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I am reading a paper by Guica et al. on Kerr/CFT correspondence (arXiv:0809.4266) and I'm not sure if I got this. They choose the boundary conditions, like a deviation of the full metric from the background Near-Horizon Extremal Kerr (NHEK) metric. Let's say that we can write that deviation like

$$\delta_\xi g_{\mu \nu}=\mathcal{L}_\xi g_{\mu\nu}=\nabla_\mu\xi_\nu+\nabla_\nu\xi_\mu$$

And the most general diffeomorphism which preserve the boundary conditions given in the text is:

$$\xi=[-r\epsilon'(\varphi)+\mathcal{O}(1)]\partial_r+\left[C+\mathcal{O}\left(\frac{1}{r^3}\right)\right]\partial_\tau+\left[\epsilon(\varphi)+\mathcal{O}\left(\frac{1}{r^2}\right)\right]\partial_\varphi+\mathcal{O}\left(\frac{1}{r}\right)\partial_\theta$$

What my mentor told me, while briefly explaining this, is that we basically need to find the most general $\xi$ such that $\mathcal{L}_\xi g_{\mu\nu}$ is within the class of the boundary conditions.

But how do I find these $\xi$?

How can you find these boundary conditions and diffeomorphisms? Or better jet, how do I find diffeomorphism using those boundary conditions? :\

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I found this paper: arxiv.org/abs/0908.0184v3, and there are lots of boundary conditions there. But I am interested into how he calculated the 4.10 formula? I have the metric (therefore all the coefficients are known), he even gives the generators of the asymptotic symmetry in equation 4.5, but I cannot reproduce the result he got :\ I used the formula $(\mathcal{L}_\xi g_{\mu\nu})^\sigma=g_{\mu\nu,\sigma}\xi^\sigma+g_{\sigma\nu}\xi^\sigma_{,\mu}+g_‌​{\mu\sigma}\xi^\sigma_{,\nu}$, and I tried going by componenets but I don't get the same result (so I must be doing something wrong :) –  dingo_d Dec 3 '12 at 11:24

1 Answer 1

Your boundary condition looks a whole lot like Killing's equation to me. Most working people I know usually use Killing's equation as a check rather than as something to solve. Note that the physicality of the equation is that $\xi^{\mu}$ is the generator of a symmetry in $g_{ab}$, so I would want to look for something that satisfies the case of being the generator of a symmetry. In the case of the Kerr metric, you know that it is axisymmetric and time-translation invariant, so you can see right away, and easily check that $\partial_{t}$ and $\partial_{\phi}$ are both candidates for $\xi^{\mu}$. I'd need to know more about what boundary condition you're looking for (usually, I think of boundary conditions as [set of fields ] = fixed value, specified on some surface, and there is no equation above.) to say much more than this, though.

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Well from what I've read, if the Lie derivative of metric tensor along some vector field is zero, then that field represents a generator of an isometry group. The boundary condition is given as a matrix of some kind of subleading terms of the deviation of the metric from the background metric. And there are two boundaries (r=$\infty$ and r=$-\infty$)... I found one paper that has some more computation, but I'm not sure how they got the $\mathcal{L}_\xi g_{\mu\nu}$ :\ –  dingo_d Dec 2 '12 at 20:24

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