The Smooth Solution:

Thirty years after Kuiper originally conjectured that the real projective plane with one handle could not be tightly immersed in three-space [More], François Haab proved that this is indeed true in the smooth case [H1]. The result requires considerable machinery, and we will not explain it in detail here, but will outline some of the highlights.

His basic idea is to consider mappings of surfaces into the plane, and to consider when these can be "factored" into an immersion of the surface in space followed by a projection into the plane. One of the important features of such projections are their fold curves. For smooth surfaces, the fold curves are formed by a collection of images of circles that are smooth immersions except at a finite number of cuspidal points.

The fold curves for tight immersions have a specific geometric form: one component is convex, and the others are locally concave (with respect to the image of the projected surface) and contain all the cusps. Haab computes strict bounds on the number of components that can exist in the fold set for a tight immersion of a given surface.

Haab then considers height functions on a tight surface and classifies the saddle points according to whether passing the saddle point changes the number of components in the level set of the height function, and he provides a bound on the number of saddle points that do not alter the number of components. He then considers the fold sets of projections of the surface, and determines that the type of saddles that occur on the fold sets can change only at points where the tangent line to the fold curve is tangent to the fold set at more than one point.

He applies this information to the case of the projective plane with one handle, and concludes that the type of saddle is constant on each component of the fold set, and that the fold set has exactly three components. He uses the fold curves and the top cycles to decompose the surface into pieces, and shows that one of these contains a cycle that separates the region, and on which the Gauss map is constant. He considers a height function in the direction of the Gauss map, and, after showing that the curve bounds a region on which the Gauss map is not constant, deduces that the height function has a local extremum inside that region. The initial decomposition ensures that this point is not a global extremum for the height function, which contradicts that fact that the immersion is tight. [More]

Haab relies heavily on the smoothness of the surface in his proof. There are several important differences that arise in the polyhedral case. [More]

[Right] The polyhedral solution
[Left] Kuiper's original question
[Up] Introduction

[TOC] [Index] [Glossary] [Mail] [Help]

9/29/94 -- The Geometry Center