Vertex Algebra Axioms and State–Field Constructions: Examples and Counterexamples

Let $V={\mathbb{C}}^2$ with basis $v_1,v_2$. Define constant fields $A(z)=E_{12}$ and $B(w)=E_{21}$, where $E_{12}{v}_2=v_1$, $E_{12}{v}_1=0$, $E_{21}{v}_1=v_2$, and $E_{21}{v}_2=0$. Their only nonzero modes are $A_{(-1)}=E_{12}$ and $B_{(-1)}=E_{21}$, so both are fields by [F1].

Their commutator is $[A(z),B(w)]=E_{11}-E_{22}\mathrm{\ne }0$. For every integer $N\ge 0$, $(z-w{)}^N(E_{11}-E_{22})$ has nonzero coefficient $z^N(E_{11}-E_{22})$, including $N=0$. Thus no locality exponent exists, although both fields are lower-truncated; the refuted implication fails.

If $a=0$, then $e^{zT}a=0$; otherwise let $d=\mathrm{deg}a$, so $T^{r}a=0$ for $r>d$ and $e^{zT}a$ is a polynomial in $z$. Hence $Y(a,z)b$ has finite support in nonnegative powers for every $b$, and its coefficients define a field. Direct substitution gives $Y(1,z)b=b$ and $Y(a,z)1=e^{zT}a$. Also $T1=0$.

For any $b\in V$, the derivation law in [A1] gives $[T,Y(a,z)]b=T((e^{zT}a)b)-(e^{zT}a)Tb=(T{e}^{zT}a)b={\mathrm{\partial }}_{z}Y(a,z)b$, so translation covariance holds.

Each $Y(a,z)$ acts by multiplication by $e^{zT}a$. Multiplication operators commute because $V$ is commutative, so $[Y(a,z),Y(c,w)]=0$ for every $a,c\in V$. Thus locality holds with exponent $N=0$. All axioms of [F1] are verified.

Since $-2\in \mathbb{C}$, [F1] gives the universal affine vertex algebra at this level; its construction does not divide by $k+h^{\vee }$.

By [F2] and [F3], invariance gives $B(h,h)=B([e,f],h)=B(e,[f,h])=2$. It also gives $B(h,e)=B(h,f)=B(e,e)=B(f,f)=0$: for example $2B(e,e)=B([h,e],e)=B(h,[e,e])=0$, and $2B(h,e)=B([h,e],h)=-B(e,[h,h])=0$; the $f$ identities follow from $[h,f]=-2f$ in the same way. Thus the $B$-dual basis to $(h,e,f)$ is $(h\mathrm{/}2,f,e)$. The adjoint Casimir is therefore ${\mathrm{\Omega }}_B=\frac{1}{2}\mathrm{ad}(h{)}^2+\mathrm{ad}(e)\mathrm{ad}(f)+\mathrm{ad}(f)\mathrm{ad}(e)$. On $h$ its last two terms give $2h+2h=4h$; on $e$ its first two nonzero terms give $2e+2e=4e$; on $f$ its first and last nonzero terms give $2f+2f=4f$. Hence ${\gamma }_B=4$ and [F4] gives $h^{\vee }=2$.

Applying [F6] to $x=e,y=f$, with $[e,f]=h$, $B(e,f)=1$, and $k=-2$, gives the stated current OPE.

At $k=-2$ and $h^{\vee }=2$, the denominator in [F5] is $2(-2+2)=0$. Therefore this normalized Sugawara expression is undefined, even though step 1.1 supplies the affine vertex algebra. No Sugawara or Virasoro structure is inferred.