String theory: Conformal invariance and Conformal Killing Vectors I am confused by the relation between the invariance of the Polyakov action under conformal transformations and the Conformal Killing Vectors (CKVs) appearing during the process of quantization.
Let me phrase my doubts in two points:

*

*The conformal gauge for the worldsheet metric can always be reached locally: this means that ignoring issues related to boundary conditions, one can choose coordinates and Weyl rescaling in such a way that the Polyakov action can be written as
\begin{align}
S=\frac{1}{2\pi\alpha'}\int d^2z\,\partial X^\mu\overline\partial X_\mu. 
\end{align}
This action is classically invariant under a subgroup of the original $\mathrm{diff}\ \times\ \mathrm{Weyl}$ given by the conformal transformations $z\to z+v(z)$. This group in two worldsheet dimensions is infinite dimensional, as it can be seen by looking at the Virasoro algebra, which has infinite generators $L_n$, $n\in\mathbb Z$.

*During the process of quantization, it becomes clear that there may not be a globally defined transformation that brings the metric into the conformal gauge (due to the existence of moduli) and that the form of the metric, once fixed, is preserved by a subgroup of the original $\mathrm{diff}\ \times\ \mathrm{Weyl}$ generated by CKVs satisfying $$(P\xi\,)_{ab}=\nabla_a\xi_b+\nabla_b\xi_a-h_{ab}\nabla_c\xi^c=0.$$ For example, the Conformal Killing Group on the sphere is $SL(2,\mathbb C)$, generated by the $L_{-1,0,1}$ subalgebra of the Virasoro algebra, whereas it is generated by the rigid translations for the torus $T^2$.

How are the above two points related?
 A: There are two potential connections to be made here: the relation between metric moduli and global gauge choice, and the relation between conformal Killing vectors and the Virasoro algebra.

*

*Firstly, while the worldsheet metric $h_{\alpha\beta}$ can always be locally set to $\eta_{\alpha\beta}$ using reparameterisation (up to a Weyl rescaling). However, this may not be possible globally. This is due to the presence of "moduli", i.e. zero modes of the operator $P^\dagger$ that sends metric deformations $t_{\alpha\beta}$ to vectors:
$$
(P^\dagger t)_\alpha\sim\nabla^\beta t_{\alpha\beta}
$$
The notation $P^\dagger$ indicates that it is the adjoint of the projector $P:(P\xi)_{\alpha\beta}=\nabla_\alpha\xi_\beta-\nabla_\beta\xi_\alpha-(\nabla\cdot\xi) h_{\alpha_\beta}$ under the natural inner product. To see the importance of moduli, consider an arbitrary infinitesimal diff-Weyl variation parameterised by $\xi$ and $\Lambda$:
$$
\delta h_{\alpha\beta}=-\underbrace{(P\xi)_{\alpha\beta}}_\text{traceless}+\underbrace{2(\Lambda -\nabla\cdot\xi)h_{\alpha\beta}}_\text{trace}
$$
Since the trace term can be set to zero with appropriate $\Lambda$, the condition for conformal gauge to hold globally is that $(P\xi)_{\alpha\beta}\overset!=t_{\alpha\beta}$ for some global vector field $\xi^\alpha$ and symmetric traceless tensor $t_{\alpha\beta}$. In other words, for a given metric deformation $t_{\alpha\beta}$, if we can find a global vector field $\xi^\alpha$ such that $t_{\alpha\beta}=(P\xi)_{\alpha\beta}$, then we can cancel the deformation with a diff-Weyl transform. However, the existence of zero modes $t^0$ of $P^\dagger$ means that
$$
(P^\dagger t^0)_\alpha=0\implies \langle P\xi, t^0\rangle=\langle\xi,P^\dagger t^0\rangle=0
$$
So $t^0_{\alpha\beta}$ is orthogonal to $(P\xi)_{\alpha\beta}$ for all global $\xi^\alpha$, thus showing that the gauge (in this case, conformal gauge) cannot hold globally on the surface. This, as the Riemann-Roch theorem shows, is actually a topological condition: the number of these moduli depends on the Euler characteristic. The name "moduli" arises from the fact that they describe the space of gauge inequivalent metrics, known as a "moduli space": moduli variations physically alter the metric.


*How does the conformal Killing group (CKG) relate to the familiar Virasoro modes? For this, we look at the conformal Killing vectors (CKV), which are the zero modes of $P$. These clearly generate the residual symmetry - in short, they consist of diffeomorphisms that are conformal, and hence can be undone by a Weyl rescaling, leaving the metric unchanged. Thus the CKG $= \mathrm{Ker} P$. In conformal gauge in particular, the condition for a CKV simplifies to being a holomorphic vector field:
$$
\bar\partial(\delta z)=\partial(\delta\bar z)=0
$$
The number of CKVs on a Riemann surface is also topological - the sphere has 3 (complex) CKVs, the torus has 1, and the higher genus surfaces have 0. Now we just need to explicitly find these linearly independent holomorphic vector fields. For the Riemann sphere $\mathbb C\cup\{\infty\}$, the most general such vector field is easily shown to be $\delta z = \epsilon_{-1}+\epsilon_0z+\epsilon_1z^2$.  Each term is an infinitesimal coordinate transformation, respectively associated with $L_{-1}, L_{0}$ and $L_{+1}$. Once exponentiated into finite, global coordinate transformations, together they form the group $SL(2,\mathbb C)$. This, up to a discrete subgroup factor, is the CKG of the sphere.
That the Virasoro modes show up again here is not a coincidence - after all, they are the Noether charges of conformal symmetry in the quantum theory, and hence generate these conformal Killing transformations! In particular, the coordinate transformations associated to $L_0$ and $L_{\pm1}$ are "simple" enough to be non-singular and globally defined (while the others are merely meromorphic but not holomorphic).
