Polchinski Equation (10.4.7) I'm having a problem to interpret eq. (10.4.7) in Polchinski:
$$
\gamma(z)\delta(\gamma(0)) = O(z), \qquad \beta(z)\delta(\gamma(0)) = O(z^{-1}).
$$
What does he mean by $\delta$? He tries to explain after but I cannot understand. What are the precise step to go from eq (10.4.3a) to (10.4.7)?
 A: The operator $\delta(\gamma(z))$ is meant to be the operator dual to the state $|0\rangle_{NS}$ according to the state-operator correspondence. One could call it $E(z)$ or anything like that.
But Polchinski uses the notation $\delta(\gamma(z))$ with this "nested" structure because the operator described in the previous paragraph may also be interpreted as the Dirac delta-function whose argument is another operator, $\gamma(z)$. Why?
Because the delta-function obeys
$$\delta(x) x = 0$$
It's only nonzero at the point where $x=0$, so if you multiply it by $x$, the product is zero everywhere. Now, substitute $\gamma(0)$ for this $x$ and you will formally get
$$\delta(\gamma(0)) \gamma(0) = 0$$
You also want to know if the values of $z$ are different.
$$\delta(\gamma(0)) \gamma(z) = ?$$
The result has to be something that is equal to $0$ for $z=0$, as argued previously. There is no branch cut etc. so the right hand side may be expanded in power series, and because it's zero at zero, it may be written as $O(z)$.
The two conditions in (10.4.7) are nothing else than translations of the two conditions in (10.4.3a) from the cylinder to the plane, i.e. according to the state-operator dictionary. As I said, replace the state $|0\rangle_{NS}$ by the $\delta$ operator. And multiply it either by the Fourier modes of $\beta$ or those of $\gamma$. The point is that the state zero or the operator $\delta$ is annihilated when multiplied by a "high enough" mode of $\beta$ or $\gamma$. The "high enough modes" are described by the inequalities in (10.4.3a).
When you translate it to the plane and take the conformal weights of $\beta$ and $\gamma$, namely $+3/2$  and $-1/2$ into account – they correspond to the "shift" of the powers when going from the cylinder to the plane – you get (10.4.7).
On the cylinder (states), $\beta_k$ modes are multiplied by $\exp(2ik\sigma)$; on the plane, they are multiplied by $z^{k+3/2}$ or something like that. Sorry if the signs are wrong, you may fix them. The claim $\beta_k$ with $k$ greater than something annihilates the $|0\rangle_{NS}$ state gets translated to the claim that a power law expansion having just $\beta(z)$ multiplied by high enough powers of $z$, when multiplied by the operator $\delta(\gamma(z))$, only yields a power law expansion with powers of $z$ greater than some constant, and you may verify that the minimum exponents are exactly what is written there.
