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For trivial coefficients in a field $k$, a Lie algebra two-coboundary is a bilinear map $\beta (x,y)=t([x,y])$ for a linear functional $t$ in the dual space ${\mathfrak{g}}^{\ast }$. Some conventions put a minus sign in this formula; the resulting vector subspace $B^2(\mathfrak{g},k)$ is the same. Each such map is a Lie algebra two-cocycle: it is an alternating bilinear form, and its cyclic expression is $t([[x,y],z]+[[y,z],x]+[[z,x],y])=0$ by the Jacobi identity. Replacing a linear section of a central extension of a Lie algebra by $s^{\prime }(x)=s(x)+t(x)z$ changes its Lie algebra two-cocycle by $-\beta$. Thus $H^2(\mathfrak{g},k)=Z^2(\mathfrak{g},k)/B^2(\mathfrak{g},k)$ in Lie algebra cohomology records exactly the obstruction to a section that is a Lie algebra homomorphism.

For trivial real coefficients, a Lie algebra two-cocycle is an alternating bilinear form satisfying $\kappa ([X,Y],Z)+\kappa ([Y,Z],X)+\kappa ([Z,X],Y)=0$. Adding a form $b([X,Y])$ does not change its Lie algebra cohomology class. The bracket $[(X,a),(Y,b)]=([X,Y],\kappa (X,Y))$ defines a central extension of a Lie algebra by $\mathbb{R}$ precisely when this cocycle identity holds. The constant discrepancy in a moment-map equivariance obstruction is such a cocycle.

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.

Choose a basis $x,y$ of the two-dimensional abelian Lie algebra $\mathfrak{a}$, and lifts with the same names. Every alternating bilinear form is determined by $c=\alpha (x,y)$, and every such form is a Lie algebra two-cocycle because every Lie bracket in $\mathfrak{a}$ is zero. Thus every central extension of a Lie algebra has a basis $x,y,z$ withIf $c=0$, all Lie brackets vanish, so this is the direct sum of Lie algebras $\mathfrak{a}\oplus kz\cong k^3$. If $c\ne 0$, replace the central basis vector by $z^{\prime }=cz$. Then $[x,y]=z^{\prime }$, which is the Heisenberg Lie algebra ${\mathfrak{h}}_3(k)$. This change of basis shows that every nonzero $c$ gives the same Lie algebra isomorphism class.

These two Lie algebras cannot be isomorphic: their derived algebras have dimensions zero and one, respectively, and a Lie algebra isomorphism preserves the derived algebra. Moreover, the Heisenberg Lie algebra cannot be a direct sum of Lie algebras $kz\oplus \mathfrak{a}$, since that direct sum of Lie algebras would have zero Lie bracket. ThereforeHere “up to isomorphism” means the Lie algebra isomorphism class of the middle Lie algebra, allowing a change of the identifications at the ends of the short exact sequence. If instead equivalence of central extensions of a Lie algebra must induce the identity on both $kz$ and $\mathfrak{a}$, the parameter $c$ cannot change: the formula in part (a) has $t([x,y])=0$. That stricter notion has one class for each $c\in k$, rather than two. This distinction explains the convention required by the question.

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.

The Witt algebra is the Lie algebra of vector fields on a circle with Laurent-polynomial coefficients. Its nontrivial central extension of a Lie algebra is the Virasoro algebra.