 Equivalence of central extensions of a Lie algebra

Two central extensions of a Lie algebra $0\to \mathfrak{z}\to E\to \mathfrak{g}\to 0$ and $0\to \mathfrak{z}\to E^{\prime }\to \mathfrak{g}\to 0$ are equivalent if a Lie algebra isomorphism $E\to E^{\prime }$ commutes with both maps in the short exact sequences, inducing the identity on the specified $\mathfrak{z}$ and $\mathfrak{g}$. This is finer than asking only that $E$ and $E^{\prime }$ be isomorphic as Lie algebras.

For $\mathfrak{z}=kz$, choose linear sections and identify the two vector spaces with $\mathfrak{g}\oplus kz$. A linear map compatible with both ends necessarily has the form $(x,uz)\mapsto (x,(u+t(x))z)$ for a linear functional $t$. If the two Lie algebra two-cocycles are $\alpha ,{\alpha }^{\prime }$, preservation of the Lie bracket is exactly ${\alpha }^{\prime }(x,y)=\alpha (x,y)+t([x,y])$. Hence equivalence means that their difference is a Lie algebra two-coboundary. If changes of the identifications at the ends are allowed, an automorphism of a Lie algebra at either end can identify further classes. For a two-dimensional abelian Lie algebra and a one-dimensional central line, the finer classes are parametrized by $k$, whereas the middle Lie algebras have only the zero and nonzero classes, represented by the abelian Lie algebra $k^3$ and the Heisenberg Lie algebra.