TITLE: Radial ordered commutation relation
QUESTION [8 upvotes]: In the book Conformal Field Theory of Francesco, Mathieu and Sénéchal, in Sec. 6.1.2, the authors state that the integral 
$$
\oint_w \mathrm{d}z~ a(z)b(w) ~=~ \oint_{C_1} \mathrm{d}z~ a(z)b(w) - \oint_{C_2} \mathrm{d} z ~b(w)a(z)\tag{6.15a} 
$$
can be seen as a commutator $$~=~[A,b(w)],\tag{6.15b} $$ where 
$$A~=~\oint a(z)\mathrm{d}z\tag{6.16}$$ 
is "the integral over space at fixed time (i.e., a contour integral of the field $a(z)$)". I feel confused about the meaning of $A$: If $A$ is a contour integral, what is the contour of the integral? The "fixed time" seems not specified in the definition of $A$.

REPLY [1 votes]: It might be useful to recall that we work here under radial ordering, where the modulus of a complex arguments is related to the "time" coordinate. So one has the following definition:
$$R(A(z)B(w))  := \left\{\begin{array}{ll}
A(z)B(w) & \textrm{for}\;|z|>|w|\\
B(w)A(z) &\textrm{for}\;|w|>|z|
\end{array}\right.$$
Then one can right
$$\oint dz [A(z),B(w)] = \oint_{|z|>|w|} dz A(z)B(w) - \oint_{|z|<|w|} dz B(w)A(z)$$
So if you observe the contours of the right side expression:
$$\oint dz [A(z),B(w)] = \oint_{C(w)}\; dz R(A(z)B(w))$$
The contour of the Integral needs to enclose $w$ but is otherwise arbitrary. It is not specified since its particular shape doesn't affect the result for holomorphic functions. You can check the details for this in the book by Blumenhagen and Plauschin "Introduction to CFT" which I believe has a nice explanation on all this.
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