The Klein bottle, though unorientable, is nevertheless the boundary of a three manifold. It is easy to see how to construct this 3 manifold. The projective plane is not a boundary of any manifold. How can we visualize this impossibility? Is there a nice embedding of the projective plane in R^4? How about an embedding of the projective plane in Euclidean space so that it has constant curvature? Same embedding question for the Klein bottle.
Hmm... not really. The Klein bottle intersects itself in R^{3}. Do you know how to represent surfaces via a quotient topology on a square? That's really the easiest way to represent these things.
take a solid cylinder and turn it around so that the two ends line up. Then instead of attaching to make a solid torus, attach by reflection in the plane of the edge discs.
The Klein bottle has Euler characteristic zero so its top Stiefel-Whitney number is zero. Its first Stiefel-Whitney class is not zero but its square is. This is easy to see because if you think of the Klein bottle as a circle bundle over the circle, the Poincare dual of the first Stiefel-Whitney class is represented by a fiber circle and its self-intersection number is zero. Thus all of the Stiefel-Whitney numbers are zero. Another way to see this is to realize that a circle bundle is always the boundary of the its disc bundle. In the case of the Klein bottle one has a flat 2 plane bundle over the circle with coordinate transformation 1 0 0 -1 (reflection in the plane) The unit disc bundle of this 2 plane bundle is a manifold whose boundary is the unit circle bundle which is the Klein bottle.
a non-orientable surface can be embedded in certain closed compact 3D manifolds. If the 3D manifold is also non-orientable, and the surface is embedded in a certain way, the surface divides the 3D manifold into two pieces, and would be the boundary of each piece. Of course these pieces can't be a subset of the real 3-space. Non-orientible surfaces cannot divide orientable 3D manifolds. Question: What happens when the projective plane is embedded in a 3D projective space? The reason I ask, is because even though this represents an "elliptic plane" in "elliptic space", the 3D projective space is an orientable manifold. So does the projective plane divide the projective space into two parts? My Guess: It just splits the 3D projective space "open" so that it becomes a bounded 3D manifold. (But what kind?)
No. If if it did then projective 2 space would be orientable. The boundary of an orientable manifold is also orientable. Intuitively if you think of P3 as the quotient of S3 by modding out opposite poles then any two hemispheres are glued together so removing the equatorial 2 sphere will not separate them. Here is another cool example: The Klein bottle is obtained from a square by making a cylindiner in one direction and a Mobius band in the other. Instead take a cube and make a solid cylinder one one direction and solid Mobius bands in the other two. This makes an orientable 3 manifold that has two transverse Klein bottles embedded in it. These two Klein bottles intersect in a circle. Neither Klein bottle separates this manifold into two parts.
There is a possibility. I've never thought of the two sided mobius as an solid enclosure resembling a Klein but in a way it does. To me it's more like a pattern that fits that of bipolar fields. Since it's an open element construct it would be difficult to make it into a solid enclosure but I suppose enough elements would merge into a bottle of sorts.
A klein bottle can be constructed from two Mobius bands by pasting them together along their boundaries (a circle)
Yep. The Klein bottle has a problem though, it violates the rubber sheet rules. When the wall area where the tube passes through is opened POOF! Klein bottle vanishes. The 3D 2 side mobius allows for a dual opening one at each pole that leads to two seperate enclosures that span the entire surfaces. Sort of a double bottle. Another aspect of the 3D 2 side mobius is it can nest a finite number of copies of itself. Using just 3 makes it possible to make a mobius capacitor and since there is 1 1/2 total twists it also behaves as a coil comprising a complete LC circuit. The frequency being controlled by the scale of construction. If none of this makes sense to you I have references available.
The Klein bottle has no problem in 4 dimensions though I am not sure if a flat Klein bottle can live in 4 dimensions.
If you can put up with watching a crusty old man for a while, he covers most all that I have mentioned and than some.
What is your point? Certainly not that a Klein bottle can't be embedded in R^3. The word "construction" in topology does not mean physical construction. It means putting a space together from pieces using identification maps and quotient topologies. By this method you construct a Klein bottle from two Mobius bands. Or conversely one says that you can "cut" a Klein bottle along a circle to make two Mobius bands.
My point is have you taped two mobius bands together along the edges and yielded a Klein bottle in so doing? Let me answer for you, no, you have not even if you tried. Physically it cannot be done. Even if you use conceptual space you must cut through boundries and even though you do the Klein bottle is imperfect and violates the rubber sheet rules and destroys itself on construction. See: When the tube passes through the wall that's ok but that area of the wall left inside the tube must be opened up to make a hole or you have no bottle at all. Make the hole, POOF!
You don't have a point. Everyone in this thread knows that you can not make a physical Klein bottle in R^3. That has nothing to do with what we were talking about.