The loop residue form satisfies the Lie two cocycle identity

For $x,y,z\in \mathfrak{g}$, invariance and symmetry give $B([x,y],z)=B(x,[y,z])=B([y,z],x)$, and the cyclic repetition gives $B([z,x],y)$ as the same scalar $b$.

For $u=x\otimes f$, $v=y\otimes q$, $w=z\otimes r$, the required cyclic sum is $b\mathrm{Res}(((fq{)}^{\mathrm{\prime }}r+(qr{)}^{\mathrm{\prime }}f+(rf{)}^{\mathrm{\prime }}q)dt)$. Expanding the three derivatives gives twice each of $f^{\mathrm{\prime }}qr,f{q}^{\mathrm{\prime }}r,fq{r}^{\mathrm{\prime }}$, so the sum is $2b\mathrm{Res}((fqr{)}^{\mathrm{\prime }}dt)$.

If $fqr={\sum }_j{a}_j{t}^j$, its derivative has $t^{-1}$ coefficient $0a_0=0$. Step 2.1 therefore vanishes. Each general input is a finite sum of pure tensors, and the cyclic expression is trilinear; distributing reduces it to these vanishing summands. Empty sums, zero arguments and repeated arguments are included.