I know that spacetime, as described by General Relativity (GR), is a pseudo-Riemannian manifold. The label "pseudo" is due to the fact that the metric of spacetime entails not only positive distances, but also negative and null distances (resulting in the three kinds of spacetime intervals: space-like, time-like, and light-like). Thus, spacetime has a very different geometry than any metric space.

My question: I wonder whether there can be an isomorphism between spacetime (defined as a pseudo--Riemannian manifold) and at least one metric space? In other words, can the GR spacetime be smoothly transformed into any metric space? If not (as my intuition says), is there proof of this fact?

  • 2
    $\begingroup$ The "pseudo" is because the metric tensor $g$ has a $-+++$ signature, allowing for positive and negative values of the norm of a vector (which, under integration over a path, leads to your "positive/negative distances"). So, yes, this smooth manifold is different than Riemannian smooth manifolds. A metric space is something different and much simpler: it is a set together with a distance mapping that assigns a positive number to any two distinct elements. A metric tensor is not a metric; distances between points of spacetime depend on path. $\endgroup$
    – Ben H
    Dec 2, 2022 at 16:59
  • $\begingroup$ Related: physics.stackexchange.com/q/562571/2451 and links therein. $\endgroup$
    – Qmechanic
    Dec 2, 2022 at 17:38
  • 1
    $\begingroup$ This is based on a confusion in terminology. A "metric space" is a fairly simple notion, a set provided with a distance function satisfying certain rules. (This distance function does define a topology on the space, the "metric topology".) A manifold is a space that already is a topological space. On a given manifold we may define an extra structure, the metric tensor, a bilinear map that takes two tangent vectors at any point to a scalar. If this map is positive definite everywhere, i.e. can be diagonalized at a given point as $({+}{+}{+}\cdots)$ then we have a "Riemannian manifold"... $\endgroup$
    – printf
    Dec 3, 2022 at 2:44
  • $\begingroup$ If however the metric tensor can be diagonalized everywhere as $({-}{+}{+}{+}\cdots)$, i.e. one minus followed by pluses, then we have what is called a "pseudo-Riemannian" (or "semi-Riemannian") manifold. [Or $({+}{-}{-}{-}\cdots)$, i.e. one plus followed by minuses, depending on your convention.] $\endgroup$
    – printf
    Dec 3, 2022 at 2:46

3 Answers 3


Yes it is, but the found structure has no physical meaning in general.

As a matter of fact, as a general result on differentiable (paracompact) manifolds, every smooth manifold $M$ (without a preferred metric) can be equipped with a smooth Riemannian metric $h$. Next, assuming $M$ is connected, you can define a distance on $M$ whose metric topology is the initial topology on $M$, so that $M$ is homeomorphic to a metric space $(M,d)$, the homeomorphism being the identity map. This distance $d(p,q)$ is nothing but the $\inf$ of the lengths of the smooth paths joining $p$ and $q$ in $M$ computed with respect to $h$:

$$d(p,q) := \inf\left\{ \left.\int_\gamma \sqrt{h(\dot{\gamma},\dot{\gamma}) }ds \:\right|\gamma \in C^\infty([a,b]; M)\:, \gamma(a)=p\:,\gamma(b)=q \right\}\:.$$

If $(M,g)$ is a time-orientable smooth Lorentzian manifold, there is an even shorter way to construct a (smooth) Riemannian metric $h$ on $M$. If $T$ is an everywhere defined timelike smooth vector field on $(M,g)$ (it exists because the manifold is time orientable) where $g$ is the (smooth) Lorentzian metric, a Riemanniam metric can be defined as $$h(X,Y):= g(X,Y) - \frac{2g(T,X)g(T,Y)}{g(T,T)}\:.$$ Indeed, $h$ is smooth by construction. Next for any $p\in T_pM$ refer to a $g$-pseudo orthonormal basis of $T_pM$, $e_0,...,e_{n-1}$, with $T$ parallel to $e_0$. Here, $X=X_p^\mu e_\mu, Y=Y_p^\nu e_\nu \in T_pM$ and the definition above yields, point by point, $$h_p(X,Y)= X^0_pY_p^0 +\sum_{a=1}^{n-1}X_p^aY_p^a$$ which is evidently a positive metric.

  • $\begingroup$ To add some other perspective on the latter part of the answer, I believe the prescription for the Riemannian metric on $M$ is known by the name of Landau-Lifshitz radar metric. $\endgroup$
    – K.T.
    Dec 3, 2022 at 17:06
  • $\begingroup$ I did not know, thank you. $\endgroup$ Dec 3, 2022 at 17:19
  • $\begingroup$ Suppose $T$ is a Killing vector for $g$. Then is $h$ the projection of $g$ onto the spatial part of the manifold? This seems like a clear enough physical meaning if it's true $\endgroup$
    – Jojo
    Dec 3, 2022 at 19:07
  • $\begingroup$ Yes, but in general there is no timelike Killing time but $h$ exists in any case. $\endgroup$ Dec 3, 2022 at 20:14
  • $\begingroup$ @Joe No, $h$ is not the projection onto the spatial part but the "inversion" through the spatial part, due to the factor of 2 in its definition. That is, $h$ equals the spatial part of $g$ with the original sign plus the temporal part of $g$ with the opposite sign. If $h$ were a projection, it would be singular and not a valid 4D metric. $\endgroup$
    – nanoman
    Dec 4, 2022 at 6:46

It is worth noting that oftentimes in mathematical analysis, metric spaces are used to define the topology of a space. Say we are considering Minkowski spacetime in $\mathbb{R}^4$. Then our Minkowski spacetime has almost all of the properties of the Wikipedia page on metric spaces, so long as we keep in mind that the distance function used to define the topology is the usual distance function $d(\vec{x},\vec{y})=\|\vec{x}-\vec{y}\|$. Convergence, limits, completeness, compactness of subsets, continuity, etc., are all defined using "the usual topology" -- the one that comes from this Euclidean metric $d$. So as far as topology is concerned, in general relativity we already have access to a metric space (the metric spaces of subsets of $\mathbb{R}^4$ in each of our coordinate charts). I bring this up because it is a concern I had when I was learning about differential geometry, the topology is not defined using the Minkowski metric.

A separate question is asking what Riemannian manifolds we can define given a manifold in general relativity. You would have to specify what structures you want to preserve on it, and what kind of isomorphism you want to talk about!


You can certainly equip a generic spacetime with a metric, if you choose to do so. For example, you could define a Riemannian metric tensor $\gamma$ from the pseudo-Riemannian metric tensor $g$ by simply flipping the relevant sign(s), and then use $\gamma$ to define a metric on the spacetime manifold. Of course, this metric will generically be physically meaningless.

  • $\begingroup$ But will this new manifold preserve the structure of the original, pseudo--Riemannian manifold? Wouldn't some of the structure (namely, the structure corresponding to the negative and null distances) be irreversibly lost during the transformation? $\endgroup$
    – Maverick
    Dec 2, 2022 at 17:52
  • $\begingroup$ We haven't lost the pseudo-Riemannian metric tensor $g$ here - we're simply using it to construct a second tensor field called $\gamma$, which we subsequently use to define a notion of distance. We are adding a (physically meaningless, but well-defined) metric structure to the spacetime. Of course, as Valter says in his answer, we certainly don't need to use $g$ to create this structure, because any smooth manifold can be equipped with a Riemannian metric tensor. $\endgroup$
    – J. Murray
    Dec 2, 2022 at 19:36
  • $\begingroup$ I think this metric defines locality? That is, which events are close and directly affect each other according to the partial differential equations of fields. $\endgroup$
    – Ryder Rude
    Dec 3, 2022 at 2:24
  • $\begingroup$ @J.Murray, is it something similar to "Riemann manifolds dual to static spacetimes", arxiv.org/abs/2004.10505, by Carolina Figueiredo, José Natário? $\endgroup$
    – JanG
    Dec 3, 2022 at 9:45

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