0
$\begingroup$

In differential geometry, Hilbert's theorem (1901) states that there exists no complete regular surface S of constant negative Gaussian curvature K immersed in $ R^3 $. This theorem answers the question for the negative case of which surfaces in $ R^3 $ can be obtained by isometrically immersing complete manifolds with constant curvature.(Wikipedia)

Among constant positive Gaussian curvature K surfaces immersed in $ R^3 $ are included the hyperbolic types (cheese tire with inward cuspidal edges), incomplete and not regular.

Let us hypothetically say someone attempts to prove irregularity of some of these positive surfaces as well. Which Lemmas should be included in proof of contradiction for cheese tires distinguishing or setting them apart from from the other regular surfaces, viz. spindles and Riemann spheres?

$\endgroup$
  • $\begingroup$ Contradiction for cheese tires to what? $\endgroup$ – Ben McKay Oct 9 '14 at 8:40
  • 1
    $\begingroup$ What is a cheese tire? $\endgroup$ – Ben McKay Oct 9 '14 at 8:40
1
$\begingroup$

May be I misunderstood your question; I reformulate it as follows: whether there exists a regular embedding of a complete surface of constant positive curvature in $R^3$ and whether this surface can have complicated topology.

The asnwer is: any complete surface of constant positive curvature is isometric to the round sphere or to its $Z_2$ quotient which is topologically the projective plane. There exists no isometric immersion of the projective plane and by the alexandrov theorem any isometric immersion of the round shpere is the standard immersion.

$\endgroup$
  • $\begingroup$ Either for K=1 or K=-1 in $ k_1 k_2 = K = constant $ $ k_1 $ goes to infinity when $ k_2 $ goes to zero producing cuspidal boundaries which demarcate periodic segments of an otherwise regular surface. Apart from above example Sievert surface also has a cuspidal edge...they are not complete regular even though isometric to round ball. $\endgroup$ – Narasimham Oct 8 '14 at 15:42
  • $\begingroup$ May be missunderstanding is due to different definitions? Because for me a sufrace isometric to the round sphere is necessary complete since completeness is for me a metric property and the round sphere is complete in the metric sense which implies that anything isometric to the round sphere is also complete which contradicts what you wrote in your comment. What is your definition of completeness? $\endgroup$ – Vladimir S Matveev Oct 9 '14 at 9:09

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.