How to construct the two sheet cover of Klein Bottle?

[Fangyang Tian]

How to construct the two sheet cover of Klein Bottle??

Dear Folks:
There are four kinds of two sheet cover (up to homeomorhism) of Klein Bottle, it is easy to check torus is one of them, what are the others?? Many thanks!

 

[lavinia]

Dear Folks:
There are four kinds of two sheet cover (up to homeomorhism) of Klein Bottle, it is easy to check torus is one of them, what are the others?? Many thanks!

The fundamental group of the Klein bottle is a split extension of Z by Z,

0 -> Z -> K -> Z -> 0 where the action of a generator of the second copy of Z on the first is multiplication by -1.

This group may be represented as all pairs of integers, (n,m) with multiplication

(n,m).(a,b) = (n + (-1)[itex]^{m}[/itex]a, m + b)

I can only find two subgroups of index two in K.

They are

1) the group generated by all pairs (m,2n). This is the fundamental group of the torus.
2) the group generated by all pairs (2m,n). This is isomorphic to the fundamental group of the Klein bottle.

So the torus and the Klein bottle are two fold covers of the Klein bottle.

BTW:Also the Klein bottle is an n-fold cover of itself for any n.

 

[lavinia]

I think this is right but correct me if not.

Any two fold cover of the Klein bottle must have Euler characteristic zero. The only compact surfaces with Euler characteristic zero are the Klein bottle and the torus.

 

[Fangyang Tian]

I think this is right but correct me if not.

Any two fold cover of the Klein bottle must have Euler characteristic zero. The only compact surfaces with Euler characteristic zero are the Klein bottle and the torus.

Sorry, I misunderstood the original problem. It should be four, but not up to homeomorphism. To be more precisely, there are four kinds of covering not four kinds of covering space.

 

[lavinia]

Sorry, I misunderstood the original problem. It should be four, but not up to homeomorphism. To be more precisely, there are four kinds of covering not four kinds of covering space.

Not sure what you mean here. What do you mean by a king of covering?

 

[morphism]

Presumably "kind of covering" means covering space up to http://planetmath.org/encyclopedia/ClassificationOfCoveringSpaces.html .

But equivalence classes of covering spaces correspond to conjugacy classes of subgroups of ##\pi_1(\text{base space})##, and since an index 2 subgroup is normal, there can be only one equivalence class of covering spaces corresponding to each such subgroup. So I agree with lavinia that the torus and the Klein bottle are the only two-sheeted covers of the Klein bottle, up to covering space equivalence.

 

[lavinia]

Presumably "kind of covering" means covering space up to http://planetmath.org/encyclopedia/ClassificationOfCoveringSpaces.html .

But equivalence classes of covering spaces correspond to conjugacy classes of subgroups of ##\pi_1(\text{base space})##, and since an index 2 subgroup is normal, there can be only one equivalence class of covering spaces corresponding to each such subgroup. So I agree with lavinia that the torus and the Klein bottle are the only two-sheeted covers of the Klein bottle, up to covering space equivalence.

Hey Morphism

Your post has me a little confused. I do not know much about covering spaces.

I thought that conjugacy classes of sub-groups of the fundamental group of the base determined the same covering space i.e. the same base and total space and the same projection mapping. So it should not matter for this problem whether the subgroup is normal or not. What should matter is how many conjugacy classes of subgroups of index two there are. As you pointed out, in this example, there is only one subgroup in each conjugacy class but I think t still need to be proved that there are only two.

I think direct proof using the structure of the fundamental group of the Klein bottle should not be hard. I would start by looking at the intersection of the subgroup of index 2 with the first Z factor (the copy that is the kernel in the short exact sequence.)

But more generally I am wondering how more interesting examples work.

Maybe two non-conjugate yet isomorphic subgroups of [itex]\pi_{1}[/itex](B) can correspond to non-homeomorphic total spaces?

Or maybe the total space could be homeomorphic yet the actions of the deck transformations are not equivalent.

Is this right?

 

[morphism]

You should ignore most of my previous post. Let me try again: double covers of the Klein bottle correspond to conjugacy classes index 2 subgroups of its fundamental group ##\pi_1##. But since an index 2 subgroup is normal, we can rephrase the previous sentence as: double covers of the Klein bottle correspond to index 2 subgroups of ##\pi_1##.

Index 2 subgroups correspond to surjective maps onto Z/2Z. If we use the presentation ##\pi_1 = \langle a,b \mid aba^{-1}b \rangle##, then it seems to me that there are precisely three such maps.

 

[lavinia]

You should ignore most of my previous post. Let me try again: double covers of the Klein bottle correspond to conjugacy classes index 2 subgroups of its fundamental group ##\pi_1##. But since an index 2 subgroup is normal, we can rephrase the previous sentence as: double covers of the Klein bottle correspond to index 2 subgroups of ##\pi_1##.

Index 2 subgroups correspond to surjective maps onto Z/2Z. If we use the presentation ##\pi_1 = \langle a,b \mid aba^{-1}b \rangle##, then it seems to me that there are precisely three such maps.

I only get two. can you define them?

 

[morphism]

Let x be a generator of Z/2Z. Then we have the following three maps:
1) a->x, b->1. (ker = <a^2,b> is iso to Z^2, so corresponds to covering by torus.)
2) a->1, b->x. (ker = <a,b^2> is iso to pi_1, so corresponds to covering by Klein.)
3) a->x, b->x.

Am I missing something? Is the 3rd map in fact impossible?

Edit: I think it's fine, and I think the kernel is iso to pi_1, so this gives us another 2-sheeted covering of the Klein bottle by itself.

P.S. In determining the iso type of the kernel, I'm using your (lavinia's) observation that a double cover of the Klein bottle is homeomorphic to either a torus or the Klein bottle itself, which means that the fundamental group of such a cover is either iso to Z^2 or pi_1. But the fundamental group is iso to the kernel of the map pi_1 -> Z/2Z. Conclusion: if the kernel is abelian, it must iso to be Z^2; otherwise, it's iso to pi_1.

 

[lavinia]

Let x be a generator of Z/2Z. Then we have the following three maps:
1) a->x, b->1. (ker = <a^2,b> is iso to Z^2, so corresponds to covering by torus.)
2) a->1, b->x. (ker = <a,b^2> is iso to pi_1, so corresponds to covering by Klein.)
3) a->x, b->x.

Am I missing something? Is the 3rd map in fact impossible?

Edit: I think it's fine, and I think the kernel is iso to pi_1, so this gives us another 2-sheeted covering of the Klein bottle by itself.

i am still confused by this but now see that the three cases determine three different subgroups of index 2 in the fundamental group of the Klein bottle.

But I am having a lot of trouble understanding what makes the two Klein bottle coverings inequivalent because for both, the covering transformation seems to be translation by 180 degrees along the fiber circles.

Here is what I mean.

Represent the fundamental group as the set of transformations of the Euclidean plane generated by the standard square lattice and the transformation, (x,y) -> (x+1/2, -y). The lattice point, (0,1) is the group element b in your notation and the transformation, (x,y) -> (x+1/2, -y) is the group element,a.

The case a-> 1 in Z2

The square lattice is a subgroup of index 2 and it is the group of covering transformation of a torus. (x,y) -> (x+1/2, -y) projects to a fixed point free involution of this torus and the quotient by this involution is a Klein bottle.

The case b-> 1 in Z2

The group generated by lattice points of the form, (n,2m) and (x,y) -> (x+1/2, -y) is of index 2 and projects the plane onto a Klein bottle. The fiber circles are the projections of the straight lines perpendicular to the x axis.

The translation, (x,y) -> (x,y+1) projects to a fixed point free involution along the fiber circles with quotient space the Klein bottle.

The case a and b -> 1 in Z2

The third case that you mention I think comes from the subgroup generated by the lattice, (n,2m), and the map (x,y) -> (x+1/2, 1 - y) which I think is the group element, ba
Here again - I think - the fiber circles are the projection of the same vertical lines in the plane and the translation (0,1) again is an involution along the fiber circles whose quotient is the Klein bottle.

What is the essential difference between case 2 and case 3? Or have I made a mistake about the last case?

 

[lavinia]

I think I found the other involution of the Klein bottle.
translate 180 degrees along the fiber while reflecting the circles parallel to the base of the fibration.

the involution is covered by the map (x,y) -> (-x,y+1/2) of the Euclidean plane.

 

[mathwonk]

so i guess the "obvious way", making the same cross - identification of the ends of a cylinder that gives the klein bottle, but extending it to the boundary of a tubular neighborhood of the cylinder, gives the torus?

 

[lavinia]

so i guess the "obvious way", making the same cross - identification of the ends of a cylinder that gives the klein bottle, but extending it to the boundary of a tubular neighborhood of the cylinder, gives the torus?

By tubular neighborhood do you mean a thickened cylinder - so it has two annuli at its boundary?

I think this manifold is a solid with two Klein bottles as its boundary.

One could also do the same thing with a solid cylinder. Identify the two discs at the boundary by a reflection along some axis through their centers. This will not be a solid torus since its boundary is a Klein bottle. This construction shows that the Klein bottle - though non-orientable - is still the boundary of a 3 manifold. One also knows this because the Klein bottle has a fixed point free involution.