Kac-Moody Lie Algebras
Chapter II: Root system and the Weyl group

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

Last update: 27 August 2012

This is a typed version of I.G. Macdonald's lecture notes on Kac-Moody Lie algebras from 1983.


In this chapter, A is a Cartan matrix until further notice.

Recall that Q=i=1n αi is the lattice generated by B in 𝔥*, and that αQ is a root of 𝔤(A) if α0 and 𝔤α0. The multiplicity of α is

mα=dimgα <

(and may well be >1).

Let R denote the set of roots. By (1.7), each root is either positive (α>0) or negative (α<0). Moreover the involution ω interchanges 𝔤α and 𝔤-α, hence if α is a root, so is -α with the same multiplicity.

Let R+ denote the set of positive roots. We have

𝔫+= αR+ 𝔤α, 𝔫-= αR+ 𝔤-α.


  1. R=R+ (-R+) (disjoint union)
  2. If αR then -αR, and mα=m-α.
  3. αiR (1in) and mαi=1.
  4. If rαiR (r) then r=±1.
  5. If αR+, αB , then α-αiR+ for some i.


(i), (ii) done above; (iii) because 𝔤αi=kei; (iv) is clear.

(v) Let [ ei1 eir ] be a nonzero element of 𝔤α, so that α=αi1 ++ αir (and r2). Then [ ei2 eir ] 0 and lies in 𝔤β, where β= αi2+ + αir . Thus βR+ and β=α-αi1

The dual root system

Recall from Chapter I that if (𝔥,B,B) is a minimal realization of the matrix A, then (𝔥*,B,B) is a minimal realization of the transposed matrix At. Let

Q= i=1n hi𝔥

be the lattice in 𝔥 spanned by h1,,hn. Then Q plays the role of Q for the Lie algebra 𝔤(At) (note that At is also a Cartan matrix). The root system R of 𝔤(At) is called the dual of the root system R. The simple roots of R are h1,,hn, etc.

The Weyl group W

For 1in define wi:𝔥* 𝔥* by

wi(λ)=λ- λ(hi)αi (1)


  1. wi is an automorphism of the lattice Q.
  2. wi(αi) =-αi; wi2=1; det(wi)=-1 .

  1. We have wi(αj)=αj -αj(hi)αi =αj-aijαi Q . Hence wiQQ.

  2. In particular, wi(αi)=αi- 2αi=-αi (because aii=2).


    wi2(λ)=wi (λ)-λ(hi)w (αi)=λ-λ (hi)αi+λ (hi)αi=λ .

    Finally, wi fixes pointwise the hyperplane { λ𝔥*:λ (hi)=0 } . Hence all but one of its eigenvalues are equal to 1, and the remaining eigenvalue is -1 (from above). Hence det(wi)=-1.

Let W denote the group of automorphisms of 𝔥* generated by w1,,wn. By (1.1) it depends only on the matrix A:W=W(A) if we need to make the dependance explicit. It is called the Weyl group of 𝔤(A), or of A.

W acts contragradiently on 𝔥:

λ(w·h)= ( w-1(λ) ) (h) (h𝔥,λ𝔥*)

In particular we have

wi(h)=h- αi(h)hi (2)


λ(wih) = (wiλ)h ( sincewi= wi-1by (2.2) ) = λ(h)-λ (hi)αi (h) = λ ( h-αi(h) hi )

for all λ𝔥*.

From (1) and (2) it follows that W acting on 𝔥 is the Weyl group of At:

W(A)W (At) .

We want next to show that W permutes the roots, ie that wR=R for each wW. For this purpose we recall (1.19) that adei, adfi are locally nilpotent derivations of 𝔤(A), so that e adei , e adfi are automorphisms. Let us compute their effect on the generators of 𝔤(A):

If h𝔥 we have (adei)h= [ei,h]=- [h,ei]=-αi (h)ei , whence (adei)2 h=0 and therefore

(a) eadei (h)=h-αi (h)ei


(a) e-adfi (h)=h-αi (h)fi

Next, we have (adei) fi=hi, (adei)2 fi= [ei,hi]=- 2ei, (adei)3 fi=0 , whence

(b) eadei (fi)=fi+ hi-ei

and likewise

(b) e-adfi (ei)=ei +hi-fi.

Now define

wi =eadei e-adfi eadei (1in)

(2.3) wi(h)= wi(h) for all h𝔥 (which justifies the choice of notation)

Proof is calculation.

wi(h) = eadei e-adfi ( h-αi(h) ei ) by (a) = eadei ( h-αi(h) fi-αi(h) ( ei+hi-fi ) ) by (a), (b) = eadei ( wih-αi (h)ei ) = wih-αi (wih)ei -αi(h)ei by (a) again = wih.

(2.4) Let αR, then wi𝔤α =𝔤wiα .


Let x𝔤α,h𝔥. Then we have

[h,wix] = wi [ wi-1 h,x ] becausewi is an automorphism = wi [wih,x] by (2.3) = α(wih) wi(x) becausex𝔤α (1.7) = (wiα)(h) wi(x)

This calculation shows that wi(x) 𝔤wiα , by (1.7) again, hence that wi𝔤α 𝔤wiα . It follows that 𝔤wiα0 , i.e. that wiα is a root; and then, replacing α by wiα we have wi 𝔤wiα 𝔤α (because wi2=1). Hence

dim𝔤α dim𝔤wiα dim𝔤α

so we have equality throughout, and hence wi𝔤α= 𝔤wiα .

(2.5) If αR and wW, then wαR and mwα=mα.


Enough to prove this when w is a generator wi of W, and then it follows from (2.4).

(2.6) If αR+,α αi , then wiαR+. Thus wi permutes the set R+={αi}.


Say αj=1n mjαj , so that the mj are 0 and some mj,ji, is >0. Then wiα=α -α(hi)αi still has mj>0 as coefficient of αj (for this j), hence is a positive root.

(2.7) Let wW be such that wαi=αj. Then

  1. whi=hj
  2. wwj=wjw .

  1. Say w=wi1 wir . Let φ= w i1 w ir Aut(𝔤(A)) . By (2.3) we have φ(h)=w(h) for h𝔥. We shall apply φ to the relation [ ei,fi ] =hi . By (2.4), φ(ei) 𝔤wαi= 𝔤αj , hence

    φ(ei)= λej (someλk)


    φ(fi)= μfj (someμk)


    whi=φ (hi)= [ φei,φfi ] = [ λej,μfj ] =λμhj

    and it remains to see that λμ=1. But this is clear since

    (wαi) (whi)= αi(hi)=2

    and also

    (wαi) (whi)=αj (λμhj)= 2λμ .

  2. Let h𝔥. Then

    wwi(h)=w ( h-αi(h) hi ) =wh-αi(h) whi


    wjw(h)=wh- αj(wh)hj =wh-αi(h) hj

    so that (ii) follows from (i).

We shall next prove that W is a Coxeter group, i.e. that it is generated by w1,,wn subject only to the relations of the form

(wiwj) mij =1(ij)

where mij are positive integers (or +). The proof will depend only on the last two propositions (2.6), (2.7).

Each element wW can be written (in many ways) as a word in the generators wi (recall that wi=wi-1), say

w=wi1wi2 wir ( 1i1,,irn )

For a given w, the least value of r is called the length l(w) of w, and if r=l(w), ( wi1, ,wir ) is called a reduced word for w.

Observe that:

  1. ( wip, ,wiq ) is a reduced word, for all p,q such that 1pqr
  2. ( wir, ,wi1 ) is a reduced word for w-1 (so that l(w-1)= l(w) ).

(2.8) Let ( wi1, ,wir ) be a reduced word for wW. Then the set S(w) of positive roots α such that w-1α is negative is

S(w)= { γ1,,γr }


γp=wi1 wip-1 αip (1pr)

(so that γ1=αi1).


By induction or r=l(w). For r=1 this is (2.6). Assume r>1 and write (for convenience of notation) sp for wip (1pr), βp=αip , so that γp=s1 sp-1βp .

  1. Since ( s1,,sr-1 ) is a reduced word, we have γp>0 for 1pr-1, by the inductive hypothesis. Consider

    γr=s1 sr-1βr =s1γr say.

    Since (s2,,sr) is a reduced word, we have γr>0 by ind. hyp., hence by (2.6) either γr>0 or γr=β1. But in the latter case s2sr-1 βr=β1 , whence by (2.7) s2sr-1 sr=s1s2 sr-1 and therefore w=s1sr= s2sr-1 , contradiction. Hence γr>0.

  2. We have w-1γp= srs1s1 sp-1βp =srsp βp=-sr sp+1βp . But srsp+1 βp>0 , by (i) applied to w-1=sr s1 . Hence w-1γp>0.
  3. Conversely, suppose γ>0 and w-1γ<0. Let w=s1 sr-1 . Then w=wsr, hence srw-1 γ<0 , whence (2.6) either (a) w-1γ<0 , in which case (ind. hyp.) γ is one of γ1,, γr-1 ; or (b) w-1γ= βr , in which case γ=wβr =γr . So γ is one of γ1,,γr .

As a corollary of (2.8) we have

(2.9) If wW is such that wαi>0 (1in) , then w=1.


For then α>0wα>0, hence s(w-1) is empty, so r=0 and therefore w=1.

In particular, it follows that W acts faithfully as a group of permutations of R:

WSym(R) .


  1. S(w) is a finite set of cardinality l(w).
  2. l(wjw)= l(w)+1 αjS(w) ,
    l(wjw)= l(w)-1 αjS(w) .

  1. S(w) is finite with l(w) elements, by (2.8). It remains to show that γ1,,γr (notation of (2.8)) are all distinct. Suppose then γp=γq for some pair p<q. Then

    s1sp-1 βp=s1 sq-1βq

    so that

    βp=sp sq-1βq

    and hence (2.7) spsq-1 sq=sp·sp sq-1= sp+1sq-1 , contradiction.

  2. Suppose l(wjw)=l(w) +1 . If w=s1sr as before, then wjs1sr is reduced, hence by (2.8) w-1wjαj<0 , i.e. w-1αj>0, so that αjS(w). Suppose l(wjw)= l(w)-1 . Put w=wjw, then by what we have just proved (with w replaced by w) w-1αj>0 , hence w-1αj<0, i.e. αjS(w). But these are the only two possibilities, because clearly

    l(wjw) l(w)+1

    and hence (replacing w by wjw) l(w)l (wjw)+1 .

    But also det(w)= (-1) l(w) , det(wjw)= -det(w) so that l(wjw) l(w) . Hence

    l(wjw) -l(w) =1 .

(2.11) Exchange Lemma Let w=wi1wir be a reduced expression, and suppose that l(wjw)< l(w) . Then for some p=1,,r we have

w=wjwi1 wip wir

i.e. we can "exchange" wj with some wip.


Induction on r=l(w). If r=1 then w=wj and there is nothing to prove. Assume that r>1, and let w=wi1 wir-1 .

  1. If l(wjw) <l(w) , apply the inductive hypothesis to w.
  2. If l(wjw) >l(w) , then by (2.10) we have αjS(w), i.e., w-1αj>0 , and αjS(w), i.e. wirw-1 αj=w-1 αj<0 . By (2.6) it follows that w-1αj= αir , so that αj=wαir , whence by (2.7)

    w=wwr= wjw=wj wi1 wir-1 .

(2.12) Theorem. W=W(A) is a Coxeter group, generated by w1,,wn subject to the relations

wi2=1 (1in)
(wiwj) mij =1(ij)

where mij[2,] are given in terms of the Cartan matrix A by the following table:

aijaji 0 1 2 3 4 (ij) mij 2 3 4 6


The exchange lemma (2.11) implies that W is a Coxeter group (proof in Bourbaki, LG + LA, Chapter IV). It remains to compute the order of wiwj in W. For the moment, exclude the case aijaji=4.

Let Vi=Ker (αi)𝔥 . Then V=ViVj together with hi and hj span 𝔥. For h=λhi+μhjV iff αi(h)=αj (h)=0 , i.e. iff


and these equations have only the solution λ=μ=0 (since we are assuming aijaji4).

Now wiwj fixes V pointwise, and on the 2-plane π spanned by hi and hj it acts as follows:

wiwj (hi) = wi ( hi-aijhj ) =-hi-aij ( hj-ajihi ) wiwj (hj) = -wihj= -hj+aji hi

and therefore the matrix of wiwjπ relative to the basis (hi,hj) is

M= ( -1+aijaji -aij aji -1 )

the eigenvalues of which are the roots of the quadratic equation

λ2+ ( 2-aij aji ) λ+1=0

so we have the following table, in which ω=exp 2iπ3 :

aijaji 0 1 2 3 4 5 eigenvalues -1,-1 ω,ω2 ±i -ω,-ω2 1,1 not roots of unity mij 2 3 4 6

When aijaji>4 the eigenvalues λ,λ satisfy λ+λ>2, hence do not lie on the unit circle in , and therefore cannot be roots of unity; consequently wiwj has infinite order in this case.

Finally, when aijaji=4, the eigenvalues are both 1, but M12 (M2=M), hence wiwjπ is unipotent and therefore again of infinite order.

Remark. We may summarize the table in (2.12) as follows:

cosπmij= min ( 1,12 aijaji ) (ij)

In particular, W is "crystallographic" – it acts faithfully on the lattice Q hence embeds in GL(n,).

The Tits cone

Let (V,B,B) be a (not necessarily minimal) realization of the Cartan matrix A over . The fundamental chamber is

C= { xV: αi(x)0 (1in) } ;

the transforms wC of C under the elements of the Weyl group are the chambers; and the union of all the chambers


is called the Tits cone.

Define a partial ordering on V:

xyiffx-y= i=1nλi hiwith allλi 0 .


  1. C is a fundamental Domain for the action of W on X (and hence W acts simply transitively on the set of chambers).
  2. Let xC and let Wx= { wW:wx=x } be the isotropy group of x in WHAT?. Then Wx is generated by the fundamental reflections wi it contains.
  3. Let xV. Then xC iff xwx for all wW.
  4. Let xV. Then xX iff α(x)0 for almost all αR+ (i.e. for all but finitely many αR+). (Hence X is a convex cone).


(i) and (ii) will both follow from the

Lemma. Let wW, and let wi1wir be a reduced word for w. If xC and wxC, then wirx=x.


For we have, putting y=wx,

wαir(y)= αir(x)0 (1)

since xC; but also wαir=-wi1 wir-1 αir<0 ((2.2), (2.8)), hence

wαir(y)0 (2)

since yC. From (1) and (2) it follows that αir(x)=0, hence wirx=x.

  1. Clearly each W–orbit in X meets C, by definition. If xC and y=wxC (with w=wi1wir as above) then y=wi1 wir-1x by the lemma. By induction on r=l(w) we conclude that y=x.
  2. Let wWx, with w as above. Then by the lemma wirWx, and hence by induction on r=l(w) we have wi1,, wirWx .
  3. Suppose xC. Induction on l(w) (again). If l(w)=1 then w=wi, and x-wix=αi (x)hi0 . Now let l(w)>1, then we can write w=wwi with l(w)= l(w)-1 , and we have

    x-wx = (x-wx)x+ w(x-wix) = (x-wx)x+ αi(x)w hi.

    By (2.8) (applied to the dual root system) we have whi0, whence αi(x)w hiWHAT? also x-w(x)0 by the inductive hypothesis, hence x-wx0 as desired.

    Conversely, if xwx for all wW, then in particular xwix (1in) , whence αi(x)hi 0 therefore αi(x)0 (1in) , i.e. xC.

  4. For each xV let

    M(x)= { αR+ : α(x)<0 }

    and let X= { xV: M(x) is finite } . We have to prove that X=X

    Let xV,wW. If αM(wx), then αR+ and (w-1α) (x)=α(wx)<0. Either w-1α>0, in which case w-1αM(x), i.e. αwM(x); or else w-1α<0. i.e. αS(w) in the notation of (2.8). Thus

    M(wx)wM(x) S(w) .

    Now S(w) is finite (2.10), hence xXM(x) finite M(wx) finite wxX. Thus X is W–stable; but clearly CX, hence XX.

    Conversely, let xX. We shall show that xX by induction on r=CardM(x). If r=0 then α(x)0 for all αR+, hence xCX. If r1 we have αi(x)<0 for some index i, i.e. αiM(x). But then

    wiαM (wix) wiα>0 and α(x)<0 αR+, ααi, α(x)<0by (2.6) αM(x) ,ααi

    from which it follows that M(wix)= M(x)-1 . By the inductive hypothesis, wixX, hence xwiX=X.

(2.14) The following conditions are equivalent:

  1. W is finite;
  2. X=V;
  3. R is finite;
  4. R is finite.


(i) (ii) Let xV. The orbit Wx of x is finite; Let y=wx be a maximal element of this orbit for the ordering on V. I claim that yC; for if not, then αi(y)<0 for some i, whence wiy=y-αi(y) hi>y , impossible. Thus wxC and therefore xw-1CX. Thus X=V.

(ii) (iii) Choose xV such that αi(x)<0 (1in) . Then α(x)<0 for all xR+; but xX, hence α(x)0 for almost all αR+. It follows that R+ is finite, hence so is R.

(iii) (i) because W acts faithfully on R: WSym(R) (remark following (2.9)).

(i) (iv) because R,R have the same Weyl group.

Let J be any subset of the index set {1,2,,n}. Then if BJ= { αj:jJ } , BJ= { hj:jJ } , (V,BJ,BJ) is a realization of the principal submatrix A. Define CJ,XJ,WJ in the obvious way:

CJ := { xV:αj (x)0for jJ } WJ := wj:jJ , XJ := wWJ wCJ.

Also let

VJ:= { xV: αj(x)=0, alljJ } =VWJ .

We have then


  1. CJ=C+VJ;
  2. XJ=X+VJ.

  1. Let xCJ. There exists yC such that αj(y)= αj(x) , all jJ, so that x-yVJ; thus x=y+(x-y) C+VJ . The reverse inclusion is obvious, because VJ and C are both contained in CJ.
  2. If wWJ, then wCJ=wC+wVJ (by (a))=wC+VJ X+VJ , hence certainly XJX+VJ. Conversely, let xX,vVJ; then α(x)0 for almost all αR+, (2.13) and β(v)=0 for all βRJ+. Hence β(x+v)0 for almost all βRJ+, hence by (2.13)(iv) again x+vXJ.

(2.16) Let xX. Then xX˚ iff Wx is finite. (X˚= interior of X)


We may assume that xC, because Wwx=wWx w-1(wW) . Let J= { j:αj (x)=0 } , then Wx=WJ by (2.13)(ii), and xVJ.

Suppose xX˚. Let vV, then y=x+λvX for some λ0, hence λv=y-xX+ VJ=XJ by (2.15). It follows that XJ=J and hence by (2.14) that Wx=WJ is finite.

Conversely, suppose that Wx is finite, so that (2.14) XJ=J. Let

U= { yV:αi (wy)>0for all wWJand alliJ } .


  1. xU, because αi(wx)=αi (x)>0 (since xC and iJ);
  2. U is open, because it is a finite intersection of open half-spaces;
  3. UX. For if yU, then ywCJ for some wWJ (because XJ=V) we have z=w-1yU CJ , whence αi(z)>0 for iJ, and αj(z)0 for jJ. Thus zC and therefore y=wzX.

Hence x is an interior point of X.

Classification of Cartan matrices

Let V be a finite-dimensional real vector space. Recall that a non-empty subset K of V is a convex cone if K is closed under addition and multiplication by non-negative scalars:

x,yKx+yK ; xK,λ0λxK .


  1. Any vector subspace of V is a convex cone.
  2. Let x1,,xrV. Then the set K of all linear combinations i=1r λixi with scalars λi0 is a closed convex cone, the cone generated by x1,,xr.

Let K be a closed convex cone in V, and let V* be the dual of V. The dual cone K*V* is defined by

K*= { ξV*: ξ(x) 0allxK }

Clearly K* is a closed convex cone (it is an intersection of closed half-spaces). The basic fact we shall need is the

Dualilty theorem K is the dual of K*, i.e.

K= { xV:ξ(x) 0for allξK* } .


Let K denote the r.h.s. above. Clearly xKxK. The nontrivial part is to prove that xKxK, i.e. that if xK there exists ξK* such that ξ(x)<0; or equivalently that if xK there exists a hyperplane H=Ker(ξ) in V which seperates x and K.

Let y,z be a positive definite scalar product on V: write 𝔤= 𝔤,𝔤 12 and d(y,z)= y-z , the usual Euclidean metric. We shall show that there is a unique point zK for which d(x,z) is minimal, and that the hyperplane (through 0) perpendicular to x-z seperates x from K.

Let S be the unit sphere in V. For each tS, let φ(t) denote the shortest distance from x to the ray +t, so that by Pythagoras

φ(t)= { x2- x,t2 ifx,t0 x ifx,t0

x z φ(t) 0 t

Clearly φ is a continuous function on S. Now SK is compact (because S is compact and K is closed), hence φ attains its lower bound on SK, i.e. there exists zK for which d(x,z) is a minimum (>0, since xK). Moreover this z is unique, for if d(x,z1) and d(x,z2) are both minimal, and z1z2 then d(x,z3) is strictly smaller, where z3=12 (z1+z2)K .

Let y=x-z, then x,y= y2>0 . I claim that u,y0 for all uK. For if uK is such that u,y>0, then the angle xzu is acute; hence if z is the foot of the perpendicular form x to the segment zu, we have zK (convexity) and d(x,z)< d(x,z) , contradiction.

z z u x

Now define ξV* by ξ(u)=-u,y; then ξK* and ξ(x)<0.

We shall require the following consequence of the duality theorem:

Lemma 1 Let A=(aij) be any real m×n matrix. Consider the systems of linear inequalities (1im, 1jn)

(1) xj>0, jaij xj<0; (2) yi0, iaij yi0.

Then either (1) has a solution or (2) has a nontrivial solution.


Let e0,,en be the standard basis of n+1 and let K be the closed convex cone generated by the m+n vectors of e0+ej (1jn), e0-jaij ej(1im) . We have

e0K scalarsλi ,μj0such that e0=juj (e0+ej)+ iλi ( e0-jaij ej ) scalarsλi ,μj0such that iaij λi=μj, λi+μj=1 (2) has a nontrivial solution.

If on the other hand e0K, then by the duality theorem there is a linear form ξ on n+1 which is negative at e0 and 0 at all points of K. But then xj=ξ(ej) (1jn) satisfy the inequalities (1).

Until further notice, A=(aij) will be a real n×n matrix satisfying

() aij0if ij;aij=0 aji=0.

If A satisfies these conditions, so does its transpose At.

Assume also that A is indecomposable.

Let V=n be the space of column vectors x= (x1,,xn) t . Write x0 (resp. x>0) to mean xi0 (resp. xi>0) for 1in. Define the closed convex cones

P = { xV: x0 } (positive octant?) K = { xV: Ax0 }

The interior of P is

P˚= { xV: x>0 } .

Lemma 2 KPP˚{0}.


Let xKP; suppose xi=0 for iI, xi>0 for iJ (where I,J are complementary subsets of {1,2,,n}). Then we have

jJaij xj=j=1n aijxj0

for all i, in particular for iI. But xj>0 and aij0, whence aij=0 for all (i,j)I×J. Since A is indecomposable, it follows that either I=, in which case xP˚; or J=, in which case x=0.

As regards K and P, there are two possibilities:

  1. KP{0};
  2. KP={0}

Case (a). By Lemma 2 we have KP={0}. Now K is connected (because convex) and we have

K=(KP˚) (KP) (KP) (disjoint union)

where P is the complement of P in V, so that

K-{0}= (KP˚) (KP)

as a union of disjoint relatively open sets, of which KP˚ is not empty (by Lemma 2). Hence there are two possibilities:

(a) K-{0} is connected, in which case KP=, i.e. KP so that (Lemma 2) KP˚{0}

(a) K-{0} is not connected, in which case K is a line in V (i.e. a 1-dimWHAT? subspace). (For if x,y lie in different components of K-{0}, the line segment (xy) is contained in K, hence must pass through 0, whence y=-λx for some λ>0. If now zK-{0}, then either z and x lie in different components, whence (as above) z=-μx,μ>0; or z and y lie in different components, in which case z=-μy=λμx. Thus K=x.)

Let N= { xV:Ax=0 } be the null-space of A. Clearly NK.

In case (a), K contains no vector subspace 0 of V, because P clearly doesn't. Hence N=0, i.e. A is nonsingular. In this case we say that A is of positive type.

In case (a) we have K=N. For if xK, then Ax0; but also -xK, whence Ax0 and therefore Ax=0, i.e. xN. So KN and therefore K=N. Hence A is singular, of rank n-1. We say that A is of zero type.

In either case (a) or (a), the inequalities


have no solution. For if Ax<0 then -xK, which in case (a) implies either x=0 or -x>0, and in case (a) implies xN, i.e. Ax=0. By Lemma 1, therefore, the inequalities


have a nontrivial solution, i.e. we have


where Kt= { xV:At x0 } . It follows that the matrix At satisfies (a), and is therefore of positive or zero type according as A is (because detAt=detA). Hence conditions (a), (a) and therefore also (b) are stable under transposition.

Case (b).In this case KP={0} and therefore also KtP={0}, i.e. the inequalities


have only the trivial solution. Hence, by Lemma 1, the inequalities


have a solution. We say that A (and At) is of negative type.

To summarize:

(2.17) Theorem (Vinberg) Each indecomposable real n×n matrix A=(aij) satisfying the conditions

() aij0if ij; aij=0 aji=0

belongs to exactly one of the following there categories:

(+) A is nonsingular, and Ax0x>0 or x=0 (positive type)
(0) rank(A)=n-1, and Ax0Ax=0 (zero type)
(-) x0>0 such that Ax0<0; x0 and Ax0x=0 (negative type)

The transposed matrix At belongs to the same category as A.

Moreover, A is of positive (resp. zero, negative) type iff x0>0 such that Ax0>0 (resp. Ax0=0, Ax0<0).


Only the last sentence requires comment. Observe that K is a finite intersection of closed half-spaces and therefore its interior K˚ is the intersection of the corresponding open half-spaces, i.e. K˚= { xV:Ax>0 } , and is non empty if A is nonsingular (because the columns of A are then a basis of V).

If A is of positive type, we have KP˚{0}, hence K˚P˚, and K˚ is nonempty, whence x0>0 with Ax0>0. If A is of zero type, then N=K intersects P˚, hence x0>0 with Ax0=0. Finally, if A is of negative type, then x0>0 with Ax0<0, from above.

Conversely, let x>0. If Ax>0, then A cannot be of zero or negative type, hence is of positive type. If Ax=0, then A cannot be of negative type or of positive type (because det(A)=0), hence is of zero type. Finally, if Ax<0, the vector y=-x satisfies Ay>0,y<0, whence A is not of positive or zero type, hence of negative type.

(2.18) If A is indecomposable and of positive or zero type, then for every proper subset J of {1,2,,n} the principal submatrix AJ has all its components of positive type.


We may assume that AJ is indecomposable. By (2.17) xV such that x>0 and Ax0. Let xJ= (xj) jJ ; then xJ>0, and AJxJ has components

jJaij xj=j=1n aijxj+ jJ (-aij)xj (iJ) .

The first sum on the right is 0, and so is the second, since -aij0 for iJ, jJ, and xj>0. Moreover the second sum is 0 iff aij=0 for all jJ. Since A is indecomposable, iJ and jJ such that aij0, hence at least one of the components of the vector AJxJ is >0. Thus we have

xJ>0, AJxJ0, AJxJ0

whence by (2.17) AJ is of positive type.


A matrix A satisfying the conditions () is said to be symmetrizable if there exists a diagonal matrix D= ( d1 dn ) with di>0 (1in) such that DAD-1 is symmetric.

(a) DAD-1symmetric D2A ( =D·DAD-1·D ) symmetric AD-2symmetric

Hence A is symmetrizable iff A can be made symmetric by multiplication (on the left or on the right) by a positive diagonal matrix;

(b) If A is symmetrizable and B=DAD-1 is symmetric, then B is uniquely determined by A. For

bij=di aijdj-1= bji=dj aijidi-1

so that

bij2=aij aji

and therefore (as bij0 if ij)

bij=- aijaji (ij);bii =aii .

Call B the symmetrization of A: notation As.

(c) If A is indecomposable and symmetrizable, the diagonal matrix D above is unique up to a positive scalar multiple. For if DAD-1= DAD-1 , then with E=D-1D we have EAE-1=A, i.e. eiaij ej-1=aij and therefore ei=ej whenever aij0, i.e. whenever i and j are joined by an edge in the graph Γ of A. Since Γ is connected (1.11) it follows that all the ei are equal, i.e. E=λ1n.

Let A=(aij) be a matrix satisfying (), Γ its graph. Let

p= ( i0,i1,,ir )

be a path in Γ, i.e. i0,,ir are vertices of Γ such that is-1is (1sr) is an edge, so that is-1is and a is-1 is 0 . Define

ap= a i0i1 a i1i2 a ir-1ir

so that ap0 and has the sign of (-1)r. If p-1= ( ir,ir-1, ,i0 ) is the reverse path, then ap/ap-1 is positive.

Notice that

  1. apq=apaq whenever pq is defined, i.e. whenever the endpoint of p is the origin of q;
  2. if B=DAD-1 (D diagonal, as above), i.e. bij=di aijdj-1 , then bp=di0 apdir-1 . In particular, bp=ap if ir=i0, i.e. if p is a loop.

(2.19) A is symmetrizable ap=ap-1 for each loop p in Γ.

In particular, A is symmetrizable if Γ is a tree.


: diagonal matrix D such that B=DAD-1 is symmetric: bij=di aijdj-1 whence ap=bp= bp-1= ap-1 for any loop p in Γ (the middle equality because B is symmetric).

: Fix an index i0. For each i[1,n] there is at least one path p from i0 to i in Γ (we may assume A is indecomposable, i.e. Γ connected). If q is another such path, then pq-1 is a loop, hence

apaq-1= aaq-1= aqp-1= aqap-1

so that

ap/ap-1= aq/aq-1 .

Hence we may define unambiguously

ei=ap/ aa-1>0

(in particular, ei0=1).

I claim that eiaij= ejaji . This is trivially true if i=j or if i and j are not lined by an edge in Γ, for then both sides are 0. If (ij) is an edge in Γ, and if p= (i0,,i) is a path from i0 to i in Γ, then q= (i0,,i,j) is a path from i0 to j, and aq=apaij so that

ej=aq/ aq-1=ap aij/aji ap01=ei aij/aji .

Hence the matrix (eiaij) is symmetric, i.e. A is symmetrizable.

Terminology. For a Cartan matrix (indecomposable)

finite type = positive type affine type = zero type indefinite type = negative type

Assume from now on that A is an indecomposable Cartan matrix. Then all the previous results are applicable.

(2.20) Let A be an indecomposable Cartan matrix of finite or affine type. Then either the graph Γ of A is a tree, or else Γ is a cycle with n3 vertices and

A= ( 2 -1 -1 -1 2 -1 -1 2 -1 -1 -1 2 )

(i.e. aij=0 if i-j2; -1if i-j=1; 2ifi=j : indices mod n). This A is of affine/zero type.


Suppose Γ contains a cycle. By considering a cycle in Γ with the least possible number of vertices, we see that A has a principal submatrix of the form

B= ( 2 a12 a1m a21 2 a23 a32 2 am-1,m am1 am,m-1 2 )

for some mn (after a permutation of the indices) with aij-1 when i-j=1 (modm) . By (2.18), B is of finite or affine type, hence x>0 such that Bx0: i.e. xi>0 (1im) such that

ai,i-1 xi-1 +2xi+ ai,i+1 xi+10 (1im)

(where suffixes are read modulo m). Add these inequalities:

i=1m ( ai+1,i +2+ ai-1,i ) xi0 (1)

Now the a's are integers -1, hence

ai+1,1+2 +ai-1,i0 .

Since the xi are >0 we must have equality throughout; so the a's are all -1, and Bx=0, whence B is of affine/zero type and therefore (2.18) B=A.

(2.21) Let A be an indecomposable Cartan matrix of finite/positive or affine/zero type. Then A is symmetrizable, and its symmetrization As is positive definite or positive semidefinite (of corank 1) according as A is of finite/positive or affine/zero type.


The first assertion follows from (2.19) and (2.20). By (2.17) x>0 such that Ax0, hence if As=DAD-1 we have y=Dx>0 and AsyC, more precisely, Asy>0 if A is of positive type, Asy=0 if A is of zero type. It follows that ( A2+λ1k ) y>0 for all scalars λ>0, hence As+λ1k is of positive type for all λ>0, hence is nonsingular (2.17). Thus the eigenvalue of As (which are real, because As is a real symmetric matrix) are all 0. If A is of positive type, they are all >0, whence As is positive definite; and if A is of zero type, 0 is an eigenvalue of As with multiplicity 1, and all other eigenvalues of A are >0.

Remark. If A is a symmetrizable indecomposable Cartan matrix of negative type, then As is indefinite. For since aii=2 the quadratic form xtAsx takes positive values, and since by (2.17) x>0 with Ax<0, the vector y=Dx satisfies y>0 and Asy<0, whence ytAsy<0. Hence, for symmetrizable A

Ais of positive type Asis positive definite Ais of zero type Asis positive semidefinite (corank 1) Ais of negative type Asis indefinite.

(2.23) Let A be an indecomposable Cartan matrix in which all proper principal minors (i.e. detAJ for all J{1,2,,n}) are positive. Then

  1. all cofactors of A are >0
  2. A is of positive (zero, negative) type according as detA is positive (zero, negative).


(i) Let Aij be the cofactor of aij in A. We may assume that ij. The expansion of det(A) is

det(A)= wSn sym(w) a1w(1) anw(n) (1)

summed over all permutations of {1,2,,n}. Write w as a product of disjoint cycles, say w=c1c2. Now Aij is the coefficient of aji in the expansion (1), hence we have to consider all cycles c such that c(j)=i: say C= ( j,i,i1,, ir-1 ) , and correspondingly the product (-1)r aii1 ai1i2 air-1j . Thus product will be zero unless p= ( i,i1,, ir-1,j ) is a simple path (i.e. with no repeated vertices) from i to j in the graph Γ of A; and then it will be equal to (-1)ra(p) =a(p)>0 , in the notation of (2.20). Thus it follows from (1) that

Aij= pa(p) det(AJ(p )) (2)

where the sum is over all simple paths p from i to j in Γ, and J(p) is the complement of the set of vertices of p. Each term in the sum (2) is >0, and the sum is not empty, because Γ is connected. Hence Aij>0.

(ii) By (2.17), A is of positive (zero, negative) type according as x>0 with Ax>0 (=0,<0) . Let y=Ax, then premultiplication by the matrix of cofactors gives

j=1n Aijyj= xidetA

Since xi>0 and Aij>0, this gives the result.

(2.24) Let A be an indecomposable Cartan matrix. Then the following are equivalent:

  1. A is of positive (resp. zero) type
  2. all proper principal minors of A are >0, and detA>0 (resp. detA=0)


(a) (b) If A is of positive type, its symmetrization A is positive definite. Hence each principal minor of A, being equal to the corresponding principal minor of A, is positive. In particular, detA>0.

If A is of zero type, we know (2.17) that det(A)=0 and (2.19) that each indecomposable proper principal submatrix AJ is of positive type, hence detAJt0 from above.

(b) (a) follows from (2.23).


positive type = finite type zero type = affine negative type = indefinite

Classification of indecomposable Cartan matrices of positive or zero type

If A=(aih) is an indecomposable Cartan matrix of positive or zero type, then for each pair i,j (ij) we have by (2.18) and (2.24)

2 aij aji 2 0

(with equality only when n=2 and A is of zero type). Thus

0aij aji4 .

Dynkin diagram

This is a fancier version of the graph Γ of A that we defined earlier. In the Dynkin diagram Δ of A the vertices i,j (ij) are connected by max ( aij, aji ) lines, with an arrow ponting towards i if aij> aji . Thus the possibilites are given in the following table:

aij aji i j
0 0
1 1
1 2
2 1
1 3
3 1
1 4
2 2
4 1

We do not attempt to define a Dynkin diagram for Cartan matrices in which aijaji>4 for some pairs i,j.

The table above shows that Δ determines A uniquely. Observe also that the Dynkin diagram of At is obtained from that of A by reversing all arrows; also that A is indecomposable iff Δ is connected.

We shall say that Δ is of finite type (resp. affine type) according as A is.

(2.25) Theorem.

  1. The connected Dynkin diagrams of finite type (resp. affine type) are exactly those listed in Table F (resp. Table A).
  2. The integers ai attached to the vertices of the diagrams in Table A are the components of the unique vector δ= ( a1,,an ) >0 such that Aδ=0 and the ai are positive relatively prime integers.


We begin by verifying the last statement. The equation Aδ=0, i.e.

j=1n aijaj =0 (1in)

can be rewritten as followss:

2ai=i mijaj

where the sum is over all the vertices j in Δ joined directly to i, and

mij= { no. of bonds joiningi andj , if there is an arrow pointing towards SOMETHING 1otherwise

[except if A= ( 2 -2 -2 2 ) , Δ= 1 1 .]

Then (ii) is easily checked diagram by diagram. It follows from (2.17) that all the diagrams in Table A are of affine type. Since each diagram in Table F occurs as a subdiagram of one in table A, it follows from (2.18) that all diagrams in Table F are of finite type.

Table F

Al (l1)
Bl (l3)
Cl (l2)
Dl (l4)
El (l=6,7,8)

(The number of vertices is n=l.)

Table A

A1 1 1 Al 1 1 1 1 1 (l2)
Bl 1 1 2 2 2 2 (l3) Bl 1 1 2 2 2 1 (l3)
Cl 1 2 2 2 1 (l2) Dl+1(2) Cl 1 1 1 1 1 (l2)
Dl 1 1 2 2 2 1 1 (l4)
E6 1 2 3 2 1 2 1
E7 1 2 3 4 3 2 1 2
E8 1 2 3 4 5 6 4 2 3
F4 1 2 3 4 2 E6(2) F4 1 2 3 2 1
G2 1 2 3 D4(3) G2 1 2 1
A2(2) BC1 1 2 A2l(2) BCl 1 2 2 2 2 (l2)
(The number of vertices is n=l+1).

If A is of (finite) type X, then det(A) is the number of 1's in the diagram X. The 1's form a single orbit under the group Aut(X).

We now have to prove the converse, namely that every connected diagram Δ of finite or affine type occurs in Table F or Table A. Let n be the number of vertices in Δ.

If n=1, the only possibility is Δ=A1.

If n=2, we have already enumerated the possibilities: A2,C2, G2 of finite type, and A1, BC1 of affine type.

If n=3, either Δ is a tree or Δ is A2 (by (2.21)). If Δ is a tree then

A= ( 2 -a 0 -b 2 -c 0 -d 2 )

where a,b,c,d are positive integers; ab3,cd3 by (2.19) and detA0, so that ab+cd4; more precisely, ab+cd=2 or 3 if Δ is of finite type, ab+cd=4 if Δ is of affine type. So the possibilities are

a b c d ab+cd
1 1 1 1 2 A3
1 1 1 2 3 B3
1 1 2 1 3 C3
1 1 1 3 4 G2
1 1 3 1 4 G2
1 2 1 2 4 BC2
1 2 2 1 4 C2
2 1 1 2 4 C2

From (2.19) we have:

() If a subdiagram of Δ occurs in Table A, then it is the whole of Δ. Hence to show that Δ Table A it is enough to show that some subdiagram of Δ is in Table A.

  1. If Δ is not a tree then by (2.20) Δ=Al (l2). So we may assume that Δ is a tree, and that n4.

  2. If Δ contains multiple bonds, they are double bonds . For otherwise Δ contains either A1 or BC1, in which case n=2 by (); or Δ contains a connected proper subdiagram of 3 nodes containing a triple bond, which (see above) can only be G2 or G2, and therefore n=3, by () again.

    If Δ contains two or more double bonds, it contains a subdiagram of type Cl or Cl or BCl (l2), hence by () this subdiagram is the whole of Δ.

    So we may assume that Δ contains at most one double bond.

  3. Suppose now that Δ has at least one branch point. If it contains a double bond as well, then it contains a subdiagram of type B2 or Bl, and again by () this subdiagram is the whole of Δ. So assume now that Δ is simple-laced (i.e. no multiple bonds). If there is more then one branch point, then Δ contains Dl (l5) as a subdiagram, hence again this is the whole of Δ. So we may assume that Δ has only one branch point. If there are 4 edges issuing from this point, then Δ contains D4 as a subdiagram, which is therefore the whole of Δ. If there are 3 edges issuing from the branch point, let p,q,r be the number of vertices of Δ on the three arms, where pqr (and p+q+r=n1).

    If p2 then Δ contains E6, hence by () Δ=E6.
    If p=1,q3 then Δ contains E7, hence Δ=E7.
    If p=1,q=2,r5 then Δ contains E8, hence Δ=E8.
    If p=1,q=2,r=2,3,4 then Δ=E6,E7,E8 respectively.
    If p=q=1,r1 then Δ=Dl (l4).

  4. Finally assume Δ has no branch points, hence is a chain. If Δ is simply-laced, then Δ=Al (l4). The remaining possibility is that Δ contains just one double bond. Suppose there are p nodes on one side of the double bond, q on the other, where pq.

    If q2 then Δ contains F4 as a proper subdiagram, contradicting ().
    If q=1,p2 then Δ contains F4 or F4 as a subdiagram. Now apply ().
    If q=1,p=1 we obtain Δ=F4; and if q=0,p0 we obtain Bl or Cl.

Explanation of notation: Let X be any of the symbols An,,G2; R a (finite) root system of type X; α1,,αl the simple roots, α0 the lowest root (i.e. ht(α0) is minimum). Let

aij= αi, αj ( 0i,jl )

Then A= (aij) 1i,jl is the Cartan matrix of type X (finite type) and A= (aij) 0i,jl is the Cartan matrix of type X (affine type).

We have α0=- 1laj αj say, with the ai positive integers, i.e.

0l ajαj=0

if we define a0=1; but then

0l ajaij= 0l ajαj (hi)=0 (0il)

showing that A is of affine type and the ai(0il) are the labels attached to the nodes of the diagram of type X in Table A.

Remark. The classification theorem (2.22) gives a complete list of the indecomposable Cartan matrices of finite or affine type. Any Cartan matrix not in this list is therefore of indefinite (i.e. negative) type, so the Cartan matrices of indefinite type are the pink-heap. However, there are two subclasses which can be explicitly classified: an indecomposable Cartan matrix A of indefinite type is said to be hyperbolic (resp. strictly hyperbolic) if every proper principal submatrix AJ has all its components of finite or affine type (resp. of finite type). All the 2×2 Cartan matrices ( 2 a12 a21 2 ) with a12a21>4 are of strictly hyperbolic type; apart from these there are only finitely many, and they all have Dynkin diagrams.

Exercise. If A is an indecomposable n×n Cartan matrix, then

  1. n5 if A = is strictly hyperbolic
  2. n10 if A is hyperbolic.

Here are two with n=10: (probably the only two)

and (I think) the only strictly hyperbolic matrix with n=5 has diagram

If A is hyperbolic and AJ is affine (J connected) then J=n-1. For otherwise we should have JK{1,,n} with K connected and both inclusions strict, and then AJ would not be either affine or finite type.

(2.26) Let A be a Cartan matrix. Then A is symmetrizable iff there exists a W–invariant symmetric bilinear form x,y on 𝔥 (with values in k), such that hi,hi is positive rational for all i. Moreover such a form is nondegenerate.


Suppose A is symmetrizable, then εj>0 such that aijεj= ajiεi for all i,j. Since the aij are integers we may assume that the εj are rational (or even positive integers).

As before, let 𝔥=1nk hi𝔥 , and let 𝔥 be a vector space complement of 𝔥 in 𝔥. Define x,y as follows:

x,hi = hi,x= εiαi (x) (x𝔥) x,y = 0if x,y𝔥

(Notice that hi,hj =εiαi (hj)= ajiεi and also hj,hi =εjαj (hi)= aijεj in particular, therefore, hi,hi =2εi>0 so that the above definition is unambiguous.) Now we have

wix,y = x,y- αi(x) hi,h = x,y- εiαi (x)αi (y) (1)

which is symmetrical in x and y, so that

wix,y = wiy,x = x,wiy

from which it follows that wx,y= x,w-1y for all wW, by induction on l(w). This x,y is W–invariant.

Conversely, if we have such a W–invariant form on 𝔥, then from (1) it follows that

αi(x) hi,y= αi(y) hix

for all x,y𝔥: taking x=hi, y=hj we obtain

2hi,hj =aji hi,hi

and therefore, putting εi=12 hi,hi >0 ,

ajiεi= hi,hj =hj,hj =aijεj

which shows that A is symmetrizable.

Finally, if h,x=0 for all x𝔥 then in particular εiαi(h)= h,hi=0 , so that hi=1n Kerαi=𝔠 𝔥 (1.10); but then h=λihi say, and

0=h,x= λiεi αi(x) (x𝔥)

so that λiεi αi=0 in 𝔥* and therefore λi==λn =0,h=0 .

From (2.26) it follows that the mapping θ:𝔥 𝔥* defined by θ(x)(y) =x,y is an isomorphism, and we can therefore transport the scalar product x,y to 𝔥:

λ,μ= θ-1(λ), θ-1(μ) .

Let αi=θ (hi)𝔥 . Then

αi(x)= x,hi= εiαi(x)

so that αi =εi ; taking x=hi we obtain (since αi(hi) =aii=2 )

2εi= hi,hi= αi, αi = εi2 αi,αi


εi=2/ αi,αi,
αi= 2αi αi,αi

and dually

αi= 2αi αi, αi .

αi is the coroot of αi.

Finally note that aij=αj (hi)= αi,αj

Action of W:

wi(h)=h- αi(h)hi

For each αi, let Hi=Ker (αi) be the hyperplane in 𝔥hi. Then wi is reflection in this hyperplane: because if m is the midpoint of h,wih then

α(m) = 12αi (h+wih) = 12 αi(h)+ wiαi (h) =0

and h=wih= αi(hhi) is a scalar multiple of αi.

So W is realized as a group generated by reflections.

We can now characterize the algebras 𝔤(A) for which A is a Cartan matrix of finite type:

(2.27) Let A be an indecomposable Cartan matrix. Then the following conditions are equivalent:

  1. A is of finite type;
  2. A is symmetrizable, and the bilinear form x,y of (2.26) (with k=) is positive definite;
  3. W is finite;
  4. R is finite;
  5. 𝔤(A) is a finite-dimensional simple Lie algebra.


(i) (ii) by (2.26) and (2.21)

(ii) (iii) (Here k=). The matrix A is nonsingular, hence h1,,hn is a basis of 𝔥 and therefore Q=qn hi is a lattice in 𝔥. Consequently End(Q) is a lattice in the real vector space End(𝔥).

Let O be the orthogonal group of the form x,y, acting on 𝔥. O is compact, hence a bounded subset of End(𝔥); W is a subgroup of O and preserves the lattice Q. i.e. WOEnd(Q), which is finite.

(iii) (ii) Let (x,y) be any positive definite scalar product on 𝔥. Then

x,y= wW (wx,wy)

is W–invariant and positive definite. Hence A is symmetrizable, by (2.23).

(iii) (iv) proved earlier (2.14)

(iv) (v) because 𝔤(A)=𝔥+ αR 𝔤α (direct sum) (1.7); and 𝔤(A) is simpe by (1.13).

Now suppose that A= (aij) 1i,jn is an indecomposable Cartan matrix of affine type. By (2.17) there is a unique vector

a= (a1,,an)t

with components ai which are mutually prime positive integers, such that

j=1n aijaj =0 (1in) (1)



Dually, the matrix At is also of affine type, hence there is a unique vector

a= ( a1,, an ) t

with components ai which are mutually prime positive integers, such that

i=1n aiaij =0 (1jn) (2)


Ata=0 .

Now A is symmetrizable (2.21), hence there is a diagonal matrix E= ( ε1 εn ) (with positive diagonal entries εi) such that AE is symmetric, i.e.

AE=EAt. (3)

But then AEa=EAta=0 by (2) and (3), so that a=λEa from (1), for some scalar λ0. Replacing E by λ-1E we have then a=Ea, i.e.,

ai=εi ai (1in) (4)

Now define

δ=j=1n ajαj 𝔥*

Then we have

δ(hi)=0 (1in) (5)

(We shall see later that δ is a root) for δ(hi)= j=1n ajaij=0 by (1). Hence W fixes δ. Dually define

c=i=1n aihi𝔥 ;

then we have

αj(c)=0 (1jn) (6)

for αj(c)= i=1n aiαj (hi)= iai aij=0 by (2). Hence W fixes c.

Recall that 𝔠=1n Kerαi is the centre of 𝔤(A), and that dim𝔠=n-l=1 he. Thus 𝔠=c (we are taking k= here): c is the canonical central element.

Next we shall construct the scalar product on 𝔥 as in (2.26). We have dim𝔥=2n-l= n+1 , so we can take as a basis of 𝔥 the elements h1,,hn and d say, where δ(d)=1. (By (5) we must have δ(d)0).

Remark. this of course does not determine d uniquely: we could add on any linear combination of h1,,hn. At this stage, however, that doesn't matter.

We have then

x,hi= εiαi(x) ;d,d =0 (7)

From (6) and (7), therefore,

c,hi=0 (1in) (8)

and in particular

c,c=0; (9)


c,d=1 (10)


c,d= d,c = aiv d,hi= aiεi αi(d) = aiαi (d)=δ(d) =1.

As in (2.26) let θ:𝔥 𝔥* be the isomorphism defined by the scalar product, so that θ(x)(y)= x,y . Then


because θ(hi)= εiαi , so that

θ(c)= aiθ(hi) =aiεi αi=aiαi =δ .

Now consider the action of the Weyl group W. We shall show that W acts as a group generated by reflections in a real Euclidean space of dimension l. Since dim𝔥=l+2, we have to cut down the number of dimensions by 2. We do this in two stages.

(1) Since W fixes c it follows that W acts (faithfully) on 𝔥=𝔥/𝔠. Each λ𝔥* such that λ(c)=0 defines a linear form on 𝔥, which we denote by λ. Thus we have α1,, αn,δ . Notice that W (in its action on 𝔥*) fixes δ. If p:𝔥𝔥 is the projection, the image C=p(C) of the fundamental chamber C is the set of h𝔥 such that αi(h) 0(1in) and the image p(X)=X of the Tits cone X is the union of the wC, wW.

(2) For each real number t0 let

Et= { x𝔥: δ(x)=t }

If t>0 this is an affine hyperplane in 𝔥, hence of dimension l. If t=0, E0=Ker(δ) =𝔥/𝔠 , where as usual 𝔥 is the subspace of 𝔥 spanned by h1,,hn. (For δ(hi)=0, 1in .) Each Et,t>0 is stable under the action of W because W fixes δ. Moreover W acts faithfully on Et(t>0). For if wW fixes Et pointwise, it fixes all points of 𝔥 not in E0, hence fixes all points of 𝔥 (for the fixed point set of w is in any case a vector subspace of 𝔥), whence w is the identity. Thus we have realized W as a group of affine-linear transformations of Et (any t>0). These actions of W are all essentially the same, so we may as well take 𝔠=1 and concentrate attention on the affine space E1.

Now the restriction to 𝔥 of the scalar product x,y is the positive semidefinite, or rank n-1 (because hi,hj =aijεj , and the matrix AE is positive semidefinite of rank n-1); also x,c=0 for all x𝔥, by (8). Hence we have a positive definite scalar product on E0=𝔥/𝔠, and therefore E1 has the structure of a Euclidean space of dimension l. The restriction of αi to E1 is an affine-linear function on this Euclidean space, and Hi= { xE1: αi(x) =0 } is an affine hyperplane in E1; these hyperplanes are the focus of an l–simplex S in E1, namely S=CE1. The generator wi of W acts on E1 as reflection in the hyperplane Hi (1in) . Thus W is realized as a group generated by reflections in the Euclidean space Esubscript?.

The transforms wS(wW) of the "fundamental alcove" S therefore fill up the space E1. Consequently the union X of the chambers wC for all wW is the open half space { x𝔥: δ(h)>0 } together with the origin. Pulling back to 𝔥, we see that the Tits cone is


where U= { h𝔥: δ(h)>0 } .

Real and imaginary roots

By (2.1) and (2.5) each αwαi (wW,1in) is a root of multiplicity mα=1. Kac calls these the real roots. In the classical case, where the Cartan matrix is of finite type, all the roots are real (proof later). In general however there will be other roots as well, whcih Kac calls imaginary roots. (The justification for this terminology will be apparent shortly.)

Let Rre,Rim denote the sets of real and imaginary roots, respectively. Also put

Rre+ = RreR+ positive real roots Rim+ = RimR+ positive imaginary roots

So by definition

Rre= i=1n Wαi .

Likewise for the dual root system R, with simple roots hi (1in) :

Rre= i=1n Whi .

Consider the real roots first. If α=wαi Rre define the coroot of α to be


This definition is justified by (2.7), because if also α=wαj , then we have αi=w-1 wαj and therefore hi=w-1 whj , i.e., whi=w hj .

We have then

α(hα)= (wαi) (whi)= αi(hi) =2 .

Next, for a real root α we define wα (acting on 𝔥 and on 𝔥*) by the formula

wα(h) = h-α(h) hα (h𝔥) wα(λ) = λ-λ(hα) α(λ𝔥)

and we verify easily that

  1. if α=wαi, then wα=wwi w-1W ;
  2. wα2=1, det(wα) =-1;
  3. wα(α)= -α,wα (hα)=- hα .

(2.28) The mapping αhα is a bijection of Rre into Rre such that

  1. hαi=hi (1in)
  2. hw.α= whα ( αRre,wW ) i.e. it is W–equivalent
  3. α>0hα>0 .


(i), (ii) are clear. As to (iii), let α=wαi, then since α>0 we have αiS (w-1) , hence by (2.10) l(wiw-1) =l(w-1)-1 , hence again by (2.10) (applied this time to R) whi>0, i.e. hα>0.

Next we consider the imaginary roots.


  1. If α is real (resp. imaginary) so is -α.
  2. αRim+ WαRim+ . Thus the set of positive imaginary roots is table under W.
  3. If α is a real root, the only multiples of α which are roots are ±α.
  4. If α is an imaginary root, then rα is an (imaginary) root for all integers r0.

  1. Let α=wαi, then -α=wwiαi is real.
  2. Let αRim+. Then ααi, hence wiα>0 by (2.5), hence wiαRim+. Hence WαRim+.

    Conversely, let αRre+, say w=wαi. Then wiw-1 α=-α<0 .

  3. If α=wαi and rα is a root, then rαi=w-1 (rα) is a root, hence r=MISSING (2.1).
  4. Proof later.

The next proposition justifies the names "real" and "imaginary".

(2.30) Assume that the Cartan matrix A is symmetrizable, and let λ,μ be a W–invariant symmetric bilinear form on 𝔥*, as in (2.26). If α is a root then

  1. α is real α,α>0 .
  2. α is imaginary α,α0 .


If α is real, say α=wαi, then α,α= αi,αi >0 (by our choice of scalar product).

Conversely, suppose αRim+. By (2.29) we have wα>0 for all wW, and wα,wα =α,α . Hence we may assume that α has minimum height in its orbit Wα, i.e. that ht(α) ht(wα) for all wW.

Since wiα=α-α (hi)αi , we have

ht(wiα)= ht(α)-α (hi)

and therefore α(hi)0 (1in), i.e. α,αi 0 and therefore also α,αi 0 . But α=miαi, say, with coefficients mi0, hence

α,α= i=1n mi α,αi 0 .


(2.31) Let βR and let αi be a simple root such that β±αi. Then the set S of integers r such that β+rαi is a root is a finite interval [-p,q] in , where p,q0 and p-q=β(hi). (If β) is positive all these roots are positive.


Without loss of generality we can assume β>0, say β=mjαj with coefficients mj0. If β+rαi is a root, we must have mi+r0 (because some mj,ji, is >0, since βαi), i.e. r-mi. It follows that the set S is bounded below (and is not empty, because 0S).

S is in any case a disjoint union of intervals in . Suppose I is one of these intervals, and consider the vector space

V=rI 𝔤β+rαi .

Then V is stable under adei and adfi – for example, [ ei, 𝔤β+rαi ] 𝔤β+(r+1)αi which is either contained in V or is zero. Hence V is stable under wi= eadei e-adfi aadei . From (2.5) it follows that the set { β+rαi:r I } is stable under wi. But now, since

wi(β+rαi)= β- ( β(hi)+r ) αi

it follows that the mapping r- (r+β(hi)) maps the interval I onto itself. Since I is bounded below (because S is), I must be a finite interval with midpoint -12β(hi). Hence all the component intervals of I have the same midpoint, hence there is only one component, i.e. S=I is an interval [-p,q] with midpoint 12(-p+q)=- 12β(hi) , i.e. p-q=β(hi) (and p,q0 because OS).

The set { β+rαi:- prq } is the αi–string through β.

Corollary. If β+αi is not a root, then β(hi)0. (For q=0 in (2.31).)


  1. This result enables us to list the roots systematically (though not their multiplicities). Clearly it is enough to consider positive roots. Suppose that we have listed all the positive roots of heigh m. Each root of height m+1 is of the form β+αi, where β is a root of height m, by (2.1)(v). By assumption, the negative values of r for which β+rαi is a root are known, hence in the notation of (2.28) p is known, hence also q=p-β(hi). So for each root β of height m and each simple root αi we can decide whether or not β+αi is a root. So we could define R axiomatically in this way.
  2. (2.31) valid for any real root α: if βR, αRre, then the α–string through β is


    where p,q0 and p-q=β(hα).

    For α=wαi; now apply (2.31) to w-1β and αi: β+rα=w (w-1β+rαi) , w-1β(hi)= β(whi)=β (hα) .

The following results should have occurred in Chapter I: they are valid for any matrix A (satisfying the condition aji=0 iff aij=0, so that the graph Γ of A is defined). If J is any subset of {1,2,,n} we have the principal submatrix AJ and its graph ΓJ, which is the full subgraph of Γ obtained by deleting the vertices of Γ not belonging to J. If γJ is connected we shall say simply that J is connected.

Now let α=1nmiαi be any element of Q. The support of α, denoted by Supp(α), is defined to be the set

Supp(α)= {i:mi0} .

(2.32) Let αR. Then Supp(α) is connected.


Let J=Supp(α), then αRJ, the root system of 𝔤(AJ). If J is not connected then AJ is decomposable, say AJ= ( AJ1 0 0 AJ2 ) and hence (1.12) 𝔤(AJ)=𝔤 (AJ1)𝔤 (AJ2) . Hence the root space 𝔤α lies in either 𝔤(AJ1) or 𝔤(AJ2); in either case, Supp(α)J: contradiction.

(2.33) Suppose R is infinite. Then for each αR+u such that α+αiR+.


Suppose not, then αR+ such that α+αiR+ (1in). Let x𝔤α, x0. Then [x,ei]=0 (1in) (because 𝔤α+αi=0), from which it follows that U(𝔫+)·x=kx and therefore the ideal 𝔞=U(𝔤)·x generated by x in 𝔤(A) is

𝔞=U(𝔫-)U (𝔥)U(𝔫+)·x =U(𝔫-)·x

Hence 𝔞β=0 unless βα. But by (1.13) we have 𝔞𝔤(A) (because clearly 𝔞𝔠), hence in particular 𝔞𝔫+, i.e. 𝔞β=𝔤β for all positive roots β. Hence all βR+ satisfy βα, whence R+ (and therefore R) is finite.

Minimal imaginary roots

Let α be a positive imaginary root. By (2.29) wα is positive for all wW. We shall say that α is minimal if ht(α) ht(wα) for all wW.

This implies in particular that ht(α) ht(wiα) (1in) ; since wiα=α-α (hi)αi , it follows that α(hi)0 (1in) , i.e. that -αC, the dual fundamental chamber. If also wα is minimal, then -wαC, and therefore by (2.13)(i) (applied in the dual situation, i.e. to At) we have α=wα. Thus each W–orbit of positive imaginary roots has a unique minimal element.

Conversely, if αR+ is such that α(hi)0 (1in) , then -αC and hence by (2.13)(iii) we have -α-wα for all wW, i.e. αwα; hence wα>0, and therefore α is a minimal positive imaginary root.

If α=mjαj, then α(hi)=mj αj(hi) aijmj . Thus for al minimal positive imaginary root the vector m=(m1,,mn)t satisfies

m0,m0, Am0

from which it follows from (2.17) that the Cartan matrix A cannot be of finite type. Thus if A is of finite type there are no imaginary roots.

From these remarks and (2.32) it follows that the minimal imaginary root α satisfies

(i) Supp(α) is connected; (ii) α(hi)0 (1in) .

In fact these necessary conditions are also sufficient:

(2.34) Let αQ+,α0. Then the following conditions are equivalent:

  1. Supp(α) is connected, and α(hi)0 (1in) ;
  2. α is a minimal positive imaginary root.


We have just observed that (ii) (i).

(i) (ii). From the remarks above, it is enough to prove that α is a root.

Suppose then that α is not a root. Let J=Supp(α), so that say

α=jJkj αj

Since α(hj)0 for all j it follows as above that the matrix AJ is not of finite type.

Choose a positive root βα of maximal height, say

β=jJmj αj .

Let γ=α-β, say

γ=jJ njαj ( nj=kj -mj )

  1. Supp(β)=J. For if not we can choose jSupp(β) and iJ-Supp(β) such that aij0, because J is connected. We have mi=0,ki1, hence β+αiα, so that (by the maximality of β) β+αi is not a root, hence by (2.31) β(hi)0. But now

    β(hi)= kSupp(β) mkαk(hi)= kaikmk ;

    the mk are >0, the aik are 0 (because ik), and at least one (namely aij) is <0. Hence β(hi)<0, contradiction.

  2. Supp(γ)J. Since AJ is not of finite type, the corresponding root system RJ is infinite (2.27), hence β+αiRJ for some iJ by (2.33). Again by the maximality of β it follows that β+αiα, hence mi=ki and therefore ni=0, i.e. iSupp(γ).
  3. Let S be a connected component of Supp(γ) and set

    β=jS mjαj, β= jJ-S mjαj

    (so that β=β+β). Firstly, if iS we have ni>0, i.e. ki>mi, hence β+αiα, hence β+αiR (by the maximality of β again) hence

    β(hi)0, alliS (1)

    j (2.31). On the other hand, for all maht iS we have

    β(hi)= jSmj aij0

    and for some iS we have β(hi)<0, otherwise aij=0 for all iS and all jJ-S, impossible since J is connected:

    β (hi)0 , alliS β (hi)<0 , someiS } (2)

    From (1) and (2) it follows that

    β(hi) 0 , alliS β(hi) >0 , someiS } (3)

    Let mS denote the vector (mj)jS; then (3) says that

    mS>0m ASmS0m ASmS0

    from which we conclude (2.17) that the matrix AS is of finite type.

  4. Now consider

    γ=iS njαj .

    Since S is a component of Supp(γ), we have aij=αj (hi)=0 for all iS and jSupp(γ)-S, whence for iS, γ(hi)= γ(hi)=α (hi)-β(hi) 0 by (1):

    γ(hi) 0,alliS (4)

    Let nS denote the vector (nj)jS; then (4) says that

    nS>0, ASnS0

    and hence by (2.17) AS is not of finite type. This contradiction completes the proof.

An immediate corollary of (2.34) is the last part of (2.29): if α is an imaginary root, then so is rα for all integers r0.

For we may assume that α is positive and minimal, and then rα satisfies the conditions of (2.34) for any integer r1.

(2.35) Let A be an indecomposable Cartan matrix.

  1. If A is of finite type, there are no imaginary roots.
  2. If A is of affine type, the imaginary roots are mδ (m,m0) , where as before δ=1n aiαi (and the ai are the labels in Table A).
  3. If A is of indefinite type, there exist positive imaginary roots α=1n kiαi such that ki>0 and α(hi)<0 for 1in.

  1. already observed.
  2. We have δ(hi)=0 (1in), and Supp(δ) is connected (all the ai are >0). Hence by (2.34) δ is an imaginary root, hence so is mδ (m,m0) by (2.29)(iv). Conversely, let α be a minimal positive imaginary root; then α(hi)0 (1in), hence by (2.17) α(hi)=0 (1in), whence α is a scalar multiple of δ.
  3. By (2.17) the inequalities x>0,Ax<0 have a solution; hence the cone P(-K) has non empty interior, and therefore contains points of the integer lattice n. In other words the inequalities x>0,Ax<0 have a solution xn, say x= (k1,,kn)t . Let α=kiαi, then ki>0 and α(hi)<0 (1in) ; and by (2.31) α is a (minimal) positive imaginary root.

Let A be a symmetrizable Cartan matrix. The standard bilinear form λ,μ on 𝔥* may clearly be chosen so that the scalar products αi,αj are integers. It follows that the number

a=min { α2: αQ, α2>0 }

exists and is >0. (Notation: α2= α,α .)

(2.36) Let A be an indecomposable Cartan matrix of finite or affine or symmetrizable hyperbolic type. If αQ and α2a, then αQ+ or -αQ+.


Suppose not, then α=β1-β2, where β1,β2Q+-{0}, and the support S1,S2 of β1,β2 respectively are disjoint. We have

a β1-β22 =β12+ β22-2 β1,β2. (1)

Suppose first that all components of S1 and of S2 are of finite type. Then B12 and B22 are >0, hence a; also B1,B20 (because αi,αj0 if ij). But this contradicts (1).

Suppose then that S1 (say) has a component of affine type. Then this component is the whole of S1, and S2 consists of a single vertex j, by virtue of (2.18); moreover by connectedness we have αi,αj<0 for some iS1, whence β1,β2<0. But this time we have B120 and B22a, whence again (1) is contradicted.

(2.37) Let A be as in (2.36).

  1. Let αQ be such that α2=a. Then α is a real root, and hence

    a=min1in αi2 (short roots)

  2. Let b=max1in αi2 , and let α=miαiQ be such that α2=b. Then α is a root iff miαi2/b (1in) (long roots)
  3. Let αQ, α0. Then αRim iff α20.


(i) We have wα2=a for all wW, hence wαQ+-Q+ by (2.36). Replacing α by -α if necessary, we may assume that αQ+. Let β be of minimal height in WαQ+, say β=miαi. Then we have

a=β,β=mi αi,β,

so that αi,β>0 for some index i, i.e. β(hi)>0. But then wiβ=β-β (hi)αi has height less than ht(β), hence wiβ-Q+ and therefore mj=0 for ji, i.e. β=miαi; but then mi=1, because a=β2=mi2 αi2mi2a . Hence β=αi and therefore α is a real root, and a=αi2 min1jn αj2 .

(ii) Suppose α is a long (real) root of R. Then α is a short real root of R; but α=2α/ α2= miαi2b ·αi , so that miαi2/b for all i. Conversely if this contradiction is satisfied, then αQ and α2=4/b=a hence αR and therefore αR.

(iii) If αRim, then α20 by (2.30). Consequently, suppose αQ-{0} and α20. By (2.36) we may assume αQ+. Again choose β of minimal height in WαQ+, say β=miαi. This time

miβ,αi =β,β= α20 .

Suppose β,αj0 for some index j; then ht(wjβ)< ht(β) , whence wjβ-Q+, which as before implies that β=mjαj, whence β2>0. Consequently β,αj0, i.e. β(hj)0, for 1jn. To complete the proof, by (2.34) it is enough to show that S=Supp(β) is connected. If not, let S1 be a component of S, and S2=S-S1, and put βi=jSi mjαj (i=1,2) . Then S1,S2 are of finite or affine type, whence β120 and β220; also β1,β2=0 (because aij=0 for all (i,j)S1×S2). Hence β2= β12+ β220 , contradiction.

Hence β is a minimal imaginary root, hence αRim+.

We can now describe the affine root systems explicitly.

Assume that the Dynkin diagram Δ is not of type BCl (l1) . Then there exists an index i such that ai=ai=1, by inspection of Table A. Denote this index by 0 and the others by 1,2,,l, where l=rank(A)=n-1. Let Q0=1lαi ,R0=RQ0, Δ0 the subdiagram of Δ obtained by erasing the vertex 0 from Δ. Then Δ0 is of finite type and R0 is a root system with Δ0 as its Dynkin diagram, hence finite by (2.27). As before let

a=min0il αi2= min1il αi2
b=max0il αi2= max1il αi2

Then b/a=1,2 or 3, and α2=a or b for all real roots αR (again by inspection of Table A). (This is not true for the excluded case BCl, where there are roots of 3 lengths.) Let Rre(s)= { αRre: α2=a } , Rre(l)= { αRre: α2=b } ; likewise R0(s), R0(l) (short roots and long roots). (If a=b then Rre(s)= Rre(l)= Rre .) Finally let

k=b/α02 =1,2or3

so that k=1 if either Δ is simply-laced or α0 is a long root.

(2.38) If R is of affine type (Table A) but not of type BCl(l1), then

  1. Rre(S)= R0(S)+δ ;
  2. Rre(l)= R0(l)+kδ .


Let α=0lmi αiQ , then β=α-m0δq0 (because a0=1). We have α2= β2 and β=1lniαi , where

ni=mi-m0ai (1il) .

  1. αRre(S) α2=a β2=a β R0(S) (2.34)(i)

  2. αRre(l) α2=b andmi αi2/b (0il) (2.34)(ii) β2=b, m0kand ni αi2/b (1il) m0kand βR0(l) .

Finally, if R is of type BCl (l1), choose a0=1 (as before, but this time a0=2); then R0 is of type Bl, so that explicitly the roots β in R0 are

±εi,± εi±εj (i<j)

where εi,εj =δij , so that β2=1 or 2.

In R we have α2=1,2 or 4 (for a real root): short, medium and long. One finds

Rre(s) = R0(s)+δ Rre(m) = R0(l)+δ (empty ifl=1) Rre(l) = 2R0(s)+ (2+1)δ=2 Rre(2)+δ.

(2.39) Let A be an indecomposable Cartan matrix, X𝔥 the Tits cone (here k=). Then the closure of X (in the usual topology of h) is given by

hXα(h) 0for all αRim+ .


This is clear if A is of finite type, for then X=X=𝔥, and Rim+=. If A is of affine type we have seen earlier that hX iff either h𝔠 or δ(h)>0, so that hX iff δ(h)0; and the positive imaginary roots are positive integer multiples of δ (2.33).

So assume that A is of indefinite type, and let

X= { h𝔥:α(h) 0,allα Rim+ }

Clearly the fundamental chamber CX, and X is W–stable by virtue of (2.29)(i or ii?). Hence wCX for all wW, and therefore XX; moreover X is closed, because it is an intersection of closed half-spaces, so that XX.

Conversely, let hX and assume first that αi(h) (1in) . Choose a positive imaginary root β such that β(hi)<0 (1in) (2.35)(iii). To show that hX it is enough by (2.13)(iv) to show that there are only finitely many positive real roots α such that α(h)<0, i.e. such that α(h)-1.

For such an α we have

wα(β)=β-β (hα)α=β+rα say

where r=-β(hα) ht(hα) (because β(hi)-1). Since wα(β) is a positive imaginary root and hX, we have (β+rα)(h)0 and therefore

β(h)-rα(h) rht(hα)

So hα has height β(h), and therefore there are only finitely many possibilities for α. Hence hX. By replacing h by a rational scalar multiple of h, it follows that

hX,αi (h)Q (1in)hX .

But these h are dense in X, hence XX.

(2.33) has the following geometrical interpretation. Let Z be the positive imaginary cone, i.e. the cone in 𝔥* generated by the positive imaginary roots, i.e.,Z is the set of all finite linear combinations ciβi with ci0 and βiRim+. Then

(2.40) The cones X in 𝔥 and Z in 𝔥* are duals of each other.

When the Cartan matrix is of indefinite type, the only case (to my knowledge) in which the closure of X of the Tits cone can be explicitly described is that in which A is hyperbolic and symmetrizable.


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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