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Each independent components of Killing vector for Schwarzschild that generates isometry are from the translation along $t$ and also from the fact that it is spherically symmetric, i.e

\begin{eqnarray} K^{(0)}&=&\partial_{t}\\\ K^{(1)}&=&\sin{\theta}\partial_{\theta}+\cot{\theta}\cos{\phi}\partial_{\phi}\\\ K^{(2)}&=&\cos{\theta}\partial_{\theta}-\cot{\theta}\sin{\phi}\partial_{\phi}\\\ K^{(3)}&=&\partial_{\phi}. \end{eqnarray}

Then we know that $K^{r}=0$ (from $K=K^{\mu}\partial_{\mu}=K^{t}\partial_{t}+K^{r}\partial_{r}+K^{\theta}\partial_{\theta}+K^{\phi}\partial_{\phi} $). I think, it's because we can put the variable $r$ out of a parenthesis, i.e for Schwarzschild \begin{equation} ds^{2}= r\left\{-\left(\frac{r-2m}{r^{2}}\right)dt^{2}+\frac{1}{r-2m}+rd\Omega^{2}\right\}. \end{equation} Similar to the Kerr metric, that the each component of the metric depends on $r$ and $\theta$, so we might put that components out of parenthesis with some algebra (I didn't do it yet), and we get $K$ for Kerr metric with $K^{r}$ and $K^{\theta}$ equal to zero.

Furthermore, I also saw on this paper [eq. (3.8), (3.9), (3.10), (3.11)] that the Killing vector $K^{\theta}$, on the near horizon extremal Kerr black hole, equal to zero. I think, it was because we can put the $\theta$ variable out of the parenthesis.

Thus, I conclude that if we can put out some variables for any metrics, the component of Killing vector along that direction equal to zero. I don't know whether it is true or not. At least it worked to the metrics above, but I don't know why it does. If it's true, I think we can reduce the 10 equations from Killing equation to calculate the Killing vector for any given metric. Please, anyone explain it.

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