Cartan matrix A of ∆ determines Φ up to isomorphism. In particular, The Dynkin diagrams characterize all possible irreducible root systems.
Proof. Since ∆ is a basis of E and ∆0 is a basis of E0, the map αi 7→ α0i extends to a unique linear isomorphism of E onto E0. The equality in the theorem implies that forα, β∈∆ we have
sf(α)f(β) =f(β)− hf(β), f(α)if(α) =f(sαβ).
Since the Weyl groups are generated by these simple reflections, this implies that the map
w7→f◦w◦f−1
is an isomorphism ofW ontoW0. Everyβ ∈Φ is of the formw(α) wherew∈W andαis a simple root. Thus
f(β) =f◦w◦f−1f(α)∈Φ0
sof(Φ) = Φ0. Sincesα(β) =β−hβ, αiα, the numberhβ, αiis determined by the reflection sα acting onβ. But then the corresponding formula for Φ0 together with the fact that
sf(α)=f◦sα◦f−1 implies that
hf(β), f(α)i=hβ, αi.
QED
6.7 The classification of the possible simple Lie algebras.
Suppose that g,h, is a pair consisting of a semi-simple Lie algebra g, and a Cartan subalgebra h. This determines the corresponding Euclidean space E and root system Φ. Suppose we have a second such pair (g0,h0). We would like to show that an isomorphism of (E,Φ) with (E0,Φ0) determines a Lie algebra isomorphism of g with g0. This would then imply that the Dynkin diagrams classify all possible simple Lie algebras. We would still be left with the problem of showing that the exceptional Lie algebras exist. We will defer this until Chapter VIII where we prove Serre’s theorem with gives a direct construction of all the simple Lie algebras in terms of generators and relations determined by the Cartan matrix.
We need a few preliminaries.
Proposition 24 Every positive root can be written as a sum of simple roots αi1+· · ·αik
in such a way that every partial sum is again a root.
Proof. By induction (on say the height) it is enough to prove that for every positive root β which is not simple, there is a simple rootαsuch thatβ−αis a root. We can not have (β, α)≤0 for all α∈∆ for this would imply that the set{β} ∪∆ is independent (by the same method that we used to prove that ∆ was independent). So (β, α)>0 for someα∈∆ and soβ−αis a root. Since β is not simple, its height is at least two, and so subtractingαwill not be zero or a negative root, hence positive. QED
Proposition 25 Letg,hbe a semi-simple Lie algebra with a choice of Cartan subalgebra, Let Φ be the corresponding root system, and let∆ be a base. Then gis generated as a Lie algebra by the subspaces gα,g−α, α∈∆.
From the representation theory ofsl(2)αwe know that [gα,gβ] =gα+β ifα+β is a root. Thus from the preceding proposition, we can successively obtain all the gβ for β positive by bracketing the gα, α ∈ ∆. Similarly we can get all the gβ for β negative from the g−α. So we can get all the root spaces. But [gα,g−α] =Chα so we can get all ofh. The decomposition
g=h⊕M
γ∈Φ
gγ
then shows that we have generated all ofg.
Here is the big theorem:
Theorem 15 Letg,handg0,h0 be simple Lie algebras with choices of Cartan subalgebras, and let Φ,Φ0 be the corresponding root systems. Suppose there is an isomorphism
f : (E,Φ)→(E0,Φ0)
which is an isometry of Euclidean spaces. Extendf to an isomorphism of h∗→h0∗
via complexification. Let f :h→h0 denote the corresponding isomorphism on the Cartan subalgebras obtained by identifying hand h0 with their duals using the Killing form.
Fix a base ∆ ofΦ and∆0 ofΦ0. Choose 06=xα∈gα, α∈∆ and06=x0α0 ∈ g0α0. Extend f to a linear map
f :h⊕M
α∈∆
gα→h0⊕ M
α0∈∆0
gα0
by
f(xα) =x0α0. Thenf extends to a unique isomorphism of g→g0.
Proof. The uniqueness is easy. Givenxαthere is a uniqueyα∈g−αfor which [xα, yα] = hα so f, if it exists, is determined on the yα and hence on all of g since thexαandyαgenerategbe the preceding proposition.
6.7. THE CLASSIFICATION OF THE POSSIBLE SIMPLE LIE ALGEBRAS.111 To prove the existence, we will construct the graph of this isomorphism.
That is, we will construct a subalgebrak ofg⊕g0 whose projections onto the first and onto the second factor are isomorphisms:
Use thexαandyαas above, with the corresponding elementsx0α0 andyα00 in g0. Let
xα:=xα⊕x0α0 ∈g⊕g0 and similarly define
yα:=yα⊕y0α0, and
hα:=hα⊕h0α0.
Let β be the (unique) maximal root ofg, and choosex∈gβ. Make a similar choice of x0 ∈gβ00 where β0 is the maximal root ofg0. Set
x:=x⊕x0.
Letm⊂g⊕g0 be the subspace spanned by all the adyα
i1· · ·adyα
imx.
The element adyα
i1· · ·adyα
imxbelongs togβ−Pα
ij ⊕g0β0−Pα0
ij
so m∩(gβ⊕g0β0) is one dimensional.
In particularmis a proper subspace ofg⊕g0.
Letk denote the subalgebra ofg⊕g0 generated by thexα the yα and the hα. We claim that
[k,m]⊂m.
Indeed, it is enough to prove thatmis invariant under the adjoint action of the generators ofk. For the adyαthis follows from the definition. For the adhαwe use the fact that
[h, yα] =−α(h)yα
to move the adhα past all the adyγ at the cost of introducing some scalar multiple, while
adhαx=hβ, αixβ+hβ0, α0ix0β0=hβ, αix because f is an isomorphism of root systems.
Finally [xα1, yα2] = 0 ifα1 6=α2 ∈ ∆ sinceα1−α2 is not a root. On the other hand [xα, yα] = hα. So we can move the adxα past the adyγ at the expense of introducing an adhα every time γ =α. Now α+β is not a root, since β is the maximal root. So [xα, xβ] = 0. Thus adxαx= 0, and we have proved that [k,m]⊂m. But sincemis a proper subspace ofg⊕g0, this implies that kis a proper subalgebra, since otherwise mwould be a proper ideal, and the only proper ideals in g⊕g0 aregandg0.
Now the subalgebra kcan not contain any element of the formz⊕0, z6= 0, for it if did, it would have to contain all of the elements of the formu⊕0 since we could repeatedly apply adxα’s until we reached the maximal root space and then get all ofg⊕0, which would mean thatkwould also contain all of 0⊕g0 and hence all ofg⊕g0 which we know not to be the case. Similarlykcan not contain any element of the form 0⊕z0. So the projections ofkontogand ontog0 are linear isomorphisms. By construction they are Lie algebra homomorphisms.
Hence the inverse of the projection of kontogfollowed by the projection ofk ontog0 is a Lie algebra isomorphism ofgontog0. By construction it sendsxα
tox0α0 andhα tohα0 and so is an extension off. QED
Chapter 7
Cyclic highest weight modules.
In this chapter,gwill denote a semi-simple Lie algebra for which we have chosen a Cartan subalgebra,hand a base ∆ for the roots Φ = Φ+∪Φ− ofg.
We will be interested in describing its finite dimensional irreducible repre-sentations. If W is a finite dimensional module forg, then hhas at least one simultaneous eigenvector; that is there is a µ∈h∗ and aw6= 0∈W such that
hw=µ(h)w ∀h∈h. (7.1)
The linear function µ is called a weight and the vector v is called a weight vector. Ifx∈gα,
hxw= [h, x]w+xhw= (µ+α)(h)xw.
This shows that the space of all vectors w satisfying an equation of the type (7.1) (for varyingµ) spans an invariant subspace. IfW is irreducible, then the weight vectors (those satisfying an equation of the type (7.1)) must span all of W. Furthermore, sinceW is finite dimensional, there must be a vectorvand a linear functionλsuch that
hv=λ(h)v ∀h∈h, eαv= 0, ∀α∈Φ+. (7.2) Using irreducibility again, we conclude that
W =U(g)v.
The module is cyclic generated by v. In fact we can be more precise: Let h1, . . . , h` be the basis of h corresponding to the choice of simple roots, let ei∈gαi, fi∈g−αi whereα1, . . . , αmare all the positive roots. (We can choose them so that eacheandf generate a littlesl(2).) Then
g=n−⊕h⊕n+, 113
wheree1, . . . , emis a basis ofn+, whereh1, . . . , h`is a basis ofh, andf1, . . . , fm is a basis of n−. The Poincar´e-Birkhoff-Witt theorem says that monomials of the form
f1i1· · ·fmimhj11· · ·hj``ek11· · ·ekmm
form a basis of U(g). Here we have chosen to place all the e’s to the extreme right, with theh’s in the middle and thef’s to the left. It now follows that the elements
f1i1· · ·fmimv
spanW. Every such element, if non-zero, is a weight vector with weight λ−(i1α1+· · ·+imαm).
Recall that
µ≺λ means thatλ−µ=X
kiαi, αi>0,
where theki are non-negative integers. We have shown that every weightµ of W satisfies
µ≺λ.
So we make the definition: A cyclic highest weight module for gis a module (not necessarily finite dimensional) which has a vectorv+ such that
x+v+= 0, ∀ x+∈n+, hv+=λ(h)v+ ∀h∈h and
V =U(g)v+.
In any such cyclic highest weight module every submodule is a direct sum of its weight spaces (by van der Monde). The weight spacesVµ all satisfy
µ≺λ and we have
V =M Vµ.
Any proper submodule can not contain the highest weight vector, and so the sum of two proper submodules is again a proper submodule. Hence any suchV has a unique maximal submodule and hence a unique irreducible quotient. The quotient of any highest weight module by an invariant submodule, if not zero, is again a cyclic highest weight module with the same highest weight.