Proving that the real projective plane is not a boundary

[lavinia]

TL;DR Summary

Looking for an intuitive proof that the real projective plane is not a boundary

While all orientable 2 dimensional compact smooth manifolds are boundaries e.g. the sphere is the boundary of a solid sphere, the torus is the boundary of a cream filled doughnut - not all unorientable surfaces are boundaries. For instance, The Projective Plane is not the boundary of any 3 dimensional manifold. I know a couple proofs that use Algebraic Topology. These are nice but do not give a geometric picture. I am looking for that picture.

One might ask why such a picture could even exist since the Projective Plane can not live in 3 space but for the Klein Bottle which also can not live in 3 space, there is - I am pretty sure but haven't done a rigorous proof - a simple picture which shows that it in fact can be made into a boundary.

There are lots of proofs that the Projective plane is not a boundary. Perhaps the simplest is that it has odd Euler characteristic. The Euler characteristic of a bounding manifold must be even. But this proof leaves me a bit cold despite its power. It gives no sense of what happens when you try to fill the Projective plane up. What goes wrong? What is the picture?

Any thoughts are appreciated.

Notes:

Making a Projective Plane

The Projective plane is often defined as the quotient space of the sphere by identifying antipodal points. As a topological space, it is also a Mobius band with a disk sewn onto its edge circle. It is also a rectangle with opposite edges pasted together with a half twist - sort of a double Mobius band.

References:

Algebraic Topology, by Alan Hatcher.
Quelques proprietes globales des variétés différentiables, by Rene Thom ,Commentari Mathematici Hevetici volume 28, 17–86 (1954)
Characteristic Classes, by John Milnor
The Topology of Fiber Bundles, by Norman Steenrod

Thom's cobordism theory is developed in his original paper and is also described in Milnor's book. Not sure about Hatcher but his book is a standard text on Algebraic Topology. Milnor's book develops the theory of Stiefel-Whitney classes which are the key invariants in cobordism theory.
Steenrod's book is a classic on fiber bundles and also develops Stiefel-Whitney classes from the point of view of obstruction theory.

 

[fresh_42]

It is really hard to visualize the projective plane, even if visualized:

 

[lavinia]

Here is another picture of the Projective Plane

http://virtualmathmuseum.org/Surface/boys_bryant-kusner/boys_bryant-kusner.html

 

[lavinia]

It is really hard to visualize the projective plane, even if visualized:View attachment 261261

View attachment 261262

A possible approach might be constructive. Here is an example for the Klein bottle - that is if I did this right. Even though it is often said to be a closed surface with no inside or outside - this picture purports to illustrate why the Klein bottle actually is a boundary of a three dimensional manifold.

One way to make a Klein bottle is to start with a cylinder and paste its two edge circles together with a reflection. One imagines bending the cylinder around until the two circles match up then gluing them together. If one does this directly for instance with a cylinder made out of a piece of cloth, one gets a torus. But if before gluing, one first reflects one of the circles around an axis through its center, one gets the Klein bottle. This reflection is much like the half twist that one does when making a Mobius band out of a strip of paper. The trouble is that this can not be done in three dimensions but it can still be easily described.

Now take a solid cylinder and do exactly the same thing. That is: paste the two solid discs at its two ends together by first reflecting one of the disks around an axis through its center. The boundary cylinder becomes a Klein bottle and the solid cylinder becomes a three dimensional manifold. Note that the central axis of the solid cylinder becomes a circle but every other tube around the axis becomes a Klein bottle. The manifold looks like a tube of Klein bottles surrounding a core circle much the way a solid torus is a tube of tori surrounding a core circle.

Often with 3 manifolds one can visualize how to construct them from three dimensional pieces that do live in three space.

 

[fresh_42]

I know that your imagination capabilities are far beyond mine. I like approaches via groups, but I don't know how to translate the topology.

If it were a boundary, shouldn't there be inward pointing and outward pointing vectors at each point. So showing that there are no such vectors would be an intuitive approach. However, I would have expected the same form the Klein bottle. Hence the first question is: what is the essential difference between the two?

 

[lavinia]

I know that your imagination capabilities are far beyond mine. I like approaches via groups, but I don't know how to translate the topology.

If it were a boundary, shouldn't there be inward pointing and outward pointing vectors at each point. So showing that there are no such vectors would be an intuitive approach. However, I would have expected the same form the Klein bottle. Hence the first question is: what is the essential difference between the two?

Yes if the manifold is a boundary then there are inward pointing vectors that point into the manifold that it bounds. I guess the picture you are suggesting is that it you follow these directions from the Projective plane on the boundary into the interior - say along geodesics - then something impossible would happen. A tempting thought.

 

[zinq]

I don't know if this counts as "intuitive", but it's a quick proof:

Let P² be the real projective plane. If M were a 3-manifold-with-boundary whose boundary ∂M were topologically P², then we could take 2 copies of M and glue them together along their P² boundaries to get what's called the "double" of M, denoted by 2M, which would then be a 3-manifold (without boundary). The Euler characteristic χ(2M) must be zero, since 2M is a compact odd-dimensional manifold.*

But a simple calculation of χ(2M) shows that it must be equal to 2 χ(M) - χ(P²), and we know χ(P²) = 1. This would mean that 0 = χ(2M) = 2χ(M) - 1, which is a contradiction since 0 is even.

_____
* For a compact orientable odd-dimensional manifold, this follows from Poincaré duality. And any compact non-orientable odd-dimensional manifold W has an orientable double cover whose Euler characteristic is just twice that of W.

 

[lavinia]

I don't know if this counts as "intuitive", but it's a quick proof:

Let P² be the real projective plane. If M were a 3-manifold-with-boundary whose boundary ∂M were topologically P², then we could take 2 copies of M and glue them together along their P² boundaries to get what's called the "double" of M, denoted by 2M, which would then be a 3-manifold (without boundary). The Euler characteristic χ(2M) must be zero, since 2M is a compact odd-dimensional manifold.*

But a simple calculation of χ(2M) shows that it must be equal to 2 χ(M) - χ(P²), and we know χ(P²) = 1. This would mean that 0 = χ(2M) = 2χ(M) - 1, which is a contradiction since 0 is even.

_____
* For a compact orientable odd-dimensional manifold, this follows from Poincaré duality. And any compact non-orientable odd-dimensional manifold W has an orientable double cover whose Euler characteristic is just twice that of W.

I am aware of this nice proof. I mentioned the Euler characteristic in my OP and had this proof in mind although one can also appeal to Thom's Theorem. A nice thing about the proof is that it works for any manifold of odd Euler characteristic. That's powerful.

One might be able to use the Euler characteristic or possibly the first Stiefel-Whitney class to determine geometric properties of immersions of the Projective plane into $R^3$. I found some stuff related to Smale's theorem on immersions of spheres. Need a little time to understand it.

If you look at post #4 I sketched a proof for the Klein Bottle. Do you think it is right?

 

[Infrared]

@lavinia I think your construction for the Klein bottle is fine. This is a special case of a mapping torus: Let $f: X\to X$ be a continuous map, and define $M_f=[0,1]\times X/\sim$ where $\sim$ is the gluing $(x,1)\sim (f(x),0).$ If $X$ is a manifold (with boundary) and $f$ is a homeomorphism, then $M_f$ should also be a manifold (with boundary).

Yours is the case $X=B^2$ and $f$ is a reflection.

 

[zinq]

Yes, definitely: D² × R / ((x,y), t) ~ ((x,-y), t+1) defines a compact 3-manifold whose boundary is a Klein bottle. (In grad school, my office-mate and I used to call this the "Klein milk bottle".)

 

[lavinia]

@lavinia I think your construction for the Klein bottle is fine. This is a special case of a mapping torus: Let $f: X\to X$ be a continuous map, and define $M_f=[0,1]\times X/\sim$ where $\sim$ is the gluing $(x,1)\sim (f(x),0).$ If $X$ is a manifold (with boundary) and $f$ is a homeomorphism, then $M_f$ should also be a manifold (with boundary).

Yours is the case $X=B^2$ and $f$ is a reflection.

@Infrared Great. And thank you.

I was thinking that one gets the same manifold with boundary as the quotient of a solid torus by the involution that rotates 180° parallel to the core circle and reflects around an axis through the transversal disks.

 

[lavinia]

Here is another picture of the Projective Plane

http://virtualmathmuseum.org/Surface/boys_bryant-kusner/boys_bryant-kusner.html

@fresh_42

The picture in the link is actually an animation. It seems like one starts with a Mobius band - with three rather than 1 half twists - then thickens it until it tops off and becomes a Boy's surface.

While I still can't follow all the self intersections , the process makes sense. One can imagine modding out by the antipodal map on the 2 sphere in stages by starting with the strip bounded by the Tropics of Cancer and Capricorn and then thickening the strip towards the poles until it closes off. The projection of these strips - if thickened symmetrically - are all Mobius bands and then they close off at the poles.

BTW: This is an example of a general construction on vector bundles that is used in the proof of Thom's theorem. This Theorem describes the equivalence classes of compact smooth manifolds that are cobordant which means that their disjoint union is a boundary of a one higher dimensional smooth manifold. Start with a vector bundle and then identify all points that lie outside of a tube around the zero section. This is called the Thom space of the bundle. In the case of the Projective plane, one has a line bundle (the Mobius band) over the circle. Closing it off at the poles is topologically the same as making the Thom space.

 

[lavinia]

I know that your imagination capabilities are far beyond mine. I like approaches via groups, but I don't know how to translate the topology.

Since the Klein bottle is covered by $R^2$ much of its topology boils down to its fundamental group. A flat Klein bottle is the quotient of Euclidean space by the action of a group of isometries called a torsion free Bieberbach group (not Justin Bieberbach the famous pop star). These groups are extensions of a finite group by a lattice. For the Klein bottle one can take the lattice to be the integer lattice in the $(x,y)$ plane together with the transformation $α:(x,y)→(x+1/2, -y)$. If one thinks of first modding out by the lattice, one gets a flat torus and the projection of $α$ becomes a fixed point free involution. The quotient of the torus by this involution is a Klein bottle.

So the fundamental group looks like $0→L^2→π→Z_2→0$. This group extension is not split because $π$ is torsion free. Notice that $α^2$ is a point in the lattice since $α^2(x,y) = (x+1,y)$.

Since the Klein bottle is covered by $R^2$ it is an Eilenberg-MacClane space and this means that its cohomology is the same as the cohomology of its fundamental group. With mod 2 coefficients, there is a subalgebra generated by the pull back of the generator of the mod 2 cohomology algebra of $Z_2$. This pull back is the first Stiefel_Whitney class. Its square is the second Stiefel-Whitney class. One can determine its square by looking at the self-intersection of its dual homology cycle. This cycle is the circle that is the projection of the $y$ axis from $R^2$.

The cool thing here is that one can use topology and geometry to calculate group cohomology which from a little browsing on the web I gather is super hard in general.
-

 

[Infrared]

The cool thing here is that one can use topology and geometry to calculate group cohomology which from a little browsing on the web I gather is super hard in general.

In undergrad, I got into a bit of trouble in a number theory class. As part of a HW problem, we had to calculate the group cohomology of $\mathbb{Z}/2.$ Apparently, we weren't supposed to use the fact that $\mathbb{R}P^\infty$ is a $K(\mathbb{Z}/2,1)$...

 

[zinq]

Here's a slightly more geometrical proof of what amounts to the same reasoning. Suppose W is a compact connected 3-manifold whose boundary ∂W is equal to ℙ².

Then since W has a Morse function f : W → R, we know that it has a handle decomposition

W = h₀ + h₁¹ + ... + h₁^K + h₂¹ + ... + h₂^L

where h_j denotes a j-handle, j = 0, 1, 2,* and where K and L are the numbers of 1-handles and 2-handles, respectively. (To shorten the notation and proof, we use the fact that, since W is connected with boundary, we need only one 0-handle and no 3-handles — but this is not important.)

Now consider the effect of adding a j-handle on the Euler characteristic of the boundary for j = 0, 1, 2. Before any handles are added, we have the empty 3-manifold ∅ and χ(∂∅) = 0. For j = 0, adding a 0-handle increases χ(∂(previous stuff)) by +2 (we add a 2-sphere). For j = 1, adding a 1-handle increases χ(∂(previous stuff)) by -2 (we remove two 2-disks and add a cylinder). For j = 2, adding a 2-handle increases χ(∂(previous stuff)) by +2 (we add a cylinder and remove two 2-disks).

Thus, starting from 0, as the Morse function passes each critical point, the boundary obtained so far changes its Euler characterist by ±2, and so we can never get a boundary with an odd Euler characteristic.

_____
* A j-handle is a copy of D^j × D^3-j that is attached (denoted by +) to the previous stuff by some embedding of its attaching region, ∂D^j × D^3-j into the boundary of the previous stuff. The 0-handle is just a closed 3-ball and just gets thrown in without any attaching (there is no previous stuff, anyway).

 

[lavinia]

Here's a slightly more geometrical proof of what amounts to the same reasoning. Suppose W is a compact connected 3-manifold whose boundary ∂W is equal to ℙ².

Then since W has a Morse function f : W → R, we know that it has a handle decomposition

W = h₀ + h₁¹ + ... + h₁^K + h₂¹ + ... + h₂^L

where h_j denotes a j-handle, j = 0, 1, 2,* and where K and L are the numbers of 1-handles and 2-handles, respectively. (To shorten the notation and proof, we use the fact that, since W is connected with boundary, we need only one 0-handle and no 3-handles — but this is not important.)

Now consider the effect of adding a j-handle on the Euler characteristic of the boundary for j = 0, 1, 2. Before any handles are added, we have the empty 3-manifold ∅ and χ(∂∅) = 0. For j = 0, adding a 0-handle increases χ(∂(previous stuff)) by +2 (we add a 2-sphere). For j = 1, adding a 1-handle increases χ(∂(previous stuff)) by -2 (we remove two 2-disks and add a cylinder). For j = 2, adding a 2-handle increases χ(∂(previous stuff)) by +2 (we add a cylinder and remove two 2-disks).

Thus, starting from 0, as the Morse function passes each critical point, the boundary obtained so far changes its Euler characterist by ±2, and so we can never get a boundary with an odd Euler characteristic.

_____
* A j-handle is a copy of D^j × D^3-j that is attached (denoted by +) to the previous stuff by some embedding of its attaching region, ∂D^j × D^3-j into the boundary of the previous stuff. The 0-handle is just a closed 3-ball and just gets thrown in without any attaching (there is no previous stuff, anyway).

I love this explanation since it builds the manifold with boundary out a sequence of intuitive steps.

@zinq I got a little confused. Walk me through it. So you start out with a 3 ball then attach a bunch of 1 handles which are solid cylinders. Removing the two open 2 disks pulls the Euler characteristic of the boundary down by 2 then adding the solid cylinder on leaves the Euler characteristic unchanged? Now add a 2 handle which is a disk cross an interval which is again a solid cylinder so this would again reduce the Euler characteristic by 2? Then attach a 3 handle which seems possible only if the boundary is a sphere so one gets two closed balls glued together and this has no boundary so not an option?

Question: Start with the 3 ball, attach the first solid cylinder. If the attaching maps are both orientation preserving the result is a solid torus. If one of the attaching maps is orientation revering one gets a solid Klein bottle?If both are orientation reversing one still gets a solid torus?

 

[zinq]

Every compact connected 3-manifold (with or without boundary) has a handle decomposition, where it can be arranged that it has only one 0-handle and only one 3-manifold. (This works for an n-manifold, too, using 0-handles through n-handles.) You are right that you can only attach a j-handle when it's possible to embed its attaching region into the (current) boundary of the previous stuff. So yes, before a 3-handle is attached the boundary of the previous stuff (which does not unusally remain connected throughout this process) it must include a connected component that is a 2-sphere. (Just one handle is attached at a time.)

Yes, when you attach the first 1-handle to the 0-handle (if there is only one 0-handle), then you get either a solid torus or a Klein milk bottle — but it depends on whether the two 2-disks of the attaching region of h₁ are attached with opposite orientation or the same orientation, respectively.

(It was through the use of handle decompositions that Stephen Smale proved his generalized Poincaré conjecture, so they can be very useful.)

You wrote: "Now add a 2 handle which is a disk cross an interval which is again a solid cylinder so this would again reduce the Euler characteristic by 2?"

Not quite: The attaching region of a 2-handle = D² × D¹ is S¹ × D¹, so that changes the boundary of previous stuff by removing an annulus and then glueing one 2-disk, by its circle boundary, to each resulting boundary component of that annulus. Which subtracts 0 but then adds 2×1 = +2, net, to the Euler characteristic of the boundary.

It's useful to look at the 2D case, say of a torus T imagine din 3-space so its Morse function f : T → R is the z-coordinate and has 4 critical points.

Then notice that for each critical value c, the change from f^-1((-∞, c-ε]) to f^-1((-∞, c+ε]) is exactly what can be achieved by adding one handle of the appropriate index (0, 1, 1, 2 in that order). (Note this is not saying the handle is the set difference f^-1((-∞, c+ε]) - f^-1((-∞, c-ε]).)

 

[lavinia]

Every compact connected 3-manifold (with or without boundary) has a handle decomposition, where it can be arranged that it has only one 0-handle and only one 3-manifold. (This works for an n-manifold, too, using 0-handles through n-handles.) You are right that you can only attach a j-handle when it's possible to embed its attaching region into the (current) boundary of the previous stuff. So yes, before a 3-handle is attached the boundary of the previous stuff (which does not unusally remain connected throughout this process) it must include a connected component that is a 2-sphere. (Just one handle is attached at a time.)

Yes, when you attach the first 1-handle to the 0-handle (if there is only one 0-handle), then you get either a solid torus or a Klein milk bottle — but it depends on whether the two 2-disks of the attaching region of h₁ are attached with opposite orientation or the same orientation, respectively.

(It was through the use of handle decompositions that Stephen Smale proved his generalized Poincaré conjecture, so they can be very useful.)

You wrote: "Now add a 2 handle which is a disk cross an interval which is again a solid cylinder so this would again reduce the Euler characteristic by 2?"

Not quite: The attaching region of a 2-handle = D² × D¹ is S¹ × D¹, so that changes the boundary of previous stuff by removing an annulus and then glueing one 2-disk, by its circle boundary, to each resulting boundary component of that annulus. Which subtracts 0 but then adds 2×1 = +2, net, to the Euler characteristic of the boundary.

Interestingly it seems there is no way to twist the attaching maps up enough to make a projective plane since there seem to be only two types of attaching maps: one which adds a solid torus like handle, the other a Klein milk bottle like handle.

 

[zinq]

As I see it, this is rigorous (modulo the handle decomposition theorem). It's just the fact that regardless of anything else, for each j = 0, 1, 2 the Euler characteristic of the boundary always changes in the same manner when a j-handle is attached.

 

[lavinia]

As I see it, this is rigorous (modulo the handle decomposition theorem). It's just the fact that regardless of anything else, for each j = 0, 1, 2 the Euler characteristic of the boundary always changes in the same manner when a j-handle is attached.

Right. And the handle body argument works in any dimension one would think. I will try it. And in higher dimensions what about other characteristic numbers like the signature of a 4 manifold?

 

[mathwonk]

I am really enjoying this question and discussion. I know nothing about this and have not spent time thinking about it yet, but the brief discussion in Mo. Hirsch's book suggests that the cobordism class of a surface is zero iff it contains an odd (oops, should be even, thanks zinq!) number of disjoint mobius strips? This probably gives an euler class proof and may also give a geometric interpretation. sorry i am so far behind you guys.

 

[zinq]

I don't have Moe's book handy, but the projective plane P contains a Möbius strip M and cannot contain another disjoint one (because the square of the generator of H¹(P; Z₂) = Z₂ is non-zero, so the dual homology class must have non-zero self-intersection). Yet P is non-zero in the Z₂ cobordism algebra of (cobordism classes of) unoriented manifolds under disjoint union and cartesian product (i.e., it is not a boundary).

 

[mathwonk]

oh i stated it backwards, class should be zero iff it contains an even number of disjoint mobius strips. thanks, zinq! "obviously" zero mobius strips implies orientable hence a boundary.

 

[zinq]

"And in higher dimensions what about other characteristic numbers like the signature of a 4 manifold?"

Well, Thom proved that the element of the other most famous cobordism algebra, the Z algebra of (cobordism classes of) oriented manifolds under disjoint union and cartesian product (modulo torsion!) that an oriented manifold belongs to is determined by its Stiefel-Whitney numbers and its Pontrjagin numbers.

So by the Hirzebruch formula (for signature in terms of Pontrjagin numbers), this includes the signature of a manifold whose dimension is a multiple of 4: Two oriented cobordant manifolds of dimension 4k must have the same signature.

 

[lavinia]

"And in higher dimensions what about other characteristic numbers like the signature of a 4 manifold?"

Well, Thom proved that the element of the other most famous cobordism algebra, the Z algebra of (cobordism classes of) oriented manifolds under disjoint union and cartesian product (modulo torsion!) that an oriented manifold belongs to is determined by its Stiefel-Whitney numbers and its Pontrjagin numbers.

So by the Hirzebruch formula (for signature in terms of Pontrjagin numbers), this includes the signature of a manifold whose dimension is a multiple of 4: Two oriented cobordant manifolds of dimension 4k must have the same signature.

Yes. What I meant was using the handle body argument for characteristic numbers other than the Euler characteristic for manifolds with boundary .

 

[mathwonk]

and i am not blaming moe for my conjecture after perusing his discussion. he just gives the computation of the cobordism group Z/2Z of non orientable surfaces as an exercise.

 

[lavinia]

and i am not blaming moe for my conjecture after perusing his discussion. he just gives the computation of the cobordism group Z/2Z of non orientable surfaces as an exercise.

I didn't get the picture of the disjoint Mobius bands. Is it like in the Klein bottle where it can be broken into two Mobius bands glued together? I guess if you narrowed one of them you would get two disjoint Mobius bands.

I wish I knew how to add pictures to a post.

 

[mathwonk]

forgive me, i have not done the work, but as i recall he says that the number of disjoint mobius bands computes the genus.

 

[lavinia]

@mathwonk Ok. If you remove a disc from a projective plane you get a Mobius band. 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.

So using @zinq 's method, removing a disc from a projective plane reduces the Euler characteristic by 1 and gives zero. Two of these connected have Euler characteristic zero minus the Euler characteristic of their boundary circle which is also zero. Connecting this to a third projective plane removing a disk brings the Euler characteristic of the Klein bottle down by one to -1 and the projective plane to 0 so the sum has Euler characteristic -1. What surface is that? It has three Mobius bands and odd Euler characteristic. So if we keep on dancin' as the song goes the Euler Characteristic keeps going down by 1 each time another Mobius band is added.

All orientable surfaces have no Mobius bands and even Euler characteristic.

 

[zinq]

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.

 

[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.