The Lie derivative of a general covariant $4$-tensor is given by
$$\mathcal{L}\_{K}R\_{abcd} = X^{e}\nabla\_{e}R\_{abcd} + R\_{ebcd}\nabla\_{a}X^{e} + R\_{aecd}\nabla\_{b}X^{e} + R\_{abed}\nabla\_{c}X^{e} + R\_{abce}\nabla\_{d}X^{e},$$


where $X^{a}$ is a y smooth vector field. If the $(0,4)$ covariant tensor $R$ is the Riemann tensor (and admits all the symmetries of the same) and $X^{a}$ is a Killing vector, then by the theorem on the inheritance of symmetries, a Killing symmetry would imply an affine symmetry which in-turn would imply a curvature symmetry (assuming the no non-metricity and no torsion). Let $K$ be a vector field that preserves the Riemann tensor $R\_{abcd}$ such that we have the condition of curvature collineation to hold, i.e.,
$$\mathcal{L}\_{K}R\_{abcd} = 0.$$
I now want to use this property of curvature collineation and derive the Killing identity: $\nabla\_{a}\nabla\_{b}K\_{c} = R\_{dabc}K^{d}$ but haven't been able to do so. Any help is appreciated.


My attempt:
Replacing with the total derivatives, we have
$$\nabla\_{a}\left(R\_{ebcd}K^{e} \right) + \nabla\_{b}\left(R\_{aecd}K^{e} \right) + \nabla\_{c}\left(R\_{abed}K^{e} \right) + \nabla\_{d}\left(R\_{abce}K^{e} \right) + \left(R\_{abcd;e} - R\_{ebcd;a} - R\_{aecd;b} - R\_{abed;c} - R\_{abce;d} \right)K^{e} = 0,$$


Now, using the symmetries of the Riemann tensor, $R\_{abcd;e} = -R\_{abde;c} - R\_{abec;d}$ and the definition of the Riemann tensor $R\_{ebcd}K^{e} = \nabla\_{c}\nabla\_{d}K\_{b}$, we obtain


$$\left[\nabla\_{c},\nabla\_{a}\right]\nabla\_{b}K\_{d} + \left[\nabla\_{d},\nabla\_{b}\right]\nabla\_{a}K\_{c} - \nabla\_{b}\left(\left[\nabla\_{c}, \nabla\_{d}\right]K\_{a} \right) - \nabla\_{d}\left(\left[\nabla\_{a},\nabla\_{b}\right]K\_{c}\right) = \left(R\_{ebcd;a} - R\_{eacd;b}\right)K^{e}.$$
I am stuck here, I don't exactly know how to handle the $\left[\nabla\_{c},\nabla\_{d}\right]\nabla\_{b}K\_{d}$ type terms.



Curvature collineation and the Killing identity
Using the identity in [here](https://mathoverflow.net/questions/334060/on-equation-delta-circ-partial-partial-x-partial-partial-x-circ-delta/334696#334696), we have
$$ [\nabla\_{[a}, \mathcal{L}\_X] R\_{bc]de} = \pi\_{d[a}{}^f R\_{bc]fe} + \pi\_{e[a}{}^f R\_{bc]df} \tag{1}$$
where we used that the first order deformation tensor is symmetric in its first two indices.


The first order deformation tensor is, explicitly
$$ \pi\_{lm}{}^n = \frac12 g^{np} \left( \nabla\_l (\nabla\_m X\_p + \nabla\_p X\_m) + \nabla\_m(\nabla\_l X\_p + \nabla\_p X\_l) - \nabla\_p (\nabla\_m X\_l + \nabla\_l X\_m) \right) $$
Rearranging we get
$$ \pi\_{lm}{}^n = \frac12 g^{np}( 2 \nabla\_l \nabla\_m X\_p + [\nabla\_m ,\nabla\_l ] X\_p + [\nabla\_l, \nabla\_p]X\_m + [\nabla\_m, \nabla\_p] X\_l ) $$
or
$$ \pi\_{lm}{}^n = \frac12 g^{np}( 2 \nabla\_l \nabla\_m X\_p + (R\_{spml} + R\_{smlp} + R\_{slmp})X^s ) $$
From the first Bianchi identity we finally get
$$ \pi\_{lm}{}^n = g^{np} ( \nabla\_l \nabla\_m X\_p + R\_{slmp} X^s) $$


**Notice that the terms in the brackets are exactly the "Killing identity" terms.** (I think we use different sign convention for Riemann.)




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An immediate consequences of the above computation:


*Killing vectors satisfy the Killing identity*


For Killing vectors, the 0th order deformation tensor vanishes (Killing's equation), and since the 1st order deformation tensor is formed through the covariant derivative of the 0th order ones, it must also vanish. And thus Killing's identity must hold.




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Now let's see what this allows us to say for curvature collineation. By the second Bianchi identity, $\nabla\_{[a}R\_{bc]de} = 0$. Assuming curvature collineation, the left hand side of equation (1) vanishes. This requires
$$ \pi\_{d[a}{}^fR\_{bc]fe} + \pi\_{e[a}{}^fR\_{bc]df} = 0 $$


Let $\Gamma$ be the vector space of type $(1,2)$-tensors symmetric in the covariant indices, and define the linear mapping $\Upsilon$ from $\Gamma$ to type $(0,5)$ tensors by


$$ \gamma\_{ab}{}^c \mapsto \gamma\_{d[a}{}^fR\_{bc]fe} + \gamma\_{e[a}{}^fR\_{bc]df} $$


For your assertion to hold (that curvature collineation implies Killing's identity), it is sufficient that the Riemann curvature tensor is such that the mapping $\Upsilon$ is injective. Obviously, as Deane noted, this fails when your manifold is flat.


In low dimensions injectivity however can fail in general. In two dimensions the antisymmetry in the $a, b, c$ indices means that $\Upsilon$ is by definition the trivial map. In three dimensions the domain $\Gamma$ is 18 dimensional, while the target is 3 dimensional (isomorphic to the space of antisymmetric matrices) and so must have a non-trivial Kernel. Even in four dimensions, the dimensional counting argument is not optimistic: let $R^\*\_{abcd} = \frac12 R\_{abef} \epsilon^{ef}{}\_{cd}$ the right Hodge dual of the Riemann curvature tensor. Taking the Hodge dual w.r.t. to the antisymmetric $a,b,c$ indices in $\Upsilon(\gamma)$ we can study the equivalent linear map
$$ \Upsilon^\*: \gamma \mapsto \gamma\_{d[a]}{}^f R^\*\_{b]fc}{}^d $$
The image is a $(0,3)$ tensor that is antisymmetric in its first two indies, so $\Upsilon^\*$ is a mapping from a 40 dimensional space to a 24 dimensional one, and still must have a non-trivial kernel in general.


(Remark: in the two dimensional case, however, using that Riemann curvature has special structure, you see that curvature collineation implies that the vector field $X$ is conformal Killing, so there's still some hope of getting the desired inequality through a different route.)




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In conclusion, I doubt that there is a general statement that curvature collineation implies Killing identity, because there are too many special cases to worry about. At best (for this line of argument) you may have a statement that states that for generic metrics in sufficiently high dimension that curvature collineation implies Killing identity.


