Evaluation modules have level zero and do not extend canonically over d

F1 gives the zero central action, hence level zero in F3's scalar-action sense. A proposed operator $D$ for $d$ must satisfy $[D,\rho (x)]=0$ by F2 at mode $m=0$. At mode $m=1$ it must satisfy $[D,a\rho (x)]=a\rho (x)$. The left side is $a[D,\rho (x)]=0$, so $a\mathrm{\ne }0$ forces $\rho (x)=0$ for every $x$. Thus a nonzero finite action cannot extend.

Conversely, if $\rho =0$, all loop and central actions are zero. For every $D$ and every $m$, both sides of $[D,a^m\rho (x)]=m{a}^m\rho (x)$ vanish; also $[D,0]=0$ for $c$ and $[D,D]=0$ for $d$. The original loop relations already hold by F1, so this verifies every full-algebra relation. When $V\mathrm{\ne }0$, the operators $0$ and ${\mathrm{id}}_V$ are distinct compatible actions, proving nonuniqueness. When $V=0$ its only endomorphism is zero. These computations prove both directions of the exact extension criterion and use no AC.