# Covariant derivative of the curvature tensor

I'm confused about the covariant derivative of the Riemannian curvature tensor.

Background

Let $$\nabla$$ be the Levi-Civita connection of some Riemannian metric. Let $$R$$ be the Riemannian curvature tensor, defined by $$R(X,Y)Z = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]} Z.$$ The only definition of 'covariant derivative of a tensor' I can find is for tensors defined as maps to $$C^{\infty}(M)$$. I understand that $$R$$ is a $$(1,3)$$-tensor (in Lee's notation, i.e. it can be viewed as mapping 1 covector-field and 3 vector-fields to a smooth function). However I don't know how to use this to make sense of the LHS of equation $$(1)$$ below.

Question

Does it hold that $$\tag{1} (\nabla_W R)(X,Y)(Z) = \nabla_W (R(X,Y)Z)?$$ If so, I would be interested in seeing a derivation of this equality from the 'covariant derivative of a $$(1,3)$$-tensor'-point of view, if that makes sense.

• To make it into an $(1,3)$ tensor one uses $\omega$, an one-form and three vector fields $X,Y,Z$ to get $$\omega(R(X,Y)Z)$$ this is locally a tetra-linear transformation $$T_pM^*\times T_pM\times T_pM\times T_pM\to\mathbb R.$$ Jun 22, 2021 at 16:31

No, it does not hold. As a tensor, $$R$$ satisfies Leibniz's rule: $$\nabla_{W}\left(R(X,Y)Z\right) = \left(\nabla_WR\right)(X,Y)Z + R\left(\nabla_WX,Y\right)Z + R\left(X,\nabla_WY\right)Z + R(X,Y)\nabla_WZ,$$ and hence, $$\left(\nabla_WR\right)(X,Y)Z = \nabla_{W}\left(R(X,Y)Z\right) - R\left(\nabla_WX,Y\right)Z - R\left(X,\nabla_WY\right)Z - R(X,Y)\nabla_WZ.$$