Proving that the real projective plane is not a boundary

[zinq]

Yes, switching the endpoints of the normal D¹-bundle to Boy's immersion everts the "sphere" ... but only after you have used a regular homotopy to get the sphere from the standard embedding to Boy's surface in the first place — that's the hard part!

But yes, for a long time the most common sphere eversion was to a middle stage that is like 4/3 of the Boy's immersion, with fourfold instead of threefold symmetry. That's the one depicted in Nelson Max's groundbreaking computer graphics animation of that eversion.

 

[lavinia]

Yes, switching the endpoints of the normal D¹-bundle to Boy's immersion everts the "sphere" ... but only after you have used a regular homotopy to get the sphere from the standard embedding to Boy's surface in the first place — that's the hard part!

But yes, for a long time the most common sphere eversion was to a middle stage that is like 4/3 of the Boy's immersion, with fourfold instead of threefold symmetry. That's the one depicted in Nelson Max's groundbreaking computer graphics animation of that eversion.

 

[lavinia]

Here is a thought. Suppose the projective plane is a boundary, $P=∂M$. Remove a disk $D$ from $P$ and glue two copies of $M-D$ together along $D$ to get a 3 manifold whose boundary is a Klein bottle, $∂M_{K}=K$.

This gluing process can be reversed by slicing $M_{K}$ along a disk $D$ to get two copies of $M-D$. Then capping $M-D$ off with a disk gives a 3 manifold whose boundary is the projective plane.

So if the projective plane is a boundary then there is some 3 manifold whose boundary is the Klein bottle that can be sliced into two congruent pieces along a disk and then adding a disk to each piece gives two 3 manifolds whose boundaries are the projective plane.

What is more, the boundary of this disk is the common boundary of two Mobius bands that are connected together to form the Klein bottle.

So it must be that no 3 manifold whose boundary is a Klein bottle can be sliced in this way and this suggests a way to visualize why a projective plane is not a boundary - although indirectly. Start with such a 3 manifold and try to slice it. For instance, try it with the Klein milk bottle.

Though this is not a general proof - - since the Klein bottle can bound all sorts of manifolds - it allows a picture.

 

[lavinia]

Here is a thought. Suppose the projective plane is a boundary, $P=∂M$. Remove a disk $D$ from $P$ and glue two copies of $M-D$ together along $D$ to get a 3 manifold whose boundary is a Klein bottle, $∂M_{K}=K$.

This gluing process can be reversed by slicing $M_{K}$ along a disk $D$ to get two copies of $M-D$. Then capping $M-D$ off with a disk gives a 3 manifold whose boundary is the projective plane.

So if the projective plane is a boundary then there is some 3 manifold whose boundary is the Klein bottle that can be sliced into two congruent pieces along a disk and then adding a disk to each piece gives two 3 manifolds whose boundaries are the projective plane.

What is more, the boundary of this disk is the common boundary of two Mobius bands that are connected together to form the Klein bottle.

So it must be that no 3 manifold whose boundary is a Klein bottle can be sliced in this way and this suggests a way to visualize why a projective plane is not a boundary - although indirectly. Start with such a 3 manifold and try to slice it. For instance, try it with the Klein milk bottle.

Though this is not a general proof - - since the Klein bottle can bound all sorts of manifolds - it allows a picture.

I reply to myself. In the case of the Klein milk bottle the circle in question is not null homologous. But if it were the boundary of a disk it would be. If one contracts the Klein milk bottle radially onto its core circle then the circle in question contracts to twice the generator of the homology class of the circle. So it can not be the boundary of a disk.

A general argument can be made for an arbitrary compact three manifold whose boundary is the Klein bottle. This manifold is not orientable - since its boundary is not orientable - so its first Stiefel-Whitney class is not zero. The circle in question is Poincare dual to the first Stiefel Whitney class of the Klein bottle (here using $Z_2$ coefficients for homology). But this is back to Algebraic Topology rather than pictures.