# Understanding scalar bulk field in $\rm AdS$

I don't really understand the standard result in Holography and AdS/CFT that a source field can be expressed at $$$$\phi_0 = \lim_{z \rightarrow 0} z^{\Delta -d} \phi(z,x)$$$$ because I don't understand some steps in the derivation. $$\Delta$$ is the dimension of the corresponding field theory operator $$O$$. $$d$$ is the spatial dimension of the AdS spacetime. I have a number of beginner questions when deriving the above.

Usually, by e.g. following these notes: https://arxiv.org/abs/1908.02667, we consider the background metric

$$$$ds^2 = \frac{L^2}{z^2} \Big( dz^2 - dt^2 -dx^2 \Big).$$$$

First of all, what does it mean that the AdS boundary is located at $$z \to 0$$? I don't see any intuition behind this at this stage?

From the action for a massive scalar

$$$$S= -\frac{1}{2} \int d^{d+1}x \sqrt{-g} \Big[ g^{MN} \partial_M \phi \partial_N \phi + m^2 \phi^2 \Big].$$$$ By varying this action we get the equation of motion $$$$\frac{1}{\sqrt{-g}} \partial_M \Big( \sqrt{-g}g^{MN} \partial_M \phi \Big) - m^2\phi^2 = 0.$$$$ Now, my second question is what it means to write this explicitly on the geometric background defined by the metric above? The above notes state that this yields $$$$z^{d+1} \partial_z \Big( z^{1-d} \partial_z \phi \Big) + z^2 \delta^{\mu \nu} \partial_\mu \partial_\nu \phi - m^2 L^2 \phi =0.$$$$ Maybe it is obvious but I don't see where this equation comes from or what it means? Would love to see this explicitly and understand it.

• For your first question you can read any standard reference on AdS space, including wiki en.wikipedia.org/wiki/Anti-de_Sitter_space#Global_coordinates
– user172341
Commented Sep 3, 2021 at 12:37
• For the second one, it suffices to write down $g^{zz}$ and so on, and evaluate the equations of motion explicitly.
– user172341
Commented Sep 3, 2021 at 12:38

The boundary of AdS is a conformal boundary, i.e., a boundary that appears after applying a conformal compactification. As you can see from the line element, the $$z = 0$$ point makes the metric diverge, and such a singular point can be removed by applying a conformal transformation of the metric $$g \to \Omega^2 g$$. This is some sort of naive definition of a conformal boundary.
Regarding your second question, he is just replacing the line element of the Poincare patch (your first equation) into the field equations (your third equation) explicitly. Recall that the capital Latin indices are split as $$(z,i)$$, and $$\sqrt{-g} = (L/z)^{d+1}$$.