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Lie Algebras Lie Algebras Shlomo Sternberg March 8, 2004 2 Chapter 5 Conjugacy of Cartan subalgebras It is a standard theorem in linear algebra that any unitary matrix can be di- agonalized (by conjugation by unitary matrices). On the other hand, it is easy to check that the subgroup T ⊂ U(n) consisting of all unitary matrices is a maximal commutative subgroup: any matrix which commutes with all diagonal unitary matrices must itself be diagonal; indeed if A is a diagonal matrix with distinct entries along the diagonal, an matrix which commutes with A must be diagonal. Notice that T is a product of circles, i.e. a torus. This theorem has an immediate generalization to compact Lie groups: Let G be a compact Lie group, and let T and T 0 be two maximal tori. (So T and T 0 are connected commutative subgroups (hence necessarily tori) and each is not strictly contained in a larger connected commutative subgroup). Then there exists an element a ∈ G such that aT 0a−1 = T . To prove this, choose one parameter subgroups of T and T 0 which are dense in each. That is, choose x and x0 in the Lie algebra g of G such that the curve t 7→ exp tx is dense in T and the curve t 7→ exp tx0 is dense in T 0. If we could find a ∈ G such that the 0 −1 0 a(exp tx )a = exp t Ada x commute with all the exp sx, then a(exp tx0)a−1 would commute with all ele- ments of T , hence belong to T , and by continuity, aT 0a−1 ⊂ T and hence = T . So we would like to find and a ∈ G such that 0 [Adax , x] = 0. Put a positive definite scalar product ( , ) on g, the Lie algebra of G which is invariant under the adjoint action of G. This is always possible by choosing any positive definite scalar product and then averaging it over G. 0 Choose a ∈ G such that (Ada x , x) is a maximum. Let 0 y := Ada x . 64 5.1. DERIVATIONS. 65 We wish to show that [y, x] = 0. For any z ∈ g we have d ([z, y], x) = (Ad y, x) = 0 dt exp tz |t=0 by the maximality. But ([z, y], x) = (z, [y, x]) by the invariance of ( , ), hence [y, x] is orthogonal to all g hence 0. QED We want to give an algebraic proof of the analogue of this theorem for Lie algebras over the complex numbers. In contrast to the elementary proof given above for compact groups, the proof in the general Lie algebra case will be quite involved, and the flavor of the proof will by quite different for the solvable and semi-simple cases. Nevertheless, some of the ingredients of the above proof (choosing “generic elements” analogous to the choice of x and x0 for example) will make their appearance. The proofs in this chapter follow Humphreys. 5.1 Derivations. Let δ be a derivation of the Lie algebra g. this means that δ([y, z]) = [δ(y), z] + [y, δ(z)] ∀ y, z ∈ g. Then, for a, b ∈ C (δ − a − b)[y, z] = [(δ − a)y, z] + [y, (δ − b)z] (δ − a − b)2[y, z] = [(δ − a)2y, z] + 2[(δ − a)y, (δ − b)z] + [y, (δ − b)2z] (δ − a − b)3[y, z] = [(δ − a)3y, z] + 3[(δ − a)2y, (δ − b)z)] + 3[(δ − a)y, (δ − b)2z] + [y, (δ − b)3z] . X n (δ − a − b)n[y, z] = [(δ − a)ky, (δ − b)n−kz]. k Consequences: • Let ga = ga(δ) denote the generalized eigenspace corresponding to the k eigenvalue a, so (δ − a) = 0 on ga for large enough k. Then [ga, gb] ⊂ g[a+b]. (5.1) • Let s = s(δ) denote the diagonizable (semi-simple) part of δ, so that s(δ) = a on ga. Then, for y ∈ ga, z ∈ gb s(δ)[y, z] = (a + b)[y, z] = [s(δ)y, z] + [y, s(δ)z] so s and hence also n = n(δ), the nilpotent part of δ are both derivations. 66 CHAPTER 5. CONJUGACY OF CARTAN SUBALGEBRAS • [δ, ad x] = ad(δx)]. Indeed, [δ, ad x](u) = δ([x, u]) − [x, δ(u)] = [δ(x), u]. In particular, the space of inner derivations, Inn g is an ideal in Der g. • If g is semisimple then Inn g = Der g. Indeed, split off an invariant comple- ment to Inn g in Der g (possible by Weyl’s theorem on complete reducibil- ity). For any δ in this invariant complement, we must have [δ, ad x] = 0 since [δ, ad x] = ad δx. This says that δx is in the center of g. Hence δx = 0 ∀x hence δ = 0. • Hence any x ∈ g can be uniquely written as x = s + n, s ∈ g, n ∈ g where ad s is semisimple and ad n is nilpotent. This is known as the decomposition into semi-simple and nilpotent parts for a semi-simple Lie algebra. • (Back to general g.) Let k be a subalgebra containing g0(ad x) for some x ∈ g. Then x belongs g0(ad x) hence to k, hence ad x preserves Ng(k) (by Jacobi’s identity). We have x ∈ g0(ad x) ⊂ k ⊂ Ng(k) ⊂ g all of these subsepaces being invariant under ad x. Therefore, the charac- teristic polynomial of ad x restricted to Ng(k) is a factor of the charactris- tic polyinomial of ad x acting on g. But all the zeros of this characteristic polynomial are accounted for by the generalized zero eigenspace g0(ad x) which is a subspace of k. This means that ad x acts on Ng(k)/k without zero eigenvalue. On the other hand, ad x acts trivially on this quotient space since x ∈ k and hence [Ngk, x] ⊂ k by the definition of the normalizer. Hence Ng(k) = k. (5.2) We now come to the key lemma. Lemma 1 Let k ⊂ g be a subalgebra. Let z ∈ k be such that g0(ad z) does not strictly contain any g0(ad x), x ∈ k. Suppose that k ⊂ g0(ad z). Then g0(ad z) ⊂ g0(ad y) ∀ y ∈ k. Proof. Choose z as in the lemma, and let x be an arbitrary element of k. By hypothesis, x ∈ g0(ad z) and we know that [g0(ad z), g0(ad z)] ⊂ g0(ad z). Therefore [x, g0(ad z)] ⊂ g0(ad z) and hence ad(z + cx)g0(ad z) ⊂ g0(ad z) for all constants c. Thus g0(ad(z + cx)) acts on the quotient space g/g0(ad z). We can factor the characteristic polynomial of ad(z + cx) acting on g as Pad(z+cx)(T ) = f(T, c)g(T, c) 5.2. CARTAN SUBALGEBRAS. 67 where f is the characteristic polynomial of ad(z + cx) on g0(ad z) and g is the characteristic polynomial of ad(z + cx) on g/g0(ad z). Write r r−1 f(T, c) = T + f1(c)T + ··· fr(c) r = dim g0(ad z) n−r n−r−1 g(T, c) = T + g1(c)T + ··· + gn−r(c) n = dim g. The fi and the gi are polynomials of degree at most i in c. Since 0 is not an eigenvalue of ad z on g/g0(ad z), we see that gn−r(0) 6= 0. So we can find r + 1 values of c for which gn−r(c) 6= 0, and hence for these values, g0(ad(z + cx)) ⊂ g0(ad z). By the minimality, this forces g0(ad(z + cx)) = g0(ad z) for these values of c. This means that f(T, c) = T r for these values of c, so each of the polynomials f1, . , fr has r + 1 distinct roots, and hence is identically zero. Hence g0(ad(z + cx)) ⊃ g0(ad z) for all c. Take c = 1, x = y − z to conclude the truth of the lemma. 5.2 Cartan subalgebras. A Cartan subalgebra (CSA) is defined to be a nilpotent subalgebra which is its own normalizer. A Borel subalgebra (BSA) is defined to be a maximal solvable subalgebra. The goal is to prove Theorem 1 Any two CSA’s are conjugate. Any two BSA’s are conjugate. Here the word conjugate means the following: Define N (g) = {x| ∃y ∈ g, a 6= 0, with x ∈ ga(ad y)}. Notice that every element of N (g) is nilpotent and that N (g) is stable under Aut(g). As any x ∈ N (g) is nilpotent, exp ad x is well defined as an automor- phism of g, and we let E(g) denote the group generated by these elements. It is a normal subgroup of the group of automorphisms. Conjugacy means that there is a φ ∈ E(g) with φ(h1) = h2 where h1 and h2 are CSA’s. Similarly for BSA’s. As a first step we give an alternative characterization of a CSA. Proposition 1 h is a CSA if and only if h = g0(ad z) where g0(ad z) contains no proper subalgebra of the form g0(ad x). 68 CHAPTER 5. CONJUGACY OF CARTAN SUBALGEBRAS Proof. Suppose h = g0(ad z) which is minimal in the sense of the proposition. Then we know by (5.2) that h is its own normalizer. Also, by the lemma, h ⊂ g0(ad x) ∀x ∈ h. Hence ad x acts nilpotently on h for all x ∈ h.
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