence is harmonic in the usual sense of the word.
2. 1 is harmonic; so is $p,$ because ${\partial }_{I_i}p$ is skew of degree $<N,$ hence zero.

Let $\mathscr{H}\subseteq 𝒮$ be the vector space of $W-$harmonic polynomials. It is a graded vector space: $$\mathscr{H}=\underset{n\ge 0}{⨁}{\mathscr{H}}^n,{\mathscr{H}}^n=\mathscr{H}\cap {𝒮}^n.$$ Moreover each ${\mathscr{H}}^n$ is $W-$stable, because if $h\in {\mathscr{H}}^n$ and $w\in W, f\in {ℐ}^{+}$ we have $${\partial }_f\left(wh\right)={\partial }_{wf}\left(wh\right)=w\cdot {\partial }_{f}h=0$$ using (Prop. 5.1). So $\mathscr{H}$ is a graded $W-$module.

Recall that $𝒥=𝒮{ℐ}^{+}$ is the ideal in $𝒮$ generated by $I_1,...,I_r.$

$\mathscr{H}$ is the orthogonal complement of $𝒥$ in $𝒮,$ i.e. $\mathscr{H}={𝒥}^{\perp }.$

		Proof.
	We have only to apply the definitions: $$\begin{array}{rcl}h\in \mathscr{H}& \iff & {\partial }_{f}h=0allf\in {ℐ}^{+}\\ & \iff & ⟨g,{\partial }_{f}h⟩=0allg\in 𝒮,f\in {ℐ}^{+}\\ & \iff & ⟨fg,h⟩=0allg\in 𝒮,f\in {ℐ}^{+}by\; (Prop.\; 6.1)\\ & \iff & h\perp 𝒥.\end{array}$$ $\square$

So $𝒮=\mathscr{H}\oplus 𝒥.$ Next we have

The mapping $\phi :\mathscr{H}\otimes ℐ\to 𝒮$ defined by $\phi (h\otimes f)=hf$ is an isomorphism of $W-$modules (i.e. $\phi$ commutes with the actions of $W$).

		Proof.
	We have to show that $\phi$ is (a) surjective, (b) injective. We show by induction on $n$ that ${𝒮}^n\subseteq \mathscr{H}ℐ.$ Clear for $n=0$ (since $1\in \mathscr{H}).$ From (Prop. 7.2) we have $${𝒮}^n={\mathscr{H}}^n+{𝒥}^n={\mathscr{H}}^n+\sum _{p=1}^n{𝒮}^{n-p}{ℐ}^p.$$ By the inductive hypothesis, ${𝒮}^{n-p}\subseteq \mathscr{H}ℐ,$ hence ${𝒮}^n\subseteq \mathscr{H}ℐ.$ Hence $𝒮\subseteq \mathscr{H}ℐ\subseteq 𝒮,$ i.e. $\mathscr{H}ℐ=𝒮.$ For each $\alpha \in {𝒩}^r,$ let $I^{\alpha }=I_1^{{\alpha }_1}\cdots I_r^{{\alpha }_r}.$ The $I^{\alpha }$ form an $ℝ-$basis of $ℐ.$ Let ${\left(h_n\right)}_{n\ge 0}$ be a basis of $\mathscr{H}$ consisting of homogeneous elements. Then the $h_n\otimes I^{\alpha }$ form a basis of $\mathscr{H}\otimes ℐ$ and we have to show that their images under $\phi ,$ namely $h_n{I}^{\alpha },$ are linearly independent over $ℝ.$ So we suppose we have a nontrivial relations $$\begin{array}{ll}\sum _{n,\alpha }{c}_{n\alpha }{h}_n{I}^{\alpha }=0,c_{n\alpha }\in ℝ,c_{n\alpha }\ne 0.& (Inv\; 7)\end{array}$$ By homogeneity we may assume that $\mathrm{deg}{h}_n+\mathrm{deg}{I}^{\alpha }$ is constant for each term in the sum. Consider the monomials $I^{\alpha }$ occuring in (Inv 7), and choose one of least degree (N.B. $\mathrm{deg}{I}^{\alpha }=\sum {\alpha }_i\mathrm{deg}{I}_i=\sum {\alpha }_i{d}_i$ ). This is not in the ideal of $ℐ$ generated by the other $I^{\beta }$ occurring in (Inv 7), otherwise the $I^{\alpha }$ would be linearly dependent over $ℝ,$ contradicting (Thm. 2.2). Hence by the Lemma 2.3 we have $$h=\sum _n{c}_{n\alpha }{h}_n\in 𝒥.$$ But $h\in \mathscr{H},$ hence by (Prop. 7.2) $h=0,$ i.e. $c_{n\alpha }=0$ for each $n:$ contradiction. $\square$

1. $P(\mathscr{H},t)=\prod _{i=1}^r\frac{1-t^{d_i}}{1-t}.$
2. $\dim \mathscr{H}=\dim (𝒮/𝒥)=\left|W\right|.$

		Proof.
	From (Prop. 3.1) and (Prop. 7.4) we have $P(\mathscr{H},t)P(ℐ,t)=P(𝒮,t)$ hence $$P(\mathscr{H},t)=P(𝒮,t)/P(ℐ,t)=\prod _{i=1}^r\frac{1-t^{d_i}}{1-t}by\; (Prop.\; 3.2).$$ From (Prop. 7.2) we have $𝒮\cong \mathscr{H}\oplus 𝒥,$ hence $\mathscr{H}\cong 𝒮/𝒥$ as vector spaces, so that $$\dim (𝒮/𝒥)=\dim \mathscr{H}=\sum \dim {\mathscr{H}}^n=P(\mathscr{H},1)=\prod _{i=1}^r{d}_i=\left|W\right|,by\; (Prop.\; 3.4).$$ $\square$

(It follows that ${\mathscr{H}}^n=0$ if $n>\sum (d_i-1)=N$ and that $\dim {\mathscr{H}}^N=1.$ Since $p\in \mathscr{H}$ we have ${\mathscr{H}}^N=ℝp.$)

Suppose $M=\underset{n\ge 0}{⨁}{M}^n$ is a graded vector space over $ℝ$ such that $W$ acts on each $M^n$ (and each $M^n$ is finite dimensional). ($W$ here could be any finite group.) Then we can define the "graded character" of $M:$ if $w\in W$ and $t$ is an indeterminate, we define $${\chi }_M(w;t)=\sum _{n\ge 0}\mathrm{trace}(w,M^n)t^n\in ℝ\left[\left[t\right]\right].$$ If $N=\underset{n\ge 0}{⨁}{N}^n$ is another such, then $W$ acts on each $M^p\otimes N^q$ (by the rule $w(x\otimes y)=wx\otimes wy, x\in M^p, y\in N^q$) and we have $$\mathrm{trace}(w,M^p\otimes N^q)=\mathrm{trace}(w,M^p)\cdot \mathrm{trace}(w,N^q).$$ Hence, as in (Prop. 3.1) we have

$${\chi }_{M\otimes N}(w;t)={\chi }_M(w;t){\chi }_N(w;t)(w\in W).$$

We shall use this to prove

$\mathscr{H}$ (or equivalently $𝒮/𝒥$) affords the regular representation of $W.$

		Proof.
	Since a representation is determined by its character, it is enough to prove that $$\mathrm{trace}(w;\mathscr{H})=\{\begin{array}{ll}\left|W\right|,& ifw=1,\\ 0,& ifw\ne 1.\end{array}\phantom{\}}$$ By (Prop. 7.5) and (Prop. 7.3) we have $${\chi }_{\mathscr{H}}(w;t)=\frac{{\chi }_{𝒮}(w;t)}{{\chi }_{ℐ}(w;t)}$$ and ${\chi }_{𝒮}(w;t)=\frac{1}{\det (1-tw)}$ as in the proof of Molien's formula (Prop. 3.3). Moreover $w$ acts as the identity on $ℐ,$ hence $\mathrm{tr}(w,{ℐ}^n)=\dim {ℐ}^n$ for each $n$ and therefore $${\chi }_{ℐ}(w;t)=\sum _{n\ge 0}\left(\dim {ℐ}^n\right)t^n=P(ℐ,t)=\prod \frac{1}{1-t^{d_i}},by\; (Prop.\; 3.2).$$ Hence $$\begin{array}{ll}{\chi }_{\mathscr{H}}(w;t)=\frac{\prod (1-t^{d_i})}{\det (1-tw)},& (Inv\; 8)\end{array}$$ and $\mathrm{trace}(w,\mathscr{H})=\sum _{n\ge 0}\mathrm{trace}(w,{\mathscr{H}}^n)={\chi }_{\mathscr{H}}(w;1).$ If $w=1$ we get $\left|W\right|,$ and if $w\ne 1$ we get 0, because then ${\chi }_{\mathscr{H}}(w;t)$ vanishes at $t=1$ (the numerator of (Inv 8) is divisible by ${(1-t)}^r$ and the denominator by at most ${(1-t)}^{r-1}.$ $\square$

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

From (Prop. 7.6) it follows that the regular representation of $W$ carries a natural grading: $\mathscr{H}=\underset{n\ge 0}{⨁}{\mathscr{H}}^n$ (or on $𝒮/𝒥$). For each irreducible representation $\rho$ of $W,$ let $\mathrm{mult}(\rho ,{\mathscr{H}}^n)$ denote the number of times $\rho$ occurs in ${\mathscr{H}}^n.$ Then we can form the generating function $$F_{\rho }(W;t)=\sum _{n\ge 0}\mathrm{mult}(\rho ,{\mathscr{H}}^n)t^n\in 𝒩\left[\left[t\right]\right].$$ This polynomial is given by

$$F_{\rho }(W;t)=\frac{1}{\left|W\right|}\prod _{i=1}^r(1-t^{d_i})\cdot \sum _{w\in W}\frac{{\chi }_{\rho }\left(w\right)}{\det (1-tw)},$$ where ${\chi }_{\rho }$ is the character of the representation $\rho .$

		Proof.
	The multiplicity of $\rho$ in ${\mathscr{H}}^n$ is the scalar product (on $W$) of ${\chi }_{\rho }$ with the character of ${\mathscr{H}}^n,$ i.e., $$\mathrm{mult}(\rho ,{\mathscr{H}}^n)=\frac{1}{\left|W\right|}\sum _{w\in W}{\chi }_{\rho }\left(w\right)\overline{\mathrm{trace}(w,{\mathscr{H}}^n)}.$$ Hence $$\begin{array}{rcl}{F}_{\rho }(W;t)& =& \frac{1}{\left|W\right|}\sum _{w\in W}{\chi }_{\rho }\left(w\right)\sum _{n\ge 0}\mathrm{trace}(w,{\mathscr{H}}^n)t^n\\ & =& \frac{1}{\left|W\right|}\sum _{w\in W}{\chi }_{\rho }\left(w\right){\chi }_{\mathscr{H}}(w;t)\\ & =& \frac{1}{\left|W\right|}\prod _{i=1}^r(1-t^{d_i})\cdot \sum _{w\in W}\frac{{\chi }_{\rho }\left(w\right)}{\det (1-tw)}.\end{array}$$ $\square$

Setting $t=1$ in Proposition 7.8 gives $F_{\rho }(W;1)={\chi }_{\rho }\left(1\right)=\mathrm{deg}\left(\rho \right)$ in accordance with the fact that the regular representation of a finite group contains each irreducible representation $\rho$ as many times as its degree.

Examples.

1. $W=S_n$ acting on $V={ℝ}^n$ by permuting the coordinates. Here $$P(ℐ,t)=\prod _{i=1}^n{(1-t^i)}^{-1}.$$ The irreducible characters of $S_n$ are ${\chi }^{\lambda }, \lambda ⊢n;$ if $w\in S_n$ has cycle type $\rho ,$ then $\det (1-tw)=\prod (1-t^{{\rho }_i})$ so that $$\begin{array}{rcl}\frac{1}{\det (1-tw)}& =& \prod _{i=1}^{l\left(\rho \right)}\frac{1}{1-t^{{\rho }_i}}\\ & =& \prod _{i=1}^{l\left(\rho \right)}(1+t^{{\rho }_i}+t^{2{\rho }_i}+\cdots )\\ & =& p_{\rho }(1,t,t^2,...),\end{array}$$ and therefore by (Prop. 7.8) $$\begin{array}{rcl}{F}_{\lambda }(S_n;t)& =& \prod _{i=1}^n(1-t^i){\sum }_{\rho ⊢n}{z}_{\rho }^{-1}{\chi }_{\rho }^{\lambda }{p}_{\rho }(1,t,t^2,...)\\ & =& \left(\prod _{i=1}^n(1-t^i)\right)\cdot s_{\lambda }(1,t,t^2,...)\\ & =& \frac{t^{n\left(\lambda \right)}\prod _{i=1}^n(1-t^i)}{\prod _{x\in \lambda }(1-t^{h\left(x\right)})}\end{array}$$ Check: $$F_{\lambda }(S_n;1)=\frac{n!}{\prod _{x\in \lambda }h\left(x\right)}=degree\; of\; {\chi }^{\lambda },$$ as it should.
2. If $\rho$ is an irreducible representation, so is $\epsilon \rho ,$ with character $\epsilon {\chi }_{\rho }.$ We have $$F_{\epsilon \rho }(W;t)=t^N{F}_{\rho }(W;t^{-1})$$ from (Prop. 7.8) (use $\sum d_i=N+r,$ (Prop. 3.4)). In particular, $F_1(W;t)=1$ and $F_{\epsilon }(W;t)=t^N.$

Recall: $$\begin{array}{rcl}K& =& ℝ(I_1,...,I_r)=field\; of\; fractions\; ofℐ,\\ L& =& ℝ(x_1,...,x_r)=field\; of\; fractions\; of𝒮.\end{array}$$ $W$ acts on $𝒮,$ hence on $L:$ $w\left(\frac{f}{g}\right)=\frac{wf}{wg}, (f,g\in 𝒮).$

1. Every element of $L$ is of the form $\frac{f}{g}$ where $f\in 𝒮, g\in ℐ.$
2. $K=L^W.$

		Proof.
	Let $u\in L,$ say $u=\frac{f}{g} (f,g\in 𝒮).$ Multiply numerator and denominator by $\prod _{w\ne 1}wg,$ then $u=\frac{f^{*}}{g^{*}}$ say with $g^{*}=\prod _{w\in W}wg\in ℐ.$ Let $u\in L; u=\frac{f}{g} (f\in 𝒮,g\in ℐ).$ Then $u$ is fixed by $W$ if and only if $f$ is, i.e. if and only if $f\in ℐ.$ So $u\in L^W\iff u\in K.$ $\square$

1. $𝒮$ is a free $ℐ-$module of rank $\left|W\right|.$
2. And $ℝ-$basis of $\mathscr{H}$ is an $ℐ-$basis of $𝒮.$
3. Any $ℝ-$basis of $𝒮$ is a $K-$basis of $L.$

		Proof.
	Let $\left(h_i\right)$ be an $ℝ-$basis of $\mathscr{H}$ $(1\le i\le \left|W\right|),$ so that $\mathscr{H}=\underset{i}{⨁}ℝh_i$ and hence $\mathscr{H}\otimes ℐ=\underset{i}{⨁}{h}_i\otimes ℐ.$ It now follows from (Prop. 7.3) that $𝒮=\underset{i}{⨁}{h}_iℐ.$ This proves ii. and hence i. As to iii., let $\left(f_i\right)$ be any $ℐ-$basis of $𝒮,$ so that every $f\in 𝒮$ has a unique expression $$f=\sum f_i{g}_i,$$ with $g_i\in ℐ.$ Hence if $g\in ℐ$ $(g\ne 0)$ we have $$\frac{f}{g}=\sum f_i\cdot \frac{g_i}{g}$$ uniquely. By (Prop. 7.10(i)) this shows that $\left(f_i\right)$ is a basis of $L$ as a $K-$vector space. $\square$

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

$$𝒮/𝒥=H^{*}(G/B);G/B \; flag\; manifold.$$

Divided differences

Let $\alpha \in R, f\in 𝒮.$ Recall (Prop. 1.1) that $f-s_{\alpha }f$ is divisible by $\alpha$ in $𝒮.$ Hence we may define an operator ${\Delta }_{\alpha }:𝒮\to 𝒮$ by $${\Delta }_{\alpha }={\alpha }^{-1}(1-s_{\alpha })$$ (i.e., ${\Delta }_{\alpha }f={\alpha }^{-1}(f-s_{\alpha }f)$). ${\Delta }_{\alpha }$ has degree -1, i.e. it maps ${𝒮}^n$ into ${𝒮}^{n-1}$ $(n\ge 0).$ Clearly it kills $ℐ.$

1. ${\Delta }_{\alpha }^2=0.$
2. ${\Delta }_{\alpha }\left(fg\right)=\left({\Delta }_{\alpha }f\right)g+\left(s_{\alpha }f\right)\left({\Delta }_{\alpha }g\right)(f,g\in 𝒮).$
3. ${\Delta }_{\alpha }\left(fg\right)=\left({\Delta }_{\alpha }f\right)g \; if\; f\in 𝒮,g\in ℐ\left(i.e.{\Delta }_{\alpha }isℐ-linear\right).$
4. ${\Delta }_{\alpha }f=⟨f,{\alpha }^{\vee }⟩ \; if\; f \; is\; linear\; (i.e.\; f\in {𝒮}^1=V).$

		Proof.
	We have $s_{\alpha }{\Delta }_{\alpha }={\left(s_{\alpha }\alpha \right)}^{-1}(s_{\alpha }-s_{\alpha }^2)={(-\alpha )}^{-1}(s_{\alpha }-1)={\Delta }_{\alpha },$ hence ${\Delta }_{\alpha }^2={\alpha }^{-1}({\Delta }_{\alpha }-s_{\alpha }{\Delta }_{\alpha })=0.$ $\begin{array}{rcl}{\Delta }_{\alpha }\left(fg\right)& =& {\alpha }^{-1}(fg-\left(s_{\alpha }f\right)\left(s_{\alpha }g\right))\\ & =& {\alpha }^{-1}\{(f-s_{\alpha }f)g+s_{\alpha }f(g-s_{\alpha }g)\}\\ & =& \left({\Delta }_{\alpha }f\right)g+\left(s_{\alpha }f\right){\Delta }_{\alpha }g.\end{array}$ If $g\in ℐ$ then $s_{\alpha }g=g,$ hence ${\Delta }_{\alpha }g=0.$ If $f\in V$ then $s_{\alpha }f=f-⟨f,{\alpha }^{\vee }⟩\alpha ,$ so ${\Delta }_{\alpha }f=⟨f,{\alpha }^{\vee }⟩.$ $\square$

We shall use these operators only when $\alpha$ is a simple root of ${\alpha }_i,$ and we shall write ${\Delta }_i={\Delta }_{{\alpha }_i} (1\le i\le r).$ For any sequence $\underset{\_}{a}=(a_1,...,a_p)$ with each $a_i\in [1,r],$ let $${\Delta }_{\underset{\_}{a}}={\Delta }_{a_1}\cdots {\Delta }_{a_p}.$$ Say $\underset{\_}{a}=(a_1,...,a_p)$ is reduced if $w=s_{a_1}\cdots s_{a_p}$ is a reduced expression, i.e. if $l\left(w\right)=p.$

1. If $\underset{\_}{a}$ is reduced, then ${\Delta }_{\underset{\_}{a}}$ depends only on $w=s_{a_1}\cdots s_{a_p}$ (write ${\Delta }_w={\Delta }_{a_1}\cdots {\Delta }_{a_p}$).
2. If $\underset{\_}{a}$ is not reduced, then ${\Delta }_{\underset{\_}{a}}=0.$

It is tempting to attempt to prove (i) by observing that it is only necessary to verify that the ${\Delta }_i$ satisfy the Coxeter relations $$\begin{array}{ll}{\Delta }_i{\Delta }_j{\Delta }_i\cdots ={\Delta }_j{\Delta }_i{\Delta }_j\cdots & (Inv\; 9)\end{array}$$ ($m_{ij}$ terms on either side, where $m_{ij}=$ order of $s_i{s}_j$ in $W$). However it seems not to be easy to verify (Inv 9) by brute force, and we therefore adopt another approach (due to Bernstein, Gelfand and Gelfand).

		Proof.
	We shall prove (i) and (ii) by induction on the length of the sequence $\underset{\_}{a}.$ $p=0$ is OK, so assume $p\ge 1.$ Let $\underset{\_}{a}$ be reduced, $w=s_{a_1}\cdots s_{a_p}.$ We shall compute ${\Delta }_{\underset{\_}{a}}\left(fg\right)$ when $f,g\in 𝒮$ and $f$ is linear, using Proposition 8.1.ii: $$\begin{array}{ll}\begin{array}{rcl}{\Delta }_{\underset{\_}{a}}\left(fg\right)& =& {\Delta }_{a_1}\cdots {\Delta }_{a_p}\left(fg\right)\\ & =& s_{a_1}\cdots s_{a_p}\left(f\right)\cdot {\Delta }_{a_1}\cdots {\Delta }_{a_p}\left(fg\right)\\ & & +\sum _{i=1}^p(s_{a_1}\cdots s_{a_{i-1}}{\Delta }_{a_i}(s_{a_{i+1}}\cdots s_{a_p}f))\cdot {\Delta }_{a_1}\cdots {\hat{\Delta }}_{a_i}\cdots {\Delta }_{a_p}g.\end{array}& (Inv\; 10)\end{array}$$ Now by Proposition 8.1.iv we have $${\Delta }_{a_i}{s}_{a_{i+1}}\cdots s_{a_p}f=⟨s_{a_{i+1}}\cdots s_{a_p}f,{\alpha }_{a_i}^{\vee }⟩=⟨f,{\beta }_i^{\vee }⟩,$$ where ${\beta }_i=s_{a_p}\cdots s_{a_{i+1}}\left({\alpha }_{a_i}\right).$ By (Root Systems, 1.16) the roots ${\beta }_1,...,{\beta }_p$ are precisely the positive roots $\beta$ such that $w\beta$ is negative. Also, by the inductive hypothesis, ${\Delta }_{a_1}\cdots {\hat{\Delta }}_{a_i}\cdots {\Delta }_{a_p}=0$ unless $s_{a_1}\cdots {\hat{s}}_{a_i}\cdots s_{a_p}$ is reduced, i.e. of length $p-1;$ moveover $$s_{a_1}\cdots {\hat{s}}_{a_i}\cdots s_{a_p}=w{s}_{a_p}\cdots s_{a_i}\cdots s_{a_p}=w{s}_{{\beta }_i}.$$ So we may rewrite (Inv 10) as follows: $$\begin{array}{ll}{\Delta }_{\underset{\_}{a}}\left(fg\right)=\left(wf\right){\Delta }_{\underset{\_}{a}}\left(g\right)=\sum _{\beta }⟨f,{\beta }^{\vee }⟩{\Delta }_{w{s}_{\beta }}\left(g\right)& (Inv\; 11)\end{array}$$ summed over $\beta \in R^{+}$ such that (1) $w\beta <0$ and (2) $l\left(w{s}_{\beta }\right)=l\left(w\right)-1.$ Thus the right hand side of (Inv 11) depends only on $w$ (and $f,g$), not on the reduced word $\underset{\_}{a}$ for $w.$ Now let $\underset{\_}{b}=(b_1,...,b_p)$ be another reduced word for $w,$ and let $E={\Delta }_{\underset{\_}{a}}-{\Delta }_{\underset{\_}{b}}.$ This is an operator of degree $-p,$ so that in particular $E\left(1\right)=0.$ From (Inv 11) it follows that $$\begin{array}{ll}E\left(fg\right)=\left(wf\right)\circ E\left(g\right)(g\in 𝒮)& (Inv\; 12)\end{array}$$ if $f\in 𝒮$ is linear. But then by iteration we see that if $f_1,...,f_n$ are linear then $$E(f_1,...,f_{n}g)=\left(w{f}_1\right)E(f_2\cdots f_{n}g)=\cdots =w(f_1\cdots f_n)E\left(g\right)$$ and hence (Inv 12) is true for all $f\in 𝒮.$ Taking now $g=1$ we have $$E\left(f\right)=\left(wf\right)E\left(1\right)=0$$ i.e., $E=0$ and therefore ${\Delta }_{\underset{\_}{a}}={\Delta }_{\underset{\_}{b}}.$ This completes the induction step for Proposition 8.2.i. Suppose now $\underset{\_}{a}=(a_1,...,a_p)$ is not reduced, and let $\underset{\_}{a}\prime =(a_1,...,a_{p-1}).$ If $\underset{\_}{a}\prime$ is not reduced then ${\Delta }_{\underset{\_}{a}\prime }=0$ by the inductive hypothesis, hence ${\Delta }_{\underset{\_}{a}}={\Delta }_{\underset{\_}{a}\prime }{\Delta }_{a_p}=0.$ If $\underset{\_}{a}\prime$ is reduced, let $w=s_{a_1}\cdots s_{a_p}, v=s_{a_1}\cdots s_{a_{p-1}}$ so that $l\left(v\right)=p-1$ and $v=w{s}_{a_p};$ since $l\left(w\right)\le p-1$ it follows that $l\left(w\right)=p-2,$ and hence ${\Delta }_v={\Delta }_w{\Delta }_{a_p}.$ Consequently $${\Delta }_{\underset{\_}{a}}={\Delta }_v{\Delta }_{a_p}={\Delta }_w{\Delta }_{a_p}^2=0$$ by Proposition 8.1.i. $\square$

From Proposition 8.2 it follows that

Let $v,w\in W.$ Then $${\Delta }_v{\Delta }_w=\{\begin{array}{ll}{\Delta }_{vw},& ifl\left(vw\right)=l\left(v\right)+l\left(w\right),\\ 0,& otherwise.\end{array}\phantom{\}}$$

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

Let $w\in W.$ Then ${\Delta }_w$ is an operator of the form $${\Delta }_w=\sum _v{a}_{vw}v,$$ where $a_{vw}\in L$ (the field of fractions of $𝒮$) and in particular $$a_{ww}=\epsilon \left(w\right)\prod _{\alpha \in R\left(w\right)}{\alpha }^{-1},$$ where $R\left(w\right)=R^{+}\cap w{R}^{-}.$ [In fact $a_{vw}=0$ unless $v\le w$ (Bruhat order) ... ]

		Proof.
	Let $w=s_{a_1}\cdots s_{a_p} (p=l\left(w\right)).$ Then $${\Delta }_w={\Delta }_{a_1}\cdots {\Delta }_{a_p}={\alpha }_{a_1}^{-1}(1-s_{a_1}){\alpha }_{a_2}^{-1}(1-s_{a_2})\cdots {\alpha }_{a_p}^{-1}(1-s_{a_p})$$ in which the coefficient of $w=s_{a_1}\cdots s_{a_p}$ is $${(-1)}^p{({\alpha }_{a_1}\cdot s_{a_1}\left({\alpha }_{a_2}\right)\cdot s_{a_1}{s}_{a_2}\left({\alpha }_{a_3}\right)\cdots s_{a_1\cdots a_{p-1}}\left({\alpha }_{a_p}\right))}^{-1}=\epsilon \left(w\right)\prod _{\alpha \in R\left(w\right)}{\alpha }^{-1}$$ by (Root Systems, 1.16). $\square$

HW: Show that $p{a}_{vw}$ is a polynomial, i.e. $p{a}_{vw}\in 𝒮$ for all $v,w.$

Let $w_0$ be the longest element of $W.$ Then $${\Delta }_{w_0}=p^{-1}\sum _{w\in W}\epsilon \left(w\right)w.$$

		Proof.
	Say ${\Delta }_{w_0}=\sum _{w\in W}{a}_{w}w,$ with coefficients $a_w (=a_{w,w_0}) \in L.$ Since $l\left(s_i{w}_0\right)<l\left(w_0\right)$ for $1\le i\le r,$ it follows from Proposition 8.4 that ${\Delta }_i{\Delta }_{w_0}=0,$ i.e. ${\Delta }_{w_0}{s}_i={\Delta }_{w_0}.$ Since the $s_i$ generate $W,$ we conclude that $${\Delta }_{w_0}=v{\Delta }_{w_0}$$ for all $v\in W,$ and hence $${\Delta }_{w_0}=\sum _{w\in W}v\left(a_w\right)vw.$$ Comparing this with the previous expression, we see that $v\left(a_w\right)=a_{vw}$ for $v,w\in W^{*}.$ Hence it is enough to determine one of the coefficients $a_w;$ now by Proposition 8.5 we have $a_{w_0}=\epsilon \left(w_0\right)p^{-1}$ (since $R\left(w_0\right)=R^{+}$). Hence $$a_w=w{w}_0\left(a_{w_0}\right)=w{\left(p\right)}^{-1}=\epsilon \left(w\right)p^{-1}.$$ $\square$

The $w\in W$ are linearly independent over $L.$ (Dedekind's lemma)

Dedekind's lemma

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

		Proof.
	Suppose we have a linear dependence relation $$\sum _1^n{f}_i{\alpha }_i=0$$ with $f_i\in L, {\alpha }_i\in \mathrm{Aut}L.$ May assume $n$ minimal (hence the $f_i\ne 0$). Let $g,h\in L.$ Then we have $$\begin{array}{ll}\sum _1^n{f}_i{\alpha }_i\left(g\right)=0,& (Inv\; 13)\end{array}$$ $$\begin{array}{ll}\sum _1^n{f}_i{\alpha }_i\left(g\right){\alpha }_i\left(h\right)=0.& (Inv\; 14)\end{array}$$ Multiply (Inv 13) by ${\alpha }_1\left(h\right)$ and subtract from (Inv 14). $$\sum _2^n{f}_i({\alpha }_i\left(h\right)-{\alpha }_1\left(h\right)){\alpha }_i\left(g\right)=0,for\; all\; g.$$ Since ${\alpha }_2\ne {\alpha }_1$ there exists $h$ such that ${\alpha }_2\left(h\right)\ne {\alpha }_1\left(h\right):$ but then $$\sum _2^n{f}_i({\alpha }_i\left(h\right)-{\alpha }_1\left(h\right)){\alpha }_i=0$$ is a nontrivial linear dependence relation of length $<n.$ $\square$

Now let ${\Delta }_i^{*}$ be the adjoint of ${\Delta }_i:$ $$⟨{\Delta }_i^{*},g⟩=⟨f,{\Delta }_{i}g⟩.$$ Since ${\Delta }_i$ has degree -1, ${\Delta }_i^{*}$ has degree +1. To get an idea of what ${\Delta }_i^{*}$ looks like, we prove (although we shan't use it)

Let $f\in 𝒮.$ Then there exists $g\in 𝒮$ (not unique) such that ${\partial }_{{\alpha }_i}g=f$ (indefinite integral: $g=\int fd{\alpha }_i$). For any such $g$ we have $${\Delta }_i^{*}f=g-s_{i}g.$$

		Proof.
	Let $h\in 𝒮.$ Then $$\begin{array}{rcl}⟨g-s_{i}g,h⟩& =& ⟨g,h-s_{i}h⟩\\ & =& ⟨g,{\alpha }_i{\Delta }_{i}h⟩\\ & =& ⟨{\partial }_{{\alpha }_i}g,{\Delta }_{i}h⟩\\ & =& ⟨f,{\Delta }_{i}h⟩\\ & =& ⟨{\Delta }_i^{*}f,h⟩.\end{array}$$ Since $h$ was arbitrary it follows that $g-s_{i}g={\Delta }_i^{*}f.$ $\square$

Now let $w\in W,$ $w=s_{a_1}\cdots s_{a_p}$ a reduced expression and define $${\Delta }_w^{*}={\Delta }_{a_p}^{*}\cdots {\Delta }_{a_1}^{*}of\; degree\; p=l\left(w\right).$$ ${\Delta }_w^{*}$ is the adjoint of ${\Delta }_w$ and we have, from Proposition 8.4:

$${\Delta }_v^{*}{\Delta }_w^{*}=\{\begin{array}{ll}{\Delta }_{wv}^{*},& ifl\left(v\right)+l\left(w\right)=l\left(vw\right),\\ 0,& otherwise.\end{array}\phantom{\}}$$

(Take adjoints: ${\Delta }_v^{*}{\Delta }_w^{*}$ is the adjoint of ${\Delta }_w{\Delta }_v$ (in that order).)

1. ${\Delta }_w$ maps $𝒥$ into $𝒥.$
2. ${\Delta }_w^{*}$ maps $\mathscr{H}$ into $\mathscr{H}.$

		Proof.
	Let $f\in 𝒮, g\in {ℐ}^{+}.$ Then by Proposition 8.1.iii ${\Delta }_i\left(fg\right)=\left({\Delta }_{i}f\right)g$ so that ${\Delta }_w\left(fg\right)=\left({\Delta }_{w}f\right)g,$ which proves (i). Recall Proposition 7.2 that $\mathscr{H}$ is the orthogonal complement of $𝒥$ in $𝒮.$ Hence if $h\in \mathscr{H}, f\in 𝒥$ we have $$⟨{\Delta }_w^{*}h,f⟩=⟨h,{\Delta }_{w}f⟩=0,$$ because ${\Delta }_{w}f\in 𝒥.$ Hence ${\Delta }_w^{*}h\in {𝒥}^{\perp }=\mathscr{H}.$ $\square$

Now let $(w\in W)$ $$h_w={\Delta }_w^{*}\left(1\right).$$ By Proposition 9.4, $h_w\in \mathscr{H}$ and is homogeneous of degree $l\left(w\right).$

(BGG) $\left\{h_w\right|w\in W\}$ is an $ℝ-$basis of $\mathscr{H}.$

		Proof.
	Let us first establish that $h_{w_0}\ne 0.$ We have $$⟨h_{w_0},p⟩=⟨{\Delta }_{w_0}^{*}\left(1\right),p⟩=⟨1,{\Delta }_{w_0}p⟩$$ and by Proposition 8.6 and skew symmetry of $p$ we have $${\Delta }_{w_0}p=p^{-1}\sum _{w\in W}\epsilon \left(w\right)\cdot wp=\left|W\right|.$$ So $⟨h_{w_0},p⟩=\left|W\right|\ne 0$ and therefore $h_{w_0}\ne 0.$ (It must be a scalar multiple of $p.$) Since $\dim \mathscr{H}=\left|W\right|$ (Proposition 7.4) it is enough to show that the $h_w$ are linearly independent over $ℝ.$ So suppose we have a linear dependence relation $$\begin{array}{ll}\sum c_w{h}_w=0(c_w\in ℝ).& (Inv\; 15)\end{array}$$ Choose $v\in W$ of minimal length such that $c_v\ne 0.$ Let $u=v^{-1}{w}_0$ and apply ${\Delta }_u^{*}$ to (Inv 15). We have by Proposition 9.3 $${\Delta }_u^{*}{h}_v={\Delta }_u^{*}{\Delta }_v^{*}\left(1\right)={\Delta }_{w_0}^{*}\left(1\right)=h_{w_0}(becausevu=w_0)$$ and if $w\ne v$ $${\Delta }_u^{*}{h}_w={\Delta }_u^{*}{\Delta }_w^{*}\left(1\right)=0by\; Proposition\; 9.3$$ because $l\left(u\right)+l\left(w\right)=N-l\left(v\right)+l\left(w\right)\ge N,$ and $uw\ne w_0$ if $w\ne v.$ So we obtain $c_v{h}_{w_0}=0,$ hence $c_v=0.$ $\square$

From Theorem 9.5 it follows that for each $n\ge 0$ $$\dim {\mathscr{H}}^n=\mathrm{Card}\{w\in W|l\left(w\right)=n\}$$ and hence $$P(\mathscr{H};t)=\sum _{w\in W}{t}^{l\left(w\right)}.$$ Comparing this with Proposition 8.6 we have the polynomial identity $$\begin{array}{ll}\sum _{w\in W}{t}^{l\left(w\right)}=\prod _{i=1}^r\frac{1-t^{d_i}}{1-t}& (Inv\; 16)\end{array}$$ valid for any finite reflection group.

Instead of appealing to Proposition 8.6 we can prove directly that the $h_w$ span $\mathscr{H}.$ Indeed let ${\mathscr{H}}_0$ be the span of the $h_w.$ Then ${\mathscr{H}}_0\le \mathscr{H}$ so that ${\mathscr{H}}_0^{\perp }\ge {\mathscr{H}}^{\perp }=𝒥,$ and ${\mathscr{H}}_0=\mathscr{H} \iff {\mathscr{H}}_0=𝒥.$ So we need to prove that $$\begin{array}{ll}if\; f\in 𝒮 \; and\; ⟨f,h_w⟩=0 \; for\; all\; w\in W, \; then\; f\in 𝒥.& (Inv\; 17)\end{array}$$

		Proof of (Inv 17).
	We may assume $f\in {𝒮}^d$ and proceed by induction on $d.$ If $d=0$ we have $f=⟨f,1⟩=⟨f,h_1⟩=0,$ so $f=0\in 𝒥.$ If $d>0$ consider $$⟨h_w,{\Delta }_{i}f⟩=⟨{\Delta }_i^{*}{h}_w,f⟩=⟨{\Delta }_i^{*}{\Delta }_w^{*}\left(1\right),f⟩$$ and ${\Delta }_i^{*}{\Delta }_w^{*}$ is either ${\Delta }_{w{s}_i}^{*}$ or 0, so that $⟨h_w,{\Delta }_{i}f⟩$ if either 0 or equal to $⟨h_{w{s}_i},f⟩=0.$ So in both cases $⟨h_w,{\Delta }_{i}f⟩=0$ for all $w\in W,$ hence (induction hypothesis) ${\Delta }_{i}f\in 𝒥,$ i.e. $f\equiv s_{i}f \left(\mathrm{mod}𝒥\right)$ for all $w\in W.$ Hence finally $f\equiv a\left(f\right) \left(=\frac{1}{\left|W\right|}\sum _{w\in W}wf\mathrm{mod}𝒥\right),$ and $a\left(f\right)\in {𝒥}^{+}$ (because $d>0$) so finally $f\in 𝒥$ as required. $\square$

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

We have $h_{s_i}={\alpha }_i^{\vee }$ because if $f\in 𝒮$ is linear $$⟨h_{s_i},f⟩=⟨{\Delta }_i^{*}\left(1\right),f⟩=⟨1,{\Delta }_{i}f⟩=⟨{\alpha }_i^{\vee },f⟩$$ by Proposition 8.1.iv.

Schubert polynomials

As before let $p=\prod _{\alpha \in R^{+}}\alpha$ and define $(w\in W)$ $$S_w=\frac{1}{\left|W\right|}{\Delta }_{w^{-1}{w}_0}\left(p\right)\in 𝒮.$$ $S_w$ is homogeneous of degree $$N-l\left(w^{-1}{w}_0\right)=l\left(w\right)$$ (since $l\left(w^{-1}{w}_0\right)=l\left(w_0\right)-l\left(w^{-1}\right)=N-l\left(w\right)$).

1. $⟨h_v,S_w⟩={\delta }_{v,w}(v,w\in W).$
2. Let $\stackrel{\_}{S_w}=S_w+𝒥\in 𝒮/𝒥.$ Then ${\left(\stackrel{\_}{S_w}\right)}_{w\in W}$ is an $ℝ-$basis of $𝒮/𝒥.$

		Proof.
	We have $$⟨h_v,S_w⟩=\frac{1}{\left|W\right|}⟨{\Delta }_v^{*}\left(1\right),{\Delta }_{w^{-1}{w}_0}\left(p\right)⟩=\frac{1}{\left|W\right|}⟨1,{\Delta }_v{\Delta }_{w^{-1}{w}_0}\left(p\right)⟩.$$ Suppose this is $\ne 0.$ Then for reasons of degree we must have $$l\left(v\right)+l\left(w^{-1}{w}_0\right)=N$$ i.e., $l\left(v\right)=l\left(w\right),$ and by Proposition 8.4 $$l\left(v\right)+l\left(w^{-1}{w}_0\right)=l\left(v{w}^{-1}{w}_0\right)$$ which forces $v=w.$ So $⟨h_v,S_w⟩=0$ unless $v=w,$ and $$⟨h_w,S_w⟩=\frac{1}{\left|W\right|}⟨1,{\Delta }_{w_0}\left(p\right)⟩$$ and since by Proposition 8.6 $${\Delta }_{w_0}\left(p\right)=\frac{1}{p}\sum _{w\in W}\epsilon \left(w\right)wp=\left|W\right|$$ we have $⟨h_w,S_w⟩=1.$ Suppose $\sum _{w\in W}{a}_w{S}_w\in 𝒥(a_w\in ℝ).$ Take the scalar produce of each side with $h_v:$ since $h_v\in \mathscr{H}\perp 𝒥$ we get $\sum _w{a}_w⟨h_v,S_w⟩=0$ and hence $a_v=0$ by (i) above. Hence the $\stackrel{\_}{S_w}$ are linearly independent over $ℝ,$ and since $\dim 𝒮/𝒥=\dim \mathscr{H}=\left|W\right|,$ they are a basis. $\square$

Suppose $W=S_n$ acting on $𝒮=ℝ[x_1,...,x_n]$ by permuting the $x_i.$ Then the monomials $x^{\alpha }=x_1^{{\alpha }_1}\cdots x_n^{{\alpha }_n}$ with ${\alpha }_i\le n-i$ form a basis of $𝒮\mathrm{mod}𝒥.$ In particular $x^{\delta }=x_1^{n-1}{x}_2^{n-2}\cdots x_{n-1}$ is a generator of ${𝒮}^N\mathrm{mod}𝒥$ $(N=\frac{1}{2}n(n-1)).$ Since $W=S_n$ acts on ${(𝒮/𝒥)}^N$ as the sign representation, it follows that $x^{w\delta }=w\left(x^{\delta }\right)\equiv \epsilon \left(w\right)x^{\delta }\mathrm{mod}𝒥$ and hence $$p=\prod _{i<j}(x_i-x_j)=\sum _{w\in S_n}\epsilon \left(w\right)x^{w\delta }\equiv n!x^{\delta }\mathrm{mod}𝒥.$$ Hence $S_w=\frac{1}{n!}{\Delta }_{w^{-1}{w}_0}\left(p\right)\equiv {\Delta }_{w^{-1}{w}_0}\left(x^{\delta }\right)\mathrm{mod}𝒥$ which ties up the $S_w$ with the Schubert polynomials of Lascoux and Schutzenberger.

Coxeter elements

Let $X$ be a tree, $G$ a group and $S:X\to G$ a mapping $x\mapsto s_x\in G.$ Suppose that $s_x,s_y$ commute whenever $x,y$ are not joined by an edge in $X.$ Then all the products $\prod _{x\in X}{s}_x$ are conjugate in $G.$

		Proof.
	Remark first that if $g_1,...,g_n\in G$ then $$g_2{g}_3\cdots g_n{g}_1=g_1^{-1}(g_1{g}_2\cdots g_n)g_1$$ is conjugate to $g_1\cdots g_n.$ So cyclic permutation of a product produces conjugate elements. We prove the lemma by induction on the number $n$ of vertices in $X.$ When $n=1$ there is nothing to prove. Suppose $n>1.$ Since $X$ is a tree it has an end vertex $a,$ say, joined to just one other vertex $b.$ Consider a product $\pi$ of the $s_x,x\in G.$ Up to conjugacy in $G$ we may assume that $s_a$ comes first: $\pi =s_a\cdots s_b\cdots =s_{a}u{s}_{b}v$ say. Now $s_a$ commutes with every $s_x$ except $s_b,$ hence commutes with $u:$ $$\pi =s_{a}u{s}_{b}v=u{s}_a{s}_{b}v,$$ hence $\pi$ is conjugate to $s_a{s}_{b}vu.$ So it remains to show that all products $s_a{s}_v\cdots$ are conjugate in $G,$ and to do this we use the inductive hypothesis. Let $X\prime$ is a tree with $n-1$ vertices. Define $s\prime :X\prime \to G$ by ${s\prime }_b=s_a{s}_b, {s\prime }_x=s_x$ if $x\ne a,b.$ Now apply the inductive hypothesis to $X\prime , s\prime .$ $\square$

Now let $W$ be an ???? Weyl group (more generally, a finite reflection group), $s_1,...,s_r$ the generators of $W.$ The Dynkin diagram $D$ of $W$ is a tree hence $(X,G)=(D,W)$ satisfies the conditions of the lemma. Consequently all the products $s_{\pi \left(1\right)}\cdots s_{\pi \left(r\right)} (\pi \in S_r)$ of the generators $s_1,...,s_r$ are conjugate in $W.$ Moreover if we had taken a different basis $B\prime =wB$ of $R$ the generators would be ${s\prime }_i=w{s}_i{w}^{-1} (w\in W),$ so all the products (for all bases) are conjugate in $W.$ These are the Coxeter elements of $W;$ they form a distinguished conjugacy class.

Example. $W=S_n:$ $s_i=(i,i+1), 1\le i\le n-1.$ Then $s_1{s}_2\cdots s_{n-1}$ is an $n-$cycle, so the Coxeter elements of $S_n$ are the $n-$cycles.

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

Now consider the characteristic polynomial $$P\left(t\right)=\det (1-tc)\in \mathbb{Z}\left[t\right](if\; W \; is\; a\; Weyl\; group)$$ whose root are the eigenvalues of $c.$ Again they do not depend on which $c$ we take, and they are $h^{\mathrm{th}}$ roots of unity, say $${\zeta }_j=\exp \frac{2\pi i{e}_j}{h}(1\le j\le r)$$ where $1\le e_1\le e_2\le \cdots \le e_r<h.$ (One shows that 1 is not an eigenvalue of $c,$ i.e. that $c$ has no fixed points $\ne 0$ in $V$). Since the eigenvalues are also $\stackrel{\_}{{\zeta }_j}$ (because they are the roots of a real polynomial) it follows that the sequence $(h-e_j)$ is the sequence $\left(e_j\right)$ in reverse order, i.e. $$\begin{array}{ll}{e}_j+e_k=hif{j}_k=r+1.& (Inv\; 18)\end{array}$$ Adding up these equations, we get $$\begin{array}{ll}{e}_1+\cdots +e_r=\frac{1}{2}hr.& (Inv\; 19)\end{array}$$ The relevance of this to polynomial invariants is that if $d_1\le \cdots \le d_r$ are the degrees of $W$ then $$\begin{array}{ll}{d}_i=e_i+1(1\le i\le r)(\Rightarrow d_j+d_k=h+2ifj+k=r+1).& (Inv\; 20)\end{array}$$ Since by Proposition 3.4 $\sum _1^r(d_i-1)=N$ it follows that $N=\sum e_i=\frac{1}{2}hr:$ $$\begin{array}{ll}hr=2N=\mathrm{Card}\left(R\right).\left(soh=\frac{\mathrm{Card}\left(R\right)}{\mathrm{rank}\left(R\right)}=top\; degree\right).& (Inv\; 21)\end{array}$$ Again, since there is a polynomial invariant $I_2={\left|x\right|}^2$ of degree 2, we have $e_1=1,$ hence ${\zeta }_1=e^{\frac{2\pi i}{h}}$ is an eigenvalue of $c,$ i.e. a root of $P\left(t\right)\in \mathbb{Z}\left[t\right].$ Hence all primitive $h^{\mathrm{th}}$ roots of unity are eigenvalues of $c,$ i.e. each positive integer less than $h$ and prime to $h$ is an exponent. $(e_1=1\Rightarrow e_r=h-1\Rightarrow d_r=hsoh=\max (d_1,...,d_r).)$

Examples.

1. $W=S_n$ acting irreducibly on the hyperplane $V$ in ${ℝ}^n$ with equation $x_1+\cdots +x_n=0.$ $c$ is an $n-$cycle, so its eigenvalues are the $n^{\mathrm{th}}$ roots of unity other than 1. So $$h=nand(e_1,...,e_{n-1})=(1,2,...,n-1).$$ This agrees with the above, because we have seen earlier that $(d_1,...,d_{n-1})=(2,3,...,n).$
2. $W=E_8.$ Here $r=8$ and $\mathrm{Card}\left(R\right)=240,$ so that $h=30.$ The positive integers less than 30 and prime to 30 are $$1,7,11,13,17,19,23,29.$$ There are 8 of them, hence they are precisely the exponents of $E_8.$
3. $W=E_7:$ $r=7, \mathrm{Card}\left(R\right)=126,$ so $h=18.$ The positive integers less than 18 and prime to 18 are $$1,5,7,11,13,17.$$ So these are 6 of the exponents, and the other one must be $\frac{1}{2}h=9,$ by (1) above.
4. $W=E_6:$ $r=6,$ $\mathrm{Card}\left(R\right)=72,$ so $h=12.$ Hence 1, 5, 7, 11 are four of the exponents. In fact, the other two are 4, 8, but one needs some extra information to deduce this. For example if one knew that $\left|W\right|=2^7\cdot 3^4\cdot 5=\prod d_i=\prod (e_i+1)=2\cdot 6\cdot 8\cdot 12(e+1)(e\prime +1)$ where $e,e\prime$ are the two unknown exponents, then since $2\cdot 6\cdot 8\cdot 12=2^7\cdot 3^2$ we have $$(e+1)(e\prime +1)=3^2\cdot 5=45$$ which together with $e+e\prime =12$ gives $ee\prime =45-1-12=32$ and hence $e,e\prime =4,8.$
5. $W=F_4:$$r=4,$ $\mathrm{Card}\left(R\right)=48,$ so $h=12$ again. Hence the exponents are 1, 5, 7, 11.

In fact the eigenvalues of the Cartan matrix $A={\left(⟨{\alpha }_i,{\alpha }_j^{\vee }⟩\right)}_{1\le i,j\le r}$ are $4{\sin }^2\frac{\pi e_i}{2h} (1\le i\le r),$ by an ingenious argument due to Coxeter [Bourbaki, p.140 Ex. 3,4].

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

$G=$ Chevalley group over ${𝔽}_q:$ $\left|G\right|=|G/B|\cdot \left|B\right|.$ $$|G/B|=\sum _{w\in W}{q}^{l\left(w\right)}\left|B\right|={(q-1)}^r{q}^N,$$ so $\left|G\right|=q^N{(q-1)}^r\sum _w{q}^{l\left(w\right)}=q^N\prod (q^{d_i}-1)=q^{2N+r}+\cdots .$

Let $c=s_1\cdots s_r$ be a Coxeter element of $W$ and put $${\alpha }_{ik}=s_1\cdots s_k\left({\alpha }_i\right),0\le k\le r,1\le i\le r.$$ Thus ${\alpha }_{i0}={\alpha }_i,$ ${\alpha }_{ir}=c\left({\alpha }_i\right)={\beta }_i.$ Let ${\gamma }_i={\alpha }_{i,i-1}.$ We have $$\begin{array}{rcl}{\alpha }_{i,k-1}-{\alpha }_{ik}& =& s_1\cdots s_{k-1}({\alpha }_i-s_k{\alpha }_i)\\ & =& ⟨{\alpha }_k^{\vee },{\alpha }_i⟩s_1\cdots s_{k-1}\left({\alpha }_k\right)\\ & =& a_{ki}{\gamma }_k,\end{array}$$ where $a_{ki}=⟨{\alpha }_k^{\vee },{\alpha }_i⟩$ is the $(k,i)$ element of the Cartan matrix $A.$ Thus $${\alpha }_{i,k-1}={\alpha }_{ik}+a_{ki}{\gamma }_k$$ and therefore, summing from $k=1$ to $i-1,$ $$\begin{array}{ll}{\alpha }_i={\gamma }_i+\sum _{k<i}{a}_{ki}{\gamma }_k.& (Inv\; 23)\end{array}$$ Likewise, summing from $k=i$ to $k=r,$ $$\begin{array}{ll}{\beta }_i={\gamma }_i-\sum _{k\ge i}{a}_{ki}{\gamma }_k=-\left({\gamma }_i+\sum _{k>i}{a}_{ki}{\gamma }_k\right)& (Inv\; 24)\end{array}$$ since $a_{ii}=2.$

Let $U=\left(\begin{array}{ccc}1& & a_{ij}\\ & \ddots & \\ & 0& 1\end{array}\right),L=\left(\begin{array}{ccc}1& & \\ & \ddots & \\ a_{ij}& & 1\end{array}\right),$ so that $A=U+L.$ Then (Inv 23), (Inv 24) show that the matrix expressing the $\beta \text{'}s$ in terms of the $\gamma \text{'}s$ is $-L,$ and that expressing the $\alpha \text{'}s$ in terms of the $\gamma \text{'}s$ is $U^{-1}.$ Hence if $C$ is the matrix of the Coxeter element $c$ we have $C=-U^{-1}L$ and hence $$\begin{array}{rcl}\det ({\lambda }^2-c)& =& \det ({\lambda }^2{1}_r-C)\\ & =& \det ({\lambda }^2+U^{-1}L)\\ & =& \det ({\lambda }^{2}U+L)(because\det U=1)\\ & =& |\begin{array}{ccc}{\lambda }^2+1& & \\ & \ddots & {\lambda }^2{a}_{ij}\\ a_{ij}& & {\lambda }^2+1\end{array}| .\end{array}$$ Now in general if $X=\left(x_{ij}\right)$ is an $r\times r$ matrix, its determinant is a sum of terms $$\begin{array}{ll}\pm x_{1\sigma \left(1\right)}\cdots x_{r\sigma \left(r\right)}\sigma \in S_r.& (Inv\; 25)\end{array}$$ Express the permutation $\sigma$ as a product of disjoint cycles: each cycle $(i_1,...,i_p)$ in $\sigma$ will contribute $x_{i_1{i}_2}{x}_{i_2{i}_3}\cdots x_{i_p{i}_1}$ to the monomial (Inv 25). Apply this observation to the determinant above: the Dynkin diagram contains no cycles, hence any $\sigma \in S_r$ containing cycles of length $>2$ will give zero. In other words, any term in the expansion of this determinant that contains ${\lambda }^2{a}_{ij}(i<j)$ must also contain $a_{ji}$ (below the diagonal). In other words we have $$\begin{array}{rcl}\det ({\lambda }^2-c)& =& \left|\begin{array}{ccc}{\lambda }^2+1& & \lambda a_{ij}\\ & \ddots & \\ \lambda a_{ij}& & {\lambda }^2+1\end{array}\right|\\ & =& \det ({(\lambda -1)}^2+\lambda A)since{a}_{ii}=2\\ & =& {\lambda }^r\det (\lambda +{\lambda }^{-1}-2+A).\end{array}$$ This vanishes when ${\lambda }^2=\exp \left(\frac{2\pi i{e}_j}{h}\right)(1\le j\le r),$ i.e. when $\lambda =\pm \exp \left(\frac{\pi i{e}_j}{h}\right),$ i.e. when $$X=\lambda +{\lambda }^{-1}-2=\pm \cos \frac{\pi e_j}{h}-2.$$ Since $$-\cos \frac{\pi e_j}{h}=\cos (\pi -\frac{\pi e_j}{h})=\cos \frac{\pi }{h}(h-e_j)=\cos \frac{\pi }{h}{e}_{r+1-j}$$ it follows that $$\det (X+A)=\prod _{i=1}^r(X+2-\cos \frac{\pi e_j}{h})$$ and hence the eigenvalues of $A$ are $2(1-\cos \frac{\pi e_j}{h})=4{\sin }^2\frac{\pi e_j}{2h}.$

The roots ${\gamma }_i=s_i\cdots s_{i-1}\left({\alpha }_i\right)$ are precisely the positive roots $\gamma$ such that $c^{-1}\gamma$ is negative (Root systems, 1.18). The formula (Inv 23) shows that they are a basis of $V.$ Let ${\Omega }_i$ be the $c-$orbit of ${\gamma }_i:$ $${\Omega }_i=\{{\gamma }_i,c{\gamma }_i,c^2{\gamma }_i,...\}.$$ Then one can show that the $r$ orbits ${\Omega }_i$ are pairwise disjoint, that each contains $h$ elements, and that $$R=\bigcup _1^r{\Omega }_i.$$

References

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

Version: September 20, 2001

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