Conformal properties of the energy-momentum tensor and Schwarzian derivative Polchinski Vol. 1 (Sec. 2.4): I'm trying to understand the Eq. 2.4.26 where he shows how the stress tensor transforms under a conformal transformation ($z \rightarrow w$):
$$(\partial w)^2 T(w) = T(z) - \frac{c}{12} \{w,z\} \tag{2.4.26}$$
Firstly, what is the reason behind the appearance of a Schwarzian derivative $\{z,w\}$ and why is it sitting with the central charge term or the anomaly term $\frac{c}{12}$? 
One observation is under a fractional linear transformation $$z \rightarrow \frac{az+b}{cz+d}$$ the Schwarzian vanishes, thereby making the transformation look "normal", but then why does this transformation play a special role in a CFT? 
 A: The significance of the Möbius transformations $\mathrm{PSL}(2,\mathbb{C})$ in 2D conformal field theory is that they are the globally defined conformal transformations on the Riemann sphere.
While the infinitesimal conformal transformations form the infinite-dimensional Witt algebra spanned by the vector fields
$$ L_n = -z^{n+1}\partial_z$$
we must be mindful that those vector fields are not globally defined on the Riemann sphere $S^2 = \mathbb{C}\cup\{\infty\}$. Obviously, they are singular at $z = 0$ for $n < -1$. Changing coordinates by $z\mapsto w = \frac{1}{z}$, we get
$$ L_n = -w^{1-n}\partial_w$$
which is singular at $w=0$, i.e $z=\infty$, for $n > 1$.
Therefore, the only globally defined conformal generators are $L_{-1},L_0,L_1$. These three generate precisely the group of Möbius transformations $z\mapsto \frac{az+b}{cz+d}$.
Thus, the symmetry group of a conformal field theory on the Riemann sphere is just $\mathrm{PSL}(2,\mathbb{C})$, and we have the requirement that the stress-energy tensor also should be invariant under this symmetry group. No such requirement can be said for the infinitesimal transformation of the Witt algebra. Nevertheless, classically, the stress-energy tensor transforms with its usual conformal weight also under those, since there is no central charge.
In the course of quantization, we incur a central charge for the Witt algebra, turning it into the Virasoro algebra1. Since the energy-momentum tensor is $T(z) = \sum_n L_n z^{n-2}$, the appearance of the central charge means the classical transformation law under the infinitesimal transformations generated by the $L_n$ may change by a quantum correction - this is precisely the Schwarzian derivative term. In the classical case $c = 0$, it vanishes, as a quantum correction (in this case often interpreted as a normal ordering constant) should.
However, if this also changed its behaviour under the global transformations, then the quantum theory would become anomalous, in particular, it would break the conservation of the Noether currents associated to $L_{-1},L_0,L_1$, which are $T(z),zT(z),z^2T(z)$. That is, anomalous transformation under a $\mathrm{PSL}(2,\mathbb{C})$ transformation would break energy-momentum conservation. This is undesirable, and, in fact, does not happen (as you may convince yourself by just chugging through the calculation of the transformation behaviour of $T$).
Now, why does the Schwarzian derivative appear as the quantum correction? If you start from the requirement that the quantum correction must vanish for $c=0$ and for $\mathrm{PSL}(2,\mathbb{C})$ transformations, then it is clear that it must be proportional to $c$. Furthermore, whatever $\{z,w\}$ is, it has to respect the group composition law that two successive transformations $z\mapsto w \mapsto u$ give the same as mapping $z\mapsto u$ directly. This is equivalent to the equation
$$ \{u,z\} = \{w,z\} + \left(\frac{\mathrm{d}w}{\mathrm{d}z}\right)^2\{u,w\} \tag{1}$$
since
$$ T(u) = \left(\frac{\mathrm{d}w}{\mathrm{d}u}\right)^2 \left(T(w) + \{u,w\}\right)$$
and
$$ T(w) = \left(\frac{\mathrm{d}z}{\mathrm{d}w}\right)^2 \left(T(z) + \{w,z\}\right)$$
but also
$$ T(u) = \left(\frac{\mathrm{d}z}{\mathrm{d}u}\right)^2 \left(T(z) + \{u,z\}\right)$$
so we obtain
$$ \left(\frac{\mathrm{d}w}{\mathrm{d}u}\right)^2 \left(\left(\frac{\mathrm{d}z}{\mathrm{d}w}\right)^2 \left(T(z) + \{w,z\}\right) + \{u,w\}\right) = \left(\frac{\mathrm{d}z}{\mathrm{d}u}\right)^2 \left(T(z) + \{u,z\}\right)$$
which gives 
$$ \left(\frac{\mathrm{d}z}{\mathrm{d}u}\right)^2 \{w,z\} + \left(\frac{\mathrm{d}w}{\mathrm{d}u}\right)^2\{u,w\} = \left(\frac{\mathrm{d}z}{\mathrm{d}u}\right)^2 \{u,z\} $$
after subtracting $\left(\frac{\mathrm{d}z}{\mathrm{d}u}\right)^2 T(z)$ from both sides. Multiplying by $\left(\frac{\mathrm{d}u}{\mathrm{d}z}\right)^2$ now yields eq. (1).
It can be shown that $(1)$ together with the requirement of $\mathrm{PSL}(2,\mathbb{C})$ invariance define the Schwarzian derivative uniquely.

1Shameless self-promotion: See this Q&A of mine for why we get a central charge in the quantum theory.
