Torus 2-sheet covering of Klein bottle

[PsychonautQQ]

So I'm trying to understand how the Torus is a 2-sheet covering of the Klein bottle. I found this on math exchange: https://math.stackexchange.com/ques...ted-covering-of-the-klein-bottle-by-the-torus.

The top response add's rigor to of the OP's observation that the double Torus with a line down the middle gives rise to two klein bottles, but the top response also says that the diagram is proof enough. Can somebody help me gain some intuition as to why this is? Why can we just draw a line down the middle? Is that an elementary transformation and thus produces a topologically equivalent structure?

If that is true, does it mean that the torus is topologically equivalent to two Klein bottles with the edge in the diagram they share identified together? Then for the covering map we should just map each of these Klein bottles in the cut torus to a Klein bottle? Am i missing something here?

 

[andrewkirk]

How's this?

A torus is a tube that we bend in 3D space and stick the ends together. To be specific, align the tube along the x axis, say with its axis running from $(-\pi,0)$ to $(\pi,0)$, then bend each of the two ends up and in through 180 degrees and bring them together at a positive coordinate on the y axis, say centred on the coordinates (0,2).

To make a klein bottle we bring the tube into 4D space, again bend the tube but this time stick the ends together the other way around. To do that we have to push the tube through where the wall of another part of the tube would be - only since we're in 4D, we can divert a little through the 4th dimension so that we can get through there without having to cut the wall.

Now consider another way to make a torus. Halfway along the pipe, at (0,0), we evert the right half of the tube (turning it inside out and folding it back over the left half), so that one end of the tube is now aligned with the other end, at $x=-\pi$. Sew together the two ends, as they are aligned, and we have another torus, only here the 'big ring' is in the y-z plane rather than the x-y plane like the one above. It is a very 'tall' torus in the x direction, with its cross section in the y-z plane being a long, thin ellipse. At every point except at the two ends - at x coordinates $-\pi$ and 0 - the shape is like a double-tube. So next we paste together the two walls of that tube, so that we now have a single tube that is covered twice. That is, there is a map from the torus to the tube under which every point on the tube, except for the rings at $x=-\pi$ and $x=0$ (the former is a join location, the latter is a fold-location), has a two-point pre-image on the torus.

Now we can turn the half-length, doubled, tube into a klein bottle using the operation described above. Since every point on the tube other than the two rings had two pre-images, the same goes for every point on the klein bottle. But then when we think about it, we realize that, since the two rings have been sewn together, every point on the ring where they have been sewn together also has two pre-images.

The dashed line on the stack exchange diagram is where the fold is made at $x=0$. The diagram is oriented with y-axis up-down and x-axis left-right, ie the same as what I described above.

This is a folksy, geometric-type explanation. The mathematical explanation is much easier, and just observes that the whole diagram is that of a torus, while each half is that of a klein bottle, so there's a continuous map from a torus to two klein bottles that are sewn together on an irreducible, simple, closed curve.

 

[PsychonautQQ]

How's this?This is a folksy, geometric-type explanation. The mathematical explanation is much easier, and just observes that the whole diagram is that of a torus, while each half is that of a klein bottle, so there's a continuous map from a torus to two klein bottles that are sewn together on an irreducible, simple, closed curve.

Wow, thanks for the great response! Can you explain how the observation of seeing two klein bottles in this diagram leads to the observation to there is an n-sheeted covering map?

 

[andrewkirk]

Let $C=[-1,1]\times[0,1]$ be the rectangle in the diagram. Let T be the torus obtained by identifying (sewing together) opposite sides of $C$, with the orientations indicated, and let $t:C\to T$ be the quotient map that performs that identification..

Note that $C=C_1\cup C_2$ where $C_1=[-1,0]\times[0,1]$ and $C_2=[0,1]\times[0,1]$ and the overlap $C_1\cap C_2$ is the line segment $L$ from (0,0) to (0,1).

Let $K$ be the klein bottle obtained by identifying opposite edges of $C_1$ with orientations as shown in the diagram, and let $k_1:C_1\to K$ be the quotient map that performs that identification.

Let $K_2$ be the klein bottle obtained by identifying opposite edges of $C_2$ with orientations as shown in the diagram, and let $k':C_2\to K_2$ be the quotient map that performs that identification.

Since $K,K_2$ are both klein bottles, there is a homeomorphism $g:K_2\to K$. We then define $k_2=g\circ k':C_2\to K$ and this is a quotient map since both $k'$ and $g$ are (all homeomorphisms are also quotient maps).

So now we have quotient maps $k_1,k_2$ from two parts of $C$ to the klein bottle $K$.

Next

• define $C'$ to be the rectangle $C$ without its right-hand edge, the line segment from (1,0) to (1,1). We observe that the map $t'$, which is the restriction of $t$ to domain $C'$, is a bijection, so it has an inverse $t'^{-1}$. It is fairly straightforward to show that $t'^{-1}$ is continuous.
• define $C'_1,C'_2$ to be $C_1,C_2$ without their respective right-hand edges. We observe that the map $k'_j$, which is the restriction of $k_j$ to domain $C'_j$ is a bijection (because the right and left edges of $C_j$ are identified, and no other points are identified), and so it covers $K$. It is continuous since it is a restriction of a continuous map.

I claim that the covering map we want is
$$f\equiv t'^{-1}\circ(k'_1\cup k'_2):T\to K$$

To prove this we need to show that:

1. $f$ is continuous
2. every point $x\in K$ has an open neighbourhood $N$ such that $f^{-1}(N)$ consists of two separated, homeomorphic images of $N$.

The proofs are not too hard. See how you go.

 

[lavinia]

Wow, thanks for the great response! Can you explain how the observation of seeing two klein bottles in this diagram leads to the observation to there is an n-sheeted covering map?

One way is to show that there is a properly discontinuous action of $Z_2$ on the torus whose quotient is the Klein bottle.

Another way is to show that every point on the Klein bottle has an open neighborhood $U$ whose inverse image in the torus is two disjoint open sets each of which is homeomorphic to $U$.

For starters you would need to identify the covering map $ρ:T→K$.

 

[lavinia]

If that is true, does it mean that the torus is topologically equivalent to two Klein bottles with the edge in the diagram they share identified together? Then for the covering map we should just map each of these Klein bottles in the cut torus to a Klein bottle? Am i missing something here?

No. In the torus the two pieces are not Klein bottles. But each piece projects onto the Klein bottle under the covering projection. What are the two pieces?

Problem 1: Describe the fundamental groups of the torus and the Klein bottle and show directly (by an algebraic rather than topological argument) that there is an injective homomorphism $h: π_1(T) ↔π_1(K)$ such that $π_1(K)/h(π_1(T)) ≈Z_2$

Problem 2: Show that it is impossible to embed a Klein bottle in a torus.

Problem 3: Show that the fundamental group of the Klein bottle is a split extension $0→Z→π_1(K)→Z→0$ of $Z$ by $Z$ but is not the direct product $Z×Z$.

Note that this means that $π_1(K)$ has no elements of finite order.

 

[PsychonautQQ]

No. In the torus the two pieces are not Klein bottles. But each piece projects onto the Klein bottle under the covering projection. What are the two pieces?

Why are the two pieces not Klein bottles? It looks like the orientations of the edges are set up correctly to me.

 

[lavinia]

Why are the two pieces not Klein bottles? It looks like the orientations of the edges are set up correctly to me.

Yes but the edges are not identified in the torus. They are identified in the Klein bottle. The diagram shows how to identify the edges to turn the torus into Klein bottle.

 

[PsychonautQQ]

I understand. Is it possible to 'cut' the torus in the way depicted in the diagram and wind up with a topologically equivalent space that would then be two Klein bottles?

 

[lavinia]

I understand. Is it possible to 'cut' the torus in the way depicted in the diagram and wind up with a topologically equivalent space that would then be two Klein bottles?

No.

If one cuts the torus along the two vertical circles, the midline circle and the vertical edge circle, it becomes two congruent cylinders. Each of these cylinders becomes a Klein bottle when their edge circles are identified with a twist as indicated by the oppositely pointing arrows in the diagram.

The language in the Stack Exchange post is a little confusing. It says "There are two Klein bottles in there." But it doesn't mean that there are two Klein bottles in the torus. It means that there are two domains within the torus that cover the Klein bottle.

The OP in this Stack Exchange post is asking how to show that the torus is a two fold cover of the Klein bottle and sees that the torus can be split into two regions that can each be made into a Klein bottle. But the OP is not sure how to go from there to defining a two fold covering.

To show that the torus is a two fold cover of the Klein bottle, you need to find the covering map. The edge identifications are not enough. I think the easiest way is to define the group of covering transformations. Since this is a two fold cover, the group is isomorphic to $Z_2$, the group with two elements. If $φ$ is the non identity element of the group then the Klein bottle is obtained from the torus by identifying each pair of points $p$ and $φ(p)$.

-
$φ$ has the following properties:

$φ^2=Id$

Around each point $p$, $φ$ maps an open neighborhood homeomorphically onto an open neighborhood of $φ(p)$. These two open neighborhoods can be chosen to be disjoint.

$φ$ can be chosen to be smooth and orientation reversing.

 

[Ray Fischer]

Lavinia #6If the Klein bottle has an edge then splitting it the two halves are two Mobius Strips one counter twisted from the other.

 

[Ray Fischer]

How's this?

A torus is a tube that we bend in 3D space and stick the ends together. To be specific, align the tube along the x axis, say with its axis running from $(-\pi,0)$ to $(\pi,0)$, then bend each of the two ends up and in through 180 degrees and bring them together at a positive coordinate on the y axis, say centred on the coordinates (0,2).

To make a klein bottle we bring the tube into 4D space, again bend the tube but this time stick the ends together the other way around. To do that we have to push the tube through where the wall of another part of the tube would be - only since we're in 4D, we can divert a little through the 4th dimension so that we can get through there without having to cut the wall.

Now consider another way to make a torus. Halfway along the pipe, at (0,0), we evert the right half of the tube (turning it inside out and folding it back over the left half), so that one end of the tube is now aligned with the other end, at $x=-\pi$. Sew together the two ends, as they are aligned, and we have another torus, only here the 'big ring' is in the y-z plane rather than the x-y plane like the one above. It is a very 'tall' torus in the x direction, with its cross section in the y-z plane being a long, thin ellipse. At every point except at the two ends - at x coordinates $-\pi$ and 0 - the shape is like a double-tube. So next we paste together the two walls of that tube, so that we now have a single tube that is covered twice. That is, there is a map from the torus to the tube under which every point on the tube, except for the rings at $x=-\pi$ and $x=0$ (the former is a join location, the latter is a fold-location), has a two-point pre-image on the torus.

Now we can turn the half-length, doubled, tube into a klein bottle using the operation described above. Since every point on the tube other than the two rings had two pre-images, the same goes for every point on the klein bottle. But then when we think about it, we realize that, since the two rings have been sewn together, every point on the ring where they have been sewn together also has two pre-images.

The dashed line on the stack exchange diagram is where the fold is made at $x=0$. The diagram is oriented with y-axis up-down and x-axis left-right, ie the same as what I described above.

This is a folksy, geometric-type explanation. The mathematical explanation is much easier, and just observes that the whole diagram is that of a torus, while each half is that of a klein bottle, so there's a continuous map from a torus to two klein bottles that are sewn together on an irreducible, simple, closed curve.

mobiusstrip16


If the Klein bottle has an edge then splitting it the two halves are two Mobius Strips one counter twisted from the other. That is what I got from the model I made! Is there something wrong with my animation?

 

[lavinia]

Lavinia #6If the Klein bottle has an edge then splitting it the two halves are two Mobius Strips one counter twisted from the other.

Not sure what your mean. It is true that the Klein bottle can be split into two Mobius bands but it does not have an edge.

In the diagram these edges are circles in the torus. They are the edges of two cylinders. To make a Klein bottle out of each of these cylinders, the opposite edge circles are identified with a reflection as indicated by the oppositely pointing arrows.

 

[Ray Fischer]

Not sure what your mean. It is true that the Klein bottle can be split into two Mobius bands but it does not have an edge.

In the diagram these edges are circles in the torus. They are the edges of two cylinders. To make a Klein bottle out of each of these cylinders, the opposite edge circles are identified with a reflection as indicated by the oppositely pointing arrows.

I show an animation (video) of forming a Klein bottle from a Torus on my Facebook page (Ray Fischer), it has an edge just like a Mobius Band. Splitting that geometry gives two Mobius Bands that are counter twisted. I can give you other data that has more information to shows why the Klein bottle should have an edge. My email is rftorbus@gmail.com . Thanks for you thoughts!

 

[lavinia]

I show an animation (video) of forming a Klein bottle from a Torus on my Facebook page (Ray Fischer), it has an edge just like a Mobius Band. Splitting that geometry gives two Mobius Bands that are counter twisted. I can give you other data that has more information to shows why the Klein bottle should have an edge. My email is rftorbus@gmail.com . Thanks for you thoughts!

I agree that one can put two Mobius bands together to get a Klein bottle. But a Klein bottle has no edge.

If you want us to look at your animation post it here.

Note that one can not make a Klein bottle in three dimensions.

 

[Ray Fischer]

I agree that one can put two Mobius bands together to get a Klein bottle. But a Klein bottle has no edge.

If you want us to look at your animation post it here.

Note that one can not make a Klein bottle in three dimensions.

So that is why I made a Mobius cup (or Torbus cup), both have edges, out of two Mobius Bands then cut it in half and found I still had two Mobius bands. Thanks?
Check out Klein Bottle on Wikipedia ! Maybe I posted a video.

 

[lavinia]

So that is why I made a Mobius cup (or Torbus cup), both have edges, out of two Mobius Bands then cut it in half and found I still had two Mobius bands. Thanks?
Check out Klein Bottle on Wikipedia ! Maybe I posted a video.

I am not sure of your question.

 

[Ray Fischer]

So that is why I made a Mobius cup (or Torbus cup), both have edges, out of two Mobius Bands then cut it in half and found I still had two Mobius bands. Thanks?
Check out Klein Bottle on Wikipedia ! Maybe I posted a video.

I am not sure of your question.

If I cannot make a Klein bottle in 3d space; how can two Mobius Bands in 3d space make a Klein bottle? What would you call the geometric figure I made? Is a Klein bottle a real geometric figure or an illusion? How can an edge be attached to a surface? As you can tell I do not know much about “Manifolds” or the logic dealing with them!

 

[lavinia]

If I cannot make a Klein bottle in 3d space; how can two Mobius Bands in 3d space make a Klein bottle?

They cannot

What would you call the geometric figure I made? Is a Klein bottle a real geometric figure or an illusion? How can an edge be attached to a surface? As you can tell I do not know much about “Manifolds” or the logic dealing with them!

You are making "immersions" of the Klein bottle in three space not actual Klein bottles. These immersed Klein bottles have self-intersections. The real Klein bottle has no self-intersections. If you take for instance the standard immersion - the tube that passes through itself then connects the edge circles - this surface intersects itself in a circle. With four dimensions this intersection circle can be avoided. As the tube bends around and nears the intersection, it is bent up into the fourth dimension. After that, it can be brought back down. Looking at it in three dimensions it would disappear near the intersection then reappear inside the tube.

It is a theorem that no non-orientable closed surface - e.g. the Klein bottle or the Projective plane - can "live" in three dimensions. The best one can do is an immersion. However, all non-orientable closed surfaces can be made in four dimensions.

BTW: Note that when the tube bends around and passes through itself the edge circle is reflected and then attached to the other edge circle. This reflection differentiates the Klein bottle from a torus. In order to make this reflection in three dimensions there is no choice but to pass the tube through itself. In four dimensions, this is not necessary.

 

[Ray Fischer]

They cannot



You are making "immersions" of the Klein bottle in three space not actual Klein bottles. These immersed Klein bottles have self-intersections. The real Klein bottle has no self-intersections. If you take for instance the standard immersion - the tube that passes through itself then connects the edge circles - this surface intersects itself in a circle. With four dimensions this intersection circle can be avoided. As the tube bends around and nears the intersection, it is bent up into the fourth dimension. After that, it can be brought back down. Looking at it in three dimensions it would disappear near the intersection then reappear inside the tube.

It is a theorem that no non-orientable closed surface - e.g. the Klein bottle or the Projective plane - can "live" in three dimensions. The best one can do is an immersion. However, all non-orientable closed surfaces can be made in four dimensions.

BTW: Note that when the tube bends around and passes through itself the edge circle is reflected and then attached to the other edge circle. This reflection differentiates the Klein bottle from a torus. In order to make this reflection in three dimensions there is no choice but to pass the tube through itself. In four dimensions, this is not necessary.

So, only 3d space is “real” or exists; my “Torbus Cup” exists and the Klein bottle is only imagery of what might exist in whatever 4d space is?When I laminate the “Twisted rings” (not a “Mobius Band”- which is not coplanar – it has a completely different geometry) with bands cut from “the twisted rings” (strings in the pictures), I call that geometry a “Torbus” or a 720 deg. split (spiral cut kerf) “Torus” – in my terms. Joining two of these counter rotated (Torbii) forms the “Torbus Cup”. This makes a “strange” geometry. Now (in 3d space) if I pull the strings through the “Hole” (over the edge) the two strings split – at a point on “the Cup” – split and form a “third string” attached to the two strings; ( the strings never leave the twisted band surface); the new string exists off the surface of the Torbus Cup. I call this “my string theory”; I do not know how to talk about this with “Topology Logic”. Can you help me understand the morphing or topology logic this 3d object might have in 4d space logic?If a line of points are brought through the hole some kind of “imaginary surface” is formed, I think! I have another paper that talks about a “Torbus Baton” that makes a similar surface. One end travels 720 deg. while the other end only travels 360 deg.

I’m guessing that these are well known concepts. I published my “3d geometry” thoughts on https://www.academia.edu/. I am trying to get more information, such as you gave me on the imagery Klein Bottle geometry. I have made and sold some interesting puzzles cut from exotic woods with this geometry. I also have a concept of “The Logos” based on this 3d geometry; which makes “Them” not imaginary to me.Thanks for your “kind help” so far.The first animation file of making the Torbus Cup I tried to send was too large, so I took out a few frames. Maybe I can be sent now! You will need a “Windows Media Player”.

 

[lavinia]

So, only 3d space is “real” or exists; my “Torbus Cup” exists and the Klein bottle is only imagery of what might exist in whatever 4d space is?When I laminate the “Twisted rings” (not a “Mobius Band”- which is not coplanar – it has a completely different geometry) with bands cut from “the twisted rings” (strings in the pictures), I call that geometry a “Torbus” or a 720 deg. split (spiral cut kerf) “Torus” – in my terms. Joining two of these counter rotated (Torbii) forms the “Torbus Cup”. This makes a “strange” geometry. Now (in 3d space) if I pull the strings through the “Hole” (over the edge) the two strings split – at a point on “the Cup” – split and form a “third string” attached to the two strings; ( the strings never leave the twisted band surface); the new string exists off the surface of the Torbus Cup. I call this “my string theory”; I do not know how to talk about this with “Topology Logic”. Can you help me understand the morphing or topology logic this 3d object might have in 4d space logic?If a line of points are brought through the hole some kind of “imaginary surface” is formed, I think! I have another paper that talks about a “Torbus Baton” that makes a similar surface. One end travels 720 deg. while the other end only travels 360 deg.

I’m guessing that these are well known concepts. I published my “3d geometry” thoughts on https://www.academia.edu/. I am trying to get more information, such as you gave me on the imagery Klein Bottle geometry. I have made and sold some interesting puzzles cut from exotic woods with this geometry. I also have a concept of “The Logos” based on this 3d geometry; which makes “Them” not imaginary to me.Thanks for your “kind help” so far.The first animation file of making the Torbus Cup I tried to send was too large, so I took out a few frames. Maybe I can be sent now! You will need a “Windows Media Player”.

This is a diversion from the OP's question. You can start a new thread.

 

[lavinia]

@PsychonautQQ

Here is another way to think about the torus covering the Klein bottle

Start with the Euclidean plane and consider the group $L$ of translations of the plane generated by the points $(n,m)$ with integer coordinates. An element of $L$ acts on $R^2$ by $(n,m)⋅(x,y)→(x+n,y+m)$.

- $L$ is called a lattice. Show that it is isomorphic to $Z×Z$.
-Show that $L$ acts properly discontinuously on the plane.
- Show that the unit square in the positive quadrant with the origin as one of its corners is a fundamental domain for the action of $L$ on $R^2$.
- Show that the quotient space of the action of $L$ on the plane is a torus so the torus is covered by $R^2$ with covering group $Z×Z$.
- Conclude that the fundamental group of the torus is $Z×Z$

Now add another transformation $σ$ of the plane and let $K$ be the group generated by $L$ and $σ$.
Let $σ$ act on the plane by $σ⋅(x,y) = (x+1/2, -y)$

- Show that $σ^2$ is an element of $L$.
-Show that $L$ is a normal subgroup of $K$
-Show that the quotient group $K/L$ is the group with two elements.
- Show that $K$ acts properly continuously on the plane. Conclude that the quotient space $R^2/K$ is a manifold.
- Split the unit square into two rectangles with a vertical slice at $x=1/2$. Show that either of these two rectangles can be chosen to be a fundamental domain of $K$.
-Show that each of these rectangles projects onto a cylinder in the torus $R^2/L$
- Show that each of these rectangles projects onto a Klein bottle in the quotient space $R^2/K$.
- Show that the group $K/L$ acts properly discontinuously on the torus.
Conclude that the torus is a two fold cover of the Klein bottle.
-Conclude that the fundamental group of the Klein bottle is $K$.

- Redo the exercise replacing $σ$ with the isometry $(x,y)→(x+1/2, -y + 1/2)$.

 

[Ray Fischer]

@PsychonautQQ

Here is another way to think about the torus covering the Klein bottle

Start with the Euclidean plane and consider the group $L$ of translations of the plane generated by the points $(n,m)$ with integer coordinates. An element of $L$ acts on $R^2$ by $(n,m)⋅(x,y)→(x+n,y+m)$.

- $L$ is called a lattice. Show that it is isomorphic to $Z×Z$.
-Show that $L$ acts properly discontinuously on the plane.
- Show that the unit square in the positive quadrant with the origin as one of its corners is a fundamental domain for the action of $L$ on $R^2$.
- Show that the quotient space of the action of $L$ on the plane is a torus so the torus is covered by $R^2$ with covering group $Z×Z$.
- Conclude that the fundamental group of the torus is $Z×Z$

Now add another transformation $σ$ of the plane and let $K$ be the group generated by $L$ and $σ$.
Let $σ$ act on the plane by $σ⋅(x,y) = (x+1/2, -y)$

- Show that $σ^2$ is an element of $L$.
-Show that $L$ is a normal subgroup of $K$
-Show that the quotient group $K/L$ is the group with two elements.
- Show that $K$ acts properly continuously on the plane. Conclude that the quotient space $R^2/K$ is a manifold.
- Split the unit square into two rectangles with a vertical slice at $x=1/2$. Show that either of these two rectangles can be chosen to be a fundamental domain of $K$.
-Show that each of these rectangles projects onto cylinders in the torus $R^2/L$
- Show that each of these rectangles projects onto Klein bottles in the quotient space $R^2/K$.
- Show that the group $K/L$ acts properly discontinuously on the torus.
Conclude that the torus is a two fold cover of the Klein bottle.
-Conclude that the fundamental group of the Klein bottle is $K$.

- Redo the exercise replacing $σ$ with the isometry $(x,y)→(x+1/2, -y + 1/2)$.

Thanks for the additional logic. Does the tube edge still join to the Klein Bottle surface? To make the Mobius Cup only edges of the two Mobius counter rotated bands are connected. What difference does this make in the logic?