Shouldn’t squaring the (metric) tensor give a scalar?

Question

What I understand is that

$$g^{\mu\nu} g_{\mu\nu} = g^2 = \text{Id}_{4×4}$$

First, is this statement true?

Now if it is, as we are contracting two indices here, shouldn’t the result be a 0-rank tensor not a second-rank tensor?

Answer 1

That does not look as the square of the metric tensor, but like the product between the metric and its inverse. In general you use upper indices for the inverse.

Anyway, I would write $$g^{\alpha \mu}g_{\mu \beta} = \delta^{\alpha}_{\beta}.$$ for that product. The result is the identity matrix.

You write $g^{\mu \nu}g_{\mu \nu}$, so you are actually taking the trace of the identity matrix: $\delta^{\mu}_{\mu} = n$, where $n$ is the dimension of your space.