Cartan Subalgebras

Lecture 11 - Cartan Subalgebras

October 11, 2012

1 Maximal Toral Subalgebras

A toral subalgebra of a Lie algebra g is any subalgebra consisting entirely of abstractly semisimple elements.

Lemma 1.1 If t ⊂ g is any toral subalgebra, then t is abelian.

Pf. Assume x ∈ t has (ad x)|t 6= 0. Because x is semisimple, it is ad-simisimple, so there is some y ∈ t with (ad x)(y) = ay. Note that

0 = (ad y)(ad x)(y) = −(ad y)2x. (1)

However, y is also semisimple, so we can write

x = x1 + ... + xn (2) where (ad y)(xi) = λixi. Therefore

n 2 X 2 0 = (ad y) x = λi xi (3) i=1 so that each λi = 0. However this contradicts

n X 0 6= ay = (ad x)y = −(ad y)x = λixi. (4) i=1

If g is a semisimple Lie algebra, then any maximal toral subalgebra h is called a Cartan subalgebra or CSA for short. Caution: in the non-semisimple case, this is not the proper use of the term CSA.

1 Since h is a commuting subalgebra, the action of adgh is simultaneously diagonalizable. This means that g has a complete root-space decomposition: there are ﬁnitely many linear ∗ functionals α ∈ h , Λ, α : h → F so that M g = gα (5) α∈Λ where

gα = { x ∈ g | h.x = α(h) x for all h ∈ h } (6) is the (non-trivial) weight-space associated to the functional α. Note that g0 is Cg(h), the centralizer of h in g.

Proposition 1.2 Assume g is a semisimple Lie algebra. Then

∗ a) Given α, β ∈ h , we have [gα, gβ] ∈ gα+β.

b) If x ∈ gα then ad x is nilpotent.

c) If β 6= −α then Lα is perpendicular to Lβ under the Killing form.

d) The restriction of adgh to g0 is non-degenerate.

Pf. Let xα ∈ gα, xβ ∈ gβ. For (a), we have

[h, [xα, xβ]] = [[h, xα], xβ] + [xα, [h, xβ]] (7)

= α(h)[xα, xβ] + β(h)[xα, xβ] = (α + β)(h)[xα, xβ]. (8)

n For (b), note that (ad xα) : gβ → gβ+nα. Since the number of weight spaces is ﬁnite, there n is some n so (ad xα) = 0. For (c), by associativity of κ we have

0 = κ([h, xα], xβ) + κ(xα, [h, xβ]) = (α(h) + β(h)) κ(xα, xβ). (9)

For (d), we know that adgh : g → g is non-degenerate, but also that adgh = 0 unless gα α = 0. Thus adgh must be non-degenerate on h.

Proposition 1.3 We have g0 = h.

Pf. Write C = Cg(h) = g0.

Step I. C contains the abstract semisimple and nilpotent parts of all its elements. Since x ∈ C has ad x : h → 0 and (ad x)s = ad xs,(ad x)n = ad xn are given as polynomials in ad x without constant term, so also (ad x)s, (ad x)n : h → 0 meaning that xs, xn ∈ C.

Step II. All semisimple elements of C lie in h. If x is semisimple and x : h → 0 then h + Fx is both an algebra and is toral.

2 Step III. The restriction of κ to h is non-degenerate. If x ∈ C is nilpotent, then because x commutes with h we have κ(x, h) = 0. But C = h + nilpotents so if h ∈ h, its dual must be in h.

Step IV. C is nilpotent. Since C = h + nilpotents and C commutes with h, any element of C is a sum of something in h with a nilpotent, and since everything commutes, and arbitrary element is nilpotent. Thus C, being ad-nilpotent, is nilpotent.

Step V. We have h ∩ [C,C] = {0}. We have κ(h, [C,C]) = κ([C, h],C) = 0. Thus [C,C] intersects h trivially.

Step VI. C is abelian. If D = [C,C] is non-trivial, then any x ∈ D is nilpotent. Also D ∩ Z(C) is nontrivial, so we can assume x ∈ D ∩ Z(C). But then xn ∈ Z(C) is non-zero, so κ(xn,C) = 0, which is impossible because κ restricted to C is non-degenerate.

Step VII. If x ∈ C \ H then its nilpotent part xn ∈ C commutes with C, and so κ(xn,C) = 0, which is impossible because κ|C is nondegenerate.

∗ ∗ We can now identify h with h via κg. Given any linear functional α ∈ h , deﬁne tα ∈ h by

α(h) = κ(tα, h) (10) for all h ∈ h.

Proposition 1.4 Assume g is a ﬁnite dimensional simple Lie algebra over an algebraically closed ﬁeld of characteristic 0. Let Φ be the set of roots of the adjoint action of h on g. Then

a) Φ spans h∗. b) If α ∈ Φ then −α ∈ Φ.

c) If α ∈ Φ and x ∈ gα, y ∈ g−α, then [x, y] = κ(x, y) tα (recall tα is the κ-dual of α).

d) If α ∈ Φ then [gα, g−α] ⊆ h is 1-dimensional.

e) If α ∈ Φ then α(tα) = κ(tα, tα) 6= 0.

f) If α ∈ Φ and xα ∈ gα, there is some yα ∈ g−α so that setting hα = [xα, yα] we have

spanF{xα, yα, hα} ≈ sl(2, F).

2tα g) Given xα ∈ gα, the choice of yα in (f) leads to hα = . In addition hα = −h−α. κ(tα,tα)

Pf. (a). If not, there is some h ∈ h so that α(h) = 0 for all α ∈ Φ. But then [h, gα] = 0 for all α ∈ Φ, so h ∈ Z(g), an impossibility.

(b). If α ∈ Φ but −α 6∈ Φ then [gα, g] = {0}, meaning gα ∈ Z(g), again an impossibility.

3 (c). Given h ∈ h, x ∈ gα, y ∈ g−α, the associativity of κ implies

κ(h, [x, y]) = κ([h, x], y) = α(h) κ(x, y) (11) = κ(tα, h) κ(x, y)

= κ(tακ(x, y), h) but then ([x, y] − κ(x, y)tα) ∈ h and h ⊥ ([x, y] − κ(x, y)tα) for all h ∈ h, forcing [x, y] = κ(x, y)tα.

(d). Follows directly from (c).

(e). Assume α(tα) = κ(tα, tα) = 0. Then s = spanC{x, y, tα) is a nilpotent Lie algebra. Consider its ad-representation on g.

2 tα (f) and (g). Given any xα ∈ Lα, pick yα ∈ L−α so that κ(xα, yα) = hα. κ(tα,tα) , Then [hα, xα] = α(hα)xα = 2xα and [hα, yα] = −α(hα)yα = −2yα, so we have our copy of sl(2, C).