What is an explicit topological embedding of the Klein bottle into $\mathbb{R}^4$ whose projection, of some sort, down to $\mathbb{R}^3$ gives the usual "beer-bottle" immersed surface (https://upload.wikimedia.org/wikipedia/commons/8/8a/Surface_of_Klein_bottle_with_traced_line.svg), and then how does that embedding and the projection yield an explicit "topological immersion" of the Klein bottle into $\mathbb{R}^3$?

By a "topological immersion" I mean at least a continuous local homeomorphism.

What I’m looking for, more precisely, is a map $f: [a, b] \times [c, d] \rightarrow \mathbb{R}^4$ with the following properties:

  • $[a, b] = [c, d] = [0, 1]$, or perhaps more conveniently, $[a, b] = [c, d] = [0, 2 \pi]$;
  • $f$ is constant on equivalence classes under the usual equivalence relation on the rectangle that identifies horizontal edges going in the same direction but identifies vertical edges in opposite directions;
  • by passing to the quotient, $f$ induces an embedding $f^{\ast}$ of the Klein bottle $K$ into $\mathbb{R}^4$;
  • the composite $g = p \circ f^{\ast}$ is an immersion of $K$ into $\mathbb{R}^3$ whose image is that “beer-bottle” surface $S$, where the "projection" $p$ has a form such as $p(x_1, x_2, x_3, x_4) = (x_1, x_2, x_3 \cos \alpha + x_4 \sin \alpha)$; and
  • the composite $q \circ g$ is therefore a parameterization of that surface $S$, where $q : [a, b] \times [c, d] \to K$ is the quotient map.
  • 1
    $\begingroup$ I guess that the usual schematic “embedding” $f$ of the Klein bottle into $\Bbb R^3$ is not a local homeomorphism (for instance, if $f(x)$ is a crossing point then I guess that for a sufficiently small neighborhood $U$ of $x$ the image $f(U)$ would be non-open, contradicting to a definition of a local homeomorphism). $\endgroup$ – Alex Ravsky Aug 27 '19 at 15:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.