I know by Whitney\'s Embedding Theorem that a Klein Bottle, which is a smooth co
ID: 3195344 • Letter: I
Question
I know by Whitney's Embedding Theorem that a Klein Bottle, which is a smooth continuous 2-dimensional manifold, can not be embedded into R3. This is a research question and I am having trouble proving it, I can't just say by Whitney's Embedding Theorem that its not possible. The way things are looking I will have to prove Whitney's weak embedding theorem and by consequence prove that a Klein Bottle cannot be embedded in R3. So can one prove that a Klein Bottle cannot be embedded in R3 but it can be embedded in R4?
Explanation / Answer
If you are willing to assume that the embedded surface S is polyhedral, you can prove that it is orientable by an elementary argument similar to the proof of polygonal Jordan Theorem. Of course the proof is translation of a homology/transversality/separation argument.
Fix a direction (nonzero vector) which is not parallel to any of the faces. For every point p in the complement of S, consider the ray starting at pp and goint to the chosen direction. If this ray does not intersect edges of S, count the number of intersection points of the ray and the surface. If this number is even, you say that p is black, otherwise p is white. If the ray intersects an edge of S, you paint ppthe same color as some nearby point whose ray does not intersect edges. It is easy to see that the color does not change along any path in the complement of S (it suffices to consider only polygonal paths avoiding points whose rays contain vertices of S).
Now take points p and q near the surface such that the segment pq is parallel to the chosen direction. Then they are of different colors. But if the surface is non-orientable, you can go from p to q along a Mobius strip contained in the surface. This contradicts the above fact about paths in the complement of S.