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I was going through Witten's paper on AdS and holography , and am confused in section 2.4. He starts by considering a massless scalar action in Euclidean AdS spacetime, with a boundary value $\phi_0$. He then looks for a "Green function" $K$, but says that it satisfies the following condition:

$$L K = 0,$$

where $L$ is the Laplacian for the scalar field. But in general say when we have a differential equation $$L \phi = f,$$

in order to solve it we construct a Green function $G$ for the differential operator $L$, it has to satisfy

$$L_x G_{xy} = \delta(x-y)$$

with proper boundary conditions and the solutions to $L$ are given by $\phi$ such that $\phi = \int_M G f $. The solution and not the Green function satisfies

$$L \phi =0.$$

Is there a deeper reason behind calling it a Green function? Does this have any relation to the fact that he is working in Euclidean AdS and not Lorentzian AdS?

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  • $\begingroup$ Minor comment to the post (v1): In the future please link to abstract pages rather than pdf files. $\endgroup$
    – Qmechanic
    Mar 1, 2020 at 11:39

1 Answer 1

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Here $x'$ denotes a point on the boundary, and $x$ denotes a point in the bulk. The problem he is looking at is given some boundary value $\phi_0(x')$ of the scalar field $\phi(x)$, how do you solve the wave equation.

$$L \phi(x) = \phi_0(x')$$

In order to do that he is solving

$$L_x K_{xx'} = \delta(x-x').$$

One way to solve this is to just consider the solutions to the equation $L K = 0$, and see which component of this equation blows up at the boundary point $x'$, which is here given by $x' = \infty$, and show that this blowing up corresponds to a delta function. Here the blowing up is given by

$$K(x_0) = C x_0^d.$$

In equation (2.18) he shows that this blowing up corresponds to a delta function, by implementing an isometry transformation

$$x^i \to \frac{x^i}{x_0^2 + \sum_{i=1}^d x_i^2}$$

which maps $x_0'=\infty$ to the origin. He then shows that the Green function

$$K(x) = C \left( \frac{x^i}{x_0^2 + \sum_{i=1}^d x_i^2} \right)^d$$

satisfies all properties of a delta function

$$L_x K(x-x') = \delta(x-x')$$ in the limit $x_0\to 0, x_i =0$. Therefore the solution of the equation of motion is given by

$$\phi(x_0,x_i) = \int_M \phi_0(x') K(x-x')$$

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