In Polchinski's longer derivation of the Weyl anomaly, he arrives at the result (equation 3.4.19): $$ \ln{\frac{Z[g]}{Z[\delta]}} = \frac{a_1}{8\pi} \int d^2\sigma \int d^2\sigma' g^{1/2} R(\sigma) G(\sigma,\sigma') g^{1/2} R(\sigma') $$ where $Z[g]$ is the gauged-fixed path integral, and $G$ is a Green's function satisfying equation 3.4.20: $g(\sigma)^{1/2} \nabla^2 G(\sigma,\sigma') = \delta^2(\sigma-\sigma')$. Taking a perturbation expansion $g_{ab} = \delta_{ab} + h_{ab}$, he arrives at equation 3.4.21: $$ \ln{\frac{Z[\delta + h]}{Z[\delta]}} \approx \frac{a_1}{8\pi^2} \int d^2z \int d^2z' (\partial_z^2 \ln{|z-z'|^2}) h_{\bar{z}\bar{z}}(z,\bar{z}) \partial^2_{z'} h_{\bar{z}\bar{z}}(z',\bar{z}')$$
Here is my naive attempt at a derivation: With a flat background, we take Polchinski's convention for flat, complex coordinates: $d^2\sigma = \frac{1}{2} d^2z$, $\delta^2(\sigma-\sigma') = 2 \delta^2(z-z',\bar{z}-\bar{z}')$.
The integral measure $d^2\sigma d^2\sigma'$ becomes $\frac{1}{4}d^2z d^2z'$. Using the complex flat metric (equation 2.1.6),
$$ g^{1/2} \nabla^2 G = \partial_a (g^{1/2} g^{ab} \partial_b) = 2 \partial_z \Big(\frac{1}{2}\partial_{\bar{z}} G\Big) + 2\partial_{\bar{z}}\Big(\frac{1}{2}\partial_zG\Big) = 2\partial_z \partial_{\bar{z}}G = 2 \delta^2(z-z',\bar{z}-\bar{z}').$$
Using $\partial_z\partial_{\bar{z}} \ln{|z|^2} = 2\pi \delta^2(z,\bar{z})$, to zeroth order, choose $G(z,z') = \frac{1}{2\pi}\ln{|z-z'|^2}$. Taking Polchinski's suggestion for the Ricci scalar, $4\partial^2_z h_{\bar{z}\bar{z}}$, and to zeroth order, $g^{1/2} = \frac{1}{2}$, we get
$$ \frac{1}{4}d^2z d^2z'\frac{1}{4} \frac{1}{2\pi} \ln{|z-z'|^2} 4\partial^2_z h_{\bar{z}\bar{z}}(z,\bar{z}) 4\partial^2_{z'} h_{\bar{z}\bar{z}}(z',\bar{z}'). $$
This has an overall factor of $\frac{1}{2}$ off from equation 3.4.21. In this, I assumed the expansion of $g^{1/2}$ was second order in $h_{ab}$ or higher.
Questions about deriving equation 3.4.21: Where did $h_{zz}$ and $h_{z\bar{z}}$ go? What are some of the contact terms? Was my choice of $G(\sigma,\sigma')$ incorrect? Factor of $\frac{1}{2}$ off in my rough derivation?
EDIT: I think I figured out $G$ and my factor of $\frac{1}{2}$. Since the integral we care about is diff invariant, we can choose the $\sigma$ coordinates to be $\sigma^1 = z$ and $\sigma^2 = \bar{z}$. With this $d^2\sigma g^{1/2} = \frac{1}{2} d^2 z$ (using $|g|^{1/2} = \frac{1}{2})$, two of these measures brings a factor of $\frac{1}{4}$. The Green's function satisfies:
$$ g^{1/2} \nabla^2 G = 2 \partial_z \partial_{\bar{z}} G = \delta^2(\sigma-\sigma') = \delta^2(z-z',\bar{z}-\bar{z}') $$
Thus $G(z,z') = \frac{1}{4\pi} \ln{|z-z'|^2}$. Putting this into the integral:
$$ \frac{a_1}{8\pi^2} \int \frac{1}{2} d^2 z \frac{1}{2} d^2 z' \frac{1}{4\pi} \ln{|z-z'|^2} 4\partial^2_z h_{\bar{z}\bar{z}}(z,\bar{z}) 4\partial^2_{z'} h_{\bar{z}\bar{z}}(z',\bar{z}') $$
All the factors of $4$ and $2$ cancels, which gives the correct prefactor of $\frac{a_1}{8\pi^2}$.
I'm still confused about the contact terms and where $h_{zz}$ and $h_{\bar{z}z}$ went - maybe you are allowed to ignore these, just perturbing the $\bar{z}\bar{z}$ component?
