AdS/Awesome

The title is taken from a Sean Carroll tweet

We define a space with p spatial directions and q time directions. Think of a space with signature (p,q+1) $ds^2 = \sum_{i=1}^p dx_i^2 - \sum_{j=1}^{q+1} dy_j^2$ and then take a pseudosphere inside given by $\sum_{i=1}^p x_i^2 - \sum{j=1}^{q+1} y_j^2 = -L^2$. Changing this minus sign will change it to (p-1,q+1) deSitter space instead. (The hyperboloid of one sheet vs two sheets.)

If you choose q=0, this gives a model for hyperbolic p space. To make it more reasonable to handle, we might as well do a Weyl transformation to bring it to a ball model or an upper half space model, but nonetheless it is still a pseudosphere in Minkowski p+1 space.

The restriction to the pseudosphere gets rid of one of the timelike directions and makes the metric of signature (p,q). So in addition to the hyperbolic geometry of (p,0) we like (p,1) as well. We like time, we could use more time directions.

Do some conformal transformations and get the metric to a form as follows

$ds^2 = \frac{1}{y^2} (-dt^2+dy^2+\sum dx_i^2)$ where y is a radial coordinate and y=0 corresponds to the conformal boundary that you never can quite get to.

We can see that if we take constant time slices we get hyperbolic spaces in the upper half space coordinate system. We also see that as you go to the conformal boundary it looks like Minkowski p space. So if we use AdS (4,1) we get our familiar spacetime at the boundary. We could put a quantum field theory on this boundary like $\mathcal{N} =4$ Super Yang-Mills or any other theory we like. Alternatively we could put a quantum gravity theory inside the AdS. It'll mess up the AdS geometry a little but not too much on the boundary. After all the conformal boundary is really far away.

This sets up a correspondence between relevant operators in the QFT on the left and the quantum gravity theory on the right. The right hand side is not the full partition function it is the partition function where we only count fields that match the boundary condition at the boundary. Of course to take care of the fact that this is a conformal boundary not a real boundary, it is regulated by cutting of at $y=\epsilon$ and extrapolating to $y=0$

In the QFT picture the y coordinate is the renormalization scale. That is why CFT's are nicer in this correspondence. Renormalization is easy if you are at a CFT.