2
$\begingroup$

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?

$\endgroup$
0

2 Answers 2

1
$\begingroup$

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}

$\endgroup$
4
  • $\begingroup$ 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
    Commented Feb 4, 2014 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
    Commented Feb 5, 2014 at 8:44
  • 1
    $\begingroup$ @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
    Commented Feb 5, 2014 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
    Commented Feb 6, 2014 at 12:51
0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ I'll try this and see the result. $\endgroup$
    – dingo_d
    Commented Feb 5, 2014 at 14:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.