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# Lemma 221 suppose that char f 0 and f g are commuting

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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.

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