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


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.


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.

  • 1
    $\begingroup$ 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 $$T_pM^*\times T_pM\times T_pM\times T_pM\to\mathbb R.$$ $\endgroup$
    – janmarqz
    Jun 22, 2021 at 16:31

1 Answer 1


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, $$ \left(\nabla_WR\right)(X,Y)Z = \nabla_{W}\left(R(X,Y)Z\right) - R\left(\nabla_WX,Y\right)Z - R\left(X,\nabla_WY\right)Z - R(X,Y)\nabla_WZ. $$


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