The Einstein-Rosen wormhole is described by the metric $$ \begin{equation} \mathrm{d} s^{2}=\frac{u^{2}}{u^{2}+2 m} \mathrm{d} t^{2}-4\left(u^{2}+2 m\right) \mathrm{d} u^{2}-\left(u^{2}+2 m\right)^{2} \mathrm{d} \Omega^{2}, \end{equation}$$ where $\mathrm{d}\Omega^{2}=\mathrm{d}\theta^2+\sin^2\theta \mathrm{d}\phi^2$ and $-\infty<u<\infty$. It is also the $X=0$ slice of the Kruskal coordinate, $$\begin{equation} \mathrm{d} s^{2}= \frac{16 m^{2}}{r} e^{ (-r/2 m)} \mathrm{d} T^{2}-\frac{16 m^{2}}{r} e^{ (-r/2 m)} \mathrm{d} X^{2} -r^{2}\left(\mathrm{d} \theta^{2}+\sin ^{2} \theta \mathrm{d} \phi^{2}\right). \end{equation}$$

In principle, a wormhole could connect two spacetime or a single spacetime. I know that the metric doesn't determine the topology of spacetime, the manifold of a wormhole connecting two spacetime and one spacetime clearly has a different topology.

Here are my questions: Is the $X=0$ slice in the Kruskal diagram a wormhole connecting two different spacetimes or one spacetime? if it is a wormhole connecting two spacetimes, how to get a wormhole solution that connects the same spacetime?


1 Answer 1


The wormhole itself is regarded as the common boundaries of one or two manifolds when identified with each other to glue them together. Therefore the (three-dimensional) hypersurface with $r=r_\mathrm{S}$ (or $u=0$ after the shift) for the Schwarzschild wormhole (and not some random interval around them). It is not possible putting this into the formulas for the Kruskal coordinates directly, which are not defined for $r=r_\mathrm{S}$ put taking the limit $r\rightarrow r_\mathrm{S}$, you indeed get $X=0$ and $T=0$.

In your case, there are two seperate manifolds/universes glued together. Consider the Schwarzschild metric: Topologically, you remove a singularity from $\mathbb{R}\times\mathbb{R}^3$ to get $\mathbb{R}\times(\mathbb{R}^3\setminus\{0\})$ and using $\mathbb{R}^3\setminus\{0\}\cong S^2\times\mathbb{R}^+$, you get $\mathbb{R}\times S^2\times\mathbb{R}^+$ with boundary $\mathbb{R}\times S^2$, basically corresponding to the surface of the blackhole going through time. Now consider the Schwarzschild wormhole: Topologically, you take two such manifolds $\mathbb{R}\times S^2\times\mathbb{R}^+$ and glue them together at $\mathbb{R}\times S^2$ to get $\mathbb{R}^2\times S^2$.

Glueing together boundaries of one or two manifolds is described by the trace and composition respectivly in cobordism theory, if you want to look that up. Cobordisms are manifolds connecting two closed manifolds with one dimension less, so for example the cylinder $M\times[0;1]$ connecting $M$ with $M$. If there is a common boundary of one cobordism and itself (like in the example just given) or with another, then you can glue them together just like above for two copies of the Schwarzschild metric, so for example getting the torus $M\times S^1$ connecting $\emptyset$ with $\emptyset$ in the example just given.

Intra-universe wormholes connecting the same universe with itself are also possible and can be guaranteed by topology. There is a theorem in "Lorentzian spacetimes" by Matt Visser which is written down on Wikipedia to guarantee their existence. As you can see there, the boundary of the manifold goes into the theorem:

If a Minkowski spacetime contains a compact region $\Omega$, and if the topology of $\Omega$ is of the form $\Omega\cong\mathbb{R}\times\Sigma$, where $\Sigma$ is a three-manifold of the nontrivial topology, whose boundary has topology of the form $\partial\Sigma\cong S^2$, and if, furthermore, the hypersurfaces $\Sigma$ are all spacelike, then the region $\Omega$ contains a quasipermanent intrauniverse wormhole.

  • $\begingroup$ So you don't need to even find the wormhole solution, you could deduce the existence of the wormhole direct from the topology of the whole spacetime, is this right? $\endgroup$
    – David Shaw
    Commented May 23, 2023 at 15:56
  • $\begingroup$ You definitely can for intra-universe wormholes. When you take a region $\Omega$ of the universe, that contains both ends of the wormhole, it won't be simply connected; meaning that its fundamental group does not vanish (it will be $\pi_1(\Omega)\cong\mathbb{Z}$). The fundamental group is measured and described by (algebraic) topology and not geometry. For wormholes connecting two universes, this no longer applies, take the Schwarzschild wormhole $S^2\times\mathbb{R}^2$ for example, which is simply connected with $\pi_1(S^2\times\mathbb{R}^2)\cong\pi_1(S^2)\times\pi_1(\mathbb{R}^2)\cong 1$. $\endgroup$ Commented May 24, 2023 at 14:17
  • $\begingroup$ If you want to see traces of the wormhole there, you probably need to consider higher homotopy groups, in this case the second homotopy group $\pi_2(S^2\times\mathbb{R}^2)\cong\pi_2(S^2)\times\pi_2(\mathbb{R}^2)\cong\mathbb{Z}$ does not vanish and again yields the $\mathbb{Z}$ we got for the intra-universe wormhole above. $\endgroup$ Commented May 24, 2023 at 14:17
  • $\begingroup$ Thank you very much, do you have any references related to what you mentioned? $\endgroup$
    – David Shaw
    Commented May 24, 2023 at 15:34
  • 1
    $\begingroup$ I think arxiv.org/abs/gr-qc/9704082 could be best for you. In the description, you find: "Here, we show that topology is too limited a tool to accurately characterize a generic traversable wormhole." Given you asked about the Schwarzschild wormhole, you are probably interested in traversable wormholes. There are also wormholes with trivial topology, for example seen in an image in that paper connecting Minkowski spacetime with the FLRW metric: researchgate.net/figure/… $\endgroup$ Commented May 26, 2023 at 10:27

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