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Asymptotic symmetry algebra

+ 7 like - 0 dislike
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So after a lot of research, and tons and tons of papers that I've went through, I finally have some idea how to solve the equations that will give me candidates for the asymptotic symmetry group for Kerr/CFT correspondence. One has boundary conditions $h_{\mu\nu}$, and metric that behaves like $g_{\mu\nu}=\bar{g}_{\mu\nu}+h_{\mu\nu}$, where $\bar{g}_{\mu\nu}$ is the background metric (in this case it's taken as near horizon extreme Kerr metric), and $h_{\mu\nu}$ are perturbations (given).

So the task is finding diffeomorphisms $\xi$ for which this metric will transform according to

$$\bar{g}_{\mu\nu}+\mathcal{L}_\xi \bar{g}_{\mu\nu}=g_{\mu\nu}=\bar{g}_{\mu\nu}+h_{\mu\nu}$$

And one solves that by solving equations

$$\mathcal{L}_\xi g_{\mu\nu}=\mathcal(r^m)$$

where $\mathcal{O}(r^m)$ are boundary conditions, given in terms of radial coordinates. From this, and the fact that $h_{\mu\nu}=h_{\nu\mu}$ I get ten equations.

Now my issue is this: What to take as ansatz for $\xi$?

In some papers, they take this:

$$\xi^\mu=\xi^\mu_0(t,\theta,\phi)+r\xi^\mu_1(t,\theta,\phi)+\mathcal{O}(r^2)$$

But in certain papers I've found that authors take this form (see after eq. 4.4 here, or here eq 5.1)

$$\xi^\mu=\sum\limits_{n=-1}^\infty \xi^\mu_n r^{-n}$$

such that we have subleading contributions (a falloff in radial coordinate).

Now, I'm inclined to use the second one, since the diffeomorphism given in Guica et. al is given with subleading terms.

I should solve this by putting a certain few terms in, and see what the boundary condition says. For instance we have

$$\mathcal{L}_\xi g_{tt}=\mathcal{O}(r^2)$$

That means that all $\mathcal{O}(r^2)$ contributions will cancel, and only equations with $r$, or smaller $r$ contributions will survive. But the answer is drastically different if I take first or second ansatz.

So what to take? Am I on the right track? If I take second one, what n should I start from? $n=-1$ or $n=-2$?

Oh and I finally got how to solve this by looking at this article and following how they got it in appendix A

EDIT:

I've been thinking. Does the fact that in their article, the metric, and the ansatz goes with powers of $v$, means that since their equations are limited by $\mathcal{O}(v)$ onward all higher power will cancel each other? In that case, since I'm taking only subleading terms in metric, and boundary conditions (different for each of 10 terms), and if I take that my ansatz falls off as $r^{-n}$, that means that all lower order correction cancel out?

The question is still: what n to take? Since it's not the same if my $\xi$ starts with $r^2$, or $r$ :\

EDIT2:

I'll put one component to show what I'm doing. Although I think something is wrong, since I'm getting complicated equations (when things get too complicated, it's kind of a show that things might not be going in the right direction, at least that's what experience showed me)

So, my metric is $g_{\mu\nu}=\bar{g}_{\mu\nu}+h_{\mu\nu}$, as explained above. The first question that comes to my mind: do I put manually $h_{\mu\nu}$ in this expression, or do I use the given boundary conditions? I could say that $h_{\mu\nu}$ is arbitrary, and my metric has to have a power fall off perturbations, so that my $tt$ component would be

$$g_{tt}=-\Omega^2(\theta)(1+r^2(1-\Lambda^2(\theta)))+r^{-1}h_{tt}(t,\theta,\phi)+\mathcal{O}(r^{-2})\quad (\star)$$

I removed $2GJ$ since it's just a factor in front of the metric, not depending on any of the variables ($t,r,\theta,\phi$), and it doesn't contribute to diffeomorphisms.

I will try this after I finish what I did.

However, I just put the given boundary conditions. For $tt$ component I have

$$\mathcal{L}_\xi g_{tt}=\mathcal{O}(r^2)=\xi^\sigma\partial_\sigma g_{tt}+g_{\sigma t}\partial_t\xi^\sigma+g_{t\sigma}\partial_t\xi^\sigma$$

Since $g_{tt}$ depends on $r$ and $\theta$, I have two components in the first argument. In the second and third arguments are the same since $g_{t\phi}=g_{\phi t}$. So I have

$$\xi^\theta\partial_\theta g_{tt}+\xi^r\partial_r g_{tt}+2(g_{tt}\partial_t\xi^t+g_{t\phi}\partial_t \xi^\phi)=\mathcal{O}(r^2)$$

$$(\xi^\theta_{-1}r+\xi^\theta_0+\xi^\theta_1r^{-1}+\mathcal{O}(r^{-2}))(-2\Omega\Omega'+2\Omega(\Lambda\Lambda'\Omega-\Omega'(1-\Lambda^2))r^2+\mathcal{O}(r^2))+(\xi^r_{-1}r+\xi^r_0+\xi^r_1r^{-1}+\mathcal{O}(r^{-2}))(-2\Omega^2(1-\Lambda^2)r+\mathcal{O}(r))+2((-\Omega^2-\Omega^2(1-\Lambda^2)r^2+\mathcal{O}(r^2))(\partial_t\xi^t_{-1}r+\partial_t\xi^t_0+\mathcal{O}(r^{-1}))+(\Lambda^2\Omega^2r+\mathcal{O}(1))(\partial_t\xi^\phi_{-1}r+\partial_t\xi^\phi_0+\partial_t\xi^\phi_1 r+\mathcal{O}(r^{-2})))=\mathcal{O}(r^2)$$

After sorting all out, and saying that all the corrections of $\mathcal{O}(r^2)$ vanish, I end up with

$$(\Lambda\Lambda'\Omega-\Omega'(1-\Lambda^2))\xi^\theta_{-1}-\Omega(1-\Lambda^2)\partial_t\xi^t_{-1}=0$$

But, this one is relatively ok, the other ones sometimes have six or more components ($\xi^\mu$). I'll work all of them out, and then try with $(\star)$ as an assumption and see if things simplify...

EDIT 3:

So things aren't simplifying. But one thing I've learned: I need to solve asymptotic Killing equation.

My next try, the one I think is a fast and relatively good try, is to just see what the leading orders of components should be, such that they satisfy the asymptotic Killing equations, given by boundary conditions. Then I'll try to do explicit calculation.

For instance (and I hope I'm on a right track) for $tt$ component I would have

$$\mathcal{L}_\xi g_{tt}=\mathcal{O}(r^2)$$ $$\xi^\theta\partial_\theta g_{tt}+\xi^r\partial_r g_{tt}+2(g_{tt}\partial_t\xi^t+g_{t\phi}\partial_t \xi^\phi)=\mathcal{O}(r^2)$$ $$\xi^\theta\mathcal{O}(r^2)+\xi^r\mathcal{O}(r)+\mathcal{O}(r^2)\partial_t\xi^t+\mathcal{O}(r)\partial_t\xi^\phi=\mathcal{O}(r^2)$$

So for this to be true (LHS=RHS) I should have

$$\xi^r\to\mathcal{O}(r)$$

$$\xi^\theta\to\mathcal{O}(1)$$

$$\partial_t\xi^t\to\mathcal{O}(1)$$

$$\partial_t\xi^\phi\to\mathcal{O}(r)$$

I hope I'm on a right track.

I got a hint how Guica et. al did it.

First we need to guess what the symmetries are and then generate boundary conditions by acting with those diffeomorphisms, and then check that the charges are finite and integrable.

So it's more a guessing game than exact solving :S

EDIT 4:

So far (by solving asymptotic Killing equations) I only get the $\xi^\theta$ component right. That is I have

$$\xi^\theta_0=\xi^\theta_{-1}r$$

which, when I put in the ansatz I get

$$\xi^\theta=\xi^\theta_{-1}r+\xi^\theta_0+\xi^\theta_1r^{-1}+\mathcal{O}(r^{-2})$$ $$\xi^\theta=\xi^\theta_1 r^{-1}+\mathcal{O}(r^{-2})$$

which is exactly the one in the article by Guica et. al. But no luck with other components.

If only there was some info on what I need to put in asymptotic Killing equation after the equal sign...

EDIT 5:

I'll write the equations I got by putting $\mathcal{L}_\xi g_{\mu\nu}=\mathcal{O}(r^n)$ where $\mathcal{O}(r^n)$ are given by boundary conditions:

$$tt:\quad (\Lambda \Lambda' \Omega+(\Lambda^2-1)\Omega')\xi^\theta_{-1}+(\Lambda^2-1)\Omega \partial_t\xi^t_{-1}=0$$

$$t r:\quad (\Lambda^2-1)r(\xi^t_{-1}+\xi^t_0r^{-1})+(\Lambda^2-1)r^3(\xi^t_{-1}+\xi^t_0r^{-1}+\xi^t_1r^{-2}+\xi^t_2r^{-3})-$$ $$-r(\xi^t_{-1}+\xi^t_0r^{-1})+\Lambda^2\xi^\phi_{-1}+\Lambda^2r^2(\xi^\phi_{-1}+\xi^\phi_0r^{-1}+\xi^\phi_1r^{-2})+\partial_t\xi^r_{-1}=0$$

$$t\theta:\quad (\Lambda^2-1)r^2(\partial_\theta\xi^t_{-1}r+\partial_\theta\xi^t_0+\partial_\theta\xi^t_1r^{-1}+\partial_\theta\xi^t_2r^{-2})-(\partial_\theta\xi^t_{-1}r+\partial_\theta\xi^t_0)+$$ $$+\Lambda^2r(\partial_\theta\xi^\phi_{-1}r+\partial_\theta\xi^\phi_0+\partial_\theta\xi^\phi_2r^{-1})+(\partial_t\xi^\theta_{-1}r+\partial_t\xi^\theta_0)=0$$

$$t\phi:\quad (2\Lambda(\Lambda'\Omega+\Lambda\Omega'))(\xi^\theta_{-1}r+\xi^\theta_0)+\Lambda^2\Omega\xi^r_{-1}+$$ $$+(\Omega(\Lambda^2-1))(\partial_\phi\xi^t_{-1}r+\partial_\phi\xi^t_0+\partial_\phi\xi^t_1r^{-1})-\Omega\partial_\phi\xi^t_{-1}+$$ $$+\Lambda^2\Omega(\partial_\phi\xi^\phi_{-1}r+\partial_\phi\xi^\phi_0)+\Lambda^2\Omega(\partial_t\xi^\phi_{-1}r+\partial_t\xi^t_0)+\Lambda^2\Omega\partial_t\xi^\phi_{-1}=0$$

$$rr:\quad \xi^\theta_{-1}r+\xi^\theta_0=0$$

$$r\theta:\quad \partial_\theta\xi^r_{-1}+(\xi^\theta_{-1}r+\xi^\theta_0)=0$$

$$r\phi:\quad (\xi^t_{-1}r+\xi^t_0)+\xi^\phi_{-1}=0$$

$$\theta\theta:\quad \Omega'(\xi^\theta_{-1}r+\xi^\theta_0)+\Omega(\partial_\theta\xi^\theta_{-1}r+\partial_\theta\xi^\theta_0)=0$$

$$\theta\phi:\quad (\partial_\phi\xi^\theta_{-1}r+\partial_\phi\xi^\theta_0)+\Lambda^2r(\partial_\theta\xi^t_{-1}r+\partial_\theta\xi^t_0+\partial_\theta\xi^t_1r^{-1})+$$ $$+\Lambda^2(\partial_\theta\xi^\phi_{-1}r+\partial_\theta\xi^\phi_0)=0$$

$$\phi\phi:\quad (\Omega\Lambda'+\Lambda\Omega')\xi^\theta_{-1}r+\Lambda\Omega r(\partial_\phi\xi^t_{-1}r+\partial_\phi\xi^t_0)+$$ $$+\Lambda\Omega\partial_\phi\xi^\phi_{-1}r=0$$

The problem is, the only equation I can get some information, about how my vector should behave in components is $rr$, from which I get the general form of $\xi^\theta$ component of diffeomorphism.

From $r\theta$ I get that the $\xi^r_{-1}$ term is a constant that doesn't depend on $\theta$, and $\theta\theta$ equation will just confirm that $rr$ one is correct.

I guess from the fact that $\xi^r_{-1}$ is not zero I could say that the $r$ component will start at that term (like in the article).

EDIT 6:

I have added the bounty to raise awareness to this, if anyone knows if what I did so far is correct, and what to do next, please do tell :)

This post imported from StackExchange Physics at 2014-03-07 13:48 (UCT), posted by SE-user dingo_d
asked Oct 24, 2013 in Theoretical Physics by dingo_d (110 points) [ no revision ]
Is this [research-level]?

This post imported from StackExchange Physics at 2014-03-07 13:48 (UCT), posted by SE-user jinawee
Well I'm writing a thesis on Kerr/CFT correspondence, and this is related to it so I guess :\

This post imported from StackExchange Physics at 2014-03-07 13:48 (UCT), posted by SE-user dingo_d
Any help on ho to proceed next?

This post imported from StackExchange Physics at 2014-03-07 13:48 (UCT), posted by SE-user dingo_d
I think I know what I've been doing wrong. When I solve this, I'll post the result in the answer. I deleted the answer, I'll post it soon.

This post imported from StackExchange Physics at 2014-03-07 13:48 (UCT), posted by SE-user dingo_d

1 Answer

+ 1 like - 0 dislike

So the way these equations are solved is by looking what order we have at the equations, this means that when we put the ansatz inside the equations, we keep all the orders that are greater than the ones equating the equation.

Then we collect all terms with the powers of $r$ (for $tt$ equation all the powers greater than $r^2$ are that of $r^3$, for $tr$, all powers greater than $r^0$ are $r^4,\ r^3,\ r^2$ and $r$, etc.) and set them equal to zero respectively. This in turn gives us 25 equations in total. From those I got only that:

$$\xi^t_{-1}=0,\ \xi^t_0=0,\xi^t_1=C(t,\phi)$$ $$\xi^r_{-1}=C_1(t,\phi)$$ $$\xi^\theta_{-1}=0,\ \xi^\theta_0=0$$ $$\xi^\phi_{-1}=0,\ \xi^\phi_0=C_2(t,\phi)$$

From additional equations I only got that $\xi^\phi_1=-\partial_t\xi^r_{-1}$, and $\xi^t_2=\partial_t\xi^r_{-1}$, that is $\xi^\phi_1+\xi^t_2=0$.

Now in principle that gives me a relatively good solution, it's just that in original article it's not specified what those constants are. Not only that, but the $\xi^r_{-1}$ constant shouldn't depend on $t$.

That way $\xi^\phi_1=0$, and $\xi^t_2=0$, and that is ok by the article. But I have no way of proving that, so it would be kinda presumptuous of me to just assume that.

So this is only a partial answer. If anyone got anything different do tell.

Maybe I should include even higher powers of $r$? Like $r^2$ in my ansatz. That would give me more components, but would also make this work even more cumbersome.

EDIT:

I used Mathematca to simplify things, and I get the correct answer finally. The only thing that is kinda unresolved is why did the authors assumed that none of the components will depend on $t$. My guess is because $U(1)$ is enhanced to Virasoro, and $U(1)$ Killing vector doesn't have $t$ dependence.

But all in all I finally have the correct answer.

This post imported from StackExchange Physics at 2014-03-07 13:48 (UCT), posted by SE-user dingo_d
answered Nov 14, 2013 by dingo_d (110 points) [ no revision ]

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