Cotangent Vectors

Given the tangent space Tₚ, we can define the cotangent vector space Tₚ* as the space of linear maps from Tₚ to the n-dimensional real space. These maps are called covectors, normal vectors or one-forms.

• The differential df of a function f is a covector, a normal vector to the surface f = constant. If the coordinates are function on M, the associated differentials form a basis in Tₚ*, which is dual to the basis in Tₚ, $d{x}^{\nu }({\partial }_{\mu })={\partial }_{\mu }(x^{\nu })={\delta }_{\mu }^{\nu }$Then, the if the maps $\omega ={\omega }_{\mu }d{x}^{\mu }$ we have $\omega ({\partial }_{\mu })=w_{\nu }d{x}^{\nu }({\partial }_{\mu })=w_{\nu }{\delta }_{\mu }^{\nu }={\omega }_{\mu }$ We can also define a cotangent vector field, a choice of covector at each point p in M. Given a general cotangent vector field there will generally not exist a function such that the cotangent vector field is its differential. Now, under coordinate transformation, $\omega ={\omega }_{\nu }d{x}^{\nu }={\tilde{\omega }}_{\mu }d{\tilde{x}}^{\mu }={\tilde{\omega }}_{\mu }\frac{\partial {\tilde{x}}^{\mu }}{\partial x^{\nu }}d{x}^{\nu }$So the components transform as, ${\omega }_{\nu }={\tilde{\omega }}_{\mu }\frac{\partial {\tilde{x}}^{\mu }}{\partial x^{\nu }}$ What about the combination of a vector and a covector? ${\tilde{\omega }}_{\mu }{\tilde{V}}^{\mu }=\frac{\partial x^{\nu }}{\partial {\tilde{x}}^{\mu }}{\omega }_{\nu }\frac{\partial {\tilde{x}}^{\nu }}{\partial x^{\rho }}{V}_{\rho }=\frac{\partial x^{\nu }}{\partial x^{\rho }}{\omega }_{\nu }{V}^{\rho }={\delta }_{\rho }^{\nu }{\omega }_{\nu }{V}^{\rho }=w_{\nu }{V}^{\nu }$We say that the indices are contracted.