# A question about the definition of the Riemann curvature tensor.

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 $$X^L\nabla_K\nabla_L-X^L\nabla_L\nabla_K=0$$?

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.