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For a closed string, the currents $J_{\pm }=P\pm T{X}^{\prime }$ have vanishing mixed Poisson brackets. The densities ${\mathcal{H}}_{\pm }=J_{\pm }^2/(4T)$ therefore generate commuting constraint algebras. With opposite Fourier orientations their modes satisfy $\{L_m,L_n\}=-i(m-n)L_{m+n}$ and the identical relation for ${\tilde{L}}_n$. This is a direct sum of Lie algebras consisting of two Witt algebras.

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.

Choose representatives ${\lambda }_1,\dots ,{\lambda }_r$ as in part (b). We use the symplectic centralizer decompositionTo prove it, a linear map commuting with every $a\in \mathfrak{a}$ must preserve every weight space: if $v\in V_{\lambda }$, then $a(Tv)=T(av)=\lambda (a)Tv$. Conversely, preserving the weight spaces makes a linear map commute with the scalar action of every $a$ on each space.

On $V_0$, the symplectic Lie algebra condition says exactly that the restriction belongs to $\mathfrak{sp}(V_0)$. On $V_{{\lambda }_i}\oplus V_{-{\lambda }_i}$, use the perfect pairing from part (b) to identify $V_{-{\lambda }_i}$ with the dual space $V_{{\lambda }_i}^{\ast }$. The action $A_i$ on $V_{{\lambda }_i}$ can be arbitrary, but its action on $V_{-{\lambda }_i}$ must be $-A_i^{\ast }$, where $A_i^{\ast }$ is the dual map. Indeed,determines $B_i=-A_i^{\ast }$ uniquely. There are no additional conditions between different pairs because $\omega$ vanishes between different pairs. Every choice of these blocks therefore gives an element of the centralizer of a Lie subalgebra. The block correspondence preserves Lie brackets, proving the claimed direct sum of Lie algebras.

The general linear Lie algebra ${\mathfrak{gl}}_m(k)$ is an abelian Lie algebra exactly when $m\le 1$: for $m\ge 2$, the matrix units satisfy $[E_{11},E_{12}]=E_{12}\ne 0$. A nonzero symplectic vector space has a symplectic basis and contains a two-dimensional symplectic subspace. The symplectic Lie algebra acting on that plane, extended by zero on its symplectic orthogonal complement, contains the matrices $H=\mathrm{diag}(1,-1)$ and $E=E_{12}$ with $[H,E]=2E\ne 0$. Thus $\mathfrak{sp}(V_0)$ is an abelian Lie algebra exactly when $V_0=0$.

Since different summands in a direct sum of Lie algebras commute, we obtain the precise criterionEquivalently, there are $2n$ distinct nonzero weights of a representation, each with a one-dimensional weight space. The condition concerns multiplicities in the defining representation; it does not assert that $\dim \mathfrak{a}=n$ for every such $\mathfrak{a}$.

For a finite-dimensional complex Lie algebra $\mathfrak{g}$, the following three conditions are equivalent descriptions of a semisimple Lie algebra.

First, its solvable radical is zero. An ideal of a Lie algebra is a linear subspace $I$ satisfying $[\mathfrak{g},I]\subseteq I$. A solvable Lie algebra is one whose derived series, defined by $I^{(0)}=I$ and $I^{(r+1)}=[I^{(r)},I^{(r)}]$, eventually becomes zero. The solvable radical is the largest solvable ideal of a Lie algebra; its vanishing is equivalent to the absence of nonzero solvable ideals.

Second, the Killing formis a nondegenerate bilinear form: if $\kappa (u,v)=0$ for every $v$, then $u=0$. Here $\mathrm{ad}u(v)=[u,v]$ is the Adjoint representation, and $\mathrm{tr}$ is the trace. This equivalence is the Cartan criterion for semisimplicity.

Third, $\mathfrak{g}$ is a direct sum of Lie algebras that are nonabelian simple Lie algebras. A simple Lie algebra is nonabelian and has no ideals of a Lie algebra other than zero and itself. Thus the three equivalent criteria are

Let $\mathfrak{a}$ be a toral subalgebra and an abelian Lie algebra in the symplectic Lie algebra, over an algebraically closed field of characteristic zero. Its defining representation has a weight decomposition. The symplectic Lie algebra identity gives $(\lambda (a)+\mu (a))\omega (v,w)=0$ for $v\in V_{\lambda }$, $w\in V_{\mu }$, so $\omega (V_{\lambda },V_{\mu })=0$ unless $\mu =-\lambda$. Since $\omega$ is a nondegenerate bilinear form, its restrictions give a perfect pairing between $V_{\lambda }$ and $V_{-\lambda }$, and a nondegenerate bilinear form on $V_0$.

An element of the centralizer of a Lie subalgebra preserves each weight space. On $V_0$ it is any element of $\mathfrak{sp}(V_0)$. On each nonzero pair, its action $A$ on $V_{\lambda }$ is arbitrary, and the perfect pairing uniquely forces the action on $V_{-\lambda }$ to be the negative dual map $-A^{\ast }$. These independent choices preserve Lie brackets, proving the displayed direct sum of Lie algebras; the sum runs over one representative per nonzero pair.

This centralizer of a Lie subalgebra is an abelian Lie algebra precisely when $V_0=0$ and all the nonzero weight spaces are one-dimensional. Indeed, ${\mathfrak{gl}}_m$ is nonabelian for $m\ge 2$ by $[E_{11},E_{12}]=E_{12}$, while every nonzero symplectic vector space contains a two-dimensional symplectic subspace on which $[\mathrm{diag}(1,-1),E_{12}]=2E_{12}$. In the stated case the surviving summands are copies of ${\mathfrak{gl}}_1$ and commute. Repeated weights of a representation, including a zero one, are therefore exactly what obstructs this commutativity.