Two Klein Bottles Sewn Together?

Question

So, I'm pretty sure at least some of you know what a Klein Bottle is. If not, a Klein Bottle, named after German mathematician Felix Klein, is basically a shape that has one side and no edge. You basically would get this if you took 2 Mobius Loops and sew them together, which creates a shape(The Klein Bottle)that requires a 4-dimensional space to exist. Here's what I wonder though.

What would happen if you kept this going. What would 2 Klein Bottles create if you tried sewing them together? Is that even possible?

Answer 1

For each $n$, define the twisted $n$-sphere bundle to be the space

$$ S^n \times I / (x,0)\sim(\alpha_0^{(n+1)}(x),1)\,, $$ where $I$ is the standard interval and $\alpha_0^{(n+1)}(x)$ flips the first coordinate of $x$ (considered as a point in $n+1$-dimensional space), leaving the others alone. In other words, we take a hollow $n$-cylinder and glue the ends together via $\alpha_0^{(n+1)}$. As examples:

• The twisted $0$-sphere bundle is the boundary of the Möbius strip (a $0$-sphere is a pair of points, which are interchanged by the antipodal map).
• The twisted $1$-sphere is the Klein bottle.

Define the twisted $n$-disk bundle to be the space $$ D^n \times I/(x,0)\sim (\alpha_0^{(n)}(x),1)\,, $$ where $D^n$ is the closed $n$-disk. As examples:

• The twisted $1$-disk bundle is the Möbius strip (a $1$-disk is the same as the interval $I$).
• The twisted $2$-disk bundle is called the solid Klein bottle. It is a three dimensional manifold whose boundary is the Klein bottle.

More generally, it is fairly intuitive to see that the boundary of the twisted $n+1$-disk bundle is the twisted $n$-sphere bundle.

Now what happens when we glue together two twisted $n$-disk bundles along their boundaries? Since the result of gluing together two $n$-disks along their boundaries is the $n$-sphere, we know that we have an $S^n$-bundle of some sort. Moreover, the gluing map at the boundary is given by $\alpha_0^{(n)}$ on each disk, so after gluing the two disks together (taking the gluing to take place in the $n+1$-th dimension), we see that the gluing on the boundary is precisely $\alpha_0^{(n+1)}$. In other words, when we glue together two twisted $n$-disk bundles along their boundaries, we get the twisted $n$-sphere bundle.

Putting all this together:

• The twisted $0$-sphere bundle is the boundary of the Möbius strip (the twisted $1$-disk bundle).
• Gluing two Möbius strips along their boundaries yields a twisted $1$-sphere bundle (a Klein bottle).
• The Klein bottle is the boundary of the twisted $2$-disk bundle (the solid Klein bottle).
• Gluing two solid Klein bottles along their boundaries yields a twisted $2$-sphere bundle.
• The twisted $2$-sphere bundle is the boundary of the twisted $3$-disk bundle.
• Gluing two twisted $3$-disk bundles along their boundaries yields a twisted $3$-sphere bundle...

... and so on.

Answer 2

all I can assume is that if this shape follows the same rules as other shapes and when you glue one side of a shape to an other identical shape both shapes loos one side and as a Klein bottle has only one side you would end up with a shape with no sides.

In short gluing the surfaces of two Klein bottles together would create a shape with no edges, no sides, no corners and no volume. It`s only characteristics are that it is curved and looped.

I know I did not use any math to come up with my answer but I like it so pleas don`t delete it.