On the wikipedia page for Kruskal-Szekeres coordinates, it states:

[Kruskal-Szekeres] coordinates have the advantage that they cover the entire spacetime manifold of the maximally extended Schwarzschild solution and are well-behaved everywhere outside the physical singularity.

In addition, when learning about singularities, it is often said that singularities are not part of the manifold - leading to incomplete geodesics (see e.g. Wald Chapter 9).

These two statements taken together seem to imply to me that KS coordinates cover the entire manifold in question and is well behaved everywhere on the manifold. The manifold in question though is (intrinsically) curved - the Schwarzschild solution has non-zero Riemann curvature. If KS coordinates cover the entirety of the manifold though, and is "well behaved" (which I assume here to mean it remains a 1-1 smooth mapping with inverse) everywhere over the manifold, then how can the manifold be curved? A 1-1 smooth mapping with inverse of the manifold onto $\mathbb{R}^4$ would imply that the manifold is diffeomorphic to $\mathbb{R}^4$ and therefore flat wouldn't it?

A lay out of the argument would be thus:

  1. $\mathcal{M}$ is the manifold corresponding to the Schwarzschild solution.

  2. The physical singularity does not exist on $\mathcal{M}$

  3. The Kruskal-Szekeres chart $\psi:\mathcal{M}\rightarrow\mathbb{R}^4$ covers all of $\mathcal{M}$ and is well behaved everywhere except at the physical singularity.

  4. From 2+3: the chart $\psi$ covers all of $\mathcal{M}$ and gives a 1-1, smooth, mapping of $\mathcal{M}$ onto $\mathbb{R}^4$

Therefore $\mathcal{M}$ is diffeomorphic to $\mathbb{R}^4$ and is flat.

The claim is obviously false, so what's the error in the logic? Does "well behaved" not mean what I think it means? Does the fact that a physical singularity "exists" albeit not on the manifold break the argument? I'm not really sure what other possibilities there are.

  • 3
    $\begingroup$ Diffeomorphic to $\Bbb R^n$ does not imply flat, btw. Take, for instance, the Poincaré metric on the unit ball. The Riemannian structure is strictly more than simply the structure of a smooth manifold. Riemannian manifolds that are isomorphic are called isometric. $\endgroup$
    – Danu
    Sep 26, 2018 at 23:07
  • $\begingroup$ This is a very good question enumaris +1. It deals with basic concepts in differential geometry that you exemplify with the Schwarzschild manifold. Please wait and see my answer coming shortly and my comment to Elio's answer below $\endgroup$
    – magma
    Sep 28, 2018 at 12:23

3 Answers 3


Your confusion starts, I believe, with your incorrect definition of "chart". You think that it is an homeomorphism $\psi:U\rightarrow\mathbb{R}^n$ where $U$ is an open subset of $\mathcal{M}$

The correct definition instead is that it is an homeomorphism $\psi:U\rightarrow V$, where $U,V$ are open subsets of $\mathcal{M}$ and $\mathbb{R}^n$, respectively.

Big difference!

For ex. $V$ needs not necessarily be homeomorphic to $\mathbb{R}^n$. As a consequence you may have a manifold $\mathcal{M}$ - covered by a single chart - that is not homeomorphic to $\mathbb{R}^n$. Example: the punctured plane $\mathbb{R}^2_O$ ($\mathbb{R}^2$ without the origin $O$), where the single chart $\psi:\mathbb{R}^2_O\rightarrow\mathbb{R}^2_O$ is just the cartesian coordinate system with $x,y\neq 0$ and $\mathcal{M},U,V=\mathbb{R}^2_O$ .

The second point to stress is that the KS chart {$T,X,\phi,\theta$} does not really cover all the Schwarzschild manifold $\mathcal{M}$, for the simple reason that the angular part $\phi,\theta$ does not cover $S^2$ completely ($0<\phi<2\pi,0<\theta<\pi$ so the poles and $\phi=0$ meridian are not covered). Now, the rectangle $0<\phi<2\pi,0<\theta<\pi$ is homeomorphic to $\mathbb{R}^2$ (using some simple arctan transformation) so we can say that on $S^2$ there is a chart $\psi:U\rightarrow\mathbb{R}^2$ where $U$ is a proper open subset of $S^2$

So we have a manifold $\mathcal{M}$ (the maximal extension of Schwarzschild manifold) with a chart $KS:U\rightarrow V$, where $U$ is a proper open subset of $\mathcal{M}$ and $V$ is:$$-\infty <X<\infty$$ $$-\infty <T^2-X^2<1$$ $$0<\phi<2\pi$$ $$0<\theta<\pi$$ and is homeomorphic to $\mathbb{R}^4$. So only a subset of $\mathcal{M}$ is homeomorphic to $\mathbb{R}^4$ by this argument. Big deal! So by looking at the KS chart you cannot really conclude whether Schwarzschild is homeomorphic to Minkowski for the simple reason that KS does not really cover Schwarzschild manifold completely.

In reality the 2 manifolds are not homeomorphic. One has no singularities, the other has a physical singularity ie. one that cannot be eliminated by a coordinates transformation. The best you can do (with KS) is to almost wrap the singularity with a nice chart like you do in $\mathbb{R}^2$ minus the origin and the positive x axis, using polar coordinates. Using algebraic topology language/methods you can prove that $\pi _2$ are different: you cannot shrink a closed surface $S^2$ around the singularity. You can read more about this here

  • $\begingroup$ Thank you for this answer, it is generally quite clear. As I understand it (from this answer and others), a summary of the errors in my argument would be basically that 1. There's an error in my understanding of "well behaved everywhere except the singularity" - the coordinates fail at the spherical pole as well. 2. Even neglecting point 1, I only mapped $\mathcal{M}$ to a subset of $\mathbb{R}^4$ and can not claim any homeomorphism to $\mathbb{R}^4$ itself. And 3. homeomorphism (and diffeomorphism) != isometric and so my claim of "flat" is false even if I showed a diffeomorphism. $\endgroup$
    – enumaris
    Oct 2, 2018 at 23:24

I believe the fundamental flaw in your reasoning is that the definition of a manifold is a more primitive notion than that of a metric, and it is the metric that gives in general relativity the notion of curvature. The clearest way to say that it that $R^4$ is not actually flat, simply because it does not have automatically a metric associated to it; it is neither flat nor curved on its own, because it is not automatically armed with enough structure to define those things. In other words, the fact that there is a smooth 1-1 mapping between $R^4$ and the maximally extended Schwarzschild solution via the Kruskal-Szekeres coordinates simply states the fact that the Schwarzschild solution lives in a 4-dimensional spacetime. It does not require, in any way, that it also possesses the geometry of flat spacetime, because $R^4$ is not flat unless you add to it the structure of Minkowski spacetime -- which, in itself, is a choice.

  • $\begingroup$ So when Wald says "Diffeomorphic manifolds have identical manifold structure." at the end of section 2.1 on page 14, he is not including the curvature of the manifold as part of the "manifold structure"? I suppose this is a very long held misconception I've had. Do you know of any resources which talk about what properties of the manifold specifically gets mapped over and what doesn't in a diffeomorphism (e.g. the dimensions of diffeomorphic manifolds must be the same, of that I'm sure)? $\endgroup$
    – enumaris
    Sep 26, 2018 at 23:10
  • $\begingroup$ Since a diffeomorphism is just a fancy way of talking about a coordinate change, then I guess you would be technically right to say that the curvature should also be unchanged after the diffeomorphism. The point I raised, however, is a bit more subtle: I am claiming that $R^4$ by itself does not carry any intrinsic notion of curvature; it's just a set of coordinates. If you claimed that the Schwarzschild solution is diffeomorphic to Minkowski spacetime, than that would mean that the Schwarzschild solution would have to be flat, but this claim is false. $\endgroup$ Sep 26, 2018 at 23:25

You must keep distinct three levels of manifold structure:

  • differential $C^\infty$ (smooth)
  • affine connection
  • riemannian (or semiriemannian).

Diffeomorphism belongs to the first level. Curvature can be defined in the second (metric is not required). GR spacetime belongs to the third (semiriemannian). In a (semi)riemannian manifold an affine conection can be defined, compatible with the metric: the so-called Levi-Civita connection.

Therefore two manifolds may be diffeomorphic even if they differ as to curvature, like Minkowsky and Schwarzschild. There is a stronger morphism between (semi)riemannian manifolds: isometry. Two (semi)riemannian manifolds are isometric if a one-one mapping exists, preserving metric.

Note that isometry can be stated without using coordinates: it only requires that metric tensor of the first manifold goes into the metric tensor of the second. Of course if two manifolds are isometric, using corresponding coordinates in both the components of the metric tensors are the same.

As to Minkowsky and Schwarzschild, they are diffeomorphic but not isometric. Existence of incomplete geodesics in Schwarzschild geometry and not in Minkowsky's is an immediate proof.

  • $\begingroup$ Elio, Minkowsky and Schwarzschild manifolds are NOT diffeomorphic. Please see my answer coming in shortly $\endgroup$
    – magma
    Sep 28, 2018 at 12:00
  • $\begingroup$ @magma I must confess that I know no proof for my statement. I relied on my intuition, and waiting for your answer tried to understand where it led me astray. Maybe I have found, but I look forward to see a real proof. $\endgroup$
    – Elio Fabri
    Sep 28, 2018 at 19:25
  • $\begingroup$ @magma it seems that both answers (and the comment) so far seem to point to "diffeomorphic != isometric" as the problem with my argument. Whereas you are saying that my argument does not even show a diffeomorphism. This is interesting, and I look forward to your answer. I suppose my argument may be flawed in many different respects, and it would be helpful to have all of them pointed out to me. $\endgroup$
    – enumaris
    Sep 28, 2018 at 23:09
  • $\begingroup$ Elio and enumaris, thank you for your patience. I feverishly worked on the answer on Friday, but I wanted to refine some of my arguments, so I took the Saturday off. I will hopefully complete it on Sunday. $\endgroup$
    – magma
    Sep 29, 2018 at 22:37
  • $\begingroup$ @ElioFabri , enumaris Finally I posted my answer below. delay due to long airflights+no wifi airports+jet lag+...some swimming-pool-induced-laziness :-) I wanted to stress why the "single chart" argument is not really meaningful $\endgroup$
    – magma
    Oct 2, 2018 at 23:14

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