Fundamental group of Project Plane with 2 points missing

[PsychonautQQ]

edit: fixed typo's andrewkirk pointed out, oops
I can cover the projective plane with 2 open sets U,V where each of these neighborhood contains the point that is missing, and the intersection of these two neighborhoods will be simply connected.

I was then hoping to invoke the Seifert-Van-Kampen theorem to see that the fundamental group would be isomorphic to the Free product of two infinite cyclic groups, because the fundmental group of U and V would both be isomorphic to the fundmanetal group of $S^1$. This is what I thought anyway.

I am a bit concerned because this is exactly the result I arrived at for a closed disk with 2 punctured points, and it seems it would be the same thing for a sphere with 2 punctured points. Am I doing something wrong here? Thanks PF!

 

[andrewkirk]

There are a few thing in the question that do not quite make sense to me, but which may just be typos.

1. I have not come across the project plane. I assume you're referring to the projective plane.
2. When you say 'the intersection of these two points will be simply connected' I presume you mean the intersection of the sets U and V (the intersection of the two points is the empty set, as they are presumed distinct).

On my reasoning, the intersection of U and V will not be simply connected. It will consist of two components, each of which is a Moebius strip not containing either of the omitted points. Since the Moebius strip has the same fundamental group as $S^1$, which is isomorphic to the integers under addition, the fund group of $U\cap V$ will be the free product of two infinite cyclic groups.

Hence $U\cap V$ is not path connected so we cannot invoke Seifert-van Kampen.

In contrast, with a closed disc, $U\cap V$ is simply connected, so we can invoke that theorem.

If my reasoning is correct, that seems to make the concern go away.

 

[PsychonautQQ]

Hm, right. I think I'm sort of seeing it, but can you help me understand why $U\cap V$ will have two components, each of which will be a moebious strip?

I am a bit concerned that we won't be able to invoke the Seifert-van Kampen theorem, as this is the second exercise at the end of the chapter called Seifert-van Kampen theorem >.<... Perhaps I could pick my open sets $U,V$ such that the intersection would be path connected and I could use the theorem?

 

[andrewkirk]

Actually, having drawn a new diagram, I now think the intersection is a Moebius strip and an open disc, rather than two Moebius strips as I said above.

I'd like to do a diagram but I'm a bit rushed right now, so I'll try to describe it instead.

Consider the usual method of drawing the Projective Plane as a square in which opposite edges are pasted together with opposite directions/senses.
Put coordinates on the square so that (0,0) and (4,4) are the bottom left and top right corners, with the first coord being horizontal direction.
Let the omitted points be P=(1,1) and Q=(3,3)
Define U as the set that contains everything below the line L1 connecting (0,1) to (1,0) and everything above the line L2 connecting (0,3) to (3,0)
Define V as the set that contains everything above the line L3 connecting (3,4) to (4,3) and everything below the line L4 connecting (1,4) to (4,1)

Then the two sets are open and cover the square, and $P\in V,\ Q\in U$.
The intersection of U and V is $A\cup B\cup C$ where A is everything below L1, B is everything above L4 and C is everything between L2 and L3.

By suitably deforming the edges of the regions and taking account of the senses of the edges that are pasted, I think we can see that C is a Moebius strip whereas $A\cup B$ is an open disc. (A Moebius strip is obtained from a square by pasting together two opposite edges with reverse sense, and leaving the other two edges free).

If that's correct then, since the fund group of the disc is trivial, the fund group of the intersection is just $\mathbb Z$ under addition.

PS I tried to think of a two-set open cover such that the intersection of the two sets was path connected but I couldn't. My intuition says it's impossible but I have no proof. The simplest space that can be obtained by pasting opposite edges of a square is a torus and I don't think it can be done for a torus either. The intersection always seems to have two components.

 

[andrewkirk]

D'oh! Just realized that U and V above are not well-defined sets, because the borders that are pasted between the two components of do not line up. So for instance the open line segment from (1,4) to (3,4) is both in and out of U.

The good news is that, if I've got it my fix for this problem right (3rd try) this means we CAN get a two-set open cover whose intersection is path-connected.

Let the square be 6 x 6 instead of 4 x 4.
Let P=(1,1), Q=(3,3)
then
define U as the region between La the line segment from (0,3) to (3,0) and
Lb the line segment from (3,6) to (6,3)
Note that its bound match, so we have a well-defined open set after pasting.

define V as the square EXCLUDING the closed set bounded by
Lc the line segment from (0,4) to (4,0) and
Ld the line segment from (2,6) to (6,2)
Note that its bounds match, so we have a well-defined open set after pasting.

$U\cap V$ is $A\cup B$ where
A is the set of points between La and Lc and
B is the set of points between Ld and Lb.
Again the bounds match, so we have a well-defined open set after pasting.

We have $P\in V-U,\ Q\in U-V$.

The sets A and B are not well-defined when we do the pasting of edges, because their points on the boundary of the square are in the set, whereas the antipodal points with which the pasting identifies them are not.
However, $A\cup B$ is well-defined because the square-boundary points of one set are pasted to the boundary points of the other. That is,
the segment (0,3)-(0,4) in A is pasted to (6,2)-(6,3) in B, and
the segment (3,0)-(4,0) in A is pasted to (2,6)-(3,6) in B.

Hence, $A\cup B$ is path-connected. I think they form an open strip with two twists - which should be orientable (unlike the one-twist Moebius strip) and have fundamental group $\mathbb Z$.

So you can use Seifert-vanKampen after all!

I hope it's right this time.

 

[PsychonautQQ]

D'oh! Just realized that U and V above are not well-defined sets, because the borders that are pasted between the two components of do not line up. So for instance the open line segment from (1,4) to (3,4) is both in and out of U.

The good news is that, if I've got it my fix for this problem right (3rd try) this means we CAN get a two-set open cover whose intersection is path-connected.

Let the square be 6 x 6 instead of 4 x 4.
Let P=(1,1), Q=(3,3)
then
define U as the region between La the line segment from (0,3) to (3,0) and
Lb the line segment from (3,6) to (6,3)
Note that its bound match, so we have a well-defined open set after pasting.

define V as the square EXCLUDING the closed set bounded by
Lc the line segment from (0,4) to (4,0) and
Ld the line segment from (2,6) to (6,2)
Note that its bounds match, so we have a well-defined open set after pasting.

$U\cap V$ is $A\cup B$ where
A is the set of points between La and Lc and
B is the set of points between Ld and Lb.
Again the bounds match, so we have a well-defined open set after pasting.

We have $P\in V-U,\ Q\in U-V$.

The sets A and B are not well-defined when we do the pasting of edges, because their points on the boundary of the square are in the set, whereas the antipodal points with which the pasting identifies them are not.
However, $A\cup B$ is well-defined because the square-boundary points of one set are pasted to the boundary points of the other. That is,
the segment (0,3)-(0,4) in A is pasted to (6,2)-(6,3) in B, and
the segment (3,0)-(4,0) in A is pasted to (2,6)-(3,6) in B.

Hence, $A\cap B$ is path-connected. I think they form an open strip with two twists - which should be orientable (unlike the one-twist Moebius strip) and have fundamental group $\mathbb Z$.

So you can use Seifert-vanKampen after all!

I hope it's right this time.

Wow, thanks for all this! You are a boss!

$S^1$ is a deformation retract of the moebius strip, and so the moebius strip has fundamental group isomorphic to the fundamental group of $S^1$, correct?
And this $A\cup B$ is a two twist strip, and also has a fundamental group isomorphic to the fundamental group of $S^1$?

 

[lavinia]

A projective plane minus a point is a Mobius band. Pull the projective plane minus a point back to the sphere. One gets the sphere minus two opposite poles. Melt the poles until all that is left is the cylinder spanning the tropics of Cancer and Capricorn. This cylinder projects to a Mobius band.

The Projective Plane minus two points is a Mobius band minus one point. It is connected but not simply connected.