1
$\begingroup$

Inspired by the Problem 5/1964 of Miklós Schweitzer Competition, my question is as follows;

Is there a necessary and sufficient condition for a surface homoemorphic to an open disc to have two congruent curves homeomorphic to a circle?

Observation: The answer to the general question (in the given link) is No, by constructing a minimal surface which does not have such curves.

Improve this question
$\endgroup$
8
  • 4
    $\begingroup$ I don't understand the problem because I don't know what definition of 'congruent' you mean. Are you assuming that the surface has a (Riemannian) metric on it (and 'congruent' means isometric with corresponding geodesic curvatures) or that it is embedded in $3$-space (and 'congruent' means extrinsically congruent as curves in $3$-space)? $\endgroup$
    – Robert Bryant
    Commented Dec 23, 2011 at 1:26
  • $\begingroup$ I'm assuming the former definition. $\endgroup$
    – Ehsan M. Kermani
    Commented Dec 23, 2011 at 1:54
  • $\begingroup$ Here is the solution to the MS problem books.google.com/… $\endgroup$
    – Gjergji Zaimi
    Commented Dec 23, 2011 at 6:36
  • $\begingroup$ @Gjergji: Thanks for the reference. @ehsanmo: In fact, the solution described in the passage Gjergji supplied assumes the latter definition of 'congruent', rather than the former. It is a completely different problem. $\endgroup$
    – Robert Bryant
    Commented Dec 23, 2011 at 12:50
  • 2
    $\begingroup$ @ehsanmo: Take a surface $S$ in Euclidean $3$-space that has constant Gauss curvature $-1$. Then any two closed curves on $S$ with sufficiently large constant geodesic curvature $\kappa >> 0$ will close at the same length and will be congruent in the first sense, but because the intrinsic isometry that identifies them will not be an ambient isometry, the two curves won't usually be congruent in the second sense. OTOH, there is a surface with two closed curves that are ambiently congruent but don't have the same geodesic curvature in $S$, so they are not congruent in the first sense. $\endgroup$
    – Robert Bryant
    Commented Dec 24, 2011 at 2:06

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .