* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project
54 MATH 223A NOTES 2011 LIE ALGEBRAS 14. Isomorphism Theorem This section contain the important theorem that two simple Lie algebras with the same Dynkin diagram are isomorphic. The proof uses the existence of a unique maximal root (Exercise 10.2.5). It also uses another lemma which we skipped: Lemma 14.0.5. Suppose that β is a positive root which is not simple. Then there exists a simple root α so that β − α is a positive root. ! ! Proof. Let β = ki αi . Then (β, β) = ki (β, αi ) > 0 implies that (β, αi ) > 0 for some αi . But this implies that β − αi is a root (Lemma 9.4.5). And it must be a positive root since one of the other coefficients kj > 0. ! 14.1. Proof of the isomorphism theorem. Proposition 14.1.1. A semisimple Lie algebra L is generated by Lα , L−α for all α ∈ ∆. Proof. Consider the subalgebra L+ generated by Lα for all α ∈ ∆. We claim!that L+ contains all Lβ for β ∈ Φ+ . If not then take the smallest positive root β = ki αi so + that Lβ is not contained in L . Then there is some simple root α so that γ = β − α is a positive root (which is smaller than β). But then Lγ ⊆ L+ and [Lα , Lγ ] = Lβ ⊆ L+ . Similarly, the subalgebra L− generated by all L−α for simple α contains all Lβ for β ∈ Φ− . So the subalgebra L" of L generated by all Lα , L−α contains Lβ for all β ∈ Φ. On the other hand hα ∈ [Lα , L−α ]. So, L" also contains all hα . But the hα generate H since the α are a basis for H ∗ . So, L" = L. ! Theorem 14.1.2. Let L, L" be simple Lie algebras over F . Let H, H " be Cartan subalgebras for L, L" . Let Φ, Φ" be the corresponding root systems. Suppose that Φ ∼ = Φ" . " " ∼ Choose any isomorphism Φ = Φ . Let ∆, ∆ be corresponding bases. Let π : H → H " be the corresponding isomorphism (whose dual π ∗ : H "∗ → H ∗ sends ∆" to ∆). Choose any nonzero xα ∈ Lα , x"α ∈ L"α . Then there is a unique isomorphism π : L → L" sending xα to x"α so that π|H = π. In short: if Φ ∼ = Φ" then L ∼ = L" . Proof. (Uniqueness) First note there are unique elements yα ∈ L−α and yα" ∈ L"−α so that [xα , yα ] = hα and [x"α , yα" ] = h"α . Since π : H → H " sends hα to h"α , π must send yα to yα" . But the elements xα , yα generate L by the proposition above. So, π is uniquely determined. Next we consider the algebra L ⊕ L" . This is a semisimple Lie algebra with exactly two nonzero proper ideal: L, L" . Take the “diagonal” D which is the subalgebra of L ⊕ L" generated by the elements xα = (xα , x"α ) and y α = (yα , yα" ). Then the projection map L ⊕ L" → L sends D onto L and similarly projection to the second factor sends D onto L" . We will show that both of these projection maps are isomorphisms giving L ∼ =D∼ = L" . Let D− be the subalgebra of L ⊕ L" generated by the elements y α for α ∈ ∆. Let β ∈ Φ+ be the unique maximal root and choose nonzero elements z ∈ Lβ , z " ∈ L"β . Let M be the D− submodule of L ⊗ L" generated by the element z = (z, z " ) ∈ Lβ ⊕ L"β . In other words, M is the span of z and all elements of the form (14.1) [y α1 [y α2 · · · [y αk z] · · · ]] MATH 223A NOTES 2011 LIE ALGEBRAS 55 Claim 1 M ∩ (Lβ ⊕ L"β ) is one-dimensional. Pf: The generators y α of D− sends Lβ ⊕ L"β into Lβ−α ⊕ L"β−α and Lγ ⊕ L"γ into Lγ−α ⊕ L"γ−α . So, we never come back to the maximal root β. Claim 2 [D, M ] ⊆ M . Pf: We show by induction on k that adxα sends the expression 14.1 above into M . The generators xα of D act trivially on (z, z " ) ∈ Lβ ⊕ L"β . So, this statement is true for k = 0. Since the difference of two simple roots is never a root, xα commutes with y α! if α" )= α. So, if α1 )= α then [xα [y α1 [y α2 · · · [y αk z] · · · ]]] = [y α1 [xα [y α2 · · · [y αk z] · · · ]]] lies in [y α1 M ] ⊆ M by induction on k. Therefore, we may assume that α1 = α. Then [xα [y α [y α2 · · · [y αk z] · · · ]]] = [y α [xα [y α2 · · · [y αk z] · · · ]]] + [hα [y α2 · · · [y αk z] · · · ]]] where hα = (hα , h"α ) = [xα , y α ]. The first summand lies in M by induction. The second summand also lies in M since ad hα acts by multiplication by the same scalar in both coordinates of Lγ ⊕ Lγ " . Claim 3 D )= L ⊕ L" . Pf: Otherwise, M is an ideal by Claim 2 and proper by Claim 1. This is a contradiction since the only nontrivial proper ideals are L ⊕ 0 and 0 ⊕ L" . Claim 4 D ∩ L = 0 = D ∩ L" . Pf: Suppose that D ∩ L is nonzero. Then it contains some (w, 0) where w ∈ L is nonzero. But xα ∈ D acts on (w, 0) by [xα , (w, 0)] = ([xα w], 0) and similarly for y α . By the Proposition, xα , yα generate L. So, [D, (w, 0)] = ([L, w], 0) = (L, 0) which would imply that D = L ⊕ L" contradicting Claim 3. Similarly D ∩ L" = 0. This implies that the projection maps D → L, D → L" are isomorphisms as claimed. These isomorphisms send xα to xα , x"α and similarly for y α . But then the composition L → D → L" sends xα to x"α as claimed. ! 14.2. Automorphisms of L. The isomorphism theorem can also be used when L = L" . In that case it says that any automorphism of Φ extends to an automorphism of the Lie algebra L. Corollary 14.2.1. We have an epimorphism of groups: Aut(L) " Aut(Φ). There is one automorphism in particular that we can write down: Corollary 14.2.2. If L is a simple Lie algebra and xα ∈ Lα , yα ∈ L−α so that [xα , yα ] = hα then there is an automorphism σ of L so that σ(xα ) = −yα , σ(yα ) = −xα and σ(hα ) = −hα . (Consequently, σ(h) = −h for all h ∈ H and σ 2 = idL .) Exercise 14.2.3. (1) Show that Aut L contains a subgroup isomorphic to Γ. (2) Prove that Φ has a unique maximal root. (See Exercise 10.2.5.) (3) Complete the proof of Theorem 12.3.4.