A Proving that the real projective plane is not a boundary

[zinq]

"So the connected sum of two projective planes is two Mobius bands glued together and so is a Klein bottle. Add add a third projective plane to get a surface I never thought about before - the connected sum of a Klein bottle and a projective plane. It contains 3 Mobius bands."

If you think of the 2-sphere as the 0, a torus as T and the projective plane as P, then compact connected surfaces under the operation of connected sum (up to homeomorphism) form a commutative monoid generated by T and P with just one relation:

P + P + P = T + P.

 

[lavinia]

Just to be clear, there is an unoriented cobordism group for surfaces (which includes all surfaces) and an oriented one (which includes all orientable surfaces together with a choice of orientation).

But not one just for non-orientable surfaces.

One thing I don't have a feel for in the oriented cobordism group is torsion. Apparently there are oriented manifolds the are not oriented cobordant to zero but whose doubles are.

 

[zinq]

I have no feel whatsoever for that torsion, either. But I read somewhere that, indeed, the only torsion in oriented cobordism is of the sort you mention: M + M = ∂W. And that the first example occurs in 5 dimensions: The class of the "Wu manifold" SU(3)/SO(3) is not a boundary but twice it is.

 

[lavinia]

I have no feel whatsoever for that torsion, either. But I read somewhere that, indeed, the only torsion in oriented cobordism is of the sort you mention: M + M = ∂W. And that the first example occurs in 5 dimensions: The class of the "Wu manifold" SU(3)/SO(3) is not a boundary but twice it is.

For a while I thought that maybe there were orientable flat Riemannian manifolds that are torsion since their Euler class and Pontryagin classes are zero. But it turns out that they are boundaries.

 

[lavinia]

Continuing the Euler characteristic approach, @fresh_42 suggested looking at immersions of the projective plane into $R^3$ and I found some papers that relate the number of triple points of the immersion to the mod 2 Euler characteristic of the surface. Similar to @mathwonk 's conjecture, the number of triple points (a point whose preimage contains exactly 3 points) of an immersed surface equals its Euler characteristic mod 2. The immersion must be in "general position" which I gather means that all self-intersections are transversal.

Still trying to understand the proof but it does reduce the proof that the projective plane is not a boundary to counting the triple points of an immersion. Boy's surface I think has one triple point.

A cursory look at Banchoff's papers gives me the idea that he is counting singularities of Morse functions or more precisely of projections of the immersion onto the axes of a coordinate system for $R^3$.

The number of triple points is a condition that can be visualized geometrically in 3 space.

More on this as I read more carefully.

http://people.math.gatech.edu/~dmargalit7/classes/math8803Fall2013/triple.pdf

https://www.ams.org/journals/proc/1...-1974-0377897-1/S0002-9939-1974-0377897-1.pdf

Note: Interestingly the normal bundle to Boy's surface is not trivial but one can still look at the unit sphere bundle which in this case is just two points in each fiber and this is a 2 fold cover of Boy's surface. So it is an immersed 2 sphere. It seems - but not sure - that one can interchange the end points in each fiber by sliding them along the unit interval in each fiber and this everts this sphere. Midway it becomes Boy's surface. The same thing works for any immersed projective plane.

 

[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.