Cartan s Criteria Math 649, 2013 Dan Barbasch February 26
Cartan s Criteria REFERENCES: Humphreys, I.2 and I.3. Definition The Cartan-Killing form of a Lie algebra is the bilinear form B(x, y) := Tr(ad x ad y). It satisfies B(ad x(y), z) + B(x, ad y(z)) = 0 Theorem (1) If B [g,g] 0 then g is solvable Theorem (2) g is semisimple if and only if B is nondegenerate.
The first theorem follows from the following essentially equivalent formulation. Theorem Suppose that g has a finite dimensional representation (π, V ) such that 1 ker π is solvable. 2 Tr(π(x) 2 ) = 0 for all x D 1 g. Then g is solvable. Corollary g solvable Tr((ad x) 2 ) = 0 x D 1 g.
Some Remarks 1 If a g is an ideal, then B a (x, y) = B g (x, y) for any x, y, a. 2 For a symmetric bilinear quadratic form, 4B(x, y) = B(x + y, x + y) B(x y, x y). 3 If g is solvable, then it has a basis for which all ad x are upper triangular. Elements in D 1 g have zero on the diagonal, so Tr((ad x) 2 ) = 0.
Cartan Subalgebras References: Humphreys II.4, II.5, Jacobson III.1-III.6, Helgason III.3. Definition A subalgebra h g is called a Cartan subalgebra (CSA) if 1 h is nilpotent 2 h is its own normalizer. If h g is a subspace, the normalizer of h is defined to be N(h) := {X g : ad X (h) h}.
Cartan Subalgebras Lemma N(h) is a subalgebra. Proof. If h h, and x, y N(h), ad[x, y](h) = ad x ad y(h) ad y ad x(h) h.
Cartan Subalgebras For a linear functional α : h C, define g α := {x g (ad h α(h)) m x = 0 some m, h h}. Theorem If h is nilpotent, then any finite dimensional representation V has a decomposition V = α V α. In particular g = α g α. An α g such that g α 0 is called a root. This follows from Lie s theorem, but only holds for nilpotent algebras. Counterexample: Upper triangular 2 2 matrices acting on K 2.
The main step in proving the theorem (a whole chapter in Jacobson) is the fact that the generalized eigenspace V λ is invariant under h. This comes from the relation L A λ = R A λ + ad(a λ), which generalizes to L m A λ = (R A λ + ad(a λ)) m = ( m k ) ad(a) k (B)R m k A λ. If m is large enough, one of the factors in the sum is zero. For the first one to be zero, it is necessary for the algebra to be nilpotent. Write h = Kx + I with I an ideal. Then check that the decomposition into generalized eigenspaces for π(x) is stabilized by π(i).
Lemma [g α, g β ] g α+β. Proof. Recall that D End(g) is called a derivation if D([x, y]) = [Dx, y] + [x, Dy]. For a derivation, (D α β) n [x, y] = ( ) n [(D α) k x, (D β) l y]. (1) l k+l=n If x and y are generalized eigenvectors with eigenvalues α, β, then for n large enough, one of the terms in each bracket of the sum in (1) is zero. This implies that [x, y] is in the generalized eigenspace of D for the eigenvalue α + β. Apply this to D = ad h for h h to conclude the claim of the lemma.
Corollary g 0 h and any x g α, (α 0) is ad-nilpotent.
Theorem A nilpotent subalgebra h g is a CSA if and only if h = g 0. Proof. Since h is nilpotent, there is a decomposition g = g 0 + α 0 g α. The normalizer N(h) is contained in g 0. So if h = g 0, then h = N(h). On the other hand, suppose h = N(h), and h g 0, but V := g 0 /h (0). Then V is a representation of h where all elements act nilpotently. By Engel s theorem, V has a nonzero joint eigenvector. Its inverse image in g 0 is in the normalizer of h, but not in h, a contradiction.
We now show that every Lie algebra has a CSA. For an element h g, let and let g 0 (h) : {x g (ad h) m x = 0 for some m}, p(t, h) := det(ti ad h) = t n + p n 1 (h)t n 1 + + p l (h)t l. (2) Note that p(0, h) = 0 because every ad h has nontrivial kernel. So let l > 0 be the smallest integer so that p l (h) 0 for some h. If p l (h) 0, then in fact l = dim g 0 (h). Definition An element h g is called regular if dim g 0 (h) is minimal. Equivalently p l (h) 0.
Theorem Assume K is algebraically closed. If h 0 is regular, then h = g 0 (h 0 ) is a CSA. Conversely, every CSA is the generalized null space of a regular element.
Proof. For a regular element, N(g 0 (h 0 )) = g 0 (h 0 ), We need to show that if h h, then ad h h is nilpotent. Suppose not. Consider λh + µh 0 for (λ, µ) K 2 (K = C if you re uncomfortable with other fields). There is a nonempty open set {(λ, µ)} so that g 0 (λh + µh 0 ) has strictly smaller dimension than h. Indeed, decompose g = g 0 (h 0 ) + g α according to generalized eigenvalues of h 0. The subspaces g α are stabilized by λh + µh 0. The set {(λ, µ) : λh + µh 0 has no generalized eigenvalue 0 on any g α } is Zariski open, and nonempty. For a pair in this set, g 0 (λh + µh 0 ) g 0 (h 0 ). If h is not nilpotent the generalized 0-eigenspace of a(λ, µ) = ad(λh + µh 0 ) is strictly smaller than h, which contradicts the fact that h 0 was assumed regular.
(continued). Conversely, let h be a CSA, and write g = h α g α where are the nonzero roots. Since there are only finitely many, let h 0 h, be such that α(h 0 ) 0 for any α. We claim g 0 (h 0 ) = h. On the one hand, g 0 (h 0 ) h because the nonzero roots don t vanish on h 0. On the other hand, ad h 0 acts nilpotently on h because h is nilpotent. So h g 0 (h 0 ).
Theorem Suppose that g has a finite dimensional representation (π, V ) such that 1 ker π is solvable. 2 Tr(π(x) 2 ) = 0 for all x D 1 g. Then g is solvable. Corollary g solvable Tr((ad x) 2 ) = 0 x D 1 g. Proof. ad : g End(g), is the representation π, ker ad = Z(g) := {x g [x, y] = 0 y g} is an abelian ideal, therefore solvable.
Definition For any Lie algebra g, define B g (x, y) := Tr(ad x ad y). Theorem g is semisimple B g is nondegenerate. Remark: Recall that g is semisimple means g has no nonzero solvable ideal. Also this implies Z(g) = 0. Definition B g (x, y) := Tr(ad z ad y) is called the Cartan-Killing form. Lemma B(x, y) = B(y, x), B(ad x(y), z) + B(y, ad x(z)) = 0.
Proof of Theorem 1 Proof. Cartan s criteria It is enough to show g D 1 g, because the assumptions also hold for D 1 g, and we can do an induction. Suppose not, i.e. g = D 1 g. Then g = h + α 0 g α and so g = D 1 g = [g λ, g µ ]. But [g λ, g µ ] g λ+µ so h = [h, h] + [g α, g α ]. Recall that V = V λ, where V λ are generalized eigenspaces for h. We will show that π(g) is nilpotent by using Engel s theorem. We start by showing that π(h) is nilpotent. Since h is nilpotent, it is solvable and Lie s theorem applies. Thus it is enough to show that λ(h) = 0 for all h h. Since Tr π(h) Vλ = λ(h) dim V λ, we get that λ([h 1, h 2 ]) = 0.
Proof of Theorem 1 continued (continued). Let now e α g α and e α g α Write h α := [e α, e α ]. We compute Tr(π(h α )). Let ρ be such that V ρ (0). Let W := i Z V ρ+iα. Then e α, e α and h α stabilize W. Therefore, Tr(π(h α )) = Tr([e α, e α ]) = 0. On the other hand, Tr W (π(h α )) = (ρ + iα)(h α ) dim V ρ+iα, so 0 = ( dim Vρ+iα ) ρ(hα ) + ( i dim Vρ+iα )α(h α )
Proof of Theorem 1 continued (continued). Solving for ρ(h α ) we get ρ(h α ) = r ρ α(h α ), r ρ Q (3) 0 = Tr(π(h α ) 2 ) = ( λ(h α) 0 r 2 λ dim V λ) α(hα ) 2. (4) If α(h α ) = 0, then (3) shows that all λ(h α ) = 0. If not, (4) shows that r 2 λ dim V λ = 0. In either case, λ(h α ) = 0 for all λ.
Proof of Theorem 1 continued (continued). It follows that the only generalized eigenspace of h is V 0. But then since π(e α ) maps V 0 to V α and the latter is (0), we get that π(g α ) acts by zero. Thus π(g) is a nilpotent algebra and therefore also solvable. Since ker π is assumed solvable, g is solvable, contradicting g = D 1 g.
Proof of Theorem 2 Cartan s criteria Proof. ( ) Suppose g is semisimple. Let g be the radical of B, i.e. g := {x g B(x, y) = 0 y g}. We need to show g = (0). First, g is an ideal because B(ad z(x), y) = B(x, ad z(y)) = 0. Next, B(x, x) = Tr((adx) 2 ) = 0, for x g. Thus g is solvable by theorem 1, so must be zero.
Proof of Theorem 2 Cartan s criteria Proof. ( ) Suppose g is not semisimple. Let I g be a nonzero abelian ideal. Such an ideal exists since there is a nontrivial solvable ideal S g. Indeed, let n be such that D (n) S 0 and D (n+1) S = 0. Then take I to be D (n) S. Any D (i) S is an ideal by induction using ad x([y, z]) = [ad x(y), z] + [y, ad x(z)]. Choose a basis of g so that the first k vectors are in I. Then for a g, and b I, ad a = ( ) 0 ad b = We can then check that Tr(ad b ad a) = 0. So (0) I radb. ( ) 0 0 0
Derivations Corollary If g is semisimple then g = [g, g]. Exercise 1 Is the converse true? 2 Compute B g for sl(n) and show that Tr(ad x ad y) = Tr(x y).
Proof of corollary 18 Cartan s criteria Suppose h g is any ideal. Then let h := {x g B(x, y) = 0 y h}. Now h h = (0). This is because h is an ideal, and Tr[(ad x) 2 ] = 0. So by Cartan s criterion, it is solvable. Since g is semisimple, this ideal must be (0). h h = (0). Thus since dim g = dim h + dim h dim(h h ) we get g = h h. In particular, apply this to h = [g, g]. Then h radb. Indeed, B(x, [y, z]) = 0 y, z B([x, y], z) = 0 y, z This is equivalent to [x, y] = 0 y g x Z(g) = (0). So g = [g, g].
Corollary For any ideal h in a semisimple Lie algebra g, we have g = h h. Definition Let g be an (arbitrary) Lie algebra. Define Der(g) := {D End g D([x, y]) = [Dx, y] + [x, Dy]}.
Lemma Der(g) forms a subalgebra. There is a natural map ad : g Der(g) with kernel Z(g). For a semisimple algebra, the map ad : g Der(g) is an inclusion. Note: 1 ad g Der(g) = D is an ideal because [D, ad x] = ad D(x). Indeed, D ad x ad x D = ad D(x). 2 B D (x, y) = B g (x, y) for x, y g. (For any ideal I g, B I (x, y) = B g (x, y), x, y I.)
Proposition If g is semisimple, g = Der(g). Proof. Let g be the orthogonal of g in D (with respect to B D ). Then g g = (0). Indeed, if x g g, 0 = B D (x, y) = B g (x, y) y g; so x = 0. As before dim D = dim g + dim g dim(g g ), so D = g + g. Furthermore if D g, ad D(x) = [D, ad x] g g = (0). But if ad D(x) = 0, D(x) = 0 x. So D = 0. Thus g = (0), and g = Derg.
Cartan subalgebras of semisimple Lie algebras (This lecture was given by E. Klebanov in 2002) We assume that the field is algebraically closed. This is not necessary, but makes the exposition simpler. Let V = g be an arbitrary finite dimensional Lie algebra and D Der(g). We recall the following basic results. Let V be a (not necessarily finite dimensional) vector space over a field K of characteristic zero. Definition An element x End(V ) is called nilpotent, if x n = 0 for some n. It is called locally nilpotent if for any v V, there is n such that x n v = 0. The element is called semisimple if any x-invariant subspace W, has an x-invariant complement. If K is algebraically closed, semisimple is the same as diagonalizable.
Proposition Let x End(V ), where V is a finite dimensional vector space over a field K. Then x = x s + x n uniquely, where x s is semisimple and x n is nilpotent and [x s, x n ] = 0. Furthermore there are polynomials p s and p n so that x s = p s (x) and x n = p n (x). In particular if W V is x-invariant, then it is x s and x n invariant as well. For a proof, consult any standard linear algebra text or Humphreys, section 4.2.
Definition If V = g is a Lie algebra, an element x g is called nilpotent (semisimple) if ad x is nilpotent (semisimple). Proposition Let D = D s + D n be the Jordan decomposition. Then D s, D n are derivations.
Proof. It is enough to show that D s is a derivation, since D n = D D s and derivations form a vector space. Let g = α g α (5) be the generalized weight decomposition. We know that [g α, g β ] g α+β and D s v = γv for any v g γ. It is now easy to check that if v g α and w g β, then Thus D s is a derivation. D s [v, w] = [D s v, w] + [v, D s w] (6)
Corollary If g is semisimple, then x = x s + x n so that x s is semisimple, x n is nilpotent and [x s, x n ] = 0. The decomposition is unique. Proof. The result follows from the previous proposition and the fact that for a semisimple algebra ad : g Der(g) is an isomorphism.
The next theorem is the main result for CSA a of finite dimensional semisimple Lie algebras. Theorem A subalgebra h g is a CSA if and only if 1 h is maximal abelian, 2 each h h is semisimple.
Proof. =. If an algebra h is abelian, it is certainly nilpotent. Let g = g 0 + α 0 g α be the root decomposition corresponding to the adjoint action of h on g. Since h acts by generalized eigenvalue 0 on g 0, and is formed of semisimple elements, it actually commutes with all of g 0. Let h 0 g 0 be a CSA of g 0. Since it commutes with h and is its own normalizer, h h 0. Since [g 0, g α ] g α, the normalizer of h 0 in g is h 0, so h 0 is a CSA of g. If x g 0, then x s, x n g 0 as well because x normalizes h. Suppose there is a nilpotent element x g 0. Then tr(ad x ad y) = 0 for any y g (because this is true for any y g α as well as for y g 0 ). Thus x = 0, so g 0 is formed of semisimple elements only. But then h 0 is a nilpotent algebra formed of semisimple elements only. Thus h 0 is abelian, so by the maximality assumption of h, we conclude h = h 0 is a CSA. Finally g 0 also commutes with h, so is contained in its normalizer, i.e. h = g 0.
(continued). =. If x h and h is a CSA, then x s, x n h as well because h is its own normalizer. If h contains any nilpotent element x then the above argument shows that tr(ad x ad y) = 0 for any y g, so x = 0. As before, since h is formed of semisimple elements only, it is abelian. Finally, any abelian algebra containing h must be in N(h) so equals h.