Mobius Band as Bundle Over S^1: Trivialization Containing (1,0) in S^1

[WWGD]

Hi:
I am trying to show that the Mobius Band M is a
bundle over S^1.
It seems easy to find trivializations for all points
except for the trouble point (1,0).

This is what I have so far:

Let R be the reals, and I<R be the unit interval
[0,1]< R (i.e., [0,1] as a subspace of the Reals.)

We have these maps:

1) p:I-->S^1 , a qiotient map, i.e., we give S^1 the

quotient topology.

2) q: a quotient map on IxR : identify points

(0,y) with (1,-y) . The "identified space"

--i.e., the quotient of IxR by q -- will be the

top space. (this is the Mob. Band.)

3)The Projection map Pi ,from the top space in 2


down to S^1 is defined by :

Pi([s,t])= p(')

( where ' is the class of s under p:[0,1]->S^1 as above, so that the clases are:

{ {0,1}, {x} ) : 0<x<1 }


Then: *****


A trivialization for U= S^1-{(1,0)} is

U itself; Pi^-1(U)=(1,0)xR , is already a

product space. The identity map gives us a homeo.


*****

A trivialization containing the point (1,0) in S^1:

This seems to work, but seems too easy:

We consider an open set (open in S^1 ; quotient ) containing (1,0). We parametrize

S^1 by arclength , "based" at (1,0), (i.e., arclength at (1,0)=0) , and we use , e.g.,

an arc U from (0.1) , to (0,-1) .

We want to show that Pi^-1(U) is homeo. to UxR . We have:


Pi^-1(U) =[0,1.57]xR union [4.71,6.28]xR


Am I on the right track?.

Thanks.

 

[quasar987]

Pi^-1(U) =[0,1.57]xR union [4.71,6.28]xR

How can these numbers be right if the bundle is a quotient of [0,1] x R ?!

For simplicity, I suggest you consider U=S^1\(-1,0) so that Pi^1(U)=([0,1]\{½} x R )/~.

And now the goal is to remove the twist of the bundle that's "at" the fiber over (1,0).

 

[WWGD]

"Pi^-1(U) =[0,1.57]xR union [4.71,6.28]xR

How can these numbers be right if the bundle is a quotient of [0,1] x R ?!"


Right. My bad; I did not rescale by 2PI. This should be:

Pi^-1(U)=[0,1/4]xR union [3/4,1]xR


"For simplicity, I suggest you consider U=S^1\(-1,0) so that Pi^1(U)=([0,1]\{½} x R )/~.

And now the goal is to remove the twist of the bundle that's "at" the fiber over (1,0)"


O.K, thanks. I tried it, but I don't see too well how to remove the twist:



This is what I got: (R^+ is the positive reals, etc.)


Pi^-1(U)= [[0,1/2)xR union (1/2,1]xR]/~ =

0xR^+ union (0,1/2)xR union (1/2,1)xR union 1xR^+ union (0,0)


i.e., I collapsed 1xR^- with 0xR^+ , and I collapsed 0xR^- with 1xR^+ , and

(0,0)~(1,0).


How does this help, tho.?. Would you please suggest.?

 

[quasar987]

With U=S^1\(-1,0), we have Pi^1(U)=([0,1]\{½} x R )/~. Write [t,x] for the equivalence class of (t,x) under ~.

Define a diffeomorphism f:Pi^1(U) --> (½,3/2) x R by setting

f([t,x]) = (t,x) for (t,x) in (½,1] x R,
f([t,x])=(t+1,-x) for (t,x) in [0,½) x R

 

[Bacle]

Quasar 987 wrote:

" With U=S^1\(-1,0), we have Pi^1(U)=([0,1]\{½} x R )/~. Write [t,x] for the equivalence class of (t,x) under ~.

Define a diffeomorphism f:Pi^1(U) --> (½,3/2) x R by setting

f([t,x]) = (t,x) for (t,x) in (½,1] x R,
f([t,x])=(t+1,-x) for (t,x) in [0,½) x R "

Interesting. I too have been curious about the details of the Mobius Bundle as

a non-trivial bundle over the circle. Just a couple of comments, though:

It seems like a point like, e.g., [5/3xR] would be sent outside of (1/2,3/2)xR.


Also: would you give some insight into how the twisting ( at the point (1,0)) is "solved"?


Specifically, It just seems strange that we can have a diffeomorphism , given the


twisting at (1,0). Would you comment.?

 

[quasar987]

Mmh, but 5/3 >1 while points of the bundle are of the form [t,x] with t in [0,1].

The idea of the construction of f is simply this: if you have a Mobius strip made of paper and if you take scissors and cut it "vertically" (this corresponds to removing the slice [{½}xR] in post #4 above), then there is no more "twist" as you can now take the remaining piece of paper and lay it flat on a table to form a rectangle: you have locally trivialized the mobius bundle. This is just what the map f does; it take ([0,1]\{½} x R )/~ and flattens it on the table (R^2).