Unraveling the Topological Equivalence of Vertical and Horizontal Tori

[DaveC426913]

So, if you take a rectangular area XY and wrap the borders up "video game style" (i.e go off the left side, reappear on the right side, go off the top, reappear at the bottom), you get a 2D surface that can be represented by a 3D torus, right? Right.

Now, there're two ways you can make a wrap-around screen into a torus - vertical or horizontal. You can join the top/bottom THEN the sides, which gets you a horizontal torus, or you can join the sides THEN the top/bottom, which gets you a vertical torus.

This seems kind of counter-inutitive to me - that you can get to two states that are equivalent but forever distinct.

Questions:

1] Are these two shapes topologically interchangeable? Can you start with one, and get to the other without tearing the surface?

2] Is there an intermediate/generic/more symmetical shape that exists halfway between those two?

3] Are there other foldings that give the same wraparound result?

What I'm trying to get at is that, from the point of view of a Flatlander living on the torus, is it possible for him to state (or be told) that he is living on a "vertical torus", not a "horizontal torus", and ne'er the twain shall meet?

 

[matt grime]

The two constructions are of course topologically interchangable. Consider the point z before the identification, it is mapped to [z] and {z} say after each identification, then the map sending [z] to {z} springs to mind.

It is important to make sure that the order of operations does not alter the outcome, but clearly in this case it does not.

The 'inbetween' stage ie after one half of either identification is done, is a tube.

Surely you can visualize rotating one to the other like sending | to -

 

[DaveC426913]

The 'inbetween' stage ie after one half of either identification is done, is a tube.

No, I don't mean "half way towards completion", I mean...

Hm...

OK, well let's work with that.

If I really build a torus, I must pick which way I'm going to make it: [ join N/S then join E/W] or [ join E/W then join N/S ] - even if the end result is the same.

Let's say I make it of ferrous material, and it has a magentic field embedded in its surface that runs N/S from top to bottom of the flat sheet.

So, I hand the flat sheet to you and tell you to make it into a torus without me seeing. You make it, then rotate its orientation (you know, from | to -) so I can't tell which way you made it (i.e. whether you joined N/S first or E/W first).

But I can tell which way you made it.

If I detect the magnetic field running around the short "tube" perimeter of the torus, then I know you joined N/S first then joined E/W.

If I detect the magnetic field running around the long "circumference" of the torus, then I know you joined E/W first then joined N/S.

Now, while *I* can tell, as an outside observer, which way it was made, that doesn't help the ant with his compass living on the surface of the toroid. He can tell where N/S is, that doesn't tell him which way his world was built. (But that does not mean that the two type of tori really are identical).


So, my first question is (now that it has a bias embedded in it, so that I can assure the whole thing is not merely being rotated) is it possible to morph one torus into the other?

 

[Cexy]

Okay, obviously a torus with magnetic field lines (or whatever) running along its surface in one direction isn't equivalent to a torus with those lines running in the other direction. But that's because you've introduced extra structure to it - those magnetic field lines!

An unadorned torus is topologically equivalent to any other unadorned torus - in your case the morphism taking one to the other is called 'rotation' ;)

 

[DaveC426913]

An unadorned torus is topologically equivalent to any other unadorned torus - in your case the morphism taking one to the other is called 'rotation' ;)

I'm still not convinced that's true. I introduced the magnetic field to make it clear what I'm getting at, but even without it, there are two distinct ways to make a torus out of a rectangular (Euclidian) surface.

 

[matt grime]

And the only way you can tell them apart is to draw lines on the sheet of paper before you make the construction. Well, that isn't allowed, at least not as a method to distinguish them, unless the lines are canonical, and they aren't, certainly not topologically. Which part of my homeomorphism do you think fails? Seriously, I've written down a topological isomorphishm for you, why don't you think about that.

If you objection is about extra structure, then of course things might not be the same *with that extra structure*.

Note, there are infinitely many ways to create tori, all of them topogically indistinguishable (the moduli space is a point): oriented compact surfaces are determined by their genus upto homeomorphism.

If you in introduce extra structure then the moduli space (a nice way of classifying the constructions up to isomorphism) gets bigger. If we require a Riemann Metric and a conformal isomorphism then there is a one dimensional family of inequivalent tori. This approximately states that the shape of 'rectangle' affects the 'lines' you drew on it. You can create a torus from any parralellogram using this cut and paste method. If you imagine that as a subset of the compelx plane with bottom left corner at 0 and sides z and w, then the quantity z/w determinies the resulting object upto conformal equivalence, approximately.Of course, we are talking mathematically (you introduced the word topologically, I think) with our mathematical definitions of all the things involved. The 'paper' doesn't have a memory, or lines drawn on it. I could after all, erase your lines when you give it to me and draw my own on at right angles.

 

[Doodle Bob]

Now, while *I* can tell, as an outside observer, which way it was made, that doesn't help the ant with his compass living on the surface of the toroid. He can tell where N/S is, that doesn't tell him which way his world was built. (But that does not mean that the two type of tori really are identical).

The answer is: no, an element of the Flatland torus, no matter how it is situated, will not be able to tell how his/her land is situated in 3-space (if it is situated in 3-space at all, it could be embedded in 4-space or even Sl(2,C)). Keep in mind that Flatlanders have little way of detecting any spatial 3rd dimesion. Even with a specialized non-contractible circle (of which there are many more than the two you've been discussing), the toroidal Flatlanders will see the coordinates of their "space" as S^1xS^1.

 

[DaveC426913]

Hm. I need to think about this more...

Meanwhile:

Is it possible to make a toroid from a rectangular piece of rubber by first joining the NW/SE corners, then joining the NE/SW corners? Can you pick *any* arbitrary axes that are perp to each other on the sheet and bend it into a torus? That seems to be the logical extension of Matt Grime's parallelogram technique.

If you can, this would answer my initial question.

 

[matt grime]

It isn't my technique, it is the standard quotient space by a free action of a group on a variety.

And the axes don't have to be perpendicular. We really aren't thinking of actually getting a piece of paper are we? Note you cannot smoothly make a torus from a piece of paper like this, like you can a mobius strip or a cylinder; you end up 'breaking' the geometry (flat torus versus non-flat torus).

 

[DaveC426913]

It isn't my technique,

"Matt Grime's technique" = "the technique referred to by Matt Grime"

And the axes don't have to be perpendicular.

You mean the axes of folding don't have to be perpendicular to each other? Wouldn't you get some bizarre results?

We really aren't thinking of actually getting a piece of paper are we?

No. What I'm trying to avoid is the supposition that we are starting with a toroidal shape (where the shape it "was" would be meaningless). I am talking about the creation of a toriod from a flat 2D surface. This forces the question of which join is made first and which is made second.

(Also, I've started talking about a rubber sheet, rather than a paper sheet - knowing that the geometry will be distorted.)

 

[matt grime]

when i said that it isn't my technique it was supposed to make you think that perhaps, just perhaps, this was a well established idea and that you might want to think of learning about it from one of the many sources available.but you are still thinking of this as an absolute physical construction, ie doing one thing first, then the other. properly you should be thinking about quotients of the plane by lattices. there is nothing mathematical in the idea of making one join first, then the other, that is actually you thinking about physical considerations. we could just as easily ask about what colour the paper is. i repeat, mathematically, we are just thinking about a quotient space, and you appear to be talking about mathematics.

I still don't understand why yo'ure not separating the mathematical from the physical.

Instead of trying to say 'ooh this doesn't feel right', why don't you attempt to refute the (obvious) homeomorphism that exists?

 

[DaveC426913]

... it was supposed to make you think that perhaps, just perhaps, this was a well established idea and that you might want to think of learning...

...Instead of trying to say 'ooh this doesn't feel right'...

This is condescending in the extreme, and quite uncalled for.

 

[matt grime]

Your question has been answered several times in this thread: all tori are topologically equivalent. If you keep ignoring that how long do you expect people's patience to hold up?

Furthermore, those two parts you quote are not even close to referring to the samething, and are in very separate paragraphs.It is good advice that if in maths you believe a proof to be fallacious, then you dissect the proof to see where it is wrong (or where you are wrong). There is an obvious homeomorphism, if you don't accept it you need to ask yourself why.

Lemma: all quotients of the plane by a lattice (folding of the opposite sides of a parallelogram) are homeomorphic.

Proof. Let u,v a pair of vectors of spanning a lattice, the quotient space has a natural choice of basis (equivalence classes) given by (s,t) <--> [su+tv] for 0<=s,t<1, this is independent of the choice of u,v and if you follow it through you can demonstrate a homeomorphism between any two lattice quotients.

 

[Hurkyl]

I think what the OP means to ask is not if there is a homeomorphism between two torii, but if one can be smoothly deformed into the other (in 3-space), without having any self-intersections on the way.

 

[matt grime]

I can't believe that since they are obviously just rotations of each other.

 

[DaveC426913]

MG, you have stated your case, and are apparently satisfied with that. If you do have more to add, then it behooves you to be civil about it, otherwise you should step back. I'm sure I'm entitled to more than a single page of dialogue in the forum before being told to go educate myself.

I am not attempting to refute anything. I have no doubt that what I'm being told is correct - but accepting the answer on authority is not the same as understanding the answer.

What I am trying to do is:
a] ensure that the question I'm trying ask is being properly conveyed despite my clumsy informal vocabulary. (Hurkyl seems to have gotten it.)
b] understand where my own perconceptions are going awry. I can't do that without picking at the answers provided.

Hurkyl: Yes, that is what I'm getting at.

I see how the end result is indistinguishable. But I find it odd that you can take two divergent routes to get there. Are there *only* two routes from rectangular sheet to torus? Is there a continuum of routes between these two to get there?

 

[matt grime]

Several questions come to mind.

Why is it odd? (Are there any other things like this where you have the same feeling?)

What consitutes a 'route'?

What is allowed in this physical construction? (Which is probably a subquestion of the previous one.)

If one were thinking purely of a bit of the real plane and its topology, then there are an infinite number of ways of quotienting out the plane and getting a torus, apply any bijection that preserves the boundary setwise.

 

[DaveC426913]

Why is it odd?

Good point. My unspoken understanding* tells me that there is a single route to construction of a torus from a flat sheet, I just see two routes that I can't reconcile.

(*A result of informal- but inadequte formal- education in post 2ndary physics & math.)

What consitutes a 'route'?

By "routes" I simply mean "instructions for getting from A to B".

In forming a 3D toroidal space from a 2D flat sheet, you must pick one axis to join first, then pick the second. Since there are always two axes, that means there are a total of two routes - XY or YX (even though XY and YX are ... commutative? Yes, that's the right word.).


Say you folded a 2D flat sheet into 3D by folding both axes at the same time, you wouldn't get a torus at all - you'd get a sphere (or some shape with no hole). A sphere is not the same, correct? A flatlander could tell the difference between living on a sphere and living on a torus, could he not? A sphere can be done in one step, a toroid requires a sequence of steps.

What is allowed in this physical construction? (Which is probably a subquestion of the previous one.)

Not sure what you're getting at. But there's prob'ly enough on this plate so far.

If one were thinking purely of a bit of the real plane and its topology, then there are an infinite number of ways of quotienting out the plane and getting a torus, apply any bijection that preserves the boundary setwise.

I know that this is the primary problem: my lack of formal training. I am confident I can visualize the concepts easily enough, but my vocabulary hinders me.


OK, I recognize that, getting from flat sheet to torus, you can do it along an infinite number of pairs of axes and will always end up with the same thing.

Still trying to reduce that two-step process to one step though.

 

[matt grime]

You're still confusing me with the conflation of the idea of physically making a torus from something with the mathematical description of a genus one compact connected surface (which is just a mathematical thing).

If you're worried about some physical things, why not the colour of the paper? There is no temporal aspect (doing one thing, then another) in the mathematical description of what is simply a set of points. You aren't actually gluing, or making anything, with a physical process here.

Before you consider if different ways of making something are equivalent you should verify that it makes sense to talk of 'making' it. So, are you happy with your idea of making a torus in the first place? After you've done that you need to decide what you mean by 'topologically the same'. Mathematically speaking we've answered this question, but your question is not in the same vein as the mathematical idea, it is somehow more physical than mathematical. As a ferinstance, if I gave you two rectangles, are they, for your purposes, 'the same'?

 

[DaveC426913]

There is no temporal aspect (doing one thing, then another) in the mathematical description of what is simply a set of points. You aren't actually gluing, or making anything, with a physical process here.

Yeah, you're right. Once made, there's no difference.

Before you consider if different ways of making something are equivalent you should verify that it makes sense to talk of 'making' it.

I suppose this is strongly similar to making a cube out of its polyhedral "net" (a term we never learned, but turning up in kids' homework). Once assembled, a cube is a cube, but there are several starting nets, the "cross" being only one.

Funny, I never had a problem with that. I guess making a toroid is the same thing...


Edit: In fact, it's pretty much exactly the same thing. Wish I had thought of that long ago.

 

[Hurkyl]

I can't believe that since they are obviously just rotations of each other.

Oh bleh, let me try that again.

I think what the OP means to ask is not if there is a homeomorphism between two torii, but if one can be smoothly deformed into the other (in 3-space), without having any self-intersections on the way, and furthermore, the map from "flatland" into the original 3-D torus followed by the deformation equals the map from "flatland" into the new 3-D torus.

 

[StatusX]

If you're talking about physically folding and gluing a piece of paper to make a torus, there's no ambiguity. You have to glue the longer sides first. :)

But the point is that the torus is really an equivalence class of topological spaces. Two members of this class are the donut shape embedded in 3D space and a square with edges identified in the proper way. The image of physically gluing edges is only supposed to give you an idea why these spaces are homeomorphic. The "gluing" (ie, identification of points on the boundary) certainly doesn't take place over time, so there's no meaing to the question of what sides are glued first.

But I do see the source your confusion. This might only add to it, but a more direct mainfestation of this kind of problem is a sphere with poles identified. This can be done either by stretching one pole around the outside to the other to form a deformed looking torus, or by pushing it straight through to form a torus with no hole in the middle. In fact, this is directly related to your question, since this is the same space you'd get if you identified a circle on a torus to a point, and whether that circle runs the long or short way around determines which shape you get. It seems that the shape yu get depends on how you do the identifications, but that only determines the way the space is embedded in R³. The space is abstract, and homeomorphic to both embeddings, but it isn't to be confused with these embeddings.

 

[matt grime]

There is yet another thing I don't get now:

why is it 'obvious' that making a cube from a net has none of the problems (whatever they may be) that making a torus from a rectangle does?

Have you even tried to glue a rectangle into a torus? It doesn't work smoothly. Why that isn't a problem for the OP also bothers me.

 

[DaveC426913]

why is it 'obvious' that making a cube from a net has none of the problems (whatever they may be) that making a torus from a rectangle does?

I didn't say 'obvious', I said I've never had a problem with it - I've not questioned it b ecasue I learned it when younger. (In retrospect, it's not obvious at all, now that I think about it.)

But coming up with the cube folding has helped me reconcile a "consistent" view the universe - when I find general rules, rather than exceptions. The torus seemed like a one-off.

Have you even tried to glue a rectangle into a torus? It doesn't work smoothly. Why that isn't a problem for the OP also bothers me.

Of course you can't - with a rigid rectangle - but this isn't about rigid rectangles. It's not a problem for me because I'm familiar with 2D and 3D topology. (I don't think you're giving me enough credit. An incomplete understanding is different from being completely ignorant.) The whole thread was just one bit I didn't get.



Thinking about the cube made me realize that there is an equivalent but more satisfactory way to understand this issue: reverse the procedure.

If you start with a cube, and you want to flatten it, you're going to have to cut it. *How* you cut it is arbitrary: you can cut along this edge, then that edge then that edge, and you'll get a "cross" net. Cut it a different way and you'll get an "L".

It now becomes obvious that the 3D shape is perfect in its symmetry, and it is only when you try to flatten it that you must choose an arbitrary way to cut it up.

Same with the torus. The torus starts off pristine - no messy, arbitrary joins. It is only when you try to flatten it that you must choose how to slice it up.

As stated, I realize this is exactly equivalent to the reverse, but it makes more intuitive sense now that the 3D shape is the initial "pristine" shape and the net is the arbitrary mangling.

 

[Doodle Bob]

Oh bleh, let me try that again.

I think what the OP means to ask is not if there is a homeomorphism between two torii, but if one can be smoothly deformed into the other (in 3-space), without having any self-intersections on the way, and furthermore, the map from "flatland" into the original 3-D torus followed by the deformation equals the map from "flatland" into the new 3-D torus.

Just to add to this a little bit:

Basically, the OP has asked, if we choose two loops, a and b, on an embedded torus in R^3 such that they generate the fundamental group of the torus, then how strong of a homeomorphism f:embedded torus --> embedded torus can we find such that f_*(a)=b and f_*(b)=a?

If a and b are nice enough (such as not being dense in the torus like those so-called NS and EW loops being mentioned before), then clearly there is a homeomorphism since we can use those loops to generate S^1 x S^1 coordinates on the embedded torus.

But the stronger question that the OP raised was whether this homeomorphism can be found by somehow mutating the embedded image of the torus -- as Hurkyl described. In other words, can f be extended to all of R^3?


This doesn't seem to me quite possible, since it seems to me that it would amount to a deformation of all of 3-space such that the inner tube of the embedded torus (homeomorphic to S^1 x open disk) would somehow be deformed to the outer tube (not homeomorphic to S^1 x open disk).

 

[DaveC426913]

Yeah. What he said.