Replacing the contravariant Minkowski metric tensor by its inverse

Question

From Wikipedia, the Minkowski metric is defined (using (- + + +) signature) as : $$\eta_{\mu \nu} = \eta^{\mu \nu} = \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$$

I am a beginner in tensor calculus, and feel uneasy that a covariant vector is being equated to a contravariant vector. Is it then possible to simply replace the covariant Minkowski metric tensor by its contravariant counterpart withing a calculation ?

For example, given a four vector $k$, is it correct that :

$k \cdot k = \eta_{\alpha \mu}k^{\mu} k^{\alpha} = \eta^{\alpha \mu}k^{\mu} k^{\alpha}$

Are there any problems with summing over all upper indices ?

Answer 1

It's important to understand that the statement

$$\eta_{\mu \nu} = \eta^{\mu \nu} = \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$$

Does not say that the metric is equal to the inverse metric. Such a statement doesn't even make sense, because the metric and inverse metric are tensors which eat different kinds of objects (vectors in the former case, covectors in the latter).

What is says is that, in the particular basis which you have chosen, the components of the metric (which are just real numbers) are equal to the components of the inverse metric (which are also real numbers).

[I] feel uneasy that a covariant vector is being equated to a contravariant vector

A covariant vector is never equal to a contravariant vector, because those two objects live in different spaces. The components of a covariant vector (in a particular choice of covariant basis) may or may not be equal to the components of a contravariant vector (in a particular choice of contravariant basis), but that is a vastly different statement (which of course depends on which bases we intend to use).

Answer 2

You can do it in this particular case, because that matrix is its own inverse; the components of $\eta_{\mu\nu}$ and $\eta^{\mu\nu}$ are the same. This doesn't happen in general, so it's basically the only case where you could replace downstairs indices by upstairs indices (though you shouldn't, because it's confusing).

Answer 3

Hint: Inverse of a diagonal matrix is inverse of each entry along the diagonals. Can you see why the components of inverse metric is just inverse of components metric here i.e. $\eta_{\mu\nu} = \eta^{\mu\nu}$?

Given a four vector you may think of $k^{\mu}$ as a column vector, and $k_{\mu}$ as a row vector. I think that will help you see why $\eta_{\alpha \mu}k^{\mu} k^{\alpha} \ne \eta^{\alpha \mu}k^{\mu} k^{\alpha}$.

What is true is the following: \begin{align} k_\mu \cdot k^{\mu} &= \eta_{\mu\alpha}k^{\alpha}k^{\mu} \\ &= \eta_{\mu\alpha}(\eta^{\alpha c}k_c) (\eta^{\mu d} k_d)\\ &= \eta^{\mu d} \eta_{\mu \alpha} \eta^{\alpha c} k_c k_d\\ &= \eta^{\mu d} \delta_\mu^c k_c k_d \qquad \qquad (\because \eta_{\mu\alpha}\eta^{\alpha c} = \delta_\mu^c)\\ &= \eta^{c d} k_c k_d \end{align}