Lie algebra extension

Remarks

The image $i(\mathfrak a)$ is then an ideal of $\mathfrak e$, and $p$ induces an isomorphism $\mathfrak e / i(\mathfrak a) \cong \mathfrak g$. The extension is central if $i(\mathfrak a)$ lies in the center of $\mathfrak e$. It is split if there is a Lie algebra homomorphism $s \colon \mathfrak g \to \mathfrak e$ with $p \circ s = \id$; then $\mathfrak e$ is a semidirect product of $i(\mathfrak a)$ and $s(\mathfrak g)$.

Example: let $\mathfrak e$ be the Heisenberg algebra with basis $x, y, z$ and $[x, y] = z$, the other brackets being $0$. Let $\mathfrak a = k$ with $i(1) = z$, and let $\mathfrak g = k^2$ abelian with $p(ax + by + cz) = (a, b)$. Then $p$ is a homomorphism because it kills every bracket and $\mathfrak g$ is abelian, and $\ker p = k z = \im i$. This is a central extension that does not split. Any elements of $\mathfrak e$ mapping to $(1, 0)$ and $(0, 1)$ have the form $x + cz$ and $y + c'z$, and their bracket is $z \ne 0$, while a splitting needs bracket $0$.

Non-example: the inclusion $k z \to \mathfrak e$ followed by the zero map $\mathfrak e \to k^2$ is not an extension, since the zero map is not surjective and its kernel is not $k z$.