Is General Relativityy described by Pseudo-Riemannian manifold or Riemannian manifold? I cannot understand the vast difference between the two manifolds. In books, General Relativity is looked as a pseudo-Riemannian manifold, though I am not sure after reading some threads on the web which confused me.
Now checking wikipedia, it says here:
After Riemannian manifolds, Lorentzian manifolds form the most important subclass of pseudo-Riemannian manifolds. They are important in applications of general relativity.
A principal basis of general relativity is that spacetime can be modeled as a 4-dimensional Lorentzian manifold of signature (3, 1) or, equivalently, (1, 3). Unlike Riemannian manifolds with positive-definite metrics, a signature of (p, 1) or (1, q) allows tangent vectors to be classified into timelike, null or spacelike.
This is the best I've gotten searching for an answer and it is still confusing and not clear.
Best Answer
In relativity (both special and general) one of the key quantities is the proper length given by:
$$ ds^2 = g_{\alpha\beta}dx^\alpha dx^\beta \tag{1} $$
where $g_{\alpha\beta}$ is the metric tensor. The physical significance of this is that if we have a small displacement in spacetime $(dx^0, dx^1, dx^2, dx^3)$ then $ds$ is the total distance moved. You can think of it as a spacetime equivalent of Pythagoras' theorem. The quantity $ds$ is an invariant i.e. all observers in any frame of reference will agree on the value of $ds$.
A metric is positive definite if $ds^2$ is always positive, and Riemannian manifolds have a metric that is positive definite.
However in relativity $ds^2$ can be positive, zero or negative, which correspond to timelike, lightlike and spacelike intervals respectively. It is because $ds^2$ can have different signs that manifolds in GR are not Riemannian but only pseudo-Riemannian.
Lorentzian manifolds are a special case of pseudo-Riemannian manifolds where the signature of the metric is $(3,1)$ (or $(1,3)$ depending on your sign convention).
If we take the metric tensor that corresponds to special relativity (i.e. flat spacetime) equation (1) becomes:
$$ ds^2 = -dt^2 + dx^2 + dy^2 + dz^2 $$
That minus sign on the $dt^2$ term means $ds^2$ can be negative as well as positive, so the manifold is pseudo-Riemannian, and the one negative and three positive signs on the right hand side make the signature $(3,1)$ so the manifold is Lorentzian.