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I am referring to this paper.

I guess that in this paper one is trying to relate the massless spin $s$ gauge fields in $AdS_4$ to conformal spin $s$ theory on $S^3$.

  • So am I right that the operator $K$ that has been defined here in $2.8$ is something in the boundary? How does one derive the explicit expression for $K$ as given in $2.12$?

    Is it solely through this particular choice of $K$ in section $2.12$ that one is implementing in section $3$ the fact that the spin-$s$ theory on the boundary is conformal?

  • In section $3$ they seem to be solely focussed on symmetric traceless rank $s$ tensors (to represent spin-$s$ on the boundary $S^3$). But why is this enough? I would think that the spin-s fields to be considered are the fields on $S^3$ which lie in those representation of $SO(4)$ which when restricted to $S0(3)$ become its highest weight $s$ representation and these are not just symmetric and traceless but also have to be transverse and also satisfy some harmonic wave equation. What about these two conditions? (This was the definition of spin-$s$ as was discussed here.)

    But when considering spin-$s$ fields on the bulk $AdS$ in equation $5.1$ the condition of transversality and the wave-equation condition seem to be back!

    I basically don't understand equations $3.1$ and $3.6$. It would be great if someone could help explain these two.

  • Is there a value of $m^2$ (in equation 5.1) at which this spin-s field on $AdS$ will be conformally coupled? (...in this paper they are focussed at the massless case ($m^2 =0$) which I would think is not necessarily conformal..)

  • With reference to the discussion below equation 5.6,

    When the bulk spin-$s$ field is massless, there are two possible dimensions of the boundary spin-$s$ current, $J_{(s)}$ - at the UV fixed point it has dimensions, $\Delta_{-} = 2-s $ and at the IR fixed point it has dimensions, $\Delta_{+} = s+d-2$

    Here two things are not being very clear to me,

    (1) How does one see the claim that at the IR fixed point the value of $\Delta_{+}$ somehow implies that now $J_{(s)}$ is a conserved current and hence the spin-s field in the boundary is now a gauge field?

    (2) Is it also being claimed that at the UV fixed point the value of $\Delta_{-}$ is precisely the same as the dimension of a spin-$s$ gauge field? What theory is this? How do we understand this? I can't wrap my head around the fact that this $J_{s}$ which I thought of as the conserved current spin-s current till now happens to have the same dimension as a gauge field!?

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I will answer one by one. K is indeed defined in 2.8. as a two-point fuction of a spin-s operator - it knows nothing about bulk, purely boundary CFT object. All two point functions are fixed by conformal symmetry, so 2.12 is the unique expression in particular coordinates. As they say, they perturb a CFT with J^2 and then flow to a fixed-point, which meand they are again in CFT –  John Sep 13 '13 at 20:02
    
I have to warn you that the questions you asked are complicated and suprising answers to some of them have been obtained quite recently. So I am not sure if the scope of the forum allows one to answer them comprehensively. –  John Sep 13 '13 at 22:18
    
@John: Wait, what are you saying? Do you mean to say that this is non-mainstream? It is fine (And good) to bring up modern mainstream research in questions. –  Dimensio1n0 Sep 14 '13 at 15:34
    
I have in mind a question on the relation between gauge invariace and being conformall. The paper I refer to is just 5 years old. The rest of the questions have pretty standard answers. –  John Sep 14 '13 at 15:51
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up vote 5 down vote accepted

On the last question, I am not sure how good you are at the representation theory, but the following fact is true: take so(d,2) (we need so(3,2) for this work), use the conformal base, i.e. Lorentz generators $L_{ab}$, translations $P_a$, conformal boosts $K_a$ and dilatation $D$, $a,b=1..d$. $P$ and $K$ behave as raising/lowering generators with respect to $D$, $[D,P]=+P$, $[D,K]=-K$. Take the vacuum to carry a spin-s representation of the Lorentz algebra and a weight $\Delta$ with respect to $D$, i.e. $|\Delta\rangle^{a_1...a_s}$. When $\Delta=d+s-2$, there is a singular vector, $P_m|\Delta\rangle^{ma_2...a_s}$. This is a standard representation theory: finding raising/lowering operators, defining vacuum, looking for singular vectors. Actually, singular vectors are exactly the conformally-invariant equations one can impose.

On the field language this means that $\partial_m J^{m a_2...a_s}=0$ is a conformally invariant equation iff the conformal dimension of $J$ is $\Delta=d+s-2$. Despite the fact that $J^{a_1...a_s}$ is a good conformal operator for any value of the conformal dimension, only for $d+s-2$ its divergence decouples. (Perhaps you have seen $L_{-2}+\alpha L_{-1}^2$ as a singular vector in the Virasoro algebra, now it is replaced with $P_m$ or $\partial_m$).

Now, having $J^{a_1..a_s}$ of weight $\Delta$ we can consider its contragradient representation or on the field language couple it via $\int \phi_{a_1..a_s}J^{a_1...a_s}$ to some other field $\phi$. That we need a conformally invariant coupling implies $\Delta_\phi=d-\Delta_J=s-2$. Not surprisingly something special must happen for $\Delta_J=d+s-2$.

$$\int (\phi_{a_1...a_s}+\partial_{a_1}\xi_{a_2...a_s})J^{a_1...a_s}=\int \phi J-\int \phi_{a_1...a_s}\partial_m J^{ma_2..a_s}=\int \phi J$$ we see that a statement that is dual to the conservation of $J$ is the gauge invariance of $\phi$.

I have not read the paper yet, but as far as I can see they play with the dimension of $J$ and for $d+s-2$ and $2-s$ it describes a conserved tensor and a gauge field just because of representation theory of the conformal group (decoupling of certain null states). At any given moment of time in the paper $J$ has some fixed dimension and is either a conserved tensor, a gauge field or just a spin-s conformal field of generic dimension $\Delta$.

On the last but one, you are right in that gauge invariance has a little to do with conformallity. The answer is spin and dimension dependent. For $s=0$ there is $m^2$ for which the scalar is conformal. For $s=1$ and certain $m^2$ the Maxwell field is a gauge field but the Maxwell equation is conformal in $d=4$ only. Beyond $d=4$ a gauge spin-one field is not conformal, or a spin-one conformal field is not a gauge field. For $s\geq2$ the situation is even more tricky: in $AdS_4$ the gauge fields are conformal, but in Minkowski space they are not conformal (in terms of gauge potentials $\phi_{\mu_1...\mu_s}$). You may have a look at http://arxiv.org/abs/0707.1085

On the second, first of all the transversality is on the right place in 5.1. Secondly, your confusion (inspired by my answer to another question) is that there are two different classes of fields people are interested in. First is the class of usual particles, where we talk about representations of the Poincare algebra $iso(d-1,1)$ if we are in $d$-dimensional Minkowski space or $so(d-1,2)$ and $so(d,1)$ if we are in anti de Sitter ot de Sitter (there we need harmonicity, tracelessness, transversality). Conformal fields are in the second class. Conformal means that it must be a representation of the conformal group $so(d,2)$ for Minkowski-$d$, note that $iso(d-1,1)\in so(d,2)$. The conformal group of anti de Sitter-$d$ is also $so(d,2)$. Note that the symmetry algebra of AdS-$(d+1)$ is exactly the conformal group of Minkowski-$d$. So when we talk about conformal fields we are interested in reprsetations of $so(d,2)$ (the signature can vary depending on the problem, it is some real form of $so(d+2)$). I would like to stress that conformal fields in d-dimensions are in one-to-one correspondence with usual fields in $AdS_{d+1}$, for the algebra is the same, which is at the core of AdS/CFT correspondence.

For example, a spin-$0$ field in Minkowski space obeys $\square \phi=0$. It gives rise to an irreducible representaion of $iso(d-1,1)$. Coincidentally, the same representation turns out to be an irreducible representation of a bigger algebra, $so(d,2)$, the conformal algebra. It is a coincidence. There exists also a spin-$0$ conformal field of weight $\Delta$, say $\phi_\Delta(x)$. Without imposing any equations it is an irreducible representation of $so(d,2)$. As a representation of its subalgebra $iso(d-1,1)$ it decomposes into an intergral of representations (Fourier) and is highly reducible. There is a special weight $\Delta=(d-2)/2$ for which $\phi_\Delta(x)$ is reducible and the decoupling of null states is achived via $\square \phi=0$ (analogous to the conservation of $J$ above). Note that $J$ above is an irreducible representation of $so(d,2)$ but it is highly reducible under $iso(d-1,1)$. For special weight $d+s-2$ we have to impose the conservation condition in order to project out the null states, but again the conserved tensor is an irreducible of $so(d,2)$ and reducible under $iso(d-1,1)$. So your confusion is because the fields are conformal, these are representations of a bigger algebra, they are more 'fat' and require less equations (even no at all) to project onto an irreducible.

$S^3$ is the analog of Minkowski-$3$ (compactified and Euclidian), then $so(4)$ is the analog of $iso(3,1)$ and they are interested in normalizable functions, these are the spherical harmonics or polynomials depending on coordinates. Then they discuss labelling of these representations using $so(4)\sim su(2)\oplus su(2)$ and proceed to doing some integrals.

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Firstly for the $d+1$-Minkowski space-time the conformal group is $SO(d+1,2)$ (..but the isometry group of the "bulk" $AdS_{d+2}$ is $O(d+1,2)$..) Now whenever the conformal group is $SO(n,2)$ are you saying that the conformal spin-s fields will be just the symmetric traceless rank-s tensors? But if the conformality condition is relaxed and we want just spin-s fields then extra conditions need to be imposed i.e of being transverse and wave equation solutions? (...can you give a reference where this is proven?..) –  user6818 Sep 16 '13 at 18:49
    
Regarding conformal spin-s fields can you then explain the counting that is being done in equation 2.3 and 2.11 of this paper, arxiv.org/abs/1309.0785 - the comment below 2.3 seems to suggest to me that they need symmetric transverse traceless rank-s tensors to get conformal spin-s fields. How come? (..it still confuses me as to why then wave equation condition needs to be imposed when conformality condition is removed!..) –  user6818 Sep 16 '13 at 18:54
    
And can you give a reference for this "decoupling" of the divergence of the "good conformal operator" ($J_{(s)}$) at scaling dimension, $d+s-2$. –  user6818 Sep 16 '13 at 18:56
    
I do not understand the first part of the comment, you shifted the dimension by 1. What is the point in $SO$ vs $O$. Firstly, all considerations are local, so it is Lie algebra that matters. Secondly, you can prohibit reflections. The answer is yes, conformal spin-s field is just symmetric and traceless at generic conformal dimension. You may have a look at arxiv.org/abs/1107.3554 for d-dimensional conformal field theory definitions. If the field is just field, not a conformal one, than you need more conditions: wave equation, transversality, gauge symmetry in massless case. –  John Sep 16 '13 at 23:29
    
We need to know that a traceless, symmetric rank-s tensor in $d=4$ has $(s+1)^2$ components. Arkady deals with $\delta\phi_{a_1...a_s}=\partial_{a_1}\xi_{a_2...a_s}+perm$ that obeys an equation of order $2s$, $\square^s\phi+...$ and there is a Bianchi identity - the equations of motion are transverse (like Maxwell, which is a particular case). Then (2.3)=field-gauge params=(s+1)^2-s^2=2s+1. Counting degrees of freedom is more complicated. The general formula is (8) in arxiv.org/pdf/1210.6821.pdf. It gives 2s(s+1)^2-s^2-(2s+1)s^2=2s(s+1) (we need to take half of it) this is 2.11 –  John Sep 16 '13 at 23:34
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