Sugawara current commutators on every restricted module

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.

Exchange $\alpha ,\beta$ in the term with coefficient $f_n(\beta )$. By the same invariant tensor, $A_{\beta ,\alpha }=-{\sum }_a{J}_{e_a}(\beta )J_{[x,e_a]}(\alpha )$. Hence the noncentral part is $\frac{1}{2}{\sum }_{\alpha +\beta =p}{f}_n(\alpha ){\sum }_a[J_{[x,e_a]}(\alpha ),J_{e_a}(\beta )]\mathrm{.}$ Here $B([x,e_a],e_a)=B(x,[e_a,e_a])=0$, so there is no additional central term. Moreover ${\sum }_a[[x,e_a],e_a]={\sum }_a[e_a,[e_a,x]]=2h^{\vee }x$. As ${\sum }_{\alpha \in \mathbb{Z}}{f}_n(\alpha )=n$, including $n=0$ and negative $n$, [F2] reduces the displayed finite sum to $n{h}^{\vee }{J}_x(p)$. Together with step 2.1, this gives $[J_x(n),T_m]=n(k+h^{\vee })J_x(p)$. Divide by $q\mathrm{\ne }0$ and reverse the commutator to obtain the claim.