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I know that if we have a Killing vector then it's straightforward to show the identity:

$$\nabla_a \nabla_b K_c = R_{cba}^k K_d$$

I'm now trying to show the following identity for a $(0,2)$ Killing tensor:

$$\nabla_{(a}\nabla_b K_{c)d} = - R_{d(ab}^{e}K_{c)e} $$

I know that I should use the fact that a Killing tensor $K_{ab}$ is symmetric and satisfies $\nabla_{(a}K_{bc)}=0$.

I've tried writing out $$(\nabla_a \nabla_b - \nabla_b\nabla_a)K_{cd} = R_{abc}^eK_{ce} + R_{and}^eK_{de}$$ and taking various permutations over a, b and c but haven't been able to make much progress with this and would appreciate some insight into this!

I also wondered if such identities generalise? Do we always have a relation for a $(0,n)$ Killing tensor that relates it's second derivatives to an algebraic relation with the Riemann tensor?

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  • $\begingroup$ The calculations and references provided in this paper (arxiv.org/abs/0907.5470) by Cook and Dray might be helpful. See theorem 1 on page 4 in particular. $\endgroup$ – Arthur Suvorov May 22 '16 at 9:31
  • $\begingroup$ Are you sure this statement is true? $\nabla_{(a}\nabla_bK_{cd)} = \nabla_{a}\nabla_{(b}K_{cd)} + \nabla_{b}\nabla_{(a}K_{cd)} + \nabla_{c}\nabla_{(a}K_{bd)} + \nabla_{d}\nabla_{(a}K_{bc)} $. By killing equation, each term on RHS is a derivative of an everywhere zero function, hence LHS is 0. So looks like your LHS should identically vanish. $\endgroup$ – zzz May 24 '16 at 1:35
  • $\begingroup$ @bianchira Yes, thanks for spotting that! The symmetrisation should only be around a,b and c, I've edited now. $\endgroup$ – Wooster May 24 '16 at 8:25
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First of all we need the equation: \begin{equation}\require{Amsmath} 2\nabla_{[a} \nabla_{b]} K_{cd} = R_{abc}{}^e K_{ed} + R_{abd}{}^e K_{ce} = 2 R_{ab(c} K_{d)e}\tag{1}.\label{eq:KT} \end{equation} We have: $$R_{d(ba}{}^e K_{c)e} = \frac{1}{3}(R_{db(a}{}^e K_{c)e} + R_{da(b}{}^e K_{c)e} + R_{dc(b}{}^e K_{a)e}) = \frac{1}{3} (\nabla_{[d} \nabla_{b]} K_{ac} + \nabla_{[d} \nabla_{a]} K_{bc} + \nabla_{[a} \nabla_{c]} K_{ba}),$$ where we used equation \eqref{eq:KT}. Now we express some terms so that index $d$ is removed from the covariant derivatives using $\nabla_{(a} K_{bc)} = 0$, thus the previous expression becomes: $$\frac{1}{3!} (\nabla_d \nabla_b K_{ac} + \nabla_b \nabla_a K_{dc} + \nabla_b \nabla_c K_{ad} + \nabla_d \nabla_a K_{bc} + \nabla_a \nabla_b K_{dc} + \nabla_a \nabla_c K_{db} + \nabla_d \nabla_c K_{ba} + \nabla_c \nabla_b K_{da} + \nabla_c \nabla_a K_{bd}) = \nabla_d \nabla_{(a} K_{bc)} + \nabla_{(a} \nabla_b K_{c)d},$$ which proves the identity.

Regarding the question about generalisation, one indication that it is possible is that if $K^a$ is a Killing vector, one can form a Killing tensor by writing $K^a K^b$ etc, and induction shows that an equation involving double covariant derivatives and Riemann tensors exists.

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