 Solution

Although ${\nabla }_{\beta }f={\partial }_{\beta }f$ for a scalar field, its first derivative is a covector field. ThereforeCommuting partial derivatives and using the torsion-free connection gives $\boxed{[{\nabla }_{\alpha },{\nabla }_{\beta }]f=0}$. For a vector field, differentiate the $(1,1)$ tensor ${\nabla }_{\beta }{V}^{\mu }$ with both its indices included:Subtract the expression with $\alpha ,\beta$ exchanged. Second derivatives of $V$ cancel; the terms multiplying first derivatives cancel pairwise, and the final lower-index connection terms cancel by symmetry. The remaining coefficient of $V^{\nu }$ isThis is the curvature commutator. On a covariant index the sign reverses, giving the curvature commutator on a covariant tensor used in the next part.