Question

${\mathit{V}}_{\mathbf{j}}$If role="math" localid="1659267226224" ${\mathit{V}}_{\mathit{i}}\mathit{=}{\mathit{g}}_{\mathit{i}\mathit{j}}\mathit{,}{\mathit{V}}^{\mathit{j}}$is a contravariant vector and is a covariant vector, show that${\mathit{U}}^{\mathbf{i}}{\mathit{V}}_{\mathbf{j}}$is a${\mathbf{2}}^{\mathbf{n}\mathbf{d}}$ -rank mixed tensor. Hint:Write the transformation equations for U and V and multiply them.

Short answer

The results are proved in the solution.

Step-by-step solution

Given information.

Co-variant and contra-variant components are defined.

Definition of 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.

Write the definitions of the covariant and contravariant vectors.

Write the definitions of the vectors.

$$\begin{gathered} {\mathrm{U}}_{\mathrm{i}}^{\text{'}}=\frac{\partial {\mathrm{x}}_{\mathrm{i}}^{\text{'}}}{\partial {\mathrm{x}}_{\mathrm{k}}}{\mathrm{U}}^{\mathrm{k}} \\ {\mathrm{V}}_{\mathrm{i}}^{\text{'}}=\frac{\partial {\mathrm{x}}_{\mathrm{j}}}{\partial {\mathrm{x}}_{\mathrm{l}}^{\text{'}}}{\mathrm{V}}_{\mathrm{l}} \end{gathered}$$

Multiply the two equations.

Multiply the above equations.

${\mathrm{U}}_{\mathrm{i}}^{\text{'}}{\mathrm{V}}_{\mathrm{j}}^{\text{'}}=\frac{\partial {\mathrm{x}}_{\mathrm{i}}^{\text{'}}}{\partial {\mathrm{x}}_{\mathrm{k}}}\frac{\partial {\mathrm{x}}_{\mathrm{j}}}{\partial {\mathrm{x}}_{\mathrm{l}}^{\text{'}}}{\mathrm{U}}^{\mathrm{k}}{\mathrm{V}}_{\mathrm{l}}$

It can be seen that this is a second-order mixed tensor.