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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.

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    $\begingroup$ If you look at the Wikipedia article for GR, it states clearly in the section about definitions that we consider a pseudo-Riemannian manifold. $\endgroup$ – ACuriousMind Oct 29 '15 at 19:19
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    $\begingroup$ Er, doesn't it say rather explicitly in the quote you give? $\endgroup$ – Kyle Kanos Oct 29 '15 at 19:36
  • $\begingroup$ Part of the confusion may also be because sometimes people are lazy, and say "Riemannian" when they mean that it could be Riemannian or pseudo-Riemannian, or even when they really mean pseudo-Riemannian. This is especially true of "some threads on the web". $\endgroup$ – Mike May 14 '19 at 15:36
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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.

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John has already given a nice intuitive answer, I'd like to write it in a bit more mathematical way. First of all metric means a bilinear defined over a product (vector) space: $g(v,w): V \times V \rightarrow \mathbb{R}$. Where the vectors $v,w$ lie in the vector space $V$. This bilinear should be smooth, symmetric $g(v,w) = g(w,v)$, and positive definite , which means for any $v,w$ it should be positive. This allows one to measure length of the vectors in $V$. And now if we weaken the condition of positive definiteness to 'non-degenrate' then we call it a pseudo-Riemannian metric. Non-degenerate means $g(v,w)=0, \forall w \in V \Rightarrow v = 0$.

The reason why in physics we usually don't go into such language is the following fact: metric is a bilinear so you can always construct the $g(v,w)$ for any pair of vectors if you know what $g(e^\mu,e^\nu)$ are, where $\{ e^\mu \}$ is a basis of the space $V$. This is nothing but the usual $g^{\mu \nu}$. For instance, take $\mathbb{R}^2$ as our $V$, you can define a metric $g^{\mu\nu} = \text{diag}(1,1)$, which is a valid (Riemannian) metric and means nothing but $g(\hat{x}, \hat{x}) = 1 = g(\hat{y}, \hat{y})$. You can check with such a metric (as John pointed out) you will always get $ds^2 > 0$.

If we try to think all these at the level of matrix $g^{\mu \nu}$ it is important to note (Gram-Schmidt theorem) at least locally one can always find a suitable orthonormal basis such that $$g^{\mu \nu} = \eta^{\mu \nu} = \text{diag} \big(-1, \cdots (s \,\, \text{times}), +1, \cdots (r \,\, \text{times}) \big) $$ Hence Riemannian metric would mean $s=0$ (because we want positive definite) and pseduo-Riemannian metric will have $s \neq 0$. In fact the 'signature' $(r,s)$ is always sufficient to tell apart between different metrics, no matter what basis you choose (Sylvester's law of inertia). For example metric which have a signature $(\text{dim}(V)-1,1)$ are called Lorentzian metrics. (Some people might exchange $r$ and $s$).

As of yet I've not talked about manifolds, be it Riemannian or pseudo-Riemannian. We can endow a 'metric structure' to a manifold also. The vector space over which we define the metric is $T_pM$, the tangent space of manifold $M$ at point $p$ on it. In fact we usually define it over the co-tangent space $T^{^*}_pM$ (which has the basis $\{dx^\mu\}$) and that's why we write it as $ds^2 = g_{\mu \nu} dx^\mu dx^\nu = g(dx^\mu, dx^\nu)$. So if you can write a (pseudo-) Riemannian metric on $T_pM$ then $M$ is said to be a (pseudo-) Riemannian manifold.

Hope it might be clear that in special relativity we restrict ourselves only to Lorentzian metrics, but in general relativity we consider all possible pseudo-Riemannian metrics (such as an anti-de Sitter metric has $s=2$).

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A pseudo Riemannian manifold is a manifold equiped with a metric of signature $(p,q)$, $p$ indicating the number of positive eigenvalues and $q$ the negative eigenvalues. For a Riemannian manifold, $q = 0$.

A spacetime of dimension $n$ is defined by a pseudo-Riemannian manifold of signature $(1, n-1)$, or alternatively $(n-1,1)$, also called a Lorentzian manifold.

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  • $\begingroup$ So Lorentzian manifold and pseudo-Riemannian manifold are the same? $\endgroup$ – Beyond-formulas Oct 29 '15 at 18:00
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    $\begingroup$ A Lorentzian manifold is a type of pseudo-Riemannian manifold, but not vice versa. For instance a metric (--++) would be pseudo-Riemannian but not Lorentzian. $\endgroup$ – Slereah Oct 29 '15 at 18:06

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