Hilbert space of quantum gravity: bulk $\otimes$ horizon

Question

I was reading a paper dealing with the Hilbert space of quantum gravity (or more precisely what should it look like considering what we know from QM and GR) ref: http://arxiv.org/abs/1205.2675 and the author writes the following: $${\cal{H}_M} = {\cal{H_{M,\,\textrm{bulk}}}}\otimes{\cal{H_{M,\,\textrm{horizon}}}}$$ for a specific manifold $\cal{M}$. I know very little about the holographic principle and the AdS-CFT correspondence but isn't it a redundant description? If there is a duality between the gravitational theory in the bulk and the CFT on the boundary, knowing one means knowing the other, so why can't we restrict ourselves to one of the Hilbert spaces? Moreover, the author writes, a couple of lines after this first element, that the two Hilbert spaces have the same dimension ( $\textrm{exp}({\frac{{\cal{Area}}}{4}})$ ) so they are totally equivalent, as a complex Hilbert space is only defined by its dimension.

Answer 1

Yes it is redundant. This is exactly what AdS/CFT is not. The degrees of freedom of the bulk are the degrees of freedom of the horizon. This is also why condensed matter analogs are rare--- the most common idea of identifying AdS/CFT boundary theories with condensed matter boundary theories is wrong, because in tranditional condensed matter systems, the boundary degrees of freedom are in addition to the bulk degrees of freedom, they are not dual to these degrees of freedom, as in AdS/CFT. The exception, where the condensed matter analog is right, is where the bulk theory is topological, like the Chern-Simons theory for the quantum hall fluid, where you can consider edge-states as describing interior physics. There might be more analogs of this sort. One has to be careful, because a lot of people have this wrong picture of AdS/CFT in the head, that it's boundary stuff in addition to bulk stuff.