4
$\begingroup$

EDIT: The well known Jordan curve theorem says: let $C\subset S^2$ be a closed simple curve on the 2-sphere. Then its complement $S^2\backslash C$ consists of two connected components, both homeomorphic to discs (in fact it is known that the closure of each component is homeomorphic to the closed disk by Jordan-Schoenflies theorem).

Is there a version of the Jordan theorem for closed simple curves in real projective plane $\mathbb{R}\mathbb{P}^2$? (The curve might be assumed to be smoothly imbedded.)

A reference would be helpful.

ADDED: Given the comment by HenrikRüping below, I realized that for my purposes it suffices to assume that the homology class of $C$ vanishes in $H_1(\mathbb{R}\mathbb{P}^2,\mathbb{Z}/2\mathbb{Z})$.

$\endgroup$
  • 1
    $\begingroup$ of course if we look at a closed simple curve representing the generator of $\pi_1$, the complement should have only one connected component homeomorphic to a disc. $\endgroup$ – HenrikRüping Dec 15 '19 at 18:40
  • $\begingroup$ @HenrikRüping: Many thanks. Corrected. $\endgroup$ – MKO Dec 15 '19 at 18:43
4
$\begingroup$

If C is null-homologous, then the complement of C has two components: a disk and a Möbius strip (as one sees since the preimage of C in the 2-sphere is 2 disjoint Jordan curves). If C is not null-homologous, then the complement of C is a single disk.

$\endgroup$
4
$\begingroup$

Well, if you take the double cover, under your assumptions the lift is two simple closed curves in $S^2,$ the complement of which will be two disks and an annulus, so the original curve bounds a disk on one side.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.