The double-trace deformation effect in AdS/CFT

Question

Let me use this paper as the reference for this.

I want to understand better the argument at the bottom of page 6.

If the bulk $AdS$ metric is written as $\frac{1}{r^2}(dr^2 + A(r)ds_{boundary}(x)^2)$ then the free massive scalars fields on it are of the form, $\phi = r^{\Delta_{+}}(\alpha(x)) + r^{\Delta_{-}}(\beta(x))$,near $r=0$. (...and the numbers $\Delta_{\pm}$ depend on the dimensionality of the AdS and the mass of the field..)

• As per their notation their "regular" boundary condition in the bulk corresponds to setting $\beta(x)$ (the coefficient of $r^{\Delta_{-}}$) to zero on the boundary and "regular" boundary condition is setting $\alpha(x)$ (the coefficient of $r^{\Delta_{+}}$) to zero.

  How is this compatible with the fact that on using the regular boundary condition the dual CFT necessarily needs to have a term of the form $\int d^d x \beta(x)O(x)$ where $O$ has dimension $\Delta_{+}$ and $\alpha = \langle O \rangle$?

  I thought that they themselves said that in the regular scenario the $\beta$ is being set to $0$ at the boundary - then what is this $\beta(x)$ that is featuring in the boundary action?

• Below equation 3.19 they argue that the when the double-trace deformation of the boundary CFT is switched off it is seeing the "irregular" scenario and when it is turned to infinity it is seeing the regular scenario.

  But how does one argue that the regular scenario is the the IR fixed point and the irregular scenario is the UV fixed point for the boundary CFT? Where is that argument?

• Rethinking the section 4 of this paper in a different way - Suppose one wants to calculate the determinant of the bulk theory $det(-\nabla^2 + m^2)$ by taking product of eigenvalues. One wants to say calculate the determinant when the bulk has been quantized with the $\Delta_{+}$ boundary condition. (...one can presumably ask the same question with $\Delta_{-}$ as well..)

  If the bulk is $AdS_{d+1}$ then one can see that the small-r asymptotics ($r=0$ is the AdS boundary in the Poincare patch) of the harmonics are of the form, $# r^{a} + # r^{b}$ for some values $a$ and $b$ (which depend on the eigenvalue and $d$).

  Now knowing the above small-r asymptotics of the harmonics how does one pick out which of these will contribute to the above determinant with say the $\Delta_{+}$ boundary condition? Can someone schematically sketch how the calculation looks like?

This post imported from StackExchange Physics at 2014-03-07 13:48 (UCT), posted by SE-user user6818

Comments

I agree that the sentence "For general masses, the field must be quantized with the boundary condition $\beta = 0$, the so-called “regular” choice of boundary conditions." is not clear (at least for me). $\beta = \beta (x)$ is a function of $x$, not a function of $r$, so at first glance, it seems a nonsense to speak about a boundary condition for $\beta$. Is it a different boundary ? In one of the cited paper, there is more information (see equation $2.9$ page $6$), but nothing definitive.

This post imported from StackExchange Physics at 2014-03-07 13:48 (UCT), posted by SE-user Trimok

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hmm..thanks :) It would be great if you can help with the other 2 points as well..:)

This post imported from StackExchange Physics at 2014-03-07 13:48 (UCT), posted by SE-user user6818

Answer 1

A possible hint :

Following equations $2.5 \to 2.7$), we may define Kernels - Fourier transform of the $2$-point function - for the limiting cases ($f=0, f = +\infty$) :

$$G_{\pm}(k) \sim \int d^dk ~e^{ik.x} \frac{1}{x^{2 \Delta_\pm}}$$

We have then : $G_{\pm}(k) \sim ~ k^{\pm2\nu}$, where $\nu > 0$

We see, that in the UV, the kernel $G^+$ diverges, so it is not relevant in the UV, but converges in the IR. In the same manner, in the IR, the kernel $G^-$ diverges, so it is not relevant for the IR, but converges in the UV, so it would seem natural to associate the conformal dimension $\Delta^-$ ($f=0$), with the UV and the conformal dimension $\Delta^+$($f=+\infty$) with the IR.

We would have a RG flow which begins with $f=0$ in the UV, to finish at $f=+\infty$ in the IR

Finally, a list of the terms employed in the paper, which are not always clear:

$$\begin {matrix} UV & IR \\ f=0 & f=+\infty\\ "irregular" quantization & "regular" quantization\\ \Delta^- & \Delta^+\\ "irregular" boundary\, value & "regular" boundary \, value\\ \alpha = source & \beta = source\\ \beta = \langle O\rangle & \alpha = \langle O\rangle\\ \gamma = - \Delta^- &\gamma= + \infty\\ \end{matrix}$$

This post imported from StackExchange Physics at 2014-03-07 13:48 (UCT), posted by SE-user Trimok

Comments

Haven't you gotten your table completely messed up? :) I thought what was stated in page 6 was that the IR is the regular scenario when $\beta$ is switched off but is also sourcing the boundary operator $O$ whose vacuum expectation value is $\alpha$? :)

This post imported from StackExchange Physics at 2014-03-07 13:48 (UCT), posted by SE-user user6818

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The argument between 2.4-2.9 is completely opaque to me! There are too many things there that don't make sense to me. (1) For 2.3 and 2.1 to be equal they need to have, $\sqrt{ det (-\frac{I}{f})}\int D\sigma e^{\int \frac{(\sigma + fO )^2 }{2f} } = 1$ How is this true!? One can think of doing the Gaussian integral substituting the field $y = \frac{ i(\sigma + fO )}{\sqrt{f}}$ but that path integral would evaluate to $\frac{i}{\sqrt{f det(I)}}$

This post imported from StackExchange Physics at 2014-03-07 13:48 (UCT), posted by SE-user user6818

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(2)To use the large-N factorization 2.4 into 2.3 don't they need that $<e^{\int [\frac{\sigma^2}{2f } + (\sigma +J)O ] } >_0 = < e^{\int [\frac{\sigma^2}{2f }] }>_0 <e^{\int [(\sigma +J)O] } >_0 $ And why should this be true!?

This post imported from StackExchange Physics at 2014-03-07 13:48 (UCT), posted by SE-user user6818

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(3) And even if I assume everything said before it, I don't see how 2.9 follows from all that!

This post imported from StackExchange Physics at 2014-03-07 13:48 (UCT), posted by SE-user user6818

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a) I made an error with regular/irregular quantization => corrected. But the other elements of the table are correct. Of course, I don't understand the $\beta=0$ switch off.

This post imported from StackExchange Physics at 2014-03-07 13:48 (UCT), posted by SE-user Trimok

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b) The passage from $2.3$ to $2.1$ is simple (up to a constant ): Imagine $\sigma$ as a one-dimensional variable. You have : $\int d\sigma e^{-{\frac{1}{2} (-f^{-1})\sigma^2}+ \sigma O} = \sqrt{\frac{2\pi}{det(-f^{-1})}} e^{\frac{1}{2}(-f)O^2}$. So, now, with basic algebra (and up to a global renormalization constant), you got $2.1$

This post imported from StackExchange Physics at 2014-03-07 13:48 (UCT), posted by SE-user Trimok

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c) Relatively to the operator $O(x)$ (you may see $O(x)$ also as a random variable), $\sigma(x)$ or $J(x)$ appear as "constants" (there are not random variables or operators, there are simple functions ), so yes, your factorization in your comment $(2)$ is correct.

This post imported from StackExchange Physics at 2014-03-07 13:48 (UCT), posted by SE-user Trimok

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d) For your last point $(3)$, you have to use $2.4$, and use the standard 2_point function. The idea is to find the new 2 point-function ($2.9$),$2.10$. If you ask a new PSE question, I will look at it, there is a little work to do, I think

This post imported from StackExchange Physics at 2014-03-07 13:48 (UCT), posted by SE-user Trimok

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Why are you integrating, $\int D\sigma e^{-\frac{1}{2}(-f^{-1})\sigma^2 + \sigma O }$? This form of the $\sigma$ dependence is only a consequence of inserting into the original $2.1$ a factor of the form $\int D\sigma e^{\int \frac{(\sigma + fO )^2 } { 2f} }$ - and this factor can be inserted (with the prefactor adjustment) only if $\sqrt{Det(-\frac{I}{f})}\int D\sigma e^{\int \frac{(\sigma + fO )^2 } { 2f} } =1$ - and this identity is not looking true unless one agrees to arbitrarily change the prefactors.

This post imported from StackExchange Physics at 2014-03-07 13:48 (UCT), posted by SE-user user6818

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For a general function $G(\sigma(x), O(x))$, you have : $\int D\sigma ~e^{\int ~dx ~G(\sigma(x), O(x))} = \int \Pi_x d\sigma(x) ~e^{\int ~dx ~G(\sigma(x), O(x))} = \int \Pi_x d\sigma(x) ~ \Pi_x e^{G(\sigma(x), O(x))} = \Pi_x \int d\sigma(x) e^{G(\sigma(x), O(x))}$

This post imported from StackExchange Physics at 2014-03-07 13:48 (UCT), posted by SE-user Trimok