The residue cocycle depends on invariant form normalization

F1 gives ${\omega }_a=a\omega$. Choose $x,y$ with $B(x,y)=1$. For modes $x_1,y_{-1}$ the old scalar cocycle is $1$, while the new one is $a$. In particular $a=2$ is an explicit failure of invariance of the numerical cocycle. Nondegeneracy permits the finite pair choice and scalar normalization.

Let $c_a$ denote the central generator for $B_a$. A map fixing all loop modes and sending $c_a$ to $a^{-1}c$ preserves brackets, since the image of $[x_m,y_n{]}_a=[x,y{]}_{m+n}+am{\delta }_{m,-n}B(x,y)c_a$ is $[x,y{]}_{m+n}+m{\delta }_{m,-n}B(x,y)c$. It is invertible, with inverse $c\mapsto a{c}_a$, so is a Lie isomorphism. This central image is forced: apply any such map to the bracket in step 1.1 and subtract its fixed loop component to get $a\phi (c_a)=c$.

Pulling a level-$k$ module for the old extension back through this isomorphism makes $c_a$ act as $a^{-1}k$. Equivalently the transported weight has ${\mathrm{\Lambda }}_a(c_a)=a^{-1}\mathrm{\Lambda }(c)$ by F3. At $a=2$ this is $k\mathrm{/}2$, differing from $k$ whenever $k\mathrm{\ne }0$. The case $k=0$ stays zero and $a=1$ gives the identity normalization; $a=0$ is excluded because both nondegeneracy and the inverse would fail. Thus rescaling requires exactly the stated adjustment of the central generator and numerical levels. No AC is used.