Mobius Band as Bundle Over S^1: Trivialization Containing (1,0) in S^1by WWGD Tags: band, bundle, mobius, trivialization 

#1
Feb2510, 01:19 PM

P: 391

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([s]') ( where [s]' 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. 



#2
Feb2510, 05:53 PM

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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). 



#3
Feb2810, 11:04 PM

P: 391

"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.? 



#4
Mar110, 07:40 AM

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



#5
Mar3110, 12:59 AM

P: 662

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 nontrivial 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.? 



#6
Mar3110, 11:57 AM

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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). 


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