# Charged CFT observables and AdS/CFT

I have a simple question regarding the holographic dictionary when mapping operators on the CFT side to those in AdS.

One piece of the dictionary is that a global symmetry maps onto a gauge symmetry in the bulk. So if I have operators charged under the global symmetry in the CFT then naively I would assume these are operators charged under the gauge symmetry in the bulk. However, since gauge symmetry is a redundant description we should also restrict to gauge singlet operators, at least to define observables in correlation functions. It seems theres an asymmetry here, all the charged operators on the CFT side are good observables, but only those in the singlet representation map to gauge invariant operators in the bulk.

Is it that the operators in the bulk are also all charged under a global symmetry (assuming we're working in the classical limit on the gravity side so it remains unbroken)?

Let me work in the usual limit where a classical theory of gravity maps to a strongly interacting conformal field theory (CFT) with some color like parameter $N$ that is taken to be very large. In this limit, a central statement of the correspondence is that a generating function for correlation functions on the CFT side is given by the on-shell gravitational action evaluated with certain boundary conditions for the fields. These boundary conditions are interpreted as sources for operators on the dual CFT side. In other words, on the CFT side, I've added something like $\int \phi_0 O$ to the action where $O$ is my operator for which I want to compute correlation functions and $\phi_0$ is a source. On the gravity side, $\phi_0$ is then interpreted as a boundary condition for some gravity field $\phi$. I use this boundary condition to solve Einstein's equations and then use that solution to evaluate the action. I get a schematic relation of the form $$S_{\rm grav}(\phi_0) = W_{\rm CFT}(\phi_0)$$ where $${\delta^n W_{\rm CFT} \over \delta \phi_0^n} = \langle O^n \rangle.$$

From this point of view, there is then no obvious impediment to considering an operator $O$ that is charged under some global symmetry on the CFT side. Indeed, that symmetry is gauged in the bulk, and there will be some corresponding field $\phi$ that is also charged under that gauge symmetry. Nevertheless, I can solve Einstein plus Maxwell or Einstein plus Yang-Mills equations in the bulk with the specified boundary value $\phi_0$. Even if $\phi$ transforms under the gauge symmetry, we usually restrict to local gauge transformations that fall to zero at the boundary. Thus $\phi_0$ will be a well defined quantity.

Added: Following the comments below, it seems important to add the following. While the gauge transformations that fall off at the boundary are redundant, the gauge transformations that don't fall off at the boundary are not gauged and are not redundant. They form an asymptotic symmetry group of the gravity theory. They act to change the state of the CFT and also (at a classical level) to change the solution to Einstein's equations. There is a matching between the global symmetry group of the CFT and the asymptotic symmetries on the gravity side.

One well cited example of an AdS/CFT system with a field that transforms under a global symmetry is the holographic superconductor: http://arxiv.org/abs/0810.1563.

• Why do you say the Hilbert space should be much larger? We should restrict to gauge transformations that vanish on the boundary, where the field theory "lives". The "large" gauge transformations which do affect boundary quantities generate the global symmetry of the CFT and are not redundant in the same way that the "small" gauge transformations that vanish at the boundary are. – user2309840 Sep 25 '15 at 4:12
• One more comment: you said in your question that you would naively assume charged operators in the CFT correspond to charged operators in gravity. I think that's not right. The boundary values of the gravity fields act as sources for the operators in the CFT. – user2309840 Sep 25 '15 at 4:14
• 1) The large gauge transformations are not redundant in the bulk. If they were, one would find unpleasant things, e.g. two CFTs at different chemical potential are gauge equivalent. 2) I stated that the boundary value of a gravity field sources an operator in the CFT. I said nothing about quantum operators on the gravity side. If you would like, you could promote the classical field to a quantum operator on the gravity side. Even then, one needs to make a careful distinction between the boundary value of this quantum operator and its bulk behavior. – user2309840 Sep 25 '15 at 18:12
• But if you agree about the large gauge transformations, then I think you also agree that the global symmetries match? – user2309840 Sep 27 '15 at 3:04
• I realize now that I'm perhaps using "large gauge transformation" in a nonstandard way. In the path integral for the gravity theory (whatever that may be), I always understood that one should fix boundary conditions for the gauge field at the boundary of AdS such that what I'm calling the "large gauge transformations" are a copy of the gauge group acting on the boundary theory. In the holographic superconductor context, for example, a U(1) gauge transformation that doesn't fall off at the boundary would act to change the phase of the condensate. – user2309840 Sep 28 '15 at 2:04