Dual-number vertex algebra that is not of CFT type

Every polynomial modulo $(e^2)$ reduces to $a+be$. If $a+be=0$ in the quotient, its degree-at-most-one representative lies in the ideal $(e^2)$ only when $a=b=0$. Thus $1,e$ are a basis, and the displayed multiplication is well-defined, commutative, associative, and unital with $e^2=0$.

Bilinearity of the product makes $a\mapsto M_a$ linear, so $Y$ is a linear state–field map. For each $a\in V$, $Y(a,z)$ is multiplication by $a$, independent of $z$. Its only possibly nonzero mode is $a_{(-1)}=M_a$, while $a_{(n)}=0$ for every $n\mathrm{\ne }-1$. Hence for every $b\in V$ all modes with $n\ge 0$ annihilate $b$, so $Y(a,z)$ is a field.

Multiplication by $1$ is the identity and $T1=0$. Since $T=0$, $e^{zT}a=a$, while $Y(a,z)1=a\cdot 1=a$. This verifies the vacuum and creation axioms.

Both $[T,Y(a,z)]$ and ${\mathrm{\partial }}_{z}Y(a,z)$ vanish because $T=0$ and $Y(a,z)$ is constant. Thus translation covariance holds.

For any $a,b\in V$, the multiplication operators commute: $Y(a,z)Y(b,w)=M_{ab}=M_{ba}=Y(b,w)Y(a,z)$ by commutativity of $V$. Therefore $[Y(a,z),Y(b,w)]=0$, so locality holds with exponent $N=0$. Along with the preceding field, vacuum/creation, and translation checks, this verifies every ordinary axiom in [F2].

Take $\omega =0$ and $c=0$. Then $Y(\omega ,z)=0={\sum }_{n\in \mathbb{Z}}{L}_n{z}^{-n-2}$, so coefficient uniqueness gives $L_n=0$ for all $n$. Their Virasoro relations hold because both the modes and the central charge are zero, and $L(-1)=0=T$.

Since $L(0)=0$, its zero eigenspace is $V_0=V$, and for each integer $n>0$ its $n$-eigenspace is $V_n=\{0\}$. Thus $V={⨁}_{n\ge 0}{V}_n$, $L(0)$ is diagonalizable, and all weight spaces are finite-dimensional; but $\dim V_0=2$.

The subspace $\mathbb{C}\mathbf{1}=\mathbb{C}1$ is one-dimensional, while $e$ is linearly independent of $1$ and belongs to $V_0=V$. Hence $V_0\mathrm{\ne }\mathbb{C}\mathbf{1}$, so the selected CFT-type condition in [F3] fails. No analytic or physical conclusion is used.