Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I would appreciate it if someone tells me how a cft on a compactified manifold (e.g. by means of periodic boundary conditions) can be meaningful? The global conformal invariance is broken due to the scale over which the manifold is compactified (e.g. the period). The local conformal invariance of course still classically exists, but for the simple case of a field on a cylinder for instance, the trace of the stress-energy tensor becomes non-zero (in proportion with the central charge); hence apparently compactification of the space-time manifold is in contrast with having conformal symmetry.

share|improve this question

2 Answers 2

First, if you talk about world sheet CFTs describing compactifications of string theory, as your focus on the "central charge" and "compactifications" suggests, for those we must realize that the world sheet conformal symmetry has nothing to do with the would-be spacetime conformal symmetry. Spacetime transformations are manifested as internal symmetries of the world sheet CFT so they are completely independent from the world sheet conformal transformations that act nonlocally on the world sheet. String theory introduces a preferred distance scale in the spacetime, the string scale (and derived scales dependent on the string coupling, such as the spacetime Planck scale as well), so the spacetime effective theories produced by string theory are never conformal (because they're not scale-invariant).

If you mean compactifications of the world volumes at which CFTs such as those from the AdS/CFT are defined, then it is true of course that they break the conformal symmetry in general. The conformal symmetry is only broken by the global effects, not locally, however. In particular, some compactifications such as your cylinder preserve the conformal symmetry completely. This is particularly well-known for the $N=4$ gauge theory on $S^3\times {\mathbb R}$, a multi-dimensional cylinder, which preserves all the conformal symmetry including the $SO(4,2)$ conformal group.

No, the stress-energy tensor of such a theory still exactly vanishes. The theory is locally the same regardless of the compactification.

share|improve this answer
    
Is your last example similar to the global breaking of conformal symmetry when going from the complex plane to the cylinder (compactifying and so introducing a length scale)? It gives a "vaccum Casimir energy" $E \propto \frac{c}{L^2}$ but still we have $T^{z\bar{z}}=0$, thus tracelessness and local conformal invariance? –  Learning is a mess May 22 '13 at 22:09

Let me answer in two steps: first of all, and contrary to what you claim, the trace of the stress tensor vanishes also on the cylinder not just on the plane. In fact it is proportional to the Ricci curvature of the world-sheet which vanishes both on the flat plane and on the flat cylinder. On the other hand, what becomes non-vanishing when passing from the plane to the cylinder is the vacuum expectation value of the chiral component $T_{--}$ of the stress tensor (the same is true for the anti-chiral one $T_{++}$). This is a subtle effect which I can explain if I am asked to do so...

Now about your "paradox": consider first a classical conformal field theory on the Minkowski cylinder, that is a theory invariant with respect to the conformal group $G$ of the Minkowski cylinder. (Modulo some covering and discrete factoring the group $G$ is just the direct product of two copies of the diffeomorphism group of circle - you can find an elegant explanation of this fact e.g. in Segal's paper on the axiomatic CFT). It must be stressed, that no subgroup of $G$ acts by rescaling of the standard flat coordinates of the cylinder, however! To be more precise about this crucial point: write the flat metric on the cylinder in the standard Minkowski flat coordinates $t,\sigma$: $$ds^2=dt^2-d\sigma^2.$$ For that matter, $\sigma$ is the angle of the rotation around the cylinder axis; it is not therefore a global coordinate on the cylinder (although $d\sigma$ makes sense globally). Now you observe that the range of $\sigma$ is the intervale $[0,2\pi[$ hence you cannot scale $t,\sigma$ as $\lambda t,\lambda \sigma$ and indeeed this "transformation" does not belong to the conformal group $G$! Instead, the shifts $t,\sigma\to t+t_0,\sigma +\sigma_0$ do belong to $G$ and they become the global (complex) scaling transformations when passing from the cylinder to the plane via the standard exponential map!

The moral of the story is the following: the conformal field theories on the cylinder and on the plane are close cousins in the sense that there are transformation rules which transform solution of one into the solution of the other (in quantum version of the theories e.g. the knowledge of the correlation functions on the cylinder makes possible to work out the correlation functions on the plane and vice versa). However, the conformal transformations which are shifts of the natural coordinates on the cylinder become (upon the exponential transformation) the scalings of the natural coordinates on the plane. Since we like to calculate on the plane, we call them always, and somewhat abusively, the " global scaling transformations".

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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