How is the $\text{AdS}_3$ stress tensor in the Fefferman-Graham expansion derived?

+ 2 like - 0 dislike
343 views

I have been reading "Lectures on black holes and the $\text{AdS}_3$/$\text{CFT}_2$ correspondence" by Per Kraus.

We consider a theory of gravity with Einstein-Hilbert action $$\frac{1}{16\pi G}\int \left(R-\frac{2}{\ell^2}\right)\sqrt{g}\mathrm{d}^3 x + \text{boundary terms}$$

One solution is $\text{AdS}_3$. I am struggling with deriving the $\text{AdS}_3$ stress tensor $$T_{ij} = \frac{1}{8\pi G\ell}\left(g_{ij}^{(2)} - \mathrm{Tr}(g^{(2)})g_{ij}^{(0)}\right) \tag{2.16}$$ where the metric in $\text{AdS}$-space is $\mathrm{d}s^2 = \mathrm{d}\eta^2 + g_{ij}\mathrm{d}x^i\mathrm{d}x^j$ in Gaussian normal coordinates. There's a radial coordinate introduced earlier as $$\mathrm{d}s^2 = \left(1 + \frac{r^2}{\ell^2}\right)\mathrm{d}t + \frac{\mathrm{d}r^2}{1+r^2/\ell^2} + r^2\mathrm{d}\phi^2\tag{2.2}$$ of which I don't know how to handle it or how to relate it to the "Fefferman-Graham expansion" $$g_{ij} = \mathrm{e}^{2\eta/\ell}g_{ij}^{(0)} + g_{ij}^{(2)} + \dots \tag{2.12}$$ where I am also not sure what role exactly the "conformal boundary metric" $g_{ij}^{(0)}$ plays and how to handle these metrics in the computation of $(2.16)$.

I think this confusion also spills to the second section, the Virasoro generators: Why does the stress tensor in terms of $w,\bar w$ defined by $g_{ij}^{(0)}\mathrm{d}x^i\mathrm{d}x^j = \mathrm{d}w\mathrm{d}\bar w$ $$T_{ww} = \frac{1}{8\pi G\ell} g^{(2)}_{ww}\tag{2.19}$$ not contain the conformal boundary metric $g^{(0)}_{ij}$?

This post imported from StackExchange Physics at 2015-11-13 22:28 (UTC), posted by SE-user Rev SS

edited Nov 13, 2015

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysic$\varnothing$OverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.