I Euclidean geometry and gravity

[pervect]

Is that "flat plane" two-dimensional or three-dimensional?

Two dimensional

 

[PeterDonis]

Then I read a reference to 'frame dragging' in rotational black holes. What effect, if any, will that have on the 'paper'?

If the paper is "hovering" statically, i.e., no radial or tangential motion, frame dragging has no effect.

 

[pervect]

In the side discussion of embedding a piece of a plane smoothly, isometrically, in a 3-sphere (following is part of a response to @pervect 's last post:

I thought I would make explicit the construction in that reference (which is presented it very compactly in any number of dimensions). Making it explicit for flat 2-surface section in a 3-sphere, the mechanism is simply to use a properly sized flat torus. As explained in:
https://en.wikipedia.org/wiki/Clifford_torus
a flat torus of just the right size will smoothly fit in a unit 3-sphere, both embedded in R4. But then, the embedding can be just ignored, and you are left with a smooth embedding of a piece of Euclidean plane in a 3-sphere.

The formalization and generalization of this is the general approach to embedding $E^n$ into higher dimesnsional spheres used in the reference I quoted earlier. The general achievement is a local embedding of $E^n$ into an m-sphere, with $m=2n-1$.

Thank you. Untangling this in terms I can more fully understand will be quite a problem. I'm suspecting the issue is fundamental in that I'm viewing curvature as being defined by a metric. Which isn't necessarily how mathematicains view it.

But I think the discussion has moved outside of what's helpful to the OP's question.

 

[PeterDonis]

I'm viewing curvature as being defined by a metric.

More precisely, by the Riemann curvature tensor that's derived from a metric via its Levi-Civita connection. There's no problem with that--that's how the term is being used in this discussion.

But the metric of what? Take the case of a 2-sphere embedded in Euclidean 3-space. The 2-sphere is curved. The 3-space is flat. But one is embedded in the other. How can that be? Taking what you appear to be trying to say at face value, that should be impossible.

The reason it's possible is that the 2-sphere embedded in Euclidean 3-space is a submanifold of that 3-space, and we can define a metric on the submanifold that is curved even though the metric of the 3-space overall is flat. They're simply two different metrics. Note that those metrics give different answers to questions like the distance between two points, even if we restrict the points under consideration to those on the 2-sphere.

What @PAllen has been describing is simply a case where that works in reverse: you have a curved 3-manifold, the 3-sphere, with a submanifold, the Clifford 2-torus, on which a flat metric can be defined. Again, you simply have two different metrics.

 

[PAllen]

The reason it's possible is that the 2-sphere embedded in Euclidean 3-space is a submanifold of that 3-space, and we can define a metric on the submanifold that is curved even though the metric of the 3-space overall is flat. They're simply two different metrics. Note that those metrics give different answers to questions like the distance between two points, even if we restrict the points under consideration to those on the 2-sphere.

When discussing isometric embeddings, there is a little more to it. The metric of the submanifold is taken to be induced from the manifold metric. This means that, e.g. the distance along a curve on the 2-sphere embedded in E3 is, in fact required to be that same whether you use the manifold metric or the induced metric on the 2-sphere. You get to choose manifold, its metric, and the definition of the submanifold embedded in it as a set of points. You do not then get to choose an arbitrary metric on the submanifold - it is taken to be that induced by the manifold metric.

 

[PeterDonis]

the distance along a curve on the 2-sphere embedded in E3 is, in fact required to be that same whether you use the manifold metric or the induced metric on the 2-sphere.

Yes. It's important to recognize also that the curve in question is a geodesic of the curved 2-sphere, but it is not a geodesic of the Euclidean 3-space. Considering the implications of that might help @pervect to see what's going on in such cases.

 

[Jaime Rudas]

a flat torus of just the right size will smoothly fit in a unit 3-sphere

What does "unit" mean in this context?

 

[PAllen]

What does "unit" mean in this context?

Radius of curvature = 1

 

[raagamuffin]

(This is an observation, not a complaint. )
As I examine these ideas , every insight I gain spurs more questions about relationships between entities, which (seems like it) can’t be answered by just knowing some facts. I’m realizing that to really make inferences , a good understanding of the math is unavoidable.
(It’s a shame that the universe can’t stick to algebra, some probability and trig )
Thanks for all your explanations, I may just have to consider more rigorous study.

 

[PAllen]

I can’t seem to let this go, partly because some have asked for how to make the embedding result (part of Euclidean plane in 3-sphere) more comprehensible as well as using only intrinsic metric tools. This post will attempt to address both of these goals head on.

I will note, in passing, that these concepts lead to much confusion on the internet. If I ask a couple different AIs these questions, they state that you can’t smoothly isometrically embed any part of hyperbolic plane in Euclidean 3 space without crinkling or folding, nor can you embed any part of a Euclidean plane in a 3-sphere without crinkling or folding. Both of these claims are flat out wrong, but the AIs are only partly to blame – there are a shocking number of superficially reputable websites that make these false claims. I will make no further comment on the hyperbolic plane issue except to note that the pseudo-sphere is a trivial refutation of the false claim (the true claim, proved long ago by Hilbert, is about the impossibility of a global embedding – the complete unbounded hyperbolic plane in $E^3$ (Euclidean 3-space)). The rest of this post will focus on clarifying the case of the Clifford torus in a 3-sphere, which disproves the false statements about embedding part of a Euclidean plane in a 3-sphere.

To start addressing how to make the result more intuitive, consider first, surfaces of revolution in $E^3$ . If you consider any straight line (geodesic) in the xz plane (standard cartesian coordinates) and rotate around the z axis, you get different embeddings of $E^2$ (Euclidean plane) – different from the obvious one. These are cones, inverted cones, or cylinders, all Euclidean flat. Now consider a curve in xz bent away from a straight line with either possible concavity. One of these will produce surfaces of revolution with positive intrinsic curvature, the other with negative intrinsic curvature (and the tractrix will produce a surface with constant negative curvature).

What we will show for the 3-sphere, directly using hyperspherical coordinates and the 3-sphere metric, is that there is a curve in the $\theta\phi$ ‘plane’ (surface of constant $\psi$) that when rotated by $\psi$ produces a surface with zero curvature – flat intrinsic metric. Of course, it is not a geodesic of the 3-sphere. Relative to this curve, if a curve bends more strongly one direction, you get surfaces of revolution of positive intrinsic curvature, if bending more the other way, you get surfaces of negative intrinsic curvature. Thus, for example, it is possible to smoothly isometrically embed a piece of hyperbolic plane in a 3-sphere, just as in $E^3$.

Surprisingly, it will be shown that a complete flat torus of just the right size smoothly, isometrically embeds in $S^3$ (3-sphere), even though this is impossible in $E^3$.

The hyperspherical coordinates form a ‘nearly complete’ chart of a 3-sphere with the following description (the coverage of this chart is sufficient for our purposes, so we need not worry about incompleteness):

$$ds^2=R^2[d\theta^2+\sin^2\theta~d\phi^2+\sin^2\theta\sin^2\phi~d\psi^2]$$

With theta and $\phi\in(0,\pi)$, $\theta\in(0,\pi)$,$\psi \in [0,2\pi)$.

Henceforward, we will consider only the unit 3-sphere, with R=1.

Consider a curve in a surface of constant $\psi$ defined by $\sin^2\theta\sin^2\phi=\frac 1 2$. [This is suggested by looking at the situation from joint embedding in $E^4$, but now fully divorced from that embedding]. It is easy to verify that within the stated coordinate bounds, this forms a closed curve (sort of a rounded square). We consider the surface revolution formed by taking this curve through all values of $\psi$. We will demonstrate that the intrinsic geometry of this surface, induced by the spherical metric above, is Euclidean flat. No use of embedding in $E^4$ or any tools beyond the metric will be used. This is, in fact, a complete, smooth, isometric embedding of a flat torus in $S^3$. Note, that only this one size of flat torus can be completely smoothly embedded.

From the curve definition above, we get by differentiation:
$$\sin^2\phi\cos^2\theta~d\theta^2 = \sin^2\theta\cos^2\phi~d\phi^2$$
Or
$$d\phi^2 = \tan^2\phi\cot^2\theta~d\theta^2$$
From the curve definition, we also have $\sin^2\phi=\frac 1 {2 \sin^2\theta}$. From these, after some rearrangements we get:
$$d\phi^2 =\frac {\cot^2\theta} {\sin^2\theta-\cos^2\theta} d\theta^2$$
Plugging this into the spherical metric we get
$$ds^2 = d\theta^2 + \frac {\cos^2\theta} {sin^2\theta-\cos^2\theta} d\theta^2 + \sin^2\theta\sin^2\phi~d\psi^2$$
Using the curve definition in the last term, and rearranging we get:
$$ds^2 = \frac {\sin^2\theta} {\sin^2\theta-\cos^2\theta} d\theta^2 + \frac 1 2 d\psi^2$$
This is the intrinsic metric for the torus induced from the spherical metric, using the inherited hyperspherical coordinates on the torus. However, rather than using $\theta$, it would be better to use a coordinate that is proportional to arclength along the $\theta\phi$ curve. By looking back at the embedding in $E^4$, a guess can be made to try the following simple coordinate change:
$$\cos\theta=\frac 1 {\sqrt 2 } \cos z$$
We have then, $\cos^2\theta=\frac 1 2 \cos^2 z$, and $\sin^2\theta~d\theta^2 = \frac 1 2 \sin^2 z~dz^2$. Plugging this into the prior metric and rearranging gives:
$$ds^2 = \frac 1 2 dz^2 +\frac 1 2 d\psi^2$$
QED. This is the Euclidean flat metric. Recalling that coordinate change can’t change intrinsic geometry, we have demonstrated that the indicated torus surface within the unit 3-sphere is Euclidean flat everywhere.