User joel hass - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T21:04:59Z http://mathoverflow.net/feeds/user/4803 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/18987/why-cant-the-klein-bottle-embed-in-mathbbr3/19000#19000 Answer by Joel Hass for Why can't the Klein bottle embed in \$\mathbb{R}^3\$? Joel Hass 2010-03-22T11:52:56Z 2010-03-22T12:02:13Z <p>Assuming that K is smooth or locally flat, there is a nice geometric reason based on transversality.</p> <p>The Klein bottle is not orientable. If it embedded in R3 so as to separate a small epsilon neighborhood of its image, then the neighborhood would be homeomorphic to a trivial line bundle over the Klein bottle, K times (-1,1), and therefore the neighborhood would also be non-orientable. But since any 3-dimensional submanifold of R3 is orientable this cannnot happen.</p> <p>So the Klein bottle embedding would have to be non separating in this neighborhood. But this means there is a curve C in the neighborhood that intersects the embedded Klein bottle once, transversally.</p> <p>Now C bounds a disk D in R3 since R3 is simply connected. (Bounding any surface is enough). One can make D transverse to the Klein bottle K, leaving boundary D fixed. One then see that K intersects D in a union of arcs and closed curves, with a single boundary point. But any such collection has an even number of boundary points, a contradiction.</p> <p>This argument works more generally to show a closed nonorientable surface cannot be embedded in a 3-manifold M with H1(M;Z2) = 0.</p>