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Unformatted text preview: easy to see that this is a linear automorphism of A since
it has the form 1 + η where η is nilpotent. So, the inverse is 1 − η + η 2 − · · · which is a
ﬁnite sum. The following lemma shows that exp(−δ ) is the inverse of exp δ .
Lemma 2.2.1. Suppose that char F = 0 and f, g are commuting nilpotent endomorphism of some vector space V (or, more generally, commuting elements of any associative
algebra with unity) then exp(f + g ) = exp(f )exp(g ). In particular, exp(f ) is a linear
automorphism of V with inverse exp(−f ).
Proof. Since f, g commute we have:
n (f + g ) = n
n
i=0 i f i g n−i MATH 223A NOTES 2011 LIE ALGEBRAS 7 Since char F = 0 we can divide both sides by n! to get
f i gj
(f + g )n
=
n!
i! j !
i+j =n
Note that for suﬃciently large n, all terms are zero since f, g are nilpotent. Thus we can
sum over all n ≥ 0 to get
exp(f + g ) = exp(f )exp(g ) as claimed.
To see that exp δ is a homomorphism of algebras, we can use the following trick:
δ (xy ) = δ ◦ µ...
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This document was uploaded on 03/09/2014 for the course MATH 223a at Brandeis.
 Fall '11
 K.Igusa
 Algebra

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