Math 222 Notes for Apr. 1
Math 222 notes for Apr. 1 Alison Miller 1 Real lie algebras and compact Lie groups, continued Last time we showed that if G is a compact Lie group with Z(G) finite and g = Lie(G), then the Killing form of g is negative definite, hence g is semisimple. Now we’ll do a converse. Proposition 1.1. Let g be a real lie algebra such that Bg is negative definite. Then there is a compact Lie group G with Lie algebra Lie(G) =∼ g. Proof. Pick any connected Lie group G0 with Lie algebra g. Since g is semisimple, the center Z(g) = 0, and so the center Z(G0) of G0 must be discrete in G0. Let G = G0/Z(G)0; then Lie(G) =∼ Lie(G0) =∼ g. We’ll show that G is compact. Now G0 has an adjoint action Ad : G0 GL(g), with kernel ker Ad = Z(G). Hence 0 ∼ G = Im(Ad). However, the Killing form Bg on g is Ad-invariant, so Im(Ad) is contained in the subgroup of GL(g) preserving the Killing! form. Since Bg is negative definite, this subgroup is isomorphic to O(n) (where n = dim(g). Since O(n) is compact, its closed subgroup Im(Ad) is also compact, and so G is compact. Note: I found a gap in this argument while writing it up; it assumes that Im(Ad) is closed in GL(g). This is in fact true, because Im(Ad) = Aut(g) is the group of automorphisms of the Lie algebra g, which is closed in GL(g); but proving this fact takes a bit of work.
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