Affine Current Algebras, Virasoro, and Sugawara

For $a,b\in \mathbb{Z}$, ${\mathrm{\partial }}_z(z^{a+b})=(a+b)z^{a+b-1}={\mathrm{\partial }}_z(z^a)z^b+z^a{\mathrm{\partial }}_z(z^b)$, so linearity and bilinearity prove the Leibniz rule on all Laurent polynomials. Multiplying a derivation by $-z^{m+1}$ preserves the Leibniz rule because $R$ is commutative; hence each $d_m$ belongs to ${\mathrm{Der}}_{\mathbb{C}}(R)$. For every $q\in \mathbb{Z}$, $d_m(z^q)=-q{z}^{m+q}$, so $d_m{d}_n(z^q)=q(n+q)z^{m+n+q}$ and $d_n{d}_m(z^q)=q(m+q)z^{m+n+q}$. Therefore $[d_m,d_n](z^q)=q(n-m)z^{m+n+q}=(m-n)d_{m+n}(z^q)$. Since Laurent monomials span $R$, this proves $[d_m,d_n]=(m-n)d_{m+n}$ and closure of $W$ under the commutator.

If a finite relation ${\sum }_m{a}_m{d}_m=0$ holds, applying it to $z$ gives $-{\sum }_m{a}_m{z}^{m+1}=0$. Distinct Laurent monomials are linearly independent in $R$, so every $a_m=0$; thus the $d_m$ form a basis of $W$.

For endomorphisms $A,B,C$ of any vector space, the commutator is alternating, and expanding $[A,[B,C]]+[B,[C,A]]+[C,[A,B]]$ cancels each of the six triple compositions with its negative. Thus the commutator bracket satisfies [F1] on ${\mathrm{End}}_{\mathbb{C}}(R)$; by the closure in step 1.1, its restriction makes $W$ a Lie algebra.

The adjoint representation is irreducible: an invariant subspace $U\subseteq \mathfrak{g}$ satisfies $[\mathfrak{g},U]\subseteq U$, so it is an ideal, and [F3] leaves only $0$ and $\mathfrak{g}$. By [F2], $T=\mathrm{ad}({\mathrm{\Omega }}_B)$ commutes with every $\mathrm{ad}(x)$. Write $n={\dim }_{\mathbb{C}}\mathfrak{g}>0$ by [F3], choose one finite basis, and let $A$ represent $T$. By [F7], $p(z)=\det (z{I}_n-A)$ is monic of degree $n$, hence is a nonconstant complex polynomial. By [F6] it has a root ${\gamma }_B\in \mathbb{C}$. By [F7], $\ker (T-{\gamma }_{B}I)\mathrm{\ne }0$. Since $T$ commutes with every $\mathrm{ad}(x)$, this eigenspace is invariant under $\mathfrak{g}$, hence is all of $\mathfrak{g}$. Therefore $T={\gamma }_B{\mathrm{id}}_{\mathfrak{g}}$, and this scalar is unique.

Choose a $Q=-B{\mathrm{\mid }}_{\mathfrak{k}}$-orthonormal basis $X_i$ of $\mathfrak{k}$. By [F4] and [F5], $Q$ is positive definite and the $B$-dual basis is $-X_i$, so ${\mathrm{\Omega }}_B=-{\sum }_i{X}_i^2$. Invariance of $Q$ makes each $\mathrm{ad}(X_i)$ skew-adjoint, and therefore $Q(TX,X)=Q\text{\hspace{0.17em}⁣}(-{\sum }_i[X_i,[X_i,X]],X)={\sum }_{i}Q([X_i,X],[X_i,X])\mathrm{.}$ For $X\mathrm{\ne }0$ this is positive: if it were zero, each $[X_i,X]$ would vanish, so $X$ would be central in $\mathfrak{k}$ and hence in $\mathfrak{g}$, contrary to [F3]. Since $T={\gamma }_{B}I$, dividing by $Q(X,X)>0$ shows that ${\gamma }_B$ is a positive real number.

Fix one finite basis $x_1,\dots ,x_N$ and its $B$-dual basis $x^1,\dots ,x^N$, so $B(x_i,x^j)={\delta }_i^j$. If $T{x}_i={\sum }_j{t}_{ji}{x}_j$, then $B(T{x}_i,x^i)=t_{ii}$ by duality. Therefore [F3] gives $\mathrm{tr}(T)={\sum }_{i}B(T{x}_i,x^i)$. All choices here are finite and $B$ is nondegenerate by [F1].

For arbitrary $x,y\in \mathfrak{g}$, [F2], symmetry and invariance give $B(Cx,y)={\sum }_{i}B([x_i,[x^i,x]],y)={\sum }_{i}B([x^i,x],[y,x_i])={\sum }_{i}B([x,[y,x_i]],x^i)\mathrm{.}$ Indeed the first conversion uses $B([a,b],c)=B(b,[c,a])$ with $a=x_i$, and the second uses the same identity with $a=x$, followed by symmetry. Step 1.1 identifies the final sum with $\mathrm{tr}(\mathrm{ad}(x)\mathrm{ad}(y))=K_{\mathfrak{g}}(x,y)$. Thus $B(Cx,y)=K_{\mathfrak{g}}(x,y)$, with no compact real form or spectral assertion.

By [F1], $K_{\mathfrak{g}}(x,y)=c^{-1}B(x,y)$. Step 1.2 implies $B(Cx-c^{-1}x,y)=0$ for every $y$, so nondegeneracy yields $Cx=c^{-1}x$ for every $x$. Hence $C={\gamma }_{B}I$ with ${\gamma }_B=c^{-1}>0$. The scalar is unique because $\mathfrak{g}\mathrm{\ne }0$. Its half $h^{\vee }={\gamma }_B\mathrm{/}2$ is therefore well defined and positive without a compact-form existence theorem or a choice axiom.

If a compact real form $\mathfrak{k}$ is supplied, [F1] and [F4] make $Q=-B{\mathrm{\mid }}_{\mathfrak{k}}$ positive definite. Take a finite $Q$-orthonormal real basis $(X_i)$ by [F4]; it is also a complex basis of $\mathfrak{g}$ because $\mathfrak{k}$ is a real form. Its $B$-dual basis is $(-X_i)$, since $B(X_i,-X_j)={\delta }_{ij}$. Thus [F2] gives $C=-{\sum }_i\mathrm{ad}(X_i{)}^2$, and step 2.1 identifies this with $2h^{\vee }I$. This proves the conditional compact-basis formula without constructing a compact real form or inferring its existence from its definition.

Fix a vector $v$. By [F1], each of the finitely many basis currents $x_a,x^a$ has a positive-mode cutoff on $v$; taking the maximum of these finitely many cutoffs gives one cutoff for all basis elements. For $r\ge 0$, normal ordering puts $J_{x_a}(r)$ on the right, so the summand vanishes on $v$ for all sufficiently large $r$. For $r<0$, the rightmost mode is $J_{x^a}(m-r)$, whose index tends to $+\mathrm{\infty }$ as $r\to -\mathrm{\infty }$, so those summands also vanish on $v$ outside a finite range. Thus only finitely many $r$ contribute to $Q_{m}v$.

For fixed $m,r$, the map $x\otimes y\mapsto :J_x(r)J_y(m-r):$ is bilinear, with its factor order fixed by the sign of $r$. The dual bases in the statement exist by [F5]. Therefore [F4] makes the sum over $a$ equal for every pair of $B$-dual bases, term by term in $r$. By step 1.1, on each vector only finitely many $r$ contribute, so summing these equal operators proves that $Q_m$ is independent of the dual bases.

Put $A(m)=\lambda (d_m,d_{-m})$. Alternatingness gives $A(-m)=-A(m)$ and $A(0)=0$. Apply the cocycle identity to $d_1,d_n,d_{-n-1}$ and use [F1]. The three terms are $(1-n)A(n+1)$, $-(2n+1)A(1)$, and $(n+2)A(n)$, so $(n-1)A(n+1)=(n+2)A(n)-(2n+1)A(1)\mathrm{.}$ This holds for every integer $n$.

Since $A(1)=0$, the recurrence for $n\ge 2$ becomes $(n-1)A(n+1)=(n+2)A(n)$. Set $a=A(2)\mathrm{/}6$ and $P(n)=a(n^3-n)$. Then $P(1)=0$, $P(2)=A(2)$, and direct expansion gives $(n-1)P(n+1)=(n+2)P(n)$. Induction using $n-1\mathrm{\ne }0$ for $n\ge 2$ proves $A(n)=P(n)$ for every positive $n$; oddness extends the formula to negative $n$, and $A(0)=0$ covers zero. By the support hypothesis, this is exactly $\lambda (d_m,d_n)=a(m^3-m){\delta }_{m+n,0}$.

The scalar $m^3-m$ is odd in $m$. Therefore $\omega (d_n,d_m)=-\omega (d_m,d_n)$. If $m=n$, the Kronecker condition forces $m=0$, and $0^3-0=0$; hence $\omega (d_m,d_m)=0$. Bilinearity extends both identities to all finite sums.

By trilinearity it suffices to use $d_m,d_n,d_p$. If $m+n+p\mathrm{\ne }0$, all three Kronecker factors vanish. If $p=-m-n$, twelve times the cyclic sum is $N=(m-n)((m+n{)}^3-(m+n))+(n-p)((n+p{)}^3-(n+p))+(p-m)((p+m{)}^3-(p+m))\mathrm{.}$ The linear part is $(m-n)p+(n-p)m+(p-m)n=0$. The cubic part, after substituting $p=-m-n$, is $(m-n)p^3+(n-p)m^3+(p-m)n^3=-(m-n)(m+n{)}^3+(m+2n)m^3-(2m+n)n^3=0,$ as expansion cancels the $m^4$, $m^{3}n$, $m{n}^3$, and $n^4$ terms. Thus $N=0$ in every case, proving the cocycle identity.

Among the nonzero $\mathfrak{k}$-invariant subspaces of finite-dimensional $H_0$, choose one, $V$, of least positive dimension. Its orthogonal complement is invariant: if $w\perp V$ and $v\in V$, then $⟨X(0)w,v⟩=-⟨w,X(0)v⟩=0$. Thus $V$ is an irreducible $\mathfrak{k}$-module.

Every positive current mode annihilates $V$, since it lowers the nonnegative energy below zero; zero modes preserve $V$. Any finite word in current modes can be reordered into negative modes, then zero modes, then positive modes by repeatedly using $AB=BA+[A,B]$. Each commutator term has shorter word length by [F2], so induction on word length and the finite number of inversions gives a finite reordering. Positive modes then kill $V$, and negative modes raise energy. Thus the current submodule generated by $V$ has ground space exactly $V$. It is nonzero, so irreducibility of $H$ makes it all of $H$; hence $H_0=V$.

Invariance of $Q$ gives ${\sum }_i([X,X_i]\otimes X_i+X_i\otimes [X,X_i])=0$. Applying the mode product $u\otimes v\mapsto u(a)v(b)$ shows the corresponding operator identity for every pair of mode indices $a,b$. Thus the diagonal zero-mode action on the inverse tensor vanishes, so $[X(0),S_m]=0$; the indices are unchanged and the normal-ordering split is unchanged. Also $Q$ is positive definite by [F6]–[F7], and finite dimensionality in [F8] gives an orthonormal basis.

Put $q=m+1$. Expand $[X(1),S_m]$ using [F1] and [F3]. Write ${\mathbf{1}}_P=1$ when condition $P$ holds and $0$ otherwise. For each $\alpha +\beta =q$, the noncentral terms are $A_{\alpha ,\beta }={\sum }_i[X,X_i](\alpha )X_i(\beta )$ with coefficient $a_{\alpha ,\beta }=\frac{1}{2}({\mathbf{1}}_{\alpha \le 0}+{\mathbf{1}}_{\beta \ge 0})$, and $B_{\alpha ,\beta }={\sum }_i{X}_i(\alpha )[X,X_i](\beta )$ with coefficient $b_{\alpha ,\beta }=\frac{1}{2}({\mathbf{1}}_{\alpha <0}+{\mathbf{1}}_{\beta \ge 1})$. Indeed, in the $r<0$ terms, commuting on the first factor gives $A$ with $\alpha =r+1\le 0$ and commuting on the second gives $B$ with $\alpha =r<0$; in the $r\ge 0$ terms the normal product is $J_{x^a}(m-r)J_{x_a}(r)$, so commuting on the first factor gives $A$ with $\beta =r\ge 0$ and commuting on the second gives $B$ with $\beta =r+1\ge 1$. The invariant-tensor identity in step 1.1 gives $A_{\alpha ,\beta }+B_{\alpha ,\beta }=0$, and $a_{\alpha ,\beta }-b_{\alpha ,\beta }=\frac{1}{2}({\mathbf{1}}_{\alpha =0}+{\mathbf{1}}_{\beta =0})$. Hence the current terms reduce to $\frac{1}{2}{\sum }_i([X,X_i](0)X_i(q)+[X,X_i](q)X_i(0))=\frac{1}{2}{\sum }_i{\mathrm{ad}}_{X_i}^2(X)(q)=-h^{\vee }X(q)\mathrm{.}$ For this reduction, apply the invariant-tensor identity at indices $(q,0)$ to replace the second summand by $-{\sum }_i{X}_i(q)[X,X_i](0)$; the resulting difference is $\frac{1}{2}{\sum }_i[[X,X_i](0),X_i(q)]$. The mode bracket at index zero identifies this with $\frac{1}{2}{\sum }_i{\mathrm{ad}}_{[X,X_i]}(X_i)(q)$, and skew-symmetry gives ${\mathrm{ad}}_{[X,X_i]}(X_i)={\mathrm{ad}}_{X_i}^2(X)$. Then [F2] gives the last equality. The central term occurs once from each original factor with mode $-1$; its contraction with the dual basis sums to $X$ because $(X_i)$ is $Q$-orthonormal, so [F3] and [F5] give $-kX(q)\mathrm{/}2$ from each factor, including when both factors have mode $-1$. Therefore $[X(1),S_m]=-(k+h^{\vee })X(q)$, and [F1] gives $[L_m,X(1)]=-X(q)$. These expansions are legitimate on every vector: in the $r\ge 0$ tail the rightmost mode has index $r$ and its commutator has index $r+1$, while in the $r<0$ tail the rightmost mode has index $m-r$ and its commutator has index $m-r+1$; all annihilate a fixed vector for sufficiently large absolute tail index by [F4].

Choose a $B$-orthonormal complex basis $(e_a)$. Such a basis exists by finite-dimensional diagonalization: a nonzero symmetric nondegenerate form has a vector of nonzero square (otherwise polarization makes it zero), rescale that vector using a square root in $\mathbb{C}$, and induct on its nondegenerate orthogonal complement. Invariance gives ${\sum }_a([x,e_a]\otimes e_a+e_a\otimes [x,e_a])=0$: in this basis the matrix of $\mathrm{ad}(x)$ is skew-symmetric, since $B([x,u],v)=-B(u,[x,v])$. Also ${\sum }_a[e_a,[e_a,x]]=2h^{\vee }x$ by [F3]. This choice concerns only $\mathfrak{g}$, not an inner product on $V$.

Fix $v\in V$ and integers $m,n$. Expand $[J_x(n),T_m]$ by the product rule in the original normal order. All terms are finite on $v$: for $r\to +\mathrm{\infty }$ the rightmost current or its commutator has index $r$ or $r+n$, and for $r\to -\mathrm{\infty }$ it has index $m-r$ or $m-r+n$. These kill $v$ beyond a common bound, using [F5] for the finite family $e_a,[x,e_a]$. To identify the expansion with the commutator, the original truncated quadratic sum also stabilizes on both $v$ and $J_x(n)v$ by [F1]. Thus a sufficiently large cutoff gives the actual commutator, and all following identities are identities of eventually constant finite sums on $v$. Each factor contributes $nk{J}_x(m+n)\mathrm{/}2$ from the central term in [F2], by ${\sum }_{a}B(x,e_a)e_a=x$; their sum is $nk{J}_x(m+n)$. This includes coincident indices: the two product-rule contributions are both retained.

Write $p=m+n$, $A_{\alpha ,\beta }={\sum }_a{J}_{[x,e_a]}(\alpha )J_{e_a}(\beta )$ and $D_{\alpha ,\beta }={\sum }_a{J}_{e_a}(\alpha )J_{[x,e_a]}(\beta )$ for $\alpha +\beta =p$. The noncentral expansion in step 2.1 is ${\sum }_{\alpha +\beta =p}(a_{\alpha ,\beta }{A}_{\alpha ,\beta }+b_{\alpha ,\beta }{D}_{\alpha ,\beta })$, with $a_{\alpha ,\beta }=\frac{1}{2}({\mathbf{1}}_{\alpha <n}+{\mathbf{1}}_{\beta \ge 0}),b_{\alpha ,\beta }=\frac{1}{2}({\mathbf{1}}_{\alpha <0}+{\mathbf{1}}_{\beta \ge n})\mathrm{.}$ Indeed the first normal-order half $r<0$ contributes $A$ at $\alpha =r+n<n$ and $D$ at $\alpha =r<0$; the reversed half $r\ge 0$ contributes $A$ at $\beta =r\ge 0$ and $D$ at $\beta =r+n\ge n$. The invariant tensor of step 1.1 gives $D_{\alpha ,\beta }=-A_{\alpha ,\beta }$. Set $f_n(j)={\mathbf{1}}_{j<n}-{\mathbf{1}}_{j<0}$. Then $a-b=(f_n(\alpha )+f_n(\beta ))\mathrm{/}2$. Cancellation leaves a finite sum, since $f_n$ is supported on $0\le j<n$ for $n>0$, on $n\le j<0$ with value $-1$ for $n<0$, and is zero for $n=0$. The uncancelled tails before this cancellation are legitimate too: for $\alpha$ sufficiently negative the rightmost index $\beta$ is sufficiently positive, and for $\alpha$ sufficiently positive both coefficients are zero.