# Boundary conditions for fields in Kerr/CFT

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

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.
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