I am currently working with Harvey Reall's notes on Black holes, and I have 2 questions concerning the interpretation of the Einstein-Rosen bridge. After some algebra, we are faced with the metric: $$ds^2 = \left(1+\frac{M}{2 \rho}\right)^4 (d\rho^2 + \rho^2d\Omega^2)$$ with $\rho > M/2$ and $d\Omega^2$ is the usual unit 2-sphere metric.

1) The notes claim that this defines a Riemannian 3-manifold with topology $\mathbb{R} \times S^2$ I am not sure where the Riemannian part kicks about. I think it only refers to the fact that the signature is manifestly full positive? More importantly, I see this more like an $\mathbb{R}^3$ metric with a weird conformal factor (and a truncation of the "radius")?

2) In terms of visualization, it suggested embedding the surface in 4d Euclidean space. Just to make sure, those this mean: Suppose the 4d spacetime (say with coordinates $(t,\rho,\theta,\phi)$) is flat. Take the surface $t = f(\rho,\theta,\phi)$ and pull-back the metric onto that surface. This gives a differential equation for $f$, which must be such that the resulting the metric is the one above. We can then plot $f(\rho,\theta,\phi)$. Is that correct?

  • $\begingroup$ Just something I'm wondering given that I'm quite new to GR: why is there no time on this metric? $\endgroup$ Nov 21, 2018 at 19:41
  • $\begingroup$ Could you please link to these notes? $\endgroup$
    – magma
    Nov 22, 2018 at 22:44
  • $\begingroup$ @GabrielGolfetti This is because the manifold is taken from the full 4d metric at constant time. $\endgroup$
    – Patrick.B
    Nov 22, 2018 at 22:52

2 Answers 2


That is indeed why it is called Riemannian (Riemannian just means the signature is all plus).

The topology of $\mathbb{R}^3$ with a truncation of the radius is $\mathbb{R} \times S^2$. In general, spherical coordinates only cover a manifold of the form $\mathbb{R} \times S^2$, since every point describes a position $\rho$ on $\mathbb{R^+} \cong \mathbb{R}$ and $(\theta, \phi)$ on $S^2$ (this is because the coordinates are degenerate at $\rho = 0$ and hence you only cover $\mathbb{R}^3$ minus a point).

It's easier to see in the two-dimensional case, as it is easier to visualize the transformation of $\mathbb{R}^2$ minus a disk to a cylinder, making the border of the hole the upper part of the cylinder, and the boundary at infinity the lower part.

  • $\begingroup$ Wait a second. So, let us say we take the usual flat metric in Cartesian coordinates $(t,x,y,z)$, this is topologically manifestly $\mathbb{R}^4$. If I now choose to describe it in using spherical coordinates instead $(t,r,\theta,\phi)$, do I get now $\mathbb{R} \times S^3$? Does this mean that $\mathbb{R}^4 \equiv$ $\mathbb{R} \times S^3$? Or only as a local statement? Given that I need at least 2 charts to cover a sphere anyways? $\endgroup$
    – Patrick.B
    Nov 22, 2018 at 22:56
  • 2
    $\begingroup$ As said the issue is that the coordinates don't cover properly $\rho =0$ (you can see the metric is degenerate there). $R^3$ is not equivalent to $R \times S^2$ but it is if you remove a point from it. $\endgroup$
    – Slereah
    Nov 22, 2018 at 22:58
  • $\begingroup$ @Slereah If you do not remove the point then we have got foliation of $R^3$ by $S^2$ spheres and if we do the fibration? Is that right? $\endgroup$
    – JanG
    Aug 26, 2023 at 10:09
  • $\begingroup$ @Patrick.B Another way to look at this is to say your example is a noncoordinate frame. Basically that means your basis 1-forms $e^a$ are closed but not exact i.e there doesn't exist $\phi$ such that $e^{\phi}=d\phi$ you'll see this a lot if you play with orthonormal frames on curved manifolds/spacetimes. $\endgroup$
    – R. Rankin
    Sep 9, 2023 at 8:35

First I would like to stress that you cannot infer the topology of a manifold from its metric. Reasons:

Topology is a global property, metric is a local property

You can squeeze and stretch a manifold without changing the topology, but these actions will definitely change its metric, so metric and topology are largely independent.

If you are only given a single coordinate system (with or without metric) you do not know enough about the manifold to draw any conclusion regarding its topology. You need an atlas to describe a manifold properly. This means of course that you must know the coordinate transformations linking the different coordinate systems.

So any topological claim made in the text is based on other considerations, not from the form of the metric.

Now your questions. 1- as you guessed Riemannian simply means that the metric signature is positive definite while pseudo-Riemannian means the metric is not positive definite. However some authors (eg. MTW) do not make this distinction.

2- yes you are basically right. You are given a 3d manifold $S(\rho,\phi,\theta)$ and its metric, you embed it into $R^4(y^0,y^1,y^2,y^3)$ with its standard metric. This is like a standard 2d Surface immersed into standard $R^3$ space. You have to find/guess the map $f:S(\rho,\phi,\theta) \to R^4(y^0,y^1,y^2,y^3)$ such that the standard metric on $R^4$, when pulled back gives the given metric. You can imagine that the 3d surface has an axial symmetry, so what you need is an axially symmetric 3d Surface $y^0=g(\rho)$. This is well described in general in MTW page 613 and page 837

  • $\begingroup$ "you cannot infer the topology of a manifold from its metric" true..in dimension greater than 2 $\endgroup$
    – R. Rankin
    Sep 9, 2023 at 8:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.