Why does the killing equation $$K_{\mu;\nu}+ K_{\nu;\mu} = 0$$ equal $$K_{\mu,\nu}+ K_{\nu,\mu} -2\Gamma^{\rho}_{\mu\nu}K_{\rho} = 0 $$

when in general a covariant derivative $V_{\beta;\alpha} = (\partial_\alpha V^\lambda + \Gamma_{\alpha \nu}^{\lambda}V^{\nu})g_{\lambda \beta}$? Where does the opposite sign of the affine connection come from and why is there not another affine connection?


  • $\begingroup$ Hi @AccidentalFourierTransform , sorry I just got back to this. Yes that solved it! I'll go check out your other answer!. $\endgroup$ – boson Mar 16 '16 at 6:25

Your expression for the covariant derivative is wrong: it should be with a minus sign (plus sign for vectors=upper index, and minus sign for covectors=lower index): $$ \begin{aligned} \nabla v^\alpha\sim \partial v^\alpha\color{red}+\Gamma^\alpha v\\ \nabla v_\alpha\sim \partial v_\alpha\color{red}-\Gamma_\alpha v \end{aligned} $$


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