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

where $K^{a}$ 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 $K^{a}$ 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: $\nabla_{a}\nabla_{b}K_{c} = R_{dabc}X^{d}$ but haven't been able to do so. Any help is appreciated.

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.

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

If the $(0,4)$ covariant tensor $R$ is the Riemann tensor and admits all the symmetries of the same, 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: $\nabla_{a}\nabla_{b}K_{c} = R_{dabc}X^{d}$ but haven't been able to do so. Any help is appreciated.

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

where $K^{a}$ 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 $K^{a}$ 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: $\nabla_{a}\nabla_{b}K_{c} = R_{dabc}X^{d}$ but haven't been able to do so. Any help is appreciated.