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:

  1. 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 $L_n$, $n\in\mathbb Z$.
  2. 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?

  • $\begingroup$ samario28, could you let me know if my answer is relevant to your question? $\endgroup$ Apr 30, 2022 at 4:55
  • $\begingroup$ I added a question to your answer! $\endgroup$
    – samario28
    Apr 30, 2022 at 16:08

1 Answer 1


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.

  1. 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.

  2. 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).

  • $\begingroup$ Thank you very much for your detailed answer. There is still something that I am missing: when is the string action a CFT? By this I mean the following: if the conformal gauge can be reached globally, the action is invariant under RESIDUAL holomorphic transformations of coordinates, which correspond to the conformal symmetry and whose Noether charges are ALL the Virasoro modes. If we are instead careful with global issues, it seems to me that the action remains invariant only under the CKG of the chosen Riemann surface (for the sphere only SL(2,C). What is wrong in what I am saying? $\endgroup$
    – samario28
    Apr 30, 2022 at 16:05
  • $\begingroup$ The string action is a CFT when the background metric is fixed (in this case to $\eta_{\alpha\beta}$, on each coordinate patch). Just like in normal QFTs, all the care that goes into constructing the CFT (e.g. charges, currents, modes, etc.) is all done with reference to infinitesimal symmetries, which form an algebra, not a group - Schottenloher discusses this in Chapter 2. That way we can ignore the global issues and just work on the Euclidean patches, which is easy. $\endgroup$ May 1, 2022 at 2:37
  • $\begingroup$ Much like in instanton contributions to gauge theories, the global residual transformations become relevant while performing the path integral (specifically while computing the Jacobian for the measure transformation on the space of metrics, which inherently considers the entire metric at once). So your answer is - yes, the group of global conformal transformations on a Riemann surface (i.e. $\mathrm{Aut}(R)$) is just the CKG, but we only care about the algebra of infinitesimal transformations to define the CFT. $\endgroup$ May 1, 2022 at 2:37
  • $\begingroup$ Your comments are very insightful and exactly touch the problems I am facing, so for that I would like to thank you a lot. 1) Why are we allowed to ignore global issues and treat the worldsheet as if it had the full conformal group of symmetries when deriving the spectrum, constructing the vertex operators et cetera, but when using the full path integral we are no longer able to do so? Shouldn't we be allowed to consider ONLY the CKG as the residual symmetry group? $\endgroup$
    – samario28
    May 1, 2022 at 17:58
  • $\begingroup$ 2) In this sense, I don't see how it is justified to use the full power of CFTs, if the proper approach (path integral) does not support the full symmetry. $\endgroup$
    – samario28
    May 1, 2022 at 17:59

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