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In Di francesco's Conformal Field Theory book he arrives at the equation (6.15)

$$\oint_C dz\; a(z)b(w)=[A,b(w)]\tag{6.15}$$

where $a(z)$ and $b(w)$ are $\textbf{holomorphic}$ fields, the contour $C$ circles counterclockwise around $w$, in the left hand side there is an implicit radial ordering and

\begin{equation} A=\oint dz\;a(z).\tag{6.16} \label{gay} \end{equation}

My question is, since $a(z)$ is holomorphic, how is $A$ not zero in virtue of Cauchy's theorem?

I think I'm missing something related to the operator character of this objects, for I understand that the presence of $b(w)$ in (6.15) might cause a singularity at $z=w$ (as we usually see in OPEs) thus making the integrals not vanishing but I can't reconciliate this with the expresion for $A$ alone.

For example this comes up when writing the Virasoro generators in terms of the energy-momentum tensor as

$$L_n=\frac{1}{2\pi i}\oint dz\; z^{n+1} T(z)$$ since I'm told $T(z)$ is holomorphic and following this reasoning this would make $L_n=0$ for $n\geq -1$, which is false.

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Nuclear Bidet
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1 Answers1

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  1. Briefly speaking, as OP already suggests, the operator $a(z)$ is only holomorphic/complex differentiable away from other operator insertions and branch points, which in turn, affect the topological codimension-1 operator $$A(C)~:=~\oint_{C} dz~a(z)\tag{6.16},$$ where $C$ is an integration contour.

  2. Note in particular that in the operator product $a(z)b(w)$ [which is typically singular for $z\!-\!w\to 0$], the $(z,w)$-dependence does not factorize.

  3. For more details, see e.g. this related Phys.SE post.

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