# Can $\mathbb{R}^4$ be globally equipped with a non-trivial non-singular Ricci-flat metric?

I'm self-studying general relativity. I just learned the Schwarzschild metric, which is defined on $$\mathbb{R}\times (E^3-O)$$. So I got a natural question: does there exist a nontrivial solution (other than Minkowski spacetime) to the vacuum Einstein field equation on a manifold diffeomorphic to the entire $$\mathbb{R}^4$$ (i.e., without singularity)? In other words, is there a Lorentzian metric is Ricci-flat but not Riemann-flat on $$\mathbb{R}^4$$?

• Any gravitational wave solution Commented Nov 21, 2023 at 23:57
• 'Schwarzschild metric defined on $\mathbb{R}\times (E^3-O)$'. I have thought that Schwarzschild metric (exterior solution) is defined outside of the matter sphere, i.e. in vacuum only?
– JanG
Commented Nov 22, 2023 at 16:46
• @JanG When the star gravitational collapses, it becomes a singular point, it is a vacuum except at r=0. Commented Nov 22, 2023 at 17:07
• @Victor I bag your pardon but a singular point does not make sense. It is much more a missing point in manifold what in turn contradicts manifold's definition.
– JanG
Commented Nov 22, 2023 at 17:16
• @Victor In case of black hole the static Schwarzschild metric ends on the $S^2$ sphere with curvature radius $R=r_{S}$.
– JanG
Commented Nov 22, 2023 at 17:20

Of course. This is easily achieved by what is known as transport of structure.

Consider ANY $$d=(n+1)$$-dimensional Lorentzian manifold $$(M,g)$$. For any point $$p\in M$$ by definition of smooth manifolds, we can find an open set $$U\subset M$$ containing $$p$$, and a diffeomorphism $$\Phi:U\to\Bbb{R}^{d}$$. Now, equip $$\Bbb{R}^d$$ with the Lorentzian metric $$g_{\Phi,U}:=\Phi_*(g|_U)$$, i.e the pushforward by $$\Phi$$ of the Lorentzian metric $$g|_U$$ restricted to $$U$$. Then, by definition, we have ensured that $$\Phi$$ is an isometry between $$(U,g|_U)$$ and $$(\Bbb{R}^d,g_{\Phi,U})$$. In particular, whatever curvature behavior is exhibited by $$(U,g|_U)$$, is also exhibited by $$(\Bbb{R}^d,g_{\Phi,U})$$. This is extremely general and we can cook up several examples this way. In particular, the take-away here is that simply requiring your underlying manifold be all of $$\Bbb{R}^4$$ is not a strong enough restriction to say anything about the behavior of metrics. In particular, it alone does not imply Riemann-flatness.

So, if you have an example of a Ricci-flat but not RIemann-flat Lorentzian manifold (eg Schwarzschild), then take a suitable open set there, and transfer that structure to $$\Bbb{R}^4$$.

• You have to to a bit more work to glue together your metric defined on a chart to a metric on the entire $\mathbb{R}^4$ as the curvature tensors are not linear in the metric, that's not obvious. (And not all charts can be extended to in such a way, that their range is all of $\mathbb{R}^4$). Commented Nov 22, 2023 at 0:03
• @SebastianRiese I think you’re missing the point here; I’m not gluing anything anywhere. By definition of a smooth manifold, I can always find such a diffeomorphism. With a diffeomorphism, I can always push-forward tensor fields. To push further, if I really wanted to, by transport of structure, I could define everything (a suitable topology, a smooth atlas, and a Lorentzian metric) on Cantor’s ‘middle-thirds’ set $C$. Commented Nov 22, 2023 at 0:06
• @SebastianRiese fine another intermediate step: a smooth manifold by definition gives me a chart $(U_0,\Phi_0)$ and a diffeomorphism $\Phi_0:U_0\to V_0$ for some open $V_0\subset\Bbb{R}^d$. Next, $V_0$ is open, so by definition, it contains an open ball $B$. Now, I restrict to $U=\Phi_0^{-1}(B)$ to get a diffeomorphism $\Phi_1:U\to B$. Finally, open balls are diffeomorphic to $\Bbb{R}^d$ (a standard exercise), let $\Psi:B\to\Bbb{R}^d$ be any such a diffeomorphism. So by composing I finally get my diffeomorphism $\Phi=\Psi\circ\Phi_1:U\to \Bbb{R}^d$. Commented Nov 22, 2023 at 0:12
• No it's clear what you mean – take the metric and push it forward along the diffeomorphism. The problem I see, is that a chart is not necessarily surjective as a function $M \to \mathbb{R}^4$ – so that construction doesn't give a metric on the entire $\mathbb{R}$. Commented Nov 22, 2023 at 0:12
• @safesphere I gave a very general procedure. To specialize to physics, as I said in my last sentence, just look at Ricci-flat metrics. These then become solutions of the vacuum-Einstein equations, which is General Relativity unless you’d like to disregard all the physical insight we have gained from studying vacuum solutions. Commented Nov 22, 2023 at 0:47

Yes you can have space-time which is Ricci flat but still has Weyl curvature. Imagine two Minkowski space-times $$M^+$$ and $$M^-$$ separated by the null plane $$\nu=0$$. So $$M^+$$ (resp. $$M^-$$) corresponds to the region where $$\nu>0$$ (resp. $$\nu<0$$). Say, the metric in $$M^-$$ is given by $$ds^2_-= 2dud\nu-2d\zeta d\bar{\zeta}$$ But as soon as you cross the $$\nu=0$$ plane, you shift your $$u$$ coordinate to $$u-f(\zeta,\bar{\zeta})$$ so that your metric now looks like $$ds^2_+=2d\nu(du-f(\zeta,\bar{\zeta})\delta(\nu)d\nu)-2d\zeta d\bar{\zeta}$$. You can see that space-time is Minkowskian except at $$\nu=0$$ where you have delta function type singularity. This is an example of Impulsive wave. Ricci flatness implies $$\frac{\partial^2f}{\partial\zeta\partial\bar{\zeta}}=0$$ which means $$f=g(\zeta)+\bar{g}(\bar{\zeta})$$ for some function $$g$$, while the Weyl curvature is proportional to $$\frac{\partial^2f}{\partial^2\zeta}$$ and $$\frac{\partial^2f}{\partial^2\bar{\zeta}}$$. This is now a gravitational impulsive wave. Although there is an apparent delta function in metric components, it is only an artifact of choice of coordinates. The metric is still $$\mathscr{C}^0$$ at the junction.