# Are Minkowski and Schwarzschild spacetimes diffeomorphic?

Another mathematical question, arising from GR. Some days ago I wrote, in an answer to 1, that they are. Then @magma commented they are not. He promised a proof, but none appeared. After magma's comment I have some doubts about my intuition and I would like to see a proof, in one sense or the other, or at least a reliable reference.

• Are they even homeomorphic? Naively, it seems to me that Schwarzschild is topologically $\mathbb{R}^2 \times S^2$, while Minkowski's just $\mathbb{R}^4$. Isn't this true? – coconut Oct 2 '18 at 10:29
• Intuitively, they can't be. Minkowski space has a trivial second homotopy group ($\pi_2=0$), while Schwarzchild spacetime has an entire line excluded (the singularity's worldline) and therefore $\pi_2=\mathbb Z$ – Gabriel Golfetti Oct 2 '18 at 10:33
• Hi Elio, it's a pleasure finding you here. I am travelling, I am in Frankfurt airport. My final destination is the Max Planck Institute in Laipzieg. – Valter Moretti Oct 2 '18 at 11:17
• Hi Elio, I finally managed to answer enumaris question :-) My argument is basically the same used here – magma Oct 3 '18 at 0:23
• @Valter Moretti Hi Valter, it's a pleasure for me too, and it was a surprise. As to your answer and the others', I suspected that the catch was in S^2. I see that in KS spacetime the spheres cannot be shrunk to a point. However can't clearly see how things go in Minkowski. What troubles me is that r=0 is a regular line in the manifold, but a coordinate singularity. So I can't see what happens. As you're reading, did you see my question about diagonalizing Boyer-Lindkquist metric? – Elio Fabri Oct 6 '18 at 16:08

They aren't. That is because Kruskal-Schwartzschild is diffeomorphic to $$S^2 \times R^2$$, while Minkowski is $$R^4$$. This is roughly similar to the difference between a 2D cylinder $$S^1\times R$$ and $$R^2$$. The formal reason is that pointed out by G. Golfetti in a comment above: the second homotopy groups are different, as is expected by the presence of the factor $$S^2$$. In M. space you can shrink two-spheres continuously into points, in K-S space you cannot: when you encounter the bifurcation surface (the 2D event horizon) you cannot go further and you pass from the right wedge to the left wedge. The standard global Kruskal coordinates map K-S manifold to $$S^2\times R^2$$. There is no way to make vanishing the radial coordinate. It would vanish on the singularity, but it does not belong to the manifold. At the center of the manifold it takes the Schwarzshild radius value which is strictly positive. There is no center of those spheres. Similar coordinates can be defined on Minkowski spacetime but they are not global since they do not cover the set $$r=0$$ which is part of the manifold.
• Valter I gave an answer along these same lines elsewhere. But I would like to see if we are really on the same wavelength. Given KS coord. {$T,X,\phi,\theta$},could you please specify which pair maps to $S^2$ and which one maps to $R^2$ ? – magma Oct 3 '18 at 0:18
• Dear Magma I am at a conference now. I will try to specify my answer when I am home again. I think that I referred to partially different coordinates and my radial coordinate is a coordinate smoothly matching the Schwarzschild coordinate $r$. However, everything can be recast using the coordinates you point out. The first couple maps to $R^2$ and the second one to $S^2$ (up to a small set...) – Valter Moretti Oct 3 '18 at 5:17