Does asymptotically AdS mean as $z \to 0$ or as $z \to \infty$ in Poincare metric of AdS?
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Hello mate here is your answer.
Now if I want to show that some other space time is asymptotically AdS, I should expect the metric to agree in the limit ρ→−∞ρ→−∞ which is the limit z→0z→0. Is that right?
Yes, this is correct. In this limit the leadin order of the metric components should agree with the pure AdS metric. There is however a more mathematically sound way of approaching the problem. One basically has to require that the asymptotic symmetry group is still the conformal group O(2,d−1)O(2,d−1), which is the isometry group of AdSdd. The procedure is laid out in this paper by Henneaux and Teitelbom, where the focus is on AdS44.
(Nb: I am trying to show that BTZ metric is asymptotically AdS).
Hope it helps you.
Now if I want to show that some other space time is asymptotically AdS, I should expect the metric to agree in the limit ρ→−∞ρ→−∞ which is the limit z→0z→0. Is that right?
Yes, this is correct. In this limit the leadin order of the metric components should agree with the pure AdS metric. There is however a more mathematically sound way of approaching the problem. One basically has to require that the asymptotic symmetry group is still the conformal group O(2,d−1)O(2,d−1), which is the isometry group of AdSdd. The procedure is laid out in this paper by Henneaux and Teitelbom, where the focus is on AdS44.
(Nb: I am trying to show that BTZ metric is asymptotically AdS).
Hope it helps you.
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Does asymptotically AdS mean as $z \to 0$ or as $z \to \infty$ in Poincare metric of AdS?
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