Question

Show that if${\mathit{V}}^{\mathbf{i}}$is a contravariant vector then${\mathit{V}}^{\mathbf{i}}\mathbf{=}{\mathit{g}}_{\mathbf{i}\mathbf{j}}{\mathit{V}}^{\mathbf{j}}$is a covariant vector, andthat if${\mathit{V}}^{\mathbf{i}}$is a covariant vector, then${\mathit{V}}^{\mathbf{i}}\mathbf{=}{\mathit{g}}_{\mathbf{i}\mathbf{j}}{\mathit{V}}^{\mathbf{j}}$is a contravariant vector.

Short answer

The proof is shown in the solution.

Step-by-step solution

Given information.

A covariant tensor and a contravariant tensor are given.

Definition of a covariance and contravariance.

The components of a vector relative to a tangent bundle basis are covariant in differential geometry if they change with the same linear transformation as the basis. If they change as a result of the inverse transformation, they are contravariant.

Begin by calculating the gradient.

Raise the index of$V_i^{\text{'}}$since$V^i$is contravariant

$\begin{array}{c}{V}_i^{\text{'}}=g_{ij}^{\text{'}}{V}^{\text{'}j}\\ =g_{ij}^{\text{'}}\frac{\partial x_j^{\text{'}}}{\partial x_k}{V}^k\end{array}$

Continue evaluation.

Continue evaluation considering$g_{ij}$is a second order covariant tensor.

$\begin{array}{c}{V}_i^{\text{'}}=g_{lm}\frac{\partial x_l}{\partial x_i^{\text{'}}}\frac{\partial x_m}{\partial x_j^{\text{'}}}\frac{\partial x_j^{\text{'}}}{\partial x_k}{V}^k\\ {\delta }_k^m=\frac{\partial x_m}{\partial x_j^{\text{'}}}\frac{\partial x_j^{\text{'}}}{\partial x_k}\\ V_i^{\text{'}}=\frac{\partial x_l}{\partial x_i^{\text{'}}}{g}_{lm}{V}^m\\ =\frac{\partial x_l}{\partial x_i^{\text{'}}}{V}_l\end{array}$

Recognize the covariant change in one brings a contravariant change in another.

$\begin{array}{c}V{\text{'}}^i=g^{ij}{V}_j^{\text{'}}\\ =g^{ij}\frac{\partial x_k}{\partial x_j^{\text{'}}}{V}_k\\ g{\text{'}}^{ij}=g^{lm}\frac{\partial x_i^{\text{'}}}{\partial x_l}\frac{\partial x_j^{\text{'}}}{\partial x_m}\\ V^{\text{'}i}=g^{lm}\frac{\partial x_i^{\text{'}}}{\partial x_l}\frac{\partial x_j^{\text{'}}}{\partial x_m}\frac{\partial x_k}{\partial x_j^{\text{'}}}{V}_k\end{array}$

Continue the evaluation.

$\begin{array}{c}{\delta }_m^k=\frac{\partial x_j^{\text{'}}}{\partial x_m}\frac{\partial x_k}{\partial x_j^{\text{'}}}\\ V^i=\frac{\partial x_i^{\text{'}}}{\partial x_l}{g}^{lm}{V}_m\\ =\frac{\partial x_i^{\text{'}}}{\partial x_l}{V}^l\end{array}$