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Suppose we are given that $R_{IJKL}X^L=0$. This implies that $$\nabla_I\nabla_JX_K-\nabla_J\nabla_IX_K=0.$$ However, because of the symmetric properties of the Riemann curvature tensor, we also know that $R_{IJKL}X^L=R_{KLIJ}X^L$. Does this imply that $X^L\nabla_K\nabla_L-X^L\nabla_L\nabla_K=0$?

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Yes, it does. First, I think it's good to establish that the equality

$$\nabla_I\nabla_JX_K-\nabla_J\nabla_IX_K=0.$$

is really saying that the $k$-th component of the local vector field

$$\nabla_I\nabla_J X-\nabla_J\nabla_I X = R(E_i, E_j)X$$

is equal to zero (where $\{E_i \}$ denotes the local frame induced by the coordinate system we're working on and is not necessarily orthonormal). Equivalently, since the Levi-Civita connection commutes with musical isomorphisms, we have

$$(\nabla^2 X_{\flat})_{ij}(E_k)-(\nabla^2 X_{\flat})_{ji}(E_k) = 0 $$

Now, $X^L\nabla_K\nabla_L-X^L\nabla_L\nabla_K$ is just the endomorphism

$$Z \mapsto (X^L\nabla_K\nabla_L-X^L\nabla_L\nabla_K)(Z) = \nabla_K \nabla_X Z - \nabla_X \nabla_K Z = R(K, X)Z $$

So your question is really: does $ R(E_i, E_j)X =0$ for all $i, j$ imply that $R(K, X)Z = 0$ for any $Z$? Now, since $R$ is a tensor, $R(K, X)Z = 0$ is true iff $R(K, X, Z, E_i) = 0 $ for all $i$. But from $ R(E_i, E_j)X = 0$ for all $i, j$, it's clear that $ R(E_i, E_j, X, E_{\ell}) = 0$ for all $i, j, k$. Equivalently, $R(E_{\ell}, X, E_i, E_j) = 0$ for all $i, j, k$. So clearly (again by tensoriality) $R(K, X, Z, E_i) = 0 $ for all $i$, and therefore $R(K, X)Z = 0$ indeed.

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