I Euclidean geometry and gravity

[PeterDonis]

stresses ... from change of proper geometry.

You still haven't shown the math of how you pick these out from other stresses. And, as I pointed out in a post some time ago, and reinforced with my post #100 just now, I don't think that's even possible. The proper acceleration of each worldline in the congruence that describes the object is a single thing. There's no way to pick out what part of it is due to "stresses from change of proper geometry" and what part is due to other stuff. And there can't be any "other" stresses that aren't accounted for by the proper acceleration.

 

[PAllen]

I think their claim is more limited than that. De Sitter spacetime has a timelike KVF, and a Killing motion using that KVF can certainly be Born rigid.

What their equation (34) shows is that, in a spacetime which is not exactly de Sitter, but also contains other stress-energy besides the cosmological constant, the particular congruence they describe is not rigid; it has nonzero expansion. That in itself doesn't show that it's impossible to find a rigid congruence; the rest of that section appears to be trying to show that, but I'm not sure I understand the argument being presented there.

Look later in the paper. They show a wider class of possible born rigid objects when the slices are not flat. But, in any case, homogeneity is nowhere near enough to allow a a rigid congruence.

 

[pervect]

Regarding this, I have the following question: Is it possible to embed a flat (Euclidean) two-dimensional surface in a curved, three-dimensional space?

I have not read anything definitive on that, but my current thinking is that it's not possible in general, but it may be possible in specific cases.

The usual approach is not about embeddings. Instead, one starts with some curved manifold, and specifies a pair of vectors at this point. The pair of vectors pick out a specific two dimensional subspace of the higher (in general) dimensional tangent space.

The tangent space is an important concept in differential geometry. In two dimensions, one might envision a sphere as a two dimensional manifold, then one can visualize the tangent space of the sphere at some point as a plane tangent to the sphere at that point. We can see from this example then that the tangent space is different from the manifold - points in the tangent space are not points on the manifold (except perhaps for one shared point, if one uses the diagram).

Tangent spaces exist in higher dimensions than the easy-to-visualize case of the tanget plane to a sphere, and since we are talking about a higher dimensional case, we'll need to use a higher dimensional tangent space.

My argument basically is that if you can't find a pair of vectors that makes the sectional curvature vanish at some point, you can't possibly embed a plane at that point. I believe you are doomed before you start.

Being able to find such a pair of vectors does not create an embedded tangent plane as you are requesting - it's more like an ininitesimallyu small version of such an embedded plane, it is only flat in an infinitesimal region. "Growing" it to a finite size is another problem - my intiuition suggests it won't always be possible but I don't have a good argument.

Since this is the result of my thinking (which isn't as great as it used to be sadly), don't regard this post with the same confidence as you would from reading something in a textbook. (You were already not doing that, I imagine, but I'll say it anyways0.

There is also the possibility of miscommuniation do to not-shared concepts. I am basically assuming we have a manifold, with a pre-existing metric that represents "distances" between nearby points on the manifold, a concept that we are not allowed to change. Perhaps this assumption can be eased, I don't know, I'm too used to making it to think seriously about not making it. I'm further making an assumption about not only about the metric, but what's called the "connection". I'm using what's called the Levi-Civita connection, which has the prpoerty that "straight lines" are the shortest distance between two points.

If I were more of a mathemtician, I probably wouldn't be talking about distances, I'd be talking about connections. I'm very used to computing curvature from the metric, but a sneaky little recollection is telling me that it's actually computed from the connection. But I don't want to try to explain connection.

To give a counter-example, we can succesfully make flat maps of the curved earth, but they won't be to scale. If we redefine distance between points to match what's drawn on the map, then suddenly the earth is flat, not round.

And if we are sufficiently deep in mathematics, we can talk about curvature without mentioning distances at all. But I'm not used to doing that. I'm more along the lines of "distances determine geometry".

 

[PAllen]

My argument basically is that if you can't find a pair of vectors that makes the sectional curvature vanish at some point, you can't possibly embed a plane at that point. I believe you are doomed before you start.

This can't be right. By that logic, you can't embed a 2-sphere in Euclidean 3-space since it has positive scalar curvature and every such plane will have zero curvature. Thus, this appears to be irrelevant to the problem.

I would say the obvious place to start would be the Gauss-Codazzi equations.

Also, there is the possibility of non-smooth embeddings, which are much easier. Two examples of surfaces that cannot be smoothly embedded to Euclidean 3-space but can be embedded without this restriction are the flat torus and a surface of constant negative curvature.

 

[PAllen]

No, I never assumed the object will remain rigid during the transport towards the mass, which would be the equivalent to spinning up the disc.

In fact, I say the exact opposite: The object will definitely change its proper geometry during the transport.

Ok, sorry I misunderstood that.

But once it has arrived, we can keep it hovering as a static object that remains rigid, which would be the equivalent of spinning at a constant rate.

In both scenarios the proper geometry of the object changes from Euclidean (when far from the mass / non-spinning) to negatively curved (when near the mass / spinning).

This is where I start to differ, and we may never agree. An object is a world tube, and all separations to time plus 3-geometry are a matter of convenience. Essential physics is contained in the world tube as whole.

In both scenarios we can apply smoothly distributed external forces needed for the proper acceleration of all object parts, which eliminates stresses related to the non-inertial motion. But the stresses related to the change of proper geometry will remain.

Here, I see what I think your are getting at (more below) but don't think it is useful (or clear to others) to think so much in terms of stresses induced by change of proper geometry. More on this below.

I disagree. The proper geometry of an object, measured with rulers at rest to that object, has very much a physical meaning. It's this geometry that determines how much rigid material you would need to build that object in its current state, for example: static near the big mass or spinning with high circumferential speed.

If you build an already spinning spoke wheel from pre-fabricated segments, you will need more segments for the circumference, than for a non-spinning version of the same radius. This is not a matter of convention.

Here is where I see what you are getting at, I think. I would describe it more as follows:

An initially rigid body forced to undergo a motion (either through changing spacetime or something like change of rotation) incompatible with rigidity will deform. When in equilibrium in a new rigid state, it would be the same as an identically constructed object with no history. However, it would typically not be the same as a conceptually similar object would be constructed in the new location/state. This gets fuzzy, but I will clarify by example.

I think my slowness to understand this point was thinking in terms of homogeneous materials. It became much easier for me to see when dropping this bias. Consider your example of a spoked wheel being spun up (I think the motion toward a BH is essentially similar, which is why I initially proposed the spinning disc comparison). Imagine the rim is extremely stiff and the spokes are thin malleable metal. Then, spinning up the disc will result in the rim moving inward and spokes buckling. Once spun up, this is no different from 'an identically constructed object' but it is certainly not how you normally construct a spinning spoked wheel. Alternatively, imagine the spokes are stiff and strong while the rim is a light stretched rubber band. Then, when spun up, you would find the rubber band was tightly stretched. Again you could build it this way, but if you're following a 'standard recipe' and building spinning, you wouldn't do so.

What I think is central here, is that the resulting matter distribution is 'unexpected'. But I think it would be hard to remove my scare quotes and rigorously define this. It is 'relative' to a notion of a construction recipe.

 

[A.T.]

Here, I see what I think your are getting at (more below) but don't think it is useful (or clear to others) to think so much in terms of stresses induced by change of proper geometry.

It might be clearer to look at the changes in proper geometry directly, rather than via the stresses. See below.

Alternatively, imagine the spokes are stiff and strong while the rim is a light stretched rubber band. Then, when spun up, you would find the rubber band was tightly stretched.

Very good example. Here, we could even replace the rubber circumference band with telescopic elements with markings on them, which instead of becoming stressed, merely indicate their current proper length for everyone to see. So you can spin up the wheel safely, and directly observe how its proper geometry changes.

 

[A.T.]

If you apply external forces to each part of the object sufficient to account for the proper acceleration of its worldline, there are no stresses left over. There can't be, because the overall proper acceleration of the worldline, which you've already stipulated is entirely compensated by external forces (such as a rocket attached to each part of the object), already takes into account all forces on that part of the object. There can't be any left over.

I don't think this argument holds true in general. The internal forces from stresses acting at a part can all cancel eachother, no matter how high the stress magnitude is. So there can be non-zero stresses, even if the external force accounts for all of proper-acceleration.

However, I now also think that providing all proper acceleration by external forces everywhere might not (always) be the best strategy to minimize all stresses.

The broader point tough, is that whatever strategy is used to apply external forces, the stresses cannot be all eliminated, if the object's stress-free proper geometry was Euclidean, and the object is now placed in non-Euclidean space.

 

[PeterDonis]

I don't think this argument holds true in general

It's not an argument, it's a definition of what proper acceleration is and what it means physically. You can't say that all of the proper acceleration of a worldline is accounted for by external forces, and also say there are internal stresses left over. That's simply a contradiction.

The internal forces from stresses acting at a part can all cancel eachother

Please give a specific example. With math.

whatever strategy is used to apply external forces, the stresses cannot be all eliminated, if the object's stress-free proper geometry was Euclidean, and the object is now placed in non-Euclidean space.

You keep using the term "stresses" without explaining how you differentiate them from "forces needed to hold the object static". I've said more than once now that I don't think there is any way of making such a distinction. You have not addressed that issue at all. You need to.

 

[PeterDonis]

You keep using the term "stresses" without explaining how you differentiate them from "forces needed to hold the object static". I've said more than once now that I don't think there is any way of making such a distinction. You have not addressed that issue at all. You need to.

Here, I'll even give a simple example for you to address the issue with.

I build a rod in flat spacetime, in free fall. It has proper length $L$.

I lower the rod, oriented radially, into the curved spacetime around a gravitating body. (It doesn't matter here whether it's a black hole or something else, as long as it's spherically symmetric and the spacetime geometry in the vacuum region around it is Schwarzschild.) I attach rockets to each part of the rod so that they provide the entire proper acceleration required to hold the rod static at a given altitude. I arrange the rocket thrusts so that the rod's proper length, as measured by radially oriented rulers, is still $L$.

What "stresses" are there on the rod that are due to its proper geometry being Euclidean, and it now being placed in a non-Euclidean space?

 

[PAllen]

It's not an argument, it's a definition of what proper acceleration is and what it means physically. You can't say that all of the proper acceleration of a worldline is accounted for by external forces, and also say there are internal stresses left over. That's simply a contradiction.

Well, it depends on the material model (equivalently, considering e.g. EM fields as well as gravity). A given bound material at some temp and pressure has an equilibrium stress free local density. Add or subtract material in a given volume and you have pressure or stress. If you add in EM forces, yes, this is accounted for by needing different externally applied proper acceleration to keep it bounded (total of tidal, hovering, and changed EM force). But without considering EM forces, you can deduce that some such effect must be present if local density is necessarily changed.

 

[PeterDonis]

it depends on the material model

Nothing that I said depends on the material model. The details of exactly what stresses will be present for an object described by a given congruence of worldlines will depend on the material model, yes--because the relationship between proper acceleration and stress-energy density depends on the material model. But the fact that all such stresses will be accounted for by the proper acceleration--that there can't be any stresses left over if you've accounted for all the proper acceleration--is not.

 

[PAllen]

Here, I'll even give a simple example for you to address the issue with.

I build a rod in flat spacetime, in free fall. It has proper length $L$.

I lower the rod, oriented radially, into the curved spacetime around a gravitating body. (It doesn't matter here whether it's a black hole or something else, as long as it's spherically symmetric and the spacetime geometry in the vacuum region around it is Schwarzschild.) I attach rockets to each part of the rod so that they provide the entire proper acceleration required to hold the rod static at a given altitude. I arrange the rocket thrusts so that the rod's proper length, as measured by radially oriented rulers, is still $L$.

What "stresses" are there on the rod that are due to its proper geometry being Euclidean, and it now being placed in a non-Euclidean space?

This isn't a good example. To see the issue, you need,e.g. a 'rectangle' bounded by some two value of $\theta$ and r, 'built' far away, then moved radially inward. Then, assuming a fixed number of 'atoms', one or both these values must change. If this is resisted (equivalent to rigid spokes in the spun up wheel case), you must have stress/compression. Yes, if you add in EM fields, this would still be represented by proper acceleration, but from combined gravity and EM fields.

For example, if the boundary of the 'rectangle' is rigidly forced to preserve perimeter (this is actually allowed without violating HN - search for 'quasi-local rigidity' to find papers on this), then the interior must change density, and thus pressure. It is still, of course, true that this would be represented in proper acceleration of 'atoms' if EM force is added to the picture. But it is an effect of the changing spacetime geometry - no killing motion possible. (I continue to prefer this phrasing over 'proper geometry', which is just a slicing choice). Further, it is certainly not the result of either tidal gravity or hovering force. [In some sense, you can think of it as resulting from changing tidal gravity, but it can't be accounted for just by considering the static tidal gravity of final state.]

 

[Tomas Vencl]

I build a rod in flat spacetime, in free fall. It has proper length $L$.

I lower the rod, oriented radially, into the curved spacetime around a gravitating body. (It doesn't matter here whether it's a black hole or something else, as long as it's spherically symmetric and the spacetime geometry in the vacuum region around it is Schwarzschild.) I attach rockets to each part of the rod so that they provide the entire proper acceleration required to hold the rod static at a given altitude. I arrange the rocket thrusts so that the rod's proper length, as measured by radially oriented rulers, is still $L$.

….and we can continue with the experiment. Take second rod with the same properties and length. Divide it into n smaller elements. Then lower the elements beside the original hovering rod and attach rockets to each part of the element so that they provide the entire prper acceleration required to hold the element static at given altitude of the element. Arrange the rocket thrusts so that the elements proper length …is still L/n…..

 

[PeterDonis]

This isn't a good example.

Depends on what you're trying to show. What I was trying to show is that I can take a rod built in flat spacetime with proper length $L$, and move it into Schwarzschild spacetime, radially oriented, in such a way that it still has proper length $L$. And then I've asked @A.T. to explain what difference "non-Euclidean spatial geometry" makes.

I agree that a purely radial rod doesn't show all the effects of tidal gravity, since there are tangential effects as well as radial effects (your rectangle example is set up to include the former as well as the latter). However, even in your example, this claim...

assuming a fixed number of 'atoms', one or both these values must change.

...is wrong, if I'm allowed to use external forces, like attaching a rocket to each atom, which @A.T. has stipulated is what's being done in his scenario. If I do that, I can make the rectangle cover the same angular coordinate range as it did in flat spacetime; I just need to adjust the proper accelerations of the rockets appropriately for the congruence of worldlines in Schwarzschild spacetime that does that.

Of course if I'm not allowed to attach a rocket to every atom, but instead I have to support the object against gravity by a more conventional means, like resting it on the surface of a planet or having a rocket at its bottom pushing on it, then of course its shape will change when I move it from flat spacetime to curved spacetime. And how its shape will change will depend on the details of what it's made of, how stiff its material is, etc., etc.

But @A.T. has posited a scenario where I am allowed to use external forces, because he thinks that by doing that we can somehow separate out "stresses due to the non-Euclidean spatial geometry" from everything else. And my point is that you can't do that. You can describe the object by a congruence of worldlines (which congruence can depend on how you're supporting the object against gravity), and you can compute the proper acceleration of every worldline in the congruence, and you can use your model of the material in the object to relate those proper accelerations to stress-energy density--and that's it. You can't split out a piece that's due to "non-Euclidean spatial geometry".

 

[PAllen]

...is wrong, if I'm allowed to use external forces, like attaching a rocket to each atom, which @A.T. has stipulated is what's being done in his scenario. If I do that, I can make the rectangle cover the same angular coordinate range as it did in flat spacetime; I just need to adjust the proper accelerations of the rockets appropriately for the congruence of worldlines in Schwarzschild spacetime that does that.

Ok, here I forgot to add 'preserving local density'.

 

[PeterDonis]

here I forgot to add 'preserving local density'.

I'm not sure you can do that. If you build the object in flat spacetime in an unstressed state, and then move it to hovering statically and being held up against gravity, the object will now be in a stressed state, and the local stress-energy density won't be the same--because it's now in a stressed state.

 

[PeterDonis]

here I forgot to add 'preserving local density'.

It occurred to me after posting #116 that by "local density" you might have meant something more like "local proper number density of atoms". Preserving that, if you're allowed to use external rockets on each atom, would mean preserving the radial and tangential proper distances, which I agree is not the same as preserving the radial and angular coordinate ranges. But it could be done.

 

[A.T.]

It's not an argument, it's a definition of what proper acceleration is and what it means physically. You can't say that all of the proper acceleration of a worldline is accounted for by external forces, and also say there are internal stresses left over. That's simply a contradiction.

I must be misunderstanding you because a counter example to the above seems too trivial:

A beam (as part of a larger structure) is under longitudinal compressive stress, but its proper acceleration is purely transversal due to transversal external forces applied uniformly along the beam.

So for each beam section all of its proper acceleration is accounted for by the transverse external forces acting on it, while the internal longitudinal forces exerted by neighboring beam sections cancel eachother.

 

[PAllen]

It occurred to me after posting #116 that by "local density" you might have meant something more like "local proper number density of atoms". Preserving that, if you're allowed to use external rockets on each atom, would mean preserving the radial and tangential proper distances, which I agree is not the same as preserving the radial and angular coordinate ranges. But it could be done.

Of course I meant number density and coordinate values.

I note you haven’t commented on the second part of that post, which, to me, proves a change in pressure due only to movement through changing curvature.

 

[PeterDonis]

A beam (as part of a larger structure) is under longitudinal compressive stress, but its proper acceleration is purely transversal due to transversal external forces applied uniformly along the beam

You don't seem to understand that the proper acceleration is determined by the beam's worldline. You can't adjust it by applying external forces. For an object that's held static in Schwarzschild spacetime, the proper acceleration is radially outward, and its magnitude (in geometric units where $G = c = 1$) is

$$
\frac{M}{r^2 \sqrt{1 - 2M / r}}
$$

where $r$ is the radial coordinate at which it is held static. There is simply no way to change that. The resultant of all the forces applied to the object must be such as to produce that proper acceleration.

 

[PeterDonis]

I note you haven’t commented on the second part of that post

I hadn't looked at it in detail because I was focusing on the first part and wanting to understand it.

Having now looked at it, I'm not sure about this:

it is certainly not the result of either tidal gravity or hovering force. [In some sense, you can think of it as resulting from changing tidal gravity, but it can't be accounted for just by considering the static tidal gravity of final state.]

Does the bolded part just mean that of course you can't just look at the final state, you have to look at the difference between the initial state and the final state?

 

[Jaime Rudas]

Also, there is the possibility of non-smooth embeddings, which are much easier.

Where could I find some introductory text that would help me delve deeper into this?

 

[PAllen]

I hadn't looked at it in detail because I was focusing on the first part and wanting to understand it.

Having now looked at it, I'm not sure about this:


Does the bolded part just mean that of course you can't just look at the final state, you have to look at the difference between the initial state and the final state?

Yes, since you are looking for a cause of changed pressure. The final state by itself has no memory of how it originated.

 

[PAllen]

Where could I find some introductory text that would help me delve deeper into this?

You can start here to look for references:

https://en.wikipedia.org/wiki/Nash_embedding_theorems

Also look at the Whitney embedding theorem and also

https://en.wikipedia.org/wiki/Embedding

 

[PAllen]

You don't seem to understand that the proper acceleration is determined by the beam's worldline. You can't adjust it by applying external forces. For an object that's held static in Schwarzschild spacetime, the proper acceleration is radially outward, and its magnitude (in geometric units where $G = c = 1$) is

$$
\frac{M}{r^2 \sqrt{1 - 2M / r}}
$$

where $r$ is the radial coordinate at which it is held static. There is simply no way to change that. The resultant of all the forces applied to the object must be such as to produce that proper acceleration.

But I think the claim is simply that you can’t deduce balance opposing forces from proper acceleration; that doesn’t mean they don’t exist, or the whole field of statics would vacuous.

 

[PeterDonis]

I think the claim is simply that you can’t deduce balance opposing forces from proper acceleration; that doesn’t mean they don’t exist

No, that's not the claim that was made. Here's the claim that was made:

A beam (as part of a larger structure) is under longitudinal compressive stress, but its proper acceleration is purely transversal due to transversal external forces applied uniformly along the beam.

The beam is supposed to be held static against gravity in Schwarzschild spacetime. That means its proper acceleration is purely radially outward, with the magnitude I gave. The longitudinal compressive stress is fine, since that means the longitudinal direction is the radial direction. But there is no way to have any transverse proper acceleration at all if the beam is static, much less to have the proper acceleration "purely transversal".

 

[PAllen]

No, that's not the claim that was made. Here's the claim that was made:


The beam is supposed to be held static against gravity in Schwarzschild spacetime. That means its proper acceleration is purely radially outward, with the magnitude I gave. The longitudinal compressive stress is fine, since that means the longitudinal direction is the radial direction. But there is no way to have any transverse proper acceleration at all if the beam is static, much less to have the proper acceleration "purely transversal".

I assumed the beam was perpendicular to r, supported against gravity by external transverse force all along the bottom of the beam. Meanwhile, the rest of the structure presses the beam longitudinally, producing balanced longitudinal stresses. These exist without being reflected in proper acceleration of the world lines.

 

[PeterDonis]

I assumed the beam was perpendicular to r, supported against gravity by external transverse force all along the bottom of the beam.

The structure would have to be narrow enough for the variation in $r$ for a transverse beam to be negligible. Otherwise its proper acceleration will no longer be purely transversal, because different parts of the beam will be at different $r$ and different angular coordinates, so their proper accelerations will differ both in magnitude and direction, with the difference being enough to be measurable.

 

[PAllen]

The structure would have to be narrow enough for the variation in $r$ for a transverse beam to be negligible. Otherwise its proper acceleration will no longer be purely transversal, because different parts of the beam will be at different $r$ and different angular coordinates, so their proper accelerations will differ both in magnitude and direction, with the difference being enough to be measurable.

I don’t see that as relevant to the example. It was just to point out that there can be balanced forces unrelated to, and in different directions, from proper acceleration of mass points. Therefore, that stress is not, in general, derivable from proper acceleration. And that if you supply the proper acceleration by ‘atomic rockets’, you have not necessarily eliminated stress.

 

[PeterDonis]

stress is not, in general, derivable from proper acceleration.

In the general sense of "stress", which in relativity means "the components of the stress-energy tensor", yes, I agree. I should have used the term "unbalanced forces" instead of "stress" to be more specific about what I was concerned about.

 

[PAllen]

You can start here to look for references:

https://en.wikipedia.org/wiki/Nash_embedding_theorems

Also look at the Whitney embedding theorem and also

https://en.wikipedia.org/wiki/Embedding

I found something right on point, but it is a thesis, so I don’t swear to its accuracy (I have not studied it in detail). Its results imply that a portion of a Euclidean plane can be smoothly isometrically embedded in any curved 3 space. It discusses use of the gauss-codazzi equations, as I suggested.


http://davidbrander.org/penn.pdf

 

[A.T.]

The beam is supposed to be held static against gravity in Schwarzschild spacetime. ...

This restriction wasn't part of your original statement:

It's not an argument, it's a definition of what proper acceleration is and what it means physically. You can't say that all of the proper acceleration of a worldline is accounted for by external forces, and also say there are internal stresses left over. That's simply a contradiction.

You claimed above that this is the very definition of proper acceleration, so it should apply in any spacetime, including flat spacetime.

But even assuming Schwarzschild spacetime, I don't get this:

The longitudinal compressive stress is fine, since that means the longitudinal direction is the radial direction.

How does this follow? The longitudinal compressive stress can come from the rest of the structure, that the beam is part of. So what does it have to do with the beam's orientation in Schwarzschild spacetime? Why can't the beam's orientation be perpendicular to purely radial?

 

[PeterDonis]

This restriction wasn't part of your original statement

No, but it was part of the specific scenario we were discussing as I understood it.

You claimed above that this is the very definition of proper acceleration, so it should apply in any spacetime, including flat spacetime.

Yes, that's correct.

even assuming Schwarzschild spacetime, I don't get this

See my discussion with @PAllen, who pointed out what you've pointed out.

 

[A.T.]

Its results imply that a portion of a Euclidean plane can be smoothly isometrically embedded in any curved 3 space.

So a zero thicknes flat sheet of paper could be somehow placed in Flamm's space without distorting / tearing it? I wonder if it's possible to visualize such an embedding. Would it be a relatively simple configuration, or one of those endlessly crumpled shapes?

 

[PAllen]

So a zero thicknes flat sheet of paper could be somehow placed in Flamm's space without distorting / tearing it? I wonder if it's possible to visualize such an embedding. Would it be a relatively simple configuration, or one of those endlessly crumpled shapes?

A smooth embedding is possible. Just try to think about a 2-sphere embedded in Euclidean 3-space. There is no curvature anywhere in the Euclidean space, but the sphere embeds just fine. Even a small part of hyperbolic 2-space is smoothly embeddable in Euclidean 3-space. So just imagine the 'reverse' as it were.

Note, for smooth local embeddings (a global embedding is where you run into all the topological constraints, requiring a large number of dimensions of the target space) the relevant formula is 2n-1. So a small volume of Euclidean 3-space may require a 5-dimensional curved space to embed.

 

[A.T.]

A smooth embedding is possible. Just try to think about a 2-sphere embedded in Euclidean 3-space. There is no curvature anywhere in the Euclidean space, but the sphere embeds just fine. Even a small part of hyperbolic 2-space is smoothly embeddable in Euclidean 3-space. So just imagine the 'reverse' as it were.

Curved 2D space embeded in flat 3D space is easy to visualize, but the reverse is tricky, because curved 3D space itself is hard to visualize.

I guess for an illustration one could project the 3D Flamm embedding space onto flat 3D space, but that is not isometric, so the embedded 2D flat sheet will then look curved. Just as if you had embedded curved 2D in flat 3D.

 

[PAllen]

Curved 2D space embeded in flat 3D space is easy to visualize, but the reverse is tricky, because curved 3D space itself is hard to visualize.

I guess for an illustration one could project the 3D Flamm embedding space onto flat 3D space, but that is not isometric, so the embedded 2D flat sheet will then look curved. Just as if you had embedded curved 2D in flat 3D.

From the thesis I referenced earlier is a notion of what embeddings are 'hard' versus 'easy'. 'Hard' implies you are likely only to get a local smooth isometric embedding. 'Easy' implies that if there are no topological constraints, you are likely to get a global embedding. The determining factor (when curvature is simple for both source and target) is source curvature minus target curvature. If this is negative, the extrinsic curvature of the embedding will be negative; if positive, it will be positive. The 'hard' case is negative extrinsic curvature. So plane into hyperbolic space is 'easy' (0 - negative=positive)), the reverse is 'hard'. Since Flamm space is, I think, mostly negative curvature, plane into this is actually the easy case. Note, sphere into $E^3$ is easy, while plane into 3-sphere is hard.

 

[A.T.]

From the thesis I referenced earlier is a notion of what embeddings are 'hard' versus 'easy'.

Yes, but I was referring to the ease of visualization / illustration.