# Wightman axioms to Vertex algebra, the inspiration for the infinitesimal translation operator T?

In section 1.1, 1.2 of Kac's book Vertex Algebras for Beginners, he deduces the axioms of vertex algebras (or more precisely, right chiral algebras) from the Wightman axioms for $2$d CFT.

Denote $\Phi_a(x)$'s the fields. The unitary representation of the conformal group associates the standard basis $\{ e_k \}$ with self-adjoint operators $P_k$ and $Q_k$. Here, $P_k$ associates to the translation by $e_k$, and $Q_k$ associates to the translation by $-e_k$ conjugated by the inversion map $( x \mapsto -x/|x|^2)$.

Assume the QFT is conformal, we can deduce some formulas for $[\Phi_a (x), P_k]$ and $[\Phi_a (x), Q_k]$. In the two-dimensional case, after changing to the light cone coordinate $t = x_0 - x_1$ and $\bar{t} = x_0 + x_1$, and introduce the operators $P = \frac{1}{2} P_0 - P_1$, $\bar{P} = \frac{1}{2} P_0 + P_1$ and some similar operators for $Q_0$, $Q_1$. We can deduce some simpler formulas for $[\Phi_a (x), P]$ and $[\Phi_a (x), Q]$

Now let $$T = \frac{1}{2} (P + [P,Q] - Q),$$ and with all the formulas mentioned above, we can deduce $T$ is the infinitesimal translation operator satisfying the translation covariance axiom, $$[T,Y(a,z)] = \partial Y(z,a)$$ where $Y(z,a)$ comes from the field $\Phi_a(x)$ in a natural way.

My question is why we should expect $T$ to be defined in this way or more precisely, what's the reason to consider the form $\frac{1}{2} ( P + [P,Q] - Q )$?

• In the definition of $Q_k$, you mean translation conjugated by inversion rather than the other way around? – Abdelmalek Abdesselam May 4 '17 at 13:50