# What is the radius of convergence of the Fefferman-Graham expansion?

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There is this general result that for any metric $ds^2$ that is asymptotically $AdS_{d+1}$, then there is a coordinate system in which $$ds^2 = \frac{1}{r^2}(dr^2 + g_{ij}(r,x^k)dx^i dx^j)$$ where $g_{ij}(r,x^k)$ admit the following expansion close to the (conformal) boundary of $AdS_{d+1}$: $$g_{ij}(r,x^k) = g^{(0)}_{ij}(x^k) + \frac{1}{r} g^{(1)}_{ij}(x^k) + O(r^{-2})\, ,$$ the boundary being at $r\to\infty$ and $1\leq i,j,k \leq d$.

[The above expansion can also contain some $r^dlog(r)$ when $d$ is even.]

I now assume that the metric satisfies Einstein's equations with some non-trivial stress-energy tensor.

I guess the radius of convergence of the above series depends on the stress-energy tensor.

My question is: how does the convergence radius depend on the stress-energy tensor?

This post imported from StackExchange Physics at 2014-09-09 10:54 (UCT), posted by SE-user Bru
asked Nov 19, 2013

What exactly do you mean by "radius of convergence"? Since the Fefferman Graham expansion is defined in a neighborhood of infinity, the proper distance from the boundary to any spacetime point with finite r must diverge. I assume you mean to ask something like "what is the greatest value of r for which the Fefferman Graham expansion converges?" This is a good question to which I don't know the answer. However, based on the fact what we have here is essentially a fancy Laurent series, my guess is that it is the greatest value of r at which a spacetime singularity exists.

This post imported from StackExchange Physics at 2014-09-09 10:54 (UCT), posted by SE-user Gabriel Herczeg

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