Wisconsin Bourbaki Seminar

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

Last update: 26 March 2014

Notes and References

This is an excerpt from notes of the Wisconsin Bourbaki Seminar, Fall 1993. The notes were written by Oliver Eng, Susan Hollingsworth, Mark Logan, Arun Ram and Louis Solomon.

§4

In the exercises below, (W,S) designates a Coxeter system. One supposes that S is finite and its cardinality is called the rank of (W,S). We identify W with a subgroup of GL(E) through σ.

  1. Let E0 be the orthgonal to E with respect to the form BM. Show that E0 is the radical of E, and that E/E0 is a direct sum of absolutely simple, pairwise nonisomorphic modules and that the number of them is the same as the number of connected componenets in the graph S.

    Solution. E is a real vector space with basis e1,,en. W acts on E by σs(e)=e-2BM (e,es)es, for each sS. By definition E0= { vE|BM (v,e)=0, for alleE } . We must show
    1. E0=rad(E), where rad(E) is the intersection of all maximal submodules of E.
    2. E/E0 is a direct sum of modules Lj corresponding to the connected components Γj of the graph Γ.
    3. The Lj are simple.
    4. The Lj are pairwise nonisomorphic.
    5. The Lj are absolutely simple, i.e. upon extending to they remain simple W-modules.
    1. Let xE0 and let M be a maximal submodule of E. Since xE0, σ(x)=x-2BM (x,es)es= x-0=x, for all sS, so x is fixed by W. Suppose xM. Then M+x is a submodule of E. Since M is maximal we must have that M+x=E. So for each basis element ei there exist miM and ci such that ei=mi+cix. Thus ei-ci(x)M and σi(ei-cix)=-ei-cixM. So ei-cix+(-ei-cix)=eiM. So M=E which is clearly a contradiction. So xM and we have that E0rad(E). Now let xrad(E). We will show that BM(x,e)=0 by showing that σs(x)=x for all sS. Suppose that there is some si such that σsi(x)x. Then rad(E)σsi(x)=x-2BM(x,esi)esi and BM(x,esi)0 since σsi(x)x. So esirad(E). Let Γi be the connected component of Γ containing esi If i and j are adjacent in Γi then BM(ei,ej)0 and since σsi(ei)=ei-2BM(ei,ej)ej we will get that ejrad(E). In this way we get ejrad(E) for all jΓi. If Γ is connected then Γi=Γ and we get that rad(E)=E. This is a contradiction since because E is finite dimensional there exist maximal submodules of E. So Γ is not connnected. For each k, labeling the components of Γ let Mj be the span of the basis vectors ek such that kΓj. Then Mj is a submodule of E. Let P=jiMj. Then PE. So P is contained in some maximal submodule M. All basis vectors in Γi are contained in rad(E) and hence in M. So M=E, but this is a contradiction. So we have that σsi(x)=x for all siS. So BM(ei,x)=0 for all i. So xrad(E). So rad(E)=E0.
    2. Let Γ1,,Γk be the connected components of the graph Γ. Let LjE/E0 be the span of the cosets ei+E0 for all iΓj. Then Lj is a submodule of E/E0 and it is clear that E/E0 is a direct sum of the Lj.
    3. Suppose that X is a nonzero submodule of Lj. Let x+E0X, xE0. Note that σsi(x)=x and BM(x,ei)=0 for all eiΓj. Since xE0 there exists ejΓj such that BM(x,ej)0. Then σj(x+E0)=x- 2BM(x,ej)ej +E0X. So ej+E0X. By a similar connectivity argument as above we have that eX for all Γj. It follows that X=Lj. Thus Lj is an irreducible submodule of E/E0.
    4. Suppose that ϕ:LjLk is a module homomorphism. Let ej be a vertex in Γj so that ej+E0Lj. Then ϕ(ej+E0)Lk and so σsj(ϕ(ej+E0))=ϕ(ej+E0). On the other hand σsj(ϕ(ej+E0)) =ϕ(σsj(ej+E0)) =ϕ(-ej+E0)=- ϕ(ej+E0). So ϕ(ej+E0)=E0 for each vertex ejΓj. Since BM(ej,ej)=10, ejE0 and we have that ϕ must be the zero map. Thus Lj and Lk are not isomorphic.
    5. Proposition 1, Chapt V §2, states that if G is a group, ρ an irreducible representation of G on a vector space V and if there exists gG such that ρ(g) is a pseudoreflection then ρ is absolutely irreducible. Let ej be a vertex in Γj and let ei be a vertex of Γj which is connected to ej. Since σsi(ej+E0) -(ej+E0)=-BM (ei,ej)ei+E0, and since BM(ei,ej)0 if i is connected to j in Γj it follows that σsi-10 and that σsi is a pseudo reflection on Lj. Thus it follows that Lj is absolutely irreducible.

    Notes. It seems that it would be instructive to understand this explicitly in the cases of A1 and A2.
  2. Suppose that Card(S)=3. If sS, let a(s)=m(u,v), where {u,v}=S-{s}. Let A=sS1/a(s).
    1. If A>1, show that BM is positive degenerate (in which case W is finite).
    2. If A=1, show that BM is positive degenerate.
    3. If A<1, show that BM is non degenerate, and of signature (2,1).
    Show that in case a), the order q of W is given by the formula q=4/(A-1).

    Solution. Recall that W acts on a vector space E over with basis e1,,en and that the form satisfies BM(ei,ej)= -cos(πm(si,sj)) =-cij where m(si,si)=1 and m(si,sj), m(si,sj)2, ij. Then the matrix of the form is B = ( 1 -c12 -c13 -c12 1 -c23 -c13 -c23 1 ) ,and det(B) = 1-c122- c132-c232 -2c12c13c23. The determinants of the principal minors are 1-cij2, ij and since cij1 we have that the determinants of the principal minors are all 0. Thus the form BM is positive if det(B)0.
    1. The possibilities such that A=1/a+1/b+1/c>1 are 1/2 1/2 x det(B)=1-0-0- cos2x-2·0=1- cos2x>0 1/2 1/2 1/3 det(B)=1-0-0- 1/4-2·0=3/4 1/2 1/3 1/4 det(B)=1-0-1/4- 1/2-2·0=1/4 1/2 1/3 1/5 det(B)=1-0-1/4- (1+54)2 -2·0=3-58 where x>0.
    2. The possibilities such that A=1/a+1/b+1/c=1 are 1/2 1/2 0 det(B)=1-0-0-1-2 ·0=0 1/2 1/3 1/6 det(B)=1-0- 1/4-3/4-2 ·0=0 1/2 1/4 1/4 det(B)=1-0- 1/2-1/2-2 ·0=0 1/3 1/3 1/3 det(B)=1- 1/4-1/4-1/4- 2(1/2)(1/2) (1/2)=0
    3. Given that cosx is a decreasing function for 0xπ it follows from b) that detB<0. We show that BM has signature (2,1).
      Case 1. Assume that BM(ei,ej)-1 for some i,j, ij. Without loss of generality we may assume that BM(e1,e2)=-c12-1. Then if we let v1 = e1, v2 = αe1+e2, v3 = βe1+γe2+e3, where α=c12<1 and β and γ are uniquely determined such that BM(vi,vj)=0. Since BM(v1,v1)= BM(e1,e1)=1, BM(v2,v2)= 1-α2>0, and det(B)<0 it follows that BM has signature (2,1).
      Case 2. BM(ei,ej)=-1 for all ij. Then the matrix of BM is given by B= ( 1-1-1 -11-1 -1-11 ) . One checks that ( 1-10 11-2 111 ) ( 1-1-1 -11-1 -1-11 ) ( 111 -111 0-21 ) = ( 1-10 11-2 111 ) ( 22-1 -22-1 0-4-1 ) = ( 400 0120 00-3 ) Thus BM has signature (2,1).
      We know that σsi,σsj is a dihedral group of order 2m(si,sj). Use the formula IS(-1)I |W||WI| =1, from Ex. 5 §3. Then we have that (-1)31|W| +(-1)2 12m(s1,s3) +(-1)2 12m(s2,s3) +(-1)2 12m(s1,s2) +(-1)31|W|+ (-1)12+(-1) 12+(-1)12+1= 1|W|. So (1/2)A-(3/2)+1 -1|W|=1|W|. Thus A/2-1/2=2/|W| and |W|=(A-1)/4.
    1. Let m be an integer 2 or +. Show that requiring 4cos2πm is equivalent to requiring m{2,3,4,6,+}.
    2. Let Γ be a discrete subgroup of E of rank dimE. Show that, if Γ is stable under W, then the integers m(s,s) for ss belong to the set {2,3,4,6,+}.
    3. Suppose that m(s,t){2,3,4,6,+} for stS. A family (xs)sS of positive real numbers is called radicielle if they satisfy the following conditions: m(s,t)=3 xs=xt m(s,t)=4 xs=2xt or xt=2xs m(s,t)=6 xs=3xt or xt=3xs m(s,t)=+ xs=xt or xs=2xt or xt=2xs. If (xs) is such a family, one puts αs=xses. Show that one has σs(αt) =αt-n (s,t)αs, wheren(s,t). Deduce that the discrete subgroup Γ with base (αs)sS is stable under W.
    4. With hypotheses as in c), suppose that the graph (W,S) is a forest. Show that then there exists at most one radicielle family (xs).

    Solution. a) and b) were done in class.
    1. If m(s,t)=2 then σs(αt) = αt-2BM (es,αt) es = αt-2xtBM (es,et)es = αt-2xt (-cos(π/2)) = αt and so n(s,t)=0.
      If m(s,t)=3 then σs(αt) = αt-2xtBM (es,et)es = αt-2xt (-cos(π/3)) = αt-xtes = αt+xtes = αt+αs and so n(s,t)=-1.
      If m(s,t)=4 then σs(αt) = αt-2xt (-cos(π/4)) = αt+2xtes = αt+ { 22xses (2/2) xses = αt+ { 2αs αs So n(s,t)=-2 or n(s,t)=-1.
      If m(s,t)=6 then σs(αt) = αt-2xt (-cos(π/6)) = αt+3xtes = αt+ { 33xses (3/3) xses = αt+ { 3αs αs So n(s,t)=-3 or n(s,t)=-1.
      If m(s,t)= then σs(αt) = αt-2xt (-cos(π/)) = αt+2xtes = αt+ { 2αs, ifxt=xs, αs, ifxt= (1/2)xs, 4αs, ifxt=2 xs. So n(s,t)=-2,-1, or -4. Note that in these cases we have that n(s,t)n(t,s)=4.
    2. Since the Coxeter graph is a forest we may let s0 be a pendant node. By induction we may assume that there is a radicelle family for the graph S-{s0}. Since s0 is connected to a unique node of S-{s0} we can certainly extend according to the rules above.

      Notes. Suppose that W is a finite reflection group with {es} a simple system such that |es|=1 for all sS. Define Φ=W{αs}sS where the αs are as given in c) above. We will show that Φα={±α} for all αΦ. Suppose the contrary. Then we may choose a simple system Δ such that αsΔ and such that Φαs{±αs}. So there exists wW, tS and c, |c|1, such that wαt=cαs. Then |xt|=|αt| =|c||αs| |αs|=|xs|. So t and s cannot be joined in the Coxeter graph by a sequence of edges labeled 3. Let A be the set of elements of S which canot be joined to s by edges labeled 3. Let F be the free group on S and let N be the normal closure of the set of relators for the Coxeter system so that F/NW. Define a map ϕ: F {±1} x { 1, ifxA, -1, ifxA. If xS then ϕ(x2)= ϕ(x)2=1 and if x,yS then ϕ((xy)m(x,y))= { 1, ifm(x,y)is even, 1, ifm(x,y)=3 andx,yA 1, ifm(x,y)=3 andx,yA. So ϕ factors through N and thus ϕ is a homomorphism from W to {±1}. Since φ(s)=1 and φ(t)=-1 it follows that σaαs and σαt are not conjugate in W. On the other hand if the form BM is invariant under W we have that wσαsw-1(v) = w ( w-1v-2BM (w-1v,αs) αs ) = v-2BM (v,wαs) wαs = σwαs = σαt. So σαs and σαt are conjugate which is a contradiction. Therefore Φα={±α} for all αΦ.

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