[Holography] Global symmetry in boundary corresponds to gauge symmetry in bulk?

[crackjack]

I hear the statement that global symmetries in the boundary field theory corresponds to gauge symmetries in the bulk.

1) Is this a generic statement that is expected to hold for all holography pairs? (Maldacena states this towards the end of his first lecture at PiTP2010, which was supposed to a collection of generic statements without reference to any particular pair, but I am not very sure)

2) Can some one give a (hand-waving) argument to support this statement?

 

[atyy]

Horwitz and Polchinski (p12): "In fact in most examples of duality there are gauge symmetries on both sides and these are unrelated to each other: the duality pertains only to the physical quantities."

 

[marcus]

Interesting question, crackjack! I suppose it's possible that in general the bulk geometry might not be rigorously defined independently of the boundary.
It may be in the case of some specific pairs, as you say, but not in all cases.

So what one might find is that because it is more convenient/feasible people are using the boundary to give a clear definite meaning to what is supposed to be going on in the bulk. One might not be dealing with a correspondence between two independently mathematically well-defined things.

So I hope very much we will get several good answers to your question! It would help clear up any such doubts.

 

[crackjack]

@atty: Probably Horowitz & Polchinski were referring strictly to local gauge symmetries on both sides, which is understandable. Here, it is between a global symmetry and a local symmetry. In that same PiTP-2010, there was a question on this during the 2nd lecture of Maldacena (you will find it within the first 5min of his 2nd lecture). And he says local symm in boundary don't really correspond to anything in bulk.

BTW, Maldacena states this global-local symm correspondence at around 1:21:35 during his 1st lecture here: http://video.ias.edu/pitp-2010. He seems to imply that, the conserved current vector for the global symm on the boundary should correspond to the boundary value for a bulk vector boson and hence a local symm in the bulk. While I do understand that the source currents that couple to the fields in the boundary correspond to the value of the bulk fields at the boundary, I don't understand why these source currents in the boundary action should have anything to do with its global symm. This might even be a simple QFT question, having nothing to do with holography per se.

This global-local correspondence also seems to be the basis of most bottom-up (ie. heuristic bulk action, rather than deriving from a string action) AdS/CMT papers on superconductivity/superfluidity.

@marcus: I tend to believe that the bulk and boundary theories can exist on their own, and that this is really a correspondence in its true sense :)

 

[Physics Monkey]

Dear crackjack,

It is true that a gauge symmetry in the bulk gives a global symmetry on the boundary. However, one can also realize global symmetries geometrically, as isometries as with the conformal group.

This paper http://arxiv.org/abs/1007.2184 contains a discussion of this issue in the other direction for some non-relativistic pairs. Older work had realized the number operator as an isometry of an extra null circle. In this paper they obtain it instead from the gauge field picture.

For bulk gauge fields, the gauge transformations that go to a constant at infinity are actually not gauged which is why they give a global symmetry. This is analogous to the way that bulk diffeomorphisms that do not vanish at the boundary represent actually symmetry operations rather than redundancy i.e. the gravitational Hamiltonian is not zero in AdS.

One may think of these transformation as measuring the total charge contained in the space. This charge can be finite because the space is non-compact.

Hope this helps.

 

[crackjack]

It is true that a gauge symmetry in the bulk gives a global symmetry on the boundary. However, one can also realize global symmetries geometrically, as isometries as with the conformal group.

That is true. But can we not think of the flux-compactified geometry in the bulk (that induces the global symmetry on the boundary) as a gauging of some global symmetries in an ungauged-supergravity? If that is so, the global symmetry in boundary would again correspond to some local gauge symm in bulk.

This paper http://arxiv.org/abs/1007.2184 contains a discussion of this issue in the other direction for some non-relativistic pairs. Older work had realized the number operator as an isometry of an extra null circle. In this paper they obtain it instead from the gauge field picture.

Interesting. They seem to transfer the geometric symmetry of the circle to a gauge field by compactifying on it.

For bulk gauge fields, the gauge transformations that go to a constant at infinity are actually not gauged which is why they give a global symmetry. This is analogous to the way that bulk diffeomorphisms that do not vanish at the boundary represent actually symmetry operations rather than redundancy i.e. the gravitational Hamiltonian is not zero in AdS.

One may think of these transformation as measuring the total charge contained in the space. This charge can be finite because the space is non-compact.

Unfortunately, I lost you here. Are you saying that bulk gauge fields that do not drop to zero at infinity (~ boundary), and thereby acting as the external source current for the corresponding operator in the boundary Lagrangian, will also correspond to the internal conserved current of some global symmetry in this same boundary theory?

 

[atyy]

This is analogous to the way that bulk diffeomorphisms that do not vanish at the boundary represent actually symmetry operations rather than redundancy i.e. the gravitational Hamiltonian is not zero in AdS.

Is this saying that an AdS solution is not physically equivalent to any smooth and invertible change of coordinates, but only to those that preserve something (what?) on the boundary?

 

[marcus]

BTW brief off-topic comment---Atyy I know you have expressed interest in Brian Swingle's work from time to time. He just gave a talk at Perimeter which is on video. Google "pirsa swingle".
The topic is entanglement in connection with the Fractional Quantum Hall Effect. So condensed matter and emergence---topics which interest you. A lot of XG Wen papers are cited throughout the talk, as I recall, and he was your teacher. So it is possible you would like this video lecture.

http://pirsa.org/11120042
"Entanglement spectrum in geometry"

 

[atyy]

@marcus - thank you, I shall certainly watch it!

BTW, I think one could talk about gauge symmetries in the bulk even if the bulk is not defined independently of the boundary. For example, isn't bulk gravity only a low energy thing, with the high energy degrees of freedom being stringy? But as long as it makes sense to talk about bulk gravity, it should still be generally covariant.

 

[crackjack]

Is this saying that an AdS solution is not physically equivalent to any smooth and invertible change of coordinates, but only to those that preserve something (what?) on the boundary?

what = May be the induced metric (~ Gibbons-Hawking surface term)? In the Hamiltonian formalism, this surface term gives the total energy of the (non-compact) spacetime. Ref: http://arxiv.org/abs/gr-qc/9501014

 

[atyy]

what = May be the induced metric (~ Gibbons-Hawking surface term)? In the Hamiltonian formalism, this surface term gives the total energy of the (non-compact) spacetime. Ref: http://arxiv.org/abs/gr-qc/9501014

Thanks. I'm not sure these are related, but I found a couple of references describing how a boundary breaks a gauge symmetry in the bulk (in general, not AdS/CFT).

http://relativity.livingreviews.org/Articles/lrr-2009-4/: However, the boundary S breaks the diffeomorphism invariance of the system, and hence, on the boundary the diffeomorphism gauge motions yield the observables O[N^a] and the gauge degrees of freedom give rise to physical degrees of freedom, making it possible to introduce edge states.

Wen: On a compactified space, the action S_bulk=∫d³x L_eff(δA_μ) is invariant under the gauge transformation. However, on a space with boundary, say, a disc D, S_bulk is not gauge invariant ...

 

[Physics Monkey]

Maybe the easiest place to think about the charge case is in flat space. Suppose their are some matter fields. A gauge transformation labeled by \alpha(x) is generated by \exp{(i \int d^d x J^0(x) \alpha(x))}. Now if we put \alpha equal to a constant then this transformation does nothing to the vector potential but it does rotate the state by an angle proportional to the total charge in the system. This total charge may itself be measured via the electric field at infinity (assuming the charge is compactly supported). In this sense the constant part of the gauge transformation is a like a global symmetry.

Now transport the story to AdS. Part of the dictionary in the bulk gauge field case is the statement that the asymptotic radial electric field sets the boundary charge density. This entry assures us that we can count the charge in two different ways and get the same answer. The first method simply integrates over the boundary charge density. The second method counts all the charge in the bulk. However, because the bulk charge sources a gauge field, it is possible to do that counting using only the asymptotic electric field. Hence the two methods will agree.

Moreover, even if the boundary is compact we can still have a finite charge density. This is incompatible with the boundary symmetry being gauged i.e. where would the field lines go? However, this is compatible with AdS/CFT since the AdS boundary is really a conformal boundary which the bulk gauge fields asymptotically approach.

 

[crackjack]

I am very sorry - I still do not understand it.

In this sense the constant part of the gauge transformation is a like a global symmetry.

I understand this (from usual Noether's theorem), but not the rest of the first para. For example, I don't understand what is the current vector doing on the gauge group element effecting gauge transformations and why is it integrated over all space when it is a local transformation. To be clear: The local gauge symmetry in the bulk is preserved when suitable boundary terms are added to the bulk action right? So why do we have to talk about global rigid transformations in the bulk?

Part of the dictionary in the bulk gauge field case is the statement that the asymptotic radial electric field sets the boundary charge density. The first method simply integrates over the boundary charge density.

Seems to me that the boundary value of the bulk gauge field acts as an arbitrary source term for the current operator of the global symmetry on the boundary (more on this in the footnote below)

The second method counts all the charge in the bulk. However, because the bulk charge sources a gauge field, it is possible to do that counting using only the asymptotic electric field.

I lost you here again. Which charge in the bulk? Since there is a gauge symmetry in the bulk, the bulk charge is a gauge non-invariant quantity right?


Footnote:
Witten states this global-local correspondence in his paper in the paragraph just above eq-2.13, but there is no justification. I can't seem to find anything in Maldacena's paper too that he cites in that paragraph in this context.
Correct me if I am wrong: From what Witten says, it seems that the Noether conserved current of the global symmetry of the boundary theory is the operator that corresponds to (ie. sourced by the boundary value of) the bulk gauge field of the bulk gauge symmetry that induced the boundary global symmetry. I was wrongly thinking that this Noether conserved current itself is the boundary value of the bulk gauge field that acts as an arbitrary external source current.

 

[Physics Monkey]

I am very sorry - I still do not understand it.I understand this (from usual Noether's theorem), but not the rest of the first para. For example, I don't understand what is the current vector doing on the gauge group element effecting gauge transformations and why is it integrated over all space when it is a local transformation. To be clear: The local gauge symmetry in the bulk is preserved when suitable boundary terms are added to the bulk action right? So why do we have to talk about global rigid transformations in the bulk?

Take a free scalar. The current is proportional to J^0 \propto -i( \phi^* \partial_t \phi - \partial_t \phi^* \phi )= -i(\phi^* \pi^* - \pi \phi). We furthermore have the canonical commutation relation [\phi(x),\pi(y) ] = i \delta(x-y). Now compute the commutator [J^0(x),\phi(y) ] = \delta(x-y) \phi(y) and the commutator [ \int d^d x J^0(x) \alpha(x) , \phi(y) ] = \alpha(y) \phi(y). Using these facts you should be able to conclude that the unitary operator I wrote above implements the transformation \phi(y) \rightarrow e^{i \alpha(y)} \phi(y) i.e. a local gauge transformation.

Seems to me that the boundary value of the bulk gauge field acts as an arbitrary source term for the current operator of the global symmetry on the boundary (more on this in the footnote below)


I lost you here again. Which charge in the bulk? Since there is a gauge symmetry in the bulk, the bulk charge is a gauge non-invariant quantity right?

Bulk charge is fine so long as you remember that it sources a field. Useful exercise: use the fact that the vector potential acts like a creation operator for electric fields (they are conjugate) to show that the operator defined by creating a charge and its corresponding Coulomb field is locally gauge invariant and sensitive only to the global gauge transformation. Basically you can show that creating the charge plus its field is like sending a Wilson line to infinity.

Footnote:
Witten states this global-local correspondence in his paper in the paragraph just above eq-2.13, but there is no justification. I can't seem to find anything in Maldacena's paper too that he cites in that paragraph in this context.
Correct me if I am wrong: From what Witten says, it seems that the Noether conserved current of the global symmetry of the boundary theory is the operator that corresponds to (ie. sourced by the boundary value of) the bulk gauge field of the bulk gauge symmetry that induced the boundary global symmetry. I was wrongly thinking that this Noether conserved current itself is the boundary value of the bulk gauge field that acts as an arbitrary external source current.

Not sure exactly what you meant here. The bulk gauge field has two "fall offs" near the boundary. One is the source i.e. the thing that couples to the operator J^\mu in the action. You get to choose this should you want to put your theory in some background fields. The other is the vev i.e. the expectation value of J^\mu.

 

[crackjack]

Thanks for the clarification! I still have a few questions, but that is because I lack a good intuitive understanding.

Take a free scalar. The current is proportional to J^0 \propto -i( \phi^* \partial_t \phi - \partial_t \phi^* \phi )= -i(\phi^* \pi^* - \pi \phi)... and the commutator [ \int d^d x J^0(x) \alpha(x) , \phi(y) ] = \alpha(y) \phi(y). Using these facts you should be able to conclude that the unitary operator I wrote above implements the transformation \phi(y) \rightarrow e^{i \alpha(y)} \phi(y) i.e. a local gauge transformation.

Ah ok. I see what you mean (the integrated Ward identity). But I have never gone beyond the leading term - I always though it doesn't work for the exponential map.
For example, even at the next order, there is an additional unwanted term:
[\frac{1}{2!} A^2, B] = \frac{1}{2!}* \left( 2i \alpha(y) AB + (\alpha (y))^2 B \right)
where, A = i \int d^d x J^0(x) \alpha(x) and B = \phi(y)

[STRIKE]Also, the above current is not the full current right? There should have been a term, J_\mu = \partial^\nu (F_{\mu \nu}\ \alpha(x)\ ) (from the pure-gauge term in the Lagrangian) which is the only culprit in making the charge non-gauge-invariant (and also another term, \propto 2i A_\mu |\phi|^2, but this is relatively harmless)[/STRIKE]

Bulk charge is fine so long as you remember that it sources a field. Useful exercise: use the fact that the vector potential acts like a creation operator for electric fields (they are conjugate) to show that the operator defined by creating a charge and its corresponding Coulomb field is locally gauge invariant and sensitive only to the global gauge transformation. Basically you can show that creating the charge plus its field is like sending a Wilson line to infinity.

Interesting.
[STRIKE]Here, you mean the total total charge, including the "culprit term"?[/STRIKE] Is the operator you have in mind the creation operator of something like F_{0i}(x)\ J^0(x)? (As an aside, isn't electric polarization the classical dual of electric field?)

Not sure exactly what you meant here. The bulk gauge field has two "fall offs" near the boundary. One is the source i.e. the thing that couples to the operator J^\mu in the action. You get to choose this should you want to put your theory in some background fields. The other is the vev i.e. the expectation value of J^\mu.

Is this dual "fall off" restricted only to adjoint gauge fields in the bulk? What about supersymmetric adjoint spinor fields?
ie. For example, for bulk fundamental matter fields, the corresponding boundary operator has no such obvious (as for bulk gauge fields) connections to the bulk? So, for these bulk fields, the only "fall off" is that their boundary value sources the corresponding boundary operator?

Addendum: I was just now told that the charge corresponding to local gauge symmetry is truly conserved (and gauge invariant) by itself, according to the relatively unknown Noether's 2nd theorem. I am not sure what this means for the above discussion.