Metric tensor and its inverse

1. Is it always allowed to represent the metric tensor $g_{\mu \nu}$ in General Relativity as a $4\times 4$ matrix?

2. If the last one is represented for example with a $4\times 4$ matrix $\mathbf{A}\in{\mathbb{R}^{4\times 4}}$ is it always true that its metric inverse $g^{\mu \nu}$ is the ordinary matrix inverse $\mathbf{A}^{-1}$ which one can calculate with ordinary linear algebra methods?

Maybe this question may sound silly, but I've understood that in differential geometry, also simple things become complicated...

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1) yes.

2) yes. So long as you're describing the ordinary physical 4-D space of standard general relativity. You can get weirdness when you describe null subsurfaces of 4-D space, and the ordinary inverse does not exist. I can point you to the part of my diss where I discuss that if you're especially curious.

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Short Answer: When considering the metric at a single point, it is a good old fashioned matrix we all know and love from elementary linear algebra.

Long Answer: Well, there is some subtlety to this question, because the entries of a metric tensor in local coordinates $x^{\mu}$ appear to be functions that are sufficiently smooth. Smooth functions form a ring and not a field! So the linear algebra has some intricacies.

This is flawed reasoning! Why? Because we specify the metric on a tangent space $g(-,-)_{p}:T_{p}M\times T_{p}M\to\mathbb{R}$. We just demand that as we "vary $p\in M$ smoothly" that the metric $g$ varies "smoothly" with $p$ (in some sense).

Having made a choice of coordinates $x^{\mu}$ in some neighborhood $U$ of $p$, we have some basis vectors for the tangent space $\left.\mathbf{e}_{\mu}=\partial_{\mu}\right|_{p}$ of $T_{p}M$. These enable us to determine the components of the metric $g_{\mu\nu}=g(\mathbf{e}_{\mu},\mathbf{e}_{\nu})_{p}$ which is a matrix as you noted in (1).

But this is at a single point $p\in M$. The values of $g_{\mu\nu}$ vary smoothly with $p$. Here's the counter-intuitive behaviour of the metric: we write $g_{\mu\nu}(x)$ to indicate the component is a "function" of the coordinates, and it should be smooth "in the obvious way". The problem is that invertibility isn't guaranteed for arbitrary matrices with components being smooth functions! So we work with a very special subcollection of such matrices.

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