Special conformal transformations as isometries of $\rm AdS_2$

Question

It is claimed in these lecture notes (page 87) that a continuous isometry of AdS in Poincare coordinates is the special conformal transformation, $\delta x_\mu = 2 c \cdot x x_\mu - x^2 c_\mu$ for $c_\mu = (0, c_i)$. For the case of AdS in 2 dimensions

$$ds^2 = \frac{1}{z^2} (dz^2 + dt^2) $$

this transformation corresponds to

$$\delta z = 2 c t z \\ \delta t = 2 c t^2 - |x^2|c = 2ct^2 - c \left(1 + \frac{t^2}{z^2}\right)$$

where we have used the equation for distance in AdS, $|x^2| = \frac{1}{z^2}(t^2 + z^2) = 1 + \frac{t^2}{z^2}$. The corresponding Killing vector should be

$$ V^\mu = \left(2tz, 2t^2 - \left(1+\frac{t^2}{z^2}\right)\right) \\ V_\mu = \left(\frac{2t}{z}, \frac{2t^2}{z^2} - \frac{1}{z^2}\left(1+\frac{t^2}{z^2}\right)\right).$$

However, this does not satisfy the Killing equation $\nabla_\mu V_\nu + \nabla_\nu V_\mu =0$, as can be seen from the $\mu = \nu = t$ component

$$2\nabla_t V_t = 2 \partial_t V_t - 2 \Gamma_{tt}^\mu V_\mu = 4 \frac{t}{z^2} - 2\frac{t}{z^4} - 4 \frac{t}{z} \neq 0 $$ where we have used that $\Gamma_{tt}^z = 1/z$, $\Gamma_{tt}^t = 0$. By looking at the Killing equations, I can guess that the proper Killing vector should instead be

$$V^\mu = (2 t z, 2t^2 - z^2)$$

but I don't see how this is equivalent to the initial definition of the SCT (unless you're sloppy and set $x^2 = z^2$, but I don't think this is right).

Answer 1

Using formulae from the note itself and setting $d=1$. $$ \delta z_\mu = 2 c^\alpha z_\alpha z_\mu - z^2 c_\mu, \qquad c_\mu = ( 0 , c ) . $$ where $z^\mu = (z,t)$ and the metric is $ds^2 = \frac{1}{z^2} [ dz^2 + dt^2 ]$. Then, $$ c^{-1} \delta z = 2 t z , \qquad c^{-1} \delta t = 2 t^2 - ( z^2 + t^2 ) = t^2 - z^2 . $$ Thus, the Killing vector is $V^\mu = ( 2 t z , t^2 - z^2 )$. You can verify that ${\cal L}_V g = 0$.

PS - The correct equation is $z^\mu z_\mu = z^2 + t^2$. The author defines the scalar product he is using in the second last paragraph of Page 86.