Conformal Invariance and the OPE

I am reading Polchinski's String Theory now and am having trouble with a couple of things mentioned on Page 45 under "Conformal Invariance and the OPE". Here is the paragraph I am reading. The sentences in italics are my questions.

He states

Conformal invariance puts strong constraints on the form of the OPE, and in particular on the OPEs of the energy-momentum tensor. Consider the OPE of $T$ with the general operator ${\cal A}$. Because $T(z)$ and ${\tilde T}({\bar z})$ are (anti)holomorphic except at insertions (What does this mean?), the corresponding coefficient functions must also have this property (What if the operator is not holomorphic? Are the coefficient functions still holomorphic?)

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The first thing means that

$$\partial_\bar{z} \langle T(z) O_1(z_1,\bar{z_1}) O_2(z_2,\bar{z_2}) ... O_n(z_n,\bar{z_n}) \rangle = 0$$

except when $z=z_i$ for some i (coinciding insertion point for T and some other operator), and $O_i$ is any local operator at a given point (not necessarily holomorphic).

This Ward identity is an operator equation, it is like an equation of motion, it tells you that the stand-alone operator T(z) has a zero z-bar derivative. This means it has a zero z-bar derivative in any state, and the operators $O_i$ just prepare the state, and whether they are holomorphic or anti-holomorphic, the z-bar derivative of T is zero, so the z-bar derivative of the correlation function is zero.

When there is a coincidence of positions, you have ordering issues preventing the naive equation of motion from holding, because the path integral time-orders all operators automatically. So the derivative includes delta-function terms which reorder the thing, and the equation fails. The holomorphic Ward identity is stronger than this in its actual statement: it says that the right hand side determines the holomorphic transformation properties of each of the local fields.

This implies that if you expand T(z)O(z') in an OPE, whether O is holomorphic or not, the coefficients must be holomorphic away from coinciding points, since

$$T(z) O(z') = \sum_{k=-n}^{\infty} C_k(z-z') O_k(z)$$

And the zero value of the z-bar derivative of the left-hand side means that the z-bar derivative of the right hand side, and therefore of the coefficient functions, is also zero except when z=z'.

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Thanks! That made it clear. – Prahar Sep 23 '12 at 17:01