Suppose $K$ is a Killing vector satisfying Killing's equation, $\nabla_bK_a+\nabla_aK_b=0$. As part of a larger problem, I am wondering if I have correctly shown that $$\nabla_a\nabla_bK_c + \nabla_b\nabla_cK_a + \nabla_c\nabla_aK_b = 0$$

My present proof, however, seems too easy. First, define $$\Pi_{abc} = \nabla_a\nabla_bK_c + \nabla_b\nabla_cK_a + \nabla_c\nabla_aK_b.$$ Now if we use geodesic coordinates at a certain point (wherein $\nabla_a = \partial_a$, since $\Gamma^a_{bc}=0$), the covariant derivatives commute at this point to give $$\Pi_{abc} = \nabla_b\nabla_aK_c + \nabla_c\nabla_bK_a + \nabla_a\nabla_cK_b = \Pi_{bac}.$$ Meanwhile, since $K$ is a Killing vector, $$\Pi_{abc} = -\nabla_a\nabla_cK_b - \nabla_b\nabla_aK_c - \nabla_c\nabla_bK_a = -\Pi_{bac}$$

Combining these results, we have $\Pi_{bac} = -\Pi_{abc} = -\Pi_{bac}$, meaning $\Pi_{abc}=0$, at least at the point where we have established geodesic coordinates. But since we can establish such coordinates anywhere, $\Pi_{abc}=0$ identically. Is this sound reasoning?

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    $\begingroup$ Note: if this proof is reasonable, it provides a shortcut to solving this question: math.stackexchange.com/questions/94440/… $\endgroup$
    – Doubt
    Nov 8, 2016 at 2:43
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    $\begingroup$ Why should covariant derivatives commute? Even in geodesic normal coordinates you should get a curvature term - remember that while $\Gamma = 0$ at the central point, $\partial \Gamma$ can be nonzero even there. I believe the overall result is true, but you will need to get the Bianchi identity for the curvature tensor involved. $\endgroup$ Nov 8, 2016 at 4:03

1 Answer 1


If $T_{abc}=\nabla_a\nabla_bK_c+ \nabla_b\nabla_c K_a+ \nabla_{c}\nabla_a K_b$, then $$ T_{abc} =_{Killing\ equation}-\nabla_a\nabla_cK_b+ \nabla_b\nabla_c K_a+ \nabla_{c}\nabla_a K_b =_{Ricci\ identity}R_{cab}^l K_l +\nabla_b\nabla_c K_a$$

Hence $$ T_{abc} + T_{bca} + T_{cab} = \{ R_{cab}^l +R_{abc}^l + R_{bca}^l \} K_l +T_{abc}=_{Bianchi\ identity} T_{abc} $$ so that by symmetry of $T$, $T=0$


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