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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 $\frac{1}{2} ( P + [P,Q] - Q )$?

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

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