# How to "topologically neutralize" this curious manifold?

Consider a Lorentzian manifold completely covered by the following coordinate system:

$$r\in[-\frac{L}{2},\infty) \\ \theta\in[0,\pi]\text{ when }r\ge\frac{L}{2} \\ \theta\in[0,\frac{\pi}{2}]\text{ when }r\lt\frac{L}{2} \\ \phi\in[0,2\pi) \\ t\in(-\infty,\infty)$$

Now, we identify each point $$(-\frac{L}{2},\theta,\phi,t)$$ with the point $$(\frac{L}{2},\pi-\theta,\phi,t)$$. (Remember, this only includes $$\theta\in[0,\frac{\pi}{2}]$$.)

Then, we impose the following metric. This metric is just an ansatz to come up with the basic shape, so I'm not too worried about the particular formulas -- in a second I'll explain what's really going on.

$$g_{tt} = -(1-\frac{r^2}{\alpha^2}) \\ g_{rr} = (1-\frac{r^2}{\alpha^2})^{-1} \\ g_{\theta\theta} = e^{-kr^2}\sin(\frac{5\pi}{2}e^{-r^2}) + r^2 - C \\ g_{\phi\phi} = [a(r)\sin^2{2\theta} + b(r)\sin^2{\theta}]\ g_{\theta\theta}$$

where $$k$$ and $$C$$ are chosen to give $$g_{\theta\theta}$$ two zeros at $$r=\pm\frac{L}{2}$$, and

$$a(r) = \begin{cases} 1 & r\le \frac{L}{2} \\ 1 - e^{-\frac{1}{(r-\frac{L}{2})^2}} & r\gt \frac{L}{2} \end{cases} \\ b(r) = \begin{cases} 0 & r\le \frac{L}{2} \\ e^{-\frac{1}{(r-\frac{L}{2})^2}} & r\gt \frac{L}{2} \end{cases} = 1 - a(r)$$

You can see the shape of these functions here, where I have renamed $$g_{\theta\theta}$$ to $$g_{hh}$$.

To understand what this manifold is, first note that, because $$g_{rr}$$ is only a function of $$r$$, the $$r$$ coordinate lines are also geodesics, at least with respect to a spatial slice (hypersurface of constant $$t$$). Now consider the cross-sections of these radial geodesics (surfaces of constant $$r$$ and $$t$$). As $$r \to \infty$$, such a surface becomes a sphere. But as $$r \to \frac{L}{2}^+$$, it becomes a pair of spheres, because of the $$\sin^2{2\theta}$$ in $$g_{\phi\phi}$$ -- one for $$\theta\in [0,\frac{\pi}{2}]$$ (call this $$S_1$$) and another for $$\theta\in [\frac{\pi}{2},\pi]$$ (call it $$S_2$$). And when $$r\lt\frac{L}{2}$$, it is again a single sphere, because we lose half the range of $$\theta$$. These single spheres are the natural continuation of the $$S_1$$ across $$r=\frac{L}{2}$$, but they are also the continuation of the $$S_2$$ across the identification mentioned earlier!

In other words, in spite of its strange coordinate system, this manifold is both smooth and closed. What's really going on is that the region $$r\in[-\frac{L}{2}, \frac{L}{2}]$$ is a "bridge" that connects $$\theta\in[0,\frac{\pi}{2}]$$ to $$\theta\in[\frac{\pi}{2},\pi]$$ via the point at $$r=\frac{L}{2}$$ -- which is a single point because $$g_{\theta\theta}=0$$ here -- this point is the origin of a spherical coordinate system where $$r\in[\frac{L}{2},\infty)$$. Since both ends of the bridge are at the same point, it can also be thought of as a loop.

Note that this manifold is asymptotically a de Sitter space, according to this formula. (Except for the offset of $$C$$ in $$g_{\theta\theta}$$, but I don't think that's really an issue as long as it varies like $$r^2$$.)

Now here's what I want to do to this manifold. In spite of the loop that it contains, I want to distort the metric in such a way as to make it "topologically neutral", in the sense that, when you take its Chern-Gauss-Bonnet integral, you get the same value as de Sitter space alone, if it didn't have this loop structure in it. And I want this spacetime to remain both smooth and stationary, ie. $$g_{\mu\nu}$$ should be independent of $$t$$, though I presume it will now contain off-diagonal terms. While doing this, we can allow $$L$$ to vary, but $$\alpha$$ should be taken as given.

This is inspired by the idea that you can induce a loop in a flat belt without lifting its ends, as long as there is a compensating $$360^{\circ}$$ twist along the belt. So how might the "twist" manifest, if the analogy holds? I was imagining that the eigenvectors of the Ricci tensor could twist about those radial geodesics, relative to the Levi-Civita connection (eg. if $$\vec v$$ is a Ricci eigenvector field, then $$\nabla_r{\vec v}$$ is roughly perpendicular to both $$\vec v$$ and $$\hat r$$).

But aside from such vague notions, I haven't the slightest clue how to try modifying this metric to achieve the desired "topological neutrality". As far as I'm concerned, any modification is as good a guess as any other. Yet I cannot possibly get anywhere without narrowing down the field of candidates.

I don't need a closed form solution, either; I assume this is more likely to be resolved by a numerical simulation. Also, I'm very interested in whether the condition of "topological neutrality" limits the possible solutions to a discrete set. But I also want to find the solution(s) themselves, approximate or not.

So any insight or general thoughts are welcome and greatly appreciated! And thank you for taking the time to read this!

• What exactly do you mean by "geodesics from a certain half-space"? Do you mean there is a totally-geodesic disk $D\subset M^3$ and geodesics meeting $D$ orthogonally? And what does it mean "a half-space at $x$"? Do you mean there is a totally-geodesic hypersurface containing $x$ and $D$? Feb 26, 2021 at 8:49
• @MoisheKohan Sorry. What if I said this: there is a 3D sub-manifold $N^3$, shaped like a worm. Both its head and its tail terminate at $x$, and that is the only point where it touches the rest of $M^3$. You would think $N^3$ has a boundary, but it doesn't; geodesic sections form spheroids rather than disks, thanks to curvature. There is a tangent plane $P$ at $x$, such that all geodesics passing through $x$ from the head of the worm pass to one side of $P$, and those from the tail pass to the other side. For this to work, I believe all geodesics must be parallel as they pass through $x$. Feb 26, 2021 at 15:58
• @MoisheKohan Actually, I know what you'll say: if all directions are parallel at $x$, it has no tangent planes. So instead, consider the extensions of all geodesics from the head of $N^3$ through $x$ into $M^3$. Call this $H^3$. And the extensions of all geodesics from the tail form $T^3$. Then, I want to say that, for sufficiently large distance from $x$, the boundary between $H^3$ and $T^3$ approaches a plane. $H^3$ and $T^3$ are what I was calling the half-spaces of $x$. Intuitively, I'm trying to say $N^3$ is a geodesic bridge between those two half-spaces, but only at $x$. Feb 26, 2021 at 17:11
• @MoisheKohan Sorry to spam you, but I finally realized something. The geodesics from the surface that divides the half-spaces are the ones that would have formed the boundary of $N^3$, so only these must become parallel at $x$, so that their extensions into $N^3$ coincide, thus keeping it closed. The geodesics from $T^3$ or $H^3$ do not become fully parallel, and that's how we can distinguish the "dividing plane". Feb 26, 2021 at 19:04
• I have to say, I do not understand your explanations at all. Did you read any Differential Geometry textbooks? (Part of the problem is that you are not using math terminology or using it in a nonstandard fashion.) Here is just one question: What you you mean when you say "parallel geodesics at $x$"? Do you mean, their velocity vectors (at $x$) are parallel? If so, the answer becomes very simple: The thing you are aiming for cannot happen. Feb 26, 2021 at 20:10