Root systems

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

Last update: 27 March 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

Let $V$ be a real vector space of finite dimension $n>0,$ and let $⟨x,y⟩$ be a positive definite symmetric inner product on $V.$ So we have $⟨x,x⟩\ne 0$ for all $x\ne 0$ in $V,$ and we write $$\left|x\right|={⟨x,x⟩}^{\frac{1}{2}}$$ for the length of $x.$ A linear transformation $f:V\to V$ is an isometry if it is length preserving: $\left|f\right(x\left)\right|=\left|x\right|$ for all $x\in V.$ Equivalently, $⟨fx,fy⟩=⟨xy⟩$ for all $x,y\in V.$

Example. $V={ℝ}^n$ with the standard inner product: $$If\; x=(x_1,...,x_n), y=(y_1,...,y_n) \; then\; ⟨x,y⟩=\sum _{i=1}^n{x}_i{y}_i.$$ In fact this is essentially the only example: given $V$ as above we can construct an orthonormal basis of $V,$ i.e. a basis $v_1,...,v_n,$ such that $⟨v_i,v_j⟩={\delta }_{ij} (1\le i,j\le n)$ and then if $x=\sum _{i=1}^n{x}_i{v}_i,$ $y=\sum _{i=1}^n{y}_i{v}_i,$ we have $⟨x,y⟩=\sum x_i{y}_i.$

If $x,y\in V$ are such that $⟨x,y⟩=0$ we say that $x,y$ are perpendicular (or orthogonal) and write $x\perp y.$ More generally, if $x,y\ne 0$ the angle $\theta (\in [0,\pi ])$ between the vectors $x,y$ is given by $$\cos \theta =\frac{⟨x,y⟩}{\left|x\right|\cdot \left|y\right|}.$$ One other piece of notation: if $x\in V,$ $x\ne 0$ we shall write $$x^{\vee }=\frac{2x}{{\left|x\right|}^2}$$ (you'll see why in a moment). We have $$\left|x^{\vee }\right|=\frac{2}{\left|x\right|},⟨x,x^{\vee }⟩=2,{\left(x^{\vee }\right)}^{\vee }=\frac{2x^{\vee }}{{\left|x^{\vee }\right|}^2}=x,{\left(cx\right)}^{\vee }=c^{-1}{x}^{\vee }(c\in ℝ, c\ne 0).$$

Reflections in $V$

Let $\alpha \in V,$ $\alpha \ne 0,$ and let $s_{\alpha }:V\to V$ be the orthogonal reflection in the hyperplane $H_{\alpha }=\{x\in V | ⟨x,\alpha ⟩=0\}$ perpendicular to $\alpha .$ Clearly $\left|s_{\alpha }\left(x\right)\right|=\left|x\right|$ for all $x\in V,$ i.e. $s_{\alpha }$ is an isometry.

1. $s_{\alpha }\left(x\right)=x-⟨x,{\alpha }^{\vee }⟩\alpha (=x-⟨x,\alpha ⟩{\alpha }^{\vee }).$
2. $s_{\alpha }{\left(x\right)}^{\vee }=s_{\alpha }\left(x^{\vee }\right)=x^{\vee }-⟨x^{\vee },\alpha ⟩{\alpha }^{\vee }.$
3. $s_{\alpha }^2=1.$
4. $⟨s_{\alpha }x,y⟩=⟨x,s_{\alpha }y⟩.$
5. Let $f:V\to V$ be an isometry. Then $s_{f\left(\alpha \right)}=f{s}_{\alpha }{f}^{-1}.$

		Proof.
	We have $x-s_{\alpha }x=\lambda \alpha ,$ for some $\lambda \in ℝ,$ so that $$\begin{array}{ll}⟨x-s_{\alpha }x,\alpha ⟩=\lambda {\left|\alpha \right|}^2.& (IGM\; 1)\end{array}$$ On the other hand, $$\begin{array}{ll}⟨x+s_{\alpha }x,\alpha ⟩=0& (IGM\; 2)\end{array}$$ because $\frac{1}{2}(x+s_{\alpha }x)\in H_{\alpha }.$ Adding (IGM 1) and (IGM 2) we get $$2⟨x,\alpha ⟩=\lambda {\left|\alpha \right|}^2$$ i.e., $\lambda =\frac{2⟨x,\alpha ⟩}{{\left|\alpha \right|}^2}=⟨x,{\alpha }^{\vee }⟩,$ which proves (i). $$\begin{array}{rcl}{s}_{\alpha }{\left(x\right)}^{\vee }& =& \frac{2s_{\alpha }\left(x\right)}{{\left|s_{\alpha }\left(x\right)\right|}^2}\\ & =& \frac{2s_{\alpha }\left(x\right)}{{\left|x\right|}^2}(because\; s_{\alpha } \; is\; an\; isometry)\\ & =& s_{\alpha }\left(x^{\vee }\right)\\ & =& x^{\vee }-⟨x^{\vee },\alpha ⟩{\alpha }^{\vee }by\; (i).\end{array}$$ Is obvious from the definition. $⟨s_{\alpha }x,y⟩=⟨x,y⟩-⟨x,{\alpha }^{\vee }⟩⟨y,\alpha ⟩=⟨x,y⟩-\frac{2⟨x,\alpha ⟩⟨y,\alpha ⟩}{{\left|\alpha \right|}^2}$ is symmetrical in $x$ and $y,$ hence equal to $⟨s_{\alpha }y,x⟩=⟨x,s_{\alpha }y⟩.$ Calculate, using (i): $$f{s}_{\alpha }{f}^{-1}\left(x\right)=f(f^{-1}x-\frac{2⟨f^{-1}x,\alpha ⟩}{{\left|\alpha \right|}^2}\alpha )=x-\frac{2⟨x,f\alpha ⟩}{{\left|f\alpha \right|}^2}f\alpha =s_{f\alpha }\left(x\right).$$ $\square$

Root systems

A root system in $V$ is a non-empty subset $R$ of $V-\left\{0\right\}$ satisfying the following two axioms:

1. (R1) For all $\alpha ,\beta \in R,$ $⟨{\alpha }^{\vee },\beta ⟩\in \mathbb{Z}$ (integrality).
2. (R2) For all $\alpha ,\beta \in R,$ $s_{\alpha }\left(\beta \right)\in R$ (symmetry).

1. Since $s_{\alpha }\left(\alpha \right)=-\alpha$ it follows from (R2) that $$\alpha \in R\Rightarrow -\alpha \in R.$$ Suppose on the other hand $\alpha \in R$ and $\beta =c\alpha \in R (c\in ℝ, c\ne 0).$ Then $⟨{\alpha }^{\vee },\beta ⟩=c⟨{\alpha }^{\vee },\alpha ⟩=2c,$ so that $2c\in \mathbb{Z};$ and $⟨\alpha ,{\beta }^{\vee }⟩=c^{-1}⟨\alpha ,{\alpha }^{\vee }⟩=2c^{-1},$ so that $2c^{-1}\in \mathbb{Z}.$ So the only possibilities for $c$ are $$c=\pm \frac{1}{2}, \pm 1, \pm 2.$$ If $$\alpha \in R, c\alpha \in R \Rightarrow c=\pm 1$$ we say that $R$ is reduced. But there are non reduced root systems as well (examples in a moment).
2. I don't demand that $R$ spans $V.$ The dimension of the subspace of $V$ spanned by $R$ is called the rank of $R.$ It is therefore the maximum number of linearly independent elements of $R.$
3. $R_1\subseteq V_1, R_2\subseteq V_2: R=R_1\cup R_2\subseteq V_1\oplus V_2$ (orthogonal direct sum). $R$ is reducible if it splits up in this way (decomposable would be a better word).

Examples.

1. $\mathrm{rank} R=1:$ only two possibilities $$R=\{\pm \alpha \}\left(A_1\right),R=\{\pm \alpha ,\pm 2\alpha \}\left(B{C}_1\right).$$ The first is reduced, the second isn't.
2. $\mathrm{rank} R=2,$ $R$ reduced. Draw pictures of $A_1\times A_1, A_2, B_2, G_2.$ The first of these is reducible, the others are irreducible.
3. $\mathrm{rank} R=2,$ $R$ non-reduced: draw picture of $B{C}_2.$
4. $V={ℝ}^n,$ standard basis $e_1,...,e_n$ $(⟨e_i,e_j⟩={\delta }_{ij}).$ $$\begin{array}{rcl}{A}_{n-1}& =& \{e_i-e_j | 1\le i,j\le n, i\ne j\}(rankn-1)\\ B_n& =& \{\pm e_i\pm e_j (1\le i<j\le n), \pm e_i (1\le i\le n)\},\\ C_n& =& \{\pm e_i\pm e_j (1\le i<j\le n), \pm 2e_i (1\le i\le n)\},\\ D_n& =& \{\pm e_i\pm e_j (1\le i<j\le n)\},\\ B{C}_n& =& B_n\cup C_n(not\; reduced).\end{array}$$ (Exercise: check (R1), (R2) in each case.)
   
   In fact, as we shall see later, this is almost a complete list of the irreducible root systems: apart from $A_n$ ($n\ge 1$), $B_n$ ($n\ge 2$), $C_n$ ($n\ge 3$), $D_n$ ($n\ge 4$), $B{C}_n$ ($n\ge 1$) there are just 5 others: $E_6, E_7, E_8, F_4, G_2$ (the last of which we have already met).
5. In $R$ is a root system, so is $$R^{\vee }=\left\{{\alpha }^{\vee } \right| \alpha \in R\}$$ (the dual root system). In the examples above, $A_{n-1}, B{C}_n,$ and $D_n$ are self dual; $B_n$ and $C_n$ are duals of each other.

I want now to start drawing some consequences from the integrality axiom (R1), which as we shall see restricts the possibilities very drastically. So let $R$ be a root system and $\alpha ,\beta \in R.$ Let $\theta ={\theta }_{\alpha \beta }\in [0,\pi ]$ be the angle between the vectors $\alpha ,\beta ,$ so that $$\cos \theta =\frac{⟨\alpha ,\beta ⟩}{\left|\alpha \right|\left|\beta \right|}$$ and hence $$\begin{array}{ll}4{\cos }^2\theta =\frac{4{⟨\alpha ,\beta ⟩}^2}{{\left|\alpha \right|}^2{\left|\beta \right|}^2}=⟨\alpha ,{\beta }^{\vee }⟩⟨{\alpha }^{\vee },\beta ⟩\in \mathbb{Z}& (IGM\; 3)\end{array}$$ so that $\left|⟨{\alpha }^{\vee },\beta ⟩\right|\le 4.$

Let $\alpha ,\beta \in R$ be linearly independent and assume $⟨\alpha ,\beta ⟩\ne 0.$ then

1. $⟨\alpha ,{\beta }^{\vee }⟩⟨{\alpha }^{\vee },\beta ⟩=1,2$ or $3.$
2. If $\left|\alpha \right|\ge \left|\beta \right|$ then $⟨{\alpha }^{\vee },\beta ⟩=\pm 1$ and $\frac{{\left|\alpha \right|}^2}{{\left|\beta \right|}^2}=1,2 or 3.$

		Proof.
	Follows from (IGM 3), since ${\cos }^2\theta \ne 0,1.$ $\frac{{\left|\alpha \right|}^2}{{\left|\beta \right|}^2}=\frac{⟨\alpha ,{\beta }^{\vee }⟩}{⟨{\alpha }^{\vee },\beta ⟩}$ (since $⟨\alpha ,\beta ⟩\ne 0$). Hence $\left|⟨{\alpha }^{\vee },\beta ⟩\right|\le \left|⟨\alpha ,{\beta }^{\vee }⟩\right|$ and it follows from (i) that $\left|⟨{\alpha }^{\vee },\beta ⟩\right|=1,\left|⟨\alpha ,{\beta }^{\vee }⟩\right|=1,2 \; or\; 3.$ $\square$

From the relation (IGM 3) we have $${\cos }^2\theta =0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, 1$$ giving $$\cos \theta =0, \pm \frac{1}{2}, \pm \frac{1}{\sqrt{2}}, \pm \frac{\sqrt{3}}{2}, \pm 1.$$ So the possible values of $\theta ={\theta }_{\alpha \beta }$ are $$\frac{\pi }{2}, \frac{\pi }{3}, \frac{2\pi }{3}, \frac{\pi }{4}, \frac{3\pi }{4}, \frac{\pi }{6}, \frac{5\pi }{6}, 0, \pi$$ or collectively $${\theta }_{\alpha \beta }=\frac{r\pi }{12}$$ where $0\le r\le 12$ and $r$ is not prime to 12 (i.e. $r\ne 1,5,7,11$).

$R$ is finite.

		Proof.
	$R$ spans a subspace $V^{\prime }$ of $V,$ so we can choose ${\alpha }_1,...,{\alpha }_r\in R$ forming a basis of $V^{\prime }.$ Let $v_1,...,v_r$ be the dual basis of $V^{\prime },$ defined by $⟨{\alpha }_i,v_j⟩={\delta }_{ij}(1\le i,j\le r).$ Let $\beta \in R,$ then ${\beta }^{\vee }\in V^{\prime },$ say $${\beta }^{\vee }=\sum _{i=1}^r{m}_i{v}_i$$ and the coefficients $m_i$ are given by $$m_i=⟨{\alpha }_i,{\beta }^{\vee }⟩.$$ So each $m_i$ is an integer and $\left|m_i\right|\le 4.$ So only finitely many possibilities. $\square$

Weyl group

Let $W=⟨s_{\alpha } | \alpha \in R⟩$ be the group of isometries of $V$ generated by the reflections $s_{\alpha },$ $\alpha \in R.$ By (R2) each $w\in W$ permutes the elements of $R,$ i.e. we have a homomorphism $$\begin{array}{ll}W\to \mathrm{Sym}\left(R\right)& (IGM\; 4)\end{array}$$ of $W$ into the group of permutations of $R.$ As in Proposition 3.4, let $V^{\prime }$ be the subspace of $V$ spanned by $R$ and let $V^{\prime \prime }={\left(V^{\prime }\right)}^{\perp }$ be the orthogonal complement of $V^{\prime },$ so that $V=V^{\prime }\oplus V^{\prime \prime }.$ Each $s_{\alpha } (\alpha \in R)$ fixes $V^{\prime \prime }$ pointwise (because $V^{\prime \prime }\subseteq H_{\alpha }$), hence each $w\in W$ fixes $V^{\prime \prime }$ pointwise.

Suppose $w\in W$ gives rise to the identity permutation under the homomorphism (IGM 4), i.e. $w\left(\alpha \right)=\alpha$ for all $\alpha \in R.$ Then $w$ fixes $V^{\prime }$ pointwise (because $R$ spans $V^{\prime }$) as well as $V^{\prime \prime },$ i.e. $w=1_V.$ So $W$ embeds in $\mathrm{Sym}\left(R\right)$ which is a finite group by Proposition 3.4. Hence $W$ is a finite group called the Weyl group of $R:$ notation $W=W\left(R\right).$

Examples.

1. Weyl groups of types $A,B,C,D$ (symmetric group, hyperoctahedral group, etc.).
2. $W\left(R^{\vee }\right)=W\left(R\right).$

Let $\alpha ,\beta \in R.$

1. If $⟨\alpha ,\beta ⟩>0$ (i.e. if ${\theta }_{\alpha \beta }$ is acute) then $\beta -\alpha \in R\cup \left\{0\right\}.$
2. If $⟨\alpha ,\beta ⟩<0$ (i.e. if ${\theta }_{\alpha \beta }$ is obtuse) then $\beta +\alpha \in R\cup \left\{0\right\}.$

		Proof.
	(ii) comes from (i) by replacing $\alpha$ by $-\alpha ,$ so it is enough to prove (i). First of all, if $\beta =c\alpha$ ($c\in ℝ$) then (as we have seen) $\frac{1}{2}, 1 \; or\; 2$ so that $$\beta -\alpha =(c-1)\alpha =\{\begin{array}{ll}-\frac{1}{2}\alpha =-\beta ,& if\; c=\frac{1}{2},\\ 0,& if\; c=1,\\ \alpha ,& if\; c=2.\end{array}\phantom{\}}$$ So we may assume $\alpha ,\beta$ linearly independent and then by Proposition 3.3 (i) either $⟨{\alpha }^{\vee },\beta ⟩=1$ or $⟨\alpha ,{\beta }^{\vee }⟩=1.$ If say $⟨{\alpha }^{\vee },\beta ⟩=1$ then by Proposition 2.1 we have $$s_{\alpha }\left(\beta \right)=\beta -⟨{\alpha }^{\vee },\beta ⟩\alpha =\beta -\alpha$$ and hence also $\beta -\alpha =-(\alpha -\beta )\in R.$ $\square$

Strings of roots

Let $\alpha ,\beta \in R$ be linearly independent and let $$I=\{i\in \mathbb{Z} | \beta +i\alpha \in R\}.$$

1. $I$ is an interval $[-p,q]$ of $\mathbb{Z},$ where $p,q\ge 0.$
2. $p-q=⟨\alpha ,{\beta }^{\vee }⟩.$

		Proof.
	Certainly $0\in I.$ Let $-p$ (resp. $q$) be the smallest (resp. largest) element of $I.$ Suppose $I\ne [-p,q].$ Then there exist $r,s\in I$ such that $s>r+1, r+1\notin I, s-1\notin I.$ $$\begin{array}{rrrcl}So\; we\; have& \beta +r\alpha \in R,& (\beta +r\alpha )+\alpha \notin R,& hence& ⟨\beta +r\alpha ,\alpha ⟩\ge 0,\\ also& \beta +s\alpha \in R,& (\beta +s\alpha )-\alpha \notin R,& hence& ⟨\beta +s\alpha ,\alpha ⟩\le 0,\end{array}$$ both by Proposition 4.1. Subtract and we get $(r-s){\left|\alpha \right|}^2\ge 0,$ hence $r\ge s,$ a contradiction. So $I=[-p,q].$ We have $s_{\alpha }(\beta +r\alpha )=\beta +r\alpha -⟨\beta +r\alpha ,{\alpha }^{\vee }⟩\alpha =\beta -(⟨{\alpha }^{\vee },\beta ⟩+r)\alpha .$ Hence $r\in I$ implies $-(⟨{\alpha }^{\vee },\beta ⟩+r)\in I.$ Take $r=q:$ $⟨{\alpha }^{\vee },\beta ⟩+q\le p,$ i.e. $⟨{\alpha }^{\vee },\beta ⟩\le p-q.$ Take $r=-p:$ $p-⟨{\alpha }^{\vee },\beta ⟩\le q,$ i.e. $⟨{\alpha }^{\vee },\beta ⟩\ge p-q.$ $\square$

The set of roots $\beta +i\alpha$ ($-p\le i\le q$) is called the $\alpha -$string through $\beta$. It follows from Proposition 5.1 that a string of roots has as most 4 elements: (take $q=0,$ i.e. $\beta$ at the end of the chain: $p=⟨{\alpha }^{\vee },\beta ⟩\le 3$ because $\alpha ,\beta$ linearly independent.)

Bases of $R$

A basis of $R$ is a subset $B$ of $R$ such that

1. (B1) $B$ is linearly independent.
2. For each $\alpha \in R$ we have $$\alpha =\sum _{\beta \in B}{m}_{\beta }\beta ,$$ with coefficients $m_{\beta }\in \mathbb{Z}$ and either all $m_{\beta }\ge 0$ or all $m_{\beta }\le 0.$

From (B2) it follows that $B$ spans the subspace $V^{\prime }$ of $V$ spanned by $R,$ hence (B1) $B$ is a basis of $V^{\prime }$ and therefore $\mathrm{Card}\left(B\right)=\mathrm{rank}\left(R\right).$

Examples. $$\begin{array}{rrcl}{A}_{n-1}:& B& =& \{e_i-e_{i+1}, 1\le i\le n-1\}.\\ B_n:& B& =& \{e_i-e_{i+1}, 1\le i\le n-1; e_n\}.\\ C_n:& B& =& \{e_i-e_{i+1}, 1\le i\le n-1; 2e_n\}.\\ D_n:& B& =& \{e_i-e_{i+1}, 1\le i\le n-1; e_{n-1}+e_n\}.\\ G_2:& & & Draw\; picture.\end{array}$$

Defining something gives no guarantee that it exists. However, the following construction provides bases (in fact, all of them). Say $x\in V$ is regular if $⟨\alpha ,x⟩\ne 0$ for all $\alpha \in R,$ i.e. if $x$ does not lie in any of the reflecting hyperplanes $H_{\alpha }$ ($\alpha \in R$). Let $x$ be regular and let $$R^{+}=R_x^{+}=\{\alpha \in R | ⟨\alpha ,x⟩>0\}.$$ Since $\alpha \in R$ implies that $-\alpha \in R$ it follows that $R=R_x^{+}\cup (-R_x^{+})$ (disjoint union). A root $\alpha \in R_x^{+}$ will be called (temporarily) decomposable if $\alpha =\beta +\gamma$ with $\beta ,\gamma \in R^{+};$ otherwise indecomposable. Let $B_x$ be the set of indecomposable elements of $R_x^{+}.$

$B_x$ is a basis of $R.$

		Proof.
	In several steps. Let $S=\sum _{\beta \in B_x}\mathbb{N}\beta .$ I clain that $R_x^{+}\subseteq S.$ Suppose not, and choose $\alpha \in R_x^{+}$, $\alpha \notin S$ such that $⟨\alpha ,x⟩$ is as small as possible. Certainly $\alpha \notin B_x$ (because $B_x\subseteq S$), hence $\alpha$ is decomposable, say $\alpha =\beta +\gamma (\beta ,\gamma \in R_x^{+}).$ Hence $$⟨\alpha ,x⟩=⟨\beta ,x⟩+⟨\gamma ,x⟩;$$ both $⟨\beta ,x⟩$ and $⟨\gamma ,x⟩$ are positive, hence less than $⟨\alpha ,x⟩.$ It follows that $\beta \in S$ and $\gamma \in S,$ hence (as $S$ is closed under addition) $\alpha \in S,$ contradiction. Hence $$R=R_x^{+}\cup (-R_x^{+})\subseteq S\cup (-S)$$ and so $B_x$ satisfies (B2). Let $\alpha ,\beta \in B_x, \alpha \ne \beta .$ Then $⟨\alpha ,\beta ⟩\le 0$ (i.e., ${\theta }_{\alpha \beta }\ge \frac{\pi }{2}$). Suppose $⟨\alpha ,\beta ⟩>0.$ By Proposition 4.1 we have $\beta -\alpha \in R$ and hence also $\alpha -\beta \in R.$ So either $\beta -\alpha \in R_x^{+},$ in which case $\beta =\alpha +(\beta -\alpha )$ is decomposable; or $\alpha -\beta \in R_x^{+},$ in which case $\alpha =\beta +(\alpha -\beta )$ is decomposable. Contradiction in either case. $B_x$ is linearly independent. Suppose not, then there exists a linear dependence relation which we can write in the form $$\sum _{\alpha \in B^{\prime }}{m}_{\alpha }\alpha =\sum _{\beta \in B^{\prime \prime }}{n}_{\beta }\beta (=\lambda \; say)$$ where $B^{\prime }, B^{\prime \prime }$ are disjoint subsets of $B_x,$ and the coefficients $m_{\alpha }, n_{\beta }$ are all $>0.$ By (2) above we have $${\left|\lambda \right|}^2=\sum _{\alpha ,\beta }{m}_{\alpha }{n}_{\beta }⟨\alpha ,\beta ⟩\le 0$$ and hence $\lambda =0.$ Hence $$0=⟨\lambda ,x⟩=\sum _{\alpha }{m}_{\alpha }⟨\alpha ,x⟩=\sum _{\beta }{n}_{\beta }⟨\beta ,x⟩.$$ Since $⟨\alpha ,x⟩$ and $⟨\beta ,x⟩$ are positive it follows that $B^{\prime }=B^{\prime \prime }=\varnothing .$ $\square$

Conversely, all bases $B$ of $R$ are of the form $B_x,$ where $x\in V$ is regular.

Let $B$ be any basis of $R,$ and let $$C=\{x\in V | ⟨\alpha ,x⟩>0 all \alpha \in B\}.$$ Then $C\ne \varnothing ,$ every $x\in C$ is regular and $B=B_x,$ for all $x\in C.$

		Proof.
	Let $B=\{{\alpha }_1,...,{\alpha }_r\};$ $B$ is a basis of $V^{\prime }$ (as remarked earlier), hence there exists a dual basis $\{v_1,...,v_r\}$ of $V^{\prime }$ such that $⟨{\alpha }_i,v_j⟩={\delta }_{ij}.$ Let $x\in V^{\prime },$ say $x=\sum _{i=1}^r{x}_i{v}_i;$ then $⟨{\alpha }_i,x⟩=x_i$ and hence $x\in C$ provided all the coefficients $x_i$ are $>0.$ So $C$ is certainly not empty. Now let $x\in C,$ $\alpha \in R^{+}:$ $\alpha =\sum _{i=1}^r{m}_i{\alpha }_i$ with $m_i\ge 0,$ hence $⟨\alpha ,x⟩=\sum _{i=1}^r{m}_i⟨{\alpha }_i,x⟩>0$ for all $\alpha \in R^{+},$ and likewise $⟨\alpha ,x⟩<0$ for all $\alpha \in R^{-}.$ So $x$ is regular and $R^{+}=R_x^{+},$ whence $B=B_x.$ $\square$

So the above construction provides all bases of $R.$

Let $B$ be a basis of $R;$ $\alpha ,\beta \in B, \alpha \ne \beta .$ Then $⟨\alpha ,\beta ⟩\le 0$ (i.e., ${\theta }_{\alpha \beta }\ge \frac{\pi }{2}$).

		Proof.
	By Proposition 6.3, $B=B_x$ for some regular $x\in V.$ Hence Proposition 6.4 follows from the proof of Proposition 6.2. $\square$

From now on it will be simpler (and will involve no loss of generality) to assume that $V^{\prime }=V,$ i.e. that $R$ spans $V.$ Let $$B=\{{\alpha }_1,...,{\alpha }_r\}$$ be a basis of $R$ (so $r=\mathrm{rank}\left(R\right)$) and let $R^{+}$ be the set of positive roots relative to $B;$ $R^{-}=-R^{+}$ the set of negative roots. (${\alpha }_1,...,{\alpha }_r$ also called a set of simple roots).

Let $s_i=s_{{\alpha }_i} (1\le i\le r)$ and set $$\rho =\frac{1}{2}\sum _{\alpha \in R^{+}}\alpha (half\; the\; sum\; of\; the\; positive\; roots).$$

1. $s_i$ permutes the set $R^{+}-\left\{{\alpha }_i\right\}.$
2. $s_i\rho =\rho -{\alpha }_i.$
3. $⟨\rho ,{\alpha }_i^{\vee }⟩=1.$

		Proof.
	Let $\beta \in R^{+}, \beta \ne {\alpha }_i.$ Then by Proposition 2.1 $$s_i\beta =\beta -⟨\beta ,{\alpha }_i^{\vee }⟩{\alpha }_i.$$ Now $\beta$ is of the form $\sum _{i=1}^r{m}_j{\alpha }_j$ with at least one coefficient $m_j,$ $j\ne i,$ positive (because $2{\alpha }_i$ is not a root). Hence the coefficient of ${\alpha }_j$ in $s_i\beta$ is also positive, hence $s_i\beta \in R^{+}.$ From (i) it follows that $$s_i\rho =\frac{1}{2}\sum _{\genfrac{}{}{0ex}{}{\beta \in R^{+}}{\beta \ne {\alpha }_i}}\beta -\frac{1}{2}{\alpha }_i=\rho -{\alpha }_i.$$ Follows from (ii), since $$s_i\rho =\rho -⟨\rho ,{\alpha }_i^{\vee }⟩{\alpha }_i.$$ $\square$

As in Proposition 6.3, let $$C=\{x\in V | ⟨x,{\alpha }_i⟩>0 (1\le i\le r)\}=\{x\in V | ⟨x,\alpha ⟩>0, \; all\; \alpha \in R^{+}\}.$$ $C$ is the Weyl chamber associated with the basis $B=\{{\alpha }_1,...,{\alpha }_r\}.$ It is the intersection of $r$ half spaces in $V$ ($\dim V=r$ now), so it is an open simplicial cone. Relative to the dual basis $\{v_1,...,v_r\}$ it is the positive octant.

From Proposition 6.5 (iii) it follows that $\rho \in C.$

Let $x\in V$ be regular. Then there exists $w\in W$ such that $wx\in C.$

		Proof.
	Choose $w\in W$ such that $⟨wx,\rho ⟩$ is as large as possible. Then for $1\le i\le r$ we have $$\begin{array}{rcl}⟨wx,\rho ⟩& \ge & ⟨s_{i}wx,\rho ⟩\\ & =& ⟨wx,s_i\rho ⟩\\ & =& ⟨wx,\rho -{\alpha }_i⟩by\; Proposition\; 6.5\\ & =& ⟨wx,\rho ⟩-⟨wx,{\alpha }_i⟩\end{array}$$ so that $⟨wx,{\alpha }_i⟩\ge 0;$ but also $⟨wx,{\alpha }_i⟩=⟨x,w^{-1}{\alpha }_i⟩\ne 0$ (because $x$ is regular and $w^{-1}{\alpha }_i\in R$), hence $⟨wx,{\alpha }_i⟩>0$ for $1\le i\le r,$ i.e., $wx\in C.$ $\square$

Let $B^{\prime }$ be another basis of $R.$ Then $B^{\prime }=wB$ for some $w\in W.$

		Proof.
	By Proposition 6.3 we have $B^{\prime }=B_x,$ some regular $x\in V,$ and the corresponding set of positive roots is $$R_x^{+}=\{\alpha \in R | ⟨\alpha ,x⟩>0\}.$$ By Proposition 6.6 there exists $w\in W$ such that $wx\in C$ and therefore $$\alpha \in R_x^{+}\iff ⟨\alpha ,x⟩>0\iff ⟨w\alpha ,wx⟩>0\iff w\alpha \in R^{+},$$ so that $R_x^{+}=w^{-1}{R}^{+}$ and hence $B^{\prime }=B_x{w}^{-1}=B.$ $\square$

We shall show later that $B^{\prime }=wB$ for exactly one $w\in W.$

1. Let $\alpha \in R,$ then $\alpha =w{\alpha }_i$ for some $w\in W$ and some $i$ (i.e. $R=WB$).
2. $W$ is generated by $s_1,...,s_r. (s_i=s_{{\alpha }_i})$

		Proof.
	Let $W_0$ be the subgroup of $W$ generated by $s_1,...,s_r.$ We shall show that (i) holds for some $w\in W_0.$ We may assume that $\alpha \in R^{+},$ for if $-\alpha =w{\alpha }_i$ then $\alpha =w{s}_i{\alpha }_i.$ For $\alpha \in R^{+},$ say $$\alpha =\sum _{i=1}^r{m}_i{\alpha }_i$$ define the height of $\alpha$ to be $$\mathrm{ht}\left(\alpha \right)=\sum _{i=1}^r{m}_i,$$ the sum of the coefficients. $(So\; \mathrm{ht}\left(\alpha \right)=1\iff \alpha \in B).$ We proceed by induction on $\mathrm{ht}\left(\alpha \right).$ We must have $⟨\alpha ,{\alpha }_i⟩>0$ for some $i,$ for otherwise we should have $${\left|\alpha \right|}^2=\sum _{i=1}^r{m}_i⟨\alpha ,{\alpha }_i⟩\le 0,$$ which is impossible. Hence $$s_i\alpha =\alpha -⟨\alpha ,{\alpha }_i^{\vee }⟩{\alpha }_{i}has\; height\mathrm{ht}\left(s_i\alpha \right)=\mathrm{ht}\left(\alpha \right)-⟨\alpha ,{\alpha }_i⟩<\mathrm{ht}\left(\alpha \right)$$ and hence by the inductive hypothesis $s_i\alpha =w{\alpha }_j$ for some $w\in W_0, {\alpha }_j\in B.$ So $\alpha =s_{i}w{\alpha }_j,$ and $s_{i}w\in W_0.$ Enough to show $s_{\alpha }\in W_0$ for each $\alpha \in R.$ But $\alpha =w{\alpha }_i$ with $w\in W_0,$ hence $s_{\alpha }=w{s}_i{w}^{-1}(Proposition\; 2.1)$ \in W_0.$$ $\square$

From Proposition 6.9, each $w\in W$ can be written in the form $$w=s_{a_1}\cdots s_{a_p}.$$ If (for a given $w$) the number $p$ of factors is as small as possible, then $s_{a_1}\cdots s_{a_p}$ is called a reduced expression for $w,$ and $p$ is the length of $w,$ denoted by $\ell \left(w\right)$ (relative to the generators $s_1,...,s_r$). Thus $\ell \left(1\right)=0; \ell \left(w\right)=1 \iff w=s_i; \ell \left(w\right)=\ell \left(w^{-1}\right).$

Let $w\in W,$ then $$\ell \left(w\right)>\ell \left(s_{i}w\right)\iff w^{-1}{\alpha }_i<0.(i.e., {\alpha }_i\in w{R}^{-}).$$

		Proof.
	Suppose $w^{-1}{\alpha }_i<0.$ Let $w=t_1\cdots t_p$ be a reduced expression for $w,$ where each $t_i$ is an $s_j,$ say $t_i=s_{{\beta }_i}, {\beta }_i\in B.$ Let $w_j=t_1\cdots t_j (0\le j\le p),$ so that $w_0=1$ and $w_p=w.$ So we have $w_0^{-1}{\alpha }_i={\alpha }_i>0$ and $w_p^{-1}{\alpha }_i=w^{-1}{\alpha }_i<0,$ hence there exists $j\in [1,p]$ such that $$\beta =w_{j-1}^{-1}{\alpha }_i>0,w_j^{-1}{\alpha }_i<0.$$ Now $w_j^{-1}=t_j{w}_{j-1}^{-1},$ so that we have $$\beta >0,t_j\beta <0,t_j=s_{{\beta }_j},{\beta }_j\in B.$$ By Proposition 6.5, $\beta \ne {\beta }_j \Rightarrow t_j\beta <0,$ so we must have $\beta ={\beta }_j$ and hence $${\alpha }_i=w_{j-1}\beta =w_{j-1}{\beta }_j$$ giving Proposition 2.1 $$s_i=w_{j-1}{t}_j{w}_{j-1}^{-1}=w_j{w}_{j-1}^{-1},$$ or $$w_j=s_i{w}_{j-1},$$ and therefore $$\begin{array}{rcl}{s}_{i}w& =& \left(s_i{w}_{j-1}\right)t_j{t}_{j+1}\cdots t_p\\ & =& (t_1\cdots t_j)t_j{t}_{j+1}\cdots t_p\\ & =& t_1\cdots t_{j-1}{t}_{j+1}\cdots t_p,\end{array}$$ showing that $$\ell \left(s_{i}w\right)\le p-1<\ell \left(w\right).$$ So we have proved that $$\begin{array}{ll}{w}^{-1}{\alpha }_i<0 \Rightarrow \ell \left(s_{i}w\right)<\ell \left(w\right).& (IGM\; 5)\end{array}$$ Suppose now that $w^{-1}{\alpha }_i>0,$ then ${\left(s_{i}w\right)}^{-1}{\alpha }_i=w^{-1}{s}_i{\alpha }_i=-w^{-1}{\alpha }_i<0,$ hence (replacing $w$ by $s_{i}w$ in (IGM 5)) we have $$w^{-1}{\alpha }_i<0 \Rightarrow \ell \left(w\right)<\ell \left(s_{i}w\right).$$ This completes the proof. $\square$

Suppose $w_1,w_2\in W, w_1\ne w_2.$ Then $w_{1}B\ne w_{2}B.$

		Proof.
	We have to show that $B\ne w_1^{-1}{w}_{2}B,$ i.e. $B\ne wB$ if $w\ne 1.$ So let $w=s_i\cdots$ be a reduced expression for $w.$ Then $\ell \left(s_{i}w\right)<\ell \left(w\right),$ hence $w^{-1}{\alpha }_i<0,$ hence $w^{-1}{\alpha }_i\notin B,$ i.e. ${\alpha }_i\notin wB.$ So $B\ne wB$ as required. $\square$

Example. Since $B$ is a basis of $R,$ so is $-B.$ Positive roots relative to $B$ are negative roots relative to $-B$ and vice versa. By Proposition 6.11 we have $-B=w_{0}B$ for a unique $w_0\in W.$ $w_0$ is called the longest element of $W$ (relative to the basis $B$). We have $w_0^2=1,$ because $w_0^{2}B=w_0(-B)=B.$

For each $w\in W$ let $$R\left(w\right)=\{\alpha \in R^{+} | w^{-1}\alpha \in R^{-}\}=R^{+}\cap w{R}^{-}.$$

Suppose that $\ell \left(w\right)>\ell \left(s_{i}w\right).$ Then $$R\left(w\right)=s_{i}R\left(s_{i}w\right)\cup \left\{{\alpha }_i\right\}.$$

		Proof.
	We have $$R\left(s_{i}w\right)=R^{+}\cap s_{i}w{R}^{-}$$ and therefore $$\begin{array}{ll}{s}_{i}R\left(s_{i}w\right)=s_i{R}^{+}\cap w{R}^{-}.& (IGM\; 6)\end{array}$$ Now by Proposition 6.5 $$\begin{array}{ll}{s}_i{R}^{+}=(R^{+}-\left\{{\alpha }_i\right\})\cup \{-{\alpha }_i\}& (IGM\; 7)\end{array}$$ and by Proposition 6.10 $w^{-1}{\alpha }_i<0,$ i.e. ${\alpha }_i\in w{R}^{-}$ and therefore $-{\alpha }_i\notin w{R}^{-}.$ Hence from (IGM 6) and (IGM 7) we deduce that $$\begin{array}{rcl}{s}_{i}R\left(s_{i}w\right)& =& (R^{+}-\left\{{\alpha }_i\right\})\cap w{R}^{-}\\ & =& R^{+}\cap w{R}^{-}-\left\{{\alpha }_i\right\}\\ & =& R\left(w\right)-\left\{{\alpha }_i\right\}.\end{array}$$ $\square$

[Compare Schubert polynomials, Ch. I, esp. (1.2).]

Note that ${\alpha }_i\notin s_{i}R\left(s_{i}w\right)$ (otherwise we should have $-{\alpha }_i=s_i{\alpha }_i\in R\left(s_{i}w\right)\subseteq R^{+},$ impossible).

1. Let $w=t_1\cdots t_p$ be a reduced expression, where $t_i=s_{{\beta }_i}, {\beta }_i\in B.$ Then $$\begin{array}{ll}R\left(w\right)=\{t_1\cdots t_{i-1}{\beta }_i | 1\le i\le p\}& (IGM\; 8)\end{array}$$ and these $p$ roots are all distinct.
2. $\ell \left(w\right)=\mathrm{Card} R\left(w\right).$

		Proof.
	Since $t_{1}w=t_2\cdots t_p$ it follows that $\ell \left(w\right)=p>\ell \left(t_{1}w\right),$ hence by Proposition 6.12 $$R\left(w\right)=\left\{{\beta }_1\right\}\cup t_{1}R(t_2\cdots t_p)$$ from which (IGM 8) follows by induction on $p$ [SP, (1.7)]. Suppose $$t_1\cdots t_{i-1}{\beta }_i=t_1\cdots t_{j-1}{\beta }_j$$ where $i<j.$ Then $${\beta }_i=t_i\cdots t_{j-1}{\beta }_j$$ and therefore by Proposition 2.1 $$\begin{array}{rcl}{t}_i=s_{{\beta }_i}& =& t_i\cdots t_{j-1}{s}_{{\beta }_j}{(t_i\cdots t_{j-1})}^{-1}\\ & =& t_i\cdots t_j{(t_i\cdots t_{j-1})}^{-1}\end{array}$$ from which it follows that $$t_i\cdots t_j=t_i{t}_i\cdots t_{j-1}=t_{i+1}\cdots t_{j-1}$$ and hence that $$w=t_1\cdots t_p=t_1\cdots \widehat{t_i}\cdots \widehat{t_j}\cdots t_p$$ contradicting the assumption that $t_1\cdots t_p$ is reduced. Hence $\mathrm{Card} R\left(w\right)=p=\ell \left(w\right).$ $\square$

Example. $R\left(w_0\right)=R^{+}\cap w_0{R}^{-}=R^{+},$ hence $\ell \left(w_0\right)=\mathrm{Card} R^{+}=number\; of\; reflections\; in\; W.$

$$\begin{array}{rclcrcl}\ell \left(w\right)& =& \ell \left(s_{i}w\right)+1& \iff & w^{-1}{\alpha }_i& <& 0,\\ \ell \left(w\right)& =& \ell \left(s_{i}w\right)-1& \iff & w^{-1}{\alpha }_i& >& 0.\end{array}$$

		Proof.
	We have $$\begin{array}{rclc}{w}^{-1}{\alpha }_i& \Rightarrow & \ell \left(w\right)>\ell \left(s_{i}w\right)& Proposition\; 6.10\\ & \Rightarrow & R\left(w\right)=s_{i}R\left(s_{i}w\right)\cup \left\{{\alpha }_i\right\}& Proposition\; 6.12\\ & \Rightarrow & \ell \left(w\right)=\ell \left(s_{i}w\right)+1& Proposition\; 6.13\end{array}$$ Replace $w$ by $s_{i}w:$ $$\begin{array}{rcl}{w}^{-1}{\alpha }_i& \Rightarrow & {\left(s_{i}w\right)}^{-1}{\alpha }_i=w^{-1}{s}_i{\alpha }_i=-w^{-1}{\alpha }_i<0\\ & \Rightarrow & \ell \left(s_{i}w\right)=\ell \left(w\right)+1.\end{array}$$ $\square$

(Exchange lemma) Let $w=t_1\cdots t_p=u_1\cdots u_p$ be two reduced expressions for $w,$ where $t_i=s_{{\beta }_i}, u_i=s_{{\gamma }_i}$ with ${\beta }_i, {\gamma }_i\in B.$ Then for some $i\in [1,p]$ we have $$w=u_1{t}_1\cdots \widehat{t_i}\cdots t_p$$ (i.e. we can exchange $u_1$ with one of the $t_i$) [SP, (1.8)].

		Proof.
	By Proposition 6.13 we have ${\gamma }_1\in R\left(w\right),$ hence ${\gamma }_1=t_1\cdots t_{i-1}{\beta }_i$ for some $i\in [1,p].$ Hence by Proposition 2.1 $$\begin{array}{rcl}{u}_1=s_{{\gamma }_1}& =& t_1\cdots t_{i-1}{s}_{{\beta }_i}{(t_1\cdots t_{i-1})}^{-1}\\ & =& (t_1\cdots t_{i-1}{t}_i){(t_1\cdots t_{i-1})}^{-1}\end{array}$$ and therefore $$t_1\cdots t_i=u_1{t}_1\cdots t_{i-1}$$ giving $w=t_1\cdots t_i\cdots t_p=u_1{t}_1\cdots t_{i-1}{t}_{i+1}\cdots t_p.$ $\square$

We shall next deduce from this exchance lemma that the Weyl group $W$ is a Coxeter group (definition later). Consider two generators $s_i,s_j$ of $W$ ($i\ne j$) and let $$\begin{array}{rcl}{m}_{ij}& =& order\; of\; s_i{s}_j \; in\; W\\ & =& order\; of\; s_j{s}_i \; in\; W\end{array}$$ (because $s_j{s}_i={\left(s_i{s}_j\right)}^{-1}$). Then we have $$\begin{array}{ll}{s}_i{s}_j{s}_i\cdots =s_j{s}_i{s}_j\cdots & (IGM\; 9)\end{array}$$ where there are $m_{ij}$ ($\ge 2$) terms on either side.

Let $w\in W,$ of length $\ell \left(w\right)=p.$ A reduced word for $w$ is a sequence $\underset{\_}{t}=(t_1,...,t_p)$ where each $t_i$ is one of the $s_j,$ and $w=t_1\cdots t_p.$ Let $S\left(w\right)$ denote the set of all reduced words for $w.$ We make $S\left(w\right)$ into a graph as follows: let $u_{ij}$ denote the word $$u_{ij}=(s_i,s_j,s_i,...)of\; length\; m_{ij}.(i\ne j)$$ Suppose $\underset{\_}{t}\in S\left(w\right)$ contains $u_{ij}$ as a subword and let $\underset{\_}{t^{\prime }}\in S\left(w\right),$ and we join $\underset{\_}{t},\underset{\_}{t^{\prime }}$ by an edge.

The graph $S\left(w\right)$ is connected.

		Proof.
	Induction on $\ell \left(w\right).$ When $\ell \left(w\right)=1, w=s_i$ and $S\left(w\right)$ has just one element. Let $\underset{\_}{t}=(t_1,...,t_p), \underset{\_}{u}=(u_1,...,u_p)\in S\left(w\right), (p=\ell \left(w\right)).$ We shall write $\underset{\_}{t}\equiv \underset{\_}{u}$ if $\underset{\_}{t}, \underset{\_}{u}$ are in the same connected component of $S\left(w\right).$ The inductive hypothesis assures us that $$\begin{array}{ll}\underset{\_}{t}\equiv \underset{\_}{u} \; if\; either\; t_1=u_1 \; or\; t_p=u_p.& (IGM\; 10)\end{array}$$ For is $w\prime =t_{1}w=u_{1}w$ then $\ell (w\prime )=p-1$ and hence $(t_2,...,t_p)\equiv (u_2,...,u_p).$ We want to prove that $\underset{\_}{t}\equiv \underset{\_}{u}.$ If $t_1=u_1$ we are through, by (IGM 10). If $t_2\ne u_1,$ then (exchange) there exists $i\in [1,p]$ such that $$\underset{\_}{a}=(u_1,t_1,...,\widehat{t_i},...,t_p)\in S\left(w\right).$$ Suppose $i\ne p.$ Then $$\underset{\_}{t}\equiv \underset{\_}{a}\equiv \underset{\_}{u}$$ by (IGM 10), and therefore $\underset{\_}{t}\equiv \underset{\_}{u}.$ Suppose $i=p.$ Let $m$ be the order of $t_1{u}_1$ in $W.$ If $m=2$ then $$\underset{\_}{a\prime }=(t_1,u_1,t_2,...,t_{p-1})\in S\left(w\right)$$ and $$\underset{\_}{t}\equiv \underset{\_}{a\prime }\equiv \underset{\_}{a}\equiv \underset{\_}{u}$$ so again $\underset{\_}{t}\equiv \underset{\_}{u}.$ Suppose $i=p$ and $m>2.$ We have $$\begin{array}{rcl}\underset{\_}{a}& =& (u_1,t_1,...,t_{p-1})\in S\left(w\right),\\ \underset{\_}{t}& =& (t_1,t_2,...,t_p),\end{array}$$ hence (exchange) there exists $i\in [1,p-1]$ such that $$\underset{\_}{b}=(t_1,u_1,t_1,...,\widehat{t_i},...,t_{p-1})\in S\left(w\right).$$ Suppose $i\ne p-1.$ Then we have $$\underset{\_}{t}\equiv \underset{\_}{b}\equiv \underset{\_}{a}\equiv \underset{\_}{u}$$ by (IGM 10), and hence $\underset{\_}{t}\equiv \underset{\_}{u}.$ Suppose $i=p-1$ and $m=3.$ Then $$\underset{\_}{b\prime }=(u_1,t_1,u_1,t_2,...,t_{p-2})\in S\left(w\right)$$ and $\underset{\_}{t}\equiv \underset{\_}{b}\equiv \underset{\_}{b\prime }\equiv \underset{\_}{u},$ so again we are through. Suppose $i=p-1$ and $m>3.$ Then we have $$\begin{array}{rcl}\underset{\_}{b}& =& (t_1,u_1,t_1,t_2,...,t_{p-2})\in S\left(w\right),\\ \underset{\_}{u}& =& (u_1,u_2,...,u_{p-1})\in S\left(w\right),\end{array}$$ so by exchange there exists $i\in [1,p-2]$ such that $$\underset{\_}{c}=(u_1,t_1,u_1,t_1,...,\widehat{t_i},...,t_{p-2})\in S\left(w\right).$$ Suppose $i\ne p-2.$ Then $$\underset{\_}{t}\equiv \underset{\_}{b}\equiv \underset{\_}{c}\equiv \underset{\_}{u}$$ and again $\underset{\_}{t}\equiv \underset{\_}{u}.$ Suppose $i=p-2$ and $m=4.$ Then $$\underset{\_}{c\prime }=(t_1,u_1,t_1,u_1,t_2,...,t_{p-2})\in S\left(w\right)$$ and $$\underset{\_}{t}\equiv \underset{\_}{c\prime }\equiv \underset{\_}{c}\equiv \underset{\_}{u}$$ so again $\underset{\_}{t}\equiv \underset{\_}{u}.$ Suppose $i=p-2$ and $m>4.$ Repeat the argument: eventually we shall get $\underset{\_}{t}\equiv \underset{\_}{u},$ as required. $\square$

The generators $s_i (1\le i\le r)$ and relations $$s_i^2=1,{\left(s_i{s}_j\right)}^{m_{ij}}=1(i\ne j)$$ form a presentation of $W.$

		Proof.
	What this means is the following: given a group $G$ and elements $g_i\in G (1\le i\le r)$ satisfying $g_i^2=1, {\left(g_i{g}_j\right)}^{m_{ij}}=1 (i\ne j),$ there exists a homomorphism $f:W\to G$ (necessarily unique) such that $$f\left(s_i\right)=g_i(1\le i\le r).$$ Let $w\in W$ and let $(t_1,...,t_p)=\underset{\_}{t}\in S\left(w\right).$ Since $w=t_1\cdots t_p$ we must have $$f\left(w\right)=f\left(t_1\right)\cdots f\left(t_p\right)=F\left(\underset{\_}{t}\right)say.$$ So we have to show that $F\left(\underset{\_}{t}\right)=F\left(\underset{\_}{u}\right)$ if $\underset{\_}{t},\underset{\_}{u}\in S\left(w\right).$ Now in $G$ we have $$g_i{g}_j{g}_i\cdots =g_j{g}_i{g}_j\cdots$$ ($m_{ij}$ terms on either side), i.e. $$f\left(s_i\right)f\left(s_j\right)f\left(s_i\right)\cdots =f\left(s_j\right)f\left(s_i\right)f\left(s_j\right)\cdots .$$ Hence $F\left(\underset{\_}{t}\right)=F\left(\underset{\_}{u}\right)$ if $\underset{\_}{t},\underset{\_}{u}$ are joined by an edge in $S\left(w\right).$ By Proposition 6.16 it follows that $F\left(\underset{\_}{t}\right)=F\left(\underset{\_}{u}\right)$ for all $\underset{\_}{t},\underset{\_}{u}\in S\left(w\right),$ as required. So $f$ is well defined and it remains to check that it is a homomorphism. Consider $f\left(s_{i}w\right):$ suppose first that $\ell \left(s_{i}w\right)=\ell \left(w\right)+1.$ If $w=t_1\cdots t_p$ is a reduced expression, then $s_{i}w=s_i{t}_1\cdots t_p$ is also reduced, hence $$f\left(s_{i}w\right)=f\left(s_i\right)f\left(t_1\right)\cdots f\left(t_p\right)=f\left(s_i\right)f\left(w\right).$$ If on the other hand $\ell \left(s_{i}w\right)=\ell \left(w\right)-1$ Proposition 6.14, replace $w$ by $s_{i}w:$ $$f\left(w\right)=f\left(s_i\right)f\left(s_{i}w\right)$$ and hence $$f\left(s_{i}w\right)=f{\left(s_i\right)}^{-1}f\left(w\right)=f\left(s_i\right)f\left(w\right)$$ since $f\left(s_i\right)=g_i=g_i^{-1}.$ So we have $$\begin{array}{ll}f\left(s_{i}w\right)=f\left(s_i\right)f\left(w\right)& (IGM\; 11)\end{array}$$ in all cases. Hence if $v\in W,$ $v=u_1\cdots u_q$ reduced, $$f\left(vw\right)=f(u_1{u}_2\cdots u_{q}w)=f\left(u_1\right)f(u_2\cdots u_{q}w)=\cdots =f\left(u_1\right)\cdots f\left(u_q\right)f\left(w\right)=f\left(v\right)f\left(w\right).$$ $\square$

Weyl chamber

$R, B$ etc. as before. Recall that the Weyl chamber associated with $B$ is $$C=\{x\in V | ⟨x,{\alpha }_i⟩>0 (1\le i\le r)\}.$$ It is an open simplicial cone and its closure in $V$ is $$\stackrel{\_}{C}=\{x\in V | ⟨x,{\alpha }_i⟩\ge 0 (1\le i\le r)\}.$$

$\stackrel{\_}{C}$ is a fundamental domain for the action of $W$ on $V$ (i.e. every $W-$orbit in $V$ meets $\stackrel{\_}{C}$ in exactly one point.)

		Proof.
	(cf. Proposition 6.6) Let $x\in V,$ let $\rho =\frac{1}{2}\sum _{\alpha >0}\alpha ,$ and choose $w\in W$ so that $⟨wx,\rho ⟩$ is as large as possible. Then for $i\in [1,r]$ we have $$\begin{array}{rcl}⟨wx,\rho ⟩& \ge & ⟨s_{i}wx,\rho ⟩\\ & =& ⟨wx,s_i\rho ⟩\\ & =& ⟨wx,\rho -{\alpha }_i⟩Proposition\; 6.5\\ & =& ⟨wx,\rho ⟩-⟨wx,{\alpha }_i⟩,\end{array}$$ so that $⟨wx,{\alpha }_i⟩\ge 0$ and hence $wx\in \stackrel{\_}{C}.$ So each $W-$orbit meets $\stackrel{\_}{C}.$ Remains to prove that if $x\in \stackrel{\_}{C}$ and $y=wx\in \stackrel{\_}{C}$ then $x=y$ (but it doesn't follow necessarily that $w=1$). We proceed by induction on $\ell \left(w\right).$ If $\ell \left(w\right)=0$ then $w=1,$ so $y=x.$ If $\ell \left(w\right)>1$ we can write $w=s_{i}w\prime$ with $\ell (w\prime )=\ell \left(w\right)-1$ (take a reduced word $w=s_i\cdots$). Then $w\prime =s_{i}w,$ so that $\ell \left(w\right)=\ell \left(s_{i}w\right)+1,$ hence $w^{-1}{\alpha }_i\in R^{-}$ by Proposition 6.14. It follows that $$⟨{\alpha }_i,y⟩=⟨{\alpha }_i,wx⟩=⟨w^{-1}{\alpha }_i,x⟩\le 0,(becausex\in \stackrel{\_}{C})$$ but also $⟨{\alpha }_i,y⟩\ge 0$ (because $y\in \stackrel{\_}{C}$). Hence $⟨{\alpha }_i,y⟩=0,$ i.e. $s_{i}y=y$ and therefore $w\prime x=s_{i}wx=s_{i}y=y.$ By the induction hypothesis we conclude that $x=y.$ $\square$

The set $$V_{\mathrm{reg}}=V-\bigcup _{\alpha }{H}_{\alpha }$$ is an open dense subset of $V.$ By Lemma 6.15 $$V_{\mathrm{reg}}=\bigcup _{w\in W}wCandV=\stackrel{\_}{V_{\mathrm{reg}}}=\bigcup _{w\in W}w\stackrel{\_}{C},$$ by taking closures.

It follows from Proposition 7.1 that $V_{\mathrm{reg}}$ is the disjoint union of the chambers $wC$ (i.e. they don't overlap). For if $x\in C, y=wx\in C$ where $w\ne 1,$ the proof of Proposition 7.1 shows that $⟨{\alpha }_i,y⟩=0$ for some $i,$ which contradicts $y\in C.$ So if $w\ne 1$ we have $C\cap w^{-1}C=\varnothing .$

Hence the chambers $wC$ $(w\in W)$ are the connected components of the topological space $V_{}:$ each $wC$ is a cone, hence convex, hence connected, also open.

The basis $B$ corresponding to $C$ may be described as follows:

Let $\alpha \in R^{+}.$ Then $\alpha \in B \iff \stackrel{\_}{C}\cap H_{\alpha }$ spans $H_{\alpha }.$

		Proof.
	Let $\alpha \in R^{+},$ say $\alpha =\sum _{i=1}^r{m}_i{\alpha }_i.$ Let $I=\left\{i \right| m_i\ne 0\}.$ We have $$\begin{array}{rclcrcl}x\in \stackrel{\_}{C}\cap H_{\alpha }& \iff & ⟨{\alpha }_i,x⟩\ge 0 (1\le i\le r)& and& ⟨\alpha ,x⟩& =& \sum _{i\in I}{m}_i⟨{\alpha }_i,x⟩=0\\ & \iff & ⟨{\alpha }_i,x⟩\ge 0 (1\le i\le r)& and& ⟨{\alpha }_i,x⟩& =& 0(i\in I).\end{array}$$ It follows that $\stackrel{\_}{C}\cap H_{\alpha }\subseteq \bigcap _{i\in I}{H}_{{\alpha }_i},$ of dimension $r-\left|I\right|.$ Hence $\stackrel{\_}{C}\cap H_{\alpha }$ spans $H_{\alpha } \iff \left|I\right|=1 \iff \alpha \in B.$ $\square$

As a corollary:

Let $B$ be a basis of $R.$ Then $B^{\vee }$ is a basis of $R^{\vee }.$

		Proof.
	Follows from Proposition 7.1a, since $H_{\alpha }=H_{{\alpha }^{\vee }}.$ $\square$

Let $(v_1,...,v_r)$ be the basis of $V$ dual to $B=({\alpha }_1,...,{\alpha }_r):$ $$⟨{\alpha }_i,v_j⟩={\delta }_{ij}.$$ If $x\in V$ we have $$x=\sum _{i=1}^r⟨x,{\alpha }_i⟩v_i$$ so that $\stackrel{\_}{C}$ is the cone consisting of all nonnegative linear combinations of the dual basis vectors $v_i.$

The dual cone $\stackrel{\_}{C^{*}}$ consists of the nonnegative linear combinations of the ${\alpha }_i,$ and we have $$x\in \stackrel{\_}{C^{*}} \iff ⟨x,v_i⟩\ge 0 (1\le i\le r).$$ (acute cone and obtuse cone: pictures for $A_2,B_2,G_2$). We make use of $\stackrel{\_}{C^{*}}$ to define a partial order on $V:$ if $x,y\in V$ then $x\ge y$ means that $x-y\in \stackrel{\_}{C^{*}},$ i.e. $$x-y=\sum _{i=1}^r{c}_i{\alpha }_{i}with\; c_i\in ℝ, c_i\ge 0,$$ or equivalently $⟨x-y,v_i⟩\ge 0 (1\le i\le r).$

Example. Suppose $R$ is of type $A_{n-1},$ ${\alpha }_i=e_i-e_{i+1} (1\le i\le n-1);$ $V\subseteq {ℝ}^n$ is the hyperplane perpendicular to $e=\frac{1}{n}(e_1+\cdots +e_n).$ We have $⟨e_i,e⟩=\frac{1}{n}=⟨e,e⟩,$ so that ${e\prime }_i=e_i-e\in V.$ The dual basis is $(v_1,...,v_{n-1})$ where $$v_i={e\prime }_1+\cdots +{e\prime }_i=e_1+\cdots +e_i-\frac{i}{n}e=\frac{1}{n}\left({\displaystyle \underset{i}{\underbrace{n-i,...,n-i,}}}{\displaystyle \underset{n-i}{\underbrace{-i,...,-i}}}\right).$$

Let $x=\sum _{i=1}^n{x}_i{e}_i, y=\sum _{i=1}^n{y}_i{e}_i\in V.$ Then $⟨x,v_i⟩=x_1+\cdots +x_i,$ hence $$x\ge y \iff x_1+\cdots +x_i\ge y_1+\cdots +y_i(1\le i\le n-1)$$ (note that $x_1+\cdots +x_n=y_1+\cdots +y_n=0$) the dominance partial order.

Let $x\in V.$ Then the following are equivalent:

1. $x\ge wx,$ all $w\in W,$
2. $x\ge s_{i}x (1\le i\le r),$
3. $x\in \stackrel{\_}{C}.$

		Proof.
	(i) $\Rightarrow$ (ii): obvious. (ii) $\Rightarrow$ (iii): We have $x-s_{i}x=⟨{\alpha }_i^{\vee },x⟩{\alpha }_i$ from Proposition 2.1, hence $x\ge s_{i}x$ means that $⟨{\alpha }_i^{\vee },x⟩\ge 0$ or equivalently $⟨{\alpha }_i,x⟩\ge 0 (1\le i\le r),$ i.e. $x\in \stackrel{\_}{C}.$ (iii) $\Rightarrow$ (i): Let $x\in \stackrel{\_}{C}, w\in W.$ Induction on $\ell \left(w\right).$ $\ell \left(w\right)=0$ implies $w=1,$ OK. Suppose $\ell \left(w\right)\ge 1.$ Then $w=w\prime s_i$ for some $i\in [1,r]$ and $\ell (w\prime )=\ell \left(w\right)-1$ (take a reduced expression for $w$ ending with $s_i$). We have $$x-wx=(x-w\prime x)+w\prime (x-s_{i}x).$$ Now $x-w\prime x\ge 0$ (induction hypothesis), and $$w\prime (x-s_{i}x)=w(s_{i}x-x)=-⟨{\alpha }_i^{\vee },x⟩w{\alpha }_i;$$ by Proposition 6.14 (with $w$ replaced by $w^{-1}$) we have $w{\alpha }_i<0,$ hence $⟨{\alpha }_i^{\vee },x⟩w{\alpha }_i\ge 0.$ So $x-wx\ge 0$ as required. $\square$

Let $x\in V$ and let $$\begin{array}{rcl}{R}_x& =& \{\alpha \in R | ⟨x,\alpha ⟩=0\}\\ W_x& =& \{w\in W | w\left(x\right)=x\}=isotropy\; group\; ofx inW.\end{array}$$ (So $R_x=\varnothing$ if and only if $x$ is regular.)

If $x\in V$ is not regular, then $R_x$ is a root system and $W_x$ is its Weyl group.

		Proof.
	Let $\alpha ,\beta \in R_x,$ then $⟨{\alpha }^{\vee },\beta ⟩\in \mathbb{Z}$ and $x\in H_{\alpha },$ so that $$⟨s_{\alpha }\beta ,x⟩=⟨\beta ,s_{\alpha }x⟩=⟨\beta ,x⟩=0,$$ so that $s_{\alpha }\beta \in R_x.$ So $R_x$ is a root system. Let ${W\prime }_x=⟨s_{\alpha } | ⟨\alpha ,x⟩=0⟩.$ Clearly ${W\prime }_x$ is a subgroup of $W_x$ and we have to show ${W\prime }_x=W_x.$ If $y=ux$ ($u\in W)$ then $⟨\alpha ,y⟩=0 \iff ⟨u^{-1}\alpha ,x⟩=0,$ and hence ${W\prime }_y$ is generated by the $s_{u^{-1}\alpha }=u^{-1}{s}_{\alpha }u$ where $\alpha \in R_x,$ so that ${W\prime }_y=u^{-1}{W\prime }_{x}u$ and likewise $W_y=u^{-1}{W}_{x}u.$ Choose $u\in W$ such that $y=ux\in \stackrel{\_}{C}.$ Enough to show ${W\prime }_y=W_y.$ So let $w\in W_y,$ i.e. $y=wy.$ The proof of Proposition 7.5 shows that if $w\ne 1$ then $w=s_{i}w\prime$ with $\ell (w\prime )<\ell \left(w\right)$ and $s_i\left(y\right)=0,$ so that $$w\prime y=s_{i}wy=s_{i}y=y$$ i.e. $w\prime \in W_y.$ By induction on $\ell \left(w\right)$ we may assume $w\prime \in {W\prime }_y$ and then $w=s_{i}w\prime \in {W\prime }_y.$ $\square$

(So the isometry group of $x\in V$ is generated by the reflections it contains.)

If $x\in \stackrel{\_}{C}, R_x=R_I$ with basis $B_I=\left\{{\alpha }_i \right| ⟨x,{\alpha }_i⟩=0\}.$

References

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

Version: October 30, 2001

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