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I'm trying to find Killing vectors of the metric,

$$ds^2 = \frac{1}{x^2}(dt^2-dx^2)$$

i.e. the half-space on anti-de Sitter space. The Ricci scalar is a constant and thus motivates one to consider a full general vector, $\vec v = (v_t, v_x)$, for which the Killing equations give,

$$\partial_t v_t + \frac{1}{x}v_x= 0, \quad \partial_x v_ x + \frac{1}{x}v_x = 0,$$ $$\partial_t u_x + \partial_x v_t + \frac{2}{x}v_t = 0.$$

The second is easy to solve, we have,

$$v_x = \frac{f(t)}{x}$$

for arbitrary $f(t)$ and then we have,

$$v_t = -\frac{1}{x^2}\int f(t) \, dt + g(x).$$

for arbitrary $g(x)$. The third equation gives the constraint, denoting $F(t) = \int f(t) \, dt$:

$$F''(t) + \left( 1-\frac{2}{x^2}\right)F(t) + xg'(x) + 2g(x) = 0.$$

I'm unsure how to proceed from here, as I don't think there's a way to solve for $F$ and $g$ separately. Though clearly if $F = 0$, one solution is $g(x) = c/x^2$. Any idea how to proceed?

1 Answers 1

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with $\quad F(t) = \int f(t) \, dt$: $$v_x = \frac{f(t)}{x}= \frac{F'(t)}{x}$$ $$v_t = -\frac{1}{x^2}\int f(t) \, dt + g(x) = -\frac{1}{x^2}F(t) + g(x)$$

$\partial_t v_x + \partial_x v_t + \frac{2}{x}v_t = 0 \quad\to\quad \frac{F''(t)}{x} + \frac{2}{x^3}F(t) + g'(x) + \frac{2}{x}\left(-\frac{1}{x^2}F(t) + g(x)\right) = 0$

$$F''(t) + xg'(x) + 2 g(x) = 0$$ This implies $\quad F''(t)=-xg'(x)-2g(x)=\lambda \quad$ with $\lambda=$constant. $$\begin{cases} F(t)=\frac{\lambda}{2}t^2+c_1t+c_0 \quad\to\quad f(t)=\lambda t+c_1\\ g(x)=\frac{c_2}{x^2}-\frac{\lambda}{2} \end{cases}$$ $$v_x = \frac{\lambda t+c_1}{x}$$ $$v_t = -\frac{\frac{\lambda}{2}t^2+c_1t+c_3}{x^2}-\frac{\lambda}{2}$$ $c_3=c_0-c_2$