2

The complement of a simple closed curve in the Riemann sphere has two connected components (Jordan). Schoenflies theorem implies that each of these components is homeomorphic to a disk and hence each has trivial fundamental group. Is there a direct (and simple) way to see that each component has trivial fundamental group without invoking Schoenflies?

Michael Albanese
  • 18,817
Chris Judge
  • 624
  • 1
    The Wikipedia article lists four "elementary proofs." Have you looked at these? https://en.wikipedia.org/wiki/Schoenflies_problem – Christian Remling Sep 03 '16 at 19:12
  • @Christian Remming: Thanks for your question. "Elementary" is not the same as "direct" or "simple". I have read the Thomassen proof and have looked at the Moise and Bing proofs.They are elementary but I wouldn't call them "simple". In fact, this question was motivated by a desire to flesh out the proof that uses the Carath'eodory extension theorem. The missing link for me is between the output of the Jordan curve theorem and the input of the Riemann mapping theorem. – Chris Judge Sep 03 '16 at 21:57
  • (continued) I think that I now know how to bridge that gap, but it's not simple: Jordan-Brouwer implies that first homology of a connected component is trivial and then since each curve is null-homologous one can construct a complex logarithm function. Then the usual proof (e.g. Ahlfors) of the Riemann mapping goes through. – Chris Judge Sep 03 '16 at 21:57
  • Chris, you are essentially using Schoenflies in your proof, as you are giving a pretty standard proof of the Schoenflies theorem with that argument. – Ryan Budney Sep 04 '16 at 03:57
  • Perhaps Borsuk's separation theorem + https://www.jstor.org/stable/2036506?seq=1#page_scan_tab_contents should lead to the required proof. The two mentioned starter theorems are simple/easy when compared to Schoenflies. I hope that you give it a try. – Włodzimierz Holsztyński Sep 04 '16 at 05:31
  • Oops, here is a better link: http://www.ams.org/journals/proc/1971-027-03/S0002-9939-1971-0271949-X/S0002-9939-1971-0271949-X.pdf – Włodzimierz Holsztyński Sep 04 '16 at 06:23
  • 1
    If you'll grant that Schoenflies is easy for a piecewise linear Jordan curve, try this: Show that each component of the complement of any compact connected set in the plane has trivial fundamental group. Do this by reducing to the case when the compact set is a piecewise linear $2$-manifold. (Any given loop in the complement of the compact set is in the complement of some neighborhood, therefore in the complement of some PL $2$-manifold neighborhood.) – Tom Goodwillie Sep 04 '16 at 12:48
  • @ Ryan Budney. Yes, I am trying to flesh out the "pretty standard" proof of Jordan-Schoenflies using Carath'eodory's theorem. A reference to complete details would be appreciated. – Chris Judge Sep 04 '16 at 13:28
  • @ Tom. Thanks! That seems like a good way to go as I agree that PL-Schoenflies is not difficult. – Chris Judge Sep 04 '16 at 13:32
  • Actually, this follows directly from Alexander duality that each component of the complement is acylic. Now, use the fact that each 2-dimensional manifold is homotopy-equivalent to a graph and hence has free fundamental group, see http://mathoverflow.net/questions/18454/fundamental-groups-of-noncompact-surfaces. – Misha Sep 05 '16 at 15:23
  • @Misha. Yes, but not really simple or direct.For example, on http://mathoverflow.net/questions/18454/fundamental-groups-of-noncompact-surfaces one learns of techinical combinatorial approaches including triangulations. Existence of triangulations is my ultimate aim...In the end, I chose to define the complex logarithm in the proof of surjectivity of Riemann mapping theorem via integrating piecewise vertical/horizontal curves and using closedness of the form $(f'(z)/f(z))dz$. Integrating over any piecewise vertical/horizontal loop in the complementary region gives zero. – Chris Judge Sep 07 '16 at 16:58

0 Answers0