The rank-one Heisenberg vacuum space is a conformal VOA of central charge one

On every polynomial, differentiation and multiplication give $[a_m,a_n]=m{\delta }_{m,-n}I$. All sufficiently large positive modes kill that polynomial. Thus $a(z)$ is a field, and its commutator is ${\mathrm{\partial }}_w\delta (z,w)$, obtained by summing $m{z}^{-m-1}{w}^{m-1}$. Multiplying this distribution by $(z-w{)}^2$ gives zero (differentiate $(z-w)\delta (z,w)=0$), so the generator is local with itself. Directly $[T,a_n]=-n{a}_{n-1}$ and $T1=0$. Also $a(z)1={\sum }_{r\ge 1}{x}_r{z}^{r-1}=e^{zT}{x}_1$, since $T^j{x}_1=j!x_{j+1}$. The negative modes are the multiplication operators $x_r$, whose monomials span $P$. All hypotheses of [F2] hold.

The reconstruction has precisely the stated normal fields. Indeed the two binomial expansions in [F3], with index $-r$, give $a(z{)}_{(-r)}B(z)=:b_r(z)B(z):$: taking the residue separates the modes $a_{-r-j}$, with coefficient $\left(\genfrac{}{}{0px}{}{r+j-1}{j}\right)z^j$, from the modes $a_j$, with coefficient $(-1{)}^{r-1}\left(\genfrac{}{}{0px}{}{r+j-1}{j}\right)z^{-r-j}$. These are the divided derivatives of the negative and nonnegative parts of $a$. By [F2]'s vacuum-state transport, the state of that product is $a_{-r}v=x_{r}v$ if $B=Y(v,z)$. Induction on the number of factors gives the displayed monomial formula. The same-sign modes commute, so nesting these normal products introduces no extra contractions. This proves the candidate state fields coincide with the reconstructed ones.

Since $Y(\omega ,z)=\frac{1}{2}:a(z{)}^2:$, its coefficient is the quadratic sum stated. For a fixed $n$ and polynomial, only finitely many terms survive: modes of large positive index kill the polynomial after normal ordering, and the sum of two negative indices is fixed. The oscillator commutator gives $[L(n),a_m]=-m{a}_{n+m}$ by the two terms $k=-m$ and $n-k=-m$ in that sum. Directly the sums at $n=0,-1$ give $L(0)={\sum }_{r\ge 1}{a}_{-r}{a}_r$ and $L(-1)={\sum }_{r\ge 1}{a}_{-r-1}{a}_r=T$, since $a_0=0$.

Let $C_{m,n}=[L(m),L(n)]-(m-n)L(m+n)$. Step 3.1 and the operator commutator identity show that $C_{m,n}$ commutes with every $a_r$. Any operator commuting with all multiplication operators and all positive derivatives is scalar: its value on $1$ is killed by every partial derivative and hence is constant, and commuting with multiplication determines its value on every monomial. Each $L(n)$ has degree $-n$, so this scalar is zero unless $m+n=0$. For $m>0$, $L(m)1=L(0)1=0$ and $L(-m)1=\frac{1}{2}{\sum }_{r=1}^{m-1}{x}_r{x}_{m-r}$. Applying $L(m)$ with step 3.1 gives $C_{m,-m}1=\frac{1}{2}{\sum }_{r=1}^{m-1}r(m-r)1=(m^3-m)1\mathrm{/}12$. Antisymmetry covers negative $m$, and $m=0$ is zero. Thus $[L(m),L(n)]=(m-n)L(m+n)+(m^3-m){\delta }_{m,-n}I\mathrm{/}12$.

The explicit Euler operator in step 3.1 has weight $d$ on $P_d$. For fixed $d$, only $x_1,\dots ,x_d$ can occur, with $m_r\le d\mathrm{/}r$, so its monomial basis is finite. There are no negative degrees, and degree zero consists of constants. Along with $L(-1)=T$ and the Virasoro relations of step 4.1, these verify every requirement in [F4]. No choice principle or analytic completion is used.