I am trying to solve the Killing equations$(\nabla_{\mu}X_{\nu}+\nabla_{\nu}X_{\mu}=0)$ for the AdS3 metric $$ds^2=\frac{-dt^2+dx^2+dz^2}{z^2}$$
The Killing equation gives me $$1)~~\partial_tX_{t}+{X_z\over z}=0,~2)~~\partial_xX_{x}-{X_z\over z}=0,~~3)~~\partial_zX_{z}+{X_z\over z}=0\\4)~~\partial_xX_{t}+\partial_tX_{x}=0,~5)~~\partial_zX_{t}+\partial_tX_{z}+{2X_t\over z}=0,~6)~~\partial_zX_{x}+\partial_xX_{z}+{2X_x\over z}=0$$
Equation 3) gives me $X_z={f(x,t)\over z}$ Plugging this in 1) and 2) gives $$X_{t}=-{1\over z^2}\int f(x,t)dt+K_1(x,z),~~X_{x}={1\over z^2}\int f(x,t)dx+K_2(t,z)$$ substituting this in 5) and 6) gives $$z\partial_zK_1(x,z)+2K_1(x,z)=-\partial_tf(x,t),~~z\partial_zK_2(t,z)+2K_1(t,z)=-\partial_xf(x,t)$$ . Solving for $K_1$ and $K_2$ gives $$K_1(x,z)=-{1\over 2}\partial_t f(x,t)+{g(x)\over z^2},~~K_2(t,z)=-{1\over 2}\partial_x f(x,t)+{h(t)\over z^2}$$. Now using this and the expression for $X_t,X_x$ in 4) we get $${1\over z^2}\left(\int \partial_t f(x,t)dx-\int \partial_x f(x,t)dt+{dg(x)\over dx}+{dh(t)\over dt}\right)-\partial_t\partial_x f(x,t)=0$$ As suggested this is of the form $A(x,t)+{B(x,t)\over z^2}=0\rightarrow A(x,t)=0,~B(x,t)=0$. Using this we get $$\partial_t\partial_x f(x,t)=0\\ \rightarrow f(x,t)=a(t)+b(x)$$ Plugging this in the second condition gives $$x{da(t)\over dt}+c_1-t{db(x)\over dx}-c_2+{dg(x)\over dx}+{dh(t)\over dt}=0$$ I am puzzled how to seperate this equation to obtain g & h in terms of a & b.