with the empty product. So we may assume that n(s) >0, so s sends some positive root to a negative root, and hence must send some simple root to a negative root. So let α ∈ ∆ be such that sα ∈ Φ−. Since µ ∈ C, we have (µ, β)≥0, ∀β ∈Φ+ and hence (µ, sα)≤0. So
0 ≥ (µ, sα)
= (s−1µ, α)
= (λ, α)
≥ 0.
So (λ, α) = 0 so sαλ =λ and hencessαλ= µ. But n(ssα) = n(s)−1 since sα = −α and sα permutes all the other positive roots. So `(ssα) = `(s)−1 and we can apply induction to conclude thats= (ssα)sαis a product of simple reflections which fixλ.
5.9 Conjugacy of Borel subalgebras
We need to prove this for semi-simple algebras since the radical is contained in every maximal solvable subalgebra.
Define a standardBorel subalgebra (relative to a choice ofCSA h and a system of simple roots, ∆) to be
b(∆) :=h⊕ M
β∈Φ+(∆)
gβ.
Define the corresponding nilpotent Lie algebra by n+(∆) := M
β∈Φ+
gβ.
Since eachsαcan be realized as (expeα)(exp−fα)(expeα) every element of W can be realized as an element of E(g). Hence all standard Borel subalgebras relative to a given Cartan subalgebra are conjugate.
Notice that ifxnormalizes a Borel subalgebra,b, then [b+Cx,b+Cx]⊂b
and so b+Cxis a solvable subalgebra containing band hence must coincide withb:
Ng(b) =b.
In particular, ifx∈bthen its semi-simple and nilpotent parts lie inb.
From now on, fix a standardBSA,b. We want to prove that any otherBSA, b0is conjugate tob. We may assume that the theorem is known for Lie algebras of smaller dimension, or for b0 with b∩b0 of greater dimension, since if dim
b∩b0 = dimb, so thatb0⊃b, we must haveb0=bby maximality. Therefore we can proceed by downward induction on the dimension of the intersection b∩b0.
Suppose b∩b0 6= 0. Let n0 be the set of nilpotent elements in b∩b0. So n0=n+∩b0.
Also [b∩b0,b∩b0] ⊂ n+∩b0 = n0 so n0 is a nilpotent ideal in b∩b0. Suppose thatn06= 0. Then sincegcontains no solvable ideals,
k:=Ng(n0)6=g.
Consider the action ofn0onb/(b∩b0). By Engel, there exists ay6∈b∩b0 with [x, y] ∈b∩b0 ∀x∈ n0. But [x, y]∈[b,b] ⊂n+ and so [x, y] ∈n0. Soy ∈k.
Thus y ∈k∩b, y 6∈b∩b0. Similarly, we can interchange the roles ofband b0 in the above argument, replacing n+ by the nilpotent subalgebra [b0,b0] of b0, to conclude that there exists ay0∈k∩b0, y06∈b∩b0. In other words, the inclusions
k∩b⊃b∩b0, k∩b0⊃b∩b0 are strict.
Both b∩k and b0∩k are solvable subalgebras of k. Let c,c0 be BSA’s containing them. By induction, there is aσ∈ E(k)⊂ E(g) withσ(c0) =c. Now letb00 be aBSAcontainingc. We have
b00∩b⊃c∩b⊃k∩b⊃b0∩b
with the last inclusion strict. So by induction there is aτ∈ E(g) withτ(b00) =b.
Hence
τ σ(c0)⊂b.
Then
b∩τ σ(b0)⊃τ σ(c0)∩τ σ(b0)⊃τ σ(b0∩k)⊃τ σ(b∩b0)
with the last inclusion strict. So by induction we can further conjugate τ σb0 intob.
So we must now deal with the case that n0 = 0, but we will still assume thatb∩b06= 0. Since any Borel subalgebra contains both the semi-simple and nilpotent parts of any of its elements, we conclude thatb∩b0 consists entirely of semi-simple elements, and so is a toral subalgebra, call itt. Ifx∈b, t∈t= b∩b0and [x, t]∈t, then we must have [x, t] = 0, since all elements of [b,b] are nilpotent. So
Nb(t) =Cb(t).
Letcbe aCSAofCb(t). Since a Cartan subalgebra is its own normalizer, we havet⊂c. So we have
t⊂c⊂Cb(t) =Nb(t)⊂Nb(c).
Lett∈t, n∈Nb(c). Then [t, n]∈cand successive brackets bytwill eventually yield 0, sincecis nilpotent. Thus (adt)kn= 0 for somek, and sincetis semi-simple, [t, n] = 0. Thusn∈Cb(t) and hencen∈csincecis its own normalizer
5.9. CONJUGACY OF BOREL SUBALGEBRAS 91 in Cb(t). Thuscis aCSAofb. We can now apply the conjugacy theorem for CSA’s of solvable algebras to conjugate cinto h.
So we may assume from now on that t⊂ h. If t =h, then decomposing b0 into root spaces underh, we find that the non-zero root spaces must consist entirely of negative roots, and there must be at least one such, since b0 6=h.
But then we can find aταwhich conjugates this into a positive root, preserving h, and thenτα(b0)∩bhas larger dimension and we can further conjugate into b.
So we may assume that
t⊂h is strict.
If
b0 ⊂Cg(t)
then since we also haveh⊂Cg(t), we can find aBSA,b00ofCg(t) containingh, and conjugateb0 tob00, since we are assuming thatt6= 0 and henceCg(t)6=g.
Sinceb00∩b⊃hhas bigger dimension thanb0∩b, we can further conjugate to bby the induction hypothesis.
If
b0 6⊂Cg(t)
then there is a common non-zero eigenvector for ad tin b0, call it x. So there is at0 ∈tsuch that [t0, x] =c0x, c06= 0. Setting
t:= 1 c0t0
we have [t, x] =x. Let Φt⊂Φ consist of those roots for whichβ(t) is a positive rational number. Then
s:=h⊕ M
β∈Φt
gβ
is a solvable subalgebra and so lies in aBSA, call itb00. Sincet⊂b00, x∈b00 we see that b00∩b0 has strictly larger dimension thanb∩b0. Alsob00∩bhas strictly larger dimension thanb∩b0 sinceh⊂b∩b00. So we can conjugateb0 to b00 and thenb00 tob.
This leaves only the caseb∩b0 = 0 which we will show is impossible. Let t be a maximal toral subalgebra of b0. We can not have t = 0, for then b0 would consist entirely of nilpotent elements, hence nilpotent by Engel, and also self-normalizing as is everyBSA. Hence it would be aCSAwhich is impossible since every CSA in a semi-simple Lie algebra is toral. So choose a CSA, h00 containing t, and then a standard BSAcontainingh00. By the preceding, we know thatb0is conjugate tob00and, in particular has the same dimension asb00. But the dimension of each standard BSA(relative to any Cartan subalgebra) is strictly greater than half the dimension of g, contradicting the hypothesis g⊃b⊕b0. QED
Chapter 6
The simple finite
dimensional algebras.
In this chapter we classify all possible root systems of simple Lie algebras. A consequence, as we shall see, is the classification of the simple Lie algebras themselves. The amazing result - due to Killing with some repair work by ´Elie Cartan - is that with only five exceptions, the root systems of the classical algebras that we studied in Chapter III exhaust all possibilities.
The logical structure of this chapter is as follows: We first show that the root system of a simple Lie algebra is irreducible (definition below). We then develop some properties of the of the root structure of an irreducible root system, in particular we will introduce its extended Cartan matrix. We then use the Perron-Frobenius theorem to classify all possible such matrices. (For the expert, this means that we first classify the Dynkin diagrams of the affine algebras of the simple Lie algebras. Surprisingly, this is simpler and more efficient than the classification of the diagrams of the finite dimensional simple Lie algebras themselves.) From the extended diagrams it is an easy matter to get all possible bases of irreducible root systems. We then develop a few more facts about root systems which allow us to conclude that an isomorphism of irreducible root systems implies an isomorphism of the corresponding Lie algebras. We postpone the the proof of the existence of the exceptional Lie algebras until Chapter VIII, where we prove Serre’s theorem which gives a unified presentation of all the simple Lie algebras in terms of generators and relations derived directly from the Cartan integers of the simple root system.
Throughout this chapter we will be dealing with semi-simple Lie algebras over the complex numbers.
93
6.1 Simple Lie algebras and irreducible root sys-tems.
We choose a Cartan subalgebrahof a semi-simple Lie algebrag, so we have the corresponding set Φ of roots and the real (Euclidean) space E that they span.
We say that Φ is irreducible if Φ can not be partitioned into two disjoint subsets
Φ = Φ1∪Φ2
such that every element of Φ1is orthogonal to every element of Φ2. Proposition 17 Ifgis simple thenΦis irreducible.
Proof. Suppose that Φ is not irreducible, so we have a decomposition as above.
Ifα∈Φ1andβ ∈Φ2then
(α+β, α) = (α, α)>0 and (α+β, β) = (β, β)>0
which means thatα+β can not belong to either Φ1or Φ2and so is not a root.
This means that
[gα,gβ] = 0.
In other words, the subalgebra g1 of g generated by all the gα, α ∈ Φ1 is centralized by all thegβ, sog1 is a proper subalgebra ofg, since ifg1=gthis would say that ghas a non-zero center, which is not true for any semi-simple Lie algebra. The above equation also implies that the normalizer ofg1contains all thegγ whereγ ranges over all the roots. But thesegγ generateg. Sog1 is a proper ideal ing, contradicting the assumption thatgis simple. QED
Let us choose a base ∆ for the root system Φ of a semi-simple Lie algebra.
We say that ∆ is irreducible if we can not partition ∆ into two non-empty mutually orthogonal sets as in the definition of irreducibility of Φ as above.
Proposition 18 Φis irreducible if and only if∆ is irreducible.
Proof. Suppose that Φ is not irreducible, so has a decomposition as above.
This induces a partition of ∆ which is non-trivial unless ∆ is wholly contained in Φ1 or Φ2. If ∆⊂Φ1say, then sinceE is spanned by ∆, this means that all the elements of Φ2are orthogonal toEwhich is impossible. So if ∆ is irreducible so is Φ. Conversely, suppose that
∆ = ∆1∪∆2
is a partition of ∆ into two non-empty mutually orthogonal subsets. We have proved that every root is conjugate to a simple root by an element of the Weyl group W which is generated by the simple reflections. Let Φ1 consist of those roots which are conjugate to an element of ∆1 and Φ2 consist of those roots which are conjugate to an element of ∆2. The reflections sβ, β∈∆2 commute with the reflectionssα, α∈∆1, and furthermore
sβ(α) =α
6.2. THE MAXIMAL ROOT AND THE MINIMAL ROOT. 95