Deriving the transformation under Weyl rescaling in Polchinski eq. (1.2.31) I have another question in Polchinski's string theory book volume 1, namely how to derive Eq. (1.2.32)?
$$(-\gamma')^{1/2} R'=(-\gamma)^{1/2} (R-2 \nabla^2 \omega)  \tag{1.2.32}$$
I have awared his Eq. (1.2.21) for the transformation of world-line metric
$$\gamma_{ab}'(\tau,\sigma)=\exp(2 \omega(\tau,\sigma )) \gamma_{ab} (\tau,\sigma)$$
, the definitions of Ricci scalar, $R:=R_{\mu \nu} g^{\mu \nu}$, Ricci, and Riemann tensors. It is still far from obvious how to construct the transformation of $R'$ from metric->Christoffel symbols->Riemann->Ricci-> (1.2.32)
My question is, how to derive Eq. (1.2.32)?
 A: As Prahar says, you just calculate the Ricci scalar $R'$ following the usual definition – the double contraction of the Riemann tensor – which is calculated from the Christoffel symbols and its derivatives where the metric $\gamma'$ expressed as $\exp(2\omega)\gamma$ is substituted everywhere. It will probably not fit one page.
However, there's a faster indirect way to establish the formula.
First, calculate $R'$ for a constant i.e. $(\tau,\sigma)$-independent $\omega$. The distances calculated by the $\gamma'$ metric are just $\exp(\omega)$ times longer than those calculated from $\gamma$. So the Ricci scalar calculated from $\gamma'$, equal to $C/a^2$ where $C$ is a constant and $a$ is some curvature radius, is $\exp(-2\omega)$ times the Ricci scalar calculated from $\gamma$. The only way how to get a formula for $\exp(-2\omega)$ in this simple rescaling by using $\gamma,\gamma'$ only is
$$ \exp(-2\omega) = \frac{\sqrt{-\gamma}}{\sqrt{-\gamma'}} $$
Therefore we have
$$ R' = R \exp(-2\omega) = R \frac{\sqrt{-\gamma}}{\sqrt{-\gamma'}}  $$
for a constant $\omega$. That agrees with your formula in your special case. Just to be sure, if we wrote $\exp(-2\omega)$ to the formula explicitly, instead of the ratio of the two square roots of determinants, it would still be valid because for the two metrics related by the rescaling, the ratio of the two determinants is given by the power of $\exp(\omega)$.
Now, consider a variable $\omega$. It's clear that $R$, the Ricci scalar, only contains up to second derivatives of $\omega$, so the expression for $R'$ can only depend on $\omega$ and the first and second derivatives of it. The dependence on $\omega$ itself has already been clarified because it's fully determined by the case of the constant $\omega$.
Now, the most general relationship for a variable $\omega$ may depend on the derivatives through $\nabla\omega\cdot \nabla \omega$ and $\nabla^2\omega$ because these are the only two covariant expressions one may construct out of the first two derivatives of $\omega$. And the relationship between $R,R'$ has to be given by a nice covariant (tensor structure respecting) formula because the relation between $\gamma,\gamma'$ is also expressed by a nice covariant (tensor structure respecting) formula.
The dependence on $\nabla\omega\cdot \nabla\omega$ actually cannot be there because it is an even function(al) of $\omega$. So you can't say that $R'$ is increased by a multiple of this expression relatively to $R$ because the opposite relationship should give a decrease but this $\nabla\omega\cdot \nabla\omega$ has the same sign in both ways.
That's why the formula only has to be modified by a multiple $\nabla^2\omega$ inserted somewhere. It's really straightforward to convince yourself that up to the ratio of the (square roots of the) determinants, the relationship has to be simply given by $\dots R-2\nabla^2\omega$. For example, you may write down the metric for the sphere in a conformally flat way and verify that if $R=0$ for a flat $\gamma$, $R'=2/a^2$, the desired Ricci scalar for a sphere of radius $a$, may be calculated from the covariant Laplacian of $\omega$.
A: I realise that this answer comes many years after the question was first asked and answered, but as I am probably not the last one struggling with this, I thought it might be worthwhile giving it anyway.
Building on Luboš explanation that by covariance the result needs to be of the form $g^{1/2} \left(R+a  \nabla \omega \cdot \nabla \omega + b \nabla^2 \omega \right)$ (with $g=-\gamma$), it is pretty straight forward to calculate the $a$ and $b$ explicitly without doing the full calculation, using symmetry arguments, or going to a specific metric. 
To determine $b$ note that $\nabla^2 \omega = \partial^2 \omega$ + terms that contain no second order derivative. We can thus find $b$ by identifying the pre-factor of $\partial^2$ in $R'$.  This can only come from the terms that have a derivative of the connection
\begin{align}
R' & = g'^{ab} R'_{ab} = g'^{ab} R'^c_{acb} = g'^{ab} \left( \partial_c \Gamma'^c_{ba} - \partial_b\Gamma'^c_{ca} + \cdots \right) \nonumber\\
\end{align} 
Now
\begin{align}
\Gamma'^{a}_{bc} &= \frac{1}{2} g'^{ad} \left( \partial_b g'_{cd} +  \partial_c g'_{bd}  -  \partial_d g'_{bc} \right)= \Gamma^{a}_{bc} + \Delta^{a}_{bc}
\end{align}
where
\begin{align}
\Delta^{a}_{bc} =  g^{ad}\left(   g_{cd}\partial_b\omega + g_{bd} \partial_c \omega - g_{bc} \partial_d \omega \right) 
\end{align}
Therefore
\begin{align}
g'^{1/2} R' & =  e^{+2\omega}g^{1/2} e^{-2\omega}  g^{ab}\left( \partial_c \Delta^c_{ba} - \partial_b\Delta^c_{ca} + \cdots \right)= - g^{1/2} d \partial^2 \omega
\end{align}
and so $b=-d=-2$.
One finds $a$ in a similar way. Because $\nabla \omega\cdot \nabla \omega =  \partial^a \omega \partial_a \omega$, $a$ follows from the coefficient of  $\partial_a\omega \partial_b \omega$ in $R'$. These terms can only come from the products of the contractions. 
\begin{align}
R' & = g'^{ab} \left( \Gamma'^{c}_{ce}\Gamma'^{e}_{ba} -  \Gamma'^{c}_{be}\Gamma'^{e}_{ca} \right) + \cdots = e^{-2w} g^{ab}  \left( \Delta^{c}_{ce}\Delta^{e}_{ba} -  \Delta^{c}_{be}\Delta^{e}_{ca} \right) + \cdots
\end{align}
Now straightforward calculation gives
\begin{align}
\Delta^{c}_{ce}\Delta^{e}_{ba} &=2d \partial_a\omega \partial_b \omega  -d  g_{ab} \partial^c\omega \partial_c \omega  \nonumber\\
\Delta^{c}_{be}\Delta^{e}_{ca} &=(d+2) \partial_a\omega \partial_b \omega  -2  g_{ab} \partial^c\omega \partial_c \omega 
\end{align}
and so $\Delta^{c}_{ce}\Delta^{e}_{ba} -  \Delta^{c}_{be}\Delta^{e}_{ca} =0$ for $d=2$. Therefore $a=0$.
