Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last update: 01 August 2012

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.


Let R be a root system in V (not necessarily reduced), R spanning V. Let V1,V2 be subspaces of V such that V = V1V2 and let Ri = RVi   (i=1,2).

In this situation the following are equivalent:

  1. V1,V2 are W-stable.
  2. RV1V2.
  3. V1V2 and R1 (resp. R2) is a root system in V1 (resp. V2), R = R1R2.

(i)⇒(ii): Let αR,  Hα = {xV | x,α=0} the hyperplane perpendicular to α. Since HαV, either V1 Hα or V2 Hα. Suppose V1 Hα, then there exists vV1 such that sαvv. We have (1.1) v-sαv = v,αα V1 (because v,sαv V1 ), hence αV1. So either αV1 or αV2, i.e., αV1V2.

(ii)⇒(iii): Let αR1,  βR2. Then α+βV1, otherwise β = (α+β)-α V1V2 = 0, impossible. Likewise α+βV2, hence α+βR. By (1.4) α,β 0. Replace β by -β and we see that α,β0. Hence α,β=0 and therefore R1R2. But R spans V, hence R1 (resp. R2) spans V1 (resp. V2). So V1V2.
Clearly R1, R2 each satisfy (R1) (integrality). As to (R2): if α,βR1 then sα(β) = β-α,βα V1R = R1. So R1, R2 are root systems.

(iii)⇒(i): Let W, W1, W2 be the Weyl groups of R, R1, R2 respectively. Then WW1×W2 (direct product). V1 is stable under W1 and fixed under W2, hence stable under W. Likewise for V2.

These equivalent conditions mean that R is reducible.

Cartan matrix

R reduced, B={α1,...,αr} a basis of R. Let aij = αi,αj ( 1i,jr ). 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 is of the form wB = { wα1,...,wαr } for some wW, and (wαi), wαj = w(αi), w(αj) = αi,αj because w is an isometry.

The matrix A satisfies the following conditions: aij; aii=2; aij0 if ij; aij=0 aji=0 (C) (aij0 because αi,αj0 if ij (1.8)).

Example. For An-1, A is the matrix 2 -1 -1 -1 -1 2 ; for G2, A= 2 -1 -3 2 detA is a positive integer.

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

Let R be another root system spanning a vector space V, and let B be a basis of R. Suppose φBB is a bijection such that φ(αi), φ(αj) = αi,αj ( 1i,jr ). Then there is a unique isomorphism f:VV such that f(R)=R and f|B=φ (i.e. f extends φ).

Since B (resp. B) is a basis of V (resp. V) there is just one isomorphism f:VV that extends φ. We have to show that f maps R onto R.

Let si = sφ(αi)   (1ir). Then we have sif (αj) = siφ (αj) = φ(αj) - φ(αi), φ(αj) φ(αi) (1.1) = f(αj) - αi,αj f(αi) = f(si(αj)) and therefore sif = fsi, i.e. si = fsif-1 (1ir).

[Don't confuse with the Coxeter matrix M=(mij),   mij =   order of   sisj: mii=1,   mij=2,3,4   or   6 if ij.]

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 α1,...,αr).
Edges: the vertices αi and αj are joined by dij = aijaji = αi,αj αi,αj edges (or bonds), together with an arrow pointing from αi to αj if |αi|>|αj| and αi,αj0, i.e. if aij>aji   (|aij|<|aji|). From (1.2) we see that dij=0,1,2 or 3. So if |αi| |αj| 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 An,Bn,Cn,Dn,G2.

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 i mij j .]

The beauty of this description of R is that

R is irreducible D is connected.

Suppose R reducible: R=R1R2. Let Bi be a basis of Ri   (i=1,2), then B=B1B2 is a basis of R, and B1B2 . Hence no vertex of D1 (the diagram of R1) is joined to any vertex of D2, the diagram of R2. So D=D1D2 is not connected.

Conversely, suppose D is not connected. Then B=B1B2 with B1B2 (and neither B1 nor B2 empty). Let Vi be the subspace spanned by Bi. Then V=V1V2 (orthogonal direct sum) and each of V1,V2 is W-stable and hence (Prop. 1.1) R is reducible.

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" φ such that φα for all αR. We have

  1. φC,
  2. |φ| |α| for all αR,
  3. φ,α = 0 or 1 for all αR+,  αφ.
  4. φ = i=1r miαi, each mi1.

Let φ be a maximal element of R for the partial order . If φ,αi<0 for some i then by (1.4) we should have either φ+αiR, contradicting the maximality of φ, or φ=-αi, contradicting φR+. So φ,αi 0 ( 1ir ). (*) Let φ = i=1r miαi. Let I = {i | mi>0},   J = {i | mi=0}. Suppose J, then there exists iI and jJ such that αi,αj < 0 (otherwise the Dynkin diagram D of R would be disconnected, contrary to (Prop. 3.1)). But then φ,αj = iI mi αi,αj < 0 (because each summand is 0 and at least one is <0) contradicting (*). So J=, i.e., mi>0 for 1ir.

Now let φ be another maximal element of R. As above we have φ,αi 0 for all i; moreover φ,αi > 0 for some i (if φ,αi = 0 for all i, then φ,x = 0 for all xV, so that φ=0). Hence φ,φ = i=1r mi φ,αi > 0. Suppose φφ. From (1.4) we infer that φ-φR. Either φ-φR+, in which case φ>φ; or φ-φR+, in which case φ>φ. Both of these are impossible, so that φ=φ.
  1. From (*) it follows that φC.
  2. Let αR; since |α| = |wα|   (wW) we may replace α by wα and hence (1.21) assume that αC (fundamental domain for W). We have φα, hence φ-α,x 0 for all xC: in particular φ-α,φ 0, φ-α,α 0 so that φ,φ φ,α α,α. Hence |φ| |α|.
  3. Let αR+. From (*) we have φ,α 0, and |φ| |α| just proved. By (1.2) it follows that φ,α = 0 or 1 if αφ. ( φ,α φ,α = 0,1,2,3 and φ,α φ,α. )

Note that (from the proof) φ = i=1r miαi, (**) with each mi a positive integer.

Let α0=-φ be the lowest root of R (relative to B). Then the relation (**) takes the form i=0r miαi = 0 where the mi are positive integers, and m0=1. We define the extended Cartan matrix (with r+1 rows and columns) A = (aij) 0i,jr (detA=0) where aij = αi,αj as before, and the extended Dynkin diagram D by the same rules as before. Each determines the other. D is connected, because from the proof of (Prop. 4.1) we have φ,αi positive and hence α0,αi negative for at least one i[1,r].


  1. r=1: R={±α1}, B={α1}, φ=α1, α0=-α1. So we have A = 2 -2 -2 2 and d01 = a01a10 =4. So D is α0 α1 (with no arrow, because |α0| = |α1| ). In all other cases the bond strengths dij   (ij) are 3 as in D.
  2. An-1   (n3): αi = ei-ei+1   (1in-1),   φ = e1-en = α1++αn-1 , (every positive root is αi++αj where ij), so α0 = en-e1. α0 α1 α2 αn-2 αn-1 (necklace) D Bn: αi = ei-ei+1   (1in-1),   αn=en,   φ = e1+e2,   α0=-e1-e2 αn αn-1 αn-2 α3 α2 α1 α0 (n3) Cn: αi = ei-ei+1   (1in-1),   αn=2en,   φ = 2e1,   α0=-2e1 α0 α1 α2 αn-2 αn-1 αn (n2) Dn: αi = ei-ei+1   (1in-1),   αn=en-1+en,   φ = e1+e2,   α0=-e1-e2 α0 α1 α2 α3 αn-3 αn-2 αn-1 αn (n4) G2: φ = 3α1+2α2,   α0 = -3α1-2α2 α1 α2 α0

Come back to the relation j=0r mjαj = 0. (Cla 1) Taking scalar products with αi we have j=0r aijmj = 0 (aij = αi,αj) (Cla 2) for each i[0,r]. Since aii=2 and aij0 for ji we can rewrite this in the form 2mi = jJ |aij|mj (Cla 3) where J = {j[0,r] | ji and aij0} gives the set of vertices of D directly linked to the vertex αi and |aij| = 1 if |αi| |αj|, i.e. i j αi αj αi αj (Cla 4) (either no arrow, or arrow pointing away from αi) whereas |aij| = dij = bond strength, if |αi| < |αj| αi αj αi αj (Cla 5) Suppose first no multiple bonds issuing from the vertex αi. Then all |aij|,  jJ are equal to 1, and (Cla 3) takes the form 2mi = jJ mj (Cla 3')


2 4 6 5 4 3 2 1 3

Next suppose that there are multiple bonds issuing from the vertex αi of 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 D can be labeled with positive integers mi in such a way that the conditions (Cla 3), (Cla 4) and (Cla 5) are satisfied. ( D,m ) 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 A and D, and define a generalized Cartan matrix to be any n×n matrix of integers 𝒜 = (aij) 1i,jn satisfying the conditions aii=2; aij0 if ij; aij=0 aji=0; aijaji4. (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 i,j   (ij) are joined by dij = aijaji   (4) edges, with an arrow pointing towards j if aij>aji (i.e. |aij| < |aji| ). As before, each of 𝒜, 𝒟 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 [1,n] into two proper subsets, I,J such that aij=0   for all   iI,  jJ I J I * 0 J 0 *

A generalized Cartan matrix 𝒜 is of affine type if

  1. 𝒜 is irreducible,
  2. There exists a vector m = m1 mn of positive integers such that 𝒜m=0
(which implies det𝒜=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 D must occur in our list, and that D (the diagram of R) is obtained from 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. An is a subdiagram of Bn, or of Cn; B2 is a subdiagram of G2). 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=(aij) Mn() satisfying the condition aij0 if ij, (Cla 7) i.e. the off-diagonal elements of A are 0. If the set [1,n] can be partitioned into two nonempty subsets I,J such that aij=0 for all (i,j) I×J, we say that A is reducible. I * 0 J 0 * I J If x = (x1,...,xn)t n is a column vector, we shall write x>0 (resp. x0) to mean that xi>0 (resp. xi0) for all i.

Let A be an irreducible matrix satisfying (Cla 7). Suppose that there exists xn such that x>0 and Ax0. Then either A is nonsingular or else Ax=0.

Suppose A is singular (detA=0). Then there exists yn,  y0 such that Ay=0. We may assume (replacing y by -y if necessary) that some yi>0. Let λ = max{ yixi | 1in} > 0 and put z = λx-y. Then Az = A(λx-y) = λAx 0. (***) Also zj = λxj-yj = xj( λ- yj xj ) is 0 for all j, and =0 for some j. Let I = {i[1,n] | zi=0}. If iI we have (Az)i = jI aijzj 0 because the aij are 0 and the zj are >0. From (***) it follows that (Az)i = 0 for all iI, i.e., jI aijzj = 0 (iI) and hence aij=0 for all iI,  jI. Since A is irreducible and I it follows that I=[1,n], so that z=0 and hence from (***) Ax = λ-1Az = 0.

If A = (aij) 1i,jn as above, and I[1,n] let Ai = (aij) i,jI . These are the principal submatrices of A, and their determinants are the principal minors.

Let A,B Mn() be two irreducible matrices satisfying (Cla 7) (off-diagonal elements 0) and such that AB (i.e., aij bij for all i,j), AB. Suppose there exists xn such that x>0 and Ax=0. Then all principal minors of B (including B itself) are nonzero.

Consider first B itself. We have Bx = (B-A)x 0 (since AB) and Bx0 (because AB). Hence detB0 by Lemma 4.4.

Now let I be a proper nonempty subset of [1,n]. If BI is reducible, there are complementary subsets J,K of I such that bjk=0 whenever (j,k) J×K, hence detBI = (detBJ) (detBK) J BJ 0 K * BK J K So we may assume BI irreducible.

Let xI = (xi) iI. If iI we have (BIxI)i = jI bijxj = j=1n bijxj + jI -bijxj which is 0 (because Bx0). Hence BIxI0. And if BIxI=0 we should have jI bijxj = 0 for all iI, whence bij=0 for iI and jI, and this is impossible because B is irreducible. So we have xI>0, BIxI0, BIxI0 and hence detBI0 by Lemma 4.4.

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

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 is an affine subdiagram of 𝒟, then is the whole of 𝒟.

Let 𝒟 be an affine diagram.

  1. Suppose 𝒟 contains a cycle (with 3 vertices). Then 𝒟 has a subdiagram of the form 1 1 1 1 (An), (n+1 vertices, n2) 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 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, n5 )
  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 pqr) 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 (n2) 1 2 2 2 2 2 Cn (n2) 1 2 2 2 2 2 Cn (n2) 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: 1 1 1 2 A1, BC1
    2. strength 3: 3 2 1 1 2 1 G2, G2
    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 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 (n1) Bn (n2) Cn (n3) C2=B2 Dn (n4) D3=A3, D2=A2 (exceptional) E6 E7 E8 F4 G2 Of these we have already encountered An,Bn,Cn,Dn and G2. 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 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 φV and c; φ 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. For f,gF we define f,g = Df,Dg; this scalar product on F is positive semidefinite (if fF is constant, i.e. Df=0, then clearly f,g = 0 for all gF). (F0 the constant functions.) If w:VV is an affine isometry then w acts on F: (wf)(x) = f(w-1x). (fF, xV) In particular, let fF be non-constant. Then Hf = {xV | f(x)=0} is an affine hyperplane in V. Let sf: VV be orthogonal reflection in this hyperplane. Then the action of sf on F is given by sf(g) = g-f,gf, 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 ( SF0= ) and satisfying the following axioms:

  1. (AR1) If a,bS then a,bS,
  2. (AR2) If a,bS then sa(b) S.
Let W = sa | aS, 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 μ: (w,x) w(x) of W×V into V is a proper map, which (since V is locally compact) means that for each compact subset K of V, μ-1(K) is compact. Since W is discrete, this in turn means that
  1. (AR3') Given any two compact subsets K, L in V, the set of wW such that wK meets L is finite.
It follows that the set {Ha | aS} 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 | aS}.


  1. R an irreducible reduced root system in V. Then S(R) = {α+k | αR, k} is an affine root system.
  2. Hence so is S(R) = {α+ 2k |α|2 | αR, k} ( 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(R).

In fact these two examples are almost exhaustive of the reduced irreducible affine root systems; there is just one other family S(BCn)   (n1) 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 S(R), R 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 η-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 A2, B2, G2.

Construction of E6,E7,E8

Let e1,...,e9 be the standard basis of 9, with the usual inner product ( ei,ej = δij ) and let e = 13 (e1++e9) so that |e|=1. Let V = {x9 | x,e=0} and let p:9V be the orthogonal projection, given by p(x) = x-x,ee. (Cla 8) Let L = p(9) , a lattice in V generated by the nine vectors fi = p(ei) = ei-13e. (1i9) (Cla 9) We have f1++f9 = 0. Finally, let R = {αL | |α|2=2} (vectors of minimal length).

R is a root system in V of type E8.

First check the axioms. R spans V because e.g. ei-ej R for ij. Since |α|2 = 2 for αR, we have α=α and hence (R1) becomes α,βR α,β . Say α=p(x) ,  β=p(y) where x,y9. Then by (Cla 8) we have α,β = x-x,ee, y-y,ee = x,y - x,e y,e , (ClaE8 1) and in particular (α=β) |x|2 - x,e2 = |α|2 = 2 so that x,e2 = |x|2-2. (ClaE8 2) Hence x,e is a rational number whose square is an integer, hence x,e and therefore (ClaE8 1) shows that α,β.

Next, if α,βR we have sα(β) = β-α,βα L and |sα(β)|2 = |β|2 = 2, hence sα(β) R, proving that (R2) holds. So R is a root system in V, clearly reduced.

Next let us list the elements of R. Suppose α=p(x) R ( x9 ). Then also α = p(x+3e) and x+3e 9. Since x+3e,e = x,e+3 we may choose xp-1(α) 9 so that x,e = 0 or ±1.
  1. If x,e=0 it follows from (ClaE8 2) that |x|2=2, so that i=19xi = 0, i=19xi2 = 2, and hence α=ei-ej = fi-fj, where 1i,j0 and ij.
  2. If x,e = ±1 we obtain likewise from (ClaE8 2) that |x|2 = 3, so that i=19xi = ±3, i=19xi2 = 3, and hence x = ±( ei+ej+ek ),   1i<j<k9, and α = p(x) = ±(ei+ej+ek-e) = ±(fi+fj+fk). To recapitulate, the elements of R are ±(fi-fj) (i<j), ±(fi+fj+fk) (i<j<k), and therefore Card(R) = 2( (92) + (93) ) = 2(36+84) = 240. Let αi = fi-fi+1 (1i7), α8 = f6+f7+f8. I claim that B = {α1,...,α8} is a basis of R. This will complete the proof, since the Dynkin diagram is α7 α6 α5 α4 α3 α2 α1 α8 ( use fi,fj = ei-13e, ej-13e = δij-29+19 = δij-19 . )
Clearly fi-fj   (1i<j8) and fi+fj+fk   (1i<j<k8) are in R+ as defined by B. So are fi-f9 = fi+ i=18fj (1i8) because this is the sum of three terms of the form fi+fj+fk   (1i<j<k8). Finally, -(fi+fj+f9) = i=19fk - (fi+fj+f9) is the sum of six of f1,...,f8. So these are the positive roots, and B is indeed a basis of R.

To obtain E7 (resp. E6) delete α1 (resp. α1,α2).

  1. For E7 the positive roots are fi-fj (2i8) (72) = 21   of these, fi+fj+fk (2i<j<k8) (73) = 35   of these, -(f1+fj+f9) (2j8) 7 of these. So 63 positive roots, 126 roots altogether.
  2. For E6 the positive roots are fi-fj (3i8) (62) = 15   of these, fi+fj+fk (3i<j<k8) (63) = 20   of these, -(f1+f2+f9) = f3++f8 1 of these. So 36 positive roots, 72 roots altogether.

Alternate description of E8

This time we work in 8, with standard basis e1,...,e8. Let L0 = {x8 | |x|2 even }. So x = i=18xiei is in L0    xi   (1i8) and xi is even (because xi2xi  mod 2). Let e = 12 (e1++e8) and L = L0+e. Then R = {αL | |α|2=2} is a root system of type E8. In this description the elements of R are ±ei±ej (1i<j8) 4(82) = 112 of these, 12( ±e1±±e8 ) with an even number of minus signs 27=128 of these. So |R|=240. The Dynkin diagram is α7 α6 α5 α4 α3 α2 α1 α8 with αi = ei-ei+1 (1i6), α7 = e7+e8-e, α8 = e6+e7 .

Construction of F4

Let L be the lattice in 4 generated by e1,...,e4 and e = 12 (e1++e4). So x = (x1,...,x4) 4 belongs to L if and only if the xi are all integers or all half-integers. Let R = {αL | |α|2=1 or 2}.

R is a root system of type F4.

If αR the coordinates of α can take only the values 0, ±12, ±1 (otherwise |α|2 (32)2 = 94 > 2 ). So R consists only of the vectors ±ei, ±ei±ej (1i<j4), 12( ±e1±e2±e3±e4 ). So Card(R) = 8+4( 42 ) + 24 = 8+24+16 = 48. Check (R1), (R2). Let α1 = e2-e3,   α2 = e3-e4,   α3 = e4,   α4 = e1-e = 12( e1-e2-e3-e4 ). Then B = {α1,...,α4} is a basis of R. The positive roots are ei, ei±ej (i<j), 12( e1±e2±e3±e4) (4+12+8=24) and the Dynkin diagram is α1 α2 α3 α4 (highest root e1+e2).

Non-reduced root systems

We have already encountered BCn = BnCn (n1). 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)).

( Rind = {αR | 12αR}. Show Rind is reduced, 2 root lengths =R1R2,   R = R1R22R1, elements of R, pairwise orthogonal ⇒ Rind type Bn, hence R = BCn. )

Coxeter groups and Weyl groups

A Coxeter group G is any group with generators say s1,...,sr and relations si2 = 1, (sisj)mij = 1 if ij where mij=mji is an integer 2, or +. Graphically it may be described by a Coxeter diagram, which is a graph with one vertex for each generator si, the vertices i,j   (ij) 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 i j by 4 and i j by 6 . This however does not exhaust the list of finite irreducible Coxeter groups: there are also the dihedral groups (order 2m) m for all m3 (which are Weyl groups only for m=3,4,6 (A2,B2,G2 respectively)) and two more with diagrams 5 5 H3 H4 H3 has order 120, H4 has order 1440. H3 is the symmetry group of an icosahedron: it contains the alternating group A5 as a subgroup of index 2, but it is not isomorphic to S5. H4 likewise is the symmetry group of a 4-dimensional regular polytope.
  2. affine, i.e. having a faithful representation in which the si 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, standard basis e1,...,er: define a scalar product by ei,ej = -cos π mij (mii=1) (π=0). We have ei,ei = 1 and ei,ej 0 if ij.

Define σi:VV by σi(x) = x-2x,eiei, and let Γ be the group of automorphisms of V generated by the σi. Then Γ is isomorphic to G (the isomorphism being σisi   (1ir) ). (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). (M=  Coxeter matrix)

The affine Weyl group

Let R be a reduced irreducible root system, spanning V. For each αR and k let Hα,k = {xV | x,α=k} (so that Hα,k = H-α,-k and Hα,0 = Hα ). Let sα,k be the orthogonal reflection in the affine hyperplane Hα,k.

If λV we denote by tλ the translation xx+λ.


  1. sα,k(x) = x-(x,α-k) α.
  2. sα,k = tkαsα.
  3. Let wW. Then wHα,k = Hwα,k.
  4. Let λP = P(R). Then tλ Hα,k = Hα,k+λ,α, tλsα,ktλ-1 = sα,k+λ,α.

  1. Let y = sα,k(x). Then x-y = cα where c, so that x-y,α=2c. (ClaA1 1) Also 12(x+y) Hα,k, so that x+y,α=2k. (ClaA1 2) Adding (ClaA1 1) and (ClaA1 2) gives x,α = c+k, so that y = x-cα = x-(x,α-k)α.
  2. sα,k(x) = x-x,αα+kα = sα(x)+kα = tkαsα(x).
  3. wsα,kw-1 = wtkαw-1 wsαw-1 = tkw(α)sw(α) = swα,k.
  4. 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).

Let Wa be the group of affine isometries of V generated by the reflections sα,k. Wa is the affine Weyl group of R. Let Q = Q(R) and P = P(R) .

(A2) Wa QW (semidirect product).

By definition, QW is the group consisting of all pairs (λ,w) with multiplication (λ,u) (μ,v) = (λ+uμ,uv). From (A1)(ii) we have tα = sα,1 sα,0 Wa, hence tλ Wa for all λQ. Define φ: QW Wa by (λ,w) tλw. Claim that φ is an isomorphism. We have φ(λ,u) φ(μ,v) = tλutμv = tλutμu-1uv = tλtuμuv = tλ+uμuv = φ( λ+uμ,uv ), so that φ is a homomorphism. It is injective because tλu=1 implies u(x)+λ = x for all xV, hence (x=0)   λ=0 and so u=1. Also φ is surjective because (A1)(ii) sα,k = φ( kα,sα ).

Let W^a PW be the group of all tλw with λP,  wW. By (A1)(iv), W^a induces an isomorphism P/Q W^a/Wa. So Wa has finite index f in W^a.


Let U = V-α,kHα,k. U is an open subset of V and its connected component are therefore also open: they are called the alcoves of Wa. Since by (A1) W^a permutes the hyperplanes Hα,k, it acts on U as a group of homomorphisms, hence permutes the alcoves.

Fix a basis B={α1,...,αr} of R and let A = {xV | x,α(0,1) for all αR+} . Let φR+ be the highest root.


  1. A is an alcove.
  2. Let xV. Then xA if and only if x,αi>0   (1ir) and x,φ <1.

  1. Certainly AU; also A is convex (if x,yA then the line segment (xy) is contained in A), hence connected. Suppose xU, xA. Then for some αR+ either x,α<0 or x,α >1, so that x is separated from A by one of the hyperplanes Hα,0, Hα,1. It follows that A is a connected component of U, i.e. an alcove.
  2. If xA, clearly x,αi>0   (1ir) and x,φ <1. Conversely if xV satisfies these inequalities, then x,α>0 for all αR+; also φα for all αR+, hence φ-α is a sum of simple roots and therefore x,φ-α 0, i.e. x,α x,φ < 1.

From (A3)(ii) it follows that A is an open Euclidean r-simplex with faces HiA   (0ir), where Hi = Hαi (1ir), H0 = Hφ,1. These r+1 hyperplanes are the walls of A. Let si= reflection in Hi (0ir).


  1. Wa permules the alcoves transitively.
  2. Wa is generated by s0,s1,...,sr.

Let W be the subgroup of Wa generated by the si (0ir). We shall show that W permutes the alcoves transitively. So let A be an alcove and let xA,  yA. The orbit Wax of x is a discrete subset of V because by (A2) it consists of the points λ+wx   ( λQ, wW ), hence so is WxWax. Choose z=wx in this orbit at minimum distance from y. If zA then for some i[0,r] y and z are on opposite sides of the hyperplane Hi. But then (see fig.) y is nearer to siz than to z, contradiction. Hence zA; but also zwA, hence A=wA. So each alcove A has walls w-1Hi   (0ir).

To complete the proof it is enough to show that sα,kW for each αR, k. Choose xU closer to Hα,k than any other reflecting hyperplane Hβ,l and let A be the alcove containing x. Then Hα,k is a wall of A, and A = w-1A, some wW. Hence Hα,k = w-1Hi for some i and therefore (A1) sα,k = w-1siw W.

Let denote the set of hyperplanes Hα,k   ( αR, k ). For each wWa let L(w) = {H | H separates A and wA}. By (A4)(ii) we can write w = sa1 sap where each ai = 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(w) = p.

(A5) Let w = sa1 sap Wa be a reduced expression. Then L(w) = {sa1sai-1Hai (1ip)} and these p hyperplanes are all distinct. Hence l(w) = CardL(w).

We proceed by induction on p=l(w). If p=0 then w=1 and L(w) is empty. Assume (A5) true for w of length p0 and consider siw of length p+1. Then we have to show that
  1. L(siw) = HisiL(w),
  2. Hi siL(w).
Consider (b) first. If HisiL(w) then Hi=siHi L(w), hence Hi = sa1sai-1Hai for some i[1,p]. From (A1) it follows that si = sa1 sai-1 sai ( sa1 sai-1 )-1 and therefore that siw = ( sa1 sai-1 sai )( sai sap ) = sa1 s^ai sap has length <p = l(w), contradiction.

Now consider (a). From above, HiL(w), i.e. Hi does not separate A and wA, hence does not separate siA and siwA. But Hi separates A and siA, hence separates A and siwA. So Hi L(siw).

Now let H,   HHi. We have HL(siw) H separates A and siwA siH separates siA and wA siH separates A and wA siH L(w) HsiL(w), (because Hi is the only hyperplane separating A and siA, hence H does not separate A and siA, hence siH does not separate siA and A).

From (A5), just as in Chapter I (Root Systems), we may deduce the exchange Lemma (1.8) for Wa, and hence that Wa is a Coxeter group with generators si   (0ir) and relations (sisj)mij = 1 where mij is the order (possibly infinite) of sisj in Wa.


  1. Wa permutes the alcoves simply transitively.
  2. A is a fundamental domain for the action of Wa on V.

  1. If wWa, w1 then l(w)0 and hence L(w) is not empty, by (A5). Hence there is at least one hyperplane Hα,k separating A and wA. Hence wAA.
  2. Let xV. Then xA for some alcove A = w-1A, hence wx wA = A. So A meets every Wa-orbit in V and it remains to show that it meets it exactly once. So suppose x and y=wx are both in A. We must show that y=x. If w=1 this is clear, so we may proceed by induction on l(w). If l(w) 1 there exists i such that l(siw) < l(w), hence (A5) HiL(w), i.e. Hi separates A and wA: say AE+,   wAE- where E+,E- are the two open half spaces with boundary Hi. But then y AwA E+E- = Hi so that siwx = siy = y. Since l(siw) < l(w) we conclude that y=x.

Minuscule weights

Let (θ1,...,θr) be the basis of V dual to (α1,...,αr): θi,αj = δij. Then θ1,...,θr is a basis of P = P(R). Also let φ = i=1r miαi be the highest root of R, so that (Prop. 4.1) all the mi are 1.

(A7) The elements of PA are 0 and the θi such that mi=1.

Suppose λP, say λ = i=1raiθi   (ai). Then λA if and only if λ,αi 0   (1ir), i.e. ai0; and λ,φ 1, i.e. 1r miai 1. So either all the ai are zero, i.e. λ=0; or miai = 1, which means that λ=θi for some i such that mi=1.

For uniformity let us define θ0=0,   m0=1,   J = {i[0,r] | mi=1}. Then (A7) says that the elements of PA are the θi, iJ. Recall that W^a is the group of all tλw   ( λP, wW ): it contains Wa as a normal subgroup of finite index, and W^a/Wa P/Q of order f.

Let G be the subgroup of gW^a such that gA=A. To describe the elements of G, we introduce the following notation: let Ri = { root system with basis αj: ji, if 1ir, R, if i=0, } and let wi be the longest element of Ri   (0ir). If i,j[1,r] and ji we have sjθi = θi-θi,αjαj = θi, hence wiθi = θi (0ir).

(A8) The elements of G are gi = tθiwiw0 (iJ) so that |G|= number of labels mi=1 in the extended Dynkin diagram of R.

If i=0, g0=t0w02, so certainly g0 maps A into A. Now let iJ,   i0; let xA and let y = gix = θi+wiw0x. We must check that yA, i.e. that y,αj>0   (1jr) and y,φ<1. Suppose first ji. Then y,αj = θi+wiw0x,αj = x,w0wiαj >0 because wiαj R- and hence w0wiαj R+.

Next, consider y,αi = 1+x,w0wiαi. We have wiαi φ, hence w0wiαi w0φ = -φ, hence 1+x,w0wiαi 1+x,-φ = 1-x,φ > 0. Finally, y,φ = θi,φ + wiw0x,φ = 1+x,w0wiφ. Now wiφ R+, because the coefficient of αi in wiφ is wiφ,θi = φ,wiθi = φ,θi = mi = 1. Hence w0wiφ R- and therefore 1+x,w0wiφ < 1. We have therefore checked that the gi do belong to G. Conversely, let gG; say g=tλw   ( λP, wW ). Then g(A) = A so that λ=g(0) AP, i.e. λ=θi for some iJ. So g(0) = gi(0) and therefore w=g-1gi fixes 0, i.e. wW and wA=A. Since AC (the Weyl chamber) it follows that ACwC, whence w=1 (because C, wC are disjoint if w1, (1.21)). Hence g=gi.

Now let w^ W^a. Then w^A is an alcove, hence w^A = wA for a unique wWa by (A6), and hence w-1w^A A, so that w-1w^G. So w^ factors uniquely as w^=wgi   ( wWa, iJ ). Hence P/Q W^a/Wa G and under this isomorphism the element giG corresponds to the coset of θi in P/Q.


  1. The θi   (iJ) are coset representatives of P/Q.
  2. f=Card(J)= number of labels mi=1 in the extended Dynkin diagram of R.

  1. Follows from the remarks above.
  2. Card(J) = |P/Q| = f by (4.3).

The θi   (iJ) are called the minuscule weights (or R).

(A10) Let λP. The λ is minuscule if and only if λ,α = 0 or 1 for all αR+.

⇒) Let αR+, so that α = i=1r θi,ααi. Since 0αφ we have 0θi,αmi for each i. If now λ0 is minuscule, λ=θi for some i such that mi=1, so that λ,α = 0 or 1.

⇐) We have λ,αi = 0 or 1 for each i, and λ,φ = i=1rmi λ,αi = 0   or   1. If λ,φ = 0 then λ,αi = 0 for each i, so that λ=0. If on the other hand λ,φ = 1, then for some i[1,r] we have λ,αi = mi = 1, and λ,αj = 0 for ji. So λ = θi is minuscule.

Finite Dynkin diagrams


Affine Dynkin diagrams



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

Version: March 12, 2012

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