Constructing a Solid Klein Bottle

[lavinia]

can one construct a solid Klein bottle - a 3 manifold whose boundary is a Klein bottle as follows.

- Start with a solid cylinder and identify the two bounding disks by a reflection.

- The boundary becomes a Klein bottle but is this a smooth manifold whose boundary is this Klein bottle?

- If so does this manifold deform onto its central circle just as a solid torus would?

- Since reflection is an isometry of the disk, can one give this manifold a flat metric?
In general if the boundaries of two Riemannian manifolds are identified by an isometry do their metrics extend?

 

[homeomorphic]

It's a disk bundle over a circle, so it does deform onto the central circle. I think the rest is true, but not completely sure.

 

[homeomorphic]

The boundary becomes a Klein bottle but is this a smooth manifold whose boundary is this Klein bottle?

I forgot to mention, 3-manifolds always have a unique smooth structure, so yes.

 

[lavinia]

Thanks homeomorphic. I have a worry that the gluing of the two solid Klein bottles can not have a flat metric. Your answer justifies the worry because it makes the computation of the homology of this manifold easy to do. The homology to me seems impossible for a manifold that is covered by a torus.

Split the 3 manifold into two solid Klein bottles with a small collar around them. Their intersection is a collar neighborhood of the bounding Klein bottle where they are glued together.

Since the solid Klein bottles deform onto a circle they have the homology of a circle so with Z2 coefficients the Meyer Vietoris sequence is

0 -> Z2 -> Z2 -> 0 -> H2(Solid K u Solid K )-> Z2 + Z2 -> Z2 + Z2 -> H2(Solid K u Solid K ) -> 0

The last H2 is by Poincare Duality.

So the Z2 homology of Solid K u Solid K is either zero of Z2 in dimensions 1 and 2.

 

[blue_raver22]

Yes, it is possible to construct a solid Klein bottle using the method described. The resulting manifold would indeed have a boundary that is a Klein bottle. However, it is important to note that this construction does not result in a smooth manifold. In order for a manifold to be smooth, it must have a smooth structure, meaning it must be locally homeomorphic to Euclidean space. In this case, the identification of the two bounding disks by reflection creates a discontinuity at the boundary, making it impossible for the manifold to have a smooth structure.

Furthermore, it is not accurate to say that the solid Klein bottle would "deform onto its central circle just as a solid torus would." The solid torus is a smooth manifold, while the solid Klein bottle, as constructed, is not. Therefore, they do not behave in the same way.

As for the metric of the solid Klein bottle, it would not be possible to give it a flat metric using this construction. The reflection isometry only affects the boundary, and does not change the interior of the solid cylinder. Therefore, the resulting manifold would still have a curved metric in the interior.

In general, if the boundaries of two Riemannian manifolds are identified by an isometry, it does not necessarily mean that their metrics will extend. This depends on the specific geometry and topology of the manifolds and how they are being identified. In the case of the solid Klein bottle, the metric of the solid cylinder would not extend to the resulting manifold due to the discontinuity at the boundary.