The AdS/Super-CFT Randall–Sundrum Brane-World Duality

By George Shiber Friday, March 11, 2016Friday, October 26, 2018 In Cosmology, Grand Unification Physics, M Theory, Mathematical Physics, Mathematics, Philosophy Of Physics, Philosophy Of Science, Physics, Super String Theory, Supersymmetry AdS/CFT, Brane-World

We are not to tell nature what she’s gotta be. She’s always got better imagination than we have ~ Richard Feynman

In this post, I will relate the AdS/CFT correspondence to the Randall-Sundrum brane-world ‘model’. The AdS/CFT correspondence ‘says’ that for every conformal field theory, there is a corresponding theory of gravity with one more dimension. More precisely, it is a correspondence between the gravitational dynamics of a $d + 1$ –dimensional anti-de Sitter spacetime $Ad{S_{d + 1}}$ , and a $d$ –dimensional conformal field theory $CF{T_d}$ , so the duality basically says that if we deform $CF{T_d}$ by source fields by adding

$\int {{d^d}} x\,J(x)\vartheta (x)$

this will be the dual to an $Ad{S_{d + 1}}$ theory with a bulk field $J$ with boundary condition:

$\underbrace {\lim }_{boundary}J\,{w^{\Delta - d + k}} = {J_{CFT}}$

with $\Delta$ the conformal dimension of the local operator $\vartheta$ and $k$ the number of covariant indices of $\vartheta$ minus the number of contravariant indices. Hence, we get a dual source field for every gauge-invariant local operator and can deduce the duality as

${\left\langle {T\left\{ {\exp \left( {\int {{d^d}x\,{J_{4D}}(x)\vartheta (x)} } \right)} \right\}} \right\rangle _{CFT}} = {\not Z_{AdS}}\left[ {\underbrace {\lim }_{boundary}J\,{w^{\Delta - d + k}} = {J_{4D}}} \right]$

or more informatively

${\left\langle {{e^{\int {{d^4}x\,{\phi _{(0)}}(x)\vartheta (x)} }}} \right\rangle _{CFT}} = {\not Z_{{\rm{String}}}}\left\langle {{r^{\Delta - d}}\phi {{(x,r)}_{\left| {_{r = 0}} \right.}} = {\phi _{(0)}}(x)} \right\rangle$

where $\phi (x,r)$ is the ‘bulk-field‘, $r$ the radial coordinate that is dual to the renormalization group in the boundary theory, with

${\phi _{(0)}}(x) \equiv {r^{\Delta - d}}\phi {(x,r)_{\left| {_{r = 0}} \right.}}$

and $r = 0$ in the CFT boundary of AdS with ${\phi _{(0)}}(x)$ coupled to $\vartheta (x)$

The left-hand-sides are the vacuum expectation value of the time-ordered exponential of the operators over CFT; the right-hand-sides are the quantum gravity generating functional with the given conformal boundary condition. So, on one side, we have a gauge theory in flat space-time at weak coupling and as the coupling increases, the theory must be described as a string-theory in curved space-time. So generally for $d = n$ , the Maldacena conjecture, based on the decoupling limit of D $n - 1$ -branes in type IIB string theory compactified on ${S^{n + 1}}$ , relates the dynamics of $Ad{S_{n + 1}}$ to an $N = n$ superconformal $U(N)$ Yang–Mills theory on its $n$ -dimensional boundary. Before I continue, keep this 5-D brane-world action in mind throughout:

$\begin{array}{c}S = \int {{d^5}} x\sqrt { - {g_{\left( 5 \right)}}} \left[ {{M^3}{R_{\left( 5 \right)}} - \Lambda } \right]\\ + \int {{d^4}x} \sqrt { - {g_{\left( 4 \right)}}} \widetilde {\not L}_{{\rm{BRANE}}}^{{\rm{Lag}}}\end{array}$

Now, different compactifications lead to other SuperCFTs on the boundary. For $d = 4$ , picking the Poincaré coordinates on $Ad{S_5}$ yields the metric

$d{s^2} = {e^{ - 2y/L}}{\left( {d{x^\mu }} \right)^2} + d{y^2}$

where in that scenario, the superconformal Yang–Mills theory would reside at the boundary $y \to - \infty$ . The Randall–Sundrum models a possible scenario for evading Kaluza–Klein compactification by localizing gravity in the presence of an uncompactified extra dimension via an insertion of a positive tension $3$ -brane, quasi-morphic to Einsteinian space-time, into $Ad{S_5}$ . The resulting Randall–Sundrum metric is then

$d{s^2} = {e^{ - 2\left| y \right|/L}}{\left( {d{x^\mu }} \right)^2} + d{y^2}$

with $y \in \left( { - \infty ,\infty } \right)$ or $y \in \left[ {0,\infty } \right)$ for a 2-sided or 1-sided Randall–Sundrum brane, respectively. So to make a Maldacena/Randall–Sundrum correspondence, let me work with the one-sided Randall–Sundrum brane, and imposing a boundary in $Ad{S_5}$ at $y = 0$ , the RS-model is hypothesized to be dual to a cut-off $CFT$ coupled to gravity, with $y = 0$ , the location of the Randall–Sundrum brane, providing the ultraviolet cut-off. Hence, this ‘super’-Maldacena conjecture reduces to the standard AdS/CFT-duality as the boundary is pushed off to $y \to - \infty$ , thus the cut-off is removed and gravity gets completely decoupled. Given that one-loop corrections to the graviton propagator induce $1/{r^3}$ corrections to the gravitational potential

$V(r) = \frac{{G{m_1}{m_2}}}{r}\left( {1 + \frac{{\alpha G}}{{{r^2}}}} \right)$

where $G$ is the four-dimensional Newton constant, one can see that the contribution of a single $CFT$ , with $\left( {{N_1},{N_{1/2}},{N_0}} \right) = \left( {{N^2},4{N^2},6{N^2}} \right)$ , is (1)

$V(r) = \frac{{G{m_1}{m_2}}}{r}\left( {1 + \frac{{2{N^2}G}}{{3\pi {r^2}}}} \right)$

with ${N_s}$ being the numbers of particle-species of spin $s$ going around the loop. Hence, via the AdS/CFT relation ${N^2} = \pi {L^3}/2{G_5}$ and the 1-sided brane-world relation $G = 2{G_5}/L$ , with ${G_5}$ being the five-dimensional Newtonian constant and $L$ the radius of $Ad{S_5}$ , it reduces to (2)

$V(r) = \frac{{G{m_1}{m_2}}}{r}\left( {1 + \frac{{2{L^2}}}{{3{r^2}}}} \right)$

Let’s deduce (1) by lowest-order quantum corrections to solutions of Einstein’s equations. Working with linearized gravity, we begin by writing the metric as

${g_{\mu \nu }} = {\eta _{\mu \nu }} + {h_{\mu \nu }}$

hence,

$\sqrt { - g} {g^{\mu \nu }} \equiv {\widetilde g^{\mu \nu }} = {\eta ^{\mu \nu }} - {\widetilde h^{\mu \nu }} + ...$

with

${\widetilde h_{\mu \nu }} = {h_{\mu \nu }} - \frac{1}{2}{\eta _{\mu \nu }}{h^\alpha }_\alpha$

Given the harmonic gauge(s)

$\left\{ {\begin{array}{*{20}{c}}{{{\not \partial }_\mu }{{\widetilde g}^{\mu \nu }} = 0}\\{{{\not \partial }_\mu }{{\widetilde h}^{\mu \nu }} = 0}\end{array}} \right.$

the linearized Einstein equation is then

${\overline {\not \Theta } _{dA}}\widetilde h_{\mu \nu }^c(x) = - 16\pi G{T_{\mu \nu }}(x)$

A Fourier transformation to momentum space yields

$\widetilde h_{\mu \nu }^c\left( p \right) = - 16\pi G{\Delta _4}\left( p \right){T_{\mu \nu }}\left( p \right)$

with

${\Delta _4}\left( p \right) = - 1/{p^2}$

being the four-dimensional massless scalar propagator. Thus, the quantum-corrected metric becomes

${\widetilde h_{\mu \nu }} = \widetilde h_{\mu \nu }^c + \widetilde h_{\mu \nu }^q$

and the quantum correction $\widetilde h_q^{\mu \nu }$ is given in momentum space by

$\begin{array}{c}\widetilde h_q^{\mu \nu }\left( p \right) = {{\not D}^{\mu \nu \alpha \beta }}\left( p \right) \cdot \\\prod {_{\mu \nu \gamma \delta }} \left( p \right)\widetilde h_c^{\gamma \delta }\left( p \right)\end{array}$

where ${\not D^{\mu \nu \alpha \beta }}$ is the graviton propagator

${\not D^{\mu \nu \alpha \beta }}\left( p \right) = \frac{1}{2}{\Delta _4}\left( p \right)\left( {{\eta ^{\mu \alpha }}{\eta ^{\mu \beta }} + {\eta ^{\mu \beta }}{\eta ^{\nu \alpha }} - {\eta ^{\mu \nu }}{\eta ^{\alpha \beta }} + ...} \right)$

and $\prod {_{\alpha \beta \gamma \delta }}$ is the one-loop graviton self-energy, which by gauge-symmetry and Lorentz invariance necessarily is

$\begin{array}{l}\prod {_{\alpha \beta \gamma \delta }} \left( p \right) = {p^4}\left[ {\prod {_1\left( {{p^2}} \right)} } \right.{\eta _{\alpha \beta }}{\eta _{\gamma \delta }} + \\\prod {_2} \left( {{p^2}} \right)\left( {{\eta _{\alpha \gamma }}{\eta _{\beta \delta }} + {\eta _{\alpha \delta }}{\eta _{\beta \gamma }}} \right) + \\\prod {_3} \left( p \right)\left( {{\eta _{\alpha \beta }}{{\widehat p}_\gamma }{{\widehat p}_\delta } + {\eta _{\gamma \delta }}{{\widehat p}_\alpha }{{\widehat p}_\beta }} \right) + \\\prod {_4} \left( {{p^2}} \right)\left( {{\eta _{\alpha \gamma }}{{\widehat p}_\beta }{{\widehat p}_\delta } + {\eta _{\alpha \delta }}{{\widehat p}_\beta }{{\widehat p}_\gamma } + {\eta _{\beta \gamma }}{{\widehat p}_\alpha }{{\widehat p}_\delta } + {\eta _{\beta \delta }}{{\widehat p}_\alpha }{{\widehat p}_\gamma }} \right) + \\\prod {_5} \left( {{p^2}} \right){\widehat p_\alpha }{\widehat p_\beta }{\widehat p_\gamma }\left. {{{\widehat p}_\delta }} \right]\end{array}$

hence, by combining, we get the quantum-corrected metric

$\begin{array}{c}h_{\mu \mu }^q\left( p \right) = - {p^2}\left[ {2\prod {_2\left( p \right)} } \right.\delta _\mu ^\alpha \delta _\nu ^\beta \\ + \prod {_1} \left( p \right){\eta _{\mu \nu }}{\eta ^{\alpha \beta }} + \\\left( {\prod {_3\left( p \right) + ...} } \right){\widehat p_\mu }{\widehat p_\nu }\left. {{\eta ^{\alpha \beta }}} \right]\widetilde h_{\alpha \beta }^c\end{array}$

which, at the linearized level, becomes

$\begin{array}{c}{h_{\mu \nu }}\left( p \right) = - 16\pi G{\Delta _4}\left( p \right) \cdot \\\left[ {{T_{\mu \nu }}\left( p \right) - \frac{1}{2}{\eta _{\mu \nu }}{T^\alpha }_\alpha \left( p \right)} \right] \cdot \\ - 16\pi G\left[ {2\prod {_2} } \right.\left( p \right){T_{\mu \nu }}\left( p \right) + \\\prod {_1} \left( p \right){\eta _{\mu \nu }}{T^\alpha }_\alpha \left. {\left( p \right)} \right]\end{array}$

Now note, after cancelling the infinities with the appropriate counterterms, the finite remainder must necessarily have the form

$\prod {_i} \left( p \right) = 32\pi G\left( {{a_i}{\rm{In}}\frac{{{p^2}}}{{{\mu ^2}}} + {b_i}} \right)$

where ${a_i}$ and ${b_i}$ $\left( {i = 1,2,3,4,5} \right)$ , are numerical coefficients and $\mu$ is an arbitrary subtraction constant having the dimensions of mass. To relate to the Newtonian potential, one Fourier-transforms back to coordinate space. For a point source, ${T_{00}}\left( x \right) = m{\delta ^3}\left( r \right)$ , we obtain to this order the following relations

$\left\{ {\begin{array}{*{20}{c}}{{g_{00}} = - \left( {1 - \frac{{2Gm}}{r} - \frac{{2\alpha {G^2}m}}{{{r^3}}}} \right)}\\{{g_{ij}} = \left( {1 + \frac{{2Gm}}{r} + \frac{{2\beta {G^2}m}}{{{r^3}}}} \right){\delta _{ij}}}\end{array}} \right.$

and since all spins contribute with the same sign as they must by general positivity arguments on the self-energy in (1) above, we get

$\begin{array}{c}\alpha = 2\beta = \frac{1}{{45\pi }}\left( {12{N_1} + 3{N_{1/2}} + {N_0}} \right)\\ = \frac{{2{N^2}}}{{3\pi }}\end{array}$

also notice that $\alpha$ determines the crucial part of the Weyl anomaly that involves the square of the Weyl tensor

${g_{\mu \nu }}\left\langle {{T^{\mu \nu }}} \right\rangle = b\left( {F + \frac{2}{3}{{\overline {\not \Theta } }_{dA}}R} \right) + b'G$

where

$\begin{array}{c}F = {C_{\mu \nu \rho \sigma }}{C^{\mu \nu \rho \sigma }} = {R_{\mu \nu \rho \sigma }}{R^{\mu \nu \rho \sigma }} - \\2{R_{\mu \nu }}{R^{\mu \nu }} + {R^2}\end{array}$

and

$\begin{array}{c}G = * {R_{\mu \nu \rho \sigma }} * {R^{\mu \nu \rho \sigma }} = {R_{\mu \nu \rho \sigma }}{R^{\mu \nu \rho \sigma }}\\ - 4{R_{\mu \nu }}{R^{\mu \nu }} + {R^2}\end{array}$

with the constants

$b = \frac{1}{{120{{\left( {4\pi } \right)}^2}}}\left[ {12{N_1} + 3{N_{1/2}} + {N_0}} \right]$

and

$b' = - \frac{1}{{720{{\left( {4\pi } \right)}^2}}}\left[ {124{N_1} + 11{N_{1/2}} + 2{N_0}} \right]$

So for the central charge, one obtains

$c = \pi {L^3}/8{G_5}$

hence, we have

$G\alpha = \frac{{G{L^3}}}{{3{G_5}}} = \frac{{2{L^2}}}{3}$

with the second equality making explicit use of the brane-world relation $G = 2{G_5}/L$ ,

making it universal and independent of which particular CFT appears in the AdS/CFT correspondence, which is precisely as desired since the Randall–Sundrum coefficient does not depend on the details of the fields propagating on the brane!

Now, on the brane-world, where the five-dimensional action has the form

with $M$ the five-dimensional Planck mass, and ${M^3} = 1/\left( {16\pi {G_5}} \right)$ , and $\Lambda$ is the cosmological constant in the bulk. Small fluctuations of the metric on the brane may be represented as

$\begin{array}{c}d{s^2} = {e^{ - 2\left| y \right|/L}}\left[ {{\eta _{\mu \nu }} + {h_{\mu \nu }}\left( {x,y} \right)} \right] \cdot \\d{x^\mu }d{x^\nu } + d{y^2}\end{array}$

with $L$ being the radius of AdS, defined by

$R_{MNPQ}^{\left( 5 \right)} = - \frac{1}{{{L^2}}}\left( {g_{MP}^{\left( 5 \right)}g_{NQ}^{\left( 5 \right)} - g_{MQ}^{\left( 5 \right)}g_{NP}^{\left( 5 \right)}} \right)$

and we have the crucial relation $\Lambda = - 12{M^3}/{L^2}$ , which yields the CFT-bane relation

$\begin{array}{l}{h_{\mu \nu }}\left( p \right) = - \frac{2}{{L{M^3}}}{\Delta _3}\left[ {{T_{\mu \nu }}\left( p \right) - \frac{1}{2}{\eta _{\mu \nu }}{T^\alpha }_\alpha \left( p \right)} \right]\\ - \frac{1}{{{M^3}}}{\Delta _{KK}}\left( p \right)\left[ {{T_{\mu \nu }}\left( p \right) - \frac{1}{3}{\eta _{\mu \nu }}{T^\alpha }_\alpha \left( p \right)} \right]\end{array}$

the above relation has physical meaning: ${\Delta _4}\left( p \right)$ , the four-dimensional massless propagator, corresponds exactly to the zero-mode graviton localized on the brane, while

${\Delta _{KK}}\left( p \right) = - \frac{1}{p}\frac{{{K_0}\left( {pL} \right)}}{{{K_1}\left( {pL} \right)}}$

is the propagator for the continuum Kaluza–Klein graviton modes, hence getting a relation between four-and-five-dimensional Newton’s constants. So, at large distances, corresponding to $pL \ll 1$ , a small-argument expansion for Bessel functions gives us

$\begin{array}{c}{\Delta _{KK}}\left( p \right) = \frac{L}{2}\left( {{\rm{In}}\frac{{{p^2}{L^2}}}{4} + 2y} \right) + \\\vartheta {\left( p \right)^2}\end{array}$

Thus, by evaluating the Fourier transform for $r \gg L$ yields the linearized metric

$\left\{ {\begin{array}{*{20}{c}}{{h_{00}} = \frac{{2Gm}}{r}\left( {1 + \frac{{2{L^2}}}{{3{r^2}}} + ...} \right)}\\{{h_{ij}} = \frac{{2Gm}}{r}\left( {1 + \frac{{{L^2}}}{{3{r^2}}} + ...} \right){\delta _{ij}}}\end{array}} \right.$

giving us a physical meaning to the Newtonian potential: therefore we have shown that the $1/{r^3}$ –corrections to Newton’s law are identical between the Maldacena ‘model’ and Randall–Sundrum ‘model’ which entails a deep classical/quantum duality expressed by

$\prod {_2} \left( p \right) + \vartheta \left( {{G^2}} \right) = \frac{L}{4}{\Delta _{KK}}\left( p \right)$

which in turn highly suggests that the ‘AdS/Super-CFT Randall–Sundrum brane-world duality’ is true and ought therefore to be undergrid by a mathematical proof: for the next post!