# Warped AdS${}_3$ and symmetry breaking

In this article it is explained how on can (in suitable coordinate basis) get a so called warped AdS${}_3$ black hole, by introducing a warping factor. The original metric in 'Euler coordinates' for AdS${}_3$ is

$$ds^2=\frac{\ell^2}{4}(-d\tau^2+d\omega^2+d\sigma^2+2\sinh \omega d\tau d\sigma)$$

which can be written as

$$ds^2=\frac{\ell^2}{4}\left(-\cosh^2\omega d\tau^2+d\omega^2+(d\sigma+\sinh \omega d\tau)^2\right)$$

This metric has $SL(2,\mathbb{R})_L\times SL(2,\mathbb{R})_R$ isometry with 3 Killing vectors for left and 3 for right sector. When we introduce a warping factor $\lambda$:

$$ds^2=\frac{\ell^2}{4}\left(-\cosh^2\omega d\tau^2+d\omega^2+\lambda(d\sigma+\sinh \omega d\tau)^2\right)$$

the original symmetry breaks to $U(1)_L\times SL(2,\mathbb{R})_R$, which manifest itself by the fact that now we only have one Killing vector for $U(1)$, and that vector should be $J_0=2\partial_\sigma$.

I have the original Killing vectors, but how do I show that warping factor will break the symmetry? Solve the Killing equation with warped metric, and show that the old Killing vectors won't solve it or? Is there an apparent way of seeing this?

• In the appendix $A$ of your reference, you have an explicit expression for the $SL(2,\mathbb R)_R$ and $SL(2,\mathbb R)_L$ killing vectors. So you can verify than $2$ of them are no more Killing vectors for the warped metrics. – Trimok Feb 6 '14 at 20:27
• So it's just putting them in Killing equation and seeing that they're no longer Killing vectors. I was thinking that there is some more profound way than that. How did they know that the warping will break the symmetry of $SL(2,\mathbb{R})$ to $U(1)$? Is it because the $\sigma$ part represents the rotation, and the wrapping factor is in front of that part of the metric which contains it? – dingo_d Feb 6 '14 at 22:20
• Maybe a possibility is to start with the original coordinates (see your previous question). The equation $(x^2+y^2)−(u^2+v^2)=−l^2$ exhibit the whole $SO(2,2)$ symmetry. Now, if we take for instance $(x^2+y^2)−(u^2+ (F_\lambda(x^2+y^2-u^2))^2 v^2)=−l^2$, ($F_\lambda$ is a function) ,we see that we have a $SO(2,1)\approx SL(2,R)$ symmetry, which is the invariance by $x^2+y^2-u^2 = Constant$, and we have a residual $SO(2) \approx U(1)$ symmetry, which is $u^2+ (F_\lambda(x^2+y^2-u^2))^2 v^2 = Constant$. – Trimok Feb 7 '14 at 9:26
• If this kind of approach (or an equivalent one) is correct, the difficult work would be to find the function $F_\lambda$, corresponding to some $\lambda$ model in the warped $\tau, \omega, \sigma$ metrics – Trimok Feb 7 '14 at 9:27
• Really? Because when I compare it to the near horizon extreme Kerr, with the $\Lambda(\theta)=1$ factor, I get the AdS${}_3$, where by just comparing I have $r\to\sinh\omega$ and $\varphi\to\sigma$ :S Plus I solved the Killing equation and got that for warped metric only $2\partial_\sigma$ is a Killing vector, and of all three of $SL(2,\mathbb{R})_R$ are too :S – dingo_d Feb 7 '14 at 9:55