Classification

Last update: 01 August 2012

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

Reducibility

Let $R$ be a root system in $V$ (not necessarily reduced), $R$ spanning $V.$ Let ${V}_{1},{V}_{2}$ be subspaces of $V$ such that $V={V}_{1}\oplus {V}_{2}$ and let

In this situation the following are equivalent:

1. ${V}_{1},{V}_{2}$ are $W-$stable.
2. $R\subseteq {V}_{1}\cup {V}_{2}.$
3. ${V}_{1}\perp {V}_{2}$ and ${R}_{1}$ (resp. ${R}_{2}$) is a root system in ${V}_{1}$ (resp. ${V}_{2}$), $R={R}_{1}\cup {R}_{2}.$

 Proof. (i)⇒(ii): Let the hyperplane perpendicular to $\alpha .$ Since ${H}_{\alpha }\ne V,$ either ${V}_{1}⊈{H}_{\alpha }$ or ${V}_{2}⊈{H}_{\alpha }.$ Suppose ${V}_{1}⊈{H}_{\alpha },$ then there exists $v\in {V}_{1}$ such that ${s}_{\alpha }v\ne v.$ We have (1.1) $v-{s}_{\alpha }v=⟨v,{\alpha }^{\vee }⟩\alpha \in {V}_{1}$ (because $v,{s}_{\alpha }v\in {V}_{1}$), hence $\alpha \in {V}_{1}.$ So either $\alpha \in {V}_{1}$ or $\alpha \in {V}_{2},$ i.e., $\alpha \in {V}_{1}\cup {V}_{2}.$ (ii)⇒(iii): Let Then $\alpha +\beta \notin {V}_{1},$ otherwise $\beta =\left(\alpha +\beta \right)-\alpha \in {V}_{1}\cap {V}_{2}=0,$ impossible. Likewise $\alpha +\beta \notin {V}_{2},$ hence $\alpha +\beta \notin R.$ By (1.4) $⟨\alpha ,\beta ⟩\ge 0.$ Replace $\beta$ by $-\beta$ and we see that $⟨\alpha ,\beta ⟩\le 0.$ Hence $⟨\alpha ,\beta ⟩=0$ and therefore ${R}_{1}\perp {R}_{2}.$ But $R$ spans $V,$ hence ${R}_{1}$ (resp. ${R}_{2}$) spans ${V}_{1}$ (resp. ${V}_{2}$). So ${V}_{1}\perp {V}_{2}.$ Clearly each satisfy (R1) (integrality). As to (R2): if $\alpha ,\beta \in {R}_{1}$ then ${s}_{\alpha }\left(\beta \right)=\beta -⟨{\alpha }^{\vee },\beta ⟩\alpha \in {V}_{1}\cap R={R}_{1}.$ So are root systems. (iii)⇒(i): Let be the Weyl groups of respectively. Then $W\cong {W}_{1}×{W}_{2}$ (direct product). ${V}_{1}$ is stable under ${W}_{1}$ and fixed under ${W}_{2},$ hence stable under $W.$ Likewise for ${V}_{2}.$ $\square$

These equivalent conditions mean that $R$ is reducible.

Cartan matrix

$R$ reduced, $B=\left\{{\alpha }_{1},...,{\alpha }_{r}\right\}$ a basis of $R.$ Let $aij = ⟨αi∨,αj⟩ ( 1≤i,j≤r ).$ The matrix $A=(aij)$ is called the Cartan matrix of $R.$ It is independent of the choice of basis $B,$ because (1.11) any other basis $B\prime$ is of the form $wB=\left\{w{\alpha }_{1},...,w{\alpha }_{r}\right\}$ for some $w\in W,$ and $⟨ (wαi)∨, wαj ⟩ = ⟨ w(αi∨), w(αj) ⟩ = ⟨αi∨,αj⟩$ because $w$ is an isometry.

The matrix $A$ satisfies the following conditions: $aij∈ℤ; aii=2; aij≤0 if i≠j; aij=0 ⇒ aji=0 (C)$ (${a}_{ij}\le 0$ because $⟨{\alpha }_{i},{\alpha }_{j}⟩\le 0$ if $i\ne j$ (1.8)).

Example. For ${A}_{n-1},$ $A$ is the matrix $2 -1 -1 ⋱ ⋱ ⋱ ⋱ -1 -1 2 ;$ for ${G}_{2},$ $A= 2 -1 -3 2$ $\mathrm{det}A$ is a positive integer.

The Cartan matrix determines the reduced root system up to isomorphism. More precisely:

Let $R\prime$ be another root system spanning a vector space $V\prime ,$ and let $B\prime$ be a basis of $R\prime .$ Suppose $\phi B\to B\prime$ is a bijection such that $⟨ φ(αi)∨, φ(αj) ⟩ = ⟨αi∨,αj⟩ ( 1≤i,j≤r ).$ Then there is a unique isomorphism $f:V\to V\prime$ such that $f\left(R\right)=R\prime$ and $f{|}_{B}=\phi$ (i.e. $f$ extends $\phi$).

 Proof. Since $B$ (resp. $B\prime$) is a basis of $V$ (resp. $V\prime$) there is just one isomorphism $f:V\to V\prime$ that extends $\phi .$ We have to show that $f$ maps $R$ onto $R\prime .$ Let Then we have $si′f (αj) = si′φ (αj) = φ(αj) - ⟨ φ(αi)∨, φ(αj) ⟩ φ(αi) (1.1) = f(αj) - ⟨αi∨,αj⟩ f(αi) = f(si(αj))$ and therefore ${s}_{i}^{\prime }\circ f=f\circ {s}_{i},$ i.e. ${s}_{i}^{\prime }=f\circ {s}_{i}\circ {f}^{-1}\phantom{\rule{.5em}{0ex}}\left(1\le i\le r\right).$ $\square$

[Don't confuse with the Coxeter matrix if $i\ne j.$]

Dynkin diagram

This is a graph that encodes the information given by the Cartan matrix. Let $B$ be a basis of $R$ (reduced) as before and let $D$ be the following graph: vertex set = $B$ (so vertices labelled ${\alpha }_{1},...,{\alpha }_{r}$).
Edges: the vertices ${\alpha }_{i}$ and ${\alpha }_{j}$ are joined by $dij = aijaji = ⟨αi∨,αj⟩ ⟨αi,αj∨⟩$ edges (or bonds), together with an arrow pointing from ${\alpha }_{i}$ to ${\alpha }_{j}$ if $|{\alpha }_{i}|>|{\alpha }_{j}|$ and $⟨{\alpha }_{i},{\alpha }_{j}⟩\ne 0,$ i.e. if From (1.2) we see that ${d}_{ij}=0,1,2$ or $3.$ So if $|{\alpha }_{i}|\ge |{\alpha }_{j}|$ the possibilities are $aij aji dij αi αj θij mij 0 0 0 i j π2 2 -1 -1 1 i j 2π3 3 -1 -2 2 i j 3π4 4 -1 -3 3 i j 5π6 6$

Examples. Dynkin diagrams of ${A}_{n},{B}_{n},{C}_{n},{D}_{n},{G}_{2}.$

This table shows that the Cartan matrix $A$ can be reconstructed unambiguously from the Dynkin diagram $D,$ hence by (1.22) determines the reduced root system $R$ up to isomorphism.

[Coxeter diagram: describes the Weyl group $\begin{array}{c} i mij j \end{array}.\right]$

The beauty of this description of $R$ is that

$R$ is irreducible $⇔$ $D$ is connected.

 Proof. Suppose $R$ reducible: $R={R}_{1}\cup {R}_{2}.$ Let ${B}_{i}$ be a basis of then $B={B}_{1}\cup {B}_{2}$ is a basis of $R,$ and ${B}_{1}\perp {B}_{2}.$ Hence no vertex of ${D}_{1}$ (the diagram of ${R}_{1}$) is joined to any vertex of ${D}_{2},$ the diagram of ${R}_{2}.$ So $D={D}_{1}\cup {D}_{2}$ is not connected. Conversely, suppose $D$ is not connected. Then $B={B}_{1}\cup {B}_{2}$ with ${B}_{1}\perp {B}_{2}$ (and neither ${B}_{1}$ nor ${B}_{2}$ empty). Let ${V}_{i}$ be the subspace spanned by ${B}_{i}.$ Then $V={V}_{1}\oplus {V}_{2}$ (orthogonal direct sum) and each of ${V}_{1},{V}_{2}$ is $W-$stable and hence (Prop. 1.1) $R$ is reducible. $\square$

So to classify irreducible root systems it suffices to list all possible connected Dynkin diagrams. To achieve this we shall first classify another class of Dynkin diagrams, called affine Dynkin diagrams (because this classification is easier). To do this we need the notion of the highest root of an (irreducible, reduced) root system.

Highest root

$R$ (reduced), $B,C$ as usual.

Suppose $R$ irreducible. Then $R$ has a "highest root" $\phi$ such that $\phi \ge \alpha$ for all $\alpha \in R.$ We have

1. $\phi \in \stackrel{—}{C},$
2. $|\phi |\ge |\alpha |$ for all $\alpha \in R,$
3. $⟨{\phi }^{\vee },\alpha ⟩=0$ or $1$ for all
4. $\phi =\sum _{i=1}^{r}{m}_{i}{\alpha }_{i},$ each ${m}_{i}\ge 1.$

 Proof. Let $\phi$ be a maximal element of $R$ for the partial order $\ge .$ If $⟨\phi ,{\alpha }_{i}⟩<0$ for some $i$ then by (1.4) we should have either $\phi +{\alpha }_{i}\in R,$ contradicting the maximality of $\phi ,$ or $\phi =-{\alpha }_{i},$ contradicting $\phi \in {R}^{+}.$ So $⟨φ,αi⟩ ≥ 0 ( 1≤i≤r ). (*)$ Let $\phi =\sum _{i=1}^{r}{m}_{i}{\alpha }_{i}.$ Let Suppose $J\ne \varnothing ,$ then there exists $i\in I$ and $j\in J$ such that $⟨{\alpha }_{i},{\alpha }_{j}⟩<0$ (otherwise the Dynkin diagram $D$ of $R$ would be disconnected, contrary to (Prop. 3.1)). But then $⟨φ,αj⟩ = ∑i∈I mi ⟨αi,αj⟩ < 0$ (because each summand is $\le 0$ and at least one is $<0$) contradicting (*). So $J=\varnothing ,$ i.e., ${m}_{i}>0$ for $1\le i\le r.$ Now let $\phi \prime$ be another maximal element of $R.$ As above we have $⟨\phi \prime ,{\alpha }_{i}⟩\ge 0$ for all $i;$ moreover $⟨\phi \prime ,{\alpha }_{i}⟩>0$ for some $i$ (if $⟨\phi \prime ,{\alpha }_{i}⟩=0$ for all $i,$ then $⟨\phi \prime ,x⟩=0$ for all $x\in V,$ so that $\phi \prime =0$). Hence $⟨φ′,φ⟩ = ∑i=1r mi ⟨φ′,αi⟩ > 0.$ Suppose $\phi \prime \ne \phi .$ From (1.4) we infer that $\phi -\phi \prime \in R.$ Either $\phi -\phi \prime \in {R}^{+},$ in which case $\phi >\phi \prime ;$ or $\phi \prime -\phi \in {R}^{+},$ in which case $\phi \prime >\phi .$ Both of these are impossible, so that $\phi \prime =\phi .$ From (*) it follows that $\phi \in \stackrel{—}{C}.$ Let $\alpha \in R;$ since we may replace $\alpha$ by $w\alpha$ and hence (1.21) assume that $\alpha \in \stackrel{—}{C}$ (fundamental domain for $W$). We have $\phi \ge \alpha ,$ hence $⟨\phi -\alpha ,x⟩\ge 0$ for all $x\in \stackrel{—}{C}:$ in particular $⟨φ-α,φ⟩ ≥ 0, ⟨φ-α,α⟩ ≥ 0$ so that $⟨φ,φ⟩ ≥ ⟨φ,α⟩ ≥ ⟨α,α⟩.$ Hence $|\phi |\ge |\alpha |.$ Let $\alpha \in {R}^{+}.$ From (*) we have $⟨\phi ,\alpha ⟩\ge 0,$ and $|\phi |\ge |\alpha |$ just proved. By (1.2) it follows that $⟨{\phi }^{\vee },\alpha ⟩=0$ or $1$ if $\alpha \ne \phi .$ $\left(⟨{\phi }^{\vee },\alpha ⟩⟨\phi ,{\alpha }^{\vee }⟩=0,1,2,3\phantom{\rule{.5em}{0ex}}and\phantom{\rule{.5em}{0ex}}⟨{\phi }^{\vee },\alpha ⟩\le ⟨\phi ,{\alpha }^{\vee }⟩.\right)$ $\square$

Note that (from the proof) $φ = ∑i=1r miαi, (**)$ with each ${m}_{i}$ a positive integer.

Let ${\alpha }_{0}=-\phi$ be the lowest root of $R$ (relative to $B$). Then the relation (**) takes the form $∑i=0r miαi = 0$ where the ${m}_{i}$ are positive integers, and ${m}_{0}=1.$ We define the extended Cartan matrix (with $r+1$ rows and columns) $A— = (aij) 0≤i,j≤r (detA—=0)$ where ${a}_{ij}=⟨{\alpha }_{i}^{\vee },{\alpha }_{j}⟩$ as before, and the extended Dynkin diagram $\stackrel{—}{D}$ by the same rules as before. Each determines the other. $\stackrel{—}{D}$ is connected, because from the proof of (Prop. 4.1) we have $⟨\phi ,{\alpha }_{i}⟩$ positive and hence $⟨{\alpha }_{0},{\alpha }_{i}⟩$ negative for at least one $i\in \left[1,r\right].$

Examples.

1. $r=1:\phantom{\rule{.5em}{0ex}}R=\left\{±{\alpha }_{1}\right\},\phantom{\rule{.5em}{0ex}}B=\left\{{\alpha }_{1}\right\},\phantom{\rule{.5em}{0ex}}\phi ={\alpha }_{1},\phantom{\rule{.5em}{0ex}}{\alpha }_{0}=-{\alpha }_{1}.$ So we have $\stackrel{—}{A}=\left(\begin{array}{cc}2& -2\\ -2& 2\end{array}\right)$ and ${d}_{01}={a}_{01}{a}_{10}=4.$ So $\stackrel{—}{D}$ is $α0 α1$ (with no arrow, because $|{\alpha }_{0}|=|{\alpha }_{1}|$). In all other cases the bond strengths are $\le 3$ as in $D.$
2. (every positive root is ${\alpha }_{i}+\cdots +{\alpha }_{j}$ where $i\le j$), so ${\alpha }_{0}={e}_{n}-{e}_{1}.$ $α0 α1 α2 αn-2 αn-1 (necklace) D—$ $αn αn-1 αn-2 α3 α2 α1 α0 (n≥3)$ $α0 α1 α2 αn-2 αn-1 αn (n≥2)$ $α0 α1 α2 α3 αn-3 αn-2 αn-1 αn (n≥4)$ $α1 α2 α0$

Come back to the relation $∑j=0r mjαj = 0. (Cla 1)$ Taking scalar products with ${\alpha }_{i}^{\vee }$ we have $∑j=0r aijmj = 0 (aij = ⟨αi∨,αj⟩) (Cla 2)$ for each $i\in \left[0,r\right].$ Since ${a}_{ii}=2$ and ${a}_{ij}\le 0$ for $j\ne i$ we can rewrite this in the form $2mi = ∑j∈J |aij|mj (Cla 3)$ where $J=\left\{j\in \left[0,r\right]\phantom{\rule{.5em}{0ex}}|\phantom{\rule{.5em}{0ex}}j\ne i\phantom{\rule{.5em}{0ex}}and\phantom{\rule{.5em}{0ex}}{a}_{ij}\ne 0\right\}$ gives the set of vertices of $\stackrel{—}{D}$ directly linked to the vertex ${\alpha }_{i}$ and $|aij| = 1 if |αi| ≥ |αj|, i.e. i j αi αj αi αj (Cla 4)$ (either no arrow, or arrow pointing away from ${\alpha }_{i}$) whereas $|aij| = dij = bond strength, if |αi| < |αj| αi αj αi αj (Cla 5)$ Suppose first no multiple bonds issuing from the vertex ${\alpha }_{i}.$ Then all are equal to 1, and (Cla 3) takes the form $2mi = ∑j∈J mj (Cla 3')$

Example.

Next suppose that there are multiple bonds issuing from the vertex ${\alpha }_{i}$ of $\stackrel{—}{D}.$ If the arrow points outwards, treat it as a single bond. If the arrow points inwards, replace $αi αj by αi αj by$

Examples. $1 2 2 1 2 3$

So the upshot is that the diagram $\stackrel{—}{D}$ can be labeled with positive integers ${m}_{i}$ in such a way that the conditions (Cla 3), (Cla 4) and (Cla 5) are satisfied. $\left(\stackrel{—}{D},m\right)$ is a labeled diagram (also called an affine Dynkin diagram). So each reduced irreducible root system gives rise to an affine Dynkin diagram.

We shall now temporarily forget about the root system $R$ that gave rise to $\stackrel{—}{A}$ and $\stackrel{—}{D},$ and define a generalized Cartan matrix to be any $n×n$ matrix of integers $𝒜 = (aij) 1≤i,j≤n$ satisfying the conditions $aii=2; aij≤0 if i≠j; aij=0 ⇒ aji=0; aijaji≤4. (Cla 6)$ We associate with $𝒜$ a diagram (or graph) $𝒟$ by the same rules as before: $𝒟$ has $n$ vertices numbered $1,2,...,n,$ and vertices are joined by edges, with an arrow pointing towards $j$ if ${a}_{ij}>{a}_{ji}$ (i.e. $|{a}_{ij}|<|{a}_{ji}|$). As before, each of determines the other.

The diagram $𝒟$ will be connected if and only if $𝒜$ is irreducible, i.e. if we cannot partition the index set $\left[1,n\right]$ into two proper subsets, $I,J$ such that

A generalized Cartan matrix $𝒜$ is of affine type if

1. $𝒜$ is irreducible,
2. There exists a vector $m=\left(\begin{array}{c}{m}_{1}\\ ⋮\\ {m}_{n}\end{array}\right)$ of positive integers such that $𝒜m=0$
(which implies $\mathrm{det}𝒜=0$). Correspondingly, the diagram $𝒟$ is of affine type if
1. $𝒟$ is connected,
2. $𝒟$ can be labeled according to the previous rules.
Finally, a subdiagram of $𝒟$ is any diagram obtained from $𝒟$ by either deleting some vertices (and the edges issuing from them) or deleting some edges (or both) (i.e. it is subgraph of $𝒟$). Then the basic fact is that

Let $𝒟$ be a diagram of affine type. Then no proper subdiagram of $𝒟$ is of affine type.

The moral of (Prop. 4.2) is that affine diagrams are rigid.

We shall use (Prop. 4.2) to classify all possible diagrams of affine type. We know that each extended diagram $\stackrel{—}{D}$ must occur in our list, and that $D$ (the diagram of $R$) is obtained from $\stackrel{—}{D}$ by deleting one vertex. So we shall also obtain a list of all possible diagrams $D.$

The diagrams $D$ do not satisfy the rigidity property (Prop. 4.2) (e.g. ${A}_{n}$ is a subdiagram of ${B}_{n},$ or of ${C}_{n};$ ${B}_{2}$ is a subdiagram of ${G}_{2}$). This is why the affine diagrams $𝒟$ are easier to classify.

The proof of (Prop. 4.2) rests on two lemmas about real matrices. We consider matrices $A=\left({a}_{ij}\right)\in {M}_{n}\left(ℝ\right)$ satisfying the condition $aij≤0 if i≠j, (Cla 7)$ i.e. the off-diagonal elements of $A$ are $\le 0.$ If the set $\left[1,n\right]$ can be partitioned into two nonempty subsets $I,J$ such that ${a}_{ij}=0$ for all $\left(i,j\right)\in I×J,$ we say that $A$ is reducible. $I * 0 J 0 * I J$ If $x={\left({x}_{1},...,{x}_{n}\right)}^{t}\in {ℝ}^{n}$ is a column vector, we shall write $x>0$ (resp. $x\le 0$) to mean that ${x}_{i}>0$ (resp. ${x}_{i}\le 0$) for all $i.$

Let $A$ be an irreducible matrix satisfying (Cla 7). Suppose that there exists $x\in {ℝ}^{n}$ such that $x>0$ and $Ax\ge 0.$ Then either $A$ is nonsingular or else $Ax=0.$

 Proof. Suppose $A$ is singular ($\mathrm{det}A=0$). Then there exists such that $Ay=0.$ We may assume (replacing $y$ by $-y$ if necessary) that some ${y}_{i}>0.$ Let $λ = max{ yixi | 1≤i≤n} > 0$ and put $z=\lambda x-y.$ Then $Az = A(λx-y) = λAx ≥ 0. (***)$ Also ${z}_{j}=\lambda {x}_{j}-{y}_{j}={x}_{j}\left(\lambda -\frac{{y}_{j}}{{x}_{j}}\right)$ is $\ge 0$ for all $j,$ and $=0$ for some $j.$ Let $I=\left\{i\in \left[1,n\right]\phantom{\rule{.5em}{0ex}}|\phantom{\rule{.5em}{0ex}}{z}_{i}=0\right\}.$ If $i\in I$ we have $(Az)i = ∑j∉I aijzj ≤ 0$ because the ${a}_{ij}$ are $\le 0$ and the ${z}_{j}$ are $>0.$ From (***) it follows that ${\left(Az\right)}_{i}=0$ for all $i\in I,$ i.e., $∑j∉I aijzj = 0 (i∈I)$ and hence ${a}_{ij}=0$ for all Since $A$ is irreducible and $I\ne \varnothing$ it follows that $I=\left[1,n\right],$ so that $z=0$ and hence from (***) $Ax = λ-1Az = 0.$ $\square$

If $A={\left({a}_{ij}\right)}_{1\le i,j\le n}$ as above, and $I\subseteq \left[1,n\right]$ let $Ai = (aij) i,j∈I .$ These are the principal submatrices of $A,$ and their determinants are the principal minors.

Let $A,B\in {M}_{n}\left(ℝ\right)$ be two irreducible matrices satisfying (Cla 7) (off-diagonal elements $\le 0$) and such that $A\le B$ (i.e., ${a}_{ij}\le {b}_{ij}$ for all $i,j$), $A\ne B.$ Suppose there exists $x\in {ℝ}^{n}$ such that $x>0$ and $Ax=0.$ Then all principal minors of $B$ (including $B$ itself) are nonzero.

 Proof. Consider first $B$ itself. We have $Bx=\left(B-A\right)x\ge 0$ (since $A\le B$) and $Bx\ne 0$ (because $A\ne B$). Hence $\mathrm{det}B\ne 0$ by Lemma 4.4. Now let $I$ be a proper nonempty subset of $\left[1,n\right].$ If ${B}_{I}$ is reducible, there are complementary subsets $J,K$ of $I$ such that ${b}_{jk}=0$ whenever $\left(j,k\right)\in J×K,$ hence $detBI = (detBJ) (detBK) J BJ 0 K * BK J K$ So we may assume ${B}_{I}$ irreducible. Let ${x}_{I}={\left({x}_{i}\right)}_{i\in I}.$ If $i\in I$ we have $(BIxI)i = ∑j∈I bijxj = ∑j=1n bijxj + ∑j∉I -bijxj$ which is $\ge 0$ (because $Bx\ge 0$). Hence ${B}_{I}{x}_{I}\ge 0.$ And if ${B}_{I}{x}_{I}=0$ we should have $∑j∈I bijxj = 0$ for all $i\in I,$ whence ${b}_{ij}=0$ for $i\in I$ and $j\notin I,$ and this is impossible because $B$ is irreducible. So we have $xI>0, BIxI≥0, BIxI≠0$ and hence $\mathrm{det}{B}_{I}\ne 0$ by Lemma 4.4. $\square$

 Proof of Proposition 4.2. Let $𝒟$ be a diagram of affine type, $𝒜$ the corresponding Cartan matrix, and let $ℰ$ be a proper subdiagram of $𝒟.$ Deleting edges from $𝒟$ corresponds to increasing off diagonal entries ${a}_{ij}$ of $𝒜$ (i.e. decreasing $|{a}_{ij}|$); deleting vertices of $𝒟$ corresponds to passing to a principal submatrix. Hence the Cartan matrix of $ℰ$ is of the form ${ℬ}_{I},$ where $ℬ\ge 𝒜$ (take $ℬ=𝒜$ outside $I$). By Lemma 4.5, ${ℬ}_{I}$ is nonsingular, hence $ℰ$ is not of affine type. $\square$

Classification of diagrams of affine type

We shall make use of Proposition 4.2 in the following form: if $𝒟$ is an affine diagram and $ℰ$ is an affine subdiagram of $𝒟,$ then $ℰ$ is the whole of $𝒟.$

Let $𝒟$ be an affine diagram.

1. Suppose $𝒟$ contains a cycle (with $\ge 3$ vertices). Then $𝒟$ has a subdiagram of the form $1 1 1 1 (An), (n+1 vertices, n≥2)$ which can be labeled, hence is affine, hence is the whole of $𝒟.$ So assume henceforth that $𝒟$ is a tree (no cycles).
2. Suppose $𝒟$ has a branch point of order $\ge 4$ (joined to 4 or more vertices). Then $𝒟$ has a subdiagram of the form $1 1 2 1 1 (D4)$ which is affine, hence is the whole of $𝒟.$
3. Suppose that $𝒟$ has 2 or more branch points of order 3. Then $𝒟$ has a subdiagram of the form $1 1 2 2 2 2 1 1 ( Dn, n≥5 )$
4. Suppose that $𝒟$ has one branch point of order 3 and at least one multiple bond. Then $𝒟$ has a subdiagram of one of the forms $2 2 2 2 2 1 1 2 2 2 2 2 1 1 (Bn) (Bn∨)$
5. Suppose that $𝒟$ has one branch point of order 3 and no multiple bonds. Let $p,q,r$ be the number of vertices on each branch (including the centre). Suppose labels at the ends of the branches are $a,b,c$ and that the label at the center is $m.$ Then $pa = qb=rc=m and 2m = (p-1)a + (q-1)b + (r-1)c = 3m- (a+b+c)$ so that $a+b+c=m$ and therefore $1p + 1q + 1r = 1,$ the solutions of which (with $p\le q\le r$) are $(p,q,r) = (3,3,3), (2,4,4), (2,3,6). (E6,E7,E8)$
6. We have now exhausted all the possibilities where $𝒟$ has a branch point. Assume now that $𝒟$ is a chain. If $𝒟$ has at least two multiple bonds it will contain one of the following diagrams, $1 2 2 2 2 2 BCn (n≥2) 1 2 2 2 2 2 Cn∨ (n≥2) 1 2 2 2 2 2 Cn (n≥2)$ each of which is labeled, hence affine. So $𝒟$ must be one of these.
7. Assume now that $𝒟$ is a chain with just one multiple bond.
1. strength 4: $\begin{array}{c} 1 1 1 2 \end{array}\phantom{\rule{2em}{0ex}}{A}_{1},\phantom{\rule{.5em}{0ex}}B{C}_{1}$
2. strength 3: $\begin{array}{c} 3 2 1 1 2 1 \end{array}\phantom{\rule{2em}{0ex}}{G}_{2},\phantom{\rule{.5em}{0ex}}{G}_{2}^{\vee }$
3. strength 2: the multiple bond can't be at an end (subdiagram of (6)). Can't contain properly either of $1 2 3 2 1 2 4 3 2 1 [fold E6] (F4∨) [fold E7] (F4)$ and can't be a subdiagram of either.
8. (No multiple bonds). Finally $\begin{array}{c} \end{array}$ is not affine, because it is a proper subdiagram of the necklace (1).
So our list of affine (i.e. labeled) diagrams is complete.

Finally, the Dynkin diagrams $D$ of irreducible reduced root systems must all occur as connected subdiagrams of these $𝒟,$ obtained by removing a vertex with label 1. Looking through our collection we obtain the list

$(classical) An (n≥1) Bn (n≥2) Cn (n≥3) C2=B2 Dn (n≥4) D3=A3, D2=A2 (exceptional) E6 E7 E8 F4 G2$ Of these we have already encountered ${A}_{n},{B}_{n},{C}_{n},{D}_{n}$ and ${G}_{2}.$ As to the other exceptional diagrams, we still have to establish that they do correspond to root systems.

Not all the affine diagrams $𝒟$ that we have encountered are of the form $\stackrel{—}{D}$ (extended Dynkin diagrams): about half of them are. The affine diagrams $𝒟$ classify (irreducible, reduced) affine root systems.

Affine root systems

$V$ as before a real vector space of finite dimension equipped with a positive definite inner product. An affine linear function on $V$ is a function of the form $f(x) = ⟨x,φ⟩ +c$ where $\phi \in V$ and $c\in ℝ;$ $\phi$ is the gradient $Df$ of $f$ (in the usual sense of elementary calculus). We may write $f = φ+c = Df+c.$ Let $F$ be the space of all these functions: it is a vector space of dimension $r+1,$ isomorphic to $V\oplus ℝ.$ For $f,g\in F$ we define $⟨f,g⟩ = ⟨Df,Dg⟩;$ this scalar product on $F$ is positive semidefinite (if $f\in F$ is constant, i.e. $Df=0,$ then clearly $⟨f,g⟩=0$ for all $g\in F$). (${F}_{0}$ the constant functions.) If $w:V\to V$ is an affine isometry then $w$ acts on $F:$ $(wf)(x) = f(w-1x). (f∈F, x∈V)$ In particular, let $f\in F$ be non-constant. Then ${H}_{f}=\left\{x\in V\phantom{\rule{.5em}{0ex}}|\phantom{\rule{.5em}{0ex}}f\left(x\right)=0\right\}$ is an affine hyperplane in $V.$ Let ${s}_{f}:V\to V$ be orthogonal reflection in this hyperplane. Then the action of ${s}_{f}$ on $F$ is given by $sf(g) = g-⟨f∨,g⟩f, where f∨ = 2f ⟨f,f⟩ ,$ just as in (1.1).

We can now define an affine root system. It is a subset $S$ of $F$ spanning $F$ and consisting of functions $\left(S\cap {F}_{0}=\varnothing \right)$ and satisfying the following axioms:

1. (AR1) If $a,b\in S$ then $⟨{a}^{\vee },b⟩\in S,$
2. (AR2) If $a,b\in S$ then ${s}_{a}\left(b\right)\in S.$
Let $W=⟨{s}_{a}\phantom{\rule{.5em}{0ex}}|\phantom{\rule{.5em}{0ex}}a\in S⟩,$ the Weyl group of $S:$ a group of isometries of $V.$
1. (AR3) $W$ (with the discrete topology) acts properly on $V.$
This means that the mapping $\mu :\left(w,x\right)↦w\left(x\right)$ of $W×V$ into $V$ is a proper map, which (since $V$ is locally compact) means that for each compact subset $K$ of $V,$ ${\mu }^{-1}\left(K\right)$ is compact. Since $W$ is discrete, this in turn means that
1. (AR3') Given any two compact subsets in $V,$ the set of $w\in W$ such that $wK$ meets $L$ is finite.
It follows that the set $\left\{{H}_{a}\phantom{\rule{.5em}{0ex}}|\phantom{\rule{.5em}{0ex}}a\in S\right\}$ of hyperplanes in $V$ is locally finite, i.e. each compact ($=$closed, bounded) subset of $V$ is intersected by only finitely many of them.

From (AR1)-(AR3) it is clear that if $S$ is an affine root system, so is $S∨ = {a∨ | a∈S}.$

Examples.

1. $R$ an irreducible reduced root system in $V.$ Then $S(R) = {α+k | α∈R, k∈ℤ}$ is an affine root system.
2. Hence so is $S{\left(R\right)}^{\vee }=\left\{{\alpha }^{\vee }+\frac{2k}{{|\alpha |}^{2}}\phantom{\rule{.5em}{0ex}}|\phantom{\rule{.5em}{0ex}}\alpha \in R,k\in ℤ\right\}$ $( because (α+k)∨ = 2(α+k) |α+k|2 = 2(α+k) |α|2 = α∨ + 2k |α|2 ).$ Warning: this is not necessarily isomorphic to $S\left({R}^{\vee }\right).$

In fact these two examples are almost exhaustive of the reduced irreducible affine root systems; there is just one other family constructed by a variant of Ex. 1.

All the main features of the finite theory (chambers, bases, Cartan matrix, Dynkin diagram) have their counterparts for affine root systems. The Dynkin diagrams are precisely the "affine" Dynkin diagrams that we have just been classifying. (In the table, the left hand column consists of the diagrams for reduced and irreducible; the right hand column consists of the others.

$reversing the arrows ↔ passing from S to S∨ ↔ transposing the Cartan matrix.$ (Affine root systems and Dedekind's $\eta -$function, Inv. Math. 14 (1972) 91-143.)
Affine (Kac-Moody) Lie algebras (see Kac's book).

The Weyl groups are affine Weyl groups (infinite discrete groups of isometries of Euclidean spaces generated by reflections). Draw pictures of

Construction of ${E}_{6},{E}_{7},{E}_{8}$

Let ${e}_{1},...,{e}_{9}$ be the standard basis of ${ℝ}^{9},$ with the usual inner product $\left(⟨{e}_{i},{e}_{j}⟩={\delta }_{ij}\right)$ and let $e = 13 (e1+⋯+e9)$ so that $|e|=1.$ Let $V = {x∈ℝ9 | ⟨x,e⟩=0}$ and let $p:{ℝ}^{9}\to V$ be the orthogonal projection, given by $p(x) = x-⟨x,e⟩e. (Cla 8)$ Let $L=p\left({ℤ}^{9}\right),$ a lattice in $V$ generated by the nine vectors $fi = p(ei) = ei-13e. (1≤i≤9) (Cla 9)$ We have ${f}_{1}+\cdots +{f}_{9}=0.$ Finally, let $R = {α∈L | |α|2=2} (vectors of minimal length).$

$R$ is a root system in $V$ of type ${E}_{8}.$

 Proof. First check the axioms. $R$ spans $V$ because e.g. ${e}_{i}-{e}_{j}\in R$ for $i\ne j.$ Since ${|\alpha |}^{2}=2$ for $\alpha \in R,$ we have ${\alpha }^{\vee }=\alpha$ and hence (R1) becomes $α,β∈R ⇒ ⟨α,β⟩ ∈ ℤ.$ Say where $x,y\in {ℤ}^{9}.$ Then by (Cla 8) we have $⟨α,β⟩ = ⟨ x-⟨x,e⟩e, y-⟨y,e⟩e ⟩ = ⟨x,y⟩ - ⟨x,e⟩ ⟨y,e⟩ , (ClaE8 1)$ and in particular $\left(\alpha =\beta \right)$ $|x|2 - ⟨x,e⟩2 = |α|2 = 2$ so that $⟨x,e⟩2 = |x|2-2. (ClaE8 2)$ Hence $⟨x,e⟩$ is a rational number whose square is an integer, hence $⟨x,e⟩\in ℤ$ and therefore (ClaE8 1) shows that $⟨\alpha ,\beta ⟩\in ℤ.$ Next, if $\alpha ,\beta \in R$ we have $sα(β) = β-⟨α,β⟩α ∈ L$ and ${|{s}_{\alpha }\left(\beta \right)|}^{2}={|\beta |}^{2}=2,$ hence ${s}_{\alpha }\left(\beta \right)\in R,$ proving that (R2) holds. So $R$ is a root system in $V,$ clearly reduced. Next let us list the elements of $R.$ Suppose $\alpha =p\left(x\right)\in R\phantom{\rule{.5em}{0ex}}\left(x\in {ℤ}^{9}\right).$ Then also $\alpha =p\left(x+3e\right)$ and $x+3e\in {ℤ}^{9}.$ Since $⟨x+3e,e⟩=⟨x,e⟩+3$ we may choose $x\in {p}^{-1}\left(\alpha \right)\cap {ℤ}^{9}$ so that $⟨x,e⟩=0$ or $±1.$ If $⟨x,e⟩=0$ it follows from (ClaE8 2) that ${|x|}^{2}=2,$ so that $∑i=19xi = 0, ∑i=19xi2 = 2,$ and hence $\alpha ={e}_{i}-{e}_{j}={f}_{i}-{f}_{j},$ where $1\le i,j\le 0$ and $i\ne j.$ If $⟨x,e⟩=±1$ we obtain likewise from (ClaE8 2) that ${|x|}^{2}=3,$ so that $∑i=19xi = ±3, ∑i=19xi2 = 3,$ and hence and $\alpha =p\left(x\right)=±\left({e}_{i}+{e}_{j}+{e}_{k}-e\right)=±\left({f}_{i}+{f}_{j}+{f}_{k}\right).$ To recapitulate, the elements of $R$ are $±(fi-fj) (i and therefore $Card(R) = 2( (92) + (93) ) = 2(36+84) = 240.$ Let $αi = fi-fi+1 (1≤i≤7), α8 = f6+f7+f8.$ I claim that $B=\left\{{\alpha }_{1},...,{\alpha }_{8}\right\}$ is a basis of $R.$ This will complete the proof, since the Dynkin diagram is $α7 α6 α5 α4 α3 α2 α1 α8$ $\left(use\phantom{\rule{.5em}{0ex}}⟨{f}_{i},{f}_{j}⟩=⟨{e}_{i}-\frac{1}{3}e,{e}_{j}-\frac{1}{3}e⟩={\delta }_{ij}-\frac{2}{9}+\frac{1}{9}={\delta }_{ij}-\frac{1}{9}.\right)$ Clearly and are in ${R}^{+}$ as defined by $B.$ So are $fi-f9 = fi+ ∑i=18fj (1≤i≤8)$ because this is the sum of three terms of the form Finally, $-(fi+fj+f9) = ∑i=19fk - (fi+fj+f9)$ is the sum of six of ${f}_{1},...,{f}_{8}.$ So these are the positive roots, and $B$ is indeed a basis of $R.$ $\square$

To obtain ${E}_{7}$ (resp. ${E}_{6}$) delete ${\alpha }_{1}$ (resp. ${\alpha }_{1},{\alpha }_{2}$).

1. For ${E}_{7}$ the positive roots are So 63 positive roots, 126 roots altogether.
2. For ${E}_{6}$ the positive roots are So 36 positive roots, 72 roots altogether.

Alternate description of ${E}_{8}$

This time we work in ${ℝ}^{8},$ with standard basis ${e}_{1},...,{e}_{8}.$ Let $L0 = {x∈ℤ8 | |x|2 even }.$ So $x=\sum _{i=1}^{8}{x}_{i}{e}_{i}$ is in and $\sum {x}_{i}$ is even (because ). Let $e = 12 (e1+⋯+e8) and L = L0+ℤe.$ Then $R=\left\{\alpha \in L\phantom{\rule{.5em}{0ex}}|\phantom{\rule{.5em}{0ex}}{|\alpha |}^{2}=2\right\}$ is a root system of type ${E}_{8}.$ In this description the elements of $R$ are $±ei±ej (1≤i So $|R|=240.$ The Dynkin diagram is $α7 α6 α5 α4 α3 α2 α1 α8$ with $αi = ei-ei+1 (1≤i≤6), α7 = e7+e8-e, α8 = e6+e7 .$

Construction of ${F}_{4}$

Let $L$ be the lattice in ${ℝ}^{4}$ generated by ${e}_{1},...,{e}_{4}$ and $e=\frac{1}{2}\left({e}_{1}+\cdots +{e}_{4}\right).$ So $x=\left({x}_{1},...,{x}_{4}\right)\in {ℝ}^{4}$ belongs to $L$ if and only if the ${x}_{i}$ are all integers or all half-integers. Let $R = {α∈L | |α|2=1 or 2}.$

$R$ is a root system of type ${F}_{4}.$

 Proof. If $\alpha \in R$ the coordinates of $\alpha$ can take only the values $0,±\frac{1}{2},±1$ (otherwise ${|\alpha |}^{2}\ge {\left(\frac{3}{2}\right)}^{2}=\frac{9}{4}>2$). So $R$ consists only of the vectors $±ei, ±ei±ej (1≤i So $\mathrm{Card}\left(R\right)=8+4\cdot \left(\genfrac{}{}{0}{}{4}{2}\right)+{2}^{4}=8+24+16=48.$ Check (R1), (R2). Let Then $B=\left\{{\alpha }_{1},...,{\alpha }_{4}\right\}$ is a basis of $R.$ The positive roots are $ei, ei±ej (i and the Dynkin diagram is $α1 α2 α3 α4$ (highest root ${e}_{1}+{e}_{2}$). $\square$

Non-reduced root systems

We have already encountered $B{C}_{n}={B}_{n}\cup {C}_{n}\phantom{\rule{.5em}{0ex}}\left(n\ge 1\right).$ In fact these are the only non-reduced irreducible root systems. I shall omit the proof, which is not difficult but not particularly instructive (Bourbaki, Groupes et Algèbres de Lie, Ch. VI §1, Props. 13 and 14 (p. 152)).

( ${R}_{\mathrm{ind}}=\left\{\alpha \in R\phantom{\rule{.5em}{0ex}}|\phantom{\rule{.5em}{0ex}}\frac{1}{2}\alpha \notin R\right\}.$ Show ${R}_{\mathrm{ind}}$ is reduced, 2 root lengths elements of $R,$ pairwise orthogonal ⇒ ${R}_{\mathrm{ind}}$ type ${B}_{n},$ hence $R=B{C}_{n}.$ )

Coxeter groups and Weyl groups

A Coxeter group $G$ is any group with generators say ${s}_{1},...,{s}_{r}$ and relations $si2 = 1, (sisj)mij = 1 if i≠j$ where ${m}_{ij}={m}_{ji}$ is an integer $\ge 2,$ or $+\infty .$ Graphically it may be described by a Coxeter diagram, which is a graph with one vertex for each generator ${s}_{i},$ the vertices being joined as follows: $mij ij 2 3 ≥4 mij$ So it is not very different from a Dynkin diagram. $G$ irreducible ⇔ diagram connected.

Classification of Coxeter groups

1. finite, irreducible. We have seen that Weyl groups are Coxeter groups. To convert the Dynkin diagram of $R$ into the Coxeter diagram of $W,$ we have only to replace PICTURE by PICTURE and $\begin{array}{c} i j \end{array}$ by $\begin{array}{c} 4 \end{array}$ and $\begin{array}{c} i j \end{array}$ by $\begin{array}{c} 6 \end{array}$. This however does not exhaust the list of finite irreducible Coxeter groups: there are also the dihedral groups (order $2m$) $m$ for all $m\ge 3$ (which are Weyl groups only for $m=3,4,6$ (${A}_{2},{B}_{2},{G}_{2}$ respectively)) and two more with diagrams $5 5 H3 H4$ ${H}_{3}$ has order 120, ${H}_{4}$ has order 1440. ${H}_{3}$ is the symmetry group of an icosahedron: it contains the alternating group ${A}_{5}$ as a subgroup of index 2, but it is not isomorphic to ${S}_{5}.$ ${H}_{4}$ likewise is the symmetry group of a 4-dimensional regular polytope.
2. affine, i.e. having a faithful representation in which the ${s}_{i}$ are reflections in hyperplanes in Euclidean space. The irreducible affine Coxeter groups are precisely the "affine Weyl groups", in one to one correspondence with the finite (reduced, irreducible) root systems. So their Coxeter diagrams are obtained from the affine Dynkin diagrams (left hand column) by replacing $by 4 by 6 α0 α1 by ∞ (infinite dihedral group).$

Geometric representation of a Coxeter group

$V={ℝ}^{r},$ standard basis ${e}_{1},...,{e}_{r}:$ define a scalar product by $⟨ei,ej⟩ = -cos π mij (mii=1)$ $\left(\frac{\pi }{\infty }=0\right).$ We have $⟨{e}_{i},{e}_{i}⟩=1$ and $⟨{e}_{i},{e}_{j}⟩\le 0$ if $i\ne j.$

Define ${\sigma }_{i}:V\to V$ by $σi(x) = x-2⟨x,ei⟩ei,$ and let $\Gamma$ be the group of automorphisms of $V$ generated by the ${\sigma }_{i}.$ Then $\Gamma$ is isomorphic to $G$ (the isomorphism being ). (Proof omitted.) $G is finite ⇔ the scalar product ⟨x,y⟩ is positive definite ⇔ the matrix M = (-cos π mij ) is positive definite, and G is affine ⇔ ⟨x,y⟩ is positive semidefinite (but not positive definite) ⇔ M is positive semidefinite (but not positive definite).$

The affine Weyl group

Let $R$ be a reduced irreducible root system, spanning $V.$ For each $\alpha \in R$ and $k\in ℤ$ let $Hα,k = {x∈V | ⟨x,α⟩=k}$ (so that ${H}_{\alpha ,k}={H}_{-\alpha ,-k}$ and ${H}_{\alpha ,0}={H}_{\alpha }$). Let ${s}_{\alpha ,k}$ be the orthogonal reflection in the affine hyperplane ${H}_{\alpha ,k}.$

If $\lambda \in V$ we denote by ${t}_{\lambda }$ the translation $x↦x+\lambda .$

(A1)

1. ${s}_{\alpha ,k}\left(x\right)=x-\left(⟨x,\alpha ⟩-k\right){\alpha }^{\vee }.$
2. ${s}_{\alpha ,k}={t}_{k{\alpha }^{\vee }}{s}_{\alpha }.$
3. Let $w\in W.$ Then $w{H}_{\alpha ,k}={H}_{w\alpha ,k}.$
4. Let $\lambda \in {P}^{\vee }=P\left({R}^{\vee }\right).$ Then $tλ Hα,k = Hα,k+⟨λ,α⟩, tλsα,ktλ-1 = sα,k+⟨λ,α⟩.$

 Proof. Let $y={s}_{\alpha ,k}\left(x\right).$ Then $x-y=c{\alpha }^{\vee }$ where $c\in ℝ,$ so that $⟨x-y,α⟩=2c. (ClaA1 1)$ Also $\frac{1}{2}\left(x+y\right)\in {H}_{\alpha ,k},$ so that $⟨x+y,α⟩=2k. (ClaA1 2)$ Adding (ClaA1 1) and (ClaA1 2) gives $⟨x,\alpha ⟩=c+k,$ so that $y = x-cα∨ = x-(⟨x,α⟩-k)α∨.$ ${s}_{\alpha ,k}\left(x\right)=x-⟨x,\alpha ⟩{\alpha }^{\vee }+k{\alpha }^{\vee }={s}_{\alpha }\left(x\right)+k{\alpha }^{\vee }={t}_{k{\alpha }^{\vee }}{s}_{\alpha }\left(x\right).$ $w{s}_{\alpha ,k}{w}^{-1}=w{t}_{k{\alpha }^{\vee }}{w}^{-1}\cdot w{s}_{\alpha }{w}^{-1}={t}_{kw\left({\alpha }^{\vee }\right)}{s}_{w\left(\alpha \right)}={s}_{w\alpha ,k}.$ $tλ sα,k tλ-1 = tλsα,k(x-λ) = tλ( (x-λ)- ( ⟨x-λ,α∨⟩-k )α∨ ) = x-( ⟨x,α∨⟩-( ⟨λ,α∨⟩+k ) )α∨ = sα,k+⟨λ,α∨⟩(x).$ $\square$

Let ${W}_{a}$ be the group of affine isometries of $V$ generated by the reflections ${s}_{\alpha ,k}.$ ${W}_{a}$ is the affine Weyl group of $R.$ Let ${Q}^{\vee }=Q\left({R}^{\vee }\right)$ and ${P}^{\vee }=P\left({R}^{\vee }\right).$

(A2) ${W}_{a}\cong {Q}^{\vee }⋊W$ (semidirect product).

 Proof. By definition, ${Q}^{\vee }⋊W$ is the group consisting of all pairs $\left(\lambda ,w\right)$ with multiplication $(λ,u) (μ,v) = (λ+uμ,uv).$ From (A1)(ii) we have ${t}_{{\alpha }^{\vee }}={s}_{\alpha ,1}{s}_{\alpha ,0}\in {W}_{a},$ hence ${t}_{\lambda }\in {W}_{a}$ for all $\lambda \in {Q}^{\vee }.$ Define $\phi :{Q}^{\vee }⋊W\to {W}_{a}$ by $\left(\lambda ,w\right)↦{t}_{\lambda }w.$ Claim that $\phi$ is an isomorphism. We have $φ(λ,u) φ(μ,v) = tλutμv = tλ⋅utμu-1⋅uv = tλ⋅tuμ⋅uv = tλ+uμ⋅uv = φ( λ+uμ,uv ),$ so that $\phi$ is a homomorphism. It is injective because ${t}_{\lambda }u=1$ implies $u\left(x\right)+\lambda =x$ for all $x\in V,$ hence and so $u=1.$ Also $\phi$ is surjective because (A1)(ii) ${s}_{\alpha ,k}=\phi \left(k{\alpha }^{\vee },{s}_{\alpha }\right).$ $\square$

Let ${\stackrel{^}{W}}_{a}\cong {P}^{\vee }⋊W$ be the group of all ${t}_{\lambda }w$ with By (A1)(iv), ${\stackrel{^}{W}}_{a}$ induces an isomorphism $P∨/Q∨ ≅ W^a/Wa.$ So ${W}_{a}$ has finite index $f$ in ${\stackrel{^}{W}}_{a}.$

Alcoves

Let $U=V-\bigcup _{\alpha ,k}{H}_{\alpha ,k}.$ $U$ is an open subset of $V$ and its connected component are therefore also open: they are called the alcoves of ${W}_{a}.$ Since by (A1) ${\stackrel{^}{W}}_{a}$ permutes the hyperplanes ${H}_{\alpha ,k},$ it acts on $U$ as a group of homomorphisms, hence permutes the alcoves.

Fix a basis $B=\left\{{\alpha }_{1},...,{\alpha }_{r}\right\}$ of $R$ and let $A = {x∈V | ⟨x,α⟩∈(0,1) for all α∈R+} .$ Let $\phi \in {R}^{+}$ be the highest root.

(A3)

1. $A$ is an alcove.
2. Let $x\in V.$ Then $x\in A$ if and only if and $⟨x,\phi ⟩<1.$

 Proof. Certainly $\varnothing \ne A\subseteq U;$ also $A$ is convex (if $x,y\in A$ then the line segment $\left(xy\right)$ is contained in $A$), hence connected. Suppose Then for some $\alpha \in {R}^{+}$ either $⟨x,\alpha ⟩<0$ or $⟨x,\alpha ⟩>1,$ so that $x$ is separated from $A$ by one of the hyperplanes It follows that $A$ is a connected component of $U,$ i.e. an alcove. If $x\in A,$ clearly and $⟨x,\phi ⟩<1.$ Conversely if $x\in V$ satisfies these inequalities, then $⟨x,\alpha ⟩>0$ for all $\alpha \in {R}^{+};$ also $\phi \ge \alpha$ for all $\alpha \in {R}^{+},$ hence $\phi -\alpha$ is a sum of simple roots and therefore $⟨x,\phi -\alpha ⟩\ge 0,$ i.e. $⟨x,\alpha ⟩\le ⟨x,\phi ⟩<1.$ $\square$

From (A3)(ii) it follows that $A$ is an open Euclidean $r-$simplex with faces where $Hi = Hαi (1≤i≤r), H0 = Hφ,1.$ These $r+1$ hyperplanes are the walls of $A.$ Let ${s}_{i}=$ reflection in ${H}_{i}$ $\left(0\le i\le r\right).$

(A4)

1. ${W}_{a}$ permules the alcoves transitively.
2. ${W}_{a}$ is generated by ${s}_{0},{s}_{1},...,{s}_{r}.$

 Proof. Let $W\prime$ be the subgroup of ${W}_{a}$ generated by the We shall show that $W\prime$ permutes the alcoves transitively. So let $A\prime$ be an alcove and let The orbit ${W}_{a}x$ of $x$ is a discrete subset of $V$ because by (A2) it consists of the points hence so is $W\prime x\subseteq {W}_{a}x.$ Choose $z=wx$ in this orbit at minimum distance from $y.$ If $z\notin A$ then for some $i\in \left[0,r\right]$ $y$ and $z$ are on opposite sides of the hyperplane ${H}_{i}.$ But then (see fig.) $y$ is nearer to ${s}_{i}z$ than to $z,$ contradiction. Hence $z\in A;$ but also $z\in wA\prime ,$ hence $A=wA\prime .$ So each alcove $A\prime$ has walls To complete the proof it is enough to show that ${s}_{\alpha ,k}\in W\prime$ for each Choose $x\in U$ closer to ${H}_{\alpha ,k}$ than any other reflecting hyperplane ${H}_{\beta ,l}$ and let $A\prime$ be the alcove containing $x.$ Then $H\alpha ,k$ is a wall of $A\prime ,$ and $A\prime ={w}^{-1}A,$ some $w\in W\prime .$ Hence ${H}_{\alpha ,k}={w}^{-1}{H}_{i}$ for some $i$ and therefore (A1) ${s}_{\alpha ,k}={w}^{-1}{s}_{i}w\in W\prime .$ $\square$

Let $ℋ$ denote the set of hyperplanes For each $w\in {W}_{a}$ let $L(w) = {H∈ℋ | H separates A and wA}.$ By (A4)(ii) we can write $w = sa1 ⋯ sap$ where each ${a}_{i}=0,1,...,r.$ Call this a reduced expression for $w$ if $p$ is as small as possible, and define the length of $w$ to be $l\left(w\right)=p.$

(A5) Let $w={s}_{{a}_{1}}\cdots {s}_{{a}_{p}}\in {W}_{a}$ be a reduced expression. Then $L(w) = {sa1⋯sai-1Hai (1≤i≤p)}$ and these $p$ hyperplanes are all distinct. Hence $l\left(w\right)=\mathrm{Card}L\left(w\right).$

 Proof. We proceed by induction on $p=l\left(w\right).$ If $p=0$ then $w=1$ and $L\left(w\right)$ is empty. Assume (A5) true for $w$ of length $p\ge 0$ and consider ${s}_{i}w$ of length $p+1.$ Then we have to show that $L\left({s}_{i}w\right)={H}_{i}\cup {s}_{i}L\left(w\right),$ ${H}_{i}\notin {s}_{i}L\left(w\right).$ Consider (b) first. If ${H}_{i}\in {s}_{i}L\left(w\right)$ then ${H}_{i}={s}_{i}{H}_{i}\in L\left(w\right),$ hence ${H}_{i}={s}_{{a}_{1}}\cdots {s}_{{a}_{i-1}}{H}_{{a}_{i}}$ for some $i\in \left[1,p\right].$ From (A1) it follows that ${s}_{i}={s}_{{a}_{1}}\cdots {s}_{{a}_{i-1}}{s}_{{a}_{i}}{\left({s}_{{a}_{1}}\cdots {s}_{{a}_{i-1}}\right)}^{-1}$ and therefore that $siw = ( sa1 ⋯ sai-1 sai )( sai ⋯ sap ) = sa1 ⋯ s^ai ⋯ sap$ has length $ contradiction. Now consider (a). From above, ${H}_{i}\notin L\left(w\right),$ i.e. ${H}_{i}$ does not separate $A$ and $wA,$ hence does not separate ${s}_{i}A$ and ${s}_{i}wA.$ But ${H}_{i}$ separates $A$ and ${s}_{i}A,$ hence separates $A$ and ${s}_{i}wA.$ So ${H}_{i}\in L\left({s}_{i}w\right).$ Now let We have $H∈L(siw) ⇔ H separates A and siwA ⇔ siH separates siA and wA ⇔ siH separates A and wA ⇔ siH ∈ L(w) ⇔ H∈siL(w),$ (because ${H}_{i}$ is the only hyperplane separating $A$ and ${s}_{i}A,$ hence $H$ does not separate $A$ and ${s}_{i}A,$ hence ${s}_{i}H$ does not separate ${s}_{i}A$ and $A$). $\square$

From (A5), just as in Chapter I (Root Systems), we may deduce the exchange Lemma (1.8) for ${W}_{a},$ and hence that ${W}_{a}$ is a Coxeter group with generators and relations ${\left({s}_{i}{s}_{j}\right)}^{{m}_{ij}}=1$ where ${m}_{ij}$ is the order (possibly infinite) of ${s}_{i}{s}_{j}$ in ${W}_{a}.$

(A6)

1. ${W}_{a}$ permutes the alcoves simply transitively.
2. $\stackrel{—}{A}$ is a fundamental domain for the action of ${W}_{a}$ on $V.$

 Proof. If then $l\left(w\right)\ne 0$ and hence $L\left(w\right)$ is not empty, by (A5). Hence there is at least one hyperplane ${H}_{\alpha ,k}$ separating $A$ and $wA.$ Hence $wA\ne A.$ Let $x\in V.$ Then $x\in \stackrel{—}{A\prime }$ for some alcove $A\prime ={w}^{-1}A,$ hence $wx\in \stackrel{—}{wA\prime }=\stackrel{—}{A}.$ So $\stackrel{—}{A}$ meets every ${W}_{a}-$orbit in $V$ and it remains to show that it meets it exactly once. So suppose $x$ and $y=wx$ are both in $\stackrel{—}{A}.$ We must show that $y=x.$ If $w=1$ this is clear, so we may proceed by induction on $l\left(w\right).$ If $l\left(w\right)\ge 1$ there exists $i$ such that $l\left({s}_{i}w\right) hence (A5) ${H}_{i}\in L\left(w\right),$ i.e. ${H}_{i}$ separates $A$ and $wA:$ say where ${E}^{+},{E}^{-}$ are the two open half spaces with boundary ${H}_{i}.$ But then $y ∈ A—∩wA— ⊆ E+—∩E-— = Hi$ so that ${s}_{i}wx={s}_{i}y=y.$ Since $l\left({s}_{i}w\right) we conclude that $y=x.$ $\square$

Minuscule weights

Let $\left({\theta }_{1},...,{\theta }_{r}\right)$ be the basis of $V$ dual to $\left({\alpha }_{1},...,{\alpha }_{r}\right):$ $⟨{\theta }_{i},{\alpha }_{j}⟩={\delta }_{ij}.$ Then ${\theta }_{1},...,{\theta }_{r}$ is a basis of ${P}^{\vee }=P\left({R}^{\vee }\right).$ Also let $φ = ∑i=1r miαi$ be the highest root of $R,$ so that (Prop. 4.1) all the ${m}_{i}$ are $\ge 1.$

(A7) The elements of ${P}^{\vee }\cap \stackrel{—}{A}$ are 0 and the ${\theta }_{i}$ such that ${m}_{i}=1.$

 Proof. Suppose $\lambda \in {P}^{\vee },$ say Then $\lambda \in \stackrel{—}{A}$ if and only if i.e. ${a}_{i}\ge 0;$ and $⟨\lambda ,\phi ⟩\le 1,$ i.e. $\sum _{1}^{r}{m}_{i}{a}_{i}\le 1.$ So either all the ${a}_{i}$ are zero, i.e. $\lambda =0;$ or $\sum {m}_{i}{a}_{i}=1,$ which means that $\lambda ={\theta }_{i}$ for some $i$ such that ${m}_{i}=1.$ $\square$

For uniformity let us define Then (A7) says that the elements of ${P}^{\vee }\cap \stackrel{—}{A}$ are the Recall that ${\stackrel{^}{W}}_{a}$ is the group of all it contains ${W}_{a}$ as a normal subgroup of finite index, and $W^a/Wa ≅ P∨/Q∨$ of order $f.$

Let $G$ be the subgroup of $g\in {\stackrel{^}{W}}_{a}$ such that $gA=A.$ To describe the elements of $G,$ we introduce the following notation: let $Ri = { root system with basis αj: j≠i, if 1≤i≤r, R, if i=0, }$ and let ${w}_{i}$ be the longest element of If $i,j\in \left[1,r\right]$ and $j\ne i$ we have ${s}_{j}{\theta }_{i}={\theta }_{i}-⟨{\theta }_{i},{\alpha }_{j}⟩{\alpha }_{j}^{\vee }={\theta }_{i},$ hence $wiθi = θi (0≤i≤r).$

(A8) The elements of $G$ are $gi = tθiwiw0 (i∈J)$ so that $|G|=$ number of labels ${m}_{i}=1$ in the extended Dynkin diagram of $R.$

 Proof. If so certainly ${g}_{0}$ maps $A$ into $A.$ Now let let $x\in A$ and let $y={g}_{i}x={\theta }_{i}+{w}_{i}{w}_{0}x.$ We must check that $y\in A,$ i.e. that and $⟨y,\phi ⟩<1.$ Suppose first $j\ne i.$ Then $⟨y,αj⟩ = ⟨θi+wiw0x,αj⟩ = ⟨x,w0wiαj⟩ >0$ because ${w}_{i}{\alpha }_{j}\in {R}^{-}$ and hence ${w}_{0}{w}_{i}{\alpha }_{j}\in {R}^{+}.$ Next, consider $⟨y,αi⟩ = 1+⟨x,w0wiαi⟩.$ We have ${w}_{i}{\alpha }_{i}\le \phi ,$ hence ${w}_{0}{w}_{i}{\alpha }_{i}\ge {w}_{0}\phi =-\phi ,$ hence $1+⟨x,w0wiαi⟩ ≥ 1+⟨x,-φ⟩ = 1-⟨x,φ⟩ > 0.$ Finally, $⟨y,φ⟩ = ⟨θi,φ⟩ + ⟨wiw0x,φ⟩ = 1+⟨x,w0wiφ⟩.$ Now ${w}_{i}\phi \in {R}^{+},$ because the coefficient of ${\alpha }_{i}$ in ${w}_{i}\phi$ is $⟨wiφ,θi⟩ = ⟨φ,wiθi⟩ = ⟨φ,θi⟩ = mi = 1.$ Hence ${w}_{0}{w}_{i}\phi \in {R}^{-}$ and therefore $1+⟨x,w0wiφ⟩ < 1.$ We have therefore checked that the ${g}_{i}$ do belong to $G.$ Conversely, let $g\in G;$ say Then $g\left(\stackrel{—}{A}\right)=\stackrel{—}{A}$ so that $\lambda =g\left(0\right)\in \stackrel{—}{A}\cap {P}^{\vee },$ i.e. $\lambda ={\theta }_{i}$ for some $i\in J.$ So $g\left(0\right)={g}_{i}\left(0\right)$ and therefore $w={g}^{-1}{g}_{i}$ fixes 0, i.e. $w\in W$ and $wA=A.$ Since $A\subseteq C$ (the Weyl chamber) it follows that $A\subseteq C\cap wC,$ whence $w=1$ (because are disjoint if $w\ne 1,$ (1.21)). Hence $g={g}_{i}.$ $\square$

Now let $\stackrel{^}{w}\in {\stackrel{^}{W}}_{a}.$ Then $\stackrel{^}{w}A$ is an alcove, hence $\stackrel{^}{w}A=wA$ for a unique $w\in {W}_{a}$ by (A6), and hence ${w}^{-1}\stackrel{^}{w}A\subseteq A,$ so that ${w}^{-1}\stackrel{^}{w}\in G.$ So $\stackrel{^}{w}$ factors uniquely as Hence $P∨/Q∨ ≅ W^a/Wa ≅ G$ and under this isomorphism the element ${g}_{i}\in G$ corresponds to the coset of ${\theta }_{i}$ in ${P}^{\vee }/{Q}^{\vee }.$

(A9)

1. The are coset representatives of ${P}^{\vee }/{Q}^{\vee }.$
2. $f=\mathrm{Card}\left(J\right)=$ number of labels ${m}_{i}=1$ in the extended Dynkin diagram of $R.$

 Proof. Follows from the remarks above. $\mathrm{Card}\left(J\right)=|{P}^{\vee }/{Q}^{\vee }|=f$ by (4.3). $\square$

The are called the minuscule weights (or ${R}^{\vee }$).

(A10) Let $\lambda \in {P}^{\vee }.$ The $\lambda$ is minuscule if and only if $⟨\lambda ,{\alpha }^{\vee }⟩=0$ or $1$ for all $\alpha \in {R}^{+}.$

 Proof. ⇒) Let $\alpha \in {R}^{+},$ so that $\alpha =\sum _{i=1}^{r}⟨{\theta }_{i},\alpha ⟩{\alpha }_{i}.$ Since $0\le \alpha \le \phi$ we have $0\le ⟨{\theta }_{i},\alpha ⟩\le {m}_{i}$ for each $i.$ If now $\lambda \ne 0$ is minuscule, $\lambda ={\theta }_{i}$ for some $i$ such that ${m}_{i}=1,$ so that $⟨\lambda ,\alpha ⟩=0$ or $1.$ ⇐) We have $⟨\lambda ,{\alpha }_{i}⟩=0$ or $1$ for each $i,$ and If $⟨\lambda ,\phi ⟩=0$ then $⟨\lambda ,{\alpha }_{i}⟩=0$ for each $i,$ so that $\lambda =0.$ If on the other hand $⟨\lambda ,\phi ⟩=1,$ then for some $i\in \left[1,r\right]$ we have $⟨\lambda ,{\alpha }_{i}⟩={m}_{i}=1,$ and $⟨\lambda ,{\alpha }_{j}⟩=0$ for $j\ne i.$ So $\lambda ={\theta }_{i}$ is minuscule. $\square$

PICTURES!

PICTURES!