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Let $\mathfrak{g}$ be a Lie algebra. Then $\operatorname{Aut}(\mathfrak{g})$ is the space of automorphisms of $\mathfrak{g}$.

I'm a little confused on what an automorphism of a Lie algebra is. Does this mean that it is an automorphism of linear algebras, i.e. an automorphism is a linear invertible map. Or do we also require such a map to preserve the bracket structure on $\mathfrak{g}$.

Bernard
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samlanader
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    You also require the map to preserve the Lie bracket. If not, the fact that $\mathfrak{g}$ is a Lie algebra (as opposed to a vector space) would be irrelevant. – Michael Albanese Dec 16 '19 at 22:24

1 Answers1

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If $\mathfrak{g} , \mathfrak{h}$ are Lie algebras, a homomorphism of Lie algebras is a map $T : \mathfrak{g} \to \mathfrak{h}$ which (a) is a linear homomorphism, and (b) satisfies $$[T(x), T(y)]_{\mathfrak{h}} = [x, y]_{\mathfrak{g}}$$ for all $x, y \in \mathfrak{g}$. A Lie algebra automorphism will be a Lie algebra homomorphism of a Lie algebra to itself which is also invertible.

AJY
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