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**1. The problem statement**

I need to prove that if two metrics are related by an overall conformal transformation of the form \overline{g}_{ab}=e^{a(x)}g_{ab} and if k^{a} is a killing vector for the metric g_{ab} then k^{a} is a conformal killing vector for the metric \overline{g}_{ab}

## Homework Equations

killing equation

killing conformal equation

## The Attempt at a Solution

i think i need to show that \overline{\nabla}_{a}k_{b}+\overline{\nabla}_{b}k_{a}=(k^{r}\nabla_{r}a(x))\overline{g}_{ab}

which as far as i understand is the killing conformal equation for the metric \overline{g}_{ab}

so using the relation \overline{\nabla}_{a}k_{b}=\nabla_{a}k_{b}-C^{r}_{ab}k_{c}

where C^{r}_{ab} are the connection coefficients relating the derivative operatrors for g_{ab} and \overline{g}_{ab}

i sustitute this in \overline{\nabla}_{a}k_{b}+\overline{\nabla}_{b}k_{a}

and using killing equation for the metric g_{ab} i obtain:

\overline{\nabla}_{a}k_{b}+\overline{\nabla}_{b}k_{a}=-k_{a}\nabla_{b}a(x)-k_{b}\nabla_{a}a(x)+g_{ab}k^{r}\nabla_{r}a(x)

which is not the conformal killing equation so im lost , can anyone help me on this?