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The isometry group of the anti-de Sitter spacetime is $SO(d-1,2)$, which has a total of $\frac{1}{2}d(d+1)$ isometries.

For the three-dimensional anti-de Sitter spacetime, these are $6$ isometries.

Would the number of isometries change for an AdS$_3$ cylinder (in global coordinates) that is radially cut-off at a finite radius?

nightmarish
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    Yes, since it is not maximally symmetric any more (you suddenly have some "border"). – ungerade Dec 11 '17 at 20:37
  • Could you elaborate? How does having a 'border' reduce the number of symmetries? – nightmarish Dec 11 '17 at 20:39
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    Maybe an example is easier: Suppose you take $R^2$. You can translate it in $x$ any $y$ direction and stay in $R^2$. As soon as you cut $R^2$ to some finite plane you can not shift anymore into these direction and stays in the plane You shift parts of you manifold into ranges that are not part of your manifold anymore. Also you have broken the translation symmetry due to the "cut-off". – ungerade Dec 11 '17 at 21:40
  • So, I suppose a border breaks all the translational, rotational and boost invariances, then? – nightmarish Dec 11 '17 at 21:53
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    I guess that depends on the spacetime under consideration. I did not provide an real answer because I do not want to check which specifically. But it is certainly not maximally symmetric anymore and therefore "looses" some symmetries. – ungerade Dec 11 '17 at 21:56
  • I suppose that, from equation (9.8) of http://www.hartmanhep.net/topics2015/9-ads3symmetries.pdf, we see that only the symmetries due to $\zeta_{0}$ and $\bar{\zeta}_{0}$ survive after the radial cut-off, but I'm not sure myself. – nightmarish Dec 11 '17 at 22:00

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