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I am going through Kerr/CFT correspondence paper again, and I am at the section where authors specify Killing vectors for near horizon extreme Kerr metric (shortly NHEK).

The metric is

$$d\bar{s}^2=2GJ\Omega^2\left(-(1+r^2)d\tau^2+\frac{dr^2}{1+r^2}+d\theta^2+\Lambda^2(d\varphi+rd\tau)^2\right)$$

Where

$$\Omega^2\equiv\frac{1+\cos^2\theta}{2},\quad\Lambda\equiv\frac{2\sin\theta}{1+\cos^2\theta}$$

It is said that the metric has enhanced $SL(2,\mathbb{R})\times U(1)$ isometry group. Now what exactly is enhanced symmetry? I only find mentioning of it in the context of string theory, so I'm not sure what to do with it.

If we ignore that for a moment, looking at the groups in question, $SL(2,\mathbb{R})$ has 3 generators ($Sl(n,\mathbb{R})$ has $n^2-1$ elements), and $U(1)$ has one.

The rotational $U(1)$ symmetry is generated by Killing vector:

$$\xi_0=-\partial_\varphi$$

While time translations become part of an enhanced $SL(2,\mathbb{R})$ isometry group generated by the Killing vectors

$$\xi_1=\frac{2r\sin\tau}{\sqrt{1+r^2}}\partial_\tau-2\sqrt{1+r^2}\cos\tau\partial_r+\frac{2\sin\tau}{\sqrt{1+r^2}}\partial_\varphi$$

$$\xi_2=-\frac{2r\cos\tau}{\sqrt{1+r^2}}\partial_\tau-2\sqrt{1+r^2}\sin\tau\partial_r-\frac{2\cos\tau}{\sqrt{1+r^2}}\partial_\varphi$$

$$\xi_3=2\partial_\tau$$

Now I wanted to try and find them, but that proved to be quite a challenge (I may try to do it in the end). So instead I wanted to check if they satisfy Killing equation. Another way of checking if they are Killing vectors is to check if Lie derivative of the metric along the Killing vectors is 0

$$\mathcal{L}_\xi g_{\mu\nu}=\xi^\sigma\partial_\sigma g_{\mu\nu}+g_{\sigma\nu}\partial_\mu\xi^\sigma+g_{\mu\sigma}\partial_\nu\xi^\sigma=0$$

So I put the two simplest ones ($\xi_0$ and $\xi_3$), and they give 0 for each component. Nice.

I try with $\xi_2$, and I get non zero components.

So what is wrong with my interpretation?

I did the calculation by hand and with RGTC package in Mathematica, using LieD which calculates Lie derivative, and I made a code that calculates Killing equation, and still got non zero result.


Edit

I'll write out what I get for $\tau\tau$ component.

So, my Killing vector $\xi_2$ has three nonzero components

$\xi^\tau=\frac{2r\sin\tau}{\sqrt{1+r^2}}$, $\xi^r=-2\sqrt{1+r^2}\cos\tau$, $\xi^\varphi=\frac{2\sin\tau}{\sqrt{1+r^2}}$

And the $\tau\tau$ part of Lie derivative is

$$\mathcal{L}_\xi g_{\tau\tau}=\xi^\sigma\partial_\sigma g_{\tau\tau}+g_{\sigma\tau}\partial_\tau\xi^\sigma+g_{\tau\sigma}\partial_\tau\xi^\sigma$$

Only nonzero metric components that can be used are $g_{\tau\tau}$, and $g_{\tau\varphi}=g_{\varphi\tau}$.

They are

$$g_{\tau\tau}=-2GJ\Omega(\theta)^2(1+r^2(1-\Lambda(\theta)^2))$$

$$g_{\varphi\tau}=4GJr\Omega(\theta)^2\Lambda(\theta)^2$$

So, for the first part of Lie derivative, since $g_{\tau\tau}$ depends only on $r$ and $\theta$ my $\xi^\sigma$ is only $\xi^r$, since $\xi^\theta=0$. And the second part we have 2 times the ($g_{\tau\tau}\partial_\tau\xi^\tau+g_{\varphi\tau}\partial_\tau\xi^\varphi$). Which, after some simplifying becomes

$$\mathcal{L}_\xi g_{\tau\tau}=\frac{8 G J r \Lambda (\theta )^2 \Omega (\theta )^2 \cos (\tau )}{\sqrt{r^2+1}}$$

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    $\begingroup$ @Qmechanic I don't think this question is considered as "homework", since it (1) arose from reading a research paper and (2) it isn't exactly about solving a problem, . $\endgroup$ – Abhimanyu Pallavi Sudhir Nov 13 '13 at 13:23
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Partial answer:

About the symmetry, the part : $-(1+r^2)d\tau^2+\frac{dr^2}{1+r^2}$ is just the metrics of AdS2, so there is a $SO(2,1)$ symmetry.

At fixed $\theta$, we have this $SO(2,1) \sim SL(2,R)$ symmetry plus the $U(1)$ symmetry corresponding to the invariance by $\phi$ translation.

The $\theta$ parameters are playing the role of geometrical terms, but do not change the nature of the symmetry.

Ref, Chapter 2, page 3

[EDIT]

Correcting OP calculus:

$\mathcal{L}_\xi g_{\tau\tau}=\xi^\sigma\partial_\sigma g_{\tau\tau}+g_{\sigma\tau}\partial_\tau\xi^\sigma+g_{\tau\sigma}\partial_\tau\xi^\sigma \tag{1}$

$g_{\tau\tau}$ depends only on $r$, and the only non-null metrics $g_{\sigma\tau} =g_{\tau\sigma}$,are $g_{\tau\tau},g_{\tau\varphi} =g_{\varphi\tau}$, so finally :

$\mathcal{L}_\xi g_{\tau\tau}=\xi^r\partial_r g_{\tau\tau}+ 2(g_{\tau\tau}\partial_\tau\xi^\tau+g_{\tau\varphi}\partial_\tau\xi^\varphi )\tag{2}$

The first term is equal to :

$$\xi^r\partial_r g_{\tau\tau} = [2 GJ \Omega^2][-2\sqrt{1+r^2}\cos\tau][2(\Lambda^2-1)r]\tag{3}$$

The second term is equal to :

$$2g_{\tau\tau}\partial_\tau\xi^\tau = [2 GJ \Omega^2]2[-1 +(\Lambda^2-1)r^2][\frac{2r\cos\tau}{\sqrt{1+r^2}}]\tag{4}$$

The third term is equal to :

$$2g_{\tau\varphi}\partial_\tau\xi^\varphi = [2 GJ \Omega^2]2[\Lambda^2r][\frac{2\cos\tau}{\sqrt{1+r^2}}]\tag{5}$$

Finally we have :

$$\mathcal{L}_\xi g_{\tau\tau}= \\ \frac{(8 GJ \Omega^2)(r \cos\tau)}{\sqrt{1+r^2}}[-(1+r^2)(\Lambda^2-1)+(-1 +(\Lambda^2-1)r^2)+ \Lambda^2]=0\tag{6}$$

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  • $\begingroup$ So that's the 'enhancing' part? $\endgroup$ – dingo_d Sep 29 '13 at 18:06
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    $\begingroup$ $SL(2,R) * U(1)$ is not the symmetry of the standard Kerr metrics, it is only correct near horizon, so one says, that, near horizon, there is an "enhanced" symmetry, which is greater than the standard symmetry for the Kerr metrics. $\endgroup$ – Trimok Sep 29 '13 at 18:10
  • $\begingroup$ Oh, that makes sense now :) I'll read the part about Killing vectors in that paper too, maybe there will be an answer there about the second part :) Thanks $\endgroup$ – dingo_d Sep 29 '13 at 18:13
  • $\begingroup$ For which $\mu, \nu$, do you get a non-zero result for $\mathcal{L}_\xi g_{\mu\nu}$ ? $\endgroup$ – Trimok Sep 29 '13 at 18:22
  • $\begingroup$ When using $\xi_2$ I have nonzero: $\tau\tau$, $\tau r$ $\tau\varphi$, $r\tau$, $r\varphi$ and $\varphi\tau$, $\varphi r$ $\endgroup$ – dingo_d Sep 29 '13 at 18:31

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