sign in sign up
<<
Home , Lie algebra

HOUSTON JOURNAL OF , Volume 1, No. 1, 1975.

A NOTE ON THE CARTAN

Morton Curtis and Alan Wiederhold

With each pair of roots of a Lie there is defined a normalized inner product which tums out to be an . These are called the Cartan Integers.In this note we show how these integers may be defined for a compact, connected, semisimpieLie without recourseto .They turn out to be intersection numbersof certain subgroupsor, alternatively,they are the degreesof certainmaps of thecircle S1 to itself.Along the way we get the roots and coroots ofa groupwithout Lie algebrasor even the . Our main motivation for doing without Lie algebrasis to be able to define and study these things for H-spaces. 1. Coroots and roots. Let G be a compact connectedLie group of dimensionn and rank r. Let T be a maximal in G, N its normalizerand W = N/T the . Thus we have P 0 >T >N • W-•-• 1, and, since T is abelian, W acts on T (by conjugation;i.e., if w E W and t E T, then w sendst to ntn -1 where n • p-l(w)). Let w • W be a reflection. Preciselythis meansthat if w: T -> T is lifted to the •: Rr • Rr, then• is reflectionin an (r-1)-hyperplaneH in Rr. Let T'= p-l(w).Since w2 = 1= p(T)we see that if x • T' thenx 2 • T. DEFINITION 1: For the reflection w E W we let V be the set of fixed points of the action of w on T.

Clearly V is a closedsubgroup of T, and since a neighborhoodof 1 in V is the image of a neighborhood of 0 in H under the covering map, we see that dim V = r-1. The identitycomponent V ø of V is thusan (r-1)-torus. Let qb'G -> G be the squaringmap

29 30 MORTON CURTIS and ALAN WIEDERHOLD •(x) = x2. Then•(T •) C V. Forif x • T• wehave x 2 • T and x(x2)x-1 = x2. DEFINITION2: Q= {tGTlxtx-1 =t -1 forxGT'}. Q is a closed subgroupof T. Clearly every element of Q • V is its own inverse; i.e., it is a square root of 1 in T. Thus Q c3V is finite and since dim V = r-1 we see that dim Q = 1. Thus Q is topologically a disjoint finite set of circles. LEMMA 1' •(T') is a componentof V. PROOF:We look at the fibers of•. If •(x) = •(y);i.e., x 2 - y2,we see that x(xy-1)x-1= x2y-lx-1= y2y-lx-1= yx-1= (xy-1)-I soxy -1 C Q. Thus the fibers of this (smooth) map • havedimension oneso •(T') isa closedconnected (r-1)-manifold in V and is thus a componentof V. Let QObe the identitycomponent of Q, sothat QO is a circlegroup. Let o• G QO be the squareroot of 1 (4= 1) in QO.Obviously o• G V. LEMMA 2: V hasat most two components. PROOF: If V hadthree or morecomponents QO and V wouldhave at leastthree pointsin common,but QOc3 V = { 1,o•} sinceeach point of thisintersection must be a squareroot of 1 in QO. We havenoted that the identitycomponent V ø of V is an(r-1)-torus. By taking an appropriateline segmentin Rr andprojecting to T by the coveringmap we canget a circle subgroupC of T suchthat C rhV o = { 1 }. Then C and V ø togetherare a "coordinatesystem" for T. In particular,any homomorphismof T to anothergroup is determinedby specifyingits restrictionto C and V ø. We are goingto define a homomorphism0' T -• S1 (a root) corresponding tothe reflection w G W. Let U = VøU •(T'). This is goingto be the kernelof the homomorphism0.

There are two cases:

(i) a G U and (ii) u. (Case(i) occurswhen the simisimple group associated to the root is S3; case (ii) occurs when it is S0(3) .) In case(i) the homomorphism is to be suchthat the combined map QOincl.• T 0 >S1 has degree 2.In case (ii) this combined mapis to have degree 1. To accomplishthese things we specify the degreeon C. In case(i) we may have A NOTE ON THE CARTAN INTEGERS 31 o•• V ø in which casethe degreeon C is to be 1 (sinceQO already"goes around twice"). If ½(T•) :# V ø then o•must be in ½(T') and then 0 is to havedegree 2 on C (sinceQO goesaround only once).Now case(ii) can occuronly if •(T •) = V ø (and o•• Vø), soin thiscase the degreeon C is to be 1. DEFINITION3:The homomorphism0:T-•S 1 just definedis the root associatedwith the reflectionw. The homomorphicinclusion QO_> T is the coroot associated with the reflection w.

Since U is the of 0 it is a subgroup of T. Let P be the centralizer of U; i.e., P= {x•Glxu=uxforallu•U).

LEMMA 3: P is connected.

PROOF: If U is connected it is a torus and the centralizer of a torus is connected

([ 1], Prop. 4.25). If Z( ) denotes"centralizer of", then we have zo(u) _c z(u) c_z(uo).

The end terms are connected and all terms are closed manifolds. Furthermore they all have dimension•-2 becauseZø(U)/U and Z(Uø)/U ø both have rank 1 and hence dimension 3 (recall dim U = r-l). Thus all three terms are equal and Z(U) is connected. LEMMA4: Letj: T->P be theinclusion and let •: S1 ->T bethe inclusion of thecoroot QO. Then in •rl(P)we have 21#I•1=o. PROOF: Let p be a path in P from 1 to a point x in T' (P is connectedand, clearly, T'C P). Then conjugationby p(t) givesa homotopyfrom the identity to conjugationby x whichis the inversemap on QO.This proves the lemma. 2. CartanIntegers. Let Wl,W2,...,wk be all of the reflectionsin W. These generateW, and,if G is semisimpiesome r of themgenerate W. For eachw i wehave definedinsection 1 a coroot13i: S1 •T anda root0i: T->S 1. Thus for each pair wi,wj we have a map S1 13i)T 0j •S1 andwe denote by dij the degree of 0j13i. Alsofor eachw i we havedefined the degreedi of [13i] in el(G). By Lemma4 di is either 1 or 2. DEFINITION4: Foreach pair wi,w j of reflections wehave an integer Cji = didij- These are called the Cartan Integers of G. 32 MORTON CURTIS and ALAN WIEDERHOLD

THEOREM:If G is semisimpieand Wl,W2..... w r generateW, thenthe Cartan Integers defined above agree with the usual Cartan Integers. PROOF: Let (,) be an inner product on œ(T)* (the dual of the Lie algebraœ(T) of T) which is invariant under the action of the Weyl group ([1], page 116). The Cartan Integers are defined by

(oi, oj) cji= 2 (0 i, 0i ) or, equivalently, Cji= 0j(ri) whereri is the ith basic translate ([2], page 41). Firstwe note that the degree dij definedabove is the same as the intersection numberofQ? and Uj which we denote by Q?91 Uj. Now it suffices to prove the theorem in case G is simple. We first prove the theoremwhen G is alsosimply connected. Then we want to showthat Cji = dij= Qi91 Uj. In œ(T)we have the line œ(Qi) and the hyperplane œ(Uj), œ(Uj) being in a familyofequally spaced parallel hyperplanes; viz.d0jl(z), with œ(Uj) =d0jl(0). _ Let œ(Qi) be the segmentof œ(Qi) from0 outjust far enoughto coverQi once underthe exponential map. We see that Qi 91Uj (atleast in absolute value) is just the numberofthe hyperplanes ind0jl(z) crossed byœ-•Qi) (including oneend of•(Qi)). (Notethat if 0i and0j areperpendicular roots then Qi c Uj andthe intersection number is zero.) Sincewe areassuming rrl(G) = 0 wehave that the basic translates { r k } generate

-- theinteger I. Soin the simplyconnected case œ(Qi ) = r i. Thuswe have IQi 91UjI = 10j(ri)l= Irjil. To getthe signswe needto considerorientations. We orient the Qi's and Ui's so thateach Qi 91Ui ispositive. Then Qi 91Uj andQj 91U i willhave the same sign (since Qi is perpendicularto Ui). Sowe orient the Qi's and Ui's by starting at anyend of the Dynkindiagram and orient each new Qj and Uj togive the right sign. Since the contains no cycles this is always possible. Now suppose•rl(G) is not necessarilyzero. Then the numberof pointsof the integerlattice lying on r i (countingone end) is just the orderof Qi in •rl(G), andwe A NOTE ON THE CARTAN INTEGERS 33 mustmultiply the intersection number by this to get 0j(r i) = Cji.

REFERENCES

1. J. F. Adams,Lectures on Lie Groups,W. A. Benjamin,Inc. (1969). 2. Hans Samelson,Notes on Lie ,van Nostrand Reinhold Mathematical Studies No. 23 (1969).

Rice University Houston, Texas ReceivedMay 2, 1975