Metric and Convariant Tensor

Question

$g_{ij}$ is the metric tensor. Show that $g^{ij}$ which satsifies $g_{ij}g^{jk}=\delta_i^k$ is a covariant tensor of rank $2$.

I am not sure how to show this? Does it instead mean to show that $g_{ij}$ is a convariant tensor of rank $2$?

Answer 1

Well, you can take the equation $$ g_{ij}g^{jk}=\delta_i^k $$ as a definition of $g^{jk}$ and establish its transformation properties.

This (tensor) equation should be valid in any basis, so $$ g'_{ij}{g'}^{jk}={\delta'}_i^k = {\delta}_i^k $$

The change of basis formula for the covariant components is

$$ g'_{ij} = \frac{\partial x_p}{\partial x'_i}\frac{\partial x_q}{\partial x'_j} g_{pq} $$

So in your new basis, $$ \frac{\partial x_p}{\partial x'_i}\frac{\partial x_q}{\partial x'_j} g_{pq} {g'}^{jk} = {\delta}_i^k = g_{ij} {g}^{jk} $$ or $$ g_{pq}\left( \delta^p_{i}\delta^q_{j} g^{jk} - \frac{\partial x_p}{\partial x'_i}\frac{\partial x_q}{\partial x'_j} {g'}^{jk} \right) = 0 \\ \delta^p_{i}\delta^q_{j} g^{jk} = \frac{\partial x_p}{\partial x'_i}\frac{\partial x_q}{\partial x'_j} {g'}^{jk} $$

Contracting both sides with $\delta^i_k$ and simplifying, you'll have

$$ g^{qp} = \frac{\partial x_p}{\partial x'_k}\frac{\partial x_q}{\partial x'_j} {g'}^{jk} $$

You can always invert this by contracting both sides with $\dfrac{\partial x'_n}{\partial x_p}\dfrac{\partial x'_m}{\partial x_q} $ to get a more familiar form in which the coordinates after a change of basis are on the left hand side.

$$ {g'}^{mn} = \frac{\partial x'_m}{\partial x_q}\frac{\partial x'_n}{\partial x_p} g^{qp} $$

which is the transformation law for the contravariant components of the 2nd order tensor.