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where µ : A × A → A is multiplication. Then δ ◦ µ = µ ◦ (δ1 + δ2 ) where
δ1 (x, y ) = (δ x, y ), δ2 (x, y ) = (x, δ y ) Since δ1 , δ2 commute, we can use the lemma to get:
exp δ (xy ) = exp δ ◦ µ(x, y ) = µ ◦ exp(δ1 + δ2 )(x, y ) = µ ◦ exp(δ1 )exp(δ2 )(x, y )
= exp δ (x) exp δ (y )
Deﬁnition 2.2.2. Suppose that x is an element of a Lie algebra L so that adx is nilpotent.
Then exp adx is called an inner automorphism of L.
Proposition 2.2.3. Suppose that L is the Lie algebra of an associative algebra A with
unity 1. Let x ∈ A be a nilpotent element. Then:
(1) exp x = xk /i! is a unit in A.
(2) adx is a nilpotent endomorphism of L.
(3) exp adx is conjugation by exp x.
Proof. By the Lemma, exp(−x) is the inverse of exp x. The other statement follow from
the following trick: Write adx = λx + ρ−x where λx is “left multiplication by x” and ρ−x
is “right multiplication by −x”: λx (y ) = xy and ρ−x (y ) = −yx. Since left and right
multiplication are commuting oper...
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This document was uploaded on 03/09/2014 for the course MATH 223a at Brandeis.
- Fall '11