I'm just brushing up on a bit of CFT, and I'm trying to understand whether conformal symmetry is local or global in the physics sense.

Obviously when the metric is viewed as dynamical then the symmetry is local, because essentially then we're dealing with a change of variables under which the metric transforms with a local scale factor $\Omega = \Omega(x)$.

Usually, however, we think of the metric as fixed. David Tong's excellent notes suggest that in this case the symmetry should be thought of as global. But I'm not sure I agree.

Say we work in 2D and have general conformal transformation given by holomorphic function $f(z)$. Under a conformal transformation $z\to w$ say, viewed actively, a general field $\Phi$ will transform to

$$(\frac{dw}{dz})^h \Phi$$

where the prefactor is clearly dependent on the spacetime point. This would suggest that the transformation is local in the physics sense.

Perhaps the distinction he is trying to make is between physical and gauge transformations. But then again I might be wrong because I thought that only global transformations had nonzero conserved quantities, and there's definitely a conserved current for conformal symmetry.

Could anyone help to clarify this for me?


1 Answer 1


Whether the conformal symmetry is local or global depends on the theory! More precisely, the symmetry that may be local is not really conformal symmetry but ${\rm diff}\times {\rm Weyl}$.

For example, in all the CFTs we use in the AdS/CFT correspondence, for example the famous ${\mathcal N}=4$ gauge theory in $d=4$, the conformal symmetry is global – and, correspondingly, it is a physical symmetry with nonzero values of generators. This is related to the fact that the CFT side of the holographic duality is a non-gravitational theory so it avoids all local symmetries related to spacetime geometry.

The previous paragraph holds even if the dimension of the CFT world volume is $d=2$. In $d=2$, it may happen that the global conformal symmetry is extended to the infinite-dimensional local symmetry where $\Omega(x)$ depends on the location. However, such an enhancement looks "automatic" only classically. Quantum mechanically, a nonzero central charge $c\neq 0$ prevents one from defining the general local conformal transformations. In all the CFTs from AdS/CFT, we have $c\geq 0$. Such a nonzero $c$ leads to the "conformal anomaly" (proportional to the world sheet Ricci scalar and $c$).

On the contrary, the world sheet $d=2$ CFT theories used to describe perturbative string theory always have a local diffeomorphism and local Weyl symmetry. This is needed to decouple all the unphysical components of the world sheet metric tensor; and a necessary condition is the incorporation of the conformal (and other) ghosts so that in the critical dimension, we have the necessary $c=0$. We say that the world sheet CFT is "coupled to gravity" as we add the world sheet metric tensor, the diff symmetry, and the Weyl symmetry. The Weyl symmetry is the symmetry under a general scaling of the world sheet metric by $\Omega(x)$ that depends on the location on the world sheet. One may gauge-fix this local Weyl symmetry along with the 2-dimensional diffeomorphism symmetry, e.g. by demanding the $\delta_{ij}$ form of the metric tensor. This gauge-fixing still preserves some residual symmetry, a subgroup of the originally infinite-dimensional "diff times Weyl" symmetry. This residual symmetry is nothing else than the infinite-dimensional conformal symmetry generated by $L_n$ and $\tilde L_n$. Because its being infinite-dimensional, we may call it a local conformal symmetry but it's really just a residual symmetry from "diff times Weyl". The global $SL(2,C)\sim SO(3,1)$ global subgroup is the Mobius group generated by $L_{0,\pm 1}$ and those with tildes, too.

As far as I know, this local conformal symmetry is a special case of some $d=2$ theories. In higher dimensions, the Weyl and diff aren't enough to kill all the components of the metric tensor and the "partially killed" theories with a dynamical metric are still inconsistent as the usual naively quantized versions of general relativity.

In all the cases above and others, it is true that the local symmetries – where the parameter $\Omega(x)$ is allowed to depend on time and space coordinates (if the latter exist) – are gauge symmetries (in the sense that the generators are obliged to annihilate physical states) while the global symmetries are always "physical" in your sense of the charge's being nonzero. These equivalences follow from some easy logical argument. When you have infinitely many generators of the (space)time-dependent symmetry transformations, it follows that all the quanta associated with these generators exactly decouple – have vanishing interactions – with the gauge-invariant degrees of freedom. So we always study the physical part of the theory only, and it's the theory composed of the gauge symmetry's singlets.

Greetings to David.

  • $\begingroup$ Hi Lubos - thanks a lot for your detailed answer. The only thing I don't understand is exactly why the symmetry can be said to be global, rather than local. Surely the fields have to transform in a position dependent way $\phi \to g(x) \phi$ which means that the symmetry should be local according to the usual definitions? I understand the distinction between the physical and gauge symmetries - I just object to calling the physical ones global just because they are physical. Is this what's going on? $\endgroup$ Dec 30, 2013 at 20:08
  • $\begingroup$ Dear Edward, I have already explained why your objection is immaterial. The definitions of local-and-gauge symmetries may superficially look different, and the same holds for global-and-physical symmetries, but one may prove that local symmetries are gauge symmetries (redundancies) while the global ones are physical, and vice versa. You may object but your objection is as unjustified as an objection against identifying 10+4 with 9+5. Well, it is the same thing! (Despite the fact that there may be people who don't understand why 10+4 and 9+5 is the same number.) $\endgroup$ Dec 31, 2013 at 10:54
  • 1
    $\begingroup$ Again, the global symmetry is a symmetry whose parameter is time-independent (which really means spacetime-independent in any theory that has symmetries mixing space and time). A local symmetry is one in which the parameters of the transformations depend on time (or spacetime), which implies an infinite number of parameters. The global symmetries's generators (conserved charges) must be allowed to have nonzero eigenvalues, otherwise one violates the clustering property. $\endgroup$ Dec 31, 2013 at 10:57
  • 1
    $\begingroup$ Dear Edward, the general holomorphic function $z\to f(z)$ is not a local symmetry on the world sheet because a general local symmetry on the world sheet is a function that arbitrarily depends on two world sheet coordinates $f(\sigma,\tau)$ i.e. $f(z,\bar z)$ and not just one - the "number" of holomorphic functions is the same as fns of 1 real var. In comparison with that, the set of holomorph. functions (even if we include those that are not one-to-one) is infinitely small. At any rate, the symmetry under general holomorphic maps is broken for CFT with $c\neq 0$ by the "conformal anomaly". $\endgroup$ Dec 31, 2013 at 11:23
  • 1
    $\begingroup$ Yup, I think I would subscribe to it. ;-) It is an infinite-dimensional global trafo by default. One may Fourier-combine the $k_n L_n$ generators into a $\int d\sigma\,k(\sigma) T_z(\sigma)$ which only contains generators that depend on the spatial coordinate $\sigma$. In string theory where a CFT is coupled to 2D gravity, the conformal symmetry is an infinite-dimensional (yet very low-dimensional, relatively to the starting group) residual symmetry from the diff x Weyl group which is a full-fledged local symmetry - generators depend on both world sheet coordinates "arbitrarily". $\endgroup$ Dec 31, 2013 at 14:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.