Commutator of covariant derivatives ($[{\nabla }_{\alpha },{\nabla }_{\beta }]V^{\mu }=R^{\mu }{}_{\nu \alpha \beta }{V}^{\nu }$)

For a torsion-free connection, the commutator of coordinate covariant derivatives vanishes on a scalar field and acts on a vector field through the displayed Riemann curvature tensor. On a covector field it instead gives $[{\nabla }_{\alpha },{\nabla }_{\beta }]A_{\mu }=-R^{\nu }{}_{\mu \alpha \beta }{A}_{\nu }$. The curvature commutator is a tensor derivation, so each index of a general tensor contributes independently with the corresponding sign. Including the lower derivative index in the second covariant derivative is essential for this identity.

For a vector field $V$, the tensorial second derivative is ${\nabla }^{2}V(X,Y)={\nabla }_X{\nabla }_{Y}V-{\nabla }_{{\nabla }_{X}Y}V$. With $R(X,Y)V={\nabla }_X{\nabla }_{Y}V-{\nabla }_Y{\nabla }_{X}V-{\nabla }_{[X,Y]}V$ and geometric torsion tensor $\mathcal{T}$, antisymmetrization givesThis follows by subtracting the two definitions and using ${\nabla }_{X}Y-{\nabla }_{Y}X=[X,Y]+\mathcal{T}(X,Y)$. By contrast, the directional operator commutator subtracts no derivative-index connection and obeys $[{\nabla }_X,{\nabla }_Y]V=R(X,Y)V+{\nabla }_{[X,Y]}V$. On scalars the directional commutator is $[X,Y]f$, while the tensorial Hessian commutator is $-\mathcal{T}(X,Y)f$. Distinguishing these objects prevents an apparent contradiction when a scalar Hessian has an antisymmetric part despite commuting coordinate vector fields.