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I have AdS${}_3$ given as a surface embedded in a 4 dimensional pseudo-Riemannian space

$$x^2+y^2-u^2-y^2=-l^2$$

With metric:

$$ds^2=dx^2+dy^2-du^2-dv^2$$

I have Killing vectors of that space given in the embedding coordinates

$$J_{ab}=x_a\frac{\partial}{\partial x^b}-x_b\frac{\partial}{\partial x^a}$$

such that for 01 component I have:

$$J_{01}=x\partial_y-y\partial_x$$

And similar for other components (since the group of AdS${}_3$ is $SO(2,2)$ there are six such vectors).

Now, I have so called 'Euler like' coordinates: $$ x=\ell\left(\cos\left(\frac{\tau}{2}\right)\cosh\left(\frac{\sigma}{2}\right)\sinh\left(\frac{\omega}{2}\right)-\sin\left(\frac{\tau}{2}\right)\sinh\left(\frac{\sigma}{2}\right)\cosh\left(\frac{\omega}{2}\right)\right) $$$$y=\ell\left(\cos\left(\frac{\tau}{2}\right)\sinh\left(\frac{\sigma}{2}\right)\cosh\left(\frac{\omega}{2}\right)+\sin\left(\frac{\tau}{2}\right)\cosh\left(\frac{\sigma}{2}\right)\sinh\left(\frac{\omega}{2}\right)\right) $$$$u=\ell\left(\cos\left(\frac{\tau}{2}\right)\cosh\left(\frac{\sigma}{2}\right)\cosh\left(\frac{\omega}{2}\right)+\sin\left(\frac{\tau}{2}\right)\sinh\left(\frac{\sigma}{2}\right)\sinh\left(\frac{\omega}{2}\right)\right) $$$$v=\ell\left(\sin\left(\frac{\tau}{2}\right)\cosh\left(\frac{\sigma}{2}\right)\cosh\left(\frac{\omega}{2}\right)-\cos\left(\frac{\tau}{2}\right)\sinh\left(\frac{\sigma}{2}\right)\sinh\left(\frac{\omega}{2}\right)\right) $$

In which my metric is:

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

This part I got easily, just differentiate and group everything and you get this (with the help of Mathematica).

What bothers me is, how do I get Killing vectors in this basis? :\

The issue is in unit vectors $\partial_x,\partial_y,\partial_u,\partial_v$. I have trouble getting vectors with unit vectors $\partial_\tau,\partial_\sigma,\partial_\omega$. I tried with differentiation, but that's not it, so I must be messing something up. I think back in uni when we transformed from Cartesian to spherical, we used Jacobian, but I cannot recall how we did this. Can anybody point me in the right direction?

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Let local coordinates $x^\mu$ and $x'^\mu$ be given with corresponding coordinate basis vectors $\partial_\mu$ and $\partial'_\mu$ respectively, then \begin{align} \partial_\mu' = \frac{\partial x^\alpha}{\partial x'^\mu}\partial_\alpha. \end{align}

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  • $\begingroup$ I did this on paper, for which I get: $\partial_x=\frac{\partial \tau}{\partial x}\partial_\tau+\frac{\partial \sigma}{\partial x}\partial_\sigma+\frac{\partial \omega}{\partial x}\partial_\omega$, but I kinda got stuck there :S Should I express $\tau,\sigma,\omega$ in terms of $x,y,u,v$ then? Is that even possible? :\ $\endgroup$ – dingo_d Feb 4 '14 at 18:43
  • $\begingroup$ I can make $\partial_\tau=\frac{\partial x}{\partial \tau}\partial_x+\frac{\partial y}{\partial \tau}\partial_y+\frac{\partial u}{\partial \tau}\partial_u+\frac{\partial v}{\partial \tau}\partial_v$, but that doesn't help much :\ $\endgroup$ – dingo_d Feb 5 '14 at 8:44
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    $\begingroup$ @dingo_d: It might be useful to treat your coordinates as a change of basis in embedding space by using $\ell$ as a fourth coordinate. In that way you get an invertible transformation. The Killing vectors you obtain should be independent of $\ell$ and restricting them to a constant $\ell$ slice gives the $AdS_3$ Killing vectors. $\endgroup$ – Olof Feb 5 '14 at 16:43
  • $\begingroup$ I should be able to change the basis with: $V^{\mu\prime}=V^\mu g_{\mu\nu} \frac{\partial x^\nu}{\partial x^{\nu\prime}} g^{\mu\prime \nu\prime}$, where primed are new coordinates ($\tau,\sigma,\phi$), and not primed are old ($x,y,u,v$), but I'm not getting a good result :\ $\endgroup$ – dingo_d Feb 6 '14 at 12:51
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Write the Jacobian matrix $J$,defined by : $^t(dx,dy,du) = J~~ ^t(d\tau,d\sigma, d\omega)$, and invert it, you have $^t(\partial_x,\partial_y,\partial_u) = J^{-1} ~~^t(\partial_\tau,\partial_\sigma,\partial_\omega)$.

The above procedure does not allow you to calculate $\partial_v$.

So repeat the same operation, but with the triplet : $^t(dx,dy,dv) = J'~~^t(d\tau,d\sigma, d\omega)$.

As a check, the expression for $\partial_x$ and $\partial_y$ must be the same in the 2 cases, otherwise something is wrong.

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  • $\begingroup$ I'll try this and see the result. $\endgroup$ – dingo_d Feb 5 '14 at 14:01

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