# Radial ordered commutation relation

Perhaps it would be more clear if the authors used the following notation: $$\begin{align}\oint_{|z-w|=\varepsilon}& \mathrm{d}z~{\cal R} a(z)b(w) \cr ~=~& \oint_{|z|=|w|+\varepsilon} \mathrm{d}z~ a(z)b(w)\cr ~-~& \oint_{|z|=|w|-\varepsilon} \mathrm{d} z ~b(w)a(z)\tag{6.15a} \cr ~=~&A(|w|\!+\!\varepsilon) b(w)-b(w)A(|w|\!-\!\varepsilon) \cr ~=:~&~[A(|w|),b(w)],\tag{6.15b} \end{align}$$ where we have defined $$A(R)~:=~\oint_{|z|=R} \mathrm{d}z~a(z).\tag{6.16}$$

In eq. (6.15a) the symbol ${\cal R}$ denotes radial ordering, $${\cal R} a(z)b(w)~:=~\left\{ \begin{array}{rcl} a(z)b(w)&{\rm for}&|z|>|w|, \cr b(w)a(z)&{\rm for}&|w|>|z|.\end{array}\right. $$

The symbol ${\cal R}$ itself is often implicitly implied in CFT texts.The non-radial-ordered OPE $a(z)b(w)$ is typically not well-defined/divergent for $|z|<|w|$. Therefore the formula $${\cal R} a(z)b(w)~=~\theta(|z|\!-\!|w|)a(z)b(w)+\theta(|w|\!-\!|z|)b(w)a(z)$$ only makes sense if we define that "zero times ill-defined is zero".

The radial-ordered OPE ${\cal R} a(z)b(w)$ is typically a meromorphic function (possible with branch cuts). Integration contours can be deformed as long as they don't cross the position of other operator insertions, cf. Cauchy's integral theorem. The commutator (6.15b) is formally singular. It is regularized via point-splitting.

*Example:*The holomorphic part of the bosonic string has non-radial-ordered OPE $$ X(z)X(w)~=~ -\frac{\alpha^{\prime}}{2}{\rm Ln} (z-w) \quad {\rm for} \quad |z|>|w|. $$ The radial-ordered OPE $${\cal R} X(z)X(w)~=~\left\{ \begin{array}{rcl} -\frac{\alpha^{\prime}}{2}{\rm Ln} (z-w)&{\rm for}&|z|>|w|, \cr -\frac{\alpha^{\prime}}{2}{\rm Ln} (w-z)&{\rm for}&|w|>|z|,\end{array}\right. $$ has a $\pm\pi i\alpha^{\prime}$ branch cut along $|z|=|w|$ because of the complex logarithm ${\rm Ln}$. This branch cut disappears when we consider derivatives of $X$.