Lie algebra automorphism

Exponential of a derivation on a Lie algebra

Let $𝔤$ be a Lie algebra over $𝕂$ ¹, and $𝑑 : 𝔤 \to 𝔤$ be a nilpotent² derivation. Then the exponential

e 𝑑 = \infty \sum 𝑖 = 0 𝑑 𝑖 𝑖! = 1 + 𝑑 + 𝑑 2 2! + \dots + 𝑑 𝑘 𝑘!

is a Lie algebra automorphism.³ lie

Proof

Since for any $𝑥, 𝑦 \in 𝔤$
$e 𝑑 (𝑥) e 𝑑 (𝑦) = (𝑛 - 1 \sum 𝑖 = 0 𝑑 𝑖 𝑥 𝑖!) (𝑛 - 1 \sum 𝑗 = 0 𝑑 𝑗 𝑦 𝑗!) = 2 𝑘 \sum 𝑛 = 0 (𝑛 \sum 𝑖 = 0 𝑑 𝑖 𝑥 𝑖! 𝑑 𝑛 - 𝑖 𝑦 (𝑛 - 𝑖)!) = 2 𝑘 \sum 𝑛 = 0 𝑑 𝑛 (𝑥 𝑦) 𝑛! = 𝑘 \sum 𝑛 = 0 𝑑 𝑛 (𝑥 𝑦) 𝑛! = e 𝑑 (𝑥 𝑦)$
it follows $e 𝑑$ is a homomorphism, and an inverse is given by $e - 𝑑$ .

tidy | en | SemBr

Footnotes

where $c h a r 𝕂 = 0$ . ↩
i.e. $𝑑 𝑘 = 0$ for some $𝑘 \in ℕ$ . ↩
1972. Introduction to Lie Algebras and Representation Theory, §2.3, pp. 8–9 ↩

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Inner Lie algebra automorphism
Lie algebra automorphism

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