[[Lie algebras MOC]]
# Lie algebra extension

Let $\mathfrak{a}, \mathfrak{b}$ be [[Lie algebra|Lie algebras]].
An **[[extension]]** of $\mathfrak{a}$ by $\mathfrak{b}$ is a Lie algebra $\mathfrak{g}$ together with a [[short exact sequence]] #m/def/lie
$$
\begin{align*}
0 \to \mathfrak{b} \hookrightarrow \mathfrak{g} \twoheadrightarrow \mathfrak{a} \to 0
\end{align*}
$$
Hence the $\mathfrak{g}$ “covers” $\mathfrak{a}$ with [[Kernel of a Lie algebra homomorphism|kernel]] $\mathfrak{b} \hookrightarrow \mathfrak{g}$. 
Note that $\mathfrak{b}$ is necessarily an [[Lie algebra ideal|ideal]],
giving the [[Quotient Lie algebra|quotient]] $\mathfrak{g} / \mathfrak{b} \cong \mathfrak{a}$ by the [[Lie algebra isomorphism theorems#First isomorphism theorem]].
Two extensions $\mathfrak{g}, \mathfrak{g}_{1}$ of $\mathfrak{a}$ by $\mathfrak{b}$ are said to be **equivalent** iff there exists an isomorphism such that the following diagram commutes ^equivalent

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## Classification

Consider an extension $0 \to \mathfrak{b} \hookrightarrow \mathfrak{g} \stackrel{p}{\twoheadrightarrow} \mathfrak{a} \to 0$.

1. Iff $\mathfrak{b}$ is [[Abelian Lie algebra|abelian]], one speaks of an **abelian extension**, ^abelian
2. Iff $\mathfrak{b} \hookrightarrow \mathfrak{g}$ is a [[central ideal]], one speaks of a **central extension**. ^central
3. Iff $\mathfrak{g} \cong \mathfrak{b} \rtimes \mathfrak{a}$ ([[Semidirect product of Lie algebras]]), one speaks of a **split extension**, equivalently $p$ is [[Split epimorphism|split epic]]. ^split
4. Iff $\mathfrak{g} \cong \mathfrak{b} \times \mathfrak{a}$ ([[Direct product of Lie algebras]]), one speaks of a **trivial extension**. ^trivial

> [!missing]- Proof of equivalence in 3.
> #missing/proof


## See also

- [[Group extension]] (the structure of that Zettel deliberately mirrors this one)


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#state/tidy | #lang/en | #SemBr