This minimal surface is so complicated that we show it in three different pictures. The surface is parametrized by the Riemann sphere and antipodal points have the same image in R^3, the surface can therefore be viewed as an image of the northern hemisphere. We begin with six images of larger and larger polar caps (drawn with decreasing scale). Notice on these propeller-shaped pieces the straight lines, they are images of three meridians which run into points (called punctures) which do not have a finite image. Near each of those punctures the surface looks somewhat like Planar

Enneper: far out it is like a plane and when moving inward we come to more and more curved parts. The three wings of the propeller already indicate the position of the three planes in R^3.

A latitude-band which touches the equator from the north also avoids the infinitely large portions near the punctures. Below we show the minimal image, a 3-fold twisted Moebius band. Observe the straight segments crossing the Moebius band, they are parts of the three straight lines which started from the pole. Finally, three meridian bands from the pole to the equator can be chosen so that they do not come too near to the punctures. The minimal image of these three bands is our last picture.

This minimal surface is the image of the 6-punctured sphere. Since antipodal points of the sphere have the same image in R^3, the surface is also the image of the 3-punctured projective plane. The punctures are so called “planar ends”, which means, the surface looks outside a large ball like three pairwise orthogonal planes. Inversion of this surface in a sphere gives the famous Boy's Surface. We show the surface in three parts. The conjugate surface does not have antipodal symmetry.

The equator band in the domain starts from the equator and extends in to one hemisphere until it approaches the three punctures. As for the polar cap, the parts nearest to the three punctures get strongly expanded and are almost flat. - For the inverted Boy surface the image of the equator doubles back onto itsself so that this part of the surface is a Moebius band. _ One can follow the bands more easily in the anaglyph views.

In the domain we can find 3 meridian bands from pole to pole which do not run through the punctures. We made them wide enough so that each comes near a puncture on both hemisspheres. For the inverted Boy surface the neigborhoods of the two domain poles have the same image: we see only one polar center. The meridian bands have therefore Moebius strips as images under the surface map. -- On the conjugate surface one can see how the (images of the) meridian bands go from polar center to polar center, each coming near a puncture on both hemisspheres.

Inverted Boys Surface

P(z) = Re ( a(z) V(z) ) where: z = exp(u + i v) a(z) = 1/(z^3 - z^{-3} + sqrt(5) ) V(z) = ( i ( z^2 + z^(-2) ) , z^2 + z^(-2) , (2 i/3) ( z^3 + z^(-3) ) )