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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.

Since the general linear Lie algebra ${\mathfrak{gl}}_1(k)$ has zero Lie bracket, identify it with $kz$. A central extension of a Lie algebra is a short exact sequenceof Lie algebra homomorphisms with $\iota (kz)$ contained in the center of a Lie algebra $Z(\tilde{\mathfrak{g}})$. Thus $\pi$ is surjective, its kernel is precisely the indicated line, and that line commutes with all of $\tilde{\mathfrak{g}}$.

Choose a linear map $s:\mathfrak{g}\to \tilde{\mathfrak{g}}$ with $\pi s=\mathrm{id}$. Such a section exists by lifting a basis. Identify $\tilde{\mathfrak{g}}$ with the direct sum of vector spaces $\mathfrak{g}\oplus kz$ by $(x,tz)\mapsto s(x)+t\iota (z)$. The difference $[s(x),s(y)]-s([x,y])$ belongs to the kernel of $\pi$, so there is a unique bilinear map $\alpha :\mathfrak{g}\times \mathfrak{g}\to k$ such thatIt is an alternating bilinear form: $\alpha (x,x)=0$, equivalently $\alpha (x,y)=-\alpha (y,x)$ in characteristic zero. Taking the central component of the Jacobi identity gives the Lie algebra two-cocycle conditionConversely, any alternating bilinear form satisfying this identity defines a Lie algebra byIndeed its Lie bracket is an alternating multilinear map and a bilinear map, and the two components of its Jacobi identity are respectively the Jacobi identity in $\mathfrak{g}$ and the displayed Lie algebra two-cocycle condition. The line $kz$ is central, and projection to $\mathfrak{g}$ gives the required central extension of a Lie algebra.

For later use, replacing $s$ by $s^{\prime }(x)=s(x)+t(x)\iota (z)$, where $t\in {\mathfrak{g}}^{\ast }$ is in the dual space, changes the Lie algebra two-cocycle toThus changes of section add a Lie algebra two-coboundary, up to the conventional sign. A section that is a Lie algebra homomorphism has ${\alpha }^{\prime }=0$ and gives a direct sum of Lie algebras.

For the Lie algebra two-cocycle ${\alpha }_c$ of part (i), the central extension of a Lie algebra has a basis $x,y,z$ and Lie bracketsPut $y^{\prime }=y+cz$. This is still a basis together with $x,z$, andConsequently $⟨x,y^{\prime }⟩$ is a Lie subalgebra isomorphic to the affine Lie algebra of the line. It commutes with $kz$ and has zero intersection with $kz$, so both summands are ideals of a Lie algebra andThis also holds for the stricter equivalence of central extensions of a Lie algebra. Indeed, choose the linear functional $t$ with $t(x)=0$, $t(y)=c$. Since $[x,y]=y$ in $\mathfrak{s}$, the change of section in part (a) gives ${\alpha }_c^{\prime }(x,y)=c-t(y)=0$. Thus every Lie algebra two-cocycle here is a Lie algebra two-coboundary, so $H^2(\mathfrak{s},k)=0$ in Lie algebra cohomology. The explicit map from the split central extension of a Lie algebra sends $x$ to $x$, $y$ to $y+cz$, and $z$ to $z$; it fixes the central line and induces the identity on the quotient Lie algebra.