Covariant Derivative

Need: coordinate-independent notion of derivatives of vector fields The covariant derivative of a vector field is a (1,1) tensor field. Linearity implies $\nabla (V+W)=\nabla V+\nabla W$ Satisfies the Leibniz rule $[\nabla (fV){]}_{\mu }^{\nu }=({\partial }_{\mu }f)V^{\nu }+f(\nabla V{)}_{\mu }^{\nu }$ A vector field is covariantly constant if the covariant derivative is zero. Let us now write the covariant derivative of a general vector in terms of the covariant derivatives of the basis vectors: $(\nabla V{)}_{\mu }^{\nu }=[\nabla (V^{\lambda }{\partial }_{\lambda }){]}_{\mu }^{\nu }={\partial }_{\mu }{V}^{\lambda }({\partial }_{\lambda }{)}^{\nu }+V^{\lambda }[\nabla ({\partial }_{\lambda }){]}_{\mu }^{\nu }={\partial }_{\mu }{V}^{\nu }+V^{\lambda }[\nabla ({\partial }_{\lambda }){]}_{\mu }^{\nu }$ We define ${\Gamma }_{\mu \lambda }^{\nu }=[\nabla ({\partial }_{\lambda }){]}_{\mu }^{\nu }$ as the connection coefficients. Consider one-forms, ${\nabla }_{\nu }(V^{\mu }{\omega }_{\mu })={\partial }_{\nu }(V^{\mu }{\omega }_{\mu })={\partial }_{\nu }{V}^{\mu }{\omega }_{\mu }+V^{\mu }{\partial }_{\nu }{\omega }_{\mu }=({\nabla }_{\nu }{V}^{\mu }){\omega }_{\mu }+V^{\mu }{\nabla }_{\nu }{\omega }_{\mu }\text{(from Leibniz rule)}$ Then, we obtain $V^{\mu }{\nabla }_{\nu }{\omega }_{\mu }=V^{\mu }{\partial }_{\nu }{\omega }_{\mu }-{\Gamma }_{\mu \lambda }^{\nu }{V}^{\lambda }{\omega }_{\mu }$ For this to be satisfied for any vector, the covariant derivative of the co-vector must satisfy ${\nabla }_{\nu }{\omega }_{\mu }={\partial }_{\nu }{\omega }_{\mu }-{\Gamma }_{\nu \mu }^{\sigma }{w}_{\sigma }$ Similarly, on a (1,1) tensor, ${\nabla }_{\nu }{T}_{\rho }^{\mu }={\partial }_{\nu }{T}_{\rho }^{\mu }+{\Gamma }_{\nu \lambda }^{\mu }{T}_{\rho }^{\lambda }-{\Gamma }_{\nu \rho }^{\lambda }{T}_{\lambda }^{\mu }$ A connection is torsion-free if the connection coefficients are symmetric in the lower indices ${\Gamma }_{\mu \nu }^{\lambda }={\Gamma }_{\nu \mu }^{\lambda }$ Such connections have the property that the covariant derivatives on a scalar commute, $[{\nabla }_{\mu },{\nabla }_{\nu }]f={\nabla }_{\mu }{\nabla }_{\nu }f-{\nabla }_{\nu }{\nabla }_{\mu }f=({\partial }_{\mu }{\partial }_{\nu }f-{\Gamma }_{\mu \nu }^{\lambda }f)-({\partial }_{\nu }{\partial }_{\mu }f-{\Gamma }_{\nu \mu }^{\lambda }f)=0$

• Levi-Civita Connection