The Sugawara mode-one commutator

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].